Transcript for:
Power Electronics Specialization - Introduction Lecture Notes

welcome to the university of colorado boulder's power electronics specialization in this course you will learn how switching power converters change electrical power from one forum to another with very high efficiency and accurate control a popular example today is the electric vehicle such as the tesla model s the heart of its power system is a power converter that changes the dc battery voltage to variable frequency and variable voltage ac to control the electric motor that drives the wheels power electronics on the vehicle also performs the function of battery charging converting the 60 hertz ac from the utility into the dc required to charge the battery with control of the current so that battery life is maximized the toyota prius a hybrid electric vehicle also contains a power electronics module under the hood it similarly controls an electric motor and battery charging another example of power electronics is the connection of a variable speed wind turbine to the ac utility the frequency and voltage of the ac produced by the electric generator vary with wind speed in this application power electronics can be used to convert the variable ac to the fixed frequency and voltage required by the utility some other examples of applications include solar power systems the power systems of mobile electronics computer server systems and spacecraft power systems in general we require high efficiency power conversion with accurate and sophisticated control to optimize energy use this specialization is an introduction to the electronic circuits used to convert electric power with high efficiency you will learn how these circuits work and how to model their losses and design their control systems welcome to the course introduction to power electronics i'm professor robert erickson from the university of colorado at boulder in the department of electrical computer and energy engineering this course is available in both non-credit and for credit versions it's the first course within the specialization on power electronics it's also the first course of our graduate certificate in power electronics and it serves also as the first course in a pathway specialization that can allow you to gain admission into our master of science degree program in this short video i'm going to discuss how the credit and non-credit courses work including enrollment and grading it's very important so i hope you'll watch the video all the way through this course uses our textbook fundamentals of power electronics however you don't have to buy the book to take the course all of the material that you need is available in in the videos and in the pdf files containing the lecture slides while we are not able to formally enforce prerequisites this course is taught at a graduate level and the basic assumption is that you are fluent in the material of circuits and electronics at the level that it's taught in a typical undergraduate electrical engineering degree so as i mentioned this course is the first course in a specialization on called introduction to power electronics if you sign up for the four credit version of this course it carries the university of colorado course number ecea 5700 and you can receive transcripted credit for this course the three other courses that follow this course in the specialization are listed here we the second chorus is on converter circuits the third is on control of converters and the fourth is on design of magnetics for converters and the intent is to uh give you the basic technical foundation to to actually build power converters in the lab for your information the specialization covers the same technical material that we teach in our regular on campus course uh that's called introduction to power electronics uh and it's numbered ecen5797 we have both four credit and non-credit versions and here's uh the difference between them so you can sign up either way if you sign up for the non-credit version uh you pay coursera the fee for the non-credit course and if you complete the assignments of the non-credit version then you can earn a coursera certificate of completion if you sign up for the far credit version then you pay university tuition and if you complete the assignments and the examination in that course then you can earn transcripted university credit that you can apply towards our campus graduate certificate in power electronics and to our master of science degree in electrical engineering both the non-credit and the four credit versions give access to the same lectures homework assignments supplementary materials and discussion forums to get access to the exams you need to sign up for the for credit version the far credit version also additionally gives you access to course facilitators also known as teaching assistants and they will hold office hours regular office hours where you can dial in remotely and and talk to the course facilitators the non-credit version is continuously available so you can start and stop at any time the four credit version is scheduled into eight week sessions so once you sign up for the four credit version you need to finish before the eight week session concludes if you sign up for the non-credit version you can transfer if you want into the four credit version uh you'll have to pay the university tuition for that for credit version to do that but when you do it then any work that you've done on the homework will be automatically transferred into the for credit version of the course we have a performance-based admission process to our ms ee degree program this involves completion of the four credit versions of the power electronics courses the four courses of this specialization with a grade point average of at least b or better throughout those four courses if you complete the far credit version then it can be applied towards that performance-based admissions process it can all the four credit version can also be applied as i mentioned to the university graduate certificates and towards the msee degree a little bit about changing your enrollment status so first of all as i mentioned you can switch from non-credit to for credit at any time one strategy might be to sign up for the non-credit version try the homeworks and see how they go and if you feel like you can handle the work then you can at that point upgrade to the four credit version so to do that you will have to pay the university tuition for the four credit version but any work that you've done in the non-credit version is automatically transferred in the far credit version of the course you can drop the course within 14 days of enrolling and receive a full tuition refund when you do this the course then does not appear at all on your university transcript after the 14 days you can still withdraw from the course at that point you won't receive a tuition refund and the course will appear on your transcript but with a grade of w which stands for withdraw the w grade does not impact your grade point average and really doesn't mean anything except that we're not going to issue you a grade you can withdraw at any time up until the point when you access the exam of the course once you access the exam then you're committed and you'll be issued a grade on your transcript at the end of the eight week session for for a four credit course your course grade is computed if you're still in the if you haven't withdrawed or withdrawn so your course credit is computed on whatever work is you've entered and you will be issued a grade i hope that if you don't complete the class that you would withdraw or drop and not just let it a grade turn in automatically into an f that's not the goal here so please be diligent in dropping or withdrawing if if needed for this course ecea 5700 there are three homework assignments and basically it's one per week or whatever pace you want to go at you're allowed unlimited attempts to these assignments they are auto graded and you can find out what your score is immediately and try again if you like they're also in video quizzes these are just for your own edification and they're not graded at the end of the four credit course there is there are some examination preparation materials these are not in the non-credit version including a practice exam this practice exam is not graded but it has the similar conditions to the the real exam and i suggest you try it and see how it goes you can withdraw from the course at any point including after you've done the practice exam then finally we have a proctored examination which is the one that counts again this is only in the four credit version of the course the proctored exam is similar to the homework assignments and has a similar format the main difference is that it's a closed book exam that it uses a proctoring service and it's a timed two-hour exam so what you will be able to do is uh have unlimited submissions to the problems within the two hour time period and get feedback on which ones are right and wrong according to the autograder but when the two hours are up then then the exam is over and it will use your last submission as the final grade so we have three homework assignments here's how they are weighted so 15 20 and 15 and then the exam is weighted 50 percent uh the this weighted average then determines the final course letter grade for this course uh and it is all really automatic or it's machine graded the non-credit course has the same materials as the four credit course except for the exam module so it has the three homework assignments it has all of the lectures and and supplementary materials to earn a coursera certificate of completion you have to successfully pass all three homework assignments and the typical passing threshold on the assignments is 70 percent so what's on the course site there are recorded lectures for the entire course there are the slides used in each lecture which you can download as pdf files and follow along there are the homework assignments and solve there are some solves sample problems for additional help there are discussion forums for each homework assignment the first assignment relates to simulation of a switching converter and so there are some simulation files you will need to be able to run lt spice which is a circuit simulation program that you can download for free the four credit version of the course has exam preparation materials and it has a link to the actual exam so for the first week of class what you should do is view the lectures for module one and then do homework assignment number one which is simulation of a boost converter i look forward to your participation in the class and i hope that you find it a useful and high quality course in this brief lecture i'm going to summarize the the technical content of the power electronics courses and specializations that we will offer for credit so the the area of power electronics is quite broad and it incorporates elements uh from a number of different subfields of electrical engineering certainly it's based on analog circuits and electronic devices we control systems and feedback systems are very important we have applications of in power systems the design of the magnetics is very important and an important part of the design of a power converter often we interface electric machines and often we will use numerical simulation so in this course introduction to power electronics ecea5700 i'm going to assume that you have some undergraduate prerequisite knowledge in analog circuits and electronics i'm not going to assume anything else and this first course will cover power converters really from the standpoint of analog circuits and electronic devices so this course one that is four weeks long plus a proctored exam at the end really is summarized by this slide in the first week we're going to do simulation of a switching converter so you will do use lt spice to look at the waveforms of a an actual switching converter we'll learn techniques of steady state converter analysis so if you have a switching converter that has switched waveforms like these how to analyze them to find things like the output voltage or the inductor currents of the converter circuit we'll then develop equivalent circuit models such as this one that model the basic conversion properties of a switching converter as well as modeling losses in the circuit and this kind of equivalent circuit then can be used to analyze a converter to predict things like its efficiency curves the follow on course two called converter circuits ecea 5701 goes into more detail in how these switching converters work we'll talk about how to realize the switches using power transistors and diodes so we'll talk some about power semiconductor devices one of the important things to learn here is how they switch and where switching loss comes from so here are some typical waveforms of the diode reverse recovery process which is one of the major sources of switching loss in a practical converter so we'll learn how to model that we'll learn about the different important power devices such as the power mosfet we're going to analyze what's known as the discontinuous conduction mode that occurs in some switching converters and we'll look at different converter circuits or circuit topologies including ones with transformer isolation here is a well-known transformer isolated converter known as the forward converter which is one of the or isolated circuits that we'll discuss the third course in this specialization is called converter control and this is an introduction to applying feedback to a switching converter generally we want to regulate an output voltage or an inductor current or some similar quantity so here's an example of a switching converter circuit known as the buck converter with a feedback loop around it to control the output voltage of this buck converter this is a pretty complex complex system and we have switched waveforms that are fairly complicated and so in course three we will learn how to model the important voltage and current components in the circuit and derive equivalent circuits that we can use to design feedback loops course three will cover an introduction then to feedback in the context of switching converters the last course ecea5704 is an introduction to magnetics for switching converters so we'll talk there about just how basic inductors and transformers work talk about their ac and dc loss mechanisms in both the core and the winding and we'll talk about how to design basic inductors and transformers for switching converters simulation is a recurring theme in this uh specialization and in the following ones as well here we're going to use lt spice in this course and in fact in the first homework this week you will use lt spice to simulate the basic waveforms of a switching converter here is actually an example from the follow-on specialization which is gets into more detail and more advanced topics in converter control but this is a an average model of a switching converter and its feedback loop that can predict things like small signal transfer functions and which can be used to design the feedback loops that follow on specialization goes into more advanced control techniques such as current mode control power factor correction input filter design uh digital control uh and and related topics we have a graduate certificate that we offer through this program the graduate certificate is comprised of this specialization the follow-on control specialization and then a capstone design project and in this project you will uh design the power conversion system for a top usb type-c interface which interfaces a computer and its battery to it to a usb cable so this involves selection of the converter circuits design of the power stage design of the magnetics design of the analog controller and then simulation and verification that your design meets requirements here is the prerequisite chain for this specialization so this course introduction to power electronics assumes a basic undergraduate knowledge of circuits and electronics but that's all it is prerequisite to the second course in the specialization called converter circuits which i mentioned before that one is prerequisite to the final two courses so the the control course and the magnetic design course both assume that you've had or gotten through the second converter circuits course so i look forward to your participation in this course i would encourage you to take advantage of the discussion forums and in the credit course we also have of course facilitators who will hold regular office hours and i urge you to take advantage of their help as well and i hope that you find this course rewarding the field of power electronics is concerned with the processing and control of electrical power using high efficiency electronic circuits the key functional element in power electronics is called the switching converter this lecture describes the basic functions performed by switching converters and it introduces how high efficiency electronic circuits can be constructed that realize these functions the switching converter generally includes two input terminals one is the power input where the where a source is connected the second input is a control input that tells the converter how to process the input power to produce the output dc to dc conversion is a common function performed by converters in which the voltage of the dc is changed and possibly regulated so we may change say a high input voltage into a low dc output voltage rectification is the conversion of ac to dc and in the process we may want to control the output voltage the dc output voltage and and maybe also the ac input current waveform the inverse of rectification is called inversion in which we have a dc input and we produce an ac output of a controllable frequency and magnitude ac to ac cyclo conversion involves changing the frequency and the amplitude of of the ac voltage in each of these the switching converter is the key electronic power processing element and it is composed of electronic uh devices as well as reactive elements and we attempt to to build this with very high efficiency control is invariably required we usually want to regulate the output voltage we may although we may want to control currents or other quantities so we're generally also building some kind of control system that adjusts the control input to the converter for example to regulate the output we can do this with either traditional analog feedback or nowadays it's economically feasible to use quite sophisticated control using digital microcontrollers high efficiency is essential in any application that involves very much power we want high efficiency nowadays of course because we want to conserve energy but in addition to that it's simply not possible and not economically feasible to process large amounts of power with low efficiency because our converter will will not be able to handle the the heat from the power lost so here's the equation for efficiency it's the ratio of output power to input power and the loss is the difference between the input and output power which we can express in terms of the output power and the efficiency in this way now our switching converters are loss limited they're limited by the amount of heat sinking that we have and the size of the enclosure and so for a loss limited converter with a given efficiency this limits the amount of output power that that the converter can produce so if we can build a converter with higher efficiency then we can either reduce the the loss and build a smaller converter with smaller heat sinks or for the same amount of heat sinking we can get more output power and build a converter that produces more power so a measure of how good the converter technology is really is the ratio of the output power to the efficiency and here it's expressed in terms of efficiency we can solve for this quantity and it turns out to be efficiency divided by one minus efficiency and it's plotted here and so you can see that even increasing the efficiency by a few percent can increase the amount of output power or reduce or increase the ratio of output power to loss substantially and and so our goal is to produce efficiencies that approach one which which makes the ratio of output power to loss approach infinity so the higher we can get on this curve the smaller our converter can be and the more input and output power it can produce this drawing here is meant to to represent a very small size converter with very large input and output converters so we have a very small converter with very high efficiency that processes a very large amount of electric power how do we build a circuit that changes the voltage with high efficiency here's a diagram of in general the different categories of electrical circuit elements that we have resistors capacitors magnetics including transformers and inductors and semiconductor devices that here are generally grouped into two two categories the first is linear mode transistors such as class a amplifiers or op amp circuits and the second is switched mode which is transistors that are operated as on off devices such as in a digital circuit where they're either all the way on or all the way off in signal processing such as you probably learned in undergraduate electronics and circuits courses we usually avoid magnetics well we don't have to but magnetics generally are more expensive they're hard to integrate on integrated circuits [Music] and most engineers don't know how to design them and so for a variety of reasons we usually ignore me or avoid magnetics but resistors and capacitors are fine linear and switched mode semiconductors are also common in power processing by contrast what we want to do is avoid using elements that consume power so for that reason we don't want to use resistors however capacitors and magnetics are fine and in fact it may be easier in a high power application to build a good inductor than it is to build a good capacitor but we commonly use inductors capacitors transformers and other elements because although they store energy they don't ideally they don't consume power we can get that stored energy back at some later time we avoid linear mode transistors for the same reason that we avoid resistors that a class a amplifier for example operates with a transistor having voltage across it and current through it at the same time and so it consumes power on the other hand switched mode transistors or transistors that are operated as on off switches are are good they don't consume power the reason for that is if we can get them to operate close to the behavior of an ideal switch then what happens is when the switch is closed there's no voltage across the switch and when the switch is open there's no current through the switch and in either state the power consumed by the switch which is the product of voltage and current is zero so to the extent that we can make a semiconductor device operate as an ideal switch like this we the the semiconductor device will be lossless so let's do a simple example of how to build a dc dc converter with high efficiency and using just inductors capacitors and switches okay so i'm going to take simple example with some round numbers here we have an input voltage i've labeled vg that is a power source and it has it's 100 volts and what we would like to do is change this 100 volts into 50 volts to supply a resistive load and just for some round numbers let's suppose the resistance is 5 ohms so that we have 10 amps flowing through our load okay so how can this converter be realized first of all let's ask how in a beginning circuits class what circuits do you know that could change 100 volts into 50 volts okay well the resistive voltage divider is one of the first circuits taught usually in a beginning circuits class and we learned that the um the output voltage of the divider is equal to this voltage divider ratio of the load resistance over the sum of the two resistances times the input voltage so if we adjust this added resistor to be the same value as our load resistor we'll get a divider ratio of a half and the output voltage will be half of the input okay the problem with this is that the same current flowing to the load also flows through this resistor and the same power in the load then is is equal to the power in this resistor and so we will have our 10 amps flowing through the resistor that has 50 volts across it so there'll be 500 watts of loss in this resistor there's 500 watts going to the load and we draw a thousand watts out of vg so the efficiency is only 50 percent at low power levels it's common to build a series pass regulator circuit that operates almost in the same way as the resistive voltage divider except instead of a resistor we put a transistor that operates in the active region [Music] and it has the same voltage across it that the resistor on the previous slide had so it dissipates the same 500 watts with the series pass regulator what the transistor allows us to do is to build a feedback loop that adjusts the control or the drive of the transistor to regulate the output voltage but in this example the series pass regulator is approximately 50 percent efficient so we need a very large transistor that can take 500 watts and that has enough heat sinking to dissipate that 500 watts which is not an easy thing to do let's consider instead how to do this function with switches so instead of an active region transistor we will use it uh semiconductors that are operated like on off switches and here i'm going to draw an ideal switch but we will realize this switch in practice with transistors and diodes that turn on and off so let's suppose we have such a switch we switch it quickly between positions one and two so that the voltage coming out of the switch network from here to here vs of t has this waveform so when the switch is in position 1 the voltage vs is equal to the input voltage vg of 100 volts when the switch is in position 2 the output voltage is 0. and we will operate the switch so half of the time it's in position one the other half it's in position two okay what do we accomplish with this well the switch changes the dc voltage level so here is the this actual waveform coming out of the switch we can uh use fourier analysis to find its dc component and you may recall from fourier analysis that the dc component of a waveform is found by integrating the waveform over one period and then dividing by the period so if you integrate this waveform the integral is the area under this curve which is uh what this width dts times the height vg so we get an area of dts vg and if you divide by the ts then what we find for the dc component is that it is d times v g where d is the duty cycle or the fraction of time that the switch is in position one okay so if the switch is in position one for half of the time then the duty cycle is a half and the out the the output voltage of the switch has a dc component that is half of the input so the switch succeeds then in changing the dc component of the voltage from 100 volts down to 50 volts generally we don't like to have that switched voltage appear across our load so we will put a low pass filter and here's an lc low pass filter that can be put in that if we design this filter correctly so the cutoff frequency is much lower than the switching frequency this filter will pass the dc component but it will reject the switching frequency and its harmonics and not let them pass through to the load so that the output voltage is smooth then essentially is just the dc component of 50 volts so this is a circuit that can change the dc voltage and it does it with components that are ideally lossless only switches inductors and capacitors so it can have efficiencies that approach 100 percent this is known as the buck converter here's one way to realize the switch using transistors and diodes we have a power mosfet and a power diode that's switched together [Music] and i've also shown here a feedback circuit say a simple analog feedback circuit that adjusts the duty cycle it actually turns the transistor on and off and adjusts its duty cycle to regulate the output voltage so this is a buck regulator it's possible to build converters that will change any voltage into any other voltage here's an example of a boost converter that can increase the voltage so we can build circuits that contain inductors capacitors and switches connected in different ways that can actually change any voltage into any other voltage as desired here's another example of a single phase inverter here we have two single pole double throw switches with the load and filter connected differentially between their outputs so that the voltage differentially between the two switch outputs looks like this can switch between plus vg when the switches are in position one or minus v v g when the switches are in position two and by changing the duty cycle we can change the average value and we can actually modulate the duty cycle sinusoidally and make a sine wave of output voltage if we like if we want to produce say a dc to ac inverter okay so in this course you're going to learn how to build these converter circuits how to analyze them and model them to for example predict their efficiencies and also how to design and model their control systems spice is a program that was originally developed at uc berkeley in the 1960s and since then there have been many commercial implementations of it and one of them lt spice is freely available now and i think works very well here's an example which we're going to talk about in this lecture one enters the schematic of a circuit in this case a buck converter and then spice can numerically calculate things of interest in the converter such as the waveforms during a in a transient analysis as well as other things like frequency analysis plots so here is a turn-on transient of a buck converter where the output starts at zero the green waveform here is the output voltage that it computes it goes through some turn-on transient and eventually settles down to a dc voltage that is approximately for the buck converter approximately equal to the input voltage multiplied by the duty cycle also in red here is a plot of the inductor current waveform during the same turn on transient to get started what you should do is follow this link to the lt spice website and from there you can download free copies of the software either for windows operating system or for the macintosh operating system from this coursera website you can also download a zip file containing the buck converter circuit files for this example so when you've done that then you can double click on this file buck.asc which will open the file in lt spice and that file contains the circuit schematic then what you do is press the run button to start the simulation when this simulation will take maybe 10 or 20 seconds to run when it's done then to display a waveform you can click on a node and it will give you the node voltage or you can click on an element and it will plot the the current through the element okay here i have opened lt spice and open to the buck converter schematic what we have here this is the buck converter power stage with the input voltage vg of 24 volts and output voltage that supplies a 5 ohm load resistor here is the lc filter and the switch in the buck converter is realized with this power mosfet and power diode the circuit has a driver and a pulse width modulator these are called behavioral models or mathematical models of these functions that we have developed for this coursera course so the pulse width modulator takes a dc input voltage here v duty that has a value of 0.4 volts dc and it produces an output logic signal c that switches on and off with a duty cycle that is determined by the duty here the duty cycle of c will be 0.4 it has a switching frequency that is set by the pulse width modulator and you can right click on the pulse width modulator and enter the switching frequency in this case the switching frequency is set to 100 kilohertz this signal c goes into the driver the driver produces an output voltage to drive the gate of the mosfet with respect to its source at the proper voltage to turn the mosfet on and off according to the signal c so to run the simulation we put the cursor on the run button the little running person there and click it it will take maybe 10 or 20 seconds for the simulation to run okay the simulation is done piece or lt spice has opened a new window here for plotting waveforms and we can move the cursor over the schematic to different places to plot different waveforms so first let's look at what the pulse width modulator is doing i will move the cursor over the pulse width modulator input the cursor changes into a little voltage probe signal and if you click there it will plot plot that voltage okay so the voltage is 400 millivolts or 0.4 volts we can look at the output voltage that the pulse width modulator makes so i'll move the cursor over c okay this is a the blue signal here is switching up and down with a frequency of 100 kilohertz let's get the magnifying glass and zoom in on part of this maybe over here so the blue waveform is c the control signal it has a switching period of 100 or switching frequency of 100 kilohertz corresponding to a switching period of 1 over 100 kilohertz or 10 microseconds and this signal is high for the duty cycle times 10 microseconds which would be 0.4 times 10 is 4 microseconds okay i'm going to erase these waveforms and now we'll look at some converter waveforms so to erase a waveform what you do is right click on the name of the waveform here on the plot so right click on the pc or you tap the track pad with two fingers on them on the mac and select delete this trace let's plot now the switch node voltage so we have to click on the window of the schematic and then when we hover over the switch node right here at the point between the transistor and diode we can click with the oscilloscope voltage probe symbol and get a plot of the switch node voltage in green so when the mosfet is on for the dts period the switch node is equal to vg like this and we have 24 volts for vg and then when the transistor is off the diode comes on and the switch node voltage is equal to ground essentially and so we have zero volts or a low voltage let's plot the inductor current also so when i hover the mouse over the inductor the the pointer turns into a little current probe to measure current in the inductor in with positive current in the direction shown by the red arrow so let's just click there and the blue trace then is a plot of the inductor current it goes up and down like this it's switching or varying between about 2.6 amps and about 1.1 amp okay finally we can plot the output voltage so we'll take the mouse and hover over the output voltage and click so the red line is the output voltage it looks like it's between 9 and 10 volts now we expect the output voltage of the buck converter to be the duty cycle times the input voltage so v should be equal to the duty cycle times vg vg is 24 volts and the duty cycle is 0.4 so 24 times 0.4 is 10 volts and what we find is the output voltage is a little bit less than 10 volts but it's close the spice simulation includes uh some of the loss mechanisms in the converter it includes the forward voltage drops of the transistor and diode here i put a little resistor in series with the inductor to model the resistance of the wire used to wind the inductor and these things all cause voltage drops that make the output voltage a little less than we would ideally expect we'll model those things in module 3 of this short course and develop some equivalent circuit models that predict more accurately what the output voltage is let's plot the input current waveform i will select the schematic window and hover the mouse over vg you can see that again it turns into a current probe symbol and the arrow is pointing from top to bottom in vg which means current is flowing from plus to minus through vg which is backwards from the direction we actually expect the current to flow so spice is going to plot this current with a negative sign or make it negative here's a plot of it when the transistor is on the current follows the inductor current again with a minus sign when the transistor is off the transistor current is zero and so v the vg current is zero also we also have this current spike here and this is caused from the reverse recovery of the diode and from the transistor and diode switching times this is another source of loss called switching loss which we're going to study in the next short course of this course we can measure voltages and currents using spice what we do is we press the control button and click on the name of the waveform and then a window comes up that tells us the average value and the rms value of the waveform so the current drawn out of vg this i of vg has an average value or dc component of here 707 milliamps so we're drawing 770 milliamps on average of current out of vg this current includes the current spike during the switching times we can use this function to measure input currents and output currents and voltages as well and calculate average powers and efficiencies of the converter likewise we can plot the current that the gate driver draws out of its power supply so what i do again is select the schematic window hover over the power supply for the gate driver and click there and we'll get the current waveform of the gate driver which looks like this the gate driver must draw current out of its power supply to turn the mosfet on and so we see these current spikes happening we so the gate driver requires some power to operate and we can measure the voltage and current of the gate driver and calculate its power as well so i will control click on that waveform and here we see that there's an average current of 14.876 milliamps drawn out of the driver that's the average value of the waveform it's actually coming in spikes that are considerably higher but the average current is the dc component of current and the average power drawn out of the gate driver supply then would be this average current multiplied by the gate driver voltage of 12 volts briefly here are a few more details regarding the models that we have provided for the pulse width modulator block and the gate driver block the functionality of this block is that it produces an output logic signal c having a duty cycle given by this formula so it's a function of vc the the input voltage it can have an offset and it can also has a gain so it's divided by some effective number v sub m you can right click on this block to set values for those things the offset voltage in vm right now vm is set to one volt the offset is set to zero so the output duty cycle is just equal numerically to the dc value vc the block also limits the duni cycle it has to be between minimum and maximum values which can be set as well and you can set the switching frequency again by right clicking on the block and typing the number in the window the gate driver block similarly is a behavioral model that represents key features of gate drivers so what you do is you apply the logic signal c to this input terminal of the driver the driver actually measures its input voltage with respect to this input reference which here is tied to ground and then it produces an output voltage here with respect to the driver vdd6 or vss signal here and what we do is we apply these two terminals to the gate and source of the mosfet to turn the mosfet on and off the output ground or reference of the driver need not be the same as the input reference we also have to supply power to the driver to make the circuit work and so here we've connected a 12 volt power supply between the input power supply vdd and its ground pin vss one last point let's look again at the complete simulation for the output voltage and inductor current so i will again plot the output voltage and we'll plot the inductor current this is a plot from t equals zero all the way to five milliseconds and you can see what happens during this turn on transit all the signals are starting at zero with the converter turned off and then when we turn on the circuit goes through some transient where the output voltage rises and there's some kind of your overshoot and ringing in both the current and the voltage and eventually the circuit settles down into steady state so here's a steady state output voltage and a steady state inductor current we can measure those values as well using spice so we zoom in at some point after the transient is done and then we can control click to measure the average voltage 9.2855 it says so again a little bit less than 10 volts we can also control click on the current to measure the average inductor current it says that is 1.8573 amps okay the homework assignment for this week is to download a boost converter file there's a zip file on the coursera site for the homework assignment you should download spice and get it to run and then run this boost simulation file on your computer and then use lt spice to answer the questions in this homework assignment which are things such as what is the steady state dc output voltage or average output voltage the inductor current calculate the system efficiency by measuring the input power and the output power and dividing and so on we previously introduced the buck converter which is a converter that is capable of stepping down the voltage ideally with very high efficiency the key steps were first to introduce a switch network here is a single pole double throw switch that in practice is built using power transistors and diodes that can switch back and forth very quickly the output voltage of the switch vs of t then is a waveform that steps up and down when the switch is in position one the output voltage vs of t is equal to the input voltage vg likewise when the switch is in position 2 the output voltage is equal to 0. we repetitively switch the the devices or the switch network with a period that is called the switching period t sub s and the switching frequency f sub s then is equal to 1 over t sub s so the switch is in position 1 for a fraction of time called the duty cycle d and the d is a number between 0 and 1 so that we're in position 1 for time d times ts then we're in position 2 for the remainder of the switching period so that would be for at time 1 minus d times ts it's traditional to use the notation that one minus d is equal to d prime and that's called the complement of d so d prime is also a number that is between 0 and 1. okay what do we accomplish then with this switch it reduces the dc component of the voltage so that vs of t has a lower dc component than does vg we previously calculated this dc component using fourier series so from fourier series the dc component is found by integrating the waveform over one period and dividing by the period and that's effectively finding the average value of the waveform so to to find the integral we would integrate the waveform or find the area under the curve that area is in fact easy to find it's simply the area of this rectangle so the area is equal to the base of the rectangle which is d times t s multiplied by the height of the rectangle which is the voltage input voltage vg so we plug that area in for the integral and then divide by the period and what we get is that our dc component of vs is equal to the duty cycle times vg and that dc component is also equal is the average of the waveform so the switch network performs the function then of changing the dc component of the voltage and it actually reduces that dc component by a factor of the duty cycle d now generally we don't like to apply this switched waveform to our load we would like to produce a nice steady output voltage without this switching so but the problem is that the switch waveform in addition to having the desired dc component also has these high frequency ac components in the fourier series and these ac components are at the switching frequency and its harmonics so how should we remove those switching components and just apply the dc component to our load well we do we remove the switching harmonics by introducing a low-pass filter and we can build a low-pass filter using elements that ideally are lossless with a circuit such as this one we can insert an inductor and a capacitor that will perform as a low-pass filter with a cutoff frequency of that we call frequency f naught or f0 [Music] so the low pass filler what the low pass filter does is to pass components in the fourier series that are at frequencies less than the cut off if not and reject the frequency components that are greater than f naught so if we choose the values of l and c to make the cutoff frequency much less than the switching frequency fs then our low pass filter will reject the switching frequency in its harmonics but it will pass the dc component so that the output voltage v of t then will will be essentially constant and equal to the dc component dvg so then we've succeeded in building a converter that has very high efficiency and uses elements that ideally are lossless and not only that but this dc output voltage is adjustable by changing simply the timing of the switching to increase or decrease the duty cycle we can change the output voltage so we can control the output voltage or make it fall of whatever waveform we like okay that is called the buck converter in the power business the term buck denotes a reduction of voltage and so the what we call the conversion ratio of the converter m of d which is defined as the ratio of the dc output voltage to the dc input voltage that m of d is equal to d for the buck converter and so the output voltage is reduced now this is a commonly used and well-known converter but there are many others the boost converter is another well-known converter and it's built by interchanging the positions of the inductor and the switch it turns out that the boost converter can increase the voltage so the output voltage is greater in magnitude than the input and it is controllable by adjustment of d another well-known converter is the buck boost converter shown below here the buck boost converter can invert the polarity of the voltage so with a positive input we get a negative output and the magnitude of the output voltage can be either increased or decreased by adjusting d in fact there are many converters known we can build a converter by some circuit containing l's and c's and embedded switches and change any voltage into any other voltage we like so we found the output voltage of the buck converter using these arguments of fourier analysis and low pass filtering but it may not be immediately obvious how to extend those arguments to find the output voltage of the boost or the buck boost it's actually possible but it may require some additional insight and cleverness and so the object in chapter two is to develop a systematic approach to solve for the voltages and currents of a switching converter okay so we're going to introduce first the key small ripple approximation that lets us solve the waveforms of the circuit with relative ease and simple equations and this small ripple approximation is valid in well-designed converters with that approximation then we will develop the principles of inductor volts second balance and capacitor amp second or charge balance that let us get a system of equations that can be solved for the voltages and currents of the converter we'll also develop some simple methods for choosing the values of inductance and capacitance in the converter and i'll illustrate the use of all of these techniques with some examples in this lecture we will develop the basic analytical techniques that allow us to solve for the voltages and currents of any arbitrary switching converter specifically we'll employ the small ripple approximation to simplify the equations and we'll derive the principles of inductor volts second balance and capacitor charge balance that give us the dc equations of the converter circuit so if you recall from the previous lectures we discussed the buck converter having a switch and having a low pass filter that removes the switching harmonics but allows the dc components or the dc component of the waveform through to the output now in a practical converter we can't build a perfect low-pass filter the low-pass filter will have attenuation but it won't completely remove the switching harmonics of the waveform and therefore the output voltage of a practical converter might look like this where there's a dc component capital v that as we previously found is equal to the duty cycle times the input voltage vg and then in addition to that we have some small ripple that is at the switching frequency and its harmonics that where a small amount of the the switching ripple gets through the filter and appears in the output and as uh sketched here this ripple is actually probably a lot larger than it would be in practice so in a practical circuit we will have some kind of specific specification on how large this switching ripple can be and generally it's a very small number so in a output voltage of say a volt or two for a computer power supply this ripple might be 10 millivolts or some very small number uh so therefore we can ex write the equation of the output voltage v of t as being equal to capital v which i here i'm using the capital letters to denote the dc components so capital v is the desired dc component equal to dvg then plus some ac variation it's called v ripple of t here that is the undesired switching harmonics that make it through the filter to the output voltage so in a well-designed converter this lc filter will have lots of attenuation and the ripple will be very small compared to the dc component so we call this the small ripple approximation where the the ripple is small compared to the desired dc component and under certain circumstances we will neglect the ripple and simply approximate the output voltage by its desired dc component okay this has the effect of decoupling the equations the differential equations of the circuit and making them very simple to solve okay so this is called the small ripple approximation to ignore this component and simply approximate v of t by capital v okay brief discussion here um as we're going to see in a minute the small ripple approximation formally can be applied only to continuous waveforms that have small ripple so we don't apply the small ripple approximation to switched waveforms in the circuit such as the switch output voltage vs of t instead we apply the small ripple approximation to continuous waveforms and specifically the inductor currents and capacitor voltages of the circuit okay so let's let's take an example here of a buck converter where here's our original circuit and we would like to solve for the waveforms such as the inductor current waveform i l of t and the output voltage waveform v of t okay so with the switch in each position the circuit reduces to a linear circuit that we can solve by standard you know sophomore circuit analysis so for example with a switch in position one the circuit reduces to this one in which the left side of the inductor is connected to vg okay this is a second order rlc circuit and we could solve it in fact in principle we can simply solve the second order differential equations of the circuit and with this being a second order circuit in general the kinds of waveforms that we get are sinusoidal or decaying exponential in nature so the actual il of t when we solve this circuit might have some waveform that looks like this for example with some initial value i l of 0 and then some solution of the second order differential equation now in a well-designed converter this low-pass filter has a cut-off frequency that is very low in frequency relative to the switching frequency and what that means is that the time constants or the time that it takes this response this ringing response to to happen is long compared to the switching period so one switching period might be just this long for example and in the small ripple approximation what we're actually doing is approximating this solution for a short time with a straight line and for that reason we also often will call the small ripple approximation instead the linear ripple approximation and any continuous waveform for a short enough time can be approximated by a straight line and so that is what we're in fact doing of course if it's a discontinuous waveform such as a switched waveform then we cannot in general approximate it with a straight line so we apply the small ripple approximation as appropriate to inductor currents and capacitor voltages that have responses such as this one that we can approximate over the switching period or a fraction of the switching period with a straight line okay we can do a similar thing with the switch in the second position as well where the inductor left side of the inductor then is connected to ground instead of vg we get another circuit to solve and we can find the waveforms for that interval as well in a similar manner so let's look at for example for the first interval finding the inductor current waveform using this small ripple approximation so from solving this circuit we have a loop right here that the inductor is connected in and the loop equation for this circuit is the inductor voltage vl is equal to the input voltage vg minus the capacitor voltage here it's called v so we get this equation for the loop that the inductor is connected in okay now v of t is the output capacitor voltage and as i we've discussed uh so far we want this capacitor voltage to have small ripple and so v of t again can be expressed as its dc component capital v plus the small ripple and here is a case where we can replace v of t by its dc component capital v to find the inductor voltage for this interval and so this is a good approximation over the short time of when the switch is in this position okay so this is a small ripple approximation knowing the inductor voltage we can now find um what the inductor current does using the well-known defining equation of an inductor that v is ldi dt in the inductor so if we plug this expression for the approximate inductor voltage into there and solve for d i dt we get this equation which says that the inductor current changes with a slope that is given by the voltage over the inductance with the voltage being equal to a constant value vg minus capital v okay so for this first interval the inductor current will start at some initial value i l of 0 and it will increase with this this slope and it will increase with that constant slope until this first interval is over and the switch is changed to position two which happens at time dts okay then during the second interval we get this circuit with the left side of the inductor connected to ground and the loop equation now around this loop where the inductor is connected says that the inductor voltage vl is equal to minus v like this we can again use the small ripple approximation to replace v of t with its dc component capital v and ignore the ripple during this interval and we can plug that voltage into the defining equation of the inductor again to find the slope of il during the second interval and what we find then is that the inductor current for this interval changes with a slope equal to minus v over l so we started at some initial value we went up during the first interval and ended at some point and then during the second interval we now change with a different slope and the slope is negative assuming the output voltage is positive so we get a slope for this interval of uh minus v over l and the current goes down until the switching period is over and we end at at time ts we end with some final value okay so the small ripple approximation makes it simple to to sketch what the inductor current waveform looks like we have straight lines that have constant slopes and are easy to solve and this greatly simplifies the equations so that we don't have to write decaying exponential type functions and have pages of algebra so then here's a summary of what the inductor current wave voltage and current waveforms look like we found the inductor voltage during each interval it was positive and equal to an approximately constant value during the first interval and during the second interval it was equal to an approximately constant negative value what this did was made the inductor current go up during the first interval and then go down with a negative slope during the second interval okay one thing we can do with this right off the right away is get an equation that helps us choose the value of inductance you can see that the actual inductor current waveform has switching ripple the current goes up and down with a triangle triangular waveform at the switching frequency and on top of that it has an average value or dc component which is the dc value of inductor current that we'll call capital i to design the inductor and choose an inductor value we generally generally will try to limit the amount of switching ripple in the inductor current how much ripple we allow is a design choice and at least for the next several weeks we will discuss converters where this ripple is limited in value to maybe 10 or 20 percent of capital i at full power okay so the ripple we commonly define the peak to average ripple as delta il and so 2 delta il is the peak-to-peak ripple which is from here to here okay and from this waveform and knowing the slope we can easily write an equation that for how large the ripple is so for example during the first sub interval with a switch in position 1 the inductor current changes from here to here and the net change is 2 delta i so we can write 2 delta il is the change in inductor current and this is simply equal to the slope during the first interval vg minus capital v over l times the length of the first interval which is dts okay so knowing the slope and the time we can find the change and we can solve then for delta il just divide both sides by two and we get this expression okay uh it's this is the equation that is commonly used to calculate the value of inductance and choose an inductor for the converter so we can solve this equation for l push l on to that side and get this equation and from this equation we can solve for the value of l that gives a given value of delta i okay there's another thing we can do besides finding the value of l we we would like to find uh the steady state voltage and current in the converter so let's consider now what the inductor current does during a turn-on transient of the converter so for example [Music] let's suppose that we start the circuit with the inductor current equal to zero and the output voltage is zero also so we'll start at 0 with the circuit turned off and at t equals zero we'll start switching with so we'll switch our switch between positions one and two with some fixed duty cycle okay we've already found what happens the inductor current will change with some slope during each interval and so during the first interval the very first switching period the inductor current will go up it with a slope that we've already found is equal to vg minus v over l okay well initially what v is the capacitor voltage vc in fact let's just call this v instead of vc so the voltage is initially zero at the output and our slope during the first interval is vg over l okay so we go up until at the end of the first interval at time dts we've reached some positive current okay what happens to the output voltage well this inductor current is small but it's a little more than zero that inductor current flows to the output and it will slightly charge the capacitor voltage up a little so v is still small but it's a tiny bit more than zero okay during the second swi interval with the switch in position two we found that the inductor current changed with a slope of minus v over l okay so we have a slope here of minus v over l well v that the output voltage is nearly zero so minus v over l is pretty close to zero and there's hardly any change in inductor current during this second interval because the slope is nearly zero okay after one switching period we end up with the inductor current at what il of ts a little bit positive it's it's a little greater than zero now we repeat the process so for the second switching period we'll go up with a slope of vg minus v over l v is a little positive now so the slope during the second interval this slope is a little bit less than that slope was last time so we don't quite increase as much okay and then during the second with the switch in position 2 for this switching period will go down with a slope of minus v over l v is a little more positive than it used to be so this slope is a little more negative than it was during the previous period and we end up with some net increase in current but it's not as much as we increased in the first period and we we repeat this process okay with the output voltage slowly charging and eventually after many switching periods we we end up at some current let's let me just draw it out here and and some output voltage where the amount we go up during the the first interval with this slope of vg minus v over l is exactly equal to the amount we decrease during the second swit second interval where the slope is minus v over l so that we end up at the same current we started the capacitor voltage does something also and it ends up at the same place it started okay and so at this point this is maybe at time nts and here's n plus 1 ts so after in switching periods where n is some large number we end up in steady state where the the waveforms become periodic each switching period has waveforms that are the same as the previous switching period and the current and voltage end up at the same place they started so you can see that that will happen in when the output voltage rises to a large enough value so v is large enough so that these slopes have adjusted and we our final value i l of n plus 1 ts equals il of nts and at that point we say that we're in steady state so the waveforms from then on um have the correct dc values of where the converter is supposed to operate we've gone through some transient that eventually has settled out and we've reached steady state okay it is possible to find a simple relationship that expresses that the circuit works in steady state and it comes from this notion that we end up at the same current that we started at uh for the inductor and we can derive it in fact from the defining relationship of the inductor the v is ldidt what we do is we integrate this equation to find the net change in inductor current over one switching period so what we can do is integrate both sides of the equation so integrate vl of t dt over one switching period so we'll integrate say from zero to ts over one switching period and here we'll integrate from i l of zero to i l of t s and when we evaluate these integrals the right hand side gives us il of ts minus il of 0 which is the net change in current if i divide both sides by l i can write the other side of the equation as 1 over l times the integral of the voltage over one period so in steady state then the net change in inductor current must be zero and therefore this integral of the voltage must be zero okay and this is the basic relationship that where we say this is volt second balance on the inductor that the integral of the voltage has dimensions of volts seconds and if you integrate the inductor voltage over one complete period that integral or area under the curve must be zero if the inductor current has no net change over what that period another way to write this equation is simply to divide both sides by the switching period ts and in that case we recognize that 1 over ts times the integral of the voltage is the dc component so another way to say this is that the dc component of the voltage applied to the adductor is zero when the circuit operates in steady state we're going to use this notation of angle brackets around a variable to mean the average value so the average value of the inductor voltage is zero in steady state okay let's try that try finding the applied volt seconds to the inductor in our buck converter we previously found this waveform was the voltage waveform applied to the inductor the inductor saw voltage of vg minus v during the first interval and minus v during the second so we have to integrate the inductor voltage and the easy way to do that is simply to find the area under the curve so during this first interval we have an area that is the base dts times the height the height is vg minus v during the second interval we have an area [Music] that is the base here d prime ts is the length of the second interval times the height which is minus v so if we add these two areas they have to add up to zero okay so here is that area and if we equate it to zero like this we get an equation and we can simplify this equation d plus d prime is in fact one i'll call d prime is one minus d so the sum of d plus d prime work works out to be one and we get dvg minus v equals zero you can solve this for the output voltage v and we find that the output voltage is dvg okay well we already knew that for the buck converter but in fact this was a way to find that the output voltage without using the arguments of fourier series and low pass filtering here we we simply draw the inductor voltage waveform and set its average value to zero and this is something we can do for any converter so we could similarly use this approach to find the output voltage of say the boost converter or the buck boost converter so volt second balance plus the small ripple approximation then gives us a way to solve for the output voltage okay capacitor charge balance or capacitor amp second balance is the dual of inductor volt second balance this is a we can apply a similar principle to a capacitor in the circuit what we do is we start with the defining equation of the capacitor i is c dv dt in the capacitor we can integrate both sides of this so integral of ic dt is the c times the integral of dvc and if you integrate over one period from say 0 to ts then this part here gives us the change in capacitor voltage over one period the net change is vc of ts minus vc of zero and that's equal to one over c times the integral of the current if we operate in steady state then there is also no net change in capacitor voltage over one period so the left-hand side of this equation is zero and this says the integral of the capacitor current must be zero when the converter operates in steady state again another way to write this we can divide by ts multiply by c and find that the dc component of capacitor current must be zero in steady state okay so we can use the principle of capacitor charge balance to get a similar equation to find the conditions under which the capacitor voltage will operate in steady state this might be more familiar than the inductor volt second balance or perhaps more intuitive if you put a dc component of current into a capacitor then the capacitor will continue to charge and if you keep putting charge on the plates of the capacitor just positive charge say then the charge will build up and the capacitor voltage will increase and it won't operate in steady state if for part of the time we put a positive current say part of the period we put a positive current on the capacitor making its voltage increase for the other part of the period we must put a negative current to bring the charge back down to have no net change in charge and therefore have no net change in capacitor voltage in this lecture we will work a boost converter example so we're going to apply the techniques of volt second balance capacitor charge balance and the small ripple approximation to find the voltages and currents of the boost converter so here i've shown a boost converter using an ideal switch so the switch is in position one for this first interval of time dts and it is in position 2 for a second interval of time d prime ts i've also labeled the inductor voltage and current as well as the capacitor voltage in current incidentally here is how we in practice normally realize the the switch using power transistors and diodes so there's a gate drive circuit that's just shown here as a voltage source but when we want the switch to be in position one this gate driver will produce a high voltage on the gate of the mosfet that will turn it on and when it's on it effectively pulls this node down to ground just like when the the switch is in position one for the switch in position two what we do is the gate driver turns off the mosfet by making its gate voltage low when the mosfet is turned off the inductor will forward bias the diode turning it on and putting the switch effectively in position two now one thing you might wonder is why does can we put a diode here and why does it automatically come on well one way to look at this is to go back to the formula for the inductor that the inductor voltage is l d i l dt which is the voltage from plus to minus across here so when the mosfet is on we build up some positive current flowing through the inductor and through the mosfet when we turn the mosfet off one way to think about it is we're effectively trying to interrupt the inductor current so the inductor current that was positive is now trying to be forced by the mosfet turning off to go immediately to zero so what is d i l dt well d i l dt is big and negative if we're shutting it off you could think of it as going heading towards minus infinity for slope but maybe in practice it isn't the idt isn't that large but it is negative and and large in magnitude so if the ildt is negative say let's call it minus infinity what that does is it will make vl become big and negative well vl is plus to minus in this direction that's the reference direction so if vl is a negative number effectively it means that the negative terminal is becoming positive and the positive terminal is going down so this node here where the mosfet is connected will head to a large positive voltage from the voltage developed in the inductor so it heads towards plus infinity but but long before it gets there it forward biases the diode so in fact when this voltage rises above the output voltage the diode will turn on and clamp this voltage to the output and once it does turn on then it provides a path for the inductor current to flow the inductor is happy and the current no longer tries to change so we can put a mosfet and a diode that will operate automatically with the mosfet sometimes in the business a diode that operates in this fashion is called a freewheeling diode okay so what we would like to do now is apply a volt second balance to the inductor and charge balance to the capacitor to solve for the voltages and currents of the converter in steady state so here's how we do it first of all on the original circuit it's important to clearly label the reference polarities of the voltages and currents and if you're not very careful about this it's very easy to make minus sign errors and get the wrong answer so to head off a lot of problems coming up soon let's label the the waveforms or the variables directly on the original converter so i've defined the inductor voltage and arbitrarily chosen its reference direction is plus on the left and minus on the right which means that if the variable vl of t is positive the voltage is in the direction shown and if vl of t is a negative number then the actual voltage will have the other polarity that's an arbitrary choice of the direction but we pick some direction for now okay once we've defined a direction for the voltage the inductor current must be defined with in a direction that flows through the inductor from the plus terminal to the minus terminal of the voltage so il of t is a quantity having a reference direction that goes from left to right through the inductor likewise here i've drawn the output voltage which coincides with the capacitor voltage as being plus on top and minus on the bottom and so the capacitor current must be defined as flowing through the capacitor from the plus terminal to the minus terminal in this direction so positive in capacitor current is flowing in this direction okay now we draw the circuit with a switch in the two positions so on the left bottom left here is the circuit that we get with the switch in position one with the right hand side of the inductor connected to ground like this and for this circuit i now carefully copy over the the directions of the voltages and currents that were labeled on the original circuit so vl of t is shown here i l of t is shown there etc and you have to keep the same polarities likewise for the switch in position 2 the right side of the inductor is connected to the output like this and i again copy the the voltage and current reference directions for the switch in position 2 and use the same reference directions as in the first position okay having done this now we can write the expressions for the voltage and current in each case so with a switch in position one we got this circuit the mosfet is on and the right side of the inductor is connected to ground so the inductor voltage here we can see is equal to the input voltage vg like this and the capacitor current we can see flows this way and it's equal to minus the load current so the load current here is v over r and from the node equation at the output node the capacitor current equals minus v over r like that these are the two things we need to know for this interval in order to apply volt second balance to the inductor and charge balance to the capacitor okay next we make the small ripple approximation and in particular we have the output capacitor voltage right here v of t which we will approximate with its dc component capital v like that so for the first interval we have an inductor voltage vl that's equal to vg and we have a capacitor current that's equal to minus v over r okay let's do the same thing for the second interval with a switch in position two the right hand side of the inductor is connected to the output node so we get this circuit and our inductor voltage now is vg minus v we'll add vg on the left minus v on the right and the capacitor current here we can express from the node equation at the output node as being equal to the inductor current il minus the load current v over r so we get these equations we again make the small ripple approximation so v of t can be replaced by its dc component capital v and also we have an inductor current term here we're going to make the small ripple approximation on it and replace the inductor current il of t with its dc component capital i or or capital il here it's on this slide it's called capital i okay let's sketch the waveforms now so the inductor voltage waveform we need to sketch in order to apply apply a volt second balance and what we found was that the inductor voltage during the first interval was equal to vg and that's for interval of length dts okay that's supposed to be a straight line equal to capital vg during the second interval the inductor voltage we found was vg minus v now where should i draw vg minus v is that a positive number or a negative number and why well it if the volt seconds are going to balance so the average vl is zero and if we have a positive voltage during the first interval then the voltage had better be negative during the second interval if we were to have any chance of making the average come out to be zero so i'm going to draw this this voltage during the second interval as a negative quantity and you can see that since the voltage has to be negative to get the volt seconds to balance v needs to be bigger than vg so therefore this is a boost converter where the output voltage v is bigger than the input voltage vg and volt seconds won't balance otherwise so this is during interval d the second interval of length d prime ts that's our inductor voltage waveform we can now apply a volt second balance to this waveform so the average vl will be well we can find the area under the curves and divide by the period but an easier way to find the average is simply to take the value during the first interval vg multiply that by the duty cycle d and add to that the voltage during the second interval vg minus v multiplied by d prime and that will give us the average so in steady state by volt second balance this must be equal to zero and now we have an equation that we can solve to find the output voltage let's let's in fact solve it let's collect terms we have vg times let's see d plus d prime then we have minus d prime v equals zero recall d plus d prime equals one so we have d prime v equals vg and solve for v we get v equals one over d prime times v g okay so that's the equation for the output voltage of the boost converter in steady state since d prime is a number less than one v is greater than vg given this voltage waveform we can also find the inductor current waveform well we don't know where it starts but in steady state it starts at some current we'll call it il of zero and for the first interval it increases because we have positive voltage so again recall the inductor voltage is ldi dt so then the slope d i l dt is the inductor voltage over l so when we have positive voltage we have positive slope in the current the current goes up with a slope of the applied voltage vg over l and during the second interval we have negative voltage so we get a negative slope and the inductor current goes back down with a slope of [Music] vg minus v over l if we're in steady state we end up at the same current that we started at okay so there's the inductor voltage waveform that i drew here's the volt second balance that we just worked out and here's a plot this is the plot of the the conversion ratio of the boost converter which is the output voltage over the input voltage and it's the one over d prime factor at d of zero we have d prime is one and the conversion ratio is one meaning that the output voltage equals the input voltage and as we increase the duty cycle this function gets larger and larger and in fact as d goes to 1 d prime goes to 0 and the ideal conversion ratio function heads to infinity at d of 1. now of course real converters won't do that they won't make infinite voltage and we'll find next week that the maximum output voltage of the boost converter is limited by losses but we haven't modeled any losses yet and so to the extent that the converter is is ideal we can get conversion ratios that approach very large numbers okay to find the dc component of the inductor current we apply charge balance to the capacitor so here is the capacitor current waveform we found that it was minus the load current during the first interval during the second interval it is the inductor current minus the load current and to get the average to be zero that second interval had better come out to be a positive current and zero is right there so we can find the average capacitor current or the dc component and by charge balance set that average current to zero in in steady state so the average capacitor current will be d times the value during the first interval minus v over r plus d prime times the value during the second interval i minus v over r and that's equal to zero in steady state so given that that's we that's this equation we can solve this equation for capital i um again this quantity here is one and when you solve for i you find that the inductor current dc component capital i is equal to the load current v over r times one over d prime so the relationship between the inductor current and the load current is the same one over d prime factor just like the the conversion ratio of the converter we may also want to you know since v is a variable we may want to replace v with its solution vg over d prime plug that in for v and you get this equation that is the solution for the inductor current dc component as a function of vg the load resistance in d prime so charge balance gives us an equation for the average inductor current and if we're designing a converter and we need to know what the currents are then we'll need this expression as well okay we can also find the current ripple we previously drew the inductor current waveform and we know the slopes so we can find delta il the peak to average inductor current ripple here's capital i the average current is the dc current we just found the ripple delta il can be found from the slope we simply take the slope [Music] which during the first interval is vg over l multiply it by the length of the interval dts that gives us the peak to peak ripple which is 2 delta il and we get this expression i can divide by 2 and find the peak to average ripple delta il and we get this equation which we can use to select the inductor to get a given ripple if we like i would also note that the peak current in our inductor which is also the peak current in our transistor and diode is the dc component capital i plus the ripple generally when we design a power converter we want to limit the peak current to a value that our components can handle and so we'll want to control the value of delta il capital i is determined by the load current but delta il is determined by the choice of inductor value so we can choose the inductance here to make the peak current be a desired value that is controlled and not too large we can do a similar thing to find the capacitor voltage ripple we found the capacitor current and from the capacitor current we can that tells us the slope and we can find the ripple on the capacitor okay so what we found was that for the first interval the capacitor current was minus the load current so the load current will discharge the capacitor and the capacitor voltage will go down with a slope that is the current divided by c okay from here so knowing the slope we can multiply the the slope which is this by the length of the first interval which is d prime ts and that's equal to the peak to peak ripple or 2 delta v so we can solve for delta v and find a value of capacitance then that will make the output voltage ripple um be sufficiently small according to whatever specification we have and the only thing i would add to this is that oftentimes in practice capacitors have significant series resistance it's called equivalent series resistance so the model for a practical capacitor especially say an electrolytic capacitor is this equivalent series resistance or esr in series with an ideal capacitor c and the ripple that we've just found is the voltage ripple on the ideal part of the capacitor but to that we have to add the voltage drop across the equivalent series resistance so if we know ic then you can multiply ic by the esr and get the voltage waveform across the esr and then add that to this voltage ripple waveform that i've just drawn which would be the voltage ripple on the ideal part of the capacitor so we've done a boost converter example and we've used the principles of inductor volt second balance and capacitor charge balance to find the dc voltage and dc currents of the converter and we've been able to derive the equation for example for the output voltage as a function of input voltage we've also been able to derive equations for the ripples in the inductor current and capacitor voltage and use them to choose the values of the capacitance we'll now briefly discuss two more finer points the first one is an example of a higher order power converter so here in section 2.4 the slides contain the full analysis of this particular converter it's an example of a converter that can contains two inductors and two capacitors it's named after its inventor dr slobodan chuck and it's called a chook converter [Music] and so here here's the converter circuit with two l's and two c's here is how the switches can be realized with a transistor and diode okay so the same analysis approaches that we've discussed so far can be applied to this converter as well but you simply have to realize that since we have two inductors we we need to apply volt second balance to each inductor and with two capacitors we need to apply charge balance to each capacitor so this will give us four equations two volts second balance equations and two charge balance equations and there are four things to solve the two capacitor dc voltages and the two inductor dc currents so we use the small ripple approximation on each inductor current and each capacitor voltage we draw their waveforms and apply the volt second balance and charge balance to get four equations now i'm not going to go through them all but for those of you who are interested you can look through the slides for this section and go through the the volt second balance and charge balance equations the result of all of that is shown here where we have volts second balance on the two inductors and charge balance on the two capacitors so we have a total of four equations we have two unknowns or four unknowns the two capacitor voltages v1 and v2 and the two inductor currents i1 and i2 so we can solve this system of equations and if we do we get this solution for the dc voltages and currents of the converter so the point of this example then is that you can apply the volt second balance and charge balance techniques to more complex converters we have to apply volt second balance on every inductor and charge balance on every capacitor to get a system of of equations that can be solved then for the the converter steady state here incidentally is the result this is the conversion ratio of the converter v2 coincides with the output voltage and it turns out to have that this converter has an inverting buck boost type characteristic we can also work out the ripples just as in the single inductor single capacitor case to find equations that help us choose values of the inductor the inductances and capacitances okay one other fine point is a case such as in the output filter of the buck converter where we want to calculate the output voltage ripple on the output capacitor and choose a value for that capacitance okay this is a case where there is no switching associated with the output node where the capacitor is connected instead we have a two pole lc filter and the current in the capacitor has ripple only and so what we've talked about so far to find delta v on this capacitor using a small ripple approximation doesn't work to illustrate that let's look at what is the capacitor current well it's the same for both intervals for both uh the first and second intervals we have the same node equation at the output node which says that the capacitor current ic is the inductor current il minus the load current vc over r so it doesn't switch and so i l is the current coming out of the inductor the inductor current as we've drawn uh looks like this uh it has a dc component and it has some switching ripple so first question is where does the dc component of the inductor current go well it can't go through the capacitor by charge balance the dc component of capacitor current is zero so all of the dc component of inductor current in steady state at least must flow through the load okay so the dc component goes this way through the load now how about the ripple well in a well-designed filter where the output capacitor actually does some filtering and then its impedance is low at high frequency we put a large enough capacitor so that it has a very low impedance at the switching frequency and its harmonics and so the switching ripple coming out of the inductor which is this triangular component of of the inductor current that switching ripple will divide between the impedances of the two branches here between the capacitor branch and the load branch and in a well-designed filter the capacitor will have a low enough impedance that nearly all of the ripple will flow through the capacitor and very little will go to the load causing ripple in the load so a good approximation is to say simply that all of the inductor current ripple flows through the capacitor so let's make that approximation and in that case the capacitor current looks like this it's simply the the same as the inductor current ripple but with no dc component okay now first of all suppose we make the small ripple approximation on the inductor current what that does is ignore the ripple and in that case the capacitor current is zero which then predicts that there's no voltage ripple so this is not a useful approximation in this case because the ripple is the only thing the current ripple is the only thing that makes voltage ripple in the capacitor so we have to include the capacitor current ripple and calculate the resulting capacitor voltage ripple okay we can do that using the defining equation of the capacitor ic is c times dvc dt so dvc dt is ic over c and that means that whenever the capacitor current is positive we charge up the capacitor and increase its voltage so here you can see for this time as drawn we have positive capacitor current positive capacitor current means the slope of the voltage is positive and the capacitor voltage increases in fact at the zero crossing of the current that's where the derivative of the voltage is zero so those are the extrema of the voltage waveform and between the this minimum and maximum of the voltage we have positive current that makes the voltage increase so over this time the voltage increases from here to here okay because of symmetry the the voltage waveform is symmetrical and we can again define the ripple delta v as being the peak to average and we have one delta v there and another delta v here so over this interval the capacitor voltage increases by a net 2 delta v okay we can find the um the relationship between this change is 2 delta v and the total charge contained in the current waveform using the defining equation for the capacitor of q equals cv and in this case over this interval where we're charging and increasing the voltage the net change in voltage again is 2 delta v and that's produced by a total charge that is the area under the current curve total charge is the integral of the current so this total charge q increases the voltage by 2 delta v and we can then apply q equals c times 2 delta v to find the relation or to find the delta v all that's left then is to calculate the area of this triangle or the total charge so the area of the triangle we can use the area formula for a triangle one half the base times the height the the height of the triangle is the inductor current ripple delta il and the base is this distance which by symmetry is half of the switching period so the total charge then is given by this equation from the triangle formula the triangle area formula we can plug that in to our q equals cv formula and solve for delta v and we get this expression for the the delta v on the the capacitor and again i would point out that if we have a capacitor with significant equivalent series resistance we must also add the the voltage drop in the equivalent resistance to to this to get the total voltage ripple on the capacitor so in cases where we have two pole output filters or higher order output filters we can't ignore the ripple and so we have to use arguments such as this in order to calculate the ripple this week we will cover chapter three concerned with a steady state equivalent circuit modeling of switching converters the object here is to develop equivalent circuits that model the important functions that the switching converter performs while ignoring uh the unimportant things such as the switching ripple so what we want to do is model the important low frequency and in this case dc components of the voltages and currents of the converter and we'll use these models to gain insight and understanding of how the circuit and converter works how it performs the important functions associated with these dc com components of the waveforms and we will use it to model the losses in particular and the efficiency of the converter we'll find that by use of equivalent circuits we can gain a lot of insight as to for example how to to develop approximations for what is important and what is not and how to handle more complex circuits for example if we want to to include all of the sources of loss in our model we can refine the model and add these loss elements which can become fairly complicated but we can handle them in a way by manipulating the circuit that lets us easily get the answers and the insight that we need later in the course we're going to refine these models to include a phenomenon known as switching loss and even later we're going to extend the models to model converter dynamics be able to find things like small signal transfer functions and say output impedances and to design the feedback loops of these converters okay so in this lecture i want to talk first about the equivalent circuit model of an ideal dc-dc converter okay what are the basic properties of an ideal dc-dc converter well first of all we're going to say that an ideal converter is 100 efficient and to begin with we won't model the losses later in the next upcoming lectures we'll refine our model and add losses in but we're going to start with the ideal case so we if we have a switching converter with an input voltage vg applied to the input port of the converter and what we'll call the input current ig and if we have an output port of our converter with a voltage v with a current that we'll label i then we can write that the input power equals the output power ideally so vg ig equals v times i i should add that in addition to this hundred percent efficiency requirement these equations also implicitly assume that the converter operates in steady state so for example if there are inductors and capacitors inside the converter that can store energy then during transients those reactive elements may store or release energy and during the time that they're doing that the input and output powers could differ but in steady state there's no net change in energy in these are stored energy in the reactive elements and then we get this first basic equation okay the second thing we found uh in the last chapter was that we could solve the converters and find that the output voltage is some function of duty cycle times the input voltage and this function of duty cycle we called the conversion ratio m and m is a function of d so for the buck converter m was equal to d and for the boost it was equal to one over d prime and so on okay there's a third equation that really comes from the first two that if you take v from equation two and substitute it into equation one then well let's do it we get v g i g on the left side of equation one is equal to v which is m times v v g times the output current i okay we can cancel the vg's and what we find is that the input current ig equals the same conversion ratio times the output current i and that's our third equation so it says that the [Music] the currents are related by the same conversion ratio except going in the opposite direction okay so those are three basic equations for the converter now what kind of equivalent circuit could we construct to go with those three equations okay well i'm going to actually draw that here so we had what vgi g equals v i was the first one second one was v equals m of d times v and the third was ig equals m of d times i okay now there's actually two different ways that we commonly model these equations with equivalent circuits the first one uses dependent sources so let's suppose let's consider our converter with an input port having vg and ig and its output port has voltage v and current i and one way to model these equations is uh with a dependent voltage source for equation two which says that the output voltage v is dependent on v g with a constant or gain of m of d so i can draw a dependent voltage source and label its value m times vg here in my notation a square source is a dependent source okay the third equation we could similarly model with a dependent source the current ig is equal to this gain m times the current i so i could write a dependent current source that produces ig and label its value m times i so this is one possible equivalent circuit that models the important low freq actually low frequency or dc voltages and currents at the terminals of the converter one can verify that equation 1 is also satisfied in fact we derive 3 using 1 but vg times mi is equal to mvg times i and we can verify from these equations that that's so this is one way to model a switching converter and in fact we often use this model in circuit simulators such as spice which have dependent sources in them that can be easily programmed to do this so that's approach number one a second approach is to recognize that these equations are actually the equations of an ideal transformer so let's draw our terminal quantities and put a transformer between them so a transformer has input power equals output power at least ideally and it also has the voltages are related by a factor which in the transformer is called the turns ratio so we could write a transformer symbol or an ideal transformer symbol and the turns ratio is the ratio of the voltages so here v is equal to the turns ratio times vg and the turns ratio apparently is the conversion ratio m so i can label the turns ratio here as 1 to m of d and we know with a transformer if we apply a voltage vg to the primary with say the plus on the dot top i'm going to put a polarity mark there where the plus sign of vg is then the voltage coming out of the transformer here will be the turns ratio times the input voltage and it's plus at the top also so the polarity mark goes on the top of the secondary winding we also know for transformers the currents follow the same rules but in the opposite direction so here the input current ig flowing into the dot is equal to the current on the secondary flowing out of the dot multiplied by the turns ratio so m of d times i okay so this is the ideal transformer model of a switching converter and it's justified because it correctly um describes the relationships between the terminal voltages and currents of the the ideal converter now i should note that real physical magnetic transformers that are constructed with turns of wire on magnetic cores can't even handle dc signals if you put a dc voltage across a trans a magnetic transformer winding the volt seconds won't balance the tree and the transformer will saturate but that's perhaps one of the good reasons to build a dc dc converter because we want that function and so here what we're going to do is we're going to make sure we all understand that this is an ideal dc transformer model by putting a straight line like this through the transformer to represent dc [Music] and we're we're at we are free to invent an equivalent circuit symbol that that models these functions and it's defined simply as being a symbol that satisfies equations one through three okay so we have two different ways to model the the terminal equations of the ideal dc dc converter and they're completely equivalent and in fact if you find that you have if you construct an equivalent circuit that has one of these you can always substitute the other for it okay so again the dc transformer model is justified because it models the important physical properties of the desired waveforms of the dc-dc converter which is conversion of the voltages and currents according to a conversion ratio m with 100 efficiency and this is a now a time invariant model that doesn't have switching so we can replace our switching converter with this transformer model and we now have we then have a circuit that can be much much more easily solved using the techniques of just basic undergraduate circuit analysis so we can do dc analysis and solve for the important quantities without worrying about what's going on with the switching and the switching ripple the only slightly strange thing here is that the turns ratio m of d is a function of duty cycle so if we change the duty cycle we can change the turns ratio at any time and if you want to turn down the output voltage just turn down the turns ratio but that's one of the additional important functions that the converter can do and it is modeled okay let's do an example of using the transformer model so here's an example of a circuit in fact let's do this on the graph paper as well so let's suppose we have a switching converter it has its input terminal vg and output terminal v and we model it with a transformer so we will draw inside the box of the switching converter a dc transformer turns ratio 1 to m and let's suppose that at the input terminals of this converter we have a power source that has some impedance associated with it so we'll model the the power source with a thevenin equivalent so i'm going to draw v1 as the thevenin equivalent voltage and some thevenin equivalent resistance that we'll call r1 that is connected to the input terminal let's also suppose we connect a load let's say a resistor r on the output and the object now is to solve for what is the output voltage well the way to do this i think the easy way to solve this is to manipulate the circuit by pushing all of the elements through to the output side of the transformer so we remove the transformer and then we then we can solve for the output voltage so if you recall from basic circuit analysis when you push a voltage source through a transformer then effectively on the output or secondary terminals of the transformer we see a voltage source of value v times the turns ratio so i'm going to write a voltage source the round source is an independent source it's a value v1 times the turns ratio m then we also need to push the resistor r through the the turns ratio and when we push a resistance through a turns ratio we must multiply by the turns ratio squared so we have an effective resistance as seen from the secondary terminals of value m squared times r1 and that's what we get between the output terminals of the converter now this is drawing or driving a load r so we'll draw r here and we have output voltage v okay now the circuit is simplified and i think it's a straightforward matter now to write the expression for the output voltage the output voltage v is equal to the volt the effective voltage source m times v1 times the divider ratio of these two resistors which would be r over r plus m squared r1 and that's the equation for the output voltage so the conversion ratio m affects not only the voltage source but also this divider ratio and the taken together we can solve for the output voltage okay so we've derived an equivalent circuit for the ideal switching converter that models the important dc components of the waveforms and the basic power conversion functions that are performed by the converter in the upcoming lectures we're going to refine this model next to add in the different kinds of types of losses in the converter and build up a model an equivalent circuit model that can be solved to find things like the efficiency i would also note that if you found the derivation of the output voltage by manipulating the ideal transformer circuit to be unfamiliar or you need if you need a brush up on how to do that there's a supplementary lecture posted to supplement this lecture that reviews the rules for manipulating ideal transformer circuits this brief video is a review of a circuit manipulations using ideal transformers the basic rules for ideal transformers in which we push an element a circuit element through a transformer to what we what we call reflecting to the other side goes as follows if we have a voltage and we push it through a transformer that has a turns ratio of n1 to n2 then as seen between the secondary terminals of the transformer effectively we have a voltage source whose value is multiplied by the turns ratio so we get n2 over n1 times v as the effective voltage source is seen by the second from the secondary terminals likewise with the current source we can push it through a transformer also but we have to divide by the turns ratio so if we have an n1 to n2 turns ratio here then as seen between the secondary terminals we effectively have a current source of value y that is divided by n2 over n1 or we can write n1 over n2 times i and the third manipulation is if we have some impedance we'll call it z and we push that through a transformer then we effectively see an impedance between the secondary terminals that is the turns ratio squared times z so we would get n2 over n1 squared z as the effective impedance seen between the secondary terminals one little case to be careful with here with z is if z is a resistor so if we have a resistor then effectively between the secondary terminals the value will be the resistance times the turns ratio squared if we have an inductor whose impedance is sl then we would multiply that impedance by the same turns ratio squared which is as if the inductance was multiplied by n2 over n1 squared so sometimes we'll just simplify this to say there's an effective inductor a value into over n one squared l if we have a capacitor recall that the capacitor the impedance of a capacitor is one over sc so the impedance here would be n2 over n1 squared times one over sc as far as the effective value of the capacitance you can see that since impedance varies inversely with capacitance this is as if we have a capacitor that gets divided by the turns ratio squared instead of multiplied so it's as if we have a capacitor value c that is n1 squared over n2 squared okay those are the rules for pushing simple elements through transformers let's do an example now suppose we have say a voltage source now let's take some larger circuit maybe with a resistor r1 some transformer with the turns ratio let's say one to in a maybe we have another resistor here the value r2 and another transformer that has the dots reversed so the primary dot is on top and the secondary dot is on the bottom of the winding and let's give this a turns ratio of in b to 1. and what we would like to do is find the effective network with respect to the output terminals and in fact maybe we'll we'll even solve for the output voltage v out so to one good way to solve this circuit is to use these circuit manipulations and push the elements through the transformers so what i'm going to do here is push them through the transformers one at a time so let's start by pushing the voltage source and r1 through the first transformer so what we will get with respect to the secondary terminals of this first transformer we'll have the voltage source pushed through and in the process it gets multiplied by n a so we have n sub a v as the voltage the effective voltage and it's in series with an effective resistor that would be the resistance r1 multiplied by the turns ratio squared so n a squared that's what we see effectively between the secondary terminals of this first transformer okay so then i'll copy the rest of the circuit after that okay next let's push the elements through the second transformer okay so here is our still our output terminals so we'll push the elements to see how they appear effectively between the secondary terminals of this second transformer so when we push r2 through the transformer we'll get a resistor on the secondary side and it gets multiplied by the turns ratio squared but here the turns ratio is 1 over nb so effectively we would get r2 over in b squared as the effect of resistance the fact that the dots are reversed doesn't affect impedances because we the dot is likely having a minus sign on the turns ratio but the impedance changes by the turns ratio squared and the minus sign cancels out okay next let's push the res r1 resistor through so we'll get an effective resistor here that is the n a squared r1 and then we have to divide by the nb squared for this impedance also okay and then finally we push the voltage source through the transformer so we have a voltage of value in a v and then we'll divide by this in b and since the dots are reversed we have to reverse the polarity of the voltage source so i'm going to draw it with minus on top and plus on the bottom okay so this is the effective network that is seen between the secondary terminals of this the last transformer with this circuit now we can solve so what is the output voltage we could just use the divider voltage divider formula so we would take the the voltage source in a over in b times v and because the polarity is reversed we have a minus sign and then that voltage gets multiplied by the divider ratio of these two resistors so it would be times the divider ratio r2 over nb squared all over what the same r2 over in b squared plus the first resistor in a squared over nb squared times r1 so that is the solution then for the output voltage so we've taken a fairly complex circuit and by manipulating it reduced it to something where we can just write the answer in the end one other thing i would point out is that you do have to be a little bit careful in manipulating here if we had wanted to solve for some other signal that's not at the output like suppose we wanted to find this current right here i'm going to call that i a um i a does not explicitly appear in this final circuit and in fact when we push the elements through the transformers we have to be careful on the second circuit ia is this current and what happens to ia on the third circuit well when we push these elements through the transformer we end up pushing ia through the transformer also so here's here's the current related to ia but ia got pushed through this turns ratio and the current on this side would actually be in b times ia okay so we could solve this circuit for that current but we would be solving for nbia and then we have to divide by in b to get back what the actual ia is as a first example of modeling loss in a converter let's model the resistance of the wire in the inductor of a boost converter the resistance of the winding or the dc copper loss of an inductor is important [Music] because it has a first order effect on the cost and efficiency of of the inductor and the converter if you don't care how large this resistance is i can wind you an inductor that is arbitrarily small and inexpensive that can give you the desired inductance uh and handle the desired current without saturating but it will use many turns of very thin wire that will have a very high resistance so the inductor then will actually have very large loss and will burn up so the resistance of the wire is important and one of the important things we have to specify and design in order to optimize the cost and efficiency of our converter design so to model this uh a common way to do it is with a lumped element model as shown here where we replace the physical inductor with a model consisting of an ideal inductor having a voltage vl in series with a resistor that is the resistance of the winding okay so between these terminals this is the model of the physical inductor and what we do is we insert this model into the converter circuit like this and then analyze this circuit so here is our inductor model okay the ideal part of the inductor with its voltage v sub l uh still must obey inductor volts second balance but now we have an added resistor in series in the circuit that has a voltage drop equal to the inductor current multiplied by uh the resistance so we apply volt second balance to vl of t in this model and as usual we apply charge balance to the capacitor current and we can get a set of equations that we can then solve for the output voltage and other quantities of the converter okay so let's do it here's our here's our circuit now to be analyzed here's the inductor voltage or the voltage of the ideal part of the inductor model and we'll have a the capacitor current there and as usual we now write the circuit with a switch in the two positions and proceed to work out the waveforms of vl of t and ic of t so with the switch in the first position we get this circuit we can identify what the inductor voltage is from this loop equation the inductor voltage vl will be equal to the input voltage vg minus the voltage drop on the resistor rl which is irl like this the capacitor current for the switch in the first position is found from the node equation at the output and as usual it is equal to the load current actually minus the load current so ic is minus v over r we now apply the small ripple approximation as usual so the output voltage v of t we replace with its dc component capital v and additionally on the inductor we will assume that the inductor has small ripple as well so we will replace i of t with its dc component capital i similar for the switch in the second position we get this circuit the inductor voltage this time is the input voltage vg minus the drop on the resistor irl minus the output voltage v and the capacitor current is the inductor current i minus the load current b over r so we get these equations and we again apply the small ripple approximation as appropriate okay so here are the waveforms then this is the inductor voltage waveform and the capacitor current waveform both of these must have zero average when the converter works in steady state the inductor voltage waveform now has some additional terms that come from the voltage drop on the the winding resistance but we can apply volt second balance to this as usual so zero or the average inductor voltage is zero and it would be d times the value during the first interval vg minus irl plus d prime times the value during the second interval which is vg minus irl minus v okay we can collect terms and equate this to zero and we get this equation for inductor volts second balance okay in the ideal case we were able to solve this for the output voltage and get an expression for the ideal output voltage but now we have an added term that comes from the inductor uh winding resistance and there's an added variable capital i uh the inductor current at this point is an unknown so we have one equation with two unknowns and we can't yet solve so to get a second equation we get that from the charge balance on the capacitor so the average capacitor current must be zero in steady state and it's we actually get the same waveform and same equation as in the ideal case for the capacitor current and when we collect terms then we get this equation which is a second equation that relates the inductor current and the output voltage okay so when taken together then we have these two equations the first is from volt second balance the second is from charge balance and we have two unknowns capital i and capital v so we can solve now to find the output voltage for example we could solve the second equation for i and plug the result into the first equation and then solve for v and if you do that here is the result v over vg i've written this equation in a nice form it has a two terms the first term one over d prime is the result for the ideal case with no loss and the second term um is we can think of as a modification of how the inductor resistance affects the output voltage and you can see that if you let rl go to zero the second term goes to one and we get the ideal expression one over d prime okay i've also written this second term in a nice form where every term is normalized and the denominator is one plus a function of rl so as rl is increased the denominator is increased and the output voltage goes down if you would like rl to have a small effect on the output voltage then we need rl to be small compared to the term it's divided by which in this case is d prime squared r so for example if rl was one percent of d prime squared r then the denominator would be 1.01 and the overall second term would be approximately 0.99 so the output voltage would decrease by one percent it's nice to write the second term in this form because we can compare see what to compare rl to and how get a good idea then of how small rl needs to be okay on the right we can see [Music] a plot of this expression v over vg for different values of r l over r the top curve is with rl equal to zero and then we get the ideal case that's one over d prime and this function goes to infinity at as d goes to one the lower curves are for different values of rl and you can see that as rl increases the output voltage goes down as we would expect but not only that there's a major qualitative change where the curve actually bends over it reaches a maximum and then it starts decreasing and at d of one the curve goes to zero so this is a pretty big difference you add a small amount of rl anything bigger than zero and at d of one the curve goes to zero instead of infinity so what's exactly going on there well we can go back and look at the original circuit what happens in the original circuit as you let d approach one well if d goes to one then the switch is always in position one and never in position two so we just connect the inductor across vg and the inductor current is large and in fact with zero rl the inductor current tends to infinity and the volt seconds don't even balance um if there is some d prime interval that's greater than zero then during the d prime interval some current will go to the output and can charge up the output voltage in the ideal case as d approaches zero the average current driving the output is infinite amps times something approaching zero duty cycle and what is the limit well the ideal equation tends to infinity at least as approached from the left but it's a non-physical answer and with any amount of rl at all the inductor current is limited to a finite value and as d goes to one in any event we we get essentially zero average current driving the output and there's no output voltage and so um the actual case with the loss resistance is the um physical answer in which the output voltage is limited and doesn't go to infinity at d of one and in fact it goes to zero moreover we can see that the maximum voltage that we can attain is limited by the loss resistance so for example suppose we wanted our converter to boost the voltage by a factor of three well if an rl over r was if you have r l over r equal to 0.05 that's not no good it's not going to work because we can't make enough output voltage r l over r equal equals 0.02 looks like it would work and .01 would be even better so the peak amount that we can boost by is is a function of rl so we've seen in this first example how we can model loss in this case we modeled the inductor winding loss with a lumped element model that that had an ideal inductor in series with an effective resistor the ideal inductor still obeys volt second balance although the the total inductor model does not because of the resistor so in general we can model losses by adding elements to our converter and then applying volt second balance to the ideal part of the inductor model and charge balance to the ideal part of the capacitor model we then get a set of equations that we can solve for things like the output voltage in the last lecture we modeled the copper loss of the inductor of a boost converter uh by inserting an extra resistor in series effectively with the inductor in a lumped element model and with that model we then calculated the effect of this resistor on the output voltage of the boost converter in this lecture we're going to extend that result and construct an equivalent circuit model to go along with with the equations so here are the equations that were found in the last lecture for this example the first equation was from inductor volts second balance and the second equation we got from capacitor charge balance and the object now is to find an equivalent circuit to go along with these equations okay the approach that we're going to take is to view these equations as the loop and node equations of our equivalent circuit model and indeed the first equation was found by finding the average voltage around the loop where the inductor is connected so what if this effectively is a loop equation in our model the second equation was found by finding the average current flowing into the capacitor from the node where the capacitor is connected namely the output node so we can view this equation as a node equation in our model as well so what we're going to do is construct equivalent circuits to go with each of these equations okay now in basic circuits classes i hope you were taught how to take a circuit and apply kirchhoff's laws to find the equations here we're going backwards we're given the equations and we want to construct a circuit well this process of going backwards turns out to not be unique there's more than one way to construct a circuit that satisfies these equations and so we have to be careful how we do it to make sure that our resulting model has physical significance and my advice here is to use the the equations in the form that they come without further manipulations so the first equation was found from inductor volt's second balance we should keep it as a sum of voltages around a loop corresponding to a kirchhoff voltage law equation and don't manipulate this now if you for example divided through by a say r then this first equation would have terms that have dimensions of current rather than voltage and you might think then that this is a node equation with with currents flowing into a node while that equation might be mathematically correct the resulting model is not physically connected to the actual converter so we don't manipulate these equations or do any kind of algebra on them we construct circuits to go along with the equations right away okay so let's do it i'm going to copy this equation that the average inductor voltage is vg minus irl minus d prime v and we will construct a circuit to go with it okay it's on the next slide but i'm going to do it actually on some graph paper here so our equation was that 0 is the average inductor voltage which is v g minus i r l minus d prime v okay so this is an equation describing the loop where the inductor is connected and it's describing actually the dc components of the voltages around that loop so i'm going to draw a little dashed inductor in that loop that has an average inductor voltage equal to zero and there's a current flowing through this inductor that is the inductor current dc component that we've been calling capital i now the average inductor voltage is zero and in fact the impedance of the inductor at dc goes to a short circuit so in fact this is a short circuit in our dc model but it's a good starting place to have a physical connection to the actual model in the actual circuit to recognize that the current in this loop is the inductor current i and the voltage is vl is the average inductor voltage that is defined with a reference direction or polarity that is consistent with the direction of the current okay so that is the voltage on the left hand side of the equation on the right hand side of the equation we have the input voltage vg so what should i draw in our loop to represent that well we'll draw a an independent voltage source of value vg and since it's on the other side of the equation then as our current goes around the loop and it's going from plus to minus through vl because vg is on the other side of the equation our current will go from minus to plus or with the other direction through vg okay the next term is i times rl that term sounds like the voltage across a resistor having current i which is the current in this loop multiplied by a resistance rl so we'll draw a resistor here and label its value rl and the voltage across it with i going from left to right would be a voltage from left to right a value irl and that polarity has the opposite polarity going around the loop as the vg term had which is consistent with a minus sign okay the last term is minus d prime v how do we handle this term well here's how v is the output voltage this isn't exactly the output voltage but it's d prime times the output voltage so i'm going to draw for now a dependent source a value d prime v so it's a voltage source that depends on the output voltage v but it's multiplied by d prime and let's see the minus sign would mean that this voltage is plus on top and minus on the bottom it has the same polarity going around the loop as the rl or the irl voltage okay so here is an equivalent circuit whose loop equation is the same as the equation that we got from volt's second balance and it is the inductor loop part of our equivalent circuit okay let's do the other other equation so the other equation was from capacitor charge balance which said that the average current through the capacitor or dc component of capacitor current is zero and what we found there was that that current was equal to d prime i minus the load current capital v over r okay this equation was from found by charge balance on the capacitor and again basically it's it was found by finding uh computing the dc components of the currents going into the node where the capacitor is connected and so we can view this as a sum of currents equaling zero which sounds like a node equation so i'm going to draw a node and this is the node where the capacitor is connected so i'll draw a dashed capacitor so the current through this capacitor our average ic is zero and we know the voltage across the capacitor is v capital v actually in our dc model the dc component of the output voltage in our model our dc model the what the impedance of our capacitor at dc is an open circuit so this will actually be an open circuit but for reference we know that v is the voltage at the node and ic is coming out of the node and flowing into the capacitor so what our equation says or charge balance equation says is that there are two other currents flowing into the node the first one is d prime i okay so this is a current that is dependent on the inductor current which was in our other circuit our earlier circuit but it's not just i it's d prime times i so i'm going to draw a dependent source with a current d prime i that flows into the node okay and then the second term is minus v over r so this is the load current and it is flowing out of the node because of the minus sign we can model that as a resistor a value r that has a voltage v across it so the resistor's in parallel with a capacitor and that way we get a current capital v over r flowing through our load okay so this is the equivalent circuit that comes from the capacitor charge balance equation the last thing to do is to combine the two circuits together so let's go back to the slides when we draw the two circuits together we get this this top circuit so the left hand side was the inductor loop equation where the inductor went right here and the right hand side was the capacitor node equation where the capacitor went right there the last thing to say about this is that we recognize we have two dependent sources that as discussed in section 3.1 are equivalent to a dc transformer so we have one dependent source this d prime v source depends on the voltage v across the other source with a factor of d prime and the dependent current source depends on the current i through the voltage source with the same factor of d prime so since the factors are the same and they depend on the voltages and currents of the corresponding sources we can combine these into the dc transformer as we've discussed previously i think the easiest way to understand what the turns ratio and polarity mark should be is to look at the voltages so we have a voltage v on the output side and d prime v on the input side of our transformer so the turns ratio is d prime to 1 which will get us the correct polarities of voltages and to look at the polarity marks we have we look at the polarities of the voltages so we have v is positive on the top on the the secondary side and d prime v is positive on the top on the primary side so the polarity marks are of the same phase we have the dots on the top in both cases as a check we should look at whether the currents work so here in our model we have a current i flowing into the dot on the primary side and coming out of the dot on the secondary side should be d prime i and that's indeed what the model says so the currents are consistent with the transformer model as well okay so we have an equivalent circuit now for our boost converter that has the ideal transformer in this case d prime to 1 is its turns ratio or if you want to call it 1 to m m is 1 over d prime and then it has an added loss resistor that comes from the inductor loss resistance okay an easy way to solve this i think is to push the elements through the transformer we've actually already done this um exercise we push vg and rl through to the secondary side if we want to solve for the output voltage so when we do that vg will be multiplied by the by m or divided by d prime rl will get multiplied by m squared or in fact divided by d prime squared so that's the effective circuit between the secondary terminals of the transformer we connect that to our load resistor and recognize that the secondary voltage is v and we can then just solve this circuit using the voltage divider formula v would be the voltage source vg over d prime multiplied by the divider ratio of these two resistors here and if you like you can divide top and bottom by r to write the expression in this form which is the way we wrote the expression in the last lecture okay so we found the same result as in the previous lecture except instead of using algebra we found it using circuits but we can do more with the equivalent circuit than just solve for the output voltage we could also for example solve for the current i on the primary side so if you say push the load resistor through the transformer to this side will get effectively between these terminals a resistor that would be d prime squared r and we can then find the current i would be vg divided by the sum of the two resistors like this another thing we can do is find the efficiency to find the efficiency we find the input power and the output power so the input power is the power going into the converter terminals here and the output power is the power coming out the output side and the efficiency is the output power over the input power okay so from the circuit we can easily write expressions for the input and output power so the input power we can write is the input voltage vg multiplied by the input current i and the output power would be the output voltage v multiplied by the output current what is the output current or the current flowing here well use the transformer we have current i flowing in the dot on the primary so the current coming out of the secondary will be d prime times i so we can write that the output power is v times d prime i okay that's shown here and now we can cancel the i's and we get that the efficiency is v over vg times d prime okay we previously found that v over vg for this converter is one over d prime times uh one over 1 plus what was it rl over d prime squared r so we have to take that v over vg and multiply it by d prime so if you multiply by d prime the the one over d prime factor goes away and we and then this added multiplying factor is in fact the efficiency here's a plot of the efficiency so for different values of r l actually for different values of r l over r we can plot a series of curves of efficiency versus duty cycle and what you can can see is that for low duty cycles the efficiency is a lot higher whereas for high duty cycles the efficiency comes to some shoulder and then drops off quickly so there's a range of duty cycles from zero up to some value depending on rl where we have high efficiencies and beyond that the the curve we get low efficiencies uh very quickly efficiency drops off quickly and this drop off coincides with the the voltage the output voltage deviating substantially from the from the ideal case the one over d prime curve so um by we can take our uh volt second balance and charge balance equations and view them as uh the loop and node equations of our equivalent circuit model and and uh therefore we can reconstruct loop equations that go with our volt second balance equations and reconstruct the equivalent circuits that go with those we can also reconstruct equivalent circuits based on node equations that correspond to the capacitor charge balance and finally we can combine these using ideal dc transformers to get a final model this final model has a physical interpretation we can plug this into a larger system and then model how the converter works in the larger system and we can solve it for things such as the efficiency or do manipulations on the circuit to find things like the voltages and currents of the converter now let's work a second example of a buck converter containing an inductor with some dc resistance so here's our converter we will model the inductor in the same way we did previously for the boost converter where we have a lumped element model of an ideal inductor in series with a an effective resistor that models the winding resistance okay we can go through the same procedure to work out volt second balance on the ideal part of the inductor model and charge balance on the capacitor and if we do this i won't go through the details but if we do it here are the equations that we get so we get this equation from volt second balance on the inductor it's like in the ideal case except that it has an additional term from the voltage drop across the winding resistance and we get this term from capacitor charge balance that is in fact the same as in the ideal case so we got an equation from inductor volt's second balance that was d times vg minus the inductor current times the inductor resistance minus the output voltage and for charge balance we got the average capacitor current which is zero was equal to the inductor current minus the load current so from the first equation we have a source a dependent source that depends on vg that we can write like this here's where our ideal inductor model goes that's effectively a short circuit with a current i flowing through this voltage we have now a term that is the voltage drop across the winding resistance and then we have the output voltage right here in the buck converter we know that the inductor is always connected to the capacitor node and so in fact we expect our inductor model to connect directly there so this would be the capacitor node and we also have our load resistor as usual connected across the capacitor so here is the equivalent circuit that we get from the two equations um this equation this model is okay as far as it goes but there's a big question of where is the dc transformer in a buck converter we would expect to have a dc transformer with a turns ratio of 1 to d equal to the conversion ratio and here we have a voltage source that is d times vg that looks like maybe it's part of the transformer but where is the rest of the transformer and for that matter where is the input voltage source vg the problem here is that we're missing an equation that describes how the converter connects to the input terminals or up to vg we need another equation to describe that connection in fact what's happening here is that in our original converter both inductor and capacitor are on the same side of the switch so we don't have a volt second balance or charge balance equation that describes the connection to the input port of the converter and so we need to manufacture another equation that describes the input port okay so we have an input port here we know there's a voltage source vg but we need to to write an equation of what the converter does what current does it draw out of vg and so the way to get that is to write an auxiliary equation or a third equation that really tells us the dc component of ig which is the current drawn out of vg or drawn out of by the converter out of the the node at the input port or input terminals or of the converter in general ig is a dependent quantity it's switched it switches when the switch changes between positions one and two it's a discontinuous waveform the small ripple approximation does not apply to it but we need to sketch what ig is and express it in terms of things that have small ripple like the inductor current and then find its dc component to give us another equation that we can use to complete our model here's our converter with our ig of t and because our switch is at the input terminal of the converter we need another equation see we called the current il so so we can sketch i g of t and basically when the switch is in position one ig is equal to il so ig follows the inductor current and when the switch is in position two ig is equal to zero so ig looks like this okay what we need to do is calculate the dc component of ig or the average value of this waveform okay so we follow il during the first interval and the first thing we will do is make the small ripple approximation on the inductor current again the inductor current has small ripple and is a continuous waveform so we can replace io of t by its dc component capital il in the usual manner and then find the average value of this waveform which is base which is just d times the capital i l plus d prime times zero so this is an equation it's an auxiliary equation for a quantity that we need to know it's a dependent quantity we need to know in order to complete our model we can construct an equivalent circuit that goes with this and then append it to the rest of the model so this this equation is uh really the equation at the node of the input terminals of the converter and what it says is that the current ig which is the current drawn out the dc component of current drawn out of vg so i'm going to draw our vg source and label the current coming out of eg as capital ig for our dc model and this is equal to d times il so it is a dependent source that is dependent on the inductor current il elsewhere in the model and we can draw a dependent source then current source and label its value dil when we append this to the rest of the model now we can get a complete equivalent circuit so here is the input port equivalent circuit that we just derived here is the equivalent circuit that we got from volt second balance and charge balance and you can see that there are now two dependent sources that can again be combined into a dc transformer and they have a turns ratio of 1 to d if you follow the voltages vg gets multiplied by d to produce the secondary voltage and the dots are or follow the plus terminals and are always on the top so here is now an equivalent circuit for the buck converter that includes the the dc transformer that we expect so in general when we have switching at one of the the ports of the converter we may need to um to write an equation for an auxiliary variable in this case the node equation at the input terminals of the converter that gives us a complete set of equations to describe the equivalent circuit so here we found the waveform of ig of t we expressed it in terms of things that had small ripple such as the inductor current and use the small ripple approximation on the inductor current and we were able to express the dc component of ig then as a function of of the dcil to get an equation that could give us the complete transformer in general we may get have a similar situation at the output if there are switched waveforms on the output side of the converter and we may have to do a similar thing there okay so in converters such as the buck or other converters even say a bug boost is another one that have switching on the input port we will in general need an equation for the average ig let's work another example in which we will include the semiconductor conduction losses in our equivalent circuit model the semiconductor conduction losses are the power losses that come from the forward voltage drops of the semiconductor devices so what we're going to do in this example is to model a boost converter we will again include the on resistance i'm sorry the the winding resistance of the inductor but now we'll also include the on resistance of the mosfet and the forward voltage drop of of the diode so for the mosfet it behaves a typical power mosfet behaves as a resistance when it is in the on state and so an equivalent circuit that we might use for the mosfet is an ideal switch in series with a resistor that we'll call r on for the diode there's multiple ways to model a diode but if we have a say a pn junction diode we can model it as a constant voltage drop that is the forward voltage drop of the diode or if we like we might further refine the model by adding an on resistance for the diode as well and in fact i'm going to call these vd and rd for the diode voltage drop and the diode on resistance typically we extract these quantities from the semiconductor data sheets for mosfets the on resistance will be specified and for the diode typically there'll be a plot of the current versus voltage curve of the diode and from this plot we can fit a curve so if we model the diode as a forward voltage plus a resistance we might approximate this curve with some straight line that's valid over the intended important operating range and so this intercept voltage would be v sub d and the slope here would give us r sub d slope on these axis is actually one over r sub d okay so for the diode then we have this model when the diode is on if we put it in series with an ideal switch like the mosfet model then we get a model that can be used in the switching converter for both intervals so what we do is we then write the as usual write the circuit when the switch is in the two positions so what we call position one is when the diode is conducting and the mosfet is off and the circuit reduces to this which now includes the on resistance of the mosfet in the the interval sub interval 1 circuit for sub interval 2 the mosfet is off or it switches off the diode is on and the circuit reduces to this where now we have the v sub d and r sub d model for the diode forward drop in our equivalent circuit in our sub circuit with these as usual we can write the um the inductor voltage and capacitor current waveforms so here as before vl of t is the voltage across the ideal inductor part of the lumped element inductor model and i sub c of t is the current in the capacitor the capacitor current waveform is the same as it has always been so far none of the losses that we've modeled have affected that but they all all of the loss elements add terms to the inductor voltage waveform but no matter we can work out the average inductor voltage anyway vl of t is d times the voltage during the first interval plus d prime times the voltage during the second interval so there are more terms but it's a straightforward matter to to write them or to write the average voltage and by volt second balance we equate this voltage to zero for the capacitor current since its waveform is the same we get the usual equation for the average or for charge balance on the capacitor so now we construct the equivalent circuit and i'm going to do this on some graph paper also so here is the inductor voltage equation that we got the average inductor voltage is 0 and it is equal to vg minus i r l when we collect terms uh minus d i r on minus d prime v d minus d prime i rd minus d prime v so there are more terms but we can write write this equation and can construct an equivalent circuit that includes an element for each term as usual okay so the first term is vg we'll write a vg voltage source the next term is i times rl we recognize that as as usual it is the voltage drop in the inductor winding resistance so we'll put a resistor value rl uh the next term is minus d times i times r on and what's new here is that there's a factor of d so the question is what should the element in our equivalent circuit b to correspond with this term and i would say there there are perhaps two choices we could consider should this be a resistor of value r on if so then that means that the current through the resistor is d times i or the other choice is that we associate d with the value of the resistor so perhaps this should be a resistor of value d times r on that has a current of i so what do you think is the correct answer well the the correct answer is that this is a resistor a value d times r on that has a current i flowing in it and the reason is that the current in this loop for this loop equation is the inductor current and if you recall in previous lectures i've drawn a little dashed inductor to model the where the inductor goes it has an average voltage that is zero and it has a current that is the inductor current i so this is the loop equation that contains the voltages around the loop where the inductor is connected and the current in that loop is in fact the inductor current i so we have to have a resistor with a value i and therefore the value of the resistor is d times r on well so now we have an element a resistor whose value varies with the duty cycle which is a kind of strange element why do we get such a thing well the reason is that the on resistance is only in the circuit when the mosfet is on which is during the first interval of length that has a duty cycle of d so for example if we let d go to zero and we never turn the mosfet on then we would not expect there to be any power loss in this resistor and the power loss that you get is a function of the duty cycle or the fraction of time that there is current flowing in this resistor making loss so this model predicts that the power loss in this resistor is the the current squared or i squared times the element value dr on now in fact what's happening is during the first interval we're getting a power loss of i squared r on but we're only getting it during the dts interval and so the average power loss with zero loss in this resistor for the second d prime interval the average loss will be the duty cycle times i squared r on and so by changing the effective value of this resistor the model correctly predicts the power loss in the mosfet okay let's keep going the next element is minus d prime v sub d okay so v sub d is the diode forward voltage drop in our diode model and it is a constant voltage i'm going to draw it as an independent source it's the value d prime v sub d um and it's plus to minus from left to right so uh the the power loss in this element also depends on the length of time that the diode conducts and the v sub d source gets multiplied by d prime the diode resistance rd similarly is an effective value of d prime rd and finally we have the output voltage multiplied by d prime this is a voltage that depends on the output capacitor voltage and we'll draw it as a dependent source the value d prime v so there's our circuit model for the inductor loop that includes all of the terms in the inductor equation okay here is the model we just constructed the capacitor node equation model is the same as it's been in the previous examples at least in the boost examples so we get this equivalent circuit as usual and we can now combine the two dependent sources into a transformer so here are the two models written together we combine these sources into a transformer we get a d prime to one boost type transformer and this is the equivalent circuit model then for the converter here is the model that we derived and let's solve the model now for the output voltage the way to do this i think that is easiest is to push all of the elements to the secondary side and combine them together so first of all i'm going to combine the two independent voltage sources i can since the v sub d voltage source is in series with these resistors we can swap the order without changing the circuit and i can combine it with vg into one source so i'm going to write one source that is a value vg minus d prime v sub d then i can combine all the resistors together by simply adding them so we have one effective resistor call it re which is equal to rl plus d r on plus d prime rd we have our d prime to 1 transformer and the output voltage finally we can push these elements through the transformer when we push the voltage source through the transformer we'll multiply by m or divide by d prime so we get an effective voltage source of value vg minus d prime v sub d all over d prime when we push the resistor through we get an effective resistance as seen from the secondary side of the resistor multiplied by m squared or divided by d prime squared so we'll get r e over d prime squared then we have our load resistance and the output voltage so we can write now the expression for the output voltage from solving the circuit it's again just a divider ratio where the reflected voltage is seen at the secondary terminals is vg over d prime minus v sub d and then times the divider ratio which is r [Music] over r plus r e over d prime squared and that's the expression for the output voltage you can see the divider ratio again is changing quite a bit with duty cycle all of these resistor values effectively are changing as the duty cycle changes as does the effect of the uh the input voltage so here is the result that i just derived on the previous slide and it's manipulated to make dimensionless terms so this is writing the result in the nice form where the first term is the ideal conversion ratio with no losses the second term is the numerator term that accounts for the losses that appear in the numerator and in this case the v sub d voltage drop of the diode appears in this term and the the final term is the denominator that comes from all the resistive volt voltage divider terms uh we can work out the efficiency as well as usual so we find the input power going into the converter we find the output power coming out and the efficiency is output power over input power the input power is vg multiplied by the current of on the input terminals which is the inductor current i and the output power is the output voltage v multiplied by this current coming out of the terminals and that current is the current i reflected through the turns ratio which gives us d prime i so v times d prime i is the output power so if you divide those then the efficiency becomes equal to uh v over vg times one over d prime and the i's cancel out so if we go back to our previous expression for the v over vg i'm sorry not times d prime not 1 over d prime so if we go back to our previous expression and multiply v over vg times d prime the 1 over d prime term cancels and the remaining terms are the efficiency and so we get this expression for the efficiency and basically we have a term in the numerator from the v sub d source and we have all the resistive terms in the denominator and this expression again because it's written in this nice normalized form tells us really the relative effects or the effect of each of the loss elements on the efficiency so here if we want the v sub d term to have a small effect on efficiency then v sub d should be small compared to v g over d prime likewise if the on resistance r on should is to have a small effect on efficiency then r on should be small compared to d prime squared r over d and so on okay uh one last little point about prediction of of the conduction losses uh using the this averaged model so we found with the mosfet on resistance that the model effectively changes the value of r on it gets multiplied by a factor of d and this makes uh makes the model correctly predict the the average power loss in the on resistance of the mosfet we should take this just a little bit further and recognize that we've also made the small ripple approximation and so these models are ignoring the effects of switching ripple and so we would expect the accuracy of the models to be good as long as the ripple is small but if the ripple gets large then the model could be inaccurate so let's consider for the case of the mosfet on resistance how does the ripple affect the on the prediction of loss okay so here's a plot of the current in the mosfet this i of t here is the mosfet current basically we have we have a mosfet that we're modeling with a switch that turns on and off as the mosfet turns on and off in series with an on resistance and there's a current i flowing in this uh switch this is i is different than the inductor current this is really the mosfet current um and so the power loss in the on resistance is the rms current so i rms squared times the on resistance and here's what that i of t looks like it's zero when the mosfet is off and when the mosfet is on the mosfet current follows the inductor current okay so when we ignore the ripple we're really drawing i of t as case a shown here where there is no ripple and with no ripple i then simply is looks like this with when it's on we have a current capital i the dc inductor current when the mosfet is off we have a current zero okay and in that case the r we can show that the rms value of i for this case a is the square root of the duty cycle times capital i okay so then the predicted power loss in the on resistance is this rms current squared so root d times i squared times the on resistance and we get d times i squared r on which is exactly the loss that that our equivalent circuit model predicts now case b and kc show examples where the ripple is greater than zero and one can show that when ripple is included the rms current is [Music] is this value root d times i and then times another factor that accounts for the ripple and that factor turns out to be one plus one third delta i over capital i squared so this extra term accounts for ripple um and here is the value then of the rms current including ripple for two cases in case b the ripple delta i is 10 of the dc current capital i and we would expect the small ripple approximation to be pretty accurate there so you can compute this term and see how big the rms current is it turns out to be 0.167 percent higher than the the ideal case with no ripple so this factor here is is this term right here and it's very very close to one so when you square this to find the power loss we find that the the computed average power loss is 0.33 percent higher than our model with the small ripple approximation predicts 0.33 percent is a pretty small number it's pretty small difference and i think the the model is very good here in fact the tolerance on the on resistance is a lot more than that case c is the extreme case shown here where the ripple is equal to capital i so we have very large we have 100 percent ripple relative to the dc current and the actual current waveform starts at zero and ramps up to twice the average so in that case where delta i equals i we can calculate this quantity and it turns out to be 15 and a half percent higher so 1.155 when you square that we and calculate the average loss we find that um there is one-third more loss than we predict so even in the extreme case of ripple equals dc component we're still only off by a factor of a third and it's not bad but when the ripple gets that large then we we know that perhaps we should account for the ripple in our computation of the loss this week we will discuss the realization of switches in our power converters using semiconductor devices such as power transistors and diodes and there's really three major topics that we need to discuss the first is when we have an ideal switch how do we decide whether to replace it with a transistor or diode or something else in order to make it work in the power converter application and i'm going to introduce in this lecture and the next several lectures the ideas of single quadrant two quadrant and four quadrant switches and how to decide which of these to use depending on the application the second topic is we need to talk a little bit about the real semiconductor devices and so i'm going to discuss power diodes power mosfets and the power igbt or insulated gate bipolar transistor which are some of the most popular and major power devices in use today and i don't want to get too heavily into semiconductor device physics but we're going to talk enough about them so that you can appreciate what are the um the major limitations and important features we need to think about from the standpoint of their use in power electronics and then the third major topic is switching loss so far we've talked about modeling conduction losses in a switching converter but what one finds in practice is that the conduction losses are not the largest uh source of power loss at least in most applications and in fact it's the switching loss that occurs during the switching transitions from the on to the off state and and vice versa that are the largest loss in the converter so what we're going to do is to refine our equivalent circuit models that we derived in chapter 3 to include switching loss and so that will have equivalent circuit models that include things like the diode reverse recovery and its effect on the converter efficiency semiconductor devices are single pole single throw type devices so they have two power terminals and a transistor or a diode really acts as an on off single pole single throw switch like this now so far in the course i've been drawing single pole double throw switches such as this one in which the switch is either in position one or in position two so the first thing we have to do is to replace the single pole double throw switch with two single pole single throw switches like this and then we can realize each of the single pole single throw switches with a semiconductor power device now this may seem like a trivial step but in fact it's not that with a single pole double throw switch there are only two possible positions whereas with the two single pole single throw switches we have two more cases the switch can be in position one or in position two if switch a is on only or switch b is on only but we also have the cases where we could turn both switches on or or we could turn both switches off and this really happens and it's something we need to consider this particular example of the buck converter if you turn both switches on at once you probably destroy your switches because you short out the power source and you can get a very large current flowing around this loop that can destroy the devices or at least cause a lot of power loss so we have to go to some lengths in our drive circuit to make sure that that never happens even for a few nanoseconds the other case where both switches are off at the same time can happen and it completely changes the characteristics of the converter when it does and so we're going to talk about that case in chapter 5 and see that it leads to the discontinuous conduction mode which is something we haven't talked about yet but we need to okay we also talk about switches with respect to how many quadrants of operation do they have to work with and this is really imposed on the switch by the rest of the converter so here i plotted what we're talking about the the horizontal axis is the off state voltage imposed on the switch by the converter when the switch is turned off so we also call this the blocking voltage or how much voltage does the switch have to block when it is off so the switch off state voltage generally is determined or given by the input voltage vg or perhaps a capacitor voltage in the converter and we have to see whether or maybe the output voltage but we have to see how large that voltage is and whether it changes polarity so perhaps it is positive that some values of duty cycle and negative at others and that affects how we realize the switch likewise the switch on state current is the current imposed on the switch by the external converter or by the converter itself when the switch is turned on generally the on state current is an inductor current or some combination of a combination of inductor currents and this can be a function of duty cycle also and perhaps it's a function of the load current so again we have to see what is the on state current and how does it vary with duty cycle is it say always positive like sketched here or can it be positive sometimes and negative other times so what's sketched here is the case where both the off state voltage and the on state current are lie and are positive and so we say that this is in the first quadrant of this plane and we call this a single quadrant switch for this particular example okay there are many different possibilities and these in different applications we may require any one or more of these so the single quadrant switch that i just uh mentioned is is one example and in fact which quadrant does it have to operate in you can see there are four possible quadrants and so there are four ways in that sense to realize a single quadrant switch the next case here is called a current bi-directional two-quadrant switch where the switch must conduct both positive and negative currents so perhaps at one duty cycle the current is positive and add another it's negative but it only has to block positive voltage when it's off and so we call this a current bi-directional two-quadrant switch there are two possibilities here depending on whether the blocking voltage is positive or negative the third example is a voltage bi-directional two-quadrant switch in which the switch has to block both polarities of voltage when it is off but only conduct one polarity of current and finally the four quadrant switch is the most general case where the switch may have to do anything it can block either positive or negative voltage when it is turned off and it can conduct either positive or negative current when it's turned on so this is the most complex case to both control and to realize with semiconductor devices but there are examples uh most notably ac to ac type converters that that need four quadrant switches okay so let's talk here about the single quadrant case so here i've defined a switch an ideal switch i've defined a voltage across this switch with some polarity and i've defined the current is flowing from the plus terminal to the negative terminal of the reference voltage of the switch and so if the the on state current and the off state voltage or of a single polarity then we have a single quadrant switch further we have what are called active switches and in these switches the switch state whether it's on or off is controlled only by a control terminal like the gate of a mosfet or igbt and so if you we know at any given time the the conducting state of the switch by the control signal that we have applied to it on the other hand a passive switch doesn't have a control terminal for example a diode and instead its switch state is controlled by the converter waveforms that are applied to the switch so if the converter voltage or current forward biases the diode then it will the switch will be on for example the silicon controlled rectifier or scr is a special case it is turned on with a control signal so the turn on transition is active but once it's on it behaves like a diode in which it won't turn off until the the external circuit or the power converter circuit reverse biases the scr okay so here's the diode and what i've drawn here is different than the on state current and off state voltage this is the voltage versus current characteristic of the diode and really you can think of this as the classic exponential characteristic of the diode that looks like this and if i i've idealized it here to ignore the forward voltage drop and put the iv characteristic right along the axis okay so when the diode is on basically it conducts positive current and when the diode is off it blocks negative voltage so in the planes that we have been drawing in the previous slides the off state voltage is negative and the on state current is positive and so the diode operates then in this quadrant in the second quadrant so if we work out these quantities by solving the converter to find its voltages and currents and see what it applies to the the switch and it turns out that if we get quadrant two then we can realize it with a diode like this incidentally suppose i had to define the voltage in the other direction i'm free to do that we could just as well define voltage upside down and this is an arbitrary choice at the outset so if we define voltage that way then we have to define current consistently as flowing from the reference positive to negative terminals and if we did that we would find that the diode would operate in the fourth quadrant instead of the second so if we do our analysis of the converter and see the applied voltages and currents on our switch and it turns out to be in this fourth quadrant then we know we need to connect a diode in the opposite direction to realize the switch okay uh power transistors such as the bipolar junction transistor or the newer insulated gate bipolar transistor our devices that have if we plot the instantaneous voltage and current they have characteristics like this depending on the gate voltage of the igbt or the the base current of the bjt and so these devices when they're on they operate with a voltage on this part of the the curve that is close to zero voltage and they conduct positive current when they're turned off by their control terminals then the characteristic is here and [Music] they can block positive voltage on the other hand practical transistors such as these uh generally are not capable of blocking significant negative voltage and if you plot their reverse characteristic it does something like this and the device will break down if you apply more than a few voltage a few volts of reverse voltage so this is a practical detail in how the devices are built so they are not capable of blocking significant reverse voltage and basically we then we can approximate the instantaneous iv characteristic of this like this as the off state is on the positive um horizontal axis and the on state is nearly on the the positive vertical axis and so we we actually have a single quadrant switch that operates in the first quadrant the mosfet is is very similar it's also a power transistor what in wide use that has similar forward characteristics so it can conduct positive current and block positive voltage in addition the power mosfet channel is symmetric and it can actually conduct reverse current as well um but in addition to that the mosfet also has um what's called a body diode that is a built-in diode that comes from shorting the substrate of the mosfet to the the source of the mosfet that adds an extra pn junction that effectively acts like a diode in parallel with the mosfet channel and so that that diode can also conduct reverse current so in that sense the mosfet is a current bi-directional device i have to add a few caveats to that though because the body diode in many mosfets is not optimized to be a good fast recovery diode we'll talk about that in a few a few lectures from now and in fact if the diode is slow enough and you switch it off quickly you can actually get such high currents flowing through that diode that it will make the device fail so there are failure mechanisms associated with turning off the slow body diode so we have to be careful if we want to operate the mosfet in the reverse direction but there are significant applications today where we do that and there are mosfets that are designed to have fast body diodes that can work in that direct can work like that so the the mosfet is widely used as a single quadrant switch just like the bjt and igbt and also there are some applications where it is used as a current bi-directional switch so here's an example here's our butt converter again and what i've done here is draw the switch in the buck converter as two single pole single throw switches and what i want to do is illustrate now by example how to go through the analysis of those switches to decide how they must be realized with transistors and diodes so what i've done is i've labeled them we have a switch a and a switch b and i've arbitrarily defined voltages across the switches so va i've just defined and taken as a reference direction that it's plus on the left and minus on the right i could just as well choose it the other direction at this point but this is how i've chosen it having chosen that we i define the direction of the switch current ia is flowing from the reference positive terminal to the reference negative terminal likewise for switch b i'm going to arbitrarily define the voltage across switch b as having a reference direction of plus on the bottom and minus on the top and so the reference direction for the current ib will flow from plus to minus through switch b now what we have to do is figure out what is the on state voltage or on state current imposed on each switch by the converter and also what is the off state voltage that the switches have to block so let's do switch a first okay so when the switch is on so we have a is on and b is off then you can see the current that switch a conducts is the inductor current so i a equals il now we've already solved this converter previously and we know that the inductor current il is the load current so il is the load current over r and i mean the load voltage over r which is the load current and v is d times vg so i get dvg over r is the current that the switch has to conduct now all of these are positive quantities we'll assume vg is positive the duty cycle is positive and r is positive it's a passive load so that the on state current ia is positive okay so here's the on state current of ia and it's a positive current when a is off it has to what voltage does it have to block well when we turn a off we turn switch b on and with a off and b on the voltage va is equal to vg and that's a positive quantity so when the switch is off it blocks a voltage vg so the on state current and the off state voltage are both positive we're in quadrant one and therefore with quadrant one we can realize the switch with a transistor say an igbt okay so with an igbt there it will block positive voltage and conduct positive current which is how it's designed to operate okay let's consider switch b next when switch b is turned on and switch a is off what current does it have to conduct well with switch b on and switch a off you can see that i b equals i l and the defined polarities of ib and il are the same in this case and so this is a positive quantity and we're conducting positive current when switch b is off then we have switch a is on and switch b has to block vg but be careful the way we define the the direction of vb uh vb is actually minus vg which is a negative quantity so i have to block negative voltage okay so in this case we have a what positive current onstate current a negative off state voltage we're in the second quadrant and that's a diode okay so switch b must be realized with a diode now you can see here also that the choice of switch isn't arbitrary i couldn't put a diode at switch a or a transistor at switch b at least not an igbt or a bjt so we have to realize the switches correctly but by doing this simple exercise we can see we can follow in a systematic way to see exactly how to realize the devices so here's a summary then i put a bjt for switch a and a diode for switch b okay in the next lecture we're going to consider two quadrant switches and see some examples where they're needed in this lecture we'll talk about current bi-directional switches so here is a realization one way to realize a current bi-directional switch so this is a single pole single throw switch that is actually a transistor connected within what we call an anti-parallel diode the combination of the two is able to conduct current in either direction so if we have our i of t switch current is positive it can flow through the transistor and if it's negative it can flow through the diode and of course we could use a bipolar transistor here or we could use a mosfet or igbt or other single quadrant switch so the combination can conduct current in either direction however it can only block voltage in one direction so when we want the switch to be off if it blocks a voltage that is positive so plus to minus in this direction both devices can be off but if we reverse the direction of the voltage you can see that the diode will turn on and short out the voltage source so this combination is capable only of blocking one polarity of voltage and so it's what we call a two quadrant switch that is current bi-directional okay so formally here is the single quadrant switch with the voltage and current directions defined the combination of transistor and diode have this composite iv characteristic and as far as the plane of off state voltage versus on state current is concerned we can operate anywhere in the first quadrant or the fourth quadrant and of course also if we want to build a current bi-directional switch that operates in only the second and third quadrants all we have to do is connect our transistor and diode up backwards just define things going in the opposite direction and we'll get the second and third quadrant the mosfet i mentioned in the last lecture is in fact a current bi-directional switch so the mosfet channel itself can conduct current in either direction and in addition it has this built-in body diode that can conduct negative current and so the mosfet by itself can be a current bi-directional two-quadrant switch i also mentioned that the body diode of practical mosfets may not be a very good diode so if we have a slow body diode we might not want to to let it conduct and have to switch it off because of its slow switching time so what one can do if you really have to is to put an external diode in series with a mosfet that only lets current flow in the positive direction and then put an external anti-parallel diode to [Music] to get a current bi-directional switch we generally don't like to have to do this and we look if we're in that situation we really have two other choices one is to buy a mosfet that has a very fast and good body diode and especially at low voltages this is a good solution or at high voltages the other option that we have is to use a different kind of device such as an igbt and nowadays you can get good igbts that are co-packaged with fast anti-parallel diodes built in here's an example of a two quadrant switch application so this is a simple inverter we have split dc input voltage so there's we have plus vg coming into our converter here this is ground or zero volts and we have minus vg there and net this is a dc input to the converter and then we have a load and we want to produce ac across this load and this node here is connected to this ground so um each of these is a two quadrant current bi-directional switch if we succeed in producing an ac output then the current in our load is ac and can go either direction and that will coincide with the inductor current so our inductor current can go either direction as well that means that our switches when they're on and conduct the inductor current must be able to conduct either polarity of current so they have to conduct the ac output current on the other hand the switches block the dc input voltage so for example if the top switch is on and the bottom switch is off then this node gets pulled up to plus vg and the voltage across the lower switch will be plus vg minus minus vg or 2vg but it's dc and since our dc input voltage doesn't change polarity we only have to block positive voltages with this switch so a current bi-directional two-quadrant switch will work and in fact current bi-directional two-quadrant switches are commonly used in [Music] inverter applications let's work out exactly how this converter works so let's sketch the voltage at the switch node so when the upper switch is turned on q1 is on then this node gets pulled up to plus vg and we leave it on for some duty cycle or some first interval length then we switch the upper device devices off turn the lower devices on and that pulls this node down to here which is minus vg and we'll leave them in the lower position for the remainder of the switching period for d prime ts and then we repeat they're supposed to be straight vertical lines okay so the average value of this waveform or dc component or at least the low frequency component is equal to d times the voltage during the first interval plus d prime times the voltage during the second interval and you can work that out then it's equal to vg times d minus d prime d prime is 1 minus d so we can also write this as vg times 2d minus 1 which is what's shown here so this us this low frequency component of this waveform gets filtered out by the lc filter and that's what's applied across our our output load here's a plot of that function you can see if d is a half the function goes to zero so we have zero net output voltage at duty cycle of a half if you increase the duty cycle above a half we get positive voltage and when we decrease the duty cycle below a half we get negative voltage and the overall conversion ratio function looks like this so what you can do is you can vary the duty cycle about a half say variant sinusoidally like this okay and if you do that what happens to the output voltage is that the output voltage will vary sinusoidally about zero so if you plug this function into here you can find that the output voltage is sinusoidal so we vary the duty cycle sinusoidally if we wanted to make say a 60 hertz output we could make this this modulation frequency of the duty cycle be 60 hertz and we would get a 60 hertz ac output the inductor current you know which is the load current will also be will also vary sinusoidally in the same way so again it's positive or negative depending on which part of the sine wave we're on and so we need current bi-directional switches here's a three-phase version of the same circuit uh each of the three phases here is one of those circuits we just talked about so it has our current bi-directional switches and an inductor driving one phase of the three-phase output so here's the three phase ac load so we vary each of these phases or vary their duty cycles sinusoidally in a three-phase manner so they're 120 and 240 degrees shifted in phase to get a three-phase ac output and with this we can get um you know by varying all three phases in this way get you know generate a three-phase ac and drive an ac load such as say a motor an ac motor or even build an inverter to connect to the utility here so again the switches have to block the dc input voltage and they have to conduct the ac output current so this is fundamentally a two quadrant current bi-directional application here's a dc dc application of the same thing this is basically a buck converter where our input voltage here vg is some bus voltage and one example is a spacecraft power bus and in this case the load resistor is replaced with a battery and there may be a filler capacitor here or not but this is basically a butt converter except that what we were calling the resistive load before now is a battery that can both store energy or supply energy so when we charge the battery we have power going this way so it comes out of the source we this actually works like a buck converter and if you just draw transistor q1 and diode d2 here and ignore the other two devices you would recognize this is a buck converter that we've already talked about in this case you don't have to turn q2 on but in fact it doesn't hurt to turn q2 on any time q1 is off so we can actually drive q2 with a complement of the gate of the base drive of q1 if we want to supply energy out of the battery back to the bus we reverse the direction of current and in that case q2 and d1 are the ones that conduct and again we can turn on q1 whenever q2 is off and just continue to drive q1 with a complement of the q2 drive signal together operated together we simply have current bi-directional switches that can conduct current in either direction depending on whether we're charging or discharging the battery this is one circuit that is popular on space some spacecraft they have say dc power supplied by solar panels but when the spacecraft is eclipsed say in earth orbit it's eclipsed by the earth and we don't have sun then we need to supply power to the spacecraft from the batteries and so we need to both charge and discharge the batteries so we can build current bi-directional switches basically as a parallel combination of a transistor and a diode we call it an anti-parallel diode this is basically taking two single quadrant switches and putting them in parallel so that they can conduct current in either direction let's continue our discussion of switch applications to talk about voltage bi-directional two-quadrant switches and four quadrant switches so another possible function that a switch must perform is the ability to block both polarities of voltage when the switch is turned off and so this is a voltage bi-directional function for a switch we can build a voltage bi-directional switch by putting two single quadrant switches in series one of which blocks positive voltage when the switches are off and the other which blocks negative voltage so here is a one well-known way to do it in which we put a diode and a transistor in series so if the voltage is negative and we want the switch to be turned off so if we have negative to positive in this case the the transistor isn't designed to be able to block any significant amount of negative voltage however the diode in series will be reverse biased when the voltage is in this direction and so the diode will turn off and even if the transistor breaks down from having too much voltage across it the diode will still be off and it will prevent any current from flowing so effectively the diode performs the function of blocking the voltage when the voltage is in this polarity with positive voltage across the switch network we can block this voltage by turning off the transistor and then it operates in the normal fashion when it's turned off to block the voltage so the diode in this case will be forward biased but still no current will flow because the transistor will block on the other hand as far as conducting current you can see that this switch really only is designed to work with positive current when it's on if the current reverses polarity of course the diode will become reverse biased and turn off and with reverse polarity the transistor in general isn't designed to conduct reverse current either so we can only have positive current and so this is a two-quadrant switch in which we can conduct either polarity i mean conduct positive current only but block either polarity of voltage another well-known device that can function in this way is the silicon-controlled rectifier or scr which is able to block reverse voltage or positive voltage if it's not triggered but it can conduct current in only one direction so then here's our ideal switch the iv curve of the the composite connection of semiconductors is this where we can be anywhere on the horizontal axis when the transistor is off and we can be on the positive vertical axis when the transistor is on so this switch will work in quadrant one and in quadrant two and of course if you would if you need a voltage bi-directional two-quadrant switch for quadrants three and four you just need to connect the the transistor and diode in the opposite direction to get it to conduct current in the other direction there are examples of the need for voltage bi-directional switches they aren't as varied as the current bi-directional case here is one that is it's a dc to three-phase ac inverter that is based on the buck boost converter and the left hand side looks like a buck boost converter where the inductor is connected to vg during the first interval by turning this transistor on during what we normally call the d prime interval of the dc-dc version of the converter here is actually divided into multiple intervals and what you do is you take the energy that was stored in the inductor during the first interval and you release it to the output by connecting different combinations of these switches uh or turning them on so for example if you turn this one on and this one on then the inductor current will flow this way it'll charge this capacitor and flow that way by controlling the relative amounts of time that the these different switches are on on the sec the output side we can control the amount of current that we send to each of the three phases and synthesize output sinusoidal waveforms in that manner i'm not going to go into that in great detail it can get complicated but basically we can build a dc to three phase ac inverter that that synthesizes output side current waveforms that are sinusoidal a uh perhaps more well-known converter that is similar to this one is called the current source inverter and we can view that one as being derived from the boost converter so basically instead of this part of the circuit we replace it simply with a dc source and an inductor and connect it here um this this inductor has a current with small ripple um and that current is again switched to the different three-phase outputs uh to synthesize output sinusoidal current waveforms so that's based on the boost and it's called the current source inverter it's a well-known inverter circuit and it requires these voltage bi-directional switches as well the reason we need voltage bi-directional switches is that these switches have to block the output line-to-line voltages and since the output voltages are ac they have to block either polarity of voltage on the other hand if if power only flows in one direction then the inductor current is always of one polarity and that current is what the switches have to conduct so a uni-directional current flow type switch is appropriate in that case okay the most general kind of switch is the four quadrant switch and here we have a switch that's capable of blocking either polarity of voltage and conducting either polarity of current so it can work at all four currents in all four quadrants this essentially has to be a a an active switch that is controlled by a control terminal there's some exceptions but in the general case it is active and you have to be careful what you wish for because it can be complicated control to control the switch to turn on and off at exactly the right time we don't have a diode that will turn on or off when in coordination with some other switch in the circuit here are some ways to realize a four quadrant switch this one this first one involves taking two voltage bi-directional switches that are two quadrant and putting them in parallel so each of these switches here can block either polarity of voltage and the the left side switch conducts positive current and the right side switch conducts negative current so together they can conduct either polarity of current the the second case here involves two current bidirectional switches that are put in series so each of these can conduct either polarity of current but the top one can only block positive voltage while the bottom one can only block negative voltage but when you put them in series they can block either polarity of voltage and then the third case is not derived from any of the previous configurations uh basically these diodes connect the transistor up in either direction as appropriate to conduct current in the right direction and when the transistor is off the the switch network will block either polarity of voltage the most commonly used version here when we need four quadrant switches is the first one and the real reason for that has to do with how we coordinate the switching turning off one four quadrant switch and turning on the next one it's easier to coordinate that switching with this network than with the other two but that's beyond the scope of what we're going to talk about but just to give you an example here is a well-known converter that uses four quadrant switches this is a called a matrix converter and it converts three phase ac into three phase ac of a different frequency and voltage and basically it takes nine four quadrant switches and these switches are able to produce output voltage waveforms of these these three nodes here that are different combinations of the input voltage the three phase input voltages and by pulse width modulating or quickly switching these devices we can actually make the average values of these output voltage waveforms um be controlled in and so we can actually synthesize switched waveforms here whose low frequency components have a different frequency and and amplitude than the input these require four quadrant switches because the switches have to block the line-to-line ac input voltage which is ac and they have to conduct the three-phase ac output current which is also ac so we need the most general type of four-quadrant switch this is a pretty interesting converter but it takes some extended time to understand how to control these switches and synthesize the waveforms and it can it can get pretty complicated okay i want to talk also just for a minute about synchronous rectifiers [Music] we've talked about the idea that a mosfet is inherently a current bi-directional switch and this property can be used in what is called a synchronous rectifier application where we replace a diode with a mosfet and take advantage of this reverse current capability so if we need a single quadrant switch that turns out to be a diode type characteristic what we can do is take the diode in this polarity and replace it with a mosfet here in which the current goes what is normally considered backwards from source to drain instead of from drain to source so with a mosfet connected backwards its iv instantaneous characteristic looks like this basically i've just flipped the voltage axis because in fact i flipped both axes because we've connected the the mosfet up backwards and what we do is we take advantage of this part and this part of the instantaneous iv characteristic to get a characteristic that works like a diode now to get this to happen you have to have some control circuit that drives the gate as needed to make the mosfet turn on and off when the diode would have turned on and off but we can build a circuit to do that and get the mosfet to work like a diode now why would you want to do that well the celebrated application of this it was the original application i'm aware of was in a low voltage computer power supplies and the trend for decades and computer power supplies has been that the voltage goes down and the current goes up as time passes and we now have processor chips with gate lengths of tens of nano meters and what happens when the the gate links are scaled down is that the voltage of the power supply has to scale down as well at the same time we're increasing the number of transistors on the chip which makes the current go up and so um what's happened in the computer business is that we need power supplies that are a fraction of a volt and tens of amps or hundreds of amps even depending on the the size of the system so how do we build a converter that can operate with high efficiency and produce such low voltages and high currents well in the chapter 3 we looked at modeling these converters and what we found was that the diode forward voltage drop appeared in our model and was a source of conduction loss and so if we have a diode with a forward voltage drop of just to pick a round number one volt and our output voltage is less than a volt then we're going to have a lot of power loss in the diode and nowadays the the diode power loss or this conduction loss may be greater than the power that goes to the load so we don't have a high efficiency system and the forward voltage drop is something that is hard to scale we can't just buy a larger diode chip and get the voltage to go down there are shocky diodes low voltage shocky diodes may have a forward drop of say 0.4 volts instead of 0.7 or 1 volt but even then we want to get the the voltage drop of our switches to be below a tenth of a volt to get good efficiency so the solution is to put a mosfet here where the forward voltage drop is dependent simply on the on resistance and the current and so if you buy a bigger mosfet you you can get a lower on resistance and if you're willing to pay you can buy as big of a mosfet as you want and get as low of an on resistance as you want and make the the forward voltage drop below so that you can get high efficiency and so that's what's done so we have our main mosfet that is the normal mosfet of the the buck converter here and then we have we replace the diode of the buck converter with a synchronous rectifier and we basically drive the synchronous rectifier with a gate drive signal that is the complement of the gate drive signal of the main fet so that when the main fed turns off the the synchronous fat turns on and vice versa there's actually a little bit of dead time there where we make sure that we don't have the case where both fits are on momentarily at the same time and shorting out vg so this is useful in low voltage high current applications and we find nowadays that eminent applications of uh tens of volts you know below 100 volts i would say the synchronous rectifier will lead to higher efficiency and so we find it used quite a bit in applications at below 100 volts where we're concerned with the efficiency for the next several lectures we're going to discuss power semiconductor devices so we will cover some of the most widely used devices currently in power electronics and specifically power diodes power mosfets and the insulated gate bipolar transistor which is an evolution of the bipolar junction transistor some of the things we're going to talk about first of all we're going to talk about this basic trade-off that's important to power electronics trade-off between the on resistance or forward voltage drop of the device versus its breakdown voltage versus its switching times and these different devices have different tradeoffs between these three figures of merit we'll also talk about minority carrier versus majority carrier devices and how they lead to different trade-offs and then the other thing we really want to understand is uh from a power electronic standpoint is switching times of these devices where the switching times come from how they work and how they lead to switching loss which is one of the important loss mechanisms in power converters that we haven't yet modeled and so another objective here is going to be to extend the equivalent circuit models of chapter 3 to include switching loss and we'll talk about how to do that as well now i don't want this to become a course in semiconductor device physics the assumption here is that you have been introduced to semiconductor devices in an electronics class my intention is to perhaps remind you of some of the introductory basics and then discuss at a high level the important things that happen in these devices from a power electronics standpoint so we want to build power switches effectively out of semiconductor devices and what this means is that the power switch must be able to change between the off state and on state and another way to say that is the off state is an insulating state in which the device does not conduct current and the on state is a conducting state where we have mobile charges that can conduct current and we need to be able to switch between the two so here's a diagram of a some kind of device where we have contacts on the two sides that are say metal contacts to the outside world these blue regions and the clear region in the middle is our switching device uh to be an insulator what that really means is that there are no mobile charges inside this insulator that can move and conduct current between the the two contacts so if we connect the voltage across this device there will be no current because we have an insulating element in the path conversely in a conductor we have a we have mobile charges in the conducting material that can easily jump from one atom to the next and conduct current so for example the cl in the classic uh metal material the electrons are thermally excited to higher energy states where they're easily able to jump from one atom to the next and conduct current so in that case if we place a voltage across the conductor then what we see is that we'll get a current to conduct so here the plus charges are are really the charges in the uh the nuclei of the atoms and the minus charges are electrons that are mobile and easily able to jump from one atom to the next so the plus charges stay fixed and the minus charges move now in this in this drawing another thing we can say is that there are no big fields coming out of the conductor the number of plus charges and minus charges is the same so that overall the conductor is charge neutral we simply have charges hopping from one uh atom to the next and every time an electron leaves the left side another one comes in on the right side so that we're charge neutral so if we say q minus is the total amount of mobile electro electron charge or the charge contained by all the minuses in this volume and we have what we call a transit time which is the time it takes your average electron to to get from one end to the other of this conducting device then we can say that the total current the average current is the total amount of charge divided by the time it takes to move across the material okay also again since the fixed positive charge is equal in magnitude to the fixed negative charge another way to say that is that the current is equal to the plus charge divided by the transit time so these are basic ideas that we call charge control ideas of of how a switch will conduct another thing to say about the conductor in a semiconductor material the basic pure silicon is uh is mostly an insulator however we can dope the silicon with impurities in the crystal lattice of the silicon that provide charges that can be become mobile so by doping the silicon to make it in-type silicon there are extra electrons in the lattice of the silicon and at room temperature they they're in thermally excited states where they are able to easily jump from one atom to the next and conduct electricity so it's similar in that case to the conductor and in p-doped silicon we actually can dope the silicon with impurities in which there are there's a deficit of electrons and actually we have what are called holes that are essentially an absence of an electron that can also jump from one atom to the next and effectively it behaves as if there is a plus charge that we call a hole that can conduct current another case yet another case is a minority carrier where if we have some in doped silicon next to say p doped silicon the in the in dope silicon the electrons can diffuse into the p-doped silicon where they are um electrons at a high energy state in the p-doped silicon and these are called minority carrier electrons and they can also conduct current likewise with we can have minority carrier holes okay let's discuss the mosfet now in the mosfet we have a source and a drain that are doped silicon and [Music] there are contacts to the source and the drain to the outside world we have a channel in between the source and drain that is where we can turn the device on and off by putting charge in and out of the channel to provide a conducting path between the source and the drain the gate in a power mosfet is polysilicon it's basically a conducting material that is a control terminal at the mosfet and separating the gate from the channel is what we call an oxide layer it's silicon dioxide that is glass and it is an insulator so basically the gate and gate oxide built on top of the channel form a capacitor and so the gate and the channel are the two plates of the capacitor and the oxide layer is the dielectric of the capacitor so right now i've illustrated the mosfet in the off state there is no charge in the channel to conduct current between the source and the drain so if we apply a voltage between the drain and the source this is an in-channel mosfet i've shown so when we put positive voltage on the drain with respect to the source we get no drain current because the channel is turned off okay to turn the mosfet on we apply positive voltage to the gate and effectively we charge up the capacitance between the gate and the channel so what's happening here is you can see the plus charges actually come in from the gate minus charges come in from the source and charges up this capacitance of the gate to the channel and the minus charges now are in the channel inhabit the channel and they're able to conduct between the source and the drain okay one i i should mention that um what what you just saw with the charges coming in was the turn on switching time the time it takes the gate driver to supply this charge through the gate and establish charge in the channel is the the turn on switching time and during that time we have a current coming in from the gate driver a gate current that is supplying this charge so it's supplying the plus charges to the gate and we get minus charges equal number of minus charges from the source and there's a current then that is the the change equal to the change in charge on this capacitance okay once the device is turned on we don't need to supply any more current we have a capacitor it's turned on and its voltage doesn't change any further then we don't need to have any gate current and the gate current can go to zero so now with the capacitor charged up these electrons can conduct current between the source and the drain so with a voltage applied to the drain we will get electrons flowing between source and drain and the drain current then we can say is equal to the total amount of minus charge in the channel divided by the transit time which is the time it takes an average charge to move across the channel okay um since the plus charge on the gate equals the minus charge in the channel another way to write the drain current is that it's equal to the plus charge on the gate divided by the transit time so the drain current is effectively controlled by the charge that we put on the control terminal or on the gate okay now let's turn it off so here i've shorted out the gate to the source which shorts out the capacitance and it draws the charge back out of the gate so that was the turn off switching time that you just saw in the turn off switching transition so we have gate current now that is extracting the the charge the plus charges out of the gate so it's a negative current the channel charges went to the drain and the drain current continued to conduct until all of the charge was gone and after that then the drain current goes to zero so the turn off switching time is the time that it takes to remove all of this charge okay a bipolar junction transistor has a similar looking structure with an emitter and a collector that are in for an npn transistor they're in doped silicon that is connected uh through contacts to the outside world and instead of a channel we have a base region that is p-type silicon there's a contact there that's the control terminal or the base terminal of the tran the bjt here effectively the base emitter junction which is a diode junction is controlled by the base terminal and in this way we can control the charge in the base and control the current so here there is no charge in the base the bjt is initially off we apply a current to the collector nothing happens we're in the off state okay to turn the bjt on we forward bias the base emitter junction so we apply a positive voltage to the base that is sufficient to forward bias this diode junction so say plus 0.7 volts or so when we do that electrons from the emitter the in region of the emitter uh will diffuse into the base and become minority carriers likewise positive charges or majority carrier holes will come in from the external base driver circuit and come into the base region okay so during this turn on switching time of the bjt we have positive current flowing into the base that is supplying the holes into the base region we likewise have elec equal number of electrons diffusing across the the diode junction to become minority carrier electrons in the base okay so we now have our electrons in the base and they're able to conduct current between the collector and emitter what i should say though is once we have supplied these minority carriers to first order we don't need any more base current just like the mosfet we've supplied the charge to the base to provide conducting or carriers that can conduct current and so to first order the base current can go to zero at this point to second order we have an effect with minority carriers called recombination where these electrons are in an a higher energy state in the base and they can actually decay out of this energy state and that's called recombination so effectively when they decay the minority carrier electron combines with a majority carrier hole and effectively the charges go away so if we want to maintain the our minority carriers in the base region we have to supply base current to to basically supply more carriers to uh and and counteract the recombination so the second order we can say that we need a current supplied from the base that is uh equal to the amount of charge divided by the lifetime where the lifetime is the average time that it takes a minority carrier to decay and go away so this is a maintenance current but it's smaller than the current that we initially had to turn the device on okay having said that if we apply a voltage to the collector now then the bjt will conduct and so we have our electrons that are able to conduct across the base into the collector and provide current so just as in the previous examples that the current that we get is equal to the total amount of electron or minus charge in the base divided by the transit time which is the time it takes to get across the base and since the minority carrier electrons are equal in magnitude to the majority carrier holes we can say that there's uh that the the current is equal to the total amount of holes or plus charge that came in from the base terminal and is still there in the base divided by the transit time okay let's turn the bjt off now so to turn the bjt off one thing we could do is simply remove the the base current and wait for the electrons to go away or recombine but that may take a long time and lead to a long switching time so better way to do it and i'll show it to you again is to reverse the direction of voltage on the base and actively remove the charge so if we actively pull the holes or these plus charges out of the base we can turn the bipolar transistor off faster so here we put a negative voltage on the base and actively remove the charge and that was the turn off switching time or the time it took to remove the charge between the base and collector again this is a charge controlled device like the other devices by pulling the plus charges out of the base we can get the minus charges to go away also and then the device stops conducting so now we're back into the off state or the insulating state where we have no minority charge in the base available to conduct current between the collector and emitter okay so i've introduced the notions that we have charge controlled power semiconductor devices and the time that it takes to insert or remove the the charge necessary to make the device conduct or or insulate is a switching time and so it takes time to do that now switching time leads to switching loss because during these switching transitions the semiconductor devices are not all the way on or all the way off but they actually have voltage and across them and current through them that leads to high instantaneous power so let's look at that here is a first attempt at explaining switching loss and we're going to refine this several times in the upcoming lectures but this is a good starting point so let's suppose we have a buck converter built here with a tran transistor that has some switching times and we have an ideal diode okay now our real diode is not ideal but when we after we talk about diodes we'll talk about how the diode affects the switching time but for now let's call it ideal and we have a transistor that has is driven by some gate driver circuit and it takes time for the gate driver to turn the transistor on and off so here's an example of the turn off transition of of the mosfet so we start out with a mosfet on right here okay so by being on the voltage across the mosfet is essentially zero and the current through the mosfet is the inductor current il okay what during this time what's happening on the diode is that the diode is off it is reverse biased and the diode voltage is minus vg like this and the diode current is zero which is shown here okay the bottom waveform here is the instantaneous power in the transistor or in the mosfet so it's the product of the mosfet voltage and current i'm not modeling forward voltage drops or conduction losses of the mosfet right here so initially with the mosfet all the way on we have zero power loss in the mosfet okay at this point right here the gate driver starts to turn off the mosfet okay so it starts with the gate driver voltage goes low or goes negative and it starts to pull charge out of the gate of the mosfet out of its capacitance and discharge the gate capacitance to to remove charge from the mosfet that takes some time the gate driver has some thevenin equivalent impedance so there's some resistance we get rc circuits effectively and it takes time to discharge the gate okay so we start to turn off the mosfet now the thing about turning off the mosfet here is that the current in the mosfet can't change until the diode turns on if you look at this this is called a clamped inductive load we have an inductor that's that's large and has small ripple its current doesn't change very quickly and the current the inductor is connected to this node and its current has to come from somewhere and if the diode is off the only place it can come from is the mosfet so what happens is that initially the voltage on the mosfet rises like this the um the gate driver is actually discharging the gate to drain capacitance of the mosfet during this time we'll talk about that more in an upcoming lecture but as long as the voltage across the mosfet is less than vg the diode is reverse biased because the diode voltage vb is the mosfet voltage minus vg and so the diode voltage is starting out at minus vg and as the mosfet voltage increases the diode voltage increases and finally at this point where the mosfet voltage equals vg we finally get that the diode voltage starts to exceed zero volts and the diode can turn on so that's what happens right here okay so the diode starts to turn on right there at that point the current from the inductor can shift from flowing through the mosfet to flowing through the diode and during this time the mosfet current will fall and how fast it it falls depends on how fast the gate driver can remove charge from the channel and the gate capacitance of the mosfet so the gate to source capacitance and so that takes some time and during that time then the diode current increases the mosfet current decreases finally right here the mosfet is all the way off with no current and with blocking the full input voltage vg and so after that we have zero current in the mosfet and the diode conducts all of the current okay so here is a plot of the instantaneous power in the mosfet and at this point it starts to rise above zero because we have voltage across the mosfet and current through the mosfet and it reaches a peak right here where the mosfet voltage is maximum and equal to vg and the mosfet current is still equal to the inductor current so we get a peak instantaneous power equal to vg times the output current or the inductor current and that's a considerable amount of power in the mosfet after that the current goes down again until at this point the power gets to zero so we have some high instantaneous power loss and in fact the energy that's lost is the integral of the power so if you find the area under this curve you get the energy in joules that is lost during the turn off switching transition as drawn here with straight lines that are idealized we could say that that energy is the area of this triangle so it's one half the base times the height of the triangle the height is vg times the inductor current here's the half and the base is the switching time it's total time from here to here okay so every time we turn the mosfet off we get that much switching loss now we multiply that by the number of times per second that we do this and we get an average power loss we also can apply the same arguments at the other transition where we turn on the transistor and really the same thing happens in reverse and we get a similar kind of expression for an energy lost during the turn on transition so we get a total switching loss according to this model that's equal to the switching frequency times the energy loss during the turn off transition plus the energy loss during the turn on transition okay this can be considerable in fact often this is larger than all the losses combined from conduction and forward voltage drops that we model in chapter three in many converters this is large the largest loss in the converter so we have to account for this and one of the things we're going to do soon is refine our models from the previous chapter to include terms that account for the switching loss the other thing i'd say is that this is highly idealized and in fact it it's not really right it's a good start but it doesn't account for what happens in the real devices so back when i was a graduate student this was what everybody taught and it led us all to try to do things to reduce switching loss and improve efficiency to and what we did was develop what were called zero current switching converters that would change the order of this switching so that the current would go to zero first and then the voltage would rise and you would think the loss would go away but people built those converters and found that they got the same exact efficiency in the same switching loss as before so in fact while this is a good start it we need to refine the models of the semiconductor devices to understand what's what the switching times are really happening what's going on and how to refine this model to correctly predict the actual switching loss so we're going to do that in the next several lectures this lecture is a short summary of what is going on inside a diode and how it switches we're going to cover reverse recovery and the basic mechanisms governing switching of diodes so here is a pn junction type diode we have p doped material on the left side of the junction and in doped material on the right side in doped material for example means that we have doped the silicon lattice with impurities that provide additional or extra electrons to the lattice that are of high energy state and easily able to conduct current by jumping from one atom to the next and we call these majority carrier electrons in the in material in a similar manner in the p material we have majority carrier holes that can easily conduct current and jump from one atom to the next when these devices are at reasonable temperatures such as at room temperature of the the thermal energy of the majority carriers makes them bounce around inside the lattice so we can think of them as having thermally induced vibrations or jumping around and bouncing around okay at the pn junction what happens then is what's called a depletion region is formed and this happens because when these majority carriers are bouncing around they tend to diffuse in directions in the direction of reduced concentration and so right here at the junction we have a high concentration of holes on the left side and a high concentration of electrons on the right and so what happens is that the the thermal energy will cause the majority carriers to diffuse in the direction of reducing concentration and so electrons will diffuse from right to left across the junction and holes will diffuse from left to right when these majority carriers diffuse across the junction they become what are called minority carriers on the opposite side and in the process of of moving across the junction they leave behind ionized atoms or net charges on the dopant atoms of the region that they left and so when electrons diffuse into the p region they leave behind plus charges or positively charged atoms in the n region and likewise holes when they diffuse into the n region they leave behind negatively ionized atoms in the p region and as a result there's an electric field between these charges and that electric field makes a voltage voltage is the integral of the electric field and so you can see there'll be a net voltage that is plus on the right side and let minus on the left side of the junction and this region here is called the depletion region or the space charge layer across this depletion region then this voltage actually opposes the diffusion so a a positive charge would have to basically diffuse uphill voltage wise and what happens is that this uh voltage across this depletion region builds up until in equilibrium the the uh the voltage completely stops these uh the further diffusion of holes and electrons across the junction and at that point we're in steady state or the device is in steady state it has this built-in voltage across the junction and no further majority carriers will diffuse here's what happens under reverse bias conditions so now we have contact attacks to the outside world and we apply a voltage that's normally we consider a negative voltage across the device so this voltage is negative with respect to the p material which is the anode and it's plus with respect to the in material which is the cathode okay when we do this we have added voltage this adds voltage across the depletion region and makes the depletion region get larger basically this entire voltage externally applied voltage is is blocked by the depletion region you can see that to increase the depletion region size and have more voltage across it you have to add charge to it and that charge actually comes from the external voltage source or the external circuit so in this sense the depletion region acts as a capacitance we have to add charge to it to increase its voltage and we generally call this the junction capacitance of the device and it's a real capacitor that can store energy in its electric field that you can get back later if you change the voltage under forward bias conditions what we do is we we increase the voltage of the p region with respect to the n region and that makes the depletion region get smaller when the depletion region is smaller it no longer stops the diffusion of majority carriers across the junction and so we we start to get a flow of charge across the junction in each direction so that holes from the p region diffuse into the n region where they become what we call minority carrier holes they're still at a high enough energy state in the in region that they can easily jump from one atom to the next and conduct current similarly electrons from the n region will diffuse into the p region where they become minority carrier electrons and are able to conduct current there under forward bias conditions then here's what happens we have current coming in the contact from the external circuit that in the p region is gives us holes those holes can do one of two things one is they can diffuse across the the junction become minority carriers in the in region and eventually they recombine with electrons majority carrier electrons in the in region and so an electron from the negative terminal may come in and recombine with one of these holes the other thing that the p m the holes can do in the p region is to recombine with an electron that has diffused in from the n region so we can have a minority carrier electron that recombines with the majority carrier hole in the p region so those are the two things that can happen and under forward bias conditions that accounts for the entire current of the diode the entire current consists of recombination on one side or the other of the junction and as long as there are minority carriers in these regions we continue to have current so what we have then is here's a plot of the minority carriers or the minority carrier concentration on the two sides of the junction so here for example we have holes that diffuse across the junction and they diffuse at some rate that depends on the slope of this concentration characteristic so they diffuse towards regions of lower concentration and they they last in this region for some time that we call the lifetime and that on average after the lifetime is over they will have recombined it they they recombine as they they diffuse and we get a concentration that reduces the farther away we get from the junction and we have that on both sides of the p-n junction so if we want to turn off the diode we have to remove these minority carriers on the two sides of the junction one by one way or another to stop the recombination and stop the current from flowing thus the diode is charge controlled by the minority carriers in the p and n regions of the diode here is the classic first order charge control model for the diode it relates first the voltage across the depletion region v to the minority carrier charge concentration at the edge of the depletion region so this q of t we can take as the charge of the minority carrier holes right here at this point or the minority carrier electrons at this point and these are related to the voltage basically the applied voltage of the diode but really the voltage across the depletion region so you're used to seeing perhaps the steady state current versus voltage characteristic of the diode that has this exponential function that's a steady state relationship and in fact the correct relationship is this one okay so the charge is charge concentration at the edge of the depletion region is this quantity given that quantity we can turns out we can find what this entire distribution of charge is and there's a total amount of charge or area under these concentration curves that is the total stored minority charge in the device in the charge control model we can make lumped element models of the device and the the simplest first order lumped element model has a single lump of charge that is this total amount of charge stored here and that charge then is is related again with some constant to the uh voltage across the depletion region with the exponential characteristic the charge control equation says that the total charge can increase if we put current more current into the diode which supplies more charge or it can decrease by recombination and so this is actually a dynamic equation of the diode that says how the stored minority charge can vary i should say also that these equations don't include the capacitance across the junction which we talked about previously that's a separate element and we're talking simply about what happens with with the holes and electrons moving across the junction okay so if we look at the charge control equation in equilibrium and the diode works in steady state dq dt is zero and then we can solve this equation equate this to zero and find that the current is equal to this stored charge over the lifetime and you can plug this stored charge equation here into there like this and if you do you find that the current follows the exponential diode relation that we're all used to which is the exponential iv curve that that does this but this is an equilibrium equation and during transients in the diode this doesn't have to happen i have to to be followed the dqdt term can be non-zero and we can actually for example have current that removes stored charge or at stored charge that deviates from this iv curve and that's in fact what happens during the switching times of the diode here in fact is a plot of what or a sketch of what really happens so here we have the diode on initially so there's some voltage like seven tenths of a volt across the diode and we're conducting some current and there's some stored charge so here i've drawn on one side of the depletion region the stored minority charge and what it looks like so again we we have a slope here that determines the rate at which the carriers diffuse and that slope is proportional to the current in the device and so we'll have charges moving this way and conducting this current now at some point t 0 then this is the initial condition at this point we start to try to switch the diode off so the external circuit will switch and as a result we will start removing charge from the diode and so what we'll observe in the external circuit is this uh this current through the diode and we can actually remove charge actively from the diode by simply having a reverse current in our charge control equation i becomes negative which makes dq dt be negative and we actively remove say the minority carrier holes backwards across the junction and minority electrons backwards across the junction that contribute to this current okay so with this negative current then say at time t1 we have negative current and we're pulling charges backwards across the junction the slope by pulling these charges we will reduce the concentration of charge uh say in the end region and with the slope going the opposite direction you'll see charge is actually moving the other direction at time t2 we've removed enough stored charge stored minority charge that the the charge right at the edge of the depletion region goes to zero and that's what happens right here okay at that point the voltage on the depletion region can start to change it's no longer this exponential function that is roughly 0.7 volts but the diode voltage starts to actually finally go negative so before this point where we still have positive minority charge at the edge of the depletion region the diode voltage remains positive and you can barely see the difference it's it's more or less 0.7 volts and it looks like the diode really is still on so at this point we start to see the voltage change and see the volt some negative voltage we continue to remove charge and what happens say at time t3 here is that we've removed this charge and the charge is now zero at this point x3 and the depletion region has actually grown and comes all the way out to x3 that which is consistent with having negative voltage across the device and finally out here at time t4 we've removed all of the storage charge there's no more minority charge left and finally the diode is all the way off and it can block the full negative voltage the circuit imposes without conducting current okay let's look a little at the power flow so power is the voltage times the current and the power in the diode at this point before we switch the diode off we had positive current flowing into the diode positive voltage across it and we had some power loss in the diode that is the conduction loss of the diode you can see where the current goes negative and the voltage is still positive over this time the diode actually supplies power it has positive voltage and negative current and we're actually some of the energy of these minority carrier charges is being extracted where we actively pull them out of the the diode now that that amount of power being supplied is not very much and in fact in a switching converter generally that power is is lost in the mosfet anyway and so it's not a very significant amount of power but then for the next part say from this time we have negative voltage that is substantial and we have negative current and this is substantial power loss that happens in the diode itself and it can make the diode hot so this is real switching loss in the diode and in fact we're going to see um in the next lecture that that we get substantial switching loss if we consider both the mosfet and the diode over this whole time and this is a significant amount of loss and it's often in what we call hard switched or conventional switching converters this is often the largest single source of loss in the circuit because you can see this is a lot of current in a typical diode this peak negative current may be several times the on state current and this is the full off state voltage so the product of these is a very high instantaneous power so this is called the reverse recovery of the diode this time here is the turn off switching time that's often called the reverse recovery time and this charge which is the minority charge that is actively removed from the device is called the recovered charge now you can also switch the diode off by simply stopping the current and waiting a long time for all the charge to recombine or we can do some combination where we maybe reduce the rate at which the current changes and goes negative which lets more of the charge recombine in the diode unless of it be actively removed but if we switch the diode off quickly this is what happens we basically actively pull all of the stored minority charge out of the diode and it becomes recovered charge the familiar iv characteristic of the diode is an equilibrium relationship and it can be violated during the switching times in particular to turn off a diode we typically have this reverse recovery transient in which the we have a large reverse current that violates the x the equilibrium exponential iv characteristic of the diode and this reverse recovery time and recovered charge can induce substantial switching loss in the mosfet and substantial loss in the diode itself as well in the next lecture we're going to calculate what the switching losses are and model them in the switching converter let's now extend our equivalent circuit models that we derived in chapter 3 to include switching loss and in particular what we're going to do is model the switching laws caused by the diode reverse recovery okay so uh to obtain tractable results anytime we're modeling switching loss where we have complicated ringing waveforms and things we do have to idealize the waveforms but we can at least include a current spike that comes from the diode switching loss and include the effects of the diode switching time and this actually gives pretty good analytical results there are many different things that cause switching loss as in some of the accompanying lectures here that are discussed what we're going to talk about here is the the loss just from the diode reverse recovery but we can include other things like the effective semiconductor output capacitances and their energy stored in a similar manner so the approach here is to correctly account for how the waveforms change and in particular when we apply inductor volts second balance or capacitor charge balance or when we calculate the average ig input current we include how the diode reverse recovery changes those waveforms and the i think the simplest way to do this is to first sketch the transistor and diode waveforms including the diode reverse recovery and then relate those waveforms to the inductor and capacitor and other waveforms so i'm going to do an example of this with a buck converter so here is a buck converter with a diode that has a reverse recovery and with a mosfet that switches quickly so we're going to neglect everything all the sources of loss except for the diode reverse recovery okay so here is the same buck converter and i've i've identified the transistor voltage and current vt and it the diode voltage and current v d and id and i've sketched ideal eyes switched waveforms for all of those here and they look like the usual rectangular waveforms except for the switching time here of the diode where we have the current spike from reverse recovery and its effect on all four of the waveforms okay when we do this there's now somewhat of an ambiguity in deciding how to define the duty cycle so if we look closely at these waveforms the switching period oops the switching period ts goes from here [Music] out to right here and what's happening is that the gate driver and the controller is telling the mosfet to turn on at this point here but then we go through the switching transition of the diode and the voltage finally changes out here okay i've also squared up this waveform this is what we call a snappy diode that where its voltage changes abruptly at the end of the the reverse recovery period to idealize that waveform so that's a somewhat pessimistic assumption as far as loss goes but anyway we have this ambiguity in what is the duty cycle so actually we can define the duty cycle however we want and we could define the duty cycle based on when the driver starts to turn on the mosfet right there or we could define it based on when the voltage of the mosfet actually changes right here and what's sketched here is the second choice it's based on the voltage in the power stage of the converter so if i extend the this light blue line which is the transistor drain to source voltage it's actually falling right there and if we extend the the pink line which is the mosfet current it has the current spike from reverse recovery that is the same as this one that is happening right before t equals 0. so what i'm going to do is define the duty cycle as is being determined by this interval from here to here so it does not include the diode switching time the justification for that is that this voltage waveform is what drives the volt second balance on the inductor and what ultimately determines the effective conversion ratio of the ideal converter and once when we add the switching loss of this converter as well if we define the duty cycle as being determined by what comes out of our controller and so this period starts here duty cycle is defined starting there we get equations that are valid but they are not as physically connected to the ideal converter we predict a different effective conversion ratio and we get a lot of extra terms around the power stage model so in effectively the transistor switching and the diode and transistor switching times change the duty cycle from what comes out of the controller and the duty cycle that the converter effectively works at is based on this voltage waveform of when the voltage at the switch node is high or when the voltage across the mosfet switches okay so with that definition dts then is this interval that starts after the diode is switched off and ends when the mosfet is switched off and then the d prime interval is the rest of the period going from here out to there okay given those definitions and given these waveforms let's now work out the inductor voltage waveform and apply volts second balance we'll work out the capacitor current waveform and apply charge balance and we'll work out the input current waveform ig and calculate its average or dc component so here's the inductor voltage waveform and with the definition that i just gave you for the duty cycle here's what it looks like and it's the same waveform we've been drawing all along this semester so that definition gives us back the original ideal inductor voltage waveform and volt second balance gives us the usual equation that the average inductor voltage is dvg minus v the reverse recovery also doesn't change the capacitor current waveform capacitors on the other side of the inductor and we get the same equation from charge balance that we always have gotten so so far no change and in fact from these two equations we could construct the usual equivalent circuit model which is shown right here with a dvg source the inductor and capacitor node and voltage loops and the load okay the only thing in the buck converter at least that is changed by the switching times is the input current waveform ig and what happens is that this uh inductor or that this uh current waveform of the transistor current is the same as ig so when so the transistor current waveform which includes the reverse recovery current spike gets drawn out of vg and changes the ig waveform so we need to work out the average value or dc component of this current and we do it in the usual way oops do it in the usual way by integrating to find its average value so the average value of ig is the area under the curve divided by the period the area under this part of this first pulse this area would be the dune cycle d times ts times the height il where il is the dc inductor current we have zero here and then we have some area here during the switching time so we have the area of this rectangle first which is the diode reverse recovery time in width and in height it's the inductor current il then we also have the area of this current spike which is the recovered charge that has area qr okay dividing through by ts gives us this expression so in the ideal case we had just one term average ig was dil coming from this area but now the reverse recovery adds two more terms and the two terms depend on the reverse recovery time and the recovered charge so an equivalent circuit to go along with that is this so ig is the current coming out of vg that's equal to d i l which is the usual um dependent source that is part of the ideal transformer then we have two more terms that i'm going to draw as independent current sources that model losses from reverse recovery so the recovered charge gives us one of the sources qr over ts is the same as qr times the switching frequency fs this is the the charge per unit switching period drawn out of the source by the recovered charge and it has an average value qr over ks and then the other term from the switching time we draw an extra amount of current equal to the inductor current uh during the switching period or the i'm sorry the reverse recovery time and that gives us another term that is this source so we put them together and combine the ideals or the in the dependent sources into a transformer and we get this equivalent circuit okay this looks really like the uh ideal buck converter with its one to d transformer turns ratio uh plus though it has these two extra sources which model the switching loss you can see that these sources consume power because they have a voltage vg across them and they draw a current out of vg and so we get a switching loss equal to vg times the sum of these currents okay this can be a considerable amount of power you have to plug in the numbers to c but it's very easy for this to be more power loss than all of the conduction losses combined it depends on the application you can also see that this power depends on the switching frequency or the switching period in fact it's directly proportional to switching frequency since fs is one over ts so if we turn up the switching frequency and switch faster we turn up this switching loss and we get more power loss which will reduce our efficiency okay we can solve the model so uh the output voltage is easy it's just the duty cycle times the input voltage as far as efficiency goes as usual we can find the input power and the output power and then divide the input power is uh the input voltage multiplied by ig and we can solve these things to find ig and it's really from the average input current equation the output power is the output voltage times the inductor current il and so we can divide to find the efficiency and when we do we get an expression like this uh it has a denominator that's one plus this uh effective switching loss or terms that depend on the switching loss and this denominator depends on the switching frequency so again efficiency will go down as the switching frequency goes up here's a plot so i'm plugging in some numbers here i think they're pretty innocent looking numbers we're 100 kilohertz switching frequency input voltage at 24 volts uh here are some numbers for the diode reverse recovery time of 75 nanoseconds and recovered charge of 70.75 micro coulombs um this is an okay diode it's not a super great diode but it's not a real bad diode either kind of an average diode and then i've plotted those equations and i haven't made any attempt to model how the recovered charge or the reverse recovery time uh change with the inductor current uh it turns out that they do change with inductor current but it's actually a pretty weak dependence and modeling exactly how they change with inductor current is fairly complicated they generally don't give you enough information on the data sheet to to have a valid model and so we'll just keep these constant but you can see when we plot this so here i'm plotting efficiency versus duty cycle what happens is the the switching loss degrades the efficiency at low duty cycle and the reason for that is that as we turn down the duty cycle we're turning down the output voltage and turning down the output power but we aren't changing the the switching loss by very much switching loss does depend on the inductor current and that is modeled but the dependence is not very strong so you can see that what happens is the efficiency becomes poor at low duty cycles and this is something we observe in practice as well and in fact at low output power the switching loss makes the the efficiency be become poor so a converter may have a good efficiency at full power but when you turn down the power the efficiency goes down for a variety of reasons but one of the major reasons is because we still have substantial switching loss without having a lot of output power to provide another example of modeling switching loss let's drive the equivalent circuit model of a boost converter including the diode reverse recovery so here's a boost converter and we're going to assume that all of the components are ideal except for the diode and its reverse recovery and we'll derive the averaged or dc equivalent circuit model with terms to incorporate the effects of reverse recovery and its loss so here is the boost converter and the transistor in diode waveforms they look uh qualitatively the same as the waveforms we drew in the last lecture for the buck converter with reverse recovery the only real difference is that the values are a little different so the voltages across the the semiconductors when they're off is equal to the output voltage v rather than the input voltage just as in the buck converter we need to correctly define the duty cycle to obtain the effective duty cycle that the power converter operates with and we define this using the voltage waveform so the voltage at this node the switching node or the voltage across the mosfet is this blue line shown here and the effective duty cycle is judged from the mosfet voltage is the time when the mosfet voltage is low so we will define this this is the interval dts and d prime ts is the remainder of the interval including the diode switching time so we have these waveforms uh with the reverse recovery uh peak peak and current and the uh and the reverse recovery time t sub r so we the first step then given these uh transistor and diode waveforms is to relate the inductor voltage and capacitor current waveforms to these waveforms so i'll define the inductor voltage here and you can see from writing the equation of this loop the inductor voltage v sub l is equal to the input voltage vg minus the transistor voltage v sub t likewise for capacitor current we can write the node equation here and the capacitor current is equal to for this converter the diode current i sub d minus the load current which is v over r okay so we can find the average values of these quantities and apply volt second balance and charge balance accounting for the effect of the reverse recovery of the diode on these waveforms one other small point we in general need to write the equation of the input current ig but for the boost converter ig is the same as il so if we find the average il or dc component of inductor current that's the same as ig and we don't have to do any further work so let's do an inductor volt second balance first so vl is vg minus vt and vt is the blue waveform right here so on the next slide that waveform is drawn again and in fact here the waveform looks the same as in the ideal case the reverse recovery time doesn't change this waveform assuming we correctly define the duty cycle as i previously mentioned so we can apply volts second balance the average inductor voltage is the duty cycle times vg for the first interval plus d prime times this vg minus v and we get the same equation as usual for the inductor volts second balance and we can even construct an equivalent circuit model from this what would we get vg would be the input voltage source it has a current capital i g and this is this voltage is equal to minus d prime v so we have a dependent source it depends on the output voltage v multiplied by d prime okay so here's our loop equation corresponding to the inductor equation and again ig is the same as the inductor current capital il which is the current in this loop okay for the capacitor we apply charge balance to the capacitor and basic basically the easy way to do this is to write the node equation at the node where the capacitor is connected which says that the diode current is equal to the capacitor current plus the load current so one way to apply charge balance is to average these currents so we can say that the dc component of diode current equals the dc component of capacitor current plus the dc component of load current now dc component of capacitor current is zero in steady state by charge balance which says that the dc component of the diode current is equal to the dc load current so we've drawn the diode current waveform already for the diode we simply find its average and equate that to the load current and that effectively gives us the equation from capacitor charge balance so here is that equation repeated and to find the average diode current we need to average this current so the average diode current is what 1 over ts times the integral of id of t over the period so we get 1 over ts times the area under the curve okay so what is the area under the curve well first we need the area of this rectangle and this rectangle has a height of il and it has a length or base that is this time this time and that time is in fact what we normally call d prime but then minus the reverse recovery time so this length here is d prime t s minus the reverse recovery time t sub r and then to that we also have to add the uh recovered charge qr and in this case it's a negative quantity we usually on data sheets express qr is a positive number so we should subtract qr okay we can now divide through by the ts and what we get is d prime il minus i l t r over t s minus q r over t s which is the value of the expression shown here okay let's construct an equivalent circuit then to go with that so this is the node equation at the output node of the capacitor and it says that the load current which is the current coming out of the node and is the well we have a voltage v across the resistor r that is our load resistor that is equal to the sum of these three terms uh the first term is d prime il which is the usual dependent source that will become part of the dc transformer second term is i l times t r over t s and it has a minus sign so it's drawn out of the node and i'll draw it as a independent source it is a loss element the value i l times t r over t s and then the last term is minus q r over t s which is also a loss term so we for the boost converter we get two extra loss terms from the diode reverse recovery and in the case of this converter these currents are actually at the output node where the capacitor is and in fact the reverse recovery causes current to be drawn out of the output capacitor as if it's loading the output so the switching loss effectively is another load on the output that is a loss element if we take the the two models together for the the inductor volt second balance and the capacitor charge balance and combine them including combining the dependent sources into the usual transformer we get this uh model now so we have the ideal boost uh transformer that is d prime to one and then we have our two uh independent current sources or sinks actually that model the loss caused by the reverse recovery and these are actual power losses the voltage across these sources is the output voltage v and they cause a power loss equal to that voltage v times the sum of the two currents like this uh here's a sample prediction of this model this sample here actually includes an inductor resistance and so the inductor resistance be if we included it would become an additional resistor in the loop where the inductor is and so with that resistor included here's a solution of v over vg for this boost converter with and without the reverse recovery and so the effect of the diode reverse recovery is to actually appear to load down the output which reduces the output voltage especially at high duty cycle okay so to summarize we can model or incorporate switching loss into our average model what we have to do is draw the actual transistor and diode voltage and current waveforms and then relate those waveforms to the inductor voltage and capacitor current waveforms and perhaps also to ig we then apply volt second balance and charge balance as usual and proceed to [Music] compute the equations of the dc conditions in the circuit including the effects of this diode reverse recovery and we can then reconstruct the equivalent circuit by a simple extension of of how we did it in chapter 3 to get equivalent circuit models that incorporate switching loss we'll conclude our discussion of power diodes with several additional important topics the first one is the trade-off between the blocking voltage or or breakdown voltage of the diode versus the forward voltage drop versus the switching speed and this is a trade-off that is inherent in any of our power semiconductor devices but it's interesting to see how it works in the pn junction type diode in any semiconductor device the breakdown voltage is related to the doping concentration of the silicon material basically silicon with heavy doping has a low on resistance but it also breaks down at a lower electric field strength than if there was light doping this is a fundamental characteristic of avalanche breakdown in semiconductor materials and so in a diode we have to include a low doping concentration high resistivity region in order to achieve a specified voltage breakdown here's a diagram that's then a more accurate description of the diode which where we have the p region on the left the n region on the right and a lightly doped region in the center here it's denoted in minus which means low concentration of doping often this is also called i for intrinsic and this is called a pin diode so we have this low doping concentration in the middle of the diode and under reverse bias constant conditions the depletion region extends far out into this in minus region uh with electric fields that the that are high in strength that the light or low doping concentration material can withstand and we get our voltage breakdown from that now the worry is that by having such a large region that is high resistivity this will increase the on resistance or the forward voltage drop of the device in a minority carrier device though such as the pn diode that doesn't happen and instead what happens is a phenomenon known as conductivity modulation basically the holes from the p region can diffuse into the lightly doped region where they become minority carriers and they actually reduce the effective resistivity of the material they provide charges that are able to conduct current even though that the doping concentration is low and in fact electrons from the n region can diffuse into this in minus region as well as well and so we get a large concentration of minority carriers in the lightly doped region that gives us a low on resistance so the good news is we can get a high voltage breakdown with a low on resistance at the same time but the bad news is we have a lot more stored minority charge in the n minus region and therefore the switching time is longer because it takes longer to remove all of that charge to turn the device off so then here are what waveforms we actually observe in a real power diode so we start out with a diode in the off state blocking negative voltage and conducting no current with this large depletion region that is blocking the voltage we then try to turn on the the diode so the whatever circuit is connected to the diode will apply positive current to the diode and raise the voltage across the diode initially this positive current charges the capacitance of the depletion region and raises the voltage at some rate eventually the voltage on the diode is positive and the diode turns on and starts to conduct current we may or may not observe some uh voltage overshoot in as shown here what's happening here is that uh if we turn the the diode on quickly it's possible that the we we initially observe the voltage of the uh lightly doped region without conductivity modulation so we have a high on resistance then and we have a large forward drop that might be 10 or 20 volts but then after a short period we get minority charge and conductivity modulation that brings the forward drop back down to something like 7 10 or 1 volt and then the diode after that conducts with the low on resistance when it's time to turn off the diode the external circuit attempts to reverse polarity or reverse the current through the diode and as we've discussed uh in the previous several lectures we get a reverse recovery transient with current flowing backwards through the diode to extract the minority charge and the only difference here now is that we have a lot more minority charge from the conductivity modulation of the the lightly doped region so we observe the reverse recovery which involves removing the recovered charge and also while the voltage is changing we're also charging the capacitance of the depletion region so as we've already discussed then this reverse recovery can induce substantial power loss in the mosfet and during the last part of the reverse recovery in the diode as well but at a minimum we see this large reverse recovery current causing a current spike in the mosfet while the mosfet is held on with height with voltage across it and so we get large instantaneous power loss induced in the mosfet by the diode reverse recovery okay here's a little terminology it's traditional in the power semiconductor field to give names to diodes according to their application a standard recovery diode typically is one that is intended for low voltage or low frequency applications such as 60 cycle or 50 cycle rectifiers often these for these diodes the reverse recovery time is not specified but they won't work in a 100 kilohertz switching converter because their reverse recovery times are too long fast recovery or ultra fast recovery or more recently other adjectives like super or hyper fast uh refer to diodes where the reverse recovery time is specified and they are intended for switching converter applications so these will have reverse recovery times for fast recovery diodes that might be between 100 nanoseconds and a microsecond or ultra fast might be 100 nanoseconds or below and nowadays we can get 600 volt pn junction rectifiers with recovery times as low as 25 or 35 nanoseconds a shocky diode we have to mention also a shocky diode has a metal semiconductor junction instead of a pn junction it is inherently a majority carrier device it does not work on minority carriers as we've been discussing so far as a result the shocky diode for silicon rectifiers doesn't have conductivity modulation nor does it have long reverse recovery times so that's good news and bad news also shocky diodes are limited for silicon devices to low voltage applications generally 100 volts and below and they're very good at say 45 volts and below they switch very fast although they do have significant junction capacitance that that can affect the circuit and cause switching loss recently we have shocky diodes built with wide bandgap semiconductors such as silicon carbide or even more recently gallium nitride these wide band gap devices can attain high high breakdown voltages of 600 or 1200 volts in a shocky diode and still have a majority carrier device they have larger forward drops maybe one and a half or two volts instead of something a little below one volt however they're very fast switching and very often in in a power converter uh you'll find that if you add a if you replace your pn diode with a schottky barrier silicon carbide diode even though the conduction loss is higher the switching loss is much lower and overall the efficiency is reduced so these have found there there are commercial devices that have found some good uh industry acceptance the major problem today with them is that they is their high cost a silicon carbide 600 volt diode might cost five or ten times as much as a comparable silicon diode but there definitely is a measurable reduction in in switching loss okay let's discuss paralleling diodes with a minority carrier or pn junction diode classically we we're not supposed to parallel diodes so for example suppose we needed a 10 amp diode but all we were able to buy or all we had in the lab were five amp diodes so you might say well i'll i'll take two five amp diodes and put them in parallel and in the process we're hoping that the 10 amps will divide evenly between the two diodes so we get five amps through each diode generally this doesn't work and the reason for this is the temperature dependence of the equilibrium iv characteristic of the diode basically when a diode gets hotter its voltage goes down for the same current or conversely for the same voltage the current will go up and this is in a minority carrier say pn junction diode so here's what happens you you put the two diodes in parallel we apply 10 amps there's there's some small difference between the diodes which makes one of the diodes take a little more current than the other and it gets a little hotter and so say that this diode 1 is hotter runs a little hotter than diode 2. that makes it take more current so instead of 5 amps it might take 6 amps and diode 2 then will take 4 amps okay that makes the first diode get hotter yet the second diode get cooler and even more of the current goes through the first diode and less through the second and so we have thermal instability where one diode hogs all the current and what will happen is that diode will hog the current it'll overheat and it'll in fact exceed its rating and so it will fail and you know once it fails if it fails as an open circuit then that makes all 10 amps go through the other diode and it fails also so in general with minority carrier devices we aren't able to parallel with majority carrier devices it the forward drop goes up as the temperature goes up and it may be possible to parallel so if you want to get minority carrier diodes to to share current we have to do what i would consider heroic measures so we can sort devices and select match devices we can package them on a common thermal substrate so they run at the same temperature and in fact if you want to build a high current module of diodes that's exactly what's done or we can try to build some external circuitry that forces the currents to balance usually that circuitry takes a lot of work to make work and generally what we would rather do is find a large enough diode and just use one one last short topic we've talked about how diode reverse recovery induces switching loss in the mosfet in a say a buck converter sometimes we have circuits where the diode is not directly connected to a transistor and here's an example of one this often happens in transformer isolated converters or in some other converter circuits that have effectively an inductor that provides current to the diode so in this this example i'm going to model the rest of the converter circuit with some switched voltage source that's shown here it's initially a positive voltage and then at some time it switches to a negative voltage and what what's shown here is an inductor and then a diode with its junction capacitance here c that's shown explicitly this is the capacitance of the depletion region so initially we have positive voltage applied that positive voltage makes positive current flow through the inductor and that current flows through the diode turning the diode on okay so the diode is on and basically this positive input voltage v1 appears across the inductor and the inductor current increases with a slope given by the voltage divided by the inductance and rises to some positive current and at this time where we switch the input this voltage negative at this point so we have negative voltage applied across the inductor the inductor current is still positive initially so this positive current continues to flow through the diode which keeps it on and so the input voltage appears across the inductor and the inductor current then ramps down from the negative applied voltage like this eventually it reaches zero and what happens then well just because the inductor current is zero doesn't mean the diode turns off we have to first remove its stored charge and that takes some time so during the reverse recovery time of the diode the diode stays on it maintains its same forward voltage drop and so we have the same negative voltage across the inductor which continues to charge the inductor current negative okay so we get some negative current in the inductor and then at some point right here finally the diode turns off because it's its stored charge has been removed by the negative current and at that point the diode switches off and what happens to the inductor current we built up some negative current in the inductor we're storing energy in the inductor one half li squared and the inductor current has to go somewhere well the only place it can go is through the capacitance now so the junction capacitance of the reverse bias diode will will form a resonance circuit in connection with the inductor that resonant circuit has some initial condition on current and so we get the inductor and capacitor forming a resonance circuit that rings like this and we get diode voltage that rings as well well eventually that ringing decays to zero by parasitic losses in the circuit so we have resistance of the inductor winding we have core loss and other losses around the circuit and something damps out the ringing and makes it decay to zero whatever that is whatever damps it dissipates the power and so it'll it will get hot and this is in fact a form of switching loss the diode induces energy storage in the inductor which is then lost after the diode turns off okay however much energy loss that is i'm going to call it w off right now as the energy loss during the diode turn off train or stored in the energy during the diode turn off transition and then lost that energy multiplied by the switching frequency is another form of switching loss and so in our power converters when we see ringing like this that decays that ringing is an additional component of switching loss in this particular case we can actually calculate how much energy there is the recovered charge which is the area under the curve is defined as the integral of the current during this this time as usual so and the recovered charge is something that is a property of the diode we can also calculate the energy stored in the inductor during this interval and basically if you we integrate the voltage on the inductor this negative voltage times the inductor current that's the integral of the power which is the total energy stored in the inductor from here to here okay well the voltage is this voltage minus v2 and the current is this current waveform but we can we can plug v2 into here and uh since v2 is a constant the the integral of the power is actually just the constant v2 times the integral of the current and the integral of the current is the recovered charge of the diode so what we get is that the the during the diode reverse recovery time energy is stored in the inductor equal to the applied voltage v2 times the diode uh stored charge or recovered charge okay that energy becomes stored in the inductor so it's equal to one half li squared at this point right here at the peak of the current and that energy then rings between the inductor and capacitor during these oscillations and eventually gets dissipated by whatever damps the ringing so this is the amount of energy we lose every time this diode switches off we multiply it by the switching frequency to get the switching loss and so the diode can induce switching loss in things the size of mosfet and this recut reverse recovery will induce ringing that that is another form of switching loss for the next two lectures we're going to cover the power mosfet in this lecture i'm going to introduce what the power mosfet is how it's constructed and some of its basic characteristics and then in the next lecture we'll talk about mosfet gate drive circuits and some practical considerations here's a cross-sectional view of a power mosfet so if you take a silicon wafer and cut it and look at from the side what what it looks like generally you see something like this so we have vertical current flow through the mosfet and really through just about every power device so for example the drain here is connects to the bottom of the silicon wafer and the source and the gate are put on the top side what i've shown here the cross-hatched regions uh represent metal contacts so we have metal contacts to the drain and to the source this shaded area here around the gate is the oxide layer that's silicon dioxide or glass that insulates the gate from the rest of the mosfet the clear area inside that is the gate that is usually a polysilicon material but it's it's a conductor that it forms the gate and then here we have the the doped silicon this is an in-channel mosfet the source is connected right here the drain is comes from the bottom at this in region but there's lightly doped in material and the drain actually is right there and so we have a mosfet that connects to the drain and source like that the p material is called the substrate and in the practical power mosfet the substrate is shorted to the source this particular structure shown here is called the dmos structure in which the gate and the source are built laterally on the the top surface of the mosfet we also have other variants uh grooved structures such as vmos or umos or sometimes known by a trade name such as trench moss are mater devices where grooves are cut into the top of the side of the silicon and the gate is actually built vertically along the side of the groove we also have the super moss structure which is a relatively recent improvement that reduces the on resistance of high voltage devices and so we have some very good 600 volt mosfets built with the super moss structure okay the the power mosfet is a highly interdigitated device meaning that we replicate the mosfet many times on the top side of of a chip and connect these all all of these little mosfets in parallel to get one large high current mosfet so here you can see there's a channel right here of a mosfet there's another one there there's another one over here another one over here and so on and so we pack these mosfets as tightly as we can together to fill the surface of a silicon wafer and build a high current device okay so then current will when the mosfet is turned on current will flow in the drain and then through the channel and out the source that's for positive positive current flow actually a better way to say it perhaps is that we have electrons coming in from the source to the in material they form a when the gate is forward biased they form an inversion layer or channel here that effectively is more in region that connects the the source to the drain and then the electrons flow this way and out the drain in the off state where the mosfet is the gate is turned off positive voltage is put on the drain with respect to the source then we have a depletion region formed that blocks the voltage and really this is the same as in the pn diode and in fact if you look at the ch the uh substrate p material and this region right here this actually looks like a pn diode as we talked about in the previous lectures for the diode we have lightly doped in region in the drain that gives the device the breakdown voltage that it requires and so in the reverse state we have an electric field really across here that has the voltage drop from drain to source okay to obtain a high voltage breakdown rating as in any semiconductor device we require lightly doped high resistivity material in the on state that again current flows this way as i mentioned before a key distinguishing feature here though is that the current has to flow through this in minus region and much of the on resistance of the device appears across that region and since it is lightly doped it has a high resistance and this is the on resistance primarily of the mosfet you do have to con you know it put that in series with the resistance of the channel and the resistance of the contacts and the leads and the package but in general this in minus region dominates the on resistance and in the mosfet we have a majority carrier device where there are no minority carriers to cause conductivity modulation the way there is in in the diode so this accounts for why mosfets tend to have a relatively high on resistance especially at high voltage levels and so a high voltage rated device will have to have an especially high resistance in minus region and the on resistance goes up quickly with voltage so we have very good mosfets at low voltage but at high voltage we start to prefer minority carrier devices that can give lower forward voltage drops okay the mosfet has this body diode which i mentioned uh last week the body diode actually is comes from this p-n junction between the substrate p material and the drain you can see there is a p-n junction there and it's a real diode so effectively we have our mosfet represented by the channel and it's in parallel with the body diode okay so this is a real diode it has reverse recovery it can conduct current if it's forward biased and that happens if we put plus on the source and minus on the drain then this diode can become forward biased and current can flow this way the diode can conduct the full rated current of the mosfet after all it is the same device however the the caveat is that the diode is not necessarily optimized for fast switching speed and it depends on the diode some mosfets are designed to have on purpose a fast recovery diode and others are not are the static iv characteristics of the mosfet this is drawn in a little different way than you might be used to seeing and it's drawn to emphasize how to control the mosfet so the vertical axis is the drain current the horizontal axis here is the gate to source voltage and we have characteristics or lines of constant drain voltage so basically if the gate voltage is below a voltage known as the threshold voltage then the mosfet is off and the gate driver has to bring the gate voltage or gate to source voltage higher than the threshold to turn the mosfet on so here as you can see we as we raise the voltage above the threshold for the first few volts the mosfet starts to turn on and when we're up here at maybe 10 or 15 volts the mosfet is all the way on and then we have you know for a given drain current we have a low drain to source voltage so we generally will operate the mosfet turned on over here where we have a low voltage and in this case the drain of source voltage is approximately proportional to the drain current and we approximate this as being saying the drain voltage is the drain current multiplied by the on resistance of the mosfet okay in minor in majority carrier devices like this we typically have a positive temperature coefficient of the on resistance which means as the device gets hot the on resistance goes up so first of all this makes it easier to parallel mosfets they don't have this mechanism this positive feedback mechanism that will make one device hog all the current on the other hand the on resistance is highest at high temperature and when we choose an on resistance we have to be cognizant of that and think about what in the worst case what is the hottest the mosfet will run and what is the on resistance under those conditions here's an equivalent circuit of the mosfet to first order showing the capacitances so the mosfet has capacitance from each terminal to each other terminal and we'll talk about what each of these is one at a time so first of all the gate to source capacitance is this ideal capacitor of the mosfet between the the gate terminal and the channel and it's drawn here with respect to the source so this is the capacitance across here from gate to channel and we charge up this gate to source capacitance to put charge in the channel and turn the mosfet on you know in addition to that you can see that there's overlap between the gate contact and the drain and this forms what's called the gate to drain capacitance okay we do what we can to make that capacitance small but in fact charging it up does help put additional charge in the drain region and somewhat reduce the on resistance typically the gate to drain capacitance is small compared to the gate to source capacitance so it's a it's a lot fewer picofarads of capacitance however it has a lot of voltage across it the gate the source capacitance might get charged up to 10 or 15 volts whereas the gate to drain capacitance gets charged up to basically the full drain voltage and so although the capacitance is small the voltage across it is large and the amount of charge in this capacitance can be large in fact in some devices it's larger than the charge in the gate to source capacitance this has some impact on considerations on how we design the gate driver circuit and we're going to talk about that in the next lecture the third capacitance is the drain to source capacitance across here and this is basically the capacitance of the the body diode it's really capacitance from here to here and this is the same capacitance that the this pn junction diode would have um and it behaves a lot like a diode capacitance it's it's non-linear we didn't talk about this with respect to the p-n junction diode but it's actually a similar kind of function here's a classic formula for a depletion capacitance and it's a non-linear device that where the capacitance varies roughly as the inverse square root of the voltage so as the voltage goes up the capacitance goes down the incremental capacitance and we get this inverse square root formula okay these output capacitances of devices such as the drain to source capacitance of the mosfet and the junction capacitance of the diode are another mechanism that can lead to switching loss basically these capacitances get charged up with energy when there's voltage on across them so for example when the mosfet is off we have vg across here there's no voltage across the mo the diode junction capacitance but as far as ac signals are going go the junction capacitance is effectively in parallel with cds and the two add together when we turn the mosfet on we short out these capacitances and any energy stored in them gets dumped into the channel of the mosfet during the turn on transition so if this is a linear capacitance we can say that these capacitances store energy equal to one half cv squared where he here's c is the effective capacitance if these were linear capacitances of the the output capacitances of the devices and vg is the voltage at the input to the converter so we lose this much energy every time we turn the mosfet on and short the capacitances out okay so this would cause a switching loss that is equal to this energy stored in these capacitances multiplied by the number of times per second that we short out the capacitances or multiplied by the switching frequency so this is another mechanism for switching loss if we want to account for the fact that the capacitances are non-linear we can integrate the power that charges up the capacitances to see how much energy the nonlinear capacitors store so we can integrate the power which is voltage times the current and plug in the expression for i as c dv dt for this capacitor and when we integrate the square root function we find that the energy that is lost is is equal to this and it has a two-thirds factor instead of a one-half factor and we evaluate the capacitance at the off state condition of the mosfet so that makes us actually lose a little markup energy instead of one half cv squared we actually have four thirds times as much or two thirds cv squared okay conclusion so the mosfet is a majority carrier device that's good news because it has very fast switching speed we don't have all the time that it takes to to remove all of this stored charge that was causing conductivity modulation in the diode and so mosfets can switch much faster and in a mosfet that is uh you know it's typical to get um switching times of a few tens of nanoseconds and in low voltage devices even faster so typical switching frequencies tens and hundreds of kilohertz are easy and we have you know low voltage computer type power supplies that run in the megahertz of switching frequency range because of the high speed of these devices the bad news though is that on resistance increases rapidly with the rated voltage because these are majority carrier devices and don't have conductivity modulation so we have good super junction mosfets at 600 volts but we can't go much higher than that before the the on resistance has increased too much to where we it's just not economical to use the device they're very easy to drive we simply switch the the gate to source voltage between 0 and 10 or 15 volts and very easy to control then when they're on and off we also have logic level mosfets that are designed to operate with a 5 volt gate voltage instead of 10 or 15. okay one other thing these devices generally are loss limited we when we design a power converter we generally don't select the mosfet based on the current rating which is a peak current rating of the device instead usually we select them on the basis of on resistance and conduction loss so often we will choose a device with a peak current rating much higher than we intend to use and instead we choose the mosfet size to get a given low on resistance as needed in our application so as a result these are very rugged devices as well that have quite a bit of margin when it comes to current foreign in this lecture we'll discuss some practical circuits for driving power mosfets and some of their uh basic considerations so what i've shown here is as an example is a synchronous buck converter in other words it's a dc to dc buck converter that has a main power transistor realized by a mosfet and the rectifier or diode is replaced with a synchronous rectifier with mosfet q2 this configuration of two mosfets uh actually is uh is found in many different applications and circuits besides this one it's found in quite a few different dc dc converters this is often called a half bridge circuit it's also found in many different dc to ac inverter circuits and the fundamentals of driving the switches really is the same in all of those circuits but here to keep things simple we'll discuss the dc to dc synchronous buck converter so we have two transistors that we have to drive so the main switch q1 and the synchronous rectifier q2 each are turned on and off by their control terminals or gates and the the voltage of the gate must be referenced to the the voltage at the source of the same transistor so to turn q2 on and off let's see its source is connected to ground so its gate voltage must be switched with respect to ground and that's relatively easy on the other hand with q1 the source of q1 is connected to the switch node which is switching high and low when q2 is on the switch node is at ground and when q1 is on the switch node is at vg so to to control q1 we must drive the gate voltage with respect to this switch node and drive what the gate voltage must be high say 12 volts above the switch node when we want to turn q1 on and it must be essentially equal to the switch node voltage when we want to turn q1 off so we need to build circuitry that is connected to the two gates here to drive the transistors in that manner here's a common way to do it called this is called a half bridge gate driver and we can buy commercial half bridge gate drivers that are implemented in high voltage integrated circuits that do this so the half bridge gate driver really comprises all of these things inside the box the half bridge gate driver contains a low side driver that drives the low side mosfet q2 with respect to ground and it contains a high side driver that drives the high side mosfet q1 and its gate with respect to its source which is the switch node each of these triangles represents really a buffer or a circuit that can produce the high currents needed to switch the gates charge the gate capacitances quickly at the proper voltage in this example 12 volts the high side driver additionally has what's called a bootstrap power supply this buffer or gate driver circuitry has to be powered and it's its effective ground is connected to the switch node and so the power supply to this uh driver circuitry must be 12 volts with respect to the switch node and so this diode and capacitor comprise what is called a bootstrap power supply when the low side mosfet q2 is turned on then the switch node here is at ground and what happens is this bootstrap diode turns on and this capacitor is charged to 12 volts through this path so we get 12 volts here okay then when q2 turns off and q1 turns on when this this gate driver turns q1 on the switch node goes high up to the input voltage and the capacitor voltage flies up with it so that the the gate drive circuit for q1 is still powered from from the 12 volt supply held on this capacitor in this at this point this diode is off and the high side driver is powered from the bootstrap capacitor so the bootstrap capacitor gets charged to 12 volts every time the low side transistor turns on something about bootstrap power supplies is they don't work unless you switch the low side transistor you have to turn it on every so often to charge up this bootstrap capacitor or it will run out of charge the gate drive circuitry generally includes what's called under voltage lockout protection in which if the power supply of the the driver circuitry is too low then the the circuitry will turn off and it will actually protect the power stage by shorting the the gate to the source and keeping the transistors turned off so if you ever let this bootstrap capacitor voltage get below the under voltage lockout of this gate driver circuit then it will just turn off and the circuit won't work so we have to keep switching q2 occasionally to make sure we charge this up now under normal operation we're switching q1 and q2 on and off at the switching frequency and that's no problem okay some more things we need in this circuit this logic input is a control signal that goes high and low and it commands uh q1 and q2 to turn on and off so when the logic signal is high typically we will turn q1 on and we'll turn q2 off so there's a block here and here that perform that function of providing the correct signals going into these buffer amplifiers to make them turn on and off in this way likewise when the logic input is low we want q1 off and q2 on okay this logic signal typically is referred to this ground and so it'll go high and low with respect to ground and the this upper block must also then perform the function of level shifting it has to level shift the signal up to this node voltage to provide the proper signals at the input of this buffer something else we have to do is we have to make sure that we never turn both q1 and q2 on at the same time if we ever do that then they'll short out our input voltage what we've been calling vg and even if they're on for a few nanoseconds at the same time during those few nanoseconds we can get a large current spike that flows through the two transistors that can be many many amps and that current spike can damage the transistors and make the circuit fail or at the very least it will cause a large amount of power loss so what we do is this logic circuitry these boxes also implement what is called uh break before make operation or non-overlapping conduction circuitry which makes sure that first first you turn the on or conducting transistor off we wait some amount of time called a dead time and then we turn the other opposite device on so both devices are off for some short amount of time that is long enough to make sure that when we account for switching times and driver delays and things we never have the case where both transistors can be on at the same time okay let's consider what happens let's suppose we had q2 on and we want to turn it off so our logic signal will start out low q2 is on then we we go to switch the logic signal high there is some dead time and then we turn q1 on so one question is what happens during the dead time which device conducts if any okay well the answer is uh during the dead time the inductor current continues to flow and we can't interrupt the inductor current let's assume here that the inductor current is positive and flowing to the output if we have a positive input and positive output voltage then the inductor current will flow in this direction if both transistors are off the inductor current has to flow somewhere well the only place it can flow then is through d2 through this body diode the inductor current will forward bias the body diode and it will conduct during the dead time and the switch volt node voltage vs will remain low with d2 conducting then right here where we turn q1 on the the upper gate driver will turn q1 on and what it does then is we have current flowing this way that will both make d2 go through a reverse recovery and cause switching loss and also it will supply the inductor current after the reverse recovery of d2 is done d2 will be off and then q1 will conduct and supply the inductor current so we have switching loss that happens even with synchronous rectifiers because of the need to have dead times we get switching loss caused by the body diodes let's consider some further details about the switching times and exactly how these devices switch if you recall from the last lecture we discussed the capacitances of the mosfet and we drew this circuit here where we explicitly showed the mosfet with its body diode and with the three capacitances gate to drain and gate to source capacitance and the output or drain to source capacitance to understand what happens during the switching times we also need to model the driver and understand how the driver circuit interacts to charge and discharge these capacitances so here's our driver symbol and what i'm going to do is represent the driver output terminals here with a simple thevenin equivalent model here in which we model the output terminals with some thevenin equivalent voltage source and some thevenin equivalent resistance okay we know the thevenin equivalent voltage is simply the open circuit output voltage which is something that will switch between zero and in this case 12 volts and r thevenin is the thevenin equivalent output resistance of the driver now real drivers are more complicated than this but this is actually a reasonably good first order model that explains all of the things we want to talk about today and in a typical driver we can think of this thevenin equivalent resistance as being the on resistance of the mosfets that are inside the driver that are driving the output terminal of the of the chip it's traditional to to rate drivers according to the peak current that they can produce at their output and so if we buy a driver that's rated at 12 volts and one amp for example then the 7 equivalent resistance would be 12 volts divided by the 1 amp or 12 ohms okay so our actual circuit then drives the gate to source of the mosfet with this thevenin equivalent so what i've done here is to insert the gate driver thevenin equivalent model in in place of the low side driver and i've replaced the mosfet with an equivalent circuit model that contains the capacitances the body diode and here we have a dependent current source that models how the drain current depends on the gate voltage let's consider the transient in which we turn off the lower mosfet wait for some dead time and then turn on the upper mosfet so we're going to draw some waveforms here's our thevenin equivalent driver voltage that will start out high so that the lower mosfet q2 is on so it's 12 volts and under these conditions the gate to source voltage of q2 this voltage here is is high also and is equal to 12 volts okay the switch node voltage vs of t is equal to ground because the lower mosfet is on and at this point what we'll do is we'll make our drivers switch low so the thevenin voltage goes low we have an rc circuit where the thevenin resistance is connected across vgs and it will pull current out of the this uh gate and discharge these capacitors so there'll be some kind of rc type decay and the voltage will go low okay when the lower mosfet is off recall the body diode is still on to conduct the inductor current and that will keep this vs switch node voltage low okay and it will stay low for the dead time so here is the dead time or perhaps even we should call this whole time here the dead time okay at this point the upper gate driver turns q1 on so what happens when it turns q1 on is the upper mosfet will try to pull this node high now as we just mentioned a few minutes ago this makes the body diode of q2 undergo a reverse recovery and so it takes some time for that reverse recovery to happen before this voltage at this node can start to rise but eventually it does rise i'm sorry that's vgs it's vs that eventually rises okay now when vs is rising what happens to the charge on this gate capacitance cgd you can see that this node here is all of a sudden going high we're trying to hold vgs low okay so um that means the voltage across this capacitance has to increase along with the drain voltage okay so if we if the voltage on this cgd increases it means current must flow through cgd according to i as c dv dt for this capacitor to charge the capacitor so we have current flowing in this direction through cgd now the big question is once that current gets to the the node here at the gate where does it go from there it has to keep has to complete the loop and keep flowing someplace okay now where can it flow can it flow this way well we hope it does we we hope the gate driver keeps pulling this node down the ground and keeps this mosfet off but the trouble is we initially the voltage here is zero so we have zero volts here and we have zero volts here coming out of this the driver or v thevenin is zero you can see right there so there's no voltage across our our thevenin and if there's no voltage across this resistor then by ohm's law there's no current through the resistor so we have zero amps of current flowing through the resistor what that means is that the only place that this current can go is this way through cgs and because of that cgs will start to charge also and we'll see vgs rise well that's not good if vgs rises above the threshold voltage of of the gate call it vthr then then this transistor will start to turn on again and when it turns on again then it tries to pull this node back down and what happens is we get oscillations where this transistor will turn on and off many times and we get all kinds of switching loss in the process so the gate to drain capacitance causes problems and it's a common problem that we have to battle when we're designing these kind of circuits it's typical for the total charge in the gate to drain capacitance can be equal or even greater than the total charge in cgs so there's a lot of charge here and it's easy for that charge if it all flows this way to start charging up cgs well the hope here is that when this voltage rises a couple of volts then we do get start to get current flowing through the thevenin resistance and through the gate driver and if the thevenin resistance is low enough we hope the gate driver can hold this node voltage low enough that the q1 or q2 doesn't turn back on another thing we can do to help the situation is to slow down the speed at which q1 turns on and so we can put a resistor in the gate of q1 that makes it turn on more slowly and makes the voltage at this node rise more slowly so dv dt is lower and since i is c dv dt in this capacitor we get less current here and it's easier for the gate driver to to pull that current out of the gate and keep the transistor turned off so you'll often see little resistors of a few tens of ohms put in series with the gate we often will put a diode in that direction so that the resistor slows down how fast the q1 turns on but the diode lets the driver turn q1 off very quickly okay so that was the switching transition where q2 turns off and then we have d2 conducting during the dead time and then q1 turns on and during this time we saw that we get switching laws caused by the reverse recovery of the body diode d2 as well as by shorting out the output capacitances of the devices and possibly but hopefully not also from oscillations and spurious switching caused by the cgd problems let's now examine the other transition where we turn q1 off and q2 on so here we have q1 initially conducting which makes the switch node high and equal to the input voltage vg when we turn q1 off what happens who conducts then well again we have positive current flowing in the inductor both transistors are off and this positive inductor current has to flow somewhere and it will flow out of d2 the body diode d2 is the only remaining diode or only remaining semiconductor device that can conduct current in that direction so we get q1 turning off then d2 turns on and q2 turns on what happens to the the charge on the output capacitances of the mosfets when we turn q1 off what happens is the inductor current can discharge these capacitors and that current inductor current will actually discharge this node down to ground so we turn q1 off here and let's assume we have a nice big gate driver that can turn q1 off quickly and in fact let's assume that the gate if we get a big enough gate driver we can turn q1 off before the inductor has current has a chance to discharge the voltage at this node by very much so q1 gets turned off right away while there's still very little voltage across q1 the voltage is held close to zero by its capacitance and so then the inductor discharges this node down to ground discharging the energy in these capacitors when we get down to ground then the diode d2 will turn on and then at some point later after the dead time is over our q2 gate driver will turn on and turn q2 on what i've just described here actually doesn't have any mechanism for switching loss q1 turns off quickly the capacitance nothing else the straight the output capacitance cds holds the voltage across q1 to be close to zero during the turn off switching time of q1 and then the inductor actually recovers the energy that was stored in these capacitors and that energy flows to the load there's no reverse recovery of a diode we have d2 turning on but no diode turning off so there is no switching loss at this transition and this switching transition of q2 i'm sorry of this one q1 d2 q2 uh is what we call a zero voltage switching transition so zero voltage switching can happen in a situation like this where the voltage across the mosfet is held close to zero during its switching transition and we have no mechanism for switching loss now the other transition in the other direction was what we call hard switched and it has switching laws from diode reverse recovery and the other things we talked about so this converter here is what we would call a hard switched converter there are converter circuits that get zero voltage switching at both transitions and those are called slot switching or zero voltage switching converters and they can have considerably higher efficiency because most of the mechanisms of switching loss have been eliminated circuits that achieve zero voltage switching at both switching transitions are more complicated than the one we're talking about here and they're beyond the scope of this course they're actually the subject of an entire different course that we teach at colorado however it's interesting that even in the basic hard switched buck converter with mosfets and good gate drivers one of the transitions is essentially lossless and it's only the other transition where we have substantial switching loss in this lecture we will discuss the bipolar junction transistor or bjt and its modern successor the insulated gate bipolar transistor or igbt here's a sketch of the bipolar junction transistor it's an npn device and we have a lightly doped in region or intrinsic region in the collector so that this base collector junction pn junction really has the same structure as the classic power diode that we've already discussed so under reverse bias conditions where the collector voltage is positive the base is more negative than the collector and so is the emitter and in fact for to turn this device off the np or pn junction between the base and emitter is reverse biased and turned off so under these conditions the device is off and the electric field is in the in minus region that is able to block the the collector voltage this is a vertical device and again this is a cross-sectional area of the transistor or of the chip like the cross-sectional areas we've been drawing uh previously for the mosfet so to turn this device on what we do is our we have a driver circuit connected to the base and emitter that will forward bias the base emitter junction so apply a positive voltage at the base with respect to the emitter to forward biases junction and when we do that electrons from the emitter are able to diffuse across this junction into the base where they become minority carriers because the collector is is more positive voltage these electrons can easily continue to to flow and they'll actually flow into the collector region so by forward biasing the base emitter junction we get the minority carriers into the base and into the n minus region that are needed to conduct to the collector to emitter current in the process you can see that these minority carrier electrons can go into the n minus region and cause conductivity modulation that effectively reduces its own resistance and so like in the diode we can this conductivity modulation gives us the best of both worlds as far as high voltage breakdown plus low on on resistance but the price we pay is that we have a lot of stored minority charge in the n minus region that has to be removed if we want to turn the transistor off okay so to turn the transistor off when we're in this situation we have two choices uh one is we can simply reverse bias the pn base emitter junction with zero base current and wait for these charges to recombine and eventually they will and once they do then the transistor will turn off but the other thing we could do is actively remove the the electrons through the base terminal actually we put we would pull the electrons backwards uh with a negative base emitter current that will uh remove the stored charge in the base in a and in minus region so so here's a sketch of what the typical waveforms look like here we have a base drive circuit that uh i'm representing with some thevenin equivalent voltage source and resistance here is the voltage of that base driver voltage source it starts out negative which reverse biases the base emitter junction and turns the transistor off at this point our driver voltage goes high to turn on the base emitter junction and you find the base emitter voltage then of the base emitter diode goes forward and goes to plus 0.7 volts and then the transistor turns on so it has low collector to emitter voltage and high collector current and it's in the on state to turn the transistor off actively what we do is we make our driver actually go to a negative voltage and with that negative voltage we actively remove the stored charge out of the base okay and so the base current that we get base current waveform will go negative and actually look a lot like the reverse recovery current of a diode it's very similar in shape so we get a large negative current to actively remove the stored charge during this time the base emitter voltage is high for at least part of the reverse recovery time eventually the voltage will go low and at that point we'll see the transistor fully turn off so in order to switch the transistor off quickly we need this large negative current and so an ideal base current waveform then looks like this we have a large positive current initially to supply the charge to the base and turn it on then we have a small maintenance current to supply recombination in the base of the minority carriers maintain the minority carriers in the base and the collector and minus region and then to turn the transistor off we have a large negative current coming out of the base to remove the stored charge quickly finally once the transistor is off we have no base current okay one thing you might ask is well why can't we just build a big big base current driver that can pull the all this storage charge out very quickly and turn the transistor off just as fast as a mosfet well first of all it takes a lot more current than a mosfet but second there are limits to how large the base current can be very large negative base currents during the turn off switching time lead to a failure mechanism that is traditionally called a second breakdown mechanism that comes from current crowding so here's a here's an example or a sketch of what happens so we have a lot of stored charge in the in minus region and in the base p region and we're going to pull current out of the base where and pull that stored charge out so you can see the way the base and emitter contacts are built a big negative current requires that we have lateral current flowing through the base region and really through the n minus region also this lateral current causes a voltage drop in the resistance of the silicon so the silicon material has some resistance and so from the direction of the current you can see that the voltage in the base will be more positive in the center than it will be at the edge and what that does is it makes the base emitter junction be more positive in the middle than it is at the edge and so during this time the collector current will tend to want to flow there in the center so we get what's called current crowding and that current crowding heats up the center part of the base emitter junction more than it heats up the edges and we get this phenomenon where the center part of the the base emitter diode junction will hog the current it can even go into thermal runaway as we've discussed previously with the case of trying to parallel diodes effectively this you can think of this as like a distributed diode and we want the current to to flow uniformly across the whole region but in fact here we're making a crowd in the center and so this part will run hotter than the rest and it will tend to hog all of the current and this can make the transistor fail so this is called current crowding if you have a transistor that fails you can actually cut it open and look at it under a microscope and see that it failed at the in the middle of the emitter and uh you know then the engineer gets told that their their base negative base current at turn off is too large you can do a similar thing at turn on where current's going the other direction and in that case it can make the current all crowd at the edges of the emitter and you can see that with a microscope also so this is called a second breakdown mechanism that causes bjts to fail and the effect is that it limits how fast we can switch the practical bjt another different characteristic of the bjt relative to the mosfet is that um the at high current the bjt runs out of gain and we're not going to go into the various reasons for this but they're well understood and so uh this is a plot a base current versus collector current for a an actual bjt power bjt that is rated um 10 amps and 600 volts and so if the bjt operates with a constant current gain beta then you would expect the base current versus collector current to be a straight line with a slope equal to that current gain but you can see that at high values of collector current the slope becomes low and the transistor runs out of current gain and this is classic for any bjt for this particular device it's rated 10 amps and you can see that you can barely get 10 amps out of this transistor at all that's really the limit and so the current rating is determined really by the this gain characteristic of the transistor with a mosfet this doesn't happen a 10 amp rated mosfet will have curves like this that extend up to 20 and 30 and much higher currents and the mosfet then is actually thermally limited so you can run a mos a 10 amp mosfet at 20 amps for very short times as long as the average is low enough and you don't overheat the device so this loss of current gain provides another mechanism but that limits the power that we can get through the bjt the darlington connected bjt is one of the classic kind of early approaches to improve the gain of high current or high voltage bjts and at high voltage it's difficult to get high current gain at the same time so a thousand volt bjt might have a current gain of only a few and the darlington connected bjt gives us two transistors worth of gain and the total gain from the uh input base current of the first transistor to the collector current of the combination now is the product of the two betas so this this is one of the traditional ways to build a high current or high voltage bjt the one problem with driving this darlington connection comes at the turn off so if we have a say another base driver we connect here and we do the same thing with it that we did with a single bjt a couple of slides ago where to turn this device off we pull in big negative current out of the base in the if we just had q1 and q2 with no diode what happens is that our base driver will turn q1 off quickly and then once q1 is off there's nothing to actively remove the stored charge in q2 so we have to wait for the the minority charge in q2 to uh to recombine and its switching time can be long uh so we have a lot of switching loss and we have slow switching times so this diode d1 is added to to fix that so that when our gate our base driver turns off q1 quickly we still have a path to remove the stored charge from q2 through the through d1 and our base driver is able to turn off q2 quickly as well okay so the bjt is one of the classic power transistors back when i was a graduate student the bjt was the workhorse of the power electronics business and then it was replaced by the mosfet which became commercially significant in the 1980s and so in low voltage applications say 600 volts and below the mosfet really is the device of choice today then at higher voltage applications the bjt it's not just is being it has been replaced by the igbt for voltages above 600 volts and so we're going to talk about the igbt next compared to the mosfet the bjt is slower switching because it's a minority carrier device but its on resistance for the same chip size is a lot lower than that of a comparable voltage mosfet okay here is the igbt really this is a combination mosfet and bjt and here's a sketch or a cross-sectional drawing of it and this actually looks exactly like the drawing for the mosfet there's only one small difference that this region right here is p instead of n so there is in fact an extra p in junction in series with what we call the drain of the mosfet that now is traditionally called the collector so the insulated gate bipolar transistor has a gate just like a mosfet the source of the mosfet is called the emitter in the igbt and the drain now is called a collector and we have this extra pn junction okay what the pn junction does is under normal conditions where the device is turned on we have current flowing this way across the pn junction and we have holes from the p material that diffuse across the pn junction and go into the in minus region and cause conductivity modulation and so these holes when the device is turned on give us minority carriers that reduce the on resistance of the device so we we get a much lower on resistance in the in minus region than we had in the comparable mosfet although we do have an extra pn junction in series that has its forward drop but what this lets us do in practice is to design igbts with very high voltages and have much lower on resistances than the mosfet would would have so above 600 volts the on resistance of the mosfet goes up very quickly but by doing this trick to get minority carrier and carrier injection we can control that mod that on resistance and so we have 1200 volt and 1700 volt devices today that are very good and and very significant commercially and we have much larger voltage devices also for for high power applications uh with with breakdown voltages of in the 2000 and 3000 and even higher range okay so effectively the igbt looks like this we have a mosfet channel in the usual place the same places in the mosfet and then we have this pn junction and to first order you might say well we have a diode in series with our mosfet but what actually happens is a little more complicated than that if you look at the the structure the substrate p material that is shorted to the what we're calling the emitter now actually gives us a pnp transistor and so a more accurate equivalent circuit for the igbt is this it has a mosfet and then the mosfet is connected to a pnp transistor and we can actually get two different components of current flow they're called i1 and i2 here so we can have current flow this way through the channel of the mosfet and out that's called i1 or we can have current flow this way just through the pnp transistor and bypassing the mosfet channel that's called i2 and effectively the mosfet drain current supplies the base current to the pnp transistor okay so the effect of this on the switching in fact the turn off switching performance is this our our gate driver can turn the mosfet off quickly which makes i one go to zero but we still have minority charge in the n minus region that will allow the pnp to continue to conduct and so i2 continues to conduct and in fact there's no way to actively remove that stored charge from the n minus region so all we can do is sit around and wait for the minority charge to recombine and once it does then finally the pnp will turn off so the igbt has this fairly slow turn off mechanism as a result and that turn off mechanism is traditionally called current tailing in the igbt so the turn off current waveform looks like this the originally when the device is on the total current flowing this way is some combination of i1 and i2 and when we turn the device off we turn off the mosfet quickly and so i1 goes away quickly and you see the the current decrease by the switching off of the mosfet to this point but after that i2 continues to flow and the fraction of the collector current that was i2 slowly decays while the minority carriers decay and eventually the device turns off and so this long time it takes the i2 component to decay is called the current tail okay so the turn off switching transition of the igbt can be significant it depends on the size and the the rating of the igbt but it may be several microseconds or in a good igbt it may be say several hundreds of nanoseconds so if we multiply the collector to emitter voltage by the collector current we get the power loss in the transistor during this turn off transition and it has some area that is the the energy loss during this turn off transition and we get switching loss then equal to the the energy loss during the turn off tail times the switching frequency and in fact there is some energy loss during the turn on transition too so we really ought to add the turn on energy loss also on data sheets they will actually usually specify values of the these turn on and turn off energy losses there are some things we can do to control this current tail one thing we could do is uh in fact by designing the circuit we can actually control the current gain of this bipolar transistor and we control can control how much of this current flows in the i2 path versus the i1 path so this is a design variable that the device designer has to play with so if we make most of the current flow through the mosfet and very little flow through the bjt then what that does is it makes this the i1 portion be large and the i2 portion be small and the current tail is smaller so we'll get less energy loss during the turn off transition and so the designers can make a faster igbt that way but the price they pay is that the forward voltage drop is larger because more of the current is flowing in the mosfet direction and less in the collector direction so another thing that that is often done is to make the opposite choice which gives us more minority charge stored charge it makes the device switch off slower but it gives us a smaller forward drop and often many manufacturers even offer two versions of their devices one that is optimized for forward drop and the other that is optimized for fast switching times and so you have to decide which one will give you more efficiency in in your application okay so the igbt really is the modern device of of choice really at voltages above 600 volts and we have very large igbt modules these modules are composed of multiple individual chips where the individual chip might be rated 50 amps or 100 amps multiple chips like that are packaged together on a common thermal substrate to make them run at the same temperature and share the current and so then we have a module of many of these chips that that module might be rated at hundreds of amps or even a thousand amps and at high voltages you know possibly several thousand volts so these are very uh good high power devices they can be controlled simply by switching their gate voltage from zero to fifteen volts and back and so you can control a lot of power um with a simple uh mosfet like gate the forward voltage drop of these devices is a diode in series with an on resistance so it's the on resistance of the n minus region in series with the base emitter or not base emitter but the pn junction voltage drop that is in the collector of the device [Music] the temperature coefficient of these devices is a combination of mosfet and bjt behavior at low current the forward voltage has a negative temperature coefficient but at high current the mosfet on resistance dominates and we have a positive temperature coefficient which means that the devices tend to share as you get close to the rated current of the device the devices are slower than a mosfet but they can be faster than a darlington or an scr or gate turn off thyristor you'll see switching frequencies for the the low uh four drop devices might be a kilohertz or several kilohertz and for fast devices sometimes they're run at tens of kilohertz and there are some you know lower voltage very fast devices there can even be run at 100 kilohertz or more here's one last brief note about switching loss and its effect on switching frequency and efficiency so we've talked about different mechanisms now for switching loss from the different devices and other mechanisms around the converter where we found that we got energy loss every time we switch and so we have different mechanisms where we could calculate the total loss during the switching transition which might come from the reverse recovery or from the switching times or from ringing an energy stored in parasitic capacitors and inductors in the converter and in each case we could find some total amount of energy that was lost during the switching transition so what we can do now for whatever our converter is is we can add up all of those losses and get a total amount of energy that is lost due to the difference switching mechanisms during each switching period so then the total switching loss will be that total energy loss multiplied by the switching frequency okay so this is uh what this loss scales linearly with frequency and as we turn up the switching frequency we turn up the switching loss now what is the effect of this on the overall efficiency well we have loss mechanisms that come from different sources back in chapter 3 we talked about conduction losses and how to model those those losses depend on say the output current but they don't depend on the switching frequency so if we say are interested in the the efficiency and the total loss maybe at full load or at some critical load power or operating point we can calculate the conduction loss at that point and it's a given value we also have what are called fixed losses that don't depend on switching frequency they don't depend on load power they're just a fixed loss that we're stuck with an example of that is the power that it takes to run the controller circuitry for our power converter and then finally we have the switching loss that uh again scales linearly with switching frequency so the total loss then is a sum of all of these three things okay now given this we can calculate the efficiency uh so the efficiency will be the output power over the input power or the output power divided by the output power plus this loss and we can plot that versus switching frequency and what we find is that the efficiency goes down of course as the frequency goes up here i've plotted the efficiency for some typical values and i have uh switching frequency varied on a logarithmic scale on the horizontal axis and so what happens is at low switching frequency we find that the conduction loss and the fixed loss dominate and the switching frequency is relatively low now if we're in this situation we may as well raise the switching frequency it's true that that will make the the efficiency go down a little but not by very much because the total loss is dominated by other things and the size of our reactive components such such as our inductors and capacitor filter you know filters in the converter and if we have a transformer in our converter power transformer also these things depend on the frequency if you raise the frequency the reactive elements get smaller so if we're down at this point we may as well raise the frequency because it makes our reactive elements get a lot smaller so we have a smaller and less expensive converter and there's very little penalty and efficiency when we do that on the other hand if we're at high frequency the switching loss term dominates the other terms and then our total loss is very sensitive to switching frequency so we probably don't want to raise the frequency too much or our switching loss will become large and our efficiency will really suffer so there's some good sweet spot of switching frequencies in the middle where the switching loss is starting to hurt but not too much yet and that's a then represents a compromise between the switching loss and its effect on efficiency and the reactive element size so then here you can even solve this formula for the critical frequency where the switching loss is equal to the other losses at that point the the total loss is twice the loss you would get at zero switching frequency and that's gives an estimate for a given technology of what kind of frequencies we can run at you may choose to run a little higher a little lower frequency depending on the details of the application but we'll probably run somewhere in this range okay so we have a fundamental tradeoff then between efficiency and switching loss versus the reactive element size and the the engineer has to choose the switching frequency accordingly so it depends on the application at high powers and high voltages we tend to run at low frequencies 10 kilohertz or one kilohertz depending on the voltage of the igbt at high voltage i mean high high frequency low voltage applications with mosfets supplying say one volt computer processor chips might run at several megahertz and this f crit may be way up here a height much higher frequency in recent years we've seen the introduction of new power semiconductor devices using wide bandgap semiconductor materials instead of silicon they use silicon carbide or gallium nitride these materials allow significant improvements in the trade-off between breakdown voltage forward voltage drop and switching speed so in this brief lecture i'm going to describe or summarize why that is so and then we'll talk about several of the commercially available devices specifically silicon carbide shocky diodes silicon carbide mosfets and gallium nitride hemp devices here is a formula that you see often quoted in the literature for the specific resistance in other words the on resistance multiplied by the active chip area for a majority carrier device such as a mosfet the ion resistance is a function of the breakdown voltage v sub b as well as the material mobility mu the material permittivity epsilon and the critical electric field at the onset of avalanche breakdown e sub c so if you want to increase the blocking voltage of your mosfet for example the on resistance will increase as the breakdown voltage squared so high voltage mosfets have much higher on resistances but the interesting thing here is the other terms these denominator terms that are a function of the semiconductor material and so it turns out that these wide band gap materials such as silicon carbide and gallium nitride have much higher critical fields e sub c that allow the a significant improvement in this trade-off here's a comparison of the parameters for these basic materials so silicon has a band gap of 1.12 electron volts silicon carbide and gallium nitride have much higher band gaps and the result of this is that the critical fields are much higher so with this significant increase in critical field then we can design the device to have a much lower on resistance at a given breakdown voltage the electron mobilities are also changed in fact it turns out that silicon carbide has lower electron mobility which is bad that means more on resistance and gallium nitride for reasons we'll discuss in a minute has higher electron mobility through what are called hot carriers that exist in the two-dimensional electron gas in the device because of the lower mobility of silicon carbide it actually is inferior at low breakdown voltages and silicon is better but when you get above 600 volts then silicon carbide begins to to exhibit significant advantages and so we see good silicon carbide mosfets at voltages starting at 600 volts and going commercially now to 10 000 volts gan doesn't have that issue and in fact gan is highly competitive with silicon it as mosfets at low voltage or at least as controlled transistors at lower voltages so we see both sub-100 volt gan devices and we see 600 volt gan devices that are very good we can also make shocky diodes with these devices and so we see high voltage shocky diodes made out especially out of silicon carbide and there are majority carrier devices that have no significant reverse recovery and we get silicon carbide shockies at 600 and 1200 volts and higher in silicon carbide we can build moss gates silicon carbide can have an oxide layer that is similar to the oxide layer in silicon and we can build mosfets with it this is not true in other materials and in particular it's not true in gam so we aren't able to build mosfets with gan and the devices that we have are junction fets instead silicon carbide mosfets then can be built with a vertical structure that's very similar to the the vertical mosfet structure seen in a silicon mosfet and it has all the same structure it has a body diode and so on again this is not the case gann is a thin film material that is deposited on top of a substrate and so we have lateral gan devices because gan devices are lateral it's somewhat more difficult to scale to higher voltages and currents although we're seeing that be being done commercially the thermal coefficient of expansion of the thin film gan device must be matched somehow to the substrate that it's deposited on and there are issues there as well that affect the structure of the gan devices so as i mentioned we have silicon carbide shocky diodes these were the first wideband gap power devices to become commercially important we see them available at 600 1200 volts and higher because the shocky diode is a majority carrier device there's no significant reverse recovery they do have a higher forward voltage drop typically one and a half to two volts but they have much lower switching loss because they're a majority carrier device what we find then is higher conduction loss lower switching loss and in most switch mode applications let's say 600 volts you see a significant gain in overall system efficiency they also cost more than silicon so there's a trade-off in the area of mosfets we have silicon mosfets at voltages up to six or 700 volts if you want to go to a higher voltage then generally we use a silica and igbt in the case of silicon carbide we have silicon carbide mosfets starting at 600 volts and commercial devices available at up to even 10 000 volts and these are real mosfets that are very similar to the silicon mosfets but with good properties at high voltage so they have much lower on resistance than a silicon mosfet they do have a pn body diode just like the silicon mosfet has in the case of silicon carbide the body diode has a forward drop of three to four volts so it's higher the the body diodes generally are built with uh good reverse recovery times below 50 nanoseconds and so we can switch these devices at a much higher frequency than we could operate in igbt so where the igbt might run at five or ten kilohertz these mosfets can run at much higher frequencies here are some sample commercial devices that you can buy today uh i've listed some from 650 volts up to 1700 volts with nice on resistances that are tens of milliohms in gan we're not able to build an oxide layer and so we can't build a moss gate and what we see for gan transistors is what is basically a junction field effect transistor where there's a source in a drain with a semiconductor material between them and a gate that is built as a diode junction on top of that material so when we negatively bias the gate we can turn off the channel when we forward bias the device with a positive gate voltage there is current that can flow between the source and the drain the traditional junction fet is a depletion mode device so to at zero gate voltage it's forward biased or turned on and to turn it off you have to apply a negative voltage to the gate early gann transistors were in fact depletion mode devices and we have cascode type circuits that combine a silicon mosfet with a depletion mode gan device to build what is effectively a an enhancement mode transistor there is another twist here in the gan devices these are what are called high electron mobility transistors or hemps they're also characterized as having a hetero junction which means that we have two different materials having different band gaps and in the gan device today we have aluminum gallium nitride which is a low band gap semiconductor material and then below it is gann which is a high band gap or wide band gap material and there is a junction between them when we place materials having different band gaps in contact at this head row junction it can form what is called a two-dimensional electron gas which is shown here and this two-dimensional electron gas has carriers having high energies that are able to conduct or jump very easily from atom to atom and effectively have very low on resistance or high mobility and this is one of the keys to the gan device of why it can achieve very low on resistance to turn this device off we negatively bias the gate which will extend its depletion region below the gate and disrupt the 2d electron gas so that the the source and drain have no conducting channel so this device can have a high breakdown field because it's a wide bandgap material and at the same time it can have very low on resistance from this two-dimensional electron gas the gan device is a junction field effect transistor that traditionally is a depletion mode device our more recent gan power transistors actually have their gate threshold is shifted positively so that they become effectively enhancement mode devices so with zero gate to source voltage the device becomes turned off and we apply a positive voltage greater than the threshold to turn the the device on however you can't apply to positive of level of a voltage or you will forward bias the gate to source diode and then you'll get significant positive current flowing into the gate which actually can damage the device so we have to limit the gate current and must apply a gate voltage that is not too high often then we'll drive these devices with a zero to five volt gate drive signal to to switch the power gann fet so in the on state the device is on with a low on resistance this is with the gate to source voltage greater than the threshold of maybe three or three and a half volts in the off state we'll generally apply zero volts between gate and source to turn the device off the device does not have a built-in body diode the way the mosfet structure has but having said that it's still possible to get the device to conduct if you apply reverse current or voltage through it let's suppose we have the gate shorted to the source and then we apply negative drain voltage in this case with reverse operation the drain can actually become the source and the source the drain so if you apply a negative enough voltage to the drain we'll have positive voltage at the gate with respect to the drain and if that positive voltage is more than the threshold the device can turn on so in that case the device will conduct and it will conduct current in this direction with a forward voltage drop between source and drain equal to the threshold or a little bit greater so it's possible to have a reverse conducting device without actually having a physical body diode so this behavior is similar to the mosfet that has a body diode except that there's no re reverse recovery because there's no real diode also the forward drop is fairly large it's it's greater than or equal to the threshold voltage of the gate so you see gan devices are actually current bi-directional devices with very fast switching times so here's a comparison of 600 volt transistors both the silicon mosfet and an equivalent gan transistor and here it's assumed that they're sized to have the same on resistance of course to achieve this on resistance the gan has a much smaller area and as a result it has much lower gate charge so because it's a smaller device the charges are smaller and it can switch much faster so we see lower gate charge lower output capacitance the only negative thing perhaps is the voltage drop when reverse conducting but then again it has or negligible reverse recovery so wide bandgap semiconductors have become commercially important while they cost more they can operate at higher voltages with higher switching speeds and with lower forward voltage drops currently the important devices are the silicon carbide shocky diode the silicon carbide mosfet and the gan hemp device homework assignment 2 is concerned with simulation of a buck converter that steps 12 volts down to 1 volt at 20 amps to supply a microprocessor you will add a synchronous rectifier to a buck converter to improve its efficiency in this lecture i'm going to illustrate how to use the schematic capture capabilities of lt spice and i'm going to work an example in which i will add a synchronous rectifier and the associated circuitry in a boost converter example here's the circuit that we'll start with this is a boost power stage having input voltage vg an inductor l1 with a mosfet and diode and then the output r and c this converter is intended to step 12 volts up to 24 volts and it operates at a duty cycle of a half here we have a voltage source to set the duty cycle to 0.5 this voltage goes into a pulse width modulator that then generates the switched waveform going to the driver which then turns the mosfet on and off to insert a synchronous rectifier in the circuit what we need to do is to take the diode and replace it with a mosfet we need to drive it with a gate driver circuit the gate driver needs a power supply to supply the gate driver and we're going to derive that power supply from the 12 volts at the input we need to generate uh complementary drive signals for the two transistors so we're going to add what's called a dead time generator that takes the pulse width modulator logic output signal and generates a second complementary signal to drive the uh synchronous rectifier fet the switching converter library that accompanies these ltspice files for this course includes what's called a dead time generator block which is shown right here this dead time generator takes the control input signal coming from the pulse width modulator c which is shown here in green and it generates two drive signals c1 and c2 so c1 is intended to drive the main power fet of the converter and c2 is intended to drive the synchronous rectifier fet so these are complementary signals when one is high the other is low and vice versa so that they switch at opposite times this dead time generator also generates what's called a dead time and here you can see the output this is the c1 output in red and the c2 output in purple and there are dead times here produced by the dead time generator block that can be set as a parameter for the block in this case it's set to 100 nanoseconds so the object here is to turn off one fet wait for the dead time and then turn on the other fet so this is called break before make operation we want to make sure that both fets are not on at the same time and the dead time you're set to 100 nanoseconds is supposed to be large enough to make sure that the switching transition of the first device being turned off can complete before the second device is told to turn on here is the lt-spice boost converter to remove the diode we use the scissors tool so i click on that tool and then i can just cut away the diode on the on the macintosh it doesn't have the toolbar like this instead you have to right click with the mouse to bring up menus that include all of these functions okay we'll insert a device here so i click on that one and let's find the mosfets okay so i will insert an nmos transistor right here okay then to stop this we hit escape i want to change this transistor into a specific device so i right click on it and i can pick a new mosfet so here's a library of parts that are built into the lt spice libraries these transistors their models are optimized so that the spice lt spice simulation will run quickly and i strongly suggest you choose one from this list okay so i'm going to choose this irf z44 in that is a 55 volt rated transistor with an on resistance of 14 milliohms uh next thing i'm going to do is add the dead time driver from our library so we go here to the component list and i need to go go to the directory that has the downloaded files from the course website and here is the block called dead time that we can insert so i'll put it right there and then we get the little pencil to draw some wires to connect this up so connect its input up to c i take the c1 output and we'll connect that to the driver input of the main transistor now we need the driver for the synchronous rectifier i can either insert a driver from the library or i can copy the one we already have let's take the duplicate or copy function copy this driver and i'll just move replace one up here lastly we need to add the bootstrap power supply for the high side driver so we derive this from the 12 volt input what we'll do is add a diode and i think i'll just use another one of these shocky diodes for that purpose so i'll copy that and place it over here okay we need to add a capacitor for this bootstrap power supply and i'll place right about there and we need to connect these components up okay let's see we need to set the value of the capacitor i'm going to make it the 10 microfarads all right i think we're ready to go let's press the run button and see how it works here i've zoomed in on the switching transition so this upper waveform is the switch node voltage right at the switch node between the two fets it's high so the synchronous rectifier is on here the gate driver turns the synchronous rectifier off right there and you can see the voltage increase a little bit what's happening is the body diode of the synchronous rectifier is being forwarded by us by the inductor current and it's turning on so we get a higher forward voltage drop during this short time during the dead time here you can see the gate drive or the gate voltage of the synchronous fed is going low right there which coincides with the jump in the switch node voltage the synchronous fed is fully turned off here and the green waveform is the gate voltage of the main fet this m1 fet in green so the gate driver starts turning it on right here and it takes there's some rise time on the gate voltage at this point here the the switch node voltage falls and we have a low voltage after that here's a magnified view of the current in the output capacitor at the time when we're turning off the synchronous rectifier so that same switch transition and what happens as we just saw was that the body diode of the synchronous rectifier is turned on during the dead time and then when we turn on the lower mosfet the body diode of the upper fet goes through a reverse recovery and that reverse recovery comes out of the capacitor and goes through the two mosfets so it goes around this loop the green waveform is that capacitor current so it's it's near zero here and it goes down to a very large negative value what is that about minus 55 amps the area of this current spike is the charge that is pulled out of the capacitor and flows through the two mosfets so this is this current is caused primarily by the reverse recovery of the body diode of the synchronous fit also some of this charge represents the change in charge on the output capacitances of the two mosfets okay so if we define this area is q you can estimate what q is it's the area of this triangle so if we find the switching time and this amplitude we can work out what q is we could also for clever figure out how to get lt spice to tell us the area of this triangle but in any event it's area we could define as some charge q and we get this charge drawn out of the capacitor once per switching period so it represents an average current that is the charge divided by the switching period we get that much average current flowing out of the capacitor as a result of this and this then represents a power loss or at least a power power flowing out of the capacitor of this current multiplied by the output voltage so q over ts times the output voltage and this average power essentially is switching loss it's the switching loss induced by the reverse recovery of the body diode primarily as well as some power loss induced by charging or discharging the output capacitances of the mosfets a little bit of that capacitive power actually is not lost but this primarily is from the reverse recovery so on the homework you will work out or estimate this power loss predicted by lt spice due to the diode reverse recovery and compare that to the total loss in your circuit so the homework assignment is to start with a basic buck converter so this circuit is given on the course website it has a main fet and a diode with the output filter and the output load again we want in this microprocessor we want to supply one volt at 20 amps so we'll have to adjust the duty cycle so that we get one volt with high accuracy and you can measure the efficiency of this circuit including both the input power the gate driver power and the output power in your calculation and then as we just did in this boost example what we want to do is in the buck replace this diode with a mosfet as a synchronous rectifier like this along with its gate driver and we'll want to add a bootstrap power supply to the upper driver and then work out the efficiency in this case and look at things like the current spike from the reverse recovery there's also some discussion questions at the end of the assignment which are optional but if you wish you may want to further investigate how to improve the efficiency of this circuit if you look here we have two different mosfets in the power electronics field we figured out that a synchronous rectifier power supplies like this can be better optimized if the mosfets are different so you might consider how this was optimized and in fact looked at whether you can find other mosfets in the ulti spice library that would further optimize it we can also optimize the dead time and the gate driver resistors to to get some improvements in efficiency as well and it's worth looking at that and trying it in lt spice if you have the time having discussed how to realize switches using semiconductor devices and to implement things such as single quadrant or two quadrant switches in a switching converter we're now in a position to discuss what what is known as the discontinuous conduction mode and chapter 5 covers the first the origin of the discontinuous conduction mode and then how to solve it and we'll do some examples of solving bach and boost converters that operate in this discontinuous conduction mode the converters that we've discussed so far in this course operate in what is called the continuous conduction mode by contrast to the mode of operation we're going to discuss now the discontinuous conduction mode occurs because there is a ripple in the current or voltage waveform that is applied to one or more of the switches inside the converter and this ripple is larger than the dc component and as a result the the current or the voltage attempts to reverse through the switch now if we realize the switch to be a uni-directional switch and we try then to reverse the polarity uh the switch doesn't behave in the way uh that it was intended and so it will switch on or off at a time that does not coincide with the driver signal switching and so we get additional uh intervals or sub intervals within the switching period as a result and this completely changes the characteristics of the converter the most traditional discontinuous mode operation occurs because the inductor current ripple is greater than the dc component of current and this causes the current through the switches to try to reverse direction the diode won't allow it the diode will turn off instead and we get a third period during the switching period and this changes the characteristics of the converter substantially so when the discontinuous mode happens what we're going to find is that the output voltage now becomes load dependent in continuous mode say for a buck converter the output voltage is the duty cycle times the input voltage any dependence on load is small and only comes from loss the effects of losses in the converters in the discontinuous mode though we have a first order dependence of the output voltage on the load current so the properties of the converter change drastically we then all of a sudden we have a high output impedance we're going to find that or it turns out that the dynamics of the converter that we're going to discuss in several weeks those dynamics change very substantially and in fact they're simpler in discontinuous mode we also find that when you remove the load the discontinuous mode causes the output voltage to not be controllable and it can do bad things so we have to account for that some converters are designed on purpose to always work in discontinuous mode one of the good things about it is that the current goes to zero before the end of the switching period the diode turns off and then there's no reverse recovery when the mosfet is next turned on so there can be less switching loss but on the other hand there are higher peak currents and so we can get more conduction loss and there's a trade often even if we don't intentionally design a converter to work in this mode many converters will operate in the discontinuous mode at low output power and so we have to be able to analyze and know what's going to happen than at those operating points so what we're going to do in this lecture is discuss the origins of the mode and find the mode boundaries the conditions under which we operate in continuous or discontinuous mode and then in the next several lectures we will analyze converters to find their output voltages and other things when they operate in the discontinuous mode so here's an example we have a basic buck converter and we've realized the switches with the conventional single quadrant switches of a transistor and a diode um here is what the inductor current waveform looks like in the continuous mode and it's the waveform we've been drawing all along this uh in this class so it has a dc component that we're labeling capital i plus it has some switching ripple that has a peak magnitude of delta i okay we previously analyzed this circuit and we know how to calculate capital i and delta i now here are the expressions so the dc component capital i is the load current v over r and delta i the the peak to average ripple is the slope times the time which we can write in terms of vg like this and the ripple depends on the duty cycle vg the inductance and the switching period okay the interesting thing about this is that the dc component depends on the load resistance or on the the current that the load decides to draw but the ripple doesn't ripple depends on all kinds of things but the one thing it doesn't depend on is the load resistance or the load current so what happens if we say increase the value of r or our load all of a sudden decides it doesn't need as much current well that will make the dc component capital i go down but it won't change the ripple now what's important here is actually the effect of that on the diode so when the diode is on it conducts the inductor current and the diode current waveform looks like this so when the diode is on it has the same capital i plus the ripple and the minimum diode current happens right there at the end of the switching period and that minimum value is the dc component of current minus the ripple okay so the diode current has to stay positive for the diode to work and when we account for the ripple the the minimum diode current is actually less than capital i so let's consider increasing the value of r to the point where the ripple equals capital i actually we reduce capital i so it's equal to the ripple then the inductor current looks like this and the diode current follows the inductor current for the second half of the period so you can see that the diode current starts out at capital i plus delta i and it ends at capital i minus delta i which for this particular load current is equal to zero what happens if we increase the load resistance even more well here is that case dc component of load current and inductor current capital i is now less than delta i so even though the average current and the load current are positive they're less than delta i and so the diode current and inductor current go to zero before the end of the switching period okay once that happens the diode will turn off and it will not allow the inductor current to continue uh in the same direction and go negative so we get a third interval now uh this uh from the point the diode turns off until the end of the switching period where uh the diode and the mosfet are off and the inductor current just sits at zero current so now our switching period is divided into three intervals there's the d1 interval the first interval which we're also going to call dts it's the d is the tren the mosfet duty cycle and d1 is the duty cycle of the first interval and they're the same here we have a d2 interval now when the diode conducts and we have a d3 interval when nobody conducts so this is the discontinuous mode of operation and this extra interval changes all the equations of the converter so first of all we need to be able to write the equations of the mode boundary and we now we have a we have a way to do that that the mode boundary happens when delta i is equal to the dc component capital i okay and in fact the conditions then are that we're in dcm or discontinuous mode when when capital i is less than delta i now at the boundary the continuous mode equations are still valid so we can plug the continuous mode equations that we had for capital i and delta i into these equations to find the boundary so here's here's the expression for capital i in terms of vg here's the expression for the delta i and [Music] so we can plug them into this and we can solve so what the vg's cancel out we can move the 2l and ts over to the left hand side and cancel the ds it looks like and we'll get this expression so this is the expression or the condition for operation in the discontinuous mode okay the quantity on the left-hand side is a function of element values in the switching period uh it's traditional to call this quantity k and so in the power electronics business we generally call capital k is 2l over rts and this 2l over rts factor keeps coming up in many of our equations over and over from here on the right hand side is the critical value of k at the boundary between modes so we call that k crit so the right hand side is a function of duty cycle it will vary with duty cycle but when k is equal to k crit then we're at the mode boundary and if k is less than k crit then we operate in the discontinuous conduction mode for the buck converter we find here that k crit is equal to d prime for different converters they'll have k crit will be a different function of duty cycle but in general we can write this expression for the mode boundary in some equation of the form k less than k crit of d for discontinuous mode so here's a plot of that k crit for the buck converter is one minus d or d prime so here's a plot of k crit versus d so k crit is one at d of zero k created 0 at d of 1 and it looks like that okay so we compare that to k so suppose k is less than 1 but the the maximum value of k crit is one so if k is less than one say some value like here then we'll be in discontinuous mode over this range where k is less than k crit then at higher duty cycles over this range will be in continuous mode on the right side here is another example where k is bigger than one and in that case k is always greater than k crit so we're always in continuous mode then for the different converters buck boost and buck boost it with their k crit expressions labeled so k crit for the buck is d prime we're going to show uh in an upcoming lecture the k crit for the boost is d times d prime squared and uh on the homework you're going to work out for the buck boost that k crit is d prime squared for the buck the maximum value of k crit for any d between zero and one is one so if k is greater than one then we're always in continuous mode but if k is less than one then we have to worry about discontinuous mode for the boost this maximum value turns out to be 4 27 we'll discuss that it's an interesting number that comes out of nature here uh the boost is less likely to run in discontinuous mode but if k is smaller than 4 27 then it can for the buck boost the maximum value of k crit is one so again with k less than one we may run in discontinuous mode okay so that's the mode boundary in the next lecture we're going to solve for the output voltage in this lecture we will discuss how to solve for the output voltage of a converter operating in discontinuous conduction mode and i'm going to explain it for the example of a buck converter what we need to do is to extend the ideas developed in chapter 2 to apply to this case where we have inductor current ripple that is large so here's what we talked about back in chapter 2 at the beginning of the course we had this technique of inductor volt second balance which said that in steady state the average voltage applied to an inductor is zero and that's true in any circuit and it doesn't matter if it's a continuous mode or discontinuous power converter or something else so it still applies and same thing with charge balance on capacitors the in steady state the average current through a capacitor is also zero and we can apply this to solve our discontinuous mode converter as well but where we have to be careful is the small ripple approximation so really by definition here in discontinuous mode we have an element where the ripple is large and we're talking in these lectures about the common example where the inductor current ripple is large compared to the dc component so we can't approximate away the inductor current ripple now in solving our discontinuous mode converter and we're going to have to use some other arguments in order to to handle the inductor current and include its ripple on the other hand we still want the output capacitor voltage to have small ripple and generally we're going to put a capacitor on our output that will be big enough to filter out the ripple and have small voltage ripple at our output so we can apply the small ripple approximation to the capacitor voltage but not to the inductor current okay so let's do this example of the buck converter and discontinuous mode and apply these arguments to solve for the output voltage here's our butt converter then and we now have three sub intervals first interval where the mosfet is on gives us this circuit for the second interval the mosfet is turned off and the diode is on so we get this circuit with the left side of the inductor connected to ground and then we now have a new third interval where both transistor and diode are off the inductor current has gone to zero then the diode is turned off and the left side of the inductor now is left hanging in space not connected anywhere with zero current flowing through the inductor so as usual we will write the circuit equations for each interval so here we have um for the first interval with the mosfet on as usual we can write the inductor voltage is vg minus v like this and the capacitor current would be from the node equation it would be the inductor current minus the load current v over r so we get that the capacitor current is il minus v over r next we apply the small ripple approximation so the output capacitor voltage v has small ripple and we can replace v in these two equations with capital v to ignore the ripple in the output so that gives us these terms but the inductor current has large ripple and we can't approximate il of t by its dc component so we're going to just leave il of t alone for now and we're going to have to handle the the ripple in il with some additional arguments in a couple of minutes for sub interval 2 the diode conducts the inductor current is positive and we have this circuit with the left side of the inductor connected to ground we can write that the inductor voltage is minus the output voltage like this the capacitor current is again the inductor current minus the load current like that and we can replace v of t by the dc component capital v using the small ripple approximation to get these equations and again the inductor current is left as il of t for sub interval 3 the inductor current has gone to zero the diode is turned off so the right hand side of the inductor is connected just or not connected it's hanging in space what is the inductor voltage during this interval well the inductor voltage is the derivative of the current we know what v l of t is l times d i l dt il is zero so the derivative of i l is zero so we get that the inductor voltage is zero during this interval so vl is zero we'll just we can simply write that i l is also zero the capacitor current is the inductor current minus the load current although the inductor current is zero so really all we need to say is the capacitor current is minus capital d over r uh where we apply the small ripple again to v of t so here's the inductor voltage waveform it's a positive voltage during the first interval a negative voltage during the second and it's zero during the third interval we can apply volt second balance to this waveform in the usual way we have a third interval the inductor voltage can be written as d1 times the voltage during the first interval plus d2 times the voltage during the second interval which is minus v plus d3 times the voltage during the third interval which is zero and by volt second balance this is equal to zero so we get this equation you can solve it for the output voltage and we get this expression for the output voltage now in continuous conduction mode without losses this equation told us what the output voltage of the buck in continuous mode was but here we have an unknown the the duty cycle d2 is not known d2 really depends on when the inductor current goes to zero and the diode turns off and we have to do further analysis to find d2 so at this point we have an equation with two unknowns v and d2 and we need to get more equations so let's keep going let's apply charge balance next here's our node where the capacitor is connected at the output terminals of the converter we can write the node equation at that point which for the buck converter the inductor is connected there so we can write that the inductor current is equal to the capacitor current plus the load current like this and what we want to do is apply charge balance to the capacitor so the hard way to apply charge balance is to take the inductor current waveform that we already drew and subtract the load current to draw the capacitor current waveform and find its average a little bit easier way to do it is to look at the original node equation at this node and compute the dc components of each term so we can say that the dc current flowing out of the inductor into this node equals the dc component of the capacitor current plus the dc component of the load current and the capacitor current by charge balance has no dc it's zero so we get that the average inductor current or dc component of inductor current equals the load current so all we really have to do is take the inductor current waveform that we've already drawn find its average value and equate that to the load current i should caution you at this point this expression of finding the average inductor current and equating it to the load is really only true in the buck converter and it's because the inductor is what's connected to the output node where the capacitor is we'll do another example next time with a boost converter where it's the diode that's connected to the output node instead of the inductor and in that case we have to draw the diode current waveform and find its dc component okay so here's the inductor current waveform we'll find its dc component and and equate it to the load here's the calculation for that the dc component of inductor current is found as usual by averaging the inductor current over one period so the average is the integral divided by the period and the integral is the area under the curve so we need to find this area okay the area and this for this waveform we have the area of a triangle so we have to use the triangle formula which would be one half the base and what the base is d1 plus d2 times ts times the height i've labeled this peak current i peak okay so that's the area or the integral if you divide by ts we get the average value okay and so here here it is divided by ts and the other thing i've done here is subtract is substitute in an expression for i peak here's how we find i peak the inductor current starts the period at zero it starts at zero because we're in discontinuous mode and we ended at zero in the last period we then we go up during the first interval with a slope given by the applied voltage over l so vg minus v over l is the slope of the inductor current during the first interval here multiply that by the length of the first interval to get the change in current and that's my peak so we can plug that expression for ip into here and finally we get we get this expression for the average inductor current uh and then lastly we equate the average inductor current to the load current and we get this expression that comes from charge balance on the capacitor so we have two equations the first one was from volt second balance the second one is from capacitor charge balance we have two unknowns the output voltage and d2 so we can solve so i'm not going to go through the algebra in this lecture i'll actually illustrate the algebra in the next example of the next lecture but basically you solve one of these equations for d2 plug it into the other equation you then get a quadratic expression that we have to find the roots of and finally from doing that we can solve and get this equation for the conversion ratio v over vg of the buck converter in discontinuous mode this equation is a function of d1 the duty cycle of the first interval it's also a function of k which is the same 2l over rts that came out of the mode boundary equation in the last lecture here's a plot so this is m which is v over vg versus duty cycle and we have now two choices for m if we're in continuous mode when k is greater than k crit then m is equal to d where whereas if we're in discontinuous mode with k less than k crit m is given by the expression from the last slide so we can plot this for different values of load resistance or different values of k and here are some different choices so for example for k equals 0.1 to plot this we first have to plot our k versus k crit formula remember k crit for the buck converter was d prime so it looked like this and if say we do the k equals 0.1 curves so here's 0.1 for k k is less than k crit for d from zero all the way up to point nine right there so we use the discontinuous formula for d between 0 and 0.9 and then for d between 0.9 and 1 we're in continuous mode and we have to use the continuous formula so here's a plot a computer drawn plot of the discontinuous formula going all the way up to d is 0.9 and after that we follow the the continuous formula of m equals d and that's what happens over the whole range of from 0 to 1 in duty cycle for the case when k is equal to 0.1 one easy way to plot this is to recognize that discontinuous mode makes the voltage rise so you can actually just plot both of these expressions and then take the largest and you'll get the right answer another thing i should point out is that what happens when you disconnect the load which is say let r go to infinity okay so k is 2l over rts so here with no load power or r is an open circuit or infinity if r goes to infinity k goes to zero and what this curve does in that case is it actually just rises all the way to one basically you get the output voltage equal to the input voltage regardless of the duty cycle it's actually the solution goes to one except right at the axis where it actually turns into a squared off function here so if say the customer disconnects the load um you will lose control of our output voltage and we have to do something about that generally our build a feedback loop that stops switching turns the converter off to keep the output voltage from rising beyond the value that we want so that we could maintain control of the output generally this is a problem in discontinuous mode when k goes to zero we tend to lose control of the output let's work an example and go through the analysis of a boost converter in discontinuous conduction mode here is our boost converter uh we know from the discussions of the previous lectures that the mode boundary between continuous and discontinuous mode occurs when the dc component of inductor current equals the uh the ripple since the diode in this converter conducts the inductor current when it's on that is in fact the condition for whether the diode is turned off by the ripple we can use our continuous mode solutions for the inductor current and its ripple from from previous analyses here's what we had before that the dc component of the inductor current was vg over d prime squared r and the ripple was given by this expression so we can plug these two into here to get the expression for the mode boundary so here it is again this is the dc component and this is the ripple so for continuous mode the dc component is greater than the ripple we can simplify this expression let's see the vg's cancel out we can put the terms that depend on duty cycle on the right side of the equation and the other terms on the left hand side and we get this expression we recognize that this is our usual k 2l over rts and this is the critical value at the boundary okay crit of d so this is what comes out of the equations for the boost converter k crit is equal to d times d prime squared here's a plot of that function the function d times d prime squared has a root at d equals zero right here and it actually has two roots at d prime equals zero or d equals one right there so it looks parabolic about d of one and so k crit is zero at both d of zero and d of one and in between it rises to a maximum this is qualitatively different than what happened in the buck and actually it's different than what happens in many other converters as well in the buck k crit was maximum at d of zero so the converter was most likely to run in discontinuous mode at low duty cycle that's not the case here here k crit goes to zero at both extremes of duty cycle for high duty cycle most converters will go continuous and the reason for that is that at high as the duny cycle approaches one the ripple goes to zero while the dc component does not so the converters uh generally run in continuous mode at high duty cycle here the boost is going continuous also at load duty cycle why is that well let's look back at the original boost converter the at low duty cycle for example in the extreme as d goes to zero we still have a path for current to flow this way through the converter if we never turn the mosfet on current will still flow out of vg through the inductor and through the load at d of zero the load voltage is vg we have load current equal to vg over r so there's substantial current flowing through our inductor however we're not if we don't switch the mosfet then there's no ripple so the the current ripple goes to zero but the dc current does not and we find that the converter must run in continuous conduction mode so the boost tends to not want to run discontinuous but certainly if k is small enough it can and one can work out the maximum of this k crit function it turns out to be at d of one third where the value is four twenty sevenths and that's roughly 0.015 so if we have k less than 0.015 then the converter will run in discontinuous mode over some intermediate range of duty cycles so here's an example with k just less than 0.1 and we run in discontinuous mode from here to here okay that's the analysis of the mode boundary now let's work out the output voltage so as usual here we will draw what the circuit reduces to with the switches in the different positions so for sub interval one the mosfet is on and the right side of the inductor is connected to ground in position two the diode conducts and in position three the diode and mosfet are both off and the inductor current is zero so we will go through volt second balance and charge balance for the two two reactive elements so in sub interval one the inductor voltage is vg and the load current or the capacitor current is minus the load current we make the small ripple approximation for the output voltage as usual and replace v of t with capital v in position two the inductor is connected to the output so the inductor voltage is vg minus v we can again make the small ripple approximation and replace v of t with capital v the capacitor current in this interval is equal to the inductor current minus the load current we do not make the small ripple approximation on i because the ripple is large in the inductor current but we can make the small ripple approximation in the voltage for the third interval the inductor current is zero and the capacitor current is minus the load current we'll make the small ripple approximation in the load current as well so we apply volt second balance here are the inductor voltages we have found for the three intervals so there we can draw the vl of t waveform to apply volt second balance we find the average inductor voltage which will be d times the our d1 times the value in the first interval vg plus d2 times the value in the second interval vg minus v plus d3 times zero and in steady state the average inductor voltage is zero we can solve this equation for the output voltage if we do this is what we get the output voltage is a function of d1 d2 and vg d2 is an unknown still at this point so we need another equation to solve for v2 for d2 so to get the second equation we apply capacitor charge balance in the case of the boost converter the elements connected to the capacitor node besides the capacitor are the load and the diode so to work out capacitor charge balance we should write the node equation so at this node the diode current is equal to the capacitor current plus the load current like this capacitor charge balance tells us that the dc component of capacitor current is zero so we at we find the dc components of these currents capacitor current has zero average and therefore the average diode current is equal to the dc load current so for the boost converter to apply charge balance we need to work out the average diode current here is our inductor current waveform that looks like the usual discontinuous mode inductor current the diode current equals the inductor current when the diode conducts and it's zero the rest of the time so this is the diode current waveform to find the dc component of this current we should integrate over one period and divide by the period the integral of the diode current then would be the area of this triangle and that area is one half the base of the triangle which is for this example d2ts times the height which is this peak current we'll call i peak we can find i peak knowing the slopes so during the first interval the inductor current increases with a slope of vg over l and i peak then would be that slope vg over l times the length of the first interval d1ts so we can take that eye peak and plug it into here okay so here is the area times 1 over ts to get the average we plug in our expression for ipeak and we get this and so if we equate the average diode current to the load current finally this is the equation that results from capacitor charge balance so we have two equations and two unknowns first equation came from volt's second balance second equation came from capacitor charge balance and the two unknowns are the output voltage v and the duty cycle d2 so we can take these two equations now eliminate d2 and solve for the output voltage the steps in that analysis are uh outlined here so what we're going to do is take the first equation and solve it for d2 so the little algebra this is what you get we'll then plug that expression for d2 into the second equation and then if we rearrange terms to try to solve for v we get this okay so this tip it turns out that this gives us a quadratic equation in the output voltage that we need to solve here is the same equation repeated so we apply the quadratic formula to solve for v and this is the answer as usual usual with the quadratic equation we get two roots so which root is correct well if you look at the um the discriminant or the quantity inside the radical that quantity is greater than one so one plus the square root of the discriminant gives us a number bigger than two dividing by two gives us a number bigger than one and the conversion ratio is bigger than one which is actually what we expect for the boost converter on the other hand the other root if we take the minus sign gives us one minus a number bigger than one so we get a negative answer well we know the boost converter gives a positive voltage but even if we didn't we could go back and look at our volt second balance equation and see that if vg is positive then v has to be positive because the duty cycles have to be positive so a negative v is actually happening because our solution is giving a negative d2 which is a non-physical answer so we select the positive sign and then this is our answer for the conversion ratio of the boost and discontinuous mode and it's valid again when k is less than k crit here's a plot of what what the function looks like so we have that m the conversion ratio is one over d prime in continuous mode and it's the function with a radical in the discontinuous mode so for for example for k equals um a number bigger than 4 27 we'll always be in continuous mode and we have this function whereas if k is say 0.1 then we'll be in discontinuous mode over some intermediate range of duty cycles and we'll follow the continuous mode result outside that range so here's a plot for that that example okay actually for large k the discontinuous mode equation can be approximated for d squared over k turns out to be much greater than one you can ignore the one take the radical and you get that m is equal to one half plus d over root k so it's actually approximately a linear function of duty cycle for small k so for example if k is 0.01 we get this function where it's nearly always in discontinuous mode with a small k and it's approximately linear function of duty cycle with a slope of 1 over root k which would be 10. okay here's a handy table of the answers for the basic converters so here's what k crit is here's the discontinuous mode m of d here's the completing the solution for d2 and then the last column is the continuous mode result so for all basic converters you can just simply go to the table and plug in the results finally here's a plot of the discontinuous mode functions the m of d functions for discontinuous mode for the basic three converters so the boost was this function that approached a linear function with a slope of one over root k it turns out that the buck boost converter exactly has a slope of 1 over root k and it is exactly linear so that's plotted here the buck boost is inverting and so the inversion is not included in this plot and then the buck turn its function turns out to be asymptotic to the buck boost line into one and so it's this function that's below those asymptotes while the boost function is above those asymptotes okay so for the boost and the buck boost we actually get functions that are almost linear in discontinuous mode one last point i would make here is for both the boost and the buck boost formulas what happens to the output voltage in this case when you remove the load well if we remove the load this is the case where r goes to infinity and that makes k go to zero so what happens here our slope is one over root k if k goes to zero our slope goes to infinity and the output voltage goes to infinity this is a real problem and it really happens if you remove the load from your booster buck boost converters its output voltage tends to infinity so physically why is that let's go back and look at the converter circuit can you explain why the output voltage of the boost converter would tend to infinity if you remove the load well the answer is if we look at just how the circuit works when the mosfet turns on the inductor current charges up to some current and some energy from vg is stored in the inductor when we turn the mosfet off that inductor current and its stored energy flows through the diode to the output and it charges up the capacitor so when it chart when that energy is transferred to the capacitor the capacitor voltage increases and its stored energy increases now there is no mechanism for discharging the capacitor if you remove the load so the only thing that can happen is every time we switch every switching period the capacitor is charged with a little more energy and its voltage rises more and more every time or during every switching period and with no mechanism for discharging the load or discharging this capacitor through the load the voltage can build up to an arbitrarily large number so it is something to watch out for we have to be careful and limit the output voltage or if our feedback loop senses this happening it needs to stop switching or turn down its duty cycle to keep this from happening but if you run an open loop converter and you say your load wire falls out then the converter will make a very large voltage it will probably exceed the voltage ratings of some of the components and and the converter will fail so we've seen that the discontinuous conduction mode is a consequence of using single quadrant switches or uni-directional switches in an application where the applied switch waveforms become bi-directional this causes the switches to to change their conducting state at times that don't coincide with the drive signal and it adds additional intervals to the converter which change the properties of the converter we found how to find the the mode boundaries by equating the dc component of the inductor current to the ripple there are some sample problems that have multiple inductors and then this gets a little more complicated but in the case with multiple inductors we have to carefully draw the diode current waveform and find under which conditions it goes to zero but in general we can find an equation of the form k is less than k crit for discontinuous mode operation we also found how to solve for a steady state in the converters we apply volt second balance and charge balance as usual to get a set of equations but we have to be careful with the small ripple approximation and in those cases where the ripple is large we have to use additional arguments this week we're going to discuss converter circuits first we're going to talk about where all these converters come from and what are the relationships uh between them so i have uh you know just brought up kind of out of the blue introduced some different converters such as the boost or the buck boost or some others that we've done in problems so i want to talk just a little bit about what the relationships are between these and how you can derive one from the other then second we're going to talk about transformer isolated converters which so far in the course have not been covered so we will talk about the important topic of how to introduce a physical magnetic isolation transformer into a power converter and discuss some of the well-known and widely used isolated converter circuits so we'll begin with uh section 6.1 called circuit manipulations and what we're going to discuss here is how to take a converter we already know and manipulate it to generate a new one and uh to perform some you know new function that we want so what we're going to do is start with a buck converter as the you know the original rudimentary converter and at the beginning of the course i introduced this converter from first principles that we have a switch network that um changes the dc component of the voltage and then this is followed by a filter that removes the switching harmonics and passes this changed dc component to the load and we know now from all of the analysis and basic arguments we've given that the conversion ratio of the buck converter m is equal to the duty cycle where this conversion ratio is the ratio of the output to the input voltages okay now something that we can do to a converter is first we can consider changing the positions of the power source and the load so here i've drawn kind of a box around the butt converter where we have input terminals labeled voltage v1 where there is a source of power connected and we have output terminals that are labeled v2 here where we've connected a load with a filter capacitor so power flows from the source to the load and we know since this is a buck converter that v2 is equal to the duty cycle times v1 now one thing that we could do is connect our power source to port 2 and our load to port 1. and here's an example of that so here i've connected the power source here as v2 the load is v1 but the relationship between v1 and v2 is the same in fact if you apply volt second balance to this inductor you find you get exactly the same relationship that v2 is still dv1 but now v2 is the power source and v1 is the load so if we solve for v1 we find that v1 is equal to v2 divided by the duty cycle and since the duty cycle is less than one the load voltage v1 is greater than the source voltage v2 so this performs the function actually of boosting the voltage and in fact if you look at this carefully you might recognize that this looks like a boost converter just drawn from right to left instead of from left to right by interchanging the source in the load what we've done is reverse the direction of power flow so this actually makes current flow the other direction through the inductor and as a result we have to look at carefully at how we realize the switches so back in chapter 4 we introduced a formal way to realize the switches and if we apply that process here what we find is that we have to put a transistor from the switching node to ground and we have to put a diode from the switching node to port one which is quite different than what we had to do in the buck converter when current flowed the other direction and something that this does is that it actually changes the definition of the duty cycle we can define the duty cycle as being uh the fraction of the time the switch is in position one or we could define it as the fraction of time the switch is in position two it's really just semantics we typically define the duty cycle according to the transistor duty cycle and here when this transistor is on the switch now is in position two not one in fact we can just define switch position one is when the transistor is on and if we do that we have to interchange d and d prime so v one we then write as one over d prime times v two instead of one over d and this then is in fact the uh the traditional equation for the boost converter that the output voltage which is now v one is one over d prime times the input voltage so the d prime comes simply because the power is flowing in the opposite direction and we have to realize the switches in a different way so the boost literally is simply a buck converter connected backwards okay a second thing we can do is we can connect converters and cascade so take your favorite converter and make call it converter number one and we know if we know its conversion ratio so the output voltage of converter one v1 is equal to this conversion ratio m1 of d times the input voltage vg and we can take that voltage v1 and use it as the input voltage of a second converter so take say your second most favorite converter and put it there is converter number two and its output voltage that here we will apply to the load is the conversion ratio m2 for the second converter times v1 okay so the overall conversion ratio of this cascade connection then is v is equal to m1 times m2 times vg so we get a conversion ratio of m1 times m2 okay well that's fine but let's let's try it so uh i think a straightforward thing to do might be to take a buck converter and put it as converter one and let's put a boost converter as converter 2. so this combination is capable of making an output voltage that is either less than or greater than vg depending on the duty cycles of the individual buck and boost converters okay and so in fact if we drive the switches of the the two converters with the same duty cycle then what we get for the cascade connection is d over one minus d which sounds like a buck boost type conversion ratio and here literally is the circuit then here's our buck converter this switch and lc filter the output of the buck becomes the input to the boost converter that has its switch and its l and c driving the load okay well that's all fine however i would say that there are some things we can do to simplify this circuit and for one thing if you look at the this inner part of this circuit we have a three pole low pass filter so there's an inductor then a capacitor and then another inductor that are filtering the ripple now you you can do this if you want but we don't really have to we can reduce the order of this low pass filter and the basic dc waveforms will be unchanged changing the order of the filler merely changes how it filters the ripple so for example we could remove c1 if we like and once we've done that we have two inductors in series and we can combine the series connection of inductors into one inductor so here's that we have the two inductors in series we combine them into a single effective inductor that's the sum of the two inductances and we get this circuit okay this is called the non-inverting buck boost converter and it's actually a popular converter it's widely used in applications such as mobile computing and cell phones where we need an output voltage that is either greater than or less than the input voltage and we actually don't have to control the two switch networks with the same duty cycle you can switch for example just one and not the other if you want and that turns out to be a good way to control them so if you want to boost we make the buck switches just always connected in position one and we operate the boost switches with some duty cycle to control the boost ratio and conversely if we want to buck we just leave the boost switches in position two and we control the the buck switches with some duty cycle accord to control the buck ratio of the converter this is a very efficient converter and especially if the output voltage is near the input voltage we have a low switching loss and low conduction loss in this converter and it can operate with very high efficiency okay there's something else we can do to this non-inverting butt boost converter if we just look at what the this circuit is with a switch in the two different positions and let's assume now that we operate both switch networks with the same duty cycle then when the switch is in position one the inductor is connected to vg and i put a dot here to to follow the polarity in which the inductor is connected so the dot is connected to vg and the non-dot side of the inductor is connected to ground so we have this circuit for sub-interval one in sub-interval two we have the dot side of the inductor connected to ground and the non-dot side connected to the output like this okay so this is what i just described and this is this non-inverting buck boost converter now one simple thing that we can do is change the polarity of the inductor during one of the intervals so for example during sub interval two if we just reverse the direction of the inductor and connect the dot side to the output voltage instead of to ground like this then that will change the polarity of the output voltage our inductor current instead of flowing in this way to the output will flow the other direction and cause a negative output voltage otherwise this is the same circuit as before so this is called the inverting buck boost converter and it turns out to reduce the number of switches necessary because if you look at this the non-dot side of the inductor now is always connected to ground and all you have to do is connect the dot side either to vg in position one or to the output voltage in position two okay so here are the two sub intervals and we can connect a switch like this as i just mentioned and what we get then is the inverting buck boost converter which is what we have talked about previously in this course so the inverting buck boost converter can be derived from the cascade connection of a buck converter and a boost converter and then by reversing the polarity of the inductor during one of the intervals we simplify this the number of switches needed and we get a minus sign so that it's an inverting converter okay so cascade connections of converters then can actually uh lead to new converters as well we cascade two converters and then simplify the result and we can generate a new converter with properties that combine the properties of the two individual original converters if you construct the equivalent circuit model of the buck boost converter what you find is that you actually get two dc transformer models in it there's a buck transformer and later there's a boost transformer in the equivalent circuit model and this actually reflects it the origins of the buck boost converter that it is in fact a cascade of a buck followed by a boost there are other cascade connections possible i just took one arbitrary one another well-known cascade connection starts with a boost and is followed by a buck so we have boost first then buck instead of buck followed by boost and if we do that connection and follow the same arguments that i applied to the buck boost converter what we get in this case is a chook converter which is named after its inventor and he actually derived this converter by cascading a boost followed by a buck now you can plug other converters in and follow them we're not going to do that but as an exercise it's an interesting thing to try to do and see what other kinds of converter circuits you can generate we saw in the last lecture how to apply transformations such as inversion of source and load or a cascade connection of converters to take a dc-dc converter and generate a new dc-dc converter that has different possibly desirable properties the approaches that were considered in that lecture will will not generate an inverter so suppose we have dc dc converters but we want to make a dc to ac or ac to dc type converter we need to be able to change the polarity of the voltage and so far in the previous lecture none of those approaches would accomplish that so in this lecture we'll talk briefly about one of the widely used ways to to produce ac and it's called the differential connection of the load so if we have two dc dc converters what we can do is connect the load differentially between their outputs so for example suppose these converters are buck converters okay so the first buck produces an output voltage that is the duty cycle times vg and it's a dc voltage the second buck all produces an output voltage that is its duty cycle times the input voltage which is also dc but we connect the load differentially between the outputs so that the voltage across the load is voltage 1 minus voltage two now if we have if both buck converters produce the same output voltage then we will get zero volts differentially across the load but if we make one of the converters produce a higher voltage than the other then we can get either positive or negative voltage across the load depending on which output which buck converter makes the higher voltage and in fact if we control the duty cycles of the two buck converters in a skillful way then we could make for example a sinusoidal voltage that appears differentially across the load so here we'll draw out the circuit here's buck converter number one with its switch and lc filter here's buck converter number two with its switch and lc filter and what we're going to do is drive buck converter number two with a complement of the duty cycle of converter one so if we drive the transistor of buck converter 1 with say the duty cycle d then buck converter 2 will have its transistor driven with d prime okay so then the voltage across the load will be v1 minus v2 or dvg minus d prime vg and we can combine the d and d prime together to get this 2d minus 1 is actually d minus d prime or d minus the quantity 1 minus d so we get 2 d minus 1. okay here's the plot of that conversion ratio 2d minus 1 versus d at d of a half the conversion ratio is zero and when d is greater than half we get a positive voltage and when d is less than a half we get a negative voltage now this circuit can be considerably simplified we have a lot of lc filters and we can again like in the last lecture simplify this filtering and the first thing we might want to do is if you look at this you see we have capacitors that are connected from the individual outputs to ground where what we might really want to do instead is directly put a capacitor across our load and filter the load voltage so here what i've done is remove these two capacitors and replace them with a single capacitor across the load once we do that you can see that now effectively we have our two inductors in series so we could take those two inductors and combine them into a single inductor like this whose value would be uh the sum of the two individual inductor values and now you can see we have simply an lc low pass filter that's taking the differential voltage coming out of the two switch networks and putting that through a two pole low pass filter to provide the load voltage okay this is the way we most often build this circuit and on the right side here is this exactly same circuit it's just redrawn in the way it's normally drawn where we have our two switch networks and the load and lc low pass filters connected differentially across the outputs this is often called the h bridge or full bridge circuit or a bridge inverter circuit it's very commonly used in single phase inverter applications where we have a dc input vg and we want to produce a sinusoidal output across the load v it's also used in dc dc applications for example in dc motor drives or servo amplifiers in control systems where we want to drive a dc motor in either direction so to drive the motor in one direction we apply a positive voltage that's dc to drive it in the opposite direction we apply a negative dc voltage and so we can do that with a circuit like this as well we might in that case actually replace the whole rlc network simply with a dc motor connected differentially between these terminals the self-inductance of the motor winding takes the place of this inductor and the inertia of the shaft replaces the capacitor if we want to make three phase ac we can do the same trick with three converters so here we might have three different buck converters one for each phase each buck converter produces an output voltage that is its duty cycle times the dc input voltage vg and we connect our three-phase load differentially across the outputs the three of the three converters okay so the line-to-line voltage from phase a to phase b will be v1 minus v2 the line to line voltage from b phase b to phase c will be v2 minus v3 and so on like this so what we do then is we control the duty cycles of the three converters to make the three individual uh ac output voltages be the desired sinusoids with no offset or no dc bias so here is that circuit here again we have three buck converters with their individual filters as before we can change the filtering and again instead of filtering the voltage from one phase to ground we might instead want to filter the ac voltage so we can move these capacitors across the ac load here the circuit is drawn with the capacitors emitted altogether and we simply have the three inductors of the three buck converters and the three switch networks one for each phase if desired we can put three-phase capacitor filters as well for example we could connect them in what's called a delta connection and make them filter each line-to-line voltage or in some applications we might omit those capacitors um if we want to build a motor drive we might replace this whole network with say a three-phase induction motor the inductors become the inductances of the windings and there will be also filtering from the inertia of the load but effectively we have a three-phase ac load that is connected differentially across the outputs of three buck converters here's another similar kind of circuit that has a boost type characteristic there's a single inductor on the input there are switches for each of the three phases and we have capacitive filters on the output and this is the current source inverter it has a boost type conversion characteristic but it has a similar three-phase bridge in it in each of these circuits i've drawn ideal switches but of course we have to realize the switches carefully according to the procedures we discussed in chapter four so for for this circuit with the buck converters our buck switches need to be current bi-directional switches because they conduct the ac output currents and in the current source inverter we need voltage bi-directional switches because they block the output voltages but we've now already discussed how to do this and we know how okay so we've seen that the differential connections of dc dc converters uh can lead to inverter type characteristics and give us a way to really synthesize and understand where some of these inverter circuits come from now there are other ways to do it as well but this differential connection is in fact the most widely used approach we previously discussed some circuit manipulations to expose the relationships between the different converters and to derive some new ones there's another way to do this we can actually derive all possible converters that have a given set of constraints such as having say two switching intervals and one inductor i'm not going to derive all the possible converters it turns out that there are an infinite number of them and they do some pretty interesting and diverse things but i do want to briefly summarize some of the basic results so what's shown here is i think of the most rudimentary class of converters where we have one inductor that is switched between the input source vg and the output load v in two different ways one way during interval one and another way during interval two now there's only a limited number of possible ways to connect an inductor between two voltages and pick one for the first interval and one for the second and we can write down all the possible ways to do that and see what converters occur and the answer is right here there are eight converters that are interesting uh in addition to a few degenerate or redundant cases that don't add anything new and we can group these converters actually into three different groups the first group are converters that i would call dc dc namely that if you have a positive input voltage then the output voltage is always of the same polarity regardless of the duty cycle there's also two converters where when you vary the duty cycle from zero to one the output voltage changes polarity in such a way that these are useful as dc to ac type inverters and then the last two are the inversion of the inverters so they're more suitable for ac to dc type applications or rectifiers so here they are the first two are the buck and the boost which we already know about the next two we've also discussed we have the the buck boost and the non-inverting buck boost so these are the four converters in this class they're all you can make with a single inductor and two intervals that are dc dc type converters okay converters five and six are ones that you might say are suitable as inverters and we've already discussed the h bridge in the last lecture this is converter number five converter number six is a pretty strange one it was invented actually the transformer isolated version of this was invented at the watkins johnson company and we call this the watkins johnson converter it has this conversion ratio 2d minus 1 divided by d it looks like this but it passes through 0 at d of a half and it can make either positive or negative output voltages although the m of d function is a non-linear one and then the last two in this class are the inverse of numbers five and six so converter number seven is what you get if you swap the source and the load of the h bridge and this one is sometimes called the current fed bridge so its conversion ratio is one over two d minus one and at d of a half the ideal conversion ratio goes to infinity so here's a plot of its conversion ratio this is a pretty strange conversion ratio but in fact if the input is a sinusoid and you want the output to be dc this is the conversion ratio that you need to perform this rectification function so when the sinusoid at the input passes through zero and the output voltage is non-zero then the ratio or the conversion ratio goes to infinity but in fact we're not making infinite output voltage we're just holding up the output voltage when the input is going through zero so if you vary the duty cycle sinusoidally about a half then you can convert an ac input voltage to a dc output and then the last one is what we get from inversion of source and load with the watkins johnson converter and its conversion ratio is the inverse which also goes to infinity at the other half so those are all the converters that are possible if you only have one inductor and you have two switching intervals we can think of other classes of converters and the next most logical class might be the class of two inductor type converters so we have two inductors and we switch them in one way between the source and load for the first interval and a different way for the second interval there are many more than eight members of this class of converters and they can do some pretty interesting different things uh here's just a couple of them the uh converter we've previously mentioned and derived this by cascade of a boost followed by a buck converter it's an inverting buck boost type converter the sepik is another converter in this class the transformer isolated version of the sepik was invented at bell labs in the 1970s and they called it a single-ended primary inductance converter whose initials are sepic this turns out to be a non-inverting buck boost converter and it's interesting actually finds quite a few applications it's interesting for one because it can its transistor has its source connected to ground we realize the switches with transistor and diode like this so it's easy to drive this transistor with a low side driver and get this non-inverting buck boost type characteristic here's another converter is the inverse of the sepik it's also a non-inverting buck boost type converter some people call this the zeta converter uh then there's a lot of other very different converters so some of them have what we call quadratic conversion ratios that are functions of the duty cycle squared here's one of those that was in an earlier homework assignment that has a conversion ratio equal to d squared and it's interesting that you can get such a conversion ratio in a converter that has three diodes and only one transistor we can get quadratic conversion ratios that have other functions also like buck boost squared or other types of conversion ratios by some of the members of this class so there are some very interesting and exotic conversion ratios that are possible in general you know if you have an arbitrary number of inductors and switches we can build are there an infinite number of converters that are possible and in this universe of converters there are many different and unusual functions of duty cycle and some of these can have practical applications so far in the course we have talked about converters containing only inductors capacitors and switches we've ignored one of the other primary elements used in switching converters namely the magnetic transformer so for the next several lectures we're going to discuss converters with transformers what the basic circuits are and how to model them and extend what we already know to understand the operation of transformer isolated converters there are a number of reasons to incorporate a transformer into a converter one of the major ones is when safety requirements require that we have transformer isolation from the power line now one way to do this is to put a 60 cycle transformer at the input of our system but 60 cycle or 50 cycle transformers elsewhere in the world these transformers are very heavy and and large and nowadays we generally try to avoid them whenever we can the size of the transformer scales more or less inversely with frequency so if we can build a transformer that operates at the converter switching frequency say 100 kilohertz instead of 60 or 50 hertz uh the size of the transformer and its weight can be re reduced by a very large margin here's an example of what an iphone charger that plugs into the 120 volt 60 cycle wall and produces a 5 volt output to drive usb connected devices and power them inside this uh charger unit is is a flyback converter which is a transformer isolated version of the buck boost converter and it contains a high frequency transformer that operates at the switching frequency another reason to use a transformer is when we have a large step up or step down of voltage suppose we wanted we had 100 volts and we wanted to make one volt we could build a buck converter and operate it at a duty cycle of one percent but such a small duty cycle may have may mean that the conduction time of the transistor is equal to the switching times of the transistors and we find then that the efficiency is very low most of the time we're switching instead of conducting so instead we could build a converter that includes a transformer that has a turns ratio and use the turns ratio to do most of the work of stepping down the voltage and that converter then operates at a much higher duty cycle where the efficiency is better yet another reason is we sometimes build multiple output converters that supply multiple voltages and these are simply additional secondary windings on the transformer inside the converter and the voltages approximately scale by the turns ratios of the windings here's a diagram of a transformer that is a physical magnetic transformer it has multiple windings each with some number of turns so winding one has n1 turns winding two as in two turns and so on and we've labeled the voltages and currents of each winding uh here the convention used with polarity marks on the windings is we're defining for now all the currents is flowing into the respective polarity mark or dot and we're defining the voltages as plus with respect to the dots to understand how a converter works that contains a transformer like this we need to first model the transformer and to write a set of equations that describe the behave its behavior what we're going to do is to employ this equivalent circuit model for the transformer and so the physical transformer on the left is replaced with this equivalent circuit model on the right where everything inside my dash lines here is internal to the model and is modeling things that are going on inside the transformer so we replace our physical transformer with this equivalent circuit model and this equivalent circuit model has elements that we can solve using standard circuit analysis so the equivalent circuit model we will use [Music] most of the time is this simple first order model that contains an ideal transformer inside these dotted lines uh and then in parallel with one of the windings of the ideal transformer is an inductor that is called the magnetizing inductance of the transformer okay so there isn't actually a physical extra inductor put on our transformer this is built into the transformer and everything inside this dashed line is an equivalent circuit that is modeling correctly the terminal equations between the voltages and currents of the windings okay the defining equations of the ideal transformer we've seen before the voltage of each winding divided by its turns is equal for all of the windings so v1 over n1 equals v2 over n2 and so on and this makes the voltages scale according to the turns ratios of the ideal transformer the equations of the currents going into the windings of the ideal transformer is this one and this is again we've seen before also the sum of the amp turns flowing into the dots of the ideal transformer is zero now in addition to the ideal transformer we're adding one more element this l sub m is called the magnetizing inductance and its current is i sub m is called the magnetizing current the magnetizing inductance um is actually real and the way to see that is to do the thought experiment of suppose we connect something up to our transformer like some ac voltage connect it up to our primary winding and suppose we disconnect all the other windings so they're winding two and winding three here are simply open circuited what do we expect this to to do well this is a real physical magnetic device it has some number of turns of wire on a iron core or ferrite core and if you have simply some turns of wire on one winding that is on a core we would expect that to behave like an inductor and the magnetizing inductance then actually models that inductance so the magnetizing inductance referred to the first winding where it's drawn here is the inductance that you would get if you simply put n1 turns of wire on the core the magnetizing inductance does behave as a real inductor it obeys the voltage as ldidt or in this case v1 is lmdimdt and it has to obey volt second balance the average value of the voltage applied to this inductor must be zero so we can apply a volt second balance to find the steady state behavior of the of circuits containing physical magnetic transformers magnetizing inductance also explains why you can't put dc through a transformer so we can't for example apply a dc voltage vg to our transformer because at dc the magnetizing inductance being an inductor has zero impedance its impedance goes to a short circuit at dc and it shorts out our dc voltage and if we apply a dc voltage volt seconds won't balance on on lm the magnetizing inductance is a physical inductor in that sense it models magnetization of the transformer core the transformer core has a bh characteristic if um you apply too many volt seconds to the transformer core it will saturate and the transformer will turn into a short circuit which generally is bad in most of our converters so if the applied volt seconds on the winding is too large what happens when you apply voltage is it we integrate so by the by faraday's law the flux in the core is given or proportional to the integral of the applied voltage as we apply voltage we integrate up the bh loop when we get into saturation then the current in the windings in the magnetizing inductance becomes very large and the magnetizing inductance essentially turns into a short circuit which is called saturation of the transformer and it's generally something we want to avoid so then we apply volt second balance to transformers when we analyze converters that contain them so the procedure here is that we replace the physical transformer with this equivalent circuit model that contains an ideal transformer plus a magnetizing inductance in parallel with one of the windings and then we apply a volt second balance to this magnetizing inductance just like we do to any other inductor in the converter and as we've been doing uh in the past and understanding how the volt seconds balance then is part of the analysis of the converter okay uh we sometimes use the terminology transformer reset and transformer reset is the mechanism by which the converter circuit makes the volt seconds balance on the transformer in many converter circuits it's not easy to add a transformer we have to add additional circuitry to make the transformer reset and get the volt second balance to happen so we're going to discuss that uh with respect to a transformer isolated version of the buck converter known as the forward converter to understand how the transformer gets reset and how we can get the volt seconds to balance one last point this equivalent circuit model is a simple first order model that is adequate for most uh things we're going to do in analysis of converters but if desired we can further refine this model to model other things so for example if we want to model the resistance of the windings like we've done in the in previous exercises for modeling losses in inductors we can model resistance of the windings of the transformer by putting resistors in series with each of these windings and in our equivalent circuit model then these resistors appear basically at the terminals of the model so in these places sometimes we model what's called core loss in the transformer and we can do that by putting a resistor in parallel with a magnetizing inductance the traditional way to do that and sometimes we model what are called leakage inductances that are effectively series inductances in series with these windings okay we're not going to model those in this course but they do exist they're a higher order effect in the transformer but there are times where we need to to include them so in the upcoming lectures we're going to use this transformer model to understand how some of the basic transformer isolated converter circuits work we're going to look at transformer isolated versions of the buck converter and the buck boost converter we will apply this model and understand the transformer reset generalizing our ideas of volts second balance and charge balance to include volt second balance on the magnetizing inductance i'll also briefly list some of the other common transformer isolated converters last time we talked about how to model a transformer and the first order model that included an ideal transformer with a magnetizing inductance in parallel with one of the windings the magnetizing inductance models magnetization of the core and it operates like a real inductor in that we have to have volt second balance on this magnetizing inductance so today we're going to talk about how to put a transformer into a buck converter and one of the common well-known and commercially important transformer isolated versions of the buck converter that is called the forward converter now putting an isolation transformer into a buck converter and getting the volt seconds to balance on the magnetizing inductance is not so easy to do here's a buck converter i've just drawn one and let's think about you know where where could we perhaps break the circuit and insert an ideal or insert an actual isolation transformer so we want to put a transformer somewhere in this circuit so perhaps we could break these wires and put the transformer here or we might try putting it over here perhaps breaking the wires right there or perhaps we could try breaking the wires right here and the question is do any of these work and could you put the transformer in any of these points and get the volt seconds to balance okay well if we put the transformer in position 1 we're connecting the primary of the transformer straight up to vg which is dc so we would be applying a dc voltage to the primary winding of the transformer and the volt seconds won't balance so that's no good likewise at position 2 we would be connecting the secondary second area of our transformer up to the output voltage v which is dc if the converter works and so we're putting dc on our secondary and so that also applies a dc voltage across the transformer the volt seconds won't balance and that's no good okay how about position three or this point three well that's a little trickier but that doesn't work either because in fact if volt seconds balance on this inductor then the average voltage across the in this filter inductor is zero and so the dc at this point at 0.3 is the same as the dc at 0.2 in fact the voltage the switch voltage at this point switches from 0 to to vg and back but it never goes negative so the average is positive and the volt seconds don't balance there either so none of these places work and it's just not so easy to do so we have to go to more work and we actually have to add extra circuitry to make the volt seconds balance and reset the transformer in any of these buck derived transformer isolated converters and so there are quite a few different schemes and different ways to do that what we're going to talk about now is one of the more popular ways that is called the forward converter here's the schematic of the forward converter we have the transformer placed in series with q1 and its output or secondary winding which is actually winding three in this diagram goes through some diodes to the output terminals and the voltage right here is the same voltage driving the the lc output filter that we have in the regular non-isolated buck converter so vs of t actually switches high and low with some duty cycle it turns out to switch high to in three over in one vg with a swit with a transistor on and it's zero with the transistor off um and so we get a dc a voltage that has a a dc average that depends on the duty cycle that gets filtered to to produce the output voltage that is very similar to the original non-isolated buck converter but then we have an extra winding here and there's some extra means for resetting the transformer and applying negative voltage across the magnetizing inductance during the d prime interval to get the average value of the transformer voltage to be zero and get the volt seconds to balance on the magnetizing inductance so we need to talk about how that works in order to understand this converter and there's actually different versions of the forward converter that have different circuits to reset the transformer this is one of the classic ones that has a winding this into winding is here only for resetting the transformer but the power actually flows through the first and third windings so to approach the analysis of a transformer isolated converter the first thing we do is replace the transformer with the equivalent circuit model so here is the equivalent circuit model we discussed in the last lecture that contains a an ideal transformer with three windings in this case and in parallel with the first winding we're going to put the magnetizing inductance the value lm and it has a magnetizing current im okay having that that model in the circuit now we can proceed and analyze the circuit in the usual way so we will work out the voltage on the inductor we'll work out the voltage on the magnetizing inductance and we'll apply volt second balance separately to each of these we also will work out the capacitor current and apply charge balance and in general we will also work out the input current ig and find its average value as well okay here are the waveforms and it turns out that with this version of the forward converter the magnetizing current operates in the discontinuous conduction mode and it has a discontinuous mode type waveform that starts at 0 rises to a peak value while the transistor is on when the transistor is off the magnetizing current decreases down to zero and then remains at zero for the rest of the switching period and in the process of going back to zero we say that the transformer has been reset and its magnetizing current has gone back to the starting point and the voltages that are applied to the induct to the magnetizing inductance that make this happen uh will then have volt second balance okay so uh for this version of the converter the magnetizing inductance must operate in discontinuous mode the output filter inductor on the other hand can operate either in continuous or discontinuous mode just like any other buck filter inductor so we need to write the [Music] write the circuit for the different uh switching intervals uh and work out the waveforms okay so let's take our equivalent circuit for the transformer and the converter and look at what happens during each interval so this is a circuit with three diodes whenever we have a lot of diodes it can be tricky to figure out which diodes conduct when so here's the explanation during the first interval the d interval we turn the mosfet on okay so q1 is on and that makes vg appear across the primary winding so here's the dot of the winding is is at the plus of v g so the windings are positive with respect to the dot we'll have vg times the turns ratio appear across this in two winding that's plus at the dot and you can see then that the voltage across this diode d1 will be vg plus this positive trans or winding two voltage which is a positive voltage and it reverse biases d1 so d1 is off and blocking voltage on the third winding we have vg times the turns ratio that is plus at the dot so there's positive voltage coming out the dot which will turn on d2 and then we'll get positive voltage after d2 and that positive voltage will turn d3 off so that's the situation during the first interval with q1 on d2 on and the other two diodes are off here's the circuit then d1 is an open circuit d3 is an open circuit q1 and d2 are short circuits okay what happens during this interval okay first of all we have the output inductor current which i'm going to call i and it'll be approximately capital i if we have small ripple that current flows out of the dot of the winding three of the transformer there's no current in winding 2 because d1 is off and so this output current which follows the the equations of the ideal transformer if we have current coming out of the dot on the the third winding we must have current going into the dot on the first winding okay recall that the sum of the currents flowing into the dots times their respective terms adds up to zero in a transformer so n1 i1 prime is this current flowing into the dot plus into i2 which is zero and then i3 times n3 is flowing into the dot but the inductor current is flowing the other direction so we could subtract 3i is the current flowing into the dot of the third winding those add up to zero and so that tells us that i1 prime the current flowing into the ideal transformer dot primary winding is equal to n3 over n1i so basically this is the reflected output inductor filler current is flowing this way okay the current ig during this interval is that current in that ideal line ideal transformer primary winding in 3 over n1i plus the magnetizing current one other thing i'll say is for forward converters we generally design the transformer with no air gap and the magnetizing inductance is very large the magnetizing current is very small so in a typical forward converter the magnetizing current im is much smaller than than the reflected load current capital i and this ig is approximately just in three over n one times i okay what happens during the next interval in the second interval we turn q2 off okay what happens when q2 is off well the magnetizing current has built up to some positive current and even though it's small it's it's not zero it's more it's positive and it has to flow somewhere before it was flowing through the mosfet but with the mosfet turned off it has to actually flow out of one of the other windings what our model here predicts is that im will flow this way and it will flow through the ideal transformer part of our primary winding model and flow out of the dot so if we have positive current flowing out of the dot on the primary then there must be current flowing into the dot of one of the other windings well it can't flow into the dot of the third winding because that would flow backwards through d2 so in fact d2 gets reverse biased and it is turned off it won't allow reverse current so the only other place it can flow is in into the dot on the secondary winding and so it flows like this and it forward biases d1 the output inductor current has to flow somewhere also if d2 is off the output inductor current will flow through d3 so we have d3 on so for the second interval here's the circuit we have q1 off d2 off d1 and d3 are on okay what happens to the transformer during this interval is that with d1 turning on the input voltage vg is is placed across the second winding like this and it is negative at the dot instead of positive so the voltage that we get across the magnetizing inductance referred to the primary is of opposite polarity negative at the dot and the value will be vg times the turns ratio so here's what we have so far for the magnetizing inductance we have a positive voltage vg applied during the first interval by the mosfet turning on when d1 turns on we have a negative voltage vg times the turns ratio n1 over n2 um and with this voltage the magnetizing current will increase during the first interval up to some peak value and then the negative voltage during the second interval will make it decrease finally when the magnetizing current gets to zero d1 turns off because it won't allow current um to flow negative through it and so we have a third interval where where d1 is off then here is the the circ the schematic during the third interval uh d3 is still on conducting the output inductor current but d1 now has turned off and the magnetizing current remains at zero for the the rest of the switching period okay so transformer reset again this is the mechanism by which the volt seconds balance and we can analyze transformer reset by writing the equations of volt second balance on the transformer so then here again was our voltage waveform the average primary voltage which is the voltage across lm will be the duty cycle d1 times positive vg plus the duty cycle d2 times this negative voltage minus n1 over n 2 vg plus d3 times 0 for the third interval and this must be zero for the converter to work here's that formula again the formula is repeated here and we can actually solve this formula to find how long d2 is so if you just solve you find that d2 is equal to the turns ratio into over n1 times d or d1 the transistor duty cycle the constraint that this imposes is that we have to reset the transformer before the end of the switching period d1 can't be or d3 the length of the last interval can't be negative so we can we can write the d3 has to be greater than or equal to zero and d3 is equal to one minus d1 minus d2 so here is that equation and if you if you require d3 to be greater than zero and you plug in our solution for d2 what we find is that d1 or the transistor duty cycle must be less than or equal to this quantity that depends on the turns ratio from n1 to n2 if we make the common or the typical choice that n1 equals into then this says that the duty cycle has to be less than a half and what's happening then is that we're applying plus vg during the the first interval we're applying minus vg during the second interval so the second interval is the same length as the first interval and to reset before the switching period is over the duty cycle then has to be less than a half if you like though you could choose other values of turns and make this limit come out to a different number so what happens if we let d be greater than a half within one equals into well here's what it looks like we our magnetizing current won't get back to zero before the switching period is over so we'll start the next switching period with some positive value of magnetizing current and then during the next switching period we do it again we don't give the transformer long enough to reset back to zero and we build up we have some net increase in magnetizing current during the next period also and this keeps happening every switching period there's a net increase in magnetizing current and we say the transformer walks up its bh loop the flux in the core will increase each switching period and eventually the transformer will saturate once it saturates it turns into a short circuit and the next time we turn on our mosfet with the transformer saturated we get a very large current run out of eg that will make the converter fail so we have to make sure to build a control circuit that doesn't let this happen and it's typical to make n1 equal to n2 and build a control circuit that limits the maximum duty cycle to be no more than a half okay we can talk about volt second balance on the output inductor and in fact this is just the same analysis as in the non-isolated buck converter we can sketch the waveform of the voltage across that inductor and it looks like this when the mosfet is on we get a voltage here equal to vg times the turns ratio from n1 to n3 that that comes out here on the right hand side of the inductor the output voltage is v so then the vl voltage during the first interval will be n3 over n1 vg minus the output voltage and for the remainder of the period when the mosfets off we have zero coming out of our or across d3 and we the voltage across vl is simply minus the output voltage so this is dts here d prime ts is there we can apply volt second balance in the usual way so d times n3 over n1 vg minus v plus d prime times minus v equals zero by volt second balance and you can solve for the output voltage v and you find that the output voltage works out to be the duni cycle times the turns ratio times vg so this is a function that looks like a buck converter with an added turns ratio of n3 over n1 okay well if the duty cycle is limited to say a half then that limits how much output voltage you can get although we can change the turns ratio to compensate for that and we will adjust the turns ratio to step the voltage down or whatever we're trying to do we also have this other constraint though that the duty cycle is limited but to this number that depends on the turns ratio of the reset winding or the n2 turns as well and so one question you might ask is well why don't i just make n2 be small if i do that that makes the maximum duty cycle be larger so if if n2 approaches a very small number then the maximum duty cycle approaches one and we can get a full range of duty cycles well the problem with that comes with the voltage applied to the transistor during the sub interval two when we're resetting the transformer diode d1 is on transistor q1 is off and we're applying voltage vg across the second winding where it's minus at the dot and plus at the non-dot side the voltage we see on the first winding then becomes what minus at the dot vg times the turns ratio in one over n2 so the voltage that is applied to the transistor q1 during this interval which is the blocking voltage of the transistor turns out to be the input voltage vg plus the voltage across the this primary winding in one over n2 vg so i get vg times the quantity one plus n one over n two okay so if we made into small and made this turns ratio in one over into d be big it means we're applying a much bigger voltage across our mosfet so a small n2 will reset the transformer quickly in a short time but it takes a lot of voltage to do that and that voltage gets blocked by q1 so we end up having a very high voltage stress on q1 and this can be a real problem in a power supply for example that must operate off the rectified 240 volt power line that we see in much of the world when you rectify 240 volts you get a voltage of what 240 root 2 that's something on the order of or something over 300 volts if we multiply that by 2 so n1 equals into then we're cl around 700 volts across the mosfet and it's hard to to buy a mosfet that that that's that large we may end up in order to get margin we may have to use say a thousand volt mosfet and those mosfets aren't very good so in fact we often want to go the other direction and make into be large not small to reduce the voltage on our mosfet but then that further limits the maximum duty cycle so we have a trade-off and the engineer has to decide so for the case where n1 equals n2 and d is less than a half then we have this factor of 2 voltage and n1 is less than n2 then d is restricted to less than some number less than a half but we have reduced voltage on the mosfet here's another version of the forward converter that's also popular and it reduces the voltage stress on the transistors this is called the two transistor forward converter we put a transistor on each side of the transformer and we have clamping diodes also as shown here and these take the place of that second winding that did the reset so what happens here is during the first interval q1 and q2 are both on both diodes are off vg is applied to the transformer and so nvg comes out the secondary this turns on d3 d4 is off and we have the usual first interval where the reflected vg is applied to the output filter during this interval magnetizing current builds up inside the transformer just as in the previous circuit okay during the second interval what we do is we turn q 1 and q 2 off this will turn d3 off and d4 on as in the previous circuit and what happens to the magnetizing current well we have magnetizing current flowing that has built up and it wants to flow this way it can't flow out the secondary winding because that would mean going into the dot and that would reverse bias d3 so instead it flows this way and it actually flows through these two primary side diodes so im will flow it will keep flowing in this direction like this it turns on d1 and d2 and effectively applies vg in the opposite direction across the transformer and that resets the transformer this is actually the same v same primary voltage winding we would get during the second interval if in the previous circuit within one equals n2 so these diodes turn on to reset the transformer the duty cycle is limited to be less than a half and the circuit works fine with no extra winding now a nice thing about this circuit is that these diodes also clamp the voltage across the transistors to be no more than vg so you can see you know what about say the drain to source voltage of q1 is here how large can that be well the biggest it can be is if when d2 is on that which puts this node at zero and this node at vg and so vg is applied across q1 same thing happens to q2 the highest this node can be is vg before d1 will turn on and clamp this voltage to vg so we have vg is the maximum voltage on q2 as well so in the previous example with 350 volts applied at the input we can use much lower voltage mosfets in fact if we use 600 volt mosfets that are very good super junction mosfet devices that we can buy nowadays we have plenty of voltage margin and so we can get a good design that's that's very reliable so this is a very popular circuit that is in commercial use today so to summarize the forward converter is a transformer isolated version of the buck converter we say it's a buck derived isolated converter we have to do some extra work to get the volt seconds to balance we saw one way to do it in which we add a second winding on the transformer that performs reset or another way is to add these extra diodes across the primary uh to reset the circuit we have two diodes on the output instead of one but with this extra circuitry then we can get the volt seconds to balance on the transformer and and get a functioning converter these are very popular and widely used converters today in this lecture we'll talk about the flyback converter which is another very popular form of transformer isolated converter the flight back converter is a transformer isolated version of the buck boost converter now it turns out that in the buck boost converter it is actually easy to add a transformer here's a diagram of the buck boost and we have a point in the middle of the buck boost converter where there is already an inductor and you could simply place a transformer at this point and the volt seconds will balance because they already balance on the inductor and in fact what we do in the flyback is is we turn the inductor into a transformer it's actually we often call it a flyback transformer and it functions in fact more like a two winding inductor so here is a way to think about it uh suppose we're in the lab winding an inductor with say some length of wire that we wind some number of turns on a magnetic cord to make an inductor and suppose instead of just winding one wire we wound two so perhaps it's like we needed a fatter wire or thicker wire to have lower resistance and we decided to just take two wires and put them together as if they were one and wind them together same number of turns okay so i could try draw that this way as is shown in this lower diagram and it actually looks like a one-to-one transformer the inductor though what we were calling the inductor in our original buck boost converter we might call now is the magnetizing inductance of this transformer so it's like we have a trans we have an inductor and then we have to a two winding ideal transformer now so far we really haven't changed anything it looks more complicated but it's it's really just functioning as an inductor with the inductor current splitting in some way between the two windings what we can do though in this case now is to now break this connection this changes how the current flows through the two windings although it doesn't change the sum of the current so if we have current in this winding call it i1 and a current in this winding we call i2 the sum of those currents adds up to the original i of the buck boost converter but the difference is that now we have i1 flowing when q1 is on and i2 flows when d1 is on so i is distributing between the two windings in a different way now the way we normally draw the flyback converter is the bottom diagram and it's nearly the same as the one above it to get a positive output voltage what i've done is reverse the polarity or reverse the dots of the winding recall that the buck boost converter inverts the polarity if we want our flyback converter to make a positive voltage then we must reverse the dots and we reverse the diode direction also so we get a positive output and then the other thing we can do if we like is now apply a turns ratio and the turns ratio will let us scale the voltage up and down and this is the conventional flyback converter there's one other small change the order of q1 and the primary winding are reversed here we usually do that to put the source of the transistor at ground and make it easy to drive with just a regular ground reference driver so the flyback converter is a buck boost converter in which we realized the inductor using two windings that effectively give us isolation and a turns ratio the flyback transformer then which is this two winding magnetic device can be modeled in the same way as a usual transformer with an ideal transformer and magnetizing inductance put in parallel but now we're using the magnetizing inductance as the actual physical inductor in our converter and this inductor is designed to store energy not only that but the winding currents look very different than the way they look in an ideal transformer we never actually have current flowing in both windings there's either current flowing in the primary when the mosfet is on or in the secondary when the diode is on but we don't have the case that the secondary current equals the turns ratio times the primary current so the magnetizing inductance is very significant its current is large and as a result we don't see what we would normally expect in an ideal transformer far from it we won't talk about magnetics design in this coursera course but i will say that the design of the flyback transformer is very different from the design of a more ideal transformer such as the one that is in the forward converter we put an air gap in this transformer that lets our inductance store substantial energy and its behavior as i mentioned is different okay having said that we can still model our converter in the usual way we replace the transformer with our transformer model having a magnetizing inductance we then solve the converter applying volt second balance to the magnetizing inductance charge balance to the capacitor and so on and we can get the equations of the converter so here's what happens during sub interval one sub interval one the mosfet is on the diode is off as is usual in the buck boost converter and with the mosfet turned on then we have the magnetizing inductance connected to vg and so we will have what vl is equal to vg we have current flowing around this path the capacitor current is simply minus the load current or minus v over r and the input current ig is equal to the magnetizing current i we'll assume we're in continuous conduction mode here and apply the small ripple approximation to v and i and get these equations for the first interval for the second interval the mosfet is off and the diode is on we now get this circuit with a diode connecting the transformer secondary to the output with the mosfet on the magnetizing current cannot flow through vg anymore in fact magnetizing current or the inductor current will flow out the secondary in our transformer equivalent circuit model the way this works is that the inductor current or magnetizing current continues to flow and it flows around this way around this path so it will flow through the ideal transformer and come out the dot on the primary side so on the secondary side we have current going into the dot equal to this primary current or i divided by the turns ratio in so now the capacitor current for this interval is that secondary current i over n minus the load current v over r and we get this equation the voltage on the magnetizing inductance vl is in fact the reflected secondary voltage with the diode on we have voltage v applied across the secondary and that voltage is negative at the dot of the secondary so it will be negative at the dot on the primary and the primary winding voltage will be v divided by the turns ratio in so the inductor voltage is minus v over n for the first interval with the mosfet off ig is equal to zero and so these are the equations for the second interval we can again apply the small ripple approximation to replace v and i with their dc components capital v and capital i so then here are those waveforms here's our vl waveform our ic waveform and our ig waveform so now we can apply volts second balance to vl so the average vl is d times this voltage plus d prime times that voltage as is shown here volt second balance says that that must be zero we can solve this then for the output voltage so what do we get v would be v g times d over d prime times the turns ratio in okay so the output voltage what is the input voltage times this but boost conversion ratio d over d prime and then also times the turns ratio in we can apply charge balance to the capacitor so the average capacitor current will be d times the value during the first interval plus d prime times the second interval value which gives us this equation we can solve that for the inductor current i or magnetizing current and what we find in that case is that the magnetizing current referred to the primary side is v over d prime r and then times the turns ratio in finally we can write the equation of ig the average ig is d times the value during the first interval i plus d prime times zero so i get the average ig is d times i so those are the equations of our dc model of the flyback as we've done in previous chapters we can construct equivalent circuit models to go with them this first one from the inductor gives us this center loop equation the second one from capacitor charge balance gives us the output side and the third equation for the average ig gives us the input port model so we get this a circuit and finally we can replace the two the ideal sources with transformers and get an equivalent circuit for the flyback the ideal flyback has a buck type transformer and then it's followed by a boost type transformer and the boost transformer additionally has a turns ratio so in the end this is a buck boost converter with an added turns ratio and the flyback then has its origins as you would expect in the buck boost so the flyback is a widely used and popular converter because it has a very low parts count it's actually known to be a cheap and inexpensive way with a very small number of parts to build a converter you can get multiple outputs for very low cost as well all we have to do is build another output so we could for example have our flyback converter with what transistor a single magnetic device flyback transformer and then one diode and one capacitor for each output and get multiple output voltages with different turns ratios and with very low parts counts small converter get a lot of different voltages one problem that is known with the flyback converter is a problem called cross regulation if we have say in two turns on this this output and in three turns on this output with their respective voltages v2 and v3 we would hope that v3 would be the turns ratio in 3 over n2 times v2 that's what we would hope and if we have good cross regulation then that's what happens it turns out with a flyback converter second order effects related to the leakage inductances these small inductances of these transformers cause the output voltages to differ substantially and it can make the output voltages highly dependent on load current so the cross regulation between the two outputs is not very good and we usually have to do some more work and more engineering to get a good design so it's inexpensive but the performance in that sense is not as good the flyback is often operated in discontinuous conduction mode that makes the magnetizing inductance very small and this transformer then can be very small and so we not only have small parts count but we have small magnetics also you can do an analysis of this converter in discontinuous mode and it turns out to be to give the the discontinuous mode buck boost equations with the an added turns ratio you