[Music] w [Music] [Music] [Music] [Music] [Music] oh [Music] greetings welcome to electronic circuits 1 this is lecture number 10 and I am bavi today we will look at uh two circuits using diodes these circuits are used in many applications one is called a halfwave rectifier and the other one the full wave rectifi uh last time we briefly talked about the halfway rectifier and I would like to refresh your memory about what we covered and now how we will continue so last time we said that uh uh in addition to looking at IV and input output characteristics of circuits we are also interested in one more type of behavior namely the time response now of course we know time response from electric circuit Theory courses but here we also would like to apply some sort of input wave maybe a sinusoidal wave to the circuit and then see what comes out of it uh this is ultimately what really the circuit does the input output characteristic and the IV characteristic are only for our own understanding and our own Insight but the actual Behavior behavior of the circuit is its response to a Time domain waveform now we said that a halfway rectifier is simply a diode D1 and the resistor R1 uh and uh it behaves like so first looked at the input output characteristic and we saw that if the input voltage is negative D1 is off uh there's no current through D1 there's no current through R1 and the output voltage is zero so the output voltage was Zero up to a certain value of the input how much voltage do we need to turn on the diode well one diode drop VD on if we use the uh uh the volt the constant voltage model so at VD on the diode turns on and from here on uh the input just goes to the output with a slope of one and the only difference is that the output has a level shift with respect to the input if the input is 2 volts the output is 2 volts minus VD on uh this was an important property we said that the output is pinned to the input meaning that as the input changes the output also changes uh in proportion except that there's a shift of VD on between the two we said the output tracks the input or the output is pinned to the input all right now we said okay let's uh uh in addition to looking at the input output characteristic let's look at the time response when the input is a sinuso it so we have a sign waveform coming in shown here in black and and we would like to see what happens at the output of the half wve rectifier well we said that if the input is small near zero then the diode is off and nothing happens this output is zero so the output is zero when the input is large enough one diode drop VD on then the diode turns on and the output is pinned to the input meaning that the output tracks the input so if the input is going up the output also goes up the difference being one diode drop VD on and then this continues until the input comes back down enough to turn off the diode and that again happens when the input is equal to one diode drop so at that point the diode turns off the current is zero now if the input comes down further and then goes and becomes negative the diode remains off there's no current the output remain remains at zero so the output remains at zero and as this input waveform repeats so does this output waveform and that's the waveform that we get okay this is what we call a half wave rectifier rectifier because it passes for example positive cycles of the input and halfwave because it only passes these half Cycles it doesn't pass the other half cycle now that may seem redundant it seems that every re has to be halfway rectifier but we will see shortly that that's not exactly true okay so this is the type of waveform that a halfway rectifier produces and now we would like to analyze this a little further and understand what it does then try to add a capacitor at the output we will see why and that has some interesting implications and once we are done with that we're comfortable with analyzing that circuit we go to a a more complete and interesting circuit called the full W rectifier and perform all those analyses again for this one so let's go back to the halfwave rectifier and uh this time uh so what I would like to remind you was that if the input is a a sine wave like this so that is VN and our circuit looks like this with an input voltage source and a resistor here and V out measured between these two two points what we saw was that the output waveform is rectified so for the output wave form we'll use a different color blue uh we saw that it's zero until V in is large enough of one diode drop to turn on the diode then it starts tracking the input with a diode level shift so the difference between these two is a constant value of vdr 700 800 molts and then it turns off again and stays off until the next cycle all right so in many applications this output is not particularly useful so imagine the charger application we said we have a charger so let's say this was a laptop charger that we had before and we said uh one side goes to the 110 volts or 20 220 volts from the wall and the other side gives us a DC that drives the laptop computer so we said first we have to reduced that very large voltage of 110 or 220 to something manageable which we use the Transformer four then we said somehow we have to go from this type of waveform to this type of waveform a rectified waveform so we have got there we have a rectified waveform but that rectified waveform can still not be applied to the laptop because it has huge fluctuations what we are interested in is the constant value so some constant value here but this keeps going up and down it goes up comes to zero goes up comes to zero so we cannot give that to the laptop or any other electronic device as the power supply so so we have to do something with that now when we studied the internal circuitry of this charger we said that we will use a filter after this point after this point to remove all these high frequencies and just extract a DC what a low pass filter does is that right a low pass filter rolls off at high frequencies so if you look at the Foria series of this waveform all those frequencies can be suppressed by a low pass filter which we would place here and we just keep the DC value well uh what does the low pass filter look like in the simplest case it consists of a resistor and a capacitor A first order low pass filter so maybe we should do that maybe we should put a resistor here and a capacitor between these two points to filter out these unwanted fluctuation well yes but what if we just do this what if we replace this resistor with a capacitor just that it's interesting let's try and see what happens the circuit doesn't bite we can always try it with that and if it doesn't work sure we'll go back to our original idea of using a complete low pass filter after this all right so uh we will construct a new circuit which we called the halfway rectifier with a smoothing capacitor so that's what it looks like we have our input a sinoid then we have our diode D1 and instead of the resistor we use the capacitor so here's a capacitor C1 and this is the output vol voltage that the circuit generates and what we would like to do is analyze the circuit and understand what this output looks like is it uh constant does it fluctuate Etc so the best test in this case again is a Time response uh so we assume V in is a sinusoid like before and try to see what V out comes out as okay so let me uh use red still for the input waveform so we have our input wave for here that's VN and then I would like to plot the output wave form as a function of time so I will go to Blue for the output wave form and start like this okay well uh remember last time when I talked about capacitors just as a quick review I said to charge a capacitor we need to push current into it right we need to put positive charge on the top plate to charge this capacitor and to discharge the capacitor we need to pull charge out of the plate this way so if the current keeps going out of the capacitor the voltage on the capacitor collapses if the current goes into the capacitor the voltage across the capacitor Rises so we just have to remember that now let's suppose that VN is zero right here then assume zero initial condition across the capacitor so zero here zero here the diode is off so nothing is happening V out is zero and again we are using the constant volt Vol AG model provided that the diode is on the input begins to go up uh the output doesn't change because the diode has not turned on yet so this side is zero this side keeps going up when this side reaches one diode drop VD on D1 turns on so somewhere around here when we are at VD on just like before uh the diode turns on when the diode turns on we have have a battery instead of D1 so when D1 is on if VN is greater than VD on then D1 turns on and the circuit reduces to this we have the input sinusoid VN then we have a battery for the diode so I from now on I just draw a battery when the diode is on I will not draw that switch that was a short circuit so we have VD on here and then we have the output so there's a capacitor here and there's a voltage that we are measuring at the output how much is V out well simple kbl this voltage minus this voltage is this voltage regardless of what we have here as long as this is an ideal voltage source and this is an ideal voltage source they override anything that place here except for another ideal voltage source so regardless of how large or small this capacitor is uh the voltage that we measure here is exactly equal to this voltage minus this voltage that's just kvl so V out is equal to V in minus V D on so again that tracking property that I mentioned earlier holds here as well okay so when the DI turns on uh the output begins to track the input with a level shift so we are here and we begin to track so we track all the way to here so far so good nothing exciting nothing that different from the original full halfway rectifier that did not have a capacitor all right so we reach this level some amount for example if the peak value of VN is v0 then the peak value that we measure here is uh let me be careful here because I need that space so uh the voltage that I get here will be v0 minus VD on so this voltage I will draw it over here sorry that I'm way out in the neverland so v0 minus v d on that's this voltage the peak voltage that I measure here now we reach V 0 at the input and now the input start going down in the previous case the output also starts going down the output still tracks the input butth how about here does it happen all right well so here's the situation we have this diode on a capacitor this voltage has reach the peak and now it's coming down the question is what happens to this voltage well there are only three possibilities this voltage can go up or it can go down or it can stay constant so if we rule out two of them the third one has to be valid all right well can this voltage go up after we go this way no because for a voltage on a capacitor to increase we have to push charge into it where where should that charge come from well that charge can only come this way through the diode and that happens only if this voltage is higher than this voltage but that cannot be true because this voltage keeps going down so we do not have enough voltage up here to be larger than this to push current through the capacitor so this voltage cannot go up can it go down well for the voltage across the capacitor to to go down what should happen is that charge must come out of the capacitor if the charge comes out of the capacitor where does it go it doesn't it cannot go that way that's open so it has to go this way but we know that the diode can conduct current only from its anode to its cathode that was our third principle so we cannot have a current going this way so only the third option is valid namely the voltage on the capacitor has to be constant after this point meaning that the diode turns off after this point so this is what will happen so here D1 is off and what we have is just an open circuit and the capacitor but the capacitor holds a voltage on it and the voltage on the capacitor happens to be V 0 minus VD on so V 0 minus VD on so this is where students often have difficulty why did the diode turn off well the best way to think about it is that when we got this to this point uh we still have VD on across the diode because it's on uh this side is sitting at this value this value and this side now wants to come down because the capacitor voltage cannot go up and it cannot go down it it stays constant and if the voltage stays constant and this voltage goes down the voltage that we have across V1 is now less than VD on and if it's less than VD on the Di has to turn off so from here on the diode turns off again assuming an a constant voltage model okay so that's good this voltage remains constant forever so it doesn't matter what happens afterwards right uh we could have negative Cycles positive Cycles this voltage always Remains the Same so that's actually very good that's exactly what we were looking for we wanted to obtain a DC a constant value that we can feed to our electronic devices uh and the value of that DC happens to be v0 the peak of the input sinusoid minus one diode drop so that's pretty good that's something we can use we don't even need to have to have to have this resistor and we don't need a complete uh low pass filter here all we need is this smoothing capacitor attached to the output of the circuit all right this is called a half wave rectifier with a smoothing capacitor because it uh carried current for this part and then stopped and from the on nothing happened all right this is the ideal world and looks pretty good but there are a few interesting issues that we have to analyze so uh what I would like to uh look at uh one quick note here is the following so let me change the color of my pen to this color all right right so reverse voltage across D1 so this D1 turns off at one point at some point in the operation so when it's off it is in possibly in Reverse bias if it's in Reverse bias it has a voltage on it remember that we can have a voltage in Reverse bias the diode is not conducting really but if the reverse bias becomes very large the diode breaks it breaks down and it begins to conduct a very large current so obviously for these things to happen to work properly you want to avoid breaking down the diode at any point in the operation so that means that we have to find the maximum reverse bias voltage across the diode and make sure that the diode that we select that we buy has has a reverse breakdown greater than what we experience here all right so how much uh reverse bias do we experience in this diode in the worst case well while the DI was turned on here so that's not a problem it turned off here so at this point we have positive more positive on this side than on this side right this much the difference but it's too little so diode is off diod is off the diod is off now the difference between the input and output is zero these two have the same voltage at this point in time okay no problem now the input voltage keeps going down whereas the output is constant this keeps going down and this is constant so now we begin to create a reverse bias across the diode and this keeps going on going on going on up to here at this point the input is very negative the output is fixed and that is the voltage drop that we have across D1 that's the reverse bias that could potentially break down the diode so this is the maximum break uh maximum reverse voltage across the diode how much is that well this value we know this value is on the capacitor V 0 minus VD on this value is minus V 0 you have a sign is so it plus v 0 here minus V 0 here so that's minus V 0 the difference between this value and this value is 2 V 0 minus VD on so that's 2 v0 minus VD on so if you buy a diode its reverse breakdown voltage must be greater than this for example if V 0 is 20 volts this is about 40 volts a little less than 40 volts so the breakdown voltage of the diode should be 50 volts 60 volts something that range to make sure it that not it does not break down okay so that was one note on that and now we are ready to move on to some other interesting points okay I would like to raise two points before we leave the halfway rectifier the first question is why did we use R1 in the original circuit so role of R1 in the original circuit this circuit here and this question is a question that you need to ask yourself in every circuit that you see for every circuit there are a bunch of devices you have to ask what is the role of this device what is the role of this device what happens if device is not there what happens if that device is not there if you ask these questions and you try to answer them you will understand the circuit much better and it's the same thing here uh we have two only two devices can I omit D1 and still have a rectifier no we saw a long time ago that diodes are necessary for this type of nonlinear operation okay so D1 is necessary but how about R1 what is R1 doing here well we can construct a circuit without R1 and see what happens if it still works great we don't need R1 we throw it away but if it doesn't then we have to put R1 back in all right so without R1 this is what the circuit looks like we have a diode and then nothing here and then this is V out and that's V in all right is there a problem here well yes there is a fundamental problem here let's suppose that for part of the operation D1 is off that's necessary for rectification right okay so if D1 is off what defines this voltage difference so if D1 is off this is what we have we have an open circuit and then we have V out it is incorrect to say that V out is zero a voltage can be zero only if we have zero current through a finite impedance not infinite impedance so what do we have here we don't know this voltage is not defined when the switch turns off we don't know what we have here so the purpose of this resistor is to make sure that when the 1 is off this voltage is in fact zero because a zero current times a zero resistor a finite resistor gives us a zero voltage so that's why that resistor has to be there now in other words there has to be something here it can be a resistor it can be a capacitor or any other component but we cannot leave it open because if we leave it open we don't know what voltage we have here when the switch when the diode has turned off so that's the fundamental reason for this okay the second point is the following for this circuit we drew the input output characteristic if you remember so we saw uh on on this uh board that the input output characteristic looks like this zero up to VD on and then takes off with the slope of one this is for one diode and one resistor the question is for the other version where I have one diode and one capacitor can I do the same thing so can we plot uh input output characteristics for the rectifier with capacitors it would be interesting to try that right uh just the way we did it for this guy we can try to do it for this guy as well well the answer to that question is uh in general no it is very difficult or not very interesting or not very useful or all of the above why well before we get tangled up by this complex circuit we can ask a simpler question uh can we plot for example something for just a capacitor can I plot I as a function of v for a capacitor for a resistor we can do it it's just a straight line for a diode we can do it it's like this or like that but for a capacitor can we do it well the problem is that here I is equal to C DV over DT so it is true that the current is a function of the voltage but it's also a function of time time in particular is a function of the derivative of the voltage with respect to time so how do we reflect this T in our in our characteristics that's why things become much more complicated and often times not very useful so because this circuit has a storage element a capacitor in it its response its input output characteristic will be time dependent will depend on the shape of the input waveform it won't be as simple as this characteristic where we just slowly change the input voltage and we observe the output voltage right it's much more complicated than this it depends on how fast the input is changing so for circuits that have a storage element in them we generally do not try to create an input output characteristic or IV characteristic in some cases it is possible in some rare cases it's also useful but for most cases it's not so we will not do that in other words for this rectifier the primary characteristic of interest is actually the time response as we have studied here very well we are now ready to go to the full wave rectifier so uh we will uh uh okay actually sorry no not yet there's one more uh idea I have to talk about before we go there let me change the color of my pen all right so in the next step we are still on the halfway rectifier in the next step we ask ourselves what happens if this rectifier drives some sort of load like you connect this adapter to your laptop or to your cell phone so then what happens is this waveform just like before or will there be any changes to it will the changes be desirable or undesirable that's why we have to analyze it all right so the next topic that we will study is half wave rectifier with cap and load and by load we mean something other than the capacitor so here's the situation I have built this nice little rectifier to generate a DC so let's say 3 volts or 20 volts and I'm happy with it I saw that it goes up and stays constant so that's great and then I come along and I connect it to something let's say I connect it to a laptop so we have a sinusid all input which is coming from the uh Outlet in the wall 110 volts or 220 volts it goes through a Transformer is reduced then it comes to the Circuit here so that's our sinusoid here and uh now previously we predicted that if we have only a capacitor and nothing else the output will be constant but now it's not exactly that we have a capacitor but then we have something else what does the laptop do in this case well the laptop needs to draw some current to operate right so let's say this is 20 volts and then there's some current here it could be several amps of current okay so then what happens we have to study that carefully uh now to study that to analyze the circuit from our basic circuit Theory knowledge we need to have some sort of model for this laptop is it a voltage source a current Source a resistor a capacitor inductor what what is it so what we will do is we'll approximate that by a resistor uh it even though it may not be a very good approximation it's still okay for our purposes so we will do this we will say here's our input going to a diode then the capacitor C1 then we model the laptop by a resistor R1 and V out is what we are measuring here here here here so that's all V out and this is D1 and this is v in okay that is the circuit that we need to analyze is the circuit different from what we saw before oh maybe what we know are two extremes if R1 is infinite R1 is not there this is what we have if C1 is zero C1 is not there but R1 is there this is what we have now it's somewhere in between we have C1 not zero we have R1 not Infinity so we have a combination so we have to see what happens all right so uh this is an interesting circuit it's not something we have seen before it takes some patience to analyze it very carefully so let's go ahead and do that well uh as usual we draw the input waveform and then try to look at the circuit in every part of the waveform and see what's going on uh let me draw the input wave form uh in Black for now so here's the input wave form V in and now we begin to draw the output wave form so for that I will choose a different color uh maybe this Orange I don't know if Orange shows hopefully it shows okay so for output we know that if the input is zero the output is zero there's nothing happening in the circuit as before for this case or for this case so the output is zero we know that the output does not change until the input is enough to turn on the diode so when the input has reached one diode the output turns on when the diode turns on the output is pinned to the input just like this it doesn't matter whether we have a resistor here or not we know that this output is exactly equal to the input minus VD on no question so the input tracks the output sorry the output tracks the input so the output goes and tracks the input all the way to here so the Orange waveform is the output waveform I will write the out here all right now the question is what happens after that Peak well when we did not have the resistor the voltage State constant now that we do have the resistor what happens okay well let's say that D1 still turns off past this point and that's a good assumption in most cases as we will see so we will make that assumption for now hopefully at some point we can check it to make sure the assumption is correct but let's me an assumption so the assumption is that D1 turns off after the peak so after this peak the DI turns off so if the diet turns off what do we have left the DI is not here we have one capacitor one resistor the voltage across the capacitor is still as before 2 V 0 minus VD on like here so this circuit can be simplified to something we are very familiar with the circuit can be simplified to one capacitor having a voltage on it equal to v0 minus VD on this voltage here uh this voltage here uh sorry this voltage V 0 minus VD on the peak value minus one diod drop and this capacitor is connected to a resistor so as you remember from basic circuit theory if we have a capacitor that is charged and a resistor is connected to it the capacitor voltage which is the same as the resistor voltage will Decay exponentially so what we would expect this voltage to do is V out I will draw it here for now and then I'll bring it back to main uh plane of waveforms as a function of time V out goes like this so that's V out of that little circuit here this is what we know from basic circuits okay so the same thing should happen here the Di has turned off we have a capacitor with that much voltage we have a resistor so from here on it's not a sinusoid anymore it's a decaying exponential so it exponentially decays I deliberately made this Decay very slow we will see why so that it just grow gradually is losing charge the capacitor through the resistor and gradually coming down okay now what is the input doing in the meantime well the input keeps going down going negative the diod still off that was our assumption the diode is off the diode is off the diode is off now let's look at this point at this point in time the output voltage has decayed the input voltage has come back up and they happen to be equal this voltage happens to be equal to this voltage does the diode turn on not quite if we use the constant voltage model we need this much voltage for the DI to turn on so not yet now we have to let the input go up uh the output will Decay until the difference between these two in this direction is sufficient to turn on the diode so when this is sufficiently POS more positive than this one one diode drop the diode turns on so diode turns on right here so D1 turns on again so maybe I should write here as well D1 turns off well I sorry I guess I wrote it before so I don't need that but anyway D1 turns off okay D1 turns on again now if D1 turns on again what does the Circuit look like here's what we have we have a volage source VD on a battery and v in a sinusoid that's going up and the capacitor again the voltage on the capacitor is about uh uh input minus VD on so at that point in time we have a situation where they inputed some amount we have VD on here and the voltage on the capacitor happens to be uh this voltage minus VD on that's when the diod turns on so now what happens well again the output is pinned to the input so the output continues to track the input we trct the input up to what point up to the peak we reach the Peak at the peak we still have V in minus VD on across the capacitor and then uh once we go past the peak the D turns off so we have another Decay afterwards in other words this waveform from here sorry this way from from here to here and from here to here repeats we Decay then we charge the capacitor again then we Decay then we charge the capacitor again and so on that is the output wave form of the halfway rectifier in the presence of a load capacitor and a load resistor all right uh the waveform that is generated here instead of the constant DC that we produced before is called a ripple so this is called Ripple we say we have Ripple on the waveform Ripple is like what you see on the ocean the waves small waves so we have a ripple here the question is is the Ripple a good thing or a bad thing if it's a good thing we want to make it larger if it's a bad think want to make it smaller well uh what this means is that what we thought was a nice constant voltage given given to our electronic device is actually not constant it fluctuates so that fluctuation is undesirable it's some cases very undesirable I'll give you a quick example let's suppose that you have some sort of audio device like an MP3 player or radio or something so imagine that you have a charger that is charging a radio okay so uh you're listening to the radio and at the same time you have pluged this into the wall and this is the DC presumably the DC this DC but when it this connects to the device the audio device because the audio device draws a current we have Ripple on this voltage so then what happens you will hear a humming noise in the background of the music that you're listening to and that humming noise is because the supply voltage that's coming to the device is not constant it keeps going up and down at a frequency of for example 60 hers in us or 50 hers in other countries so that is a problem the Ripple must be minimized well now the question we have to ask ourselves is how much is this amplitude of ripple from this peak to this value how much have we lost we have to calculate that once we calculate it we see how it depends on the various parameters of the circuit and from there we can decide how we should reduce the amount of Ripple how we should supress this effect all right so how do we calculate the Ripple this looks a little complicated because we have a nonlinear circuit we have part of a sinusoid then an exponential decay and so forth so uh doesn't seem to be straightforward uh but we can make some approximations the approximations are as follows we know that this circuit is presumably welld designed which means his Ripple has to be small okay if the Ripple is small it means that this Decay is very slow we start from here it decays very very slowly and touches here and then again so for the Riel to be small the Decay has to be very slow so that we don't lose much charge from this capacitor when the diode turns off all right so if the Ripple is slow and small then we can draw two conclusions conclusion number one if the Ripple is very slow low and small what happens is that the diode turns off here and it turns on around here it turns on very briefly and that makes sense because if we didn't take much charge out of this capacitor by the laptop we don't need to provide much charge to it in the next cycle so the diode turns on only briefly here if the Ripple is small and slow so what I can say is that this Decay lasts about one period of the input it starts from here and lasts about not completely one period but close to one period so I will write that down here that's a very important assumption that we're going to make so if Ripple is small then Decay lasts about t in and T in is the period of the input so for example it's easier if I show it here so TN is from Peak to peak one peak one positive Peak to the next positive Peak so here's T in or from one zero Crossing to another zero Crossing from here to here or from here to here Etc that's T in so we can say that from here to from here when the DAT turns off onto the di turns on again is approximately TN this is only true if the Ripple is small and the Ripple has to be small for this design to be a good design if the Ripple is large then we have lots of humming noise in the audio device example I showed you I mentioned and then it's not a good design okay so if that's the case uh we can assume that this Decay lasts about t in seconds if the frequency of this voltage is uh 50 htz in other countries then T in is 20 milliseconds if it's 60 HZ like in us then T is about 16.7 NCS 16.7 milliseconds right so those are the numbers that we can keep in mind all right so we made one assumption so far uh that Ripple is small hence the Decay is slow and therefore the time for from here to here is approximately equal to TN all right now to calculate how much the voltage has fallen from here to here I need to go back to basic circuits and analyze this type of circuit again so let's go ahead and do that so let me add a page okay here's the circuit that we analyzed assuming the diode has turned off we have a resistor and the capacitor and we have R1 and C1 there's a voltage uh some voltage here let's call it V uh one the initial voltage and uh I would like to analyze the circuit now remember from basic circuits that the response looks like this uh okay so let me call this V out uh but I will write initial voltage on C1 is equal to V1 so V out as a function of time for this very simple circuit is like this it starts at V1 goes to zero and we know that this equation is V1 x p minus t over to okay so that's what we know when a capacitor discharges through a resistor now we are making assumption that the discharge the Decay is very slow does that allow us to make an approximation for this wave for yes and this is a very good approximation to remember for your career often times is very helpful okay so let's suppose that we are observing this waveform for a Time duration much less than to so for example to is 1 second as an extreme case but we are observing this wave form for only 1 millisecond so the observation window the observation time this T so from zero to some T is much less than the time constant in other words we not interested in what happens later we are only interested in the first let's say Mond while to is 10 milliseconds 100 milliseconds 1 second very large okay so if T the observation window from here to here is much less than to I can approximate this exponential you remember exponential of a very small number is one plus that small number so we can say that if T is much less than to then V1 X of minus t/ to is approximately equal to V1 1 - t/ to and this is not that surprising uh the idea here is that if we are interested only a very small part of this waveform we can approximate it by straight line This is a straight line so we can approximate by a straight line right that's all all this says okay good so in the halfway rectifier with a capacitor and a resistor connected to out to the output if the Decay is low we can approximate the Decay by this waveform so let's go back there for a second and see what we have seen so far so what I'm saying is that this Decay from here to here which is in this simple circuit a capacitor discharging through a resistor because the diode is off can be approximated by a straight line a straight line going down and that line is given by the initial voltage whatever we have here which we know the voltage we have here is V 0 minus VD on Peak minus one diode one diode drop multiplied by 1 minus t over to assuming that this is a very slow Decay the the observation Time For Us is from here to here a approximately T in seconds the time constant of this discharge is very large much greater than our observation window so we can approximate this charact this line This Way From Here by a straight Decay and by a linear Decay so I will just remember that this voltage the initial voltage is equal to V 0 minus VD out VD on so let's go back here here and uh right uh for the half wave rectifier we have V out is equal to V 0 minus VD on multiplied by 1 - t over tow this is the Decay Behavior during the time the diode is off okay so we approximated that exponential decay by linear Decay now what we are really interested in is the amount of Decay at the end of that discharge so we go back here we are interested in this voltage here if we decay deced and decayed and decayed until the DI turned on so at this point we past this point we don't Decay anymore so how much is the voltage here if we know the voltage here so this voltage is V 0 minus VD ion we are interested in this voltage now we know the time from here to here we said that's approximately equal to TN he said if the Ripple is small then we can say the Decay lasts about t in seconds so we go back here and we replace t with t in so we say V out at tal T in is equal to V 0 minus VD on 1 - t in over tow and this is an approximation all right so this is what we have we had this Decay remember we charged up and then we Decay and what we have found is the equation for this Decay from here which I'm assuming zero to here which I'm assuming approximately TN as a function of time okay and that is the equation that we had V out all right so what I'm really interested in is this height how much we had at the beginning and how much we have after T seconds so we found the voltage at this point in time that's this we have the voltage at this time which is just v 0us v on if we subtract these two voltages we find the amplitude of the Ripple so we say Ripple amplitude is equal to V out at time equal Z minus V out at time equals T in so here we have V 0us VD on here we have all of this so difference between these two is just that so that's equal to V 0 minus VD on time T in over tow this is the Ripple amplitude Okay uh T is the time constant of the circuit equal to R1 C1 TN is the period of the input sometimes we prefer to express it as 1/ F in F in being the frequency of the input so sometimes we write this as V 0 minus VD on divided by multiplied sorry divided by R1 C1 FN FN is the input frequency 60 HZ in US 50 HZ in other countries for the line voltage so this is a simple compact equation for the amplitud of Ripple that we get here we can see that if C1 is uh smaller and smaller the Ripple becomes larger and larger why well it makes sense because if the capacitor is small the decays really fast faster and faster so the Ripple becomes larger we are here we Decay down here of course so long as the approximation of slow Decay holds this equation also holds uh this equation also says that if R1 is small the Ripple is large why well if R1 is small again the Decay is faster and that means that if you connect a heavier load by heavy May mean a load that draws a lot of current then the repel will be larger so if the charger like this charger DW drives a uh let's say a simple Bluetooth device which is it doesn't draw much current then Ripple might be only 10 molts but if the same charger has to drive something with a lot more current let's say laptop or some other device then the report will be larger and that's what this equation tells us okay so this a nice little equation uh that is very useful in trying to decide the values that we need here here this is usually given but at least these two values for a certain amount of ripple at the output of the halfwave rectifier all right well uh looks like we did not get a chance to look at the fullway rectifier uh and at this point our time is up so I will uh see you next time when we talk about the fullway rectifier and we repeat all of these analyses for it I will see you next time [Music] f