[Music] welcome to the course on design of power electronic converters today's lecture is going to be on bipolar pwm last class we had discussed edge bridge converters and we saw the different possible equivalent circuits and uh the switching combinations and there i had mentioned you about pulse width modulations that there are different types of pulse width modulations for h bridge converters and depending on your application you may choose the suitable one the two most popular ones are bipolar pwm and unipolar pwm the first one bipolar pwm is what we are going to see in this lecture so in bipolar pwm what happens is that it uses only the two switching combinations that means these two devices ta plus and tb minus the diagonal switch these two are operated together they are always turned on and turned off together and the other two diagonal devices ta minus and tb plus these are also operated together and always turned on and off together so you can consider it like a like a pair so that's like this t a plus n t b minus these always operated together and t a minus n t b plus these two devices are also always operated together so only use of diagonal switching combinations further you may recall that what will be the output voltages when your voltage is a has to be vo or vdc rather so vo whenever we want it to be equal to vdc at that time the switch da plus and db minus r on and whenever we want vo to be equal to minus v dc at that time m t a minus n t b plus i turned on and we had also seen before that what will be the corresponding dc bus current idc is going to be plus i o whenever t a plus t b minus r on and it is going to be minus i o whenever t a minus in tb plus are on ok and depending on the direction of the current either the switch will be conducting that means the mosfet will be conducting or the diode will be conducting so these things we have seen before so now uh let us look into the waveforms so this is the career waveform which is shown here which is triangular carrier is chosen and this has the the switching frequency fs okay so the time period one time period that you see here this is equal to ts the switching time period then this is the reference this one is your vo ref the output voltage reference waveform with which we do the comparison and then on comparison what we observe here is that the logic that is followed is whenever your ref is greater than v carrier at that time ta plus and tb minus are turned on and whenever the opposite is happening that means uh v carrier is greater than v or f then the other two diagonal devices tb plus nta minus are turned on okay so we can do these comparisons like this at every point and then accordingly here the output voltage wherever ta plus tv minus are on at that place the output voltage is going to be equal to plus vdc otherwise the output voltage will be equal to minus vdc so the output voltage then switches between plus vdc and minus vdc continuously and what will be the leg voltages so leg voltage will be what whenever the upper switch is on at that time it is vdc otherwise it is zero so the leg a voltage v a n here t a plus is on so it is equal to plus vdc and otherwise it this is ta minus is on so it will be equal to 0 and and so on similarly for the leg voltage vbn also we see the same thing here tb plus is on upper switch of leg b is on so this is vdc and here tb minus is on and so the leg voltage is equal to 0 and this continues okay then the load current now load current for the waveforms that is shown here here it is assumed that the load current is all throughout positive okay it's not going negative now it depends on the load don't think that that is the case always it may become positive negative then depends on the load and also the on the output voltage so ah then what will happen whenever the voltage applied is plus vdc at the output at that time the load current will increase okay and then because most of the time the load current load is going to be inductive in nature so when you apply a positive voltage across it the current will increase and then when we apply a zero voltage or a negative voltage then the current will reduce and so that is what is happening over here when this voltage is output voltage is minus vdc and this is continuing now what about your idc the dc bus current so dc mass current here we see that in this part your which are the devices conducting ta plus and tb minus okay at that time what will happen your current will be equal to idc idc will be equal to io and whenever the other two diagonal switches are on idc will be equal to minus of io and that is what we observe here that whenever t a b plus and t a minus are on so this part is opposite of this this part whenever over here your this is when t b plus and t a minus are on okay then your this device voltages now blocking voltage will be vdc vt a plus tb minus whenever they are not conducting this will be equal to vdc else whenever they are conducting for ideal case it will be 0 and similarly you can draw the other waveforms also vta minus and v t b plus okay whenever they are conducting zero not conducting blocking voltages vdc and the current that is carried by the devices now here we what we have assumed the current direction is always positive so when the current direction is all throughout positive what will happen whenever these two switches are turned on the first two diagonal switches it is ta plus and tb minus which are going to conduct and whenever the opposite is happening so at that time the current direction is like this so it flows through the diodes comes through here and it flows through the diode here as well okay so d a minus in db plus the two diodes are going to conduct and that is what we see here ita plus and itb minus carries the current io whenever they are on and else whenever these other two diagonal devices are on it is the two diodes which are carrying the current now this is uh the waveforms that is that shown here they are assuming the vo ref is a dc reference now this h bridge is also used for your single phase dc to ac conversion so there the reference can be sinusoidal so how the waveforms are going to be then so this is your vo ref which is sinusoidal here you can see there and this reference this is a triangular carrier which let's say goes from 0 to 1 so obviously this also can have maximum value as equal to 1 and not more than that if it becomes more than that so then it will be greater than the carrier waveform and then the pulses will start getting missed out okay and then this is the vo average so what this pulse says that we are seeing is what we get by doing the comparison so when we do these comparisons here so uh whenever this can is greater than the carrier your your output is going to be here equal to plus vdc else this is going to be equal to minus vdc so this is what is um your 0 over here so this is your 0 it goes from plus vdc to minus vdc and this waveform is actually the vo waveform the actual output voltage which you will see on the oscilloscope the instantaneous voltage waveform and if we take the average of that waveform if we do the average of it so that is this vo bar for every ts time period every switching time period if you take the average of v o then what you will be getting is this v o bar which is this sinusoid which you are getting and which has to resemble the reference waveform to begin with this average is also you can say is the fundamental component of this switched voltage waveform now next let us see how to obtain the reference waveform how do you obtain the equation for it so for that let us consider only one switching time period ts so let's say this is your vo waveform for one switching time period ts and this part is when um your the first two diagonal switches are on so we call it as dts and this other is your one minus dts and this waveform will have an average and that is that vo bar that we are talking about and this varies from plus v dc to minus vdc so we can write this vo bar is equal to vdc into dts plus minus vdc into 1 minus dts by ts so then further what you get from this is vdc into 2d minus 1 is what your vo bar is equal to from which you can write d as equal to vo bar by 2vdc plus half so using this equation you can obtain the duty ratio now if vo bar is something varying is what our target voltage target output voltage is a sinusoidal voltage let's say this is vo hat sine of omega 1 t now i am using this omega 1 to denote that this is the fundamental so in that case your duty ratios itself that modulating waveform is your v or ref what we can call it as this is equal to your vo hat sine of omega 1 t by 2 vdc plus half so this is the modulating waveform equation for the modulating waveform which you can use to compare with the triangular carrier and generate the gate pulses for the bipolar pwm there is another term corresponding to a pulse width modulation which is used in power electronic converters which is called as the modulation index now this modulation index it is denoted as m a many times is defined as vo hat by vdc now i am defining it as vo head the peak of the output voltage by vdc peak of the output voltage means the averaged waveform vo bar of that's the peak of that sometimes in books people also write the rms value of it so know that your value of modulation index will change accordingly so you can use either definition but you just have to be aware which definition you are using because your modulation index value will change accordingly so now what happens is that if this is your triangular career waveform and then you have this reference now your this career varies from 0 to 1 let's say now the vo ref that if that we are using it can have a maximum value as equal to one otherwise it will start crossing these career peaks okay and so then what will happen is that the comparison will not happen it will some pulses will start to miss out like for example if the reference waveform becomes like this so here till here you will actually get no pulse so then that means we we are losing over the control the desired average waveform that we were supposed to get which we are targeting that is not going to come by the by the pwm method and it will have more harmonics into it lower order harmonics may start coming as well so ah this is mostly avoided ah but although there may be situations where this rule may be violated that we would like to keep this reference wave from below one so then what that is called as over modulation so that is also done but for our purpose we would not like to violate it let's say we want to keep it below this the reference should be below the courier waveforms maximum so accordingly there is a maximum value for this modulation index which this modulation index can take for a particular method of pwm so then if if we want to write that ah we want to find out that for bipolar pwm what that maximum modulation index is going to be so your vo ref that means can be maximum equal to 1 and so your vo head by 2 vdc plus half can be maximum equal to 1 according to the equation that we found out just now so if you solve it what you get is that vo hat can be equal to vdc so that means your maximum modulation index can be equal to 1 in case of bipolar pwm so what is the use of these maximum modulation index this gives us an indication of what is the minimum vdc that you require for your converter usually when you are designing a converter then you know what is your output what is the targeted output then if you have the targeted output then you have to find out how much is the dc bus voltage that needs to be created so using the maximum modulation index for the particular pwm we are using you will be knowing that this is the minimum dc bus voltage that is required okay so maximum modulation index indicates minimum vdc needed to obtain vo hat now let us find out an expression for capacitor current this capacitor current also plays a very important role in the choice of capacitor and here to do this we will ignore the switching frequency component we will use this idc bar because your dc bus current is idc which is the actual current but we will be ignoring the switching frequency components for now and that we will be denoting as idc bar so everything taking as the average expression so v o bar is v o hat sine of omega 1 t and i o bar this is also the average of the output current the load current over one switching time period so i o hat sine of omega 1 t minus let's say phi 1 there is a lag in it in which is phi 1. so if we assume everything to be ideal and lossless we can write vdc into idc bar as equal to vo bar into io bar that's the dc power is equal to ac output power so then you substitute it vo hat sine of omega 1 t into io hat sine of omega 1 t minus of phi 1 if you solve it you have to do little trigonometry here if you solve that what you will be getting is this vo hat io head by 2vdc cos of phi 1 minus vo hat io hat by 2 vdc cos of 2 omega 1 t minus of phi 1 now what you observe here is that this component is not varying it is a dc value so this is your dc component idc of the capacitor current now usually what we will expect is that it's a dc bus current means it should be a flat dc but we have seen that it's a flat dc is not what flows through the dc bus it's a ripple it has got a switching frequency ripple which we have ignored so ah then definitely it should be only the dc component which should be there so that of course we are seeing but what we see is that there is something else to it and this is varying with a frequency twice the fundamental so this is the second harmonic of omega one okay so second harmonic component is also present in case of h bridge converter when we use this bipolar pwm and that is a disadvantage because then the second harmonic is a much lower frequency harmonic let's say if you have a 50 hertz output then you and you in the capacitor current there is a 100 hertz component so you have to choose a capacitor large enough which can withstand that 100 hertz component of current in the dc bus and still maintain the dc bus voltage so let us look into the current waveforms so this is the output voltage waveform instantaneous output voltage waveform this is your vo which varies from plus v dc to minus vdc and this is the zero of it okay instantaneous output voltage waveform and this is the vo bar the average of it what we saw over switching time periods if we average the instantaneous output voltage waveform will be getting that vo bar then what happens further is that that if we draw the load current waveform then if it is a sinusoidal waveform the load current will be how it is going to appear since it's assuming an inductive load so whenever the voltage is positive so here the voltage is positive it's plus vdc what will happen the current will increase and then whenever the voltage is minus vdc the current will decrease and it this uh nature of the current increasing decreasing it continues and it follows kind of a sinusoidal nature okay so then your this current this is your actual i o current which you will be seeing on the oscilloscope and if you again take the average of it so that is the average current this is the average current the black one is the average current io bar which we just used for calculation just now i o bar and idc waveform so this is the idc waveform you can see here that whenever your this part let's say whenever the first two diagonal switches are on at that time this is equal to io so here this is equal to io so this part it is equal to plus io and this part you can see that it is equal to minus io here and so forth it continues like this so using this waveform you can find out this idc waveform also all right and what we see is that that it is it's a very switched current waveform so it has got several components in it one component is your this dc that will be there which is your idc and then it will have also the second harmonic component so that will be your idc 2 lettuce qualities and of course the switching frequency component will be there and finally this is your actual idc waveform that we see that you are going to see on the oscilloscope so the key points of this lecture are that that when you choose a modulation strategy you have to analyze that modulation strategy different modulation strategies are there you can pick one ah for your converter you can analyze different waveforms for that modulation strategy and you have to find out the modulating waveform because when you are going to design it you have do you have a particular output you need that output so you have to get the modulating waveform which can generate that particular output so those equations are required and then you also find out the maximum modulation index that will give you an indication of the minimum dc bus voltage that is required and you should also observe the capacitor current ripples or you can find it out if equations are available for that or you can simulate and you can observe the capacitor current waveforms then that will give you an idea that how big capacitance you need to maintain the dc bus voltage thank you [Music] you