Lec 7: Unipolar PWM

Welcome to the course on Design of Power Electronic Converters. So, we were discussing the module on Analysis of Power Electronic Converters.

Electronic Converters. So, we were discussing the module on Analysis of Power Electronic Converters.

And today's lecture, we will discuss Unipolar PWM. Previously, we had discussed Bipolar PWM for H bridge converters. So, there is another PWM method for it popular PWM method there are many other PWM methods as well, the other popular method, the another one is Unipolar PWM.

So, in this one what is done is that these legs this leg A and leg B they are treated individually. In Bipolar PWM what we had seen was these two diagonal switches TA plus and TB minus

individually. In Bipolar PWM what we had seen was these two diagonal switches TA plus and TB minus and TA minus and TB plus they were always switched on and off together. But, in case of Unipolar PWM

we look into individual legs rather than looking into the whole converter together.

So, accordingly each leg A and B has got a reference waveform.

So, different reference waveforms for modulation are given.

And since, we are not using specifically your this combination of TA plus, TB minus and TA minus TB plus instead, we are looking

into individual legs. So, therefore, all four switching combinations arise in Unipolar PWM, all four switching combinations they are used.

Now, for example, let us say this upper switch is ON for leg A.

So, this one is On and for leg B also the way the reference waveform is for leg B the upper switch of leg B gets On. So, in that case TA plus TB plus is this switching combination which is going to be

used. Then it may so happen that both the lower switches are getting on together depending on the

used. Then it may so happen that both the lower switches are getting on together depending on the leg A the reference waveform and its comparison with the carrier and leg B's reference waveform in comparison with the carrier. So, all of these your TA plus TB plus

being on together TA minus TB minus being on together and TA plus TB minus and TA minus and TB plus all these four switching combinations arise in the Unipolar PWM.

And depending on that what is the direction of the current that we have seen before it depends on the direction of current. So, let us say your direction of current was positive and the upper switch of leg A and the upper switch of leg B was On.

So, now, what we can see here is that, that the current has to flow like this.

Now this upper switch is On. So, it is going to then flow through the diode and come here and flow like this. This we have discussed before that depending on the direction of the current whether the IGBT is conducting or the diode is conducting it is dependent on that.

And what is the current over here idc current over here that also depends on the switch combination and not on the direction of the load current. So, now, let us look into the waveforms. So, here you can see this is the carrier waveform, this is the triangular carrier waveform

and this will have the switching frequency fs, high switching frequency and then this is the reference waveform for leg A,

this is for leg A and this one is for leg B. Now, when the comparison takes place,

So, whenever you can see here in this part the reference we end ref the reference of leg A is greater than the carrier and so, the upper switch TA plus is On.

And then for this much area your what you can see that it is TA minus the lower switch of leg A is On and so on this continues and what will be the leg voltage V AN? Whenever the upper switch is On the voltage is going to be equal to Vdc otherwise it will be equal to 0.

And for leg B. So, leg B this is the reference and then when you compare what will happen here you can see that this part your reference is greater than the carrier. So, upper switch TB plus is On and here reference is lower than carrier. So, TB minus lower switch is On and so on it

continues and whenever the upper switch is On, the voltage the leg voltage becomes equal to Vdc.

And then we know that your output voltage Vo can be written as the difference of the leg voltages V AN minus V BN and that is what is shown here as the output voltage what you see here is that that over here this is Vdc, V AN and this is 0. So, the difference is Vdc

and here both of them are 0. So, the output voltage is 0, then further here again this is VDC and this is 0. So, this is again Vdc and this continues like this.

Then we see the load current assuming it to be inductive load. So, what will happen whenever the voltage is output voltage is positive, the current will be increasing that is what we see here the current to be increasing and then here when the voltage is 0 output voltage there the current will decrease and so forth it will continue like this.

Then we see the idc current waveform there what will happen when in this part, in this part your what is on is your TA plus and TB minus. Here your TA plus

and TB minus these two are On and you know that for this switching combination for positive direction of current if your idc equal to load current. So, what we see

here this is equal to the load current. Then in this part over here in this interval your what is on is your the lower two switches the bottom two switches TA minus and TB minus these two are On together and at that time what will happen

your DC bus current will be 0 we have seen the equivalent circuits before, the four equivalent circuits and accordingly these waveforms are getting formed.

Then, we can look into the switch voltages. So, whenever the switch is on at that time the switch voltage is 0 otherwise whenever it is off it is blocking the dc bus voltage Vdc and that is what we see here for V TA plus and similarly, for V TB plus also the voltage

across switch TB plus the upper switch of leg B we see that whenever it is off, then it is going to block Vdc and whenever it is On assuming it to be ideal the voltage is 0.

Similarly, for your V TA minus and V TB minus the lower two switches voltages are also shown here you can obtain it like that. Now, coming to the switch currents, Now, depending on the switching combination and the direction of the current this gets decided that whether the

switch is conducting that means the transistor is conducting or whether the diode is conducting.

So, what we can see here now, here we have assumed the direction of current to be positive.

So, now, the two switching combinations that we saw is that TA plus and TB minus are becoming On and the direction of the current is positive. So, it flows like this or these two lower switches are getting turned On.

So, when these lower two switches are getting turned on at that time, what will happen is that, that your current direction is still positive it flows through here and then it has to go through the diode. So, TB minus and TA minus are supposed to conduct. So, TA plus, TB minus,

TA minus, TB minus these two combinations and here TA plus TB plus the upper two switching combinations that also we can look into. So, upper two switches it will be flowing like this here and then it flows through this diode. So, accordingly

if we see this whenever the upper two switches are On that is the switching combination here your this current iTA plus and iDB plus means through the diode of TB plus and through the

IGBT TA plus this will be equal to the load current i o otherwise it will be 0 and whenever the switching combination is TA plus TB minus here in this part.

So, then the current will be equal to load current through those devices iTA plus iTB minus.

And then whenever the lower two switching combinations are arising, lower two switches being on at that time your diode DA minus and TB minus are carrying the current. So, at that time they will have the load current. So, this is what is the current waveform

current. So, at that time they will have the load current. So, this is what is the current waveform through these two iDA minus and iTB minus. So, this is how you can obtain the current waveforms through the devices through the IGBTs and through the switches. So,

it mainly depends on which of the devices are conducting depending on the switching combination which is arising and the second the direction of the load current.

Now, in this case I have shown you where your references are not varying like the references appeared to be as DC references, but it may so, happen the references may be sinusoidal

and this is what is shown here that this is your reference one of the references. So, this is for

your reference for leg A, so, V AN ref and then this one is the reference for leg B, V BN ref and depending on these the comparisons takes place and whenever this is higher

here your reference is greater than the carrier waveform, then at that time the upper switch gets On and the leg voltage becomes equal to Vdc.

So, this varies from this leg voltage varies from 0 to Vdc. Similarly, for leg B also you

can do the comparison and this also varies from 0 to Vdc and the output voltage this output voltage gets formed by the subtraction of these two V AN and V BN and so you can subtract these two

waveforms and you will see this is what you are going to get, this blue waveform and this varies from here you can see this varies from 0 to Vdc for this much period it varies from

0 to Vdc and then for another period you can see that that it varies from 0 to minus Vdc.

Because that is how the difference over here is appearing here the V BN voltages have got greater width than the V AN with the V AN switching the voltage that is appearing and so,

accordingly it varies from 0 to minus Vdc. And then if you take the average of this waveform, the switched voltage waveform so, this is that you are going to get and that is your Vo bar average means here average over one switching cycle.

And similarly, your average for these leg voltages are also shown this is your V BN bar the average over one switching time period and continuously it is done like that. So, V AN bar that is the

average for V AN, the switched voltage waveform that you are going into obtain. And you can see that this V AN bar average it resembles the V AN ref and VB N bar that resembles the V BN ref

and the subtraction of these two V AN bar and VB N bar is what you will be getting as the Vo bar.

Now, let us discuss how do you obtain the modulating waveform for this Unipolar PWM. So, as we already saw that there are two reference waveforms one for leg A and leg B. So, this is given by your V AN ref which will be equal to your Vo bar means

basically the average of the output voltage that you want to get Vo bar 2 Vdc plus half. Again, to

remind you that this average that we are telling is not over the fundamental frequency of Vo bar but over the switching time period. So, Vo bar by 2 Vdc plus half

is what is the reference that you can use for V AN ref and for V BN ref the reference waveform or the modulating waveform that you can use is minus Vo bar by 2 Vdc plus half

and this let us say your Vo bar the average that you need or the fundamental that you need of the switched output voltage waveform is equal to Vo hat Sin of Omega 1 t, Omega 1 is for to the denote that this is the fundamental.

So, then this will become equal to Vo hat Sin of omega 1 t by 2 Vdc plus half

and this will become equal to minus Vo hat by 2 Vdc Sin of Omega 1 t plus half.

So, what you see is that these two reference waveforms are shifted by half means basically there is a DC shift and they are two sinusoids which are 180 degree out of phase.

Next, let us look into the modulation index for Unipolar PWM. We had defined the modulation index ma before in bipolar PWM also. So, it this we had defined as Vo hat by Vdc and what could

be the maximum value for Unipolar PWM if we want to find out that so V AN ref what could be the max value of it. So, that is going to be equal to 1 because it is a

sinusoid and its amplitude can be maximum equal to 1 as we are taking the carrier to be maximum equal to 1. So, this carrier,

this varies from 0 to 1. So, this modulating waveform can also be max equal to 1. So,

then this will be equal to half plus Vo bar by 2 Vdc based on what we just saw.

So, half plus Vo hat by 2 Vdc because Sin Omega 1 t can have a maximum value equal to 1 and so, from here what you get this Vo hat can be equal to

Vdc this is what is possible maximum. So, that means the maximum modulation index that will be equal to your Vo hat by Vdc equal to 1

and if you cross that, then you will be doing over modulation that also is done sometimes, but if you want full control, then we should not exceed the maximum modulation index.

And so, from this what we understand is that to obtain Vo hat we have to at least have a dc bus voltage which is equal to Vdc. So, let us say if your Vo hat is equal to 500 volt, then you have to take the Vdc equal to 500 volt your dc bus had to be at least

equal to 500 volts or it has to be greater than that. Now, let us discuss the Capacitor Current this is similar to what we had discussed for Bipolar PWM.

So, if we take the average over switching cycles, So, if this is Vo bar Vo hat Sin of Omega 1 t and io bar the average Current that you get over switching cycle sin omega 1 t minus of phi 1 and this is all ignoring the switching

frequency components. So, Vdc idc bar if we equate the DC power and AC power.

frequency components. So, Vdc idc bar if we equate the DC power and AC power.

So, Vo bar io bar and what is your idc bar that is your basically you have ignored switching frequency components.

So, when you substitute this and then you solve it what you will be getting is idc bar Vo hat a Vo hat by 2 Vdc cos of phi 1 minus Vo hat io hat by 2 Vdc cos of 2 omega 1 t minus of

phi 1. So, this is the idc the dc component and this one is the second harmonic of omega 1.

phi 1. So, this is the idc the dc component and this one is the second harmonic of omega 1.

So, what we see is that if we ignore the switching frequency components, then your, the DC component and the second harmonic component for Unipolar PWM also turns out to be the same. So, that means irrespective of whether you are using bipolar, PWM Unipolar PWM, the capacitor that you choose for DC bus,

it should be able to withstand the second harmonic component of your fundamental omega 1.

And what it means that it should be able to withstand it basically means that your capacitor should be big enough that in spite of the second harmonic it is able to maintain the DC bus voltage and not higher ripples or voltage ripples are not increasing.

So, let us look into the waveforms. So, this is your output voltage waveform that we just saw. And this is the average tau of that which is your Vo bar.

And then, this one this current which is ripple that you see is your load current and whenever the voltage is positive let us say hey this is positive this part you can see here that the current is actually rising it is increasing.

And let us say over here, you can see that that the current is decreasing because the output voltage has become 0. So that switching frequency ripple is there, but otherwise it is following a sinusoidal pattern. So, when you take the average of this

rippled waveform, this black one, this is your io bar the averaged one and when you see the actual the blue one that is your io, which is what you will be seeing on the oscilloscope. And here also this one is the blue waveform is the actual waveform that you

oscilloscope. And here also this one is the blue waveform is the actual waveform that you will be seeing on the oscilloscope and that is your Vo; instantaneous voltage waveform Vo.

Then your idc waveform the DC bus current waveform, you can form it using all the switching combinations in the direction of the current. So, what we see here is that here sometimes it is equal to io sometimes it is becoming 0 because we saw that when the voltage

is positive this output voltage is positive at that time the switching combinations mostly that arise are the first two diagonal switches that means TA plus TB minus are on.

So, in this region mostly your switching combinations that are arising are TA plus TB minus and the upper two switches being on TA plus and TB plus and the lower two switches being on

TA minus and TB minus. And for these two switching combination your idc equal to 0 whereas, for this switching combination when the Current is positive idc equal to io and so, that is what you will be observing here.

So, what you see here is that this is becoming in sometimes it is following your idc which is equal to io, same as io and otherwise it is becoming equal to 0. So, idc is switching between IO and

your 0 that is what we see. And over here when the voltage is becoming negative, the average voltage is becoming negative the switching combinations that mostly arise are TA minus and TB

plus and then next is your TA minus and TB minus or your TA plus and TB plus these are the three switching combination and corresponding to this switching combination your idc equal

to minus of io and corresponding to these two switching combination your idc equal to 0.

So, accordingly you see here idc equal to minus of io that means what you just, if you flip this waveform if you invert it, so it will be following something like this

and that is what we observe here idc equal to minus of io and otherwise it is becoming equal to 0. So, whenever these combinations are arising TA minus TB minus and TA plus TB plus,

then the Current is becoming equal to 0. So, sometimes it follows plus io and 0 and other half it follows your minus io and 0. So that is your current waveform and it will have

the components. So, one component is your this idc idc you can say that the capital IDC and this

the components. So, one component is your this idc idc you can say that the capital IDC and this one is your instantaneous current waveform idc and but then you have another component which is

the second harmonic component and that will be something like this that you are going to get and you can call it is idc 2, the second harmonic component of the fundamental.

So, the key points of this lecture are the same as what we had the key points in Bipolar PWM.

You have to analyse different waveforms for any modulation strategy, whichever modulation strategy you pick up, you find out the modulating waveforms because you have to get the reference to obtain your output and then you find out what is the minimum dc bus voltage that is required

using the maximum modulation index. And then you also observe the ripples in the capacitor current

that will help you in choosing the capacitor when you design the converter. Thank you.