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Last class, we talked about a four quadrant
converter, AC to DC converter. So, the configuration,
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we start with thyristor. So, let us see, how
the four quadrant converter looks like.
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You have the diode, so this thyristor with
this diode together, it will act as a bidirectional
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switch. So, we want both voltage and current
to be
current to be changed; in both directions,
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we require that is E0 I0. So, for the other
side also, we require. So, we will put one
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more link here. This is the configuration.
So, we will be able to do the four quadrant
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operation. Both voltage and current, we can
change it. See, four quadrant operations,
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both the voltage and current, we can change
it independent of each other. At the same
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time, if you see here, the load is connected
here; this is the wave form, input side and
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this is the output side, this we are talking
about dc to dc converter, dc to dc.
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So here, this is the output; output we can
have both positive and negative, current also
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we can have positive and negative and previous
class, we also studied about the PWM mode
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of operation. So, PWM mode, we can have the
E0, we can control. So, we can control in
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such a way, a sinusoid average, sinusoidal
variation is possible, possible here, E0 side;
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this is possible.
Now, if that is possible; can we have a sinusoidal
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output voltage? We can have a sinusoidal output
voltage like this. If the sinusoidal output
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voltage is possible, we can also have a sinusoidal
output current. So, voltage and current depending
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on the load current may be leading, current
may be lagging for a inductive load. So, by
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proper control, can we have a unity power
factor also like this? It is possible, we
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will see that one. Then see, this previously,
we talked about front end converter ac to
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dc converters. Now, here if you see here,
if you see here, this E0 with PWM control
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E0 we are changing, in a PWM control, the
fundamental is we are making us; the average
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value we are varying at a sinusoidal, so here.
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Now, this is the current with unity power
factor. Then the question is can you use this
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configuration for an ac to dc converter with
a unity power factor? See, now the power flow
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can go from source to load as well as load
to the source. Then the same thing we can
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use it for an ac to dc converter with unity
power factor. Now for PWM control, why we
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require PWM? We want PWM so that the harmonics
shifted to the high frequency side; we come
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to that one soon. So, this configuration,
we can use it for an ac to dc converter with
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unity power factor.
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So, instead of thyristor, we will be talking
about here for high frequency PWM, we require;
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so we can use instead of thyristor, we can
use IGBTâ€™s or MOSFET because thyristor require
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a force computation that we have not shown
here. This thyristor require a force, so extra
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circuit is required. That we are not talking
about that one.
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Now, another thing is these two limbs; this
limb and this one - A and B, this is one leg
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- A leg, this is B leg. These are independent
of each other. See, that means the switching
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of A is not dependent on B. So, it can be
independent. So, B side, we have introduced
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to have a negative voltage and current here.
That means we want to generate, we want to
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run the system in all four quadrant operation.
Now, let us talk about the PWM operation now,
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ac to dc converter. So, how an ac to dc converter
with unity power factor; what we want? We
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come to that one now. Let us go to the next
slide now.
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AC to DC converter with unity power factor,
converter with unity power factor, power factor,
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the circuit: now ac to dc, so let us see this
is our ac side. Then, as before, we have high
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inductance is there, to drop the ripple voltage,
to drop the ripple voltage and also to reduce
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the ripple current. Then, so here, I will
use each limb as switch IGBT or MOSFET, you
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can use it. So, I will show it as a symbol
switch with freewheeling diode, freewheeling
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diode also required. This is connected here,
this side will go here, A, B and you have
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the dc side, the capacitor here. This is our
E0, this is input, our single phase ac Vin,
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this is inductance L.
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Now, we are using PWM here such that across
AB, there is a fundamental voltage, PWM, there
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is a fundamental voltage and the other frequencies,
PWM voltage. We will come to that one. But
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for the analysis, due to with PWM, PWM switching
for leg A and leg B, there will be resultant
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fundamental AB voltage which is we will be
controlling. So, for unity power factor; how
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we have to control such that we get unity
power factor? So, let us say, this is our
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Vin, Vin and let us say this is our Iin. Iin
probably, we will mark with a different colour.
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So Iin, so Iin has to be in phase with Vin,
unity power factor. So, this is Iin.
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So, this Iin flowing through L, it will have
a drop across the inductance and what is a
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drop? That drop will be L omega into Iin,
the peak value of, we are trying to draw the
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phasor diagram, the peak of value of this
current. Now, this VAB, fundamental AB will
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be, so this voltage will be perpendicular
to, this will be perpendicular to, this will
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be equal to this perpendicular to Vin, the
drop across a inductance that is L omega into
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Iin. So, the VAB fundamental will be here,
this is the VAB, VAB fundamental. This is
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when the power flow from this side to that
is source side to the dc output side.
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Now, when the power is reversed; how will
it happen, the phasor diagrams? Again, we
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will draw it here. This is Vin, this our Vin
again, power reversal with unity power factor
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means Iin has to be in this direction. So,
Iin we can draw with a different colour. Now,
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this is your Iin, Iin phasor and the Vin will
be in this direction, j omega here, it will
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be in this direction. So, this is VAB, this
vector will be L omega Iin, omega is the mains
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frequency.
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So if you see, this is for power. Here, V
into Iin is negative, Vin Iin will be power
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will be flowing in the other side that is
dc side to ac side here. In this case, it
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is going from ac to dc. So, if you take with
unity power factor as the power for a fixed
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Iin, the movement of VAB will be along this
direction for forward as well as reverse power
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flow. Now, as a Iin varies; Iin varies Vin
will also vary, Iin magnitude varies. So,
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Iin magnitude depends on the load connected
across E0. So, we have to control VAB for
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any load. Any load that means what is meant
by load here? You have a dc here, output dc
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current, so for that dc current E0 into I0
current, output dc current, there is a power.
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That power has to be match taken from the
input current, input source that is Vin into
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Iin. So, Iin will be connected, Iin will be
related to the output load dc current.
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So, depending on the load, Iin will vary.
So, for a particular Iin, for forward or reverse
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power flow, VAB has to be moved along this
direction. So, we have to control the PWM
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such that your VAB moves along this direction
for unity power factor. So, how do you do
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that one? That is more important.
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So, as I told before, we are using PWM operation.
So, let us talk about the PWM. In PWM, how
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we can control the VAB, fundamental VAB, amplitude?
And also, I told previously, these two legs
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are independent of each other, we can control
independently. So, how do you do that one?
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So, let us take only this part now that is
a converter alone and study the PW. So, let
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us go to the next slide now.
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We will talk about the converter part and
how we do the PWM. So, the input the output
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voltage E0, I will divide into 2 values of
E0 by 2, E0 by 2 and this also E0 by 2, E0
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by 2. So, this way, we will do it. Now, we
have the switch here, we have the freewheeling
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diode here, this is our A. Now, other limb,
we are not we are not talking about those.
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What we can use for A, the same thing can
be applied to B leg also. Now see, even though
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the output is not divided as E0 by 2, E0 by
2; I have divided, for analysis I have divided
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into E0 by 2 but the total output dc is still
E0.
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Now, let us take a condition when S1 is on;
S1 is on, what is the voltage at A? The voltage
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at A means we have to measure the voltage
with respect to a fixed point. So, what we
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will do? We will measure with respect to the
fictitious center of E0. Here we represent
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it as O; this point, we will mark as O. That
means only thing we have to measure the voltage
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with respect to a common reference point.
The reference point is O that is that is a
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fictitious; it is not there, for analysis,
we are doing it at the centre part of output
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E0 voltage E0.
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When switch, top switch is on that is I will
make this one as S1, this one as S2; when
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the top switch is on, the voltage A with respect
to zero, this we will say pole voltage VA0
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is equal to VA0 is equal to sorry E0 by 2.
We will assume the switching are instantaneous
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when S1 is off and S2 is on. When S2 is on,
VA0 will be minus E0 by 2, VA0 is equal to
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minus E0 by 2. In this case, we are turning
S1 on; in the second case, here S2 on. Now
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PWM operation, we have to switch on S1 and
S2 such that the average variation at point
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A with respect to O should vary sinusoidal.
So, how do you that one?
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Here, we will use a sine triangle PWM technique.
So now, let us go to the sine triangle PWM.
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In sine triangle PWM; what we will do? The
sine wave, we are comparing with a fixed triangular
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waveform, waveform of high frequency and the
sine wave is the fundamental wave what we
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want, the reference wave. Let us take a typical
example, how the sine triangle PWM, we are
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comparing.
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This is the VA0 we want. So, we generate the
VA0. This VA0 will be compared with a fixed
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triangular, triangle waveform of fixed amplitude
and fixed frequency. So, let us take triangle
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waveform. May be triangle waveform, we will
draw with a different colour so that clarity
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will be more. This is our triangle waveform,
this we can generate with in an analog circuit
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or a microcontroller or in VSP. The generation
of this sine triangle waveform is not that
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difficult but only thing is the triangle frequency
is much higher than the sine frequency. Why
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we do that one? So that the harmonic frequency;
you will know, how the harmonics are generated?
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The high amplitude harmonics will be shifted
to the high frequency side which is the frequency
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of the triangle wave form.
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So, high frequency amplitude, high amplitude
harmonics are shifted to the high frequency
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side that is at the triangle side, then the
impedance for this harmonics L omega will
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increase and the current ripple drawn from
the mains will decrease. So, the current will
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be as close as to a sinusoidal wave. It will
happen that way. So, this is the thing. So,
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we will call this waveform, the modulating
wave. That is our reference sine wave and
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the triangle wave we call as carrier wave
that is VC (t) VC (t), the amplitude of that
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one, V is equal to VC.
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Now, as I told, whenever the sine is greater
than the triangle waveform, whenever the triangle
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is greater than the sorry whenever the sine
is greater than the triangle; we will be switching
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the top switch one. Top switch is one means
the VA0 will be Vdc by 2. So now, let us use
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a different colour so that it will be like
this; this point that is from this point to
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this one, Vdc by 2. Then when the triangle
is greater than the sine wave, it will be
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minus Vdc by 2. So, as we go along the sine
triangle comparison for full one cycle of
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sine wave; the output waveform, PWM pulse
width modulated, PWM pulse with modulated
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VA0 will be like this.
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See, when it comes to the center, the Vdc,
the plus Vdc by 2 is more and the minus Vdc
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by 2 is, the width of minus Vdc by 2 is smaller.
So, you can see, the average value, if you
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see; this average value will be varying sinusoidal.
So, we will study that one; is it following
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the sinusoidal way? And we also want, the
output should be proportional to our Vm (t),
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modulating wave that is our reference wave.
So, how we can control the output VA0 fundamental
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with respect to our modulating wave? We will
study now. Let us complete this one. So, negative
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side, it will happen this way.
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So, this way, we will get the PWM waveform.
So, if you see here, the positive portion
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from here to here, this will be sorry E0 by
2 and this side will be minus E0 by 2 that
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is I will just remove this one sorry. So,
I will rewrite this one for clarity. So, this
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will be plus E0, E0 by 2 and below it is minus
E0 by 2. So, you will get this type of PWM
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wave form and the average waveform, the average
wave form will be proportional to our Vm (t).
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This we will check now but this average variation
will be proportional to our Vm (t).
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Now, let us check whether this average variation
is proportional to Vm (t). If it is proportional
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to Vm (t); the moment Vm (t) amplitude we
reduce, the average variation amplitude also
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will reduce. The Vm (t) if you can shift it
with respect to, slightly shift it with respect
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to sine triangle waveform, again the average
variation at the VA0 will also shift. Let
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us see whether it is true. So, what I stated
now, by using this PWM, sine triangle PWM
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by switching the top switch and bottom switch
with respect to the sine compared with triangle;
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we get a PWM wave form which I shown here,
pulse width waveform like this. The average
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variation will vary sinusoidal and proportional
to Vm (t). Let us see, whether it is proportional
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to Vm (t).
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One assumption also we made; we will make
it, the frequency of the triangle wave form
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is very high compared to sin wave. Why that
is required, that assumption? Let us take
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one triangle period, one triangle period.
See, triangle which is going from negative
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maximum to 0 and then positive maximum and
comes back, this one. So, if you see here,
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one triangle period, we call as the carrier
period. So, when the triangle goes from minus
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negative to positive, this duration is equal
to Tc by 2, half carrier period; similarly
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here also, Tc by 2, half carrier period.
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Now, during one triangle period, any triangle,
we are just expanded here; we are comparing
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one portion of a sine wave with respect to
the triangle. Now, we will assume, during
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this one carrier period that is starting from
here to here, Tc, we will assume the frequency
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of the triangle waveform, we will assume the
frequency of the triangle waveform is so high
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during this period. The variation of the sinusoidal
waveform, that part of the sinusoidal wave
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form during this period is constant. It is
not varying much. So, for analysis, we can
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assume, we are comparing the part of the sine
wave here.
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See, assuming high frequency; see, this is
our sine wave amplitude during one triangle
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period. So, what we said here? The frequency
of I will write here, frequency of triangle
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or carrier triangle is higher than the than
the sine waveform, sine wave. That is the
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modulating wave. See, modulating and carrier;
these terminology, we borrowed from the communication
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paper, modulating, modulating, modulating
sorry it has gone to other side modulating
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wave, relating wave.
So, how much time? Typically 9 times or 11
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times, 17 times, 21 times; I am talking about
the frequency of the triangle wave with respect
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to the modulating sine wave so that during
a sampling period sorry during a carrier period
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that is one triangle period; the sinusoidal
waveform, the part of the sinusoid which is
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compared with the triangle wave during that
period is nearly constant. So now, as I told,
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whenever the sine is greater than the triangle,
S1 is on, the top switch is on that is this
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is our, with a diode, will mark it for clarity
here; this is S1, this is S2. So, whenever
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the sine is greater than the triangle; top
switch is on.
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So, during this portion, S1 is on; this is
S1 here, S10 from here to here. Also, when
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during the negative slope of the triangle
from here to this period that is here also,
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S1 is on and during this period, when the
triangle is greater than the sine wave here,
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S2 on, this period. Now, what we have to find
out? Same like in SMPS; see if you see here,
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suppose you have a 10 volt, you want a 5 volt,
how do you use it? Using a chopper; in a period,
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part of the time, we will be switching on
the device so that the output input voltage
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of 10 volt will across a load and rest of
the period, it will be 0.
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So, if you see here, so we want the 10 volts,
so we will we doing like this, we will be
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turning the device for half the period and
half the period will be switch off. So, the
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output will be 0 to 10 during this period
and rest of the period, it will be 0. So,
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the average value will be 5 volt. That is
how you generate 5 volt here using volt second.
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That means volt second means 10 into Ton,
this is Ton divide by total period will be
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equal to 5 volt if Ton is half of T. If Ton
is 25% of T means the output voltage will
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be 25% of 10 volts that is 2.5.
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So, by varying the Ton period, we can vary
the output voltage here. The same way here,
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we are trying to turn on the device here.
S2 is turned on during this period, S1 is
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turned on this period and S1 is turned on
period. So, same like our previous SMPS case,
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we should we should find out what is the volt
second or the average value from the volt
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second, average value during one sampling
period. What is the purpose? We want to find
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out this average value; is it proportional
to the Vm (t), part of that sine wave during
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that period? So, that is what we want to find
out. Let us take S1 on period, the volt second
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during S1 on period.
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00:33:45,470 --> 00:33:52,470
Now, let us see find out the volt second during
one triangle period and find out the average
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value during a triangle period. See, if you
see here, when S1 is on; so this slopes if
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00:34:03,920 --> 00:34:10,920
you take, from here to here, it is the slope.
It will go minus Vc to Vc. So, total this
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00:34:14,770 --> 00:34:21,770
height is that is I would say O, here, O to
T sorry P, here that is this height. So, from
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00:34:30,470 --> 00:34:37,470
this point to the vertical height, O to P
that is this point is 2 Vc, 2 Vc.
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00:34:49,159 --> 00:34:56,159
So, when the vertical height goes to 2 Vc;
what is the X axis? X axis is equal to Tc
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00:35:00,609 --> 00:35:07,609
by 2. So, for 2 Vc, it is Tc by 2. Then, what
is for this height? This height is Vc plus
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00:35:15,060 --> 00:35:22,060
Vm. Vm, we are measuring from the zero that
zero sides, from the zero axis side, the zero
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00:35:22,240 --> 00:35:29,240
voltage level. So, this much distance if you
see here; this is Vm, this is Vm, this is
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00:35:32,660 --> 00:35:39,660
equal to Vm during that period and from here
to the midpoint, here, midpoint of the vertical
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00:35:40,930 --> 00:35:47,930
height is equal to Vc. So, for 2 Vc, the period
is equal to Tc Tc by 2. Then what is the period?
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00:35:50,210 --> 00:35:55,270
That is to find out the period from here to
here, S10 from this period we are finding
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00:35:55,270 --> 00:36:02,270
out, this period, from here to here. What
is the period? It is equal to so Tc by 2;
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00:36:02,830 --> 00:36:06,040
for 2 Vc, it is Tc by 2.
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00:36:06,040 --> 00:36:13,040
Then, what it is for Vc plus Vm (t), Vm (t)
at that instant? That is the amplitude of
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00:36:15,980 --> 00:36:22,980
the sine wave. During this portion, VA0, during
this period, VA0 sorry VA during this period,
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00:36:27,910 --> 00:36:34,910
we are talking about the VA0, during that
period, it is equal to VDC by 2. But if you
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00:36:38,890 --> 00:36:45,890
see here, during this period also S1 is on
from here to here and this period and this
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00:36:45,950 --> 00:36:50,500
period because of the symmetry of the wave
form and assuming the amplitude of the sine
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00:36:50,500 --> 00:36:56,850
wave is constant, this period from here to
here and this period from here to here, both
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00:36:56,850 --> 00:37:03,850
on periods are equal. So, the total on period
is equal to 2 times Tc by 2.
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00:37:07,930 --> 00:37:14,930
Now, during the S2 on period, the output voltage
will be, so what is the total area volt second?
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00:37:16,850 --> 00:37:23,850
So, this period multiplied by VDC 2 by is
the total volt second during on period. So,
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00:37:27,550 --> 00:37:34,550
S1 on, the volt second that is voltage into
time is equal to when they multiplied to,
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00:37:41,610 --> 00:37:48,610
this 2 and this 2 will get cancel. So, Tc
into Tc into Vc plus Vm (t) divide by 2 Vc
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00:38:06,930 --> 00:38:13,930
into sorry not it is, our notation is not
VDC, it is E0 sorry E0 by 2. So, this is E0
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00:38:23,619 --> 00:38:30,619
by 2. So, here also it is we have to write
E0 by 2. Here, it will be into this is the
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00:38:40,369 --> 00:38:43,320
volt second, E0 by 2.
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00:38:43,320 --> 00:38:50,320
Then what will happen when S2 is on? S2 is
on, the time period is this is during S1 is
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00:38:54,250 --> 00:38:59,290
on, this period. So, Tc minus this period
we will give you that is the period where
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00:38:59,290 --> 00:39:06,290
S1 is on that is this period, we will give
the S2 on period. So, S2 on is equal to total
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00:39:12,490 --> 00:39:19,490
Tc minus Tc. So here, let see clarity, I will
remove this diagram. So, we have more clarity
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00:39:32,650 --> 00:39:39,650
here. So, S2 on period, so Tc minus and Tc
minus, there is a total carrier period triangle
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00:39:46,540 --> 00:39:53,540
period minus the on period of S1 on period
that is what we got; Tc into Vc plus Vm (t)
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00:39:59,760 --> 00:40:06,430
that is the Vm (t) is the sine amplitude during
the triangle period during the comparison
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00:40:06,430 --> 00:40:13,430
under consideration now; Vm (t) divided by
2 Vc and this is the S2 on period. Now, the
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00:40:20,190 --> 00:40:24,450
volt second multiplied by minus E0 by 2.
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00:40:24,450 --> 00:40:29,100
So, for clarity, let us go the next page.
I have redrawn the sine triangle comparison
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00:40:29,100 --> 00:40:36,100
during one sampling period. Now, we know the
volt second, volt second when S1 on, volt
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00:40:42,320 --> 00:40:49,320
second when S1on is we got it that is Tc into
Vc plus Vm (t) divided by 2 Vc the whole multiplied
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00:41:01,570 --> 00:41:08,570
by E0 by 2 volt second S10. Now, what is the
volt second during S2 on? The period, the
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00:41:16,310 --> 00:41:22,970
time period during each S2 on multiplied by
minus Vdc by 2. We know that when S2 in on,
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00:41:22,970 --> 00:41:26,800
when the triangle is greater than the sine
wave that is here, during this portion; so
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00:41:26,800 --> 00:41:31,060
thus S2 will be on and output voltage V0 will
be minus Vdc by 2.
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00:41:31,060 --> 00:41:36,750
So, what to the S2 on period? S2 on period
is very easy to find out. We know the S1 on
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00:41:36,750 --> 00:41:43,750
period; Tc minus that one. So, S2 on is equal
to Tc minus Tc into Vc plus Vm (t) divided
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00:41:59,040 --> 00:42:06,040
by 2 Vc. This is the S2 on period multiplied
by minus E0 by 2 will give the volt second.
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00:42:11,520 --> 00:42:18,520
So, the volt second during S2 on, when S2
on is equal to Tc minus Tc into Vc plus Vm
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00:42:36,640 --> 00:42:42,940
(t). Vm (t) is a function of time at which
a triangle period, we are sampling. That is
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00:42:42,940 --> 00:42:49,940
why we have put as a function of time divide
by 2 Vc whole multiplied by minus E0 by 2.
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00:42:55,090 --> 00:43:02,090
Now, the average value is the summation of
these 2 volt second divide by Tc. So, VA0
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00:43:05,720 --> 00:43:12,720
average for a period T0, I can write it here
so that it will be clear; VA0 average, total
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00:43:20,740 --> 00:43:27,740
volt second divide by Tc is equal to total
volt second divide by Tc for triangle period
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00:43:35,810 --> 00:43:41,420
that is the average value. So, how it will
be? If you sum these 2 and multiply if you
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00:43:41,420 --> 00:43:48,420
see, it will be equal to 1 by Tc that is the
average value into Tc by 2 plus Tc into Vm
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00:44:08,930 --> 00:44:15,930
(t) divide by 2 Vc then minus Tc by 2 plus
Tc into Vm (t) divide by 2 Vc; we are taken
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00:44:35,510 --> 00:44:42,510
care of the positive and negative multiplied
by E0 by 2, here I will write the E0 by 2.
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00:44:47,570 --> 00:44:54,570
So, this will be, so if you see here, Tc by
2 Tc by 2 get cancelled. So, if I you say,
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00:44:58,660 --> 00:45:05,660
this one and this one get cancelled. Then
the whole VA0 average, I will write it here;
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00:45:10,050 --> 00:45:17,050
VA0, average value of VA0 during a sampling
period sorry during a sampling period means
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00:45:21,670 --> 00:45:28,670
during the triangle period, during the period
Tc during the, during a period, period Tc
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00:45:36,420 --> 00:45:43,420
is equal to VA0 average will be equal to from
this two equations, it will be E0 by 2 into
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00:45:50,800 --> 00:45:57,800
the value of sine wave at that instant divide
by Vc, Vc is the carrier period.
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00:45:58,530 --> 00:46:04,710
So, if you see, the average value will vary
when Vm (t) vary because these two rest of
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00:46:04,710 --> 00:46:11,710
the periods are constant. So, during assuming
high frequency sine triangle PWM such that
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00:46:12,130 --> 00:46:17,660
during a triangle period, the sine amplitude
is not varying, we are assuming constant;
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00:46:17,660 --> 00:46:24,660
so that is why we said high frequency PWM
such that during sine triangle PWM, the average
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00:46:25,850 --> 00:46:32,850
value of the output VA0 PWM wave form will
be proportional to Vm (t). So, to control
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00:46:35,750 --> 00:46:42,750
the Vm (t) or control the average value VA0
average, you wave the Vm (t) amplitude. So,
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00:46:42,930 --> 00:46:46,880
it will be proportionally varying. Now, this
way, we will be controlling.
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00:46:46,880 --> 00:46:53,880
Then, the same way, it is for leg A. In the
same way, we can control the leg B also; so
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00:46:54,200 --> 00:47:01,200
that is VB0. So finally, what to the AB? VAB
will be equal to, VAB will be equal to VA0
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00:47:04,600 --> 00:47:05,960
minus VB0.
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00:47:05,960 --> 00:47:12,960
If you refer the previous figure that is here;
so the voltage VAB, we are independently controlling
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00:47:17,810 --> 00:47:24,810
VA0. A0 means the zero with respect to the
fictitious center point of this power supply.
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00:47:24,900 --> 00:47:31,900
So, VA0 minus VB0 will give VAB. So, we have
to generate the reference waveform VAB such
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00:47:32,430 --> 00:47:39,430
that we get unity power factor, unity power
factor as I shown here. So, VAB has to be
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00:47:41,090 --> 00:47:47,700
moved in this direction sorry the tip of the
VAB should be moved in this direction depending
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00:47:47,700 --> 00:47:52,290
on IA such that unity power factor, we will
get.
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00:47:52,290 --> 00:47:58,890
Now, see average value, we got. Now, we went
for the high frequency PWM, sine triangle
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00:47:58,890 --> 00:48:05,530
PWM; how about the harmonics? See, if you
analyze this one through Fourier series, its
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00:48:05,530 --> 00:48:12,420
highly it will get into lot of complication.
But in a simple way, we can analyze how the
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00:48:12,420 --> 00:48:19,170
harmonics, how the harmonics are present during
a sine triangle PWM. See, this I am taking
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00:48:19,170 --> 00:48:24,700
from a famous book by Professor Ned Mohan.
So here, we can analyze that one.
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00:48:24,700 --> 00:48:31,700
Now see, sine triangle PWM as I told is the
fundamental waveform, the average waveform
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00:48:31,950 --> 00:48:37,859
is proportional to Vm (t), our sine wave or
whatever the waveform you are trying to compare
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00:48:37,859 --> 00:48:44,170
with a triangle wave form. Now, where is the
next higher order harmonics? Because the output
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00:48:44,170 --> 00:48:48,369
waveform is a pulse width modulated waveform,
pulse width modulated wave form, that is what
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00:48:48,369 --> 00:48:54,060
we have shown before and we have talked only
about the fundamental.
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00:48:54,060 --> 00:48:59,050
So, apart from the fundamental, if you have
the harmonics, this harmonics will be, the
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00:48:59,050 --> 00:49:03,750
high harmonics amplitude will be shifted to
the triangular frequency. So, the advantage
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00:49:03,750 --> 00:49:08,650
is if you use a high frequency triangle waveform,
if possible if your power converter permits;
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00:49:08,650 --> 00:49:15,560
the next high amplitude harmonics will be
at the triangular frequency. So, the current
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00:49:15,560 --> 00:49:22,560
drawn due to this harmonic voltage will be
much less because of the L omega, omega is
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00:49:22,670 --> 00:49:27,730
more, harmonic that is the amplitude is, that
the frequency of that one depends on the triangular
305
00:49:27,730 --> 00:49:28,140
frequency.
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00:49:28,140 --> 00:49:33,990
So, L omega will be higher, the current drawn
due to their harmonic, harmonic will be much
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00:49:33,990 --> 00:49:38,290
less. So, the ripple current is much lower
and output you will get nearly sinusoidal
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00:49:38,290 --> 00:49:44,000
current. So, in the next class, we will study
briefly, simple thumb rule for a sine triangle;
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00:49:44,000 --> 00:49:50,790
how the harmonics are placed? Now, then for
our front end ac to dc converter, how do you
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00:49:50,790 --> 00:49:57,790
use the PWM such that some of the harmonics
can be suppressed. So, as the harmonics are
311
00:49:58,750 --> 00:50:03,740
suppressed, the current drawn from the mains
or the ac side will be more sinusoidal with
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00:50:03,740 --> 00:50:10,740
unity power factor that is what we want. That
we will talk in the next class.