WEBVTT
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We discussed about the front end Ac to Dc
converter and how we can control the harmonics
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or suppose some of the low order harmonics
without resorting to high frequency PWM switching.
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So, the configuration used, I will again draw
it for a clarity again.
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You have the primary transformer, then two
secondaries, then you have the two converters,
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one is here. We will assume the switches are
bidirectional, so diodes I am not putting
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now. So, we should assume that the diode is
there. So here, the transformer either the
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transformer leakage inductance or extra inductance,
you have to put it to take care of the ripple,
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this should be there.
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Similarly the second converter, inductance
is here. So, this converters, the PWM; this
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is converter A1B1, this is A2B2. For A1B1,
we will
use the same reference wave that is the modulating,
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same sin wave. But the triangle waves for
A1B1 are 180 degree phase shift. For A2B2
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but for the switching sequence for B1 is reversed.
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So, what is for s1? We will use it for here,
the bottom switch of B1. Then here, for the
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second converter, the same modulating signal
we use but the triangular wave forms for A2B2
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are 180 degree phase shifted but A1 and A2
are 90 degree phase shifted. So, we have the
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output diode is like this; this one will also
come here, this one also, this one come here,
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so parallel operation.
So, by doing this way, what happens? You will
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have the harmonics like this; you have the
Vn here, harmonic order, amplitude, then you
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have the harmonic order here, number of harmonics.
So, you have the fundamental here. Then all
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the harmonics at Fc and its side bands that
is this one; this is Fc and its side bands
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also at 3 Fc and its side bands, this is 3
Fc and its side bands, all this will get cancelled
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because of the PWM technique used, sin triangle
PWM used for the leg A and leg B.
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So, last class we learned for A when the sin
is greater than the A leg, leg A. Sin greater
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than triangle; top switch is on, top switch
is on. Then sin less than triangle, the bottom
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switch is on that is for leg A. And for, this
is 2 for this converter also. But for leg
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B, sin greater than the triangle, the bottom
switch is on and sin less than triangle, top
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on. This we are using. By using this one and
the triangle, triangle frequency, Fc is equal
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to 11 times fm we are using. So, all this
harmonics Fc, 3 Fc odd multiple get cancelled.
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Now, since this two are parallel, the triangle
wave form A1 and A2 are 90 degree phase shifted.
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So, all the harmonics and the side bands of
f2 that is this one because of the 90 degree,
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it will get cancelled here in the transformer.
Here it may be flowing but it has opposite
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effect in the transformer core and it will
not generate any flux and it will not couple
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to the transformer primary.
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So, transformer primary will have the fundamental,
then side bands at 4 fc plus or minus 1 and
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minus 3, that way it goes. So, this is the
thing. Now, let us let us study how the typical
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wave forms look like.
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We have converter 1, converter 2, this is
switch. So, let us draw the modulating wave
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and the sin wave. So, 11 times, approximately
11 times we can draw it like this. Let us
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draw that; this how to get a feel for it that
is why this the triangle wave form, we will
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make 11 times. So, let us draw the modulating
wave. We will use the same modulating wave,
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this is the modulating wave.
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Now, for the second leg; if this is A, this
side is B. So, for B phase, we have the inverted
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one. Let us draw the inverted one with a different
colour. This is 180 degree phase shifted.
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Now, let us draw the switching pattern, switching
pattern; first one let us take for
the leg A. Leg A, which triangle we will be
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using? The blue triangle we will be using
for leg A. So, I will mark it accordingly,
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A is here, A. So, whenever the sin is greater
than the triangle, top switch is on; so here,
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for leg A.
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And, when the sin is less than the triangle
that means we are comparing the modulating
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wave and the blue triangle waveform. When
the sin is less than the triangle, the bottom
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switch is on. So, if you do, when the top
switch is on, this point, the centre point
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with respect to our DC midpoint is equal to
plus VDC by 2. So, there it will be minus
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VDC by 2. So, we can see this comparison,
we can take it this way that is the second
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one here. During this portion, it will be
triangle is greater than the sin wave.
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So, if you use that blue triangle wave form
and the, so we are comparing, all the time
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comparing with the blue triangle wave form
and the modulating sin wave. So, whenever
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sin is greater than the triangle, the top
switch is on. So, the leg voltage or the pole
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voltage A with respect to our fictitious centre
point of the output DC plus VDC by 2. Otherwise
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it is minus VD when the bottom switch is on,
it is minus VDC by 2. So, the waveform will
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look like this, this wave. All this hatched
portion shows that the top switch S1 is on,
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S1, top switch is on. The bottom one we will
say 2.
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Now, for what is the pattern, the voltage
wave form, pole voltage or the leg voltage
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B0 for the second leg? Then we have to use
the sin as well as the red triangular wave
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form. So, we will go to the red marking. So
here, red triangle waveform means here. So,
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the wave form; so here, if you see here, when
the triangle, triangle is greater than the
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sin wave, here the switching sequence is inverted
that is top switch is on.
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So, whenever the triangle is greater than
the sin wave; the top is on, we will have
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VDC by 2, the pole voltage wave form, so this
way. So whenever, here whenever the sin greater
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than the triangle; what we are doing? The
bottom switch is on. No sorry, whenever the
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sin greater than triangle, the bottom switch
is on. But it is reverse of the other phase.
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So, what you have to remember here?
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The triangle waveform 11 times, 11 times the
sin wave and the triangle waveform for the
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other one, the leg B is inverted, triangle
wave form and the switch sequence for the
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leg A when the sin is greater than the triangle,
the top switch is on; when the sin is less
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than the triangle, the bottom switch is on.
In the leg B whenever sin is greater than
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the triangle, we are turning on the bottom
switch. The logic is inverted.
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Now, what is the final wave form across the
load VAB? That is we are more interested in
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that one that is equal to VA0, VA0 minus VB0.
Let us do the VA0 minus V0. This is, the top
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one is VA0. So, this is VA0 and this one is
VB0. Now, VA0 minus VB0, we can do it like
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that, we can just subtract it. So, these type
of pulses, appear across the across the waveform.
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Then, height of this one will be, previously
it was VDC by 2, the height will be VDC. Here
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it is VDC by 2, this height, if you take it,
this height is equal to VDC by 2 that is 2
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here also. But in this case, it is VDC. So
subtraction; that minus, minus get added.
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So, to make all the wave form within the phase,
I have slightly scaled down the height. But
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the actual height is VDC here that is VAB
waveform. So, if you do field analysis of
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this wave form, it will have the fundamental,
2 times the individual leg A or individual
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B. At the same time, it will not contain the
harmonics at fc and it side bands and 3 fc
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also. Here, with one converter, the next higher
amplitude harmonics happens at the side bands
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of 2 fc. Now, with one more converter, we
will try to cancel the 2 fc side also so that
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the final high order harmonics happens, happens
at the side bands of 4 fc. So 4 fc, fc is
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11 times; 4 fc means 11 into 44 times. So,
44 time the fundamental.
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So, we by a very interesting technique without
resorting into high frequency switching, we
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have shifted the harmonics to very high, to
the very high frequency side this is VDC.
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Now, for the next converter that is the converter,
the bottom converter, again I will try to
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the draw the figure. It is like this; here
we will say, this leg is A2, this leg is B2.
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Our AC side is here. So here, with respect
to this triangle wave form, for the same A
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leg and same leg A leg here, it is 90 degree
phase shifted. So, let us draw the 90 degree
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phase shifted wave form.
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So, this is the one; 90 degree phase shifted.
This is for leg A2 sorry this is according
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to this one, if it is 90 degree, this for
leg B2, red one. Sin wave is the same, same
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modulating wave we are using. So, all the
fundamental get added. Now, for the for the
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blue triangle of this one, 90 degree triangle
we will draw there, 90 degree phase shifted
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that is this one. This is for leg A2, leg
A2.
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Now, for A01 let us draw VA20 let us draw.
VA20, when the sin is greater than the triangle,
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the top switch is on that is from here and
here, VDC by 2. So, if you repeat for other
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triangles also, you will get this pattern.
But here the height is VDC by 2 and plus VDC
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by 2 minus VDC by 2, it will go. So, this
is the pole voltage wave form of the second
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converter with 90 degree phase shifted triangle
with respect to the same leg of the converter
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1 that is this converter.
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So, I will just mark it to get the pulse width,
the change of pulse width, the feeling for
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the change of pulse width. Then for leg B,
we will change to the converter. It will be
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like this; sorry, this is the one. So, we
are comparing the red triangle with the modulating
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wave, sin wave. Here, whenever the sin is
greater than the triangle, the bottom switch
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is on; when the sin is less than the triangle,
top switch is on, the switching logic interchanged,
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same as before.
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Only the case is the triangles for the, similar
index for the second converter is 90 degree
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phase shifted with respect to the same converter,
the top converter or the converter 1. So,
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this is the case. So again, the pole voltage
wave form VA2 B2 that is VA2 B2 sorry I will,
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the VA2 B2 will be VA2 B2 that means VA20
minus VB20 will be like this. So, this height
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will be again VDC. We will go to plus VDC
here and here it is minus VDC. So, these are
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the VA0 VB0 wave form coming across the two
converters.
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Now, if you sum these two voltage, that voltage
will have a shape like this because that the
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current wave form to get a feeling of the
harmonics, in instead of summing the current
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wave form if these two voltage wave form you
can sum; this may be the replica of the voltage
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waveform appearing across the transformer
- two secondary’s and reflector on the primary.
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If you see here, transformer, primary we will
be giving the voltages and in this wave form,
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all the harmonics, individual harmonics will
dropped across the leakage inductance.
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So, why I am summing? This is not actually
happening in the converter; these two wave
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form never summed. But this wave form when
you sum, this the resultant wave form, frequency
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spectrum of the resultant wave form approximately
gives the harmonics spectrum of the suppressed
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current wave forms. We can get a feel for
that one if you do the effective of that one.
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So, this is a stepped wave form, PWM with
stepped wave form.
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This wave form will not contain all the fc
and its harmonics fc, 3 fc, 5 fc and harmonics
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will be highly suppressed and then the side
bands at 2 fc. So, this will have only harmonics,
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the zero line is here; this will have the
harmonics only at the side bands of 4 fc.
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So, you will get a stepped wave form and the
primary current will be more sinusoidal here.
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So, we will have a stimulation study of this
one later. But now the PWM technique, we know
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it.
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Now, how to generate the required Vm so that
the required VAB, VAB is generated such that
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the transformer primary always used to draw
unity current with unity power factor or the
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current drawn from the transformer, primary
side will be nearly sinuous order and in phase
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with the main. So, we will study that part
now. Before going to the study, controller
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part; see, one more thing we have to learn
here. How to design this L and C? Let us study
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the how to design the inductor for one converter.
Let us draw the converter once again.
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This is the mains, this is the inductance,
this is the switch, here; so, this is our
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A and this is B. Now, according to the phasor
diagram with unity power factor; this is VS
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and this is IS in phase with that is this
current IS in phase with VS and the drop cross
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inductance L will be, for power flow from
source to the load, it will be like this.
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Otherwise, it will be like this so. VAB will
be like this, this is the VAB.
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Now, so VS plus IS, if you take the fundamental
component of this AB, AB fundamental component
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that we will represent as Vr, Vr, Vr is the
fundamental
component of the VAB, VAB PWM signal, PWM
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wave form, wave form that
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is Vr; fundamental component of the VAB PWM
wave form.
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Now, if you assume the harmonics are highly
suppressed, due to fundamental, there is a
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drop across this inductance. So, from this
diagram, what is the relation between the
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fundamental voltage and the inductive drop
that is this one and the VAB wave form? It
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will be equal to Vr is equal to root of VS
square plus L omega; omega the mains frequency,
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this is the L into
IS square. This L omega IS is the peak value
of the sinuous order current here; L omega,
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have drawn.
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Now, this Vr is generated by modulating the
sin triangle PWM and from the previous sin
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triangle PWM and our V0 is here. This is our
V0 sorry E0. So, VAB will be for from the
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previous sin triangle analysis, what you have
done two days before; the VAB, for one leg,
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the maximum amplitude of V will be proportional
to the maximum will be equal to VDC by 2 or
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E0 by 2.
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So, when subtract A and B, the maximum value,
the fundamental maximum value of VAB will
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be equal to E, E0 that is the maximum voltage
Vr we can get. But in all practical applications,
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we will not go to the extreme modulation.
If you take a sin triangle like this, we will
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not go to this range.
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See, if you go here, the pulse would become
very narrow; the system may not respond here.
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So, we will always restrict the modulation
index that is the height of the sin wave modulation
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index m I will say is equal to our VC value
by Vm maximum value will be always theoretically
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this will be equal to 1 but we will restrict
to maximum, it will be equal to 0.8 for practical
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limitation.
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So from here, Vr will be 0.8 of the maximum
E0, the output E0. This will be equal to root
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of VA square plus L, for a 50 hertz operation
omega is equal to 2 pi into f, f is equal
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to 50 into IS maximum. How do you find out
this IS maximum here? How do you find out
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what is the maximum IS so that from these
two equations; from this equation, let us
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find out what is L. We will square this one
such that Vr is equal to as I told, in the
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modulation index, we will go upto not to 1,
it is only 0.8.
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So, 0.8 E0 square
minus VS square divided by omega square IS
square is equal to or square root of this
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one will be equal to our L value. Now, if
you see here, this L value depends on VS we
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know. How to find out this IS? Omega we know,
omega is equal to 2 pi into f, our mains frequency.
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So, how to find out this IS? IS varies with
our load that means load which is connected
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here, what load it is coming here; this is
DC voltage, so load. So, may be IS we can
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find out from the power balance.
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See, output power is equal to E0 into I0,
output power and for with unity power factor,
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we will be assuming, we will assume the efficiency
of the system is it is across the system 100%.
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So, input power should be equal to output
power. This is the output power, output power,
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this will be equal to, input power is how
much? That is VS rms into IS rms. So, E0 we
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know, I0 maximum load current we can find
out, VS rms is known; from this one, we can
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find out what is for a maximum load, what
is the IS rms? From the IS rms, we can find
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out ISP. We can square it, you will get it.
So, L this way, we can find out.
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Now, how to find out the value of this capacitor?
That is very important. So, let us find out
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the next one, how to find out the value of
the capacitor? Let us draw the converter once
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again that is only one side converter only
we will draw it.
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This is our input source here, then we have
this is the A point, A leg, this is our B
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leg and your capacitor. Capacitor has to take
care of the ripple current coming from here.
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If this is the I0, the current ripple current
coming through this current, the current which
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is coming from here, let us say this is IL,
IL contains sorry IL contains, this IL contains
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DC current plus the ripple power. The ripple
will pass through the inductance that is the
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oscillating part not the DC part. The fluctuating
part, it will go through the capacitor and
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the capacitor we should choose it such a way,
it should give a stiff DC link.
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That means due to the ripple current, fluctuating
current, the voltage ripple across this one
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should be as small as possible. Small means
how much? 5% of the DC, we should, can have
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only that much ripples. So, let us find out
what is the ripple current coming here? Here
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again, if you see here, this is VAB; fundamental
component input power of this converter should
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be also be equal to output power of this one.
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So, the input power to the converter is equal
to Vr (t) into our iS (t) that is coming from
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here, this is the iS (t). This current, input
assuming 100% efficiency VS (t) and iS (t)
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should be also be equal to this is we assuming,
this is equal to E0 into I0. Now, sorry V0
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that is sorry not I0, that current E0 into
the output power input power should be equal
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to the output power. So, output power is equal
to E0 plus IL into I0. So, we will take IL
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here.
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So, what are, what is Vr (t) into iS (t)?
Vr (t) is equal to if you take this is VS,
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this is IS, this is our fundamental Vr (t)
and this has a delta here. So, Vr (t) will
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be Vr sin omega t minus delta into IS sin
omega t. So, net current IL is equal to Vr
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IS into sin omega t minus delta into sin omega
t divide by E0. This is the current IL coming
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out of this one. Input power is equal to output
power. Now, if you see here, this can be extended
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like; Vr IS divided by E0 into cos delta minus
cos 2 omega t minus delta.
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So, IL contains a DC part that is DC, this
is the DC part that is equal to the I0 and
40:28.880 --> 40:35.880
there is a fluctuating part that is 2 times
the input frequency. This will go through
40:38.710 --> 40:45.710
this capacitor that is
this will go through this one, this part,
this one, the one which is this one. So, we
40:55.760 --> 41:01.020
have to design the capacitor for this 2 times
the frequency, there is a second harmonics
41:01.020 --> 41:07.160
current flowing through this one. So again,
once again, this is the input power, sinusoidal
41:07.160 --> 41:11.840
fundamental, sinusoidal current, input power
voltage into current, this is the output power
41:11.840 --> 41:12.720
from the converter.
41:12.720 --> 41:19.430
Now, Vr (t) from this, from this definition,
it is Vr into sin omega t minus del into IS
41:19.430 --> 41:26.080
sin omega t. So, the IL which is coming from
here, if you see here, you will get this equation
41:26.080 --> 41:30.700
but when you convert to this one, there is
a two halves to be there because sin A sin
41:30.700 --> 41:34.390
B, we are converting to cos. So, these two
halves to be there sorry there is a mistake
41:34.390 --> 41:41.390
here. So, I will correct that one now. So
this, there has to be a 2 here; so 2 is here.
41:42.680 --> 41:49.680
So, this current, this into cos two omega
t, this value into cos omega t will, cos 2
41:50.150 --> 41:57.150
omega t will flow through this one. That will
generate a ripple current across this one
41:57.710 --> 42:03.210
and let us see what is that current. If it
is cos to omega t; we will integrate approximately.
42:03.210 --> 42:10.210
This is the ripple peak, let us take the ripple
peak, let us take as IL hat, ripple peak.
42:12.180 --> 42:16.480
If we integrate divided by 2 omega c if you
do the capacitor, capacitor is equal to integral,
42:16.480 --> 42:23.480
the voltage delta V is V is equal to integral
1 by c i dt. So even that, this function if
42:26.620 --> 42:32.980
you integrate, you will get this one. So,
this will be you want is equal to 5% of the
42:32.980 --> 42:39.980
delta V, 5% of our less than or equal to 5%
of 0.5%, 0.05 of E0, this has to be there.
42:43.950 --> 42:50.950
So, E0 we know, IL peak we know, we can find
out C. So, if you see this equation, I will
42:56.570 --> 43:03.570
write it here, the C will be, from the inequality
here, C should be greater than or equal to
43:07.540 --> 43:14.540
IL hat that is the ripple peak that is from
here, IL hat divided by 2 into 0.05 E0 into
43:36.390 --> 43:43.390
omega. So, from this equation, we can find
out the value of C approximately. So, C should
43:43.590 --> 43:46.000
be greater than this value.
43:46.000 --> 43:53.000
Now, L and C we have found out. Now, what
we want? We want to control the output voltage.
43:54.920 --> 44:00.390
So, what is the control block we require?
What is the Vr we have to generate? That is
44:00.390 --> 44:07.310
fundamental of VAB, we have to generate for
all load conditions such that the input current
44:07.310 --> 44:12.270
in the phase with our mains voltage. How we
will, how we will do it, control block of
44:12.270 --> 44:19.270
the front end converter?
44:26.510 --> 44:33.510
What we want here? We want an output E0, output
E0. So, this is our reference. So, reference
44:38.600 --> 44:45.600
I will mark it as E0 star. So, this is the
reference. Whatever we want, we can fix it
44:49.670 --> 44:56.670
as reference. So, it will be, it will be compared.
That means our output voltage, we have sensed
45:03.570 --> 45:10.570
and feedback to this one. So, the moment we
are comparing the reference plus the feedback,
45:12.760 --> 45:19.590
this is the feedback; there should be a controller.
So, what type of controller, we will decide
45:19.590 --> 45:26.590
later. We will just put controller. So controller,
what it should give? It should give the IS.
45:32.760 --> 45:33.620
Why?
45:33.620 --> 45:40.620
Because any time the E varies, changes with
respect to reference value that means the
45:44.650 --> 45:50.550
there is a change in the capacitor voltage.
That shows there is a change in the output
45:50.550 --> 45:57.550
load current. So, any change in the load current
assuming 100% efficiency and the input power
46:00.800 --> 46:06.170
should be equal to the output power. So, for
the required current, there should be an input
46:06.170 --> 46:11.580
current required. So, this indicates, any
change in E0 with respect to E0 indicates
46:11.580 --> 46:15.760
that there is a current demand, change in
current demand from the mains side. So, this
46:15.760 --> 46:22.760
will indicate to the IS, load current IS which
is flowing through the inductance or which
46:23.230 --> 46:30.230
is taken from the mains, this IS. This IS,
again we have to sense it from our mains,
46:37.470 --> 46:38.940
you have to sense it from here.
46:38.940 --> 46:45.940
I will just draw it here; this is our input
that is this is the current IS. This we have
46:49.980 --> 46:56.870
to sense it using a Hall Effect sensor LEM
current sensor we have to use it, we have
46:56.870 --> 47:03.870
to properly scale and give it to here. So,
here also we need a current controller to
47:08.740 --> 47:15.740
control. What type of controller; we are not,
we do not know. So, this controller, what
47:18.180 --> 47:25.180
it should give? To get the correct IS and
IS star, IS to match, what is required is
47:33.440 --> 47:40.440
the fundamental component of the AB, AB with
PWM. It should be the required VAB or reference
47:42.880 --> 47:43.810
value you should get it.
47:43.810 --> 47:50.810
So, the VAB, the reference value for VAB is
the output from here. So, this will go to
47:54.620 --> 48:01.620
the Vr reference. So, controller we take it
with for only a one converter but the two
48:05.550 --> 48:10.850
converters, appropriate gain of two we have
to give at the various places. But the same
48:10.850 --> 48:17.850
controller is used we can use it for the both
converters. So, this is for Vr star. So, this
48:17.960 --> 48:24.960
Vr star, it will go to the PWM converter,
PWM - pulse width modulation of the converter.
48:25.460 --> 48:32.460
So, converter will give what? It will give
the correct Vr. This is the reference value
48:36.170 --> 48:41.630
and the converter will PWM will give the correct
reference value.
48:41.630 --> 48:48.630
Now, this Vr, so if you see here, this is
VS; so VS plus, VS plus L into d IS by dt
48:58.470 --> 49:05.240
is equal to Vr here. So Vr, all already we
got here, we know the VS value. So, if you
49:05.240 --> 49:12.240
want this is VS, so VS, this is plus; VS minus
Vr, VS plus L into di by dt is equal to Vr.
49:23.220 --> 49:30.220
So, VS minus Vr is equal to L into di by dt.
So, this you integrate with the gain of 1
49:39.130 --> 49:46.130
by L. Here you will get the IL. This IS, these
are actual currents; we have to bring it to
49:51.250 --> 49:57.830
the controller side. So, we may require a
gain, gain function to reduce it or if you
49:57.830 --> 50:01.280
want to filter it, you may have filter also
over there or you will put simple gain function
50:01.280 --> 50:08.280
k, current gain so that IS reference and if
it is for used, if the controller is built
50:11.100 --> 50:16.470
using analog controllers, may be IS can vary
from plus minus to plus ten only.
50:16.470 --> 50:20.260
So, actual current when we sense it and bring
it back to this one, there is a gain function
50:20.260 --> 50:27.260
is required. This is how you do it. Now, we
have, this is an inner current loop is there.
50:28.200 --> 50:33.330
So here, if you see here, again I will explain,
you have the E0 reference and feedback we
50:33.330 --> 50:39.620
have taken from the converter. If there is
any mismatch in E0 and E0, it reflects on
50:39.620 --> 50:44.940
the type of load current required. Load current,
there is a variation in the load current.
50:44.940 --> 50:49.830
So, there is a variation in the load current
that should reflect on the input current drawn
50:49.830 --> 50:55.220
IS. So, this controller output should point
to the IS. This IS and actual current IS,
50:55.220 --> 51:01.850
we will compare and this compare out, would
give a controller. So, this controller should,
51:01.850 --> 51:06.980
this IS star and IS should match, controller
should give the correct Vr reference that
51:06.980 --> 51:12.230
is the reference for or the modulating wave
for sin triangle PWM. So, if the correct waveform
51:12.230 --> 51:19.230
is given, the PWM converter, PWM converter
will give the correct Vr here.
51:21.220 --> 51:28.220
Now, this is the converter output, this Vr
or VS minus Vr, integrate 1 by L integrate
51:33.280 --> 51:40.280
will give you IS value. This IS is stepped
down, you can feedback it here. Now, IL we
51:40.680 --> 51:47.290
got. How to find out the or how do you get
the E0 value? How do you make the block diagram
51:47.290 --> 51:54.290
for the E0? So, if you see, from the power
balance with unity power factor, VS IS rms,
52:01.290 --> 52:08.290
VS IS, these are the rms values is equal to
E0 into I0, E0 into I0. VS IS are the rms
52:31.450 --> 52:38.450
values; so if you take the peak value, if
VS and IS are the peak value, then VS peak
52:43.910 --> 52:50.910
IS peak divide by root 2, divide by root 2
is equal to E0 I0. IS we already know it,
52:57.920 --> 53:04.920
so VS IS divided by 2 E0 is equal to our I0.
This is the dc value current.
53:14.140 --> 53:20.150
So, if there is no disturbance and the load
is pakka matching, this I0 will, I0 coming
53:20.150 --> 53:27.150
from here. So, we have designed the, we are
now we are worried about the DC value. We
53:27.480 --> 53:32.570
have chosen the capacitor such that the fluctuating
part will go through this one and capacitor
53:32.570 --> 53:37.280
value you have chosen such that the voltage
ripples is very minimal. So, the DC value,
53:37.280 --> 53:42.850
it will go here, this I0. So, this both already
we have to use it this way.
53:42.850 --> 53:49.850
You already know IS. So, from the IS, there
is a gain block. This gain block is equal
53:56.430 --> 54:03.430
to sorry, this is the VS not IS, this is VS
into VS peak, IS peak VS peak. So, IS is coming
54:12.990 --> 54:19.990
here. IS we know it, we can generate the IS
value here, get the IS peak value. IS into
54:22.090 --> 54:29.090
VS divide by 2 E0 is our I0. I0 at any instance,
any load change happens if I0 and the I load
54:51.940 --> 54:58.940
are same, then the difference is 0. So, there
is no discharging in the capacitor voltage
54:59.050 --> 54:59.750
wave form.
54:59.750 --> 55:04.360
But suddenly, load changes increase or decrease;
IL I load will change immediately. So, unless
55:04.360 --> 55:10.370
that is indicated to the converter I0, immediately
there is a fluctuation can happen in the capacitor
55:10.370 --> 55:17.370
voltage. So, this difference immediately goes
to the capacitor. That will generate a ripple;
55:17.900 --> 55:24.900
1 by C, this is the integral that will give
the V0 side. This V0 we will feedback to this
55:27.670 --> 55:34.670
one. This is the complete block diagram for
controlling the converter.
55:35.700 --> 55:42.700
So, again I will say how this one is. You
have the E0 reference and E0 feedback; that
55:43.650 --> 55:50.650
we are taking tapping from here with proper
sensors and we are giving it there. With proper
55:51.050 --> 55:58.050
sensors we are taking from here and we are
giving it here. Now, when the load current
56:06.050 --> 56:13.050
and the DC value of the current coming from
here, this converter, when they are equal,
56:13.260 --> 56:18.050
the capacitor voltage will not fluctuate,
capacitor voltage will be steady and the controller
56:18.050 --> 56:19.130
will be in the steady state.
56:19.130 --> 56:26.130
But due to sudden change in load that is what
in your control. This I0 can change but our
56:27.800 --> 56:33.770
controller is giving some steady id. Unless
this is modified, Vr is modified, this will
56:33.770 --> 56:38.630
not change. So, sudden change in load, the
difference will immediately go through this
56:38.630 --> 56:44.700
one and there will be a change in the capacitor
voltage V0. That change in current, difference
56:44.700 --> 56:51.700
in moments, sudden change in load I0 minus
IL load integral 1 by C integral, this difference;
56:51.900 --> 56:55.060
that will feedback to this one here.
56:55.060 --> 57:02.060
Now, the moment change happens, immediately
IS has to change. So, IS you are sensing from
57:03.880 --> 57:09.900
here, IS is given. Any change in the IS through
the controller, we will generate the correct
57:09.900 --> 57:14.630
Vr. We will give the power converter so that
the power converter, reference wave form that
57:14.630 --> 57:19.290
is the modulating wave, we will give to converter
such that this converter will give the correct
57:19.290 --> 57:20.240
Vr.
57:20.240 --> 57:27.240
So, from this equation, Vr, VS minus Vr integral
1 by L gives the current IS here. That IS,
57:32.640 --> 57:37.560
instantaneous value of IS divided by K, we
will feed it here. So here, if you see, this
57:37.560 --> 57:44.140
current is a sinusoidal current and we are
this current, we are controlling with this
57:44.140 --> 57:48.870
one and all our controller here is due to
through PWM. So, we will assume the PWM frequency
57:48.870 --> 57:55.380
is so high. So, during the control action,
we can assume this IC, the sinusoidal part
57:55.380 --> 58:02.380
of this current IC with a small ripple, it
is not changing much. So, we are using a sinusoidal
58:02.880 --> 58:07.880
current reference and sinusoidal current feedback
here.
58:07.880 --> 58:12.070
Then that is the one of the simplest way of
doing the thing. Otherwise, the other way
58:12.070 --> 58:16.450
to convert this converter is alternating current
to the DC equivalent and then do the controllers;
58:16.450 --> 58:21.490
then you require other signalling process.
So here, what we assume? The PWM action is
58:21.490 --> 58:27.720
so fast. During that small moment IS reference
and IS feedback are not, sinusoidal feedback
58:27.720 --> 58:34.720
are not changing. There is not much variation
in the IS. So, what you have to give this
58:35.350 --> 58:36.280
IS value?
58:36.280 --> 58:40.470
Peak value of the IS reference comes; so we
have to give a sinusoidal reference here.
58:40.470 --> 58:47.470
So, with that sinusoidal reference IS should
be in phase with our mains voltage VS. So,
58:48.090 --> 58:55.090
that will take care of the unity power factor.
So, this IS will break here and this controller
59:00.410 --> 59:06.900
because we know, this IS is a sinusoidal current,
so this current also has to be sinusoidal.
59:06.900 --> 59:13.900
So, this sinusoidal current we are assuming
that control is through PWM with high frequency.
59:14.670 --> 59:20.030
During that PWM operation, this IS and IS
star not, there is not much change in the
59:20.030 --> 59:26.280
IS and IS star. So, but we want unity power
factor. So, this IS, what you drawn should
59:26.280 --> 59:28.020
be in phase with VS.
59:28.020 --> 59:34.410
So, we should generate a sinusoidal reference
wave form in phase with this one with unit
59:34.410 --> 59:39.470
amplitude and we should multiply with this
IS and give it here. So, we have to modify
59:39.470 --> 59:46.470
this part. Let us see how we can do that one?
So, the controller has to be accordingly modified.
59:50.810 --> 59:57.810
So, the controller output is here, then we
will generate a sinusoidal signal which is
1:00:06.880 --> 1:00:13.880
in phase with our Vsd that is sin omega t
10, we have to generate, a sin omega t ten
1:00:14.910 --> 1:00:17.100
with unit amplitude.
1:00:17.100 --> 1:00:24.100
This we should multiply, these two we have
to multiply. Multiplication block, I will
1:00:27.510 --> 1:00:33.030
make it like this, this is the multiplication
block. We give it here and this one, you should
1:00:33.030 --> 1:00:40.010
give it here, output should be given here.
So, this sin omega t which is in phase with
1:00:40.010 --> 1:00:45.570
this one that will ensure that this current
drawn IS will be in phase with VS all the
1:00:45.570 --> 1:00:48.990
time and you have to generate Vr.
1:00:48.990 --> 1:00:55.150
So here, what is more important is how to
find out this controller, this controller
1:00:55.150 --> 1:01:00.560
and this controller? What is the function
of this controller such that input should
1:01:00.560 --> 1:01:07.560
track the output or the output should track
the input as close as possible so that any
1:01:09.080 --> 1:01:15.100
change in IS with IS, immediately we have
to generate the Vr. So, that controller has
1:01:15.100 --> 1:01:20.080
to take care of that one. So, let us study
out some of the standard controller like P
1:01:20.080 --> 1:01:23.100
controller is there, PI controller, PID controller.
1:01:23.100 --> 1:01:29.170
So, if you use this controller with how the
whole system, whole the current block; first
1:01:29.170 --> 1:01:34.720
we will start with inner current block, how
inner current block will work. That we will
1:01:34.720 --> 1:01:38.990
study in the next class. So, we will go to
the controller and what are the controllers,
1:01:38.990 --> 1:01:42.250
standard controller, what are its problems?
Then we will try to fit in one of this standard
1:01:42.250 --> 1:01:45.120
controller blocks and we will choose the correct
one.