WEBVTT
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Last class we started with the DC to AC convertor,
then we started with single phase, the basic
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configuration of half bridge converter, then
how we will be able to get a rectangular pulse
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of equal on and equal off duration and the
height depends on the DC link. Then we found
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that for DC to AC that means for AC waveform,
we have to control both frequency as well
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as the amplitude. The amplitude depends on,
what we are talking about the amplitude is
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the fundamental component and for the pervious
pulse waveform like this, this type of rectangular
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pulse, what we found is that by varying the
S1 on and S2 off, we can control the frequency
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but we will not be able to control the fundamental.
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Then from the half bridge, we introduced one
more leg here - S3 S4; then the load is connected
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between A and B, so A is called this whole
leg, leg is called the single pole double
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throw switch, so pole A and pole B. So, the
total voltage, the net voltage coming across
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the load is VA0 minus VB0.
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So here, we said, for the pole voltage A0
when it goes to minus VDC by 2 plus VDC by
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2 minus VDC by 2; the total load voltage waveform,
it will go from plus VDC to minus VDC. So,
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that way we have increased the amplitude.
But here the frequency can be controlled by
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controlling the S1 on and then S1 off here
and S2 on here.
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So, how to control the voltage? And, previously
also we talked about the fundamental and the
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harmonics; this will have fundamental third,
fifth, all harmonics and the harmonic amplitude
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will be inversely proportional to its harmonic
order.
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Then we said we found out that for controlling
the voltage as well as frequency, we require
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two degrees of freedom. So, by turning on
S1 and S2 instants, we can control the frequency.
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But independently controlling the S3 and S4
and the phase difference between that switch
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sequence of pole A with respect to pole B;
by controlling that one, we can control the
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output voltage also. Here, we have done that
way; here it is shown, how to control that.
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This is the pole VA0. So here, like the half
bridge converter, we are not splitting the
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DC link into VDC by 2 into VDC by 2. Instead,
we are measuring the AB with respect to a
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common point O. O, we assume to be at the
centre point of the DC link. So, fictitious
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centre point; that we call as VA0, it will
again it will be VDC by 2 minus VDC by 2 and
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given a delay.
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The previous circuit if you see here, we are
given full 180 degree for S1 and S3 on. When
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this S1 is on here, S3 is on here, the red
line. So, 180 degree we have given here. Now,
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instead of 180 degree, slightly less than
180 degrees we have given here, this alpha.
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So, the moment alpha is given, there will
be a period during which both S1 and S3 are
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conducting. S1 and S3 are conducting means
A and B will be shorted to this point. So,
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output will be 0. So, that zero point that
zero interval is placed symmetrically between
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this notches. So, increasing the alpha, this
notch width increase or that pulse point,
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this will slowly reduce and the output fundamental
will reduce. Then, we found out the harmonics;
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how to find out the harmonics and we found
out how the fundamental varies.
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So, we found out fundamental will be proportional
to alpha by 2. So, we got a voltage control
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by varying that overlap angle alpha but it
is not proportional to alpha, it is proportional
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to cos alpha by 2. So, for any linear control,
we want a function which is proportional to
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the output fundamental V. So, there we said,
we will talk about the sin triangle PWM. So,
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before coming to sin triangle PWM, let us
go to, this is only for single phase; how
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we can get three phase?
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Three phase inverter, we studied the basic
operation. So previously, let us see all this
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three phase operation is extended, is in proper
extension of our single phase half bridge
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converter. So, this three phase structure
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can be realised by properly extending our
single phase half bridge converter, half bridge
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inverter.
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See, let us again go back to our basic inverter
scheme, VDC by 2 VDC by 2. This is our centre
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point O, we have the switches and it is assumed
that it is bidirectional, freewheeling diode
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is also there. Now, we have the A here, pole
A and we said we can add any number of legs
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to this inverter and each leg can be independently
controlled and it is not related to the switching
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of other leg.
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So, for three phase; what is the next we should
do? We should add three more legs here. So,
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this is A, this our three more single pole,
this is C, single pole double throw switch.
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That means this pole B can be either connected
to this point positive rail or to the negative
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rail by appropriately switching the devices.
Now, where we connect the load?
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So, let us take same this point. We connected
here. This is B. What will happen to C? Again,
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C will be, we can do it this way. Then we
want a three phase inverter, three phase we
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require means we require symmetric three phase
that means the waveform should be exactly
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same, the positive half and negative cycle.
But there is amplitude and the duration. Also,
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the B and C phase should be appropriately
phase shifted, three phase should be B phase
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should be 120 phase shifted with respect to
A and C phase will be 120 degree 240 degree
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phase shifted with respect to A.
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So, that means B, the pattern the VB0 phase
shift or the switching sequence should be
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or the the switching sequence, switching sequence
should be 120 phase shifted with respect to
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A that is delay, a delay of 120 degree with
respect to VA0 waveform. Then what will be
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for phase C? VC0 phase shift 240 degree or
120 degree with respect to VBA or 250 240
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degree delay with respect to VA0. So, let
us draw the pattern. Now, let us go to the
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next page.
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So, for the reference, let us again draw our
inverter, basic schematic here. This is A,
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this is our load for A, zero, VDC by 2, then
V, then C. This is connected here and the
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load is connected here. Let us draw the waveforms.
Now VA0, this is our VA0 X axis. So, let us
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take 180 means this much. So, VDC by 2 here,
so we have 3 here, 3 here, again 3 here, 3
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here. So, minus VDC by 2, this is our VA0.
So here, S1 on or top switch is on. So, this
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is one and when the bottom switch is on, we
have minus VDC by 2. So, this is plus VDC
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by 2 from here to here and from here to here,
it is minus VDC by 2. This we have studied
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last class.
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Now, what we want? The same sequence, so we
are talking about each pole, the top switch
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and the bottom switch. Now, let us draw the
B phase. So, for clarity, we will use a different
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colour. Let this be the axis for the B phase,
so 120 degree phase shifted. We know that
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this is zero, this is pi and this is 2 pi.
So 120, this is pi means this is pi by 2.
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So, this we can divide into three divisions;
so this is 60, 60 each. So, 120 mean it will
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come here. So, after the A phase top switch
is on, 120 degree phase shifted, we should
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start the pole B operation. So, after 120
degree, again the top switch is on. So again
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here, it will go like this. So, when the bottom
switch, here, when this is turned on, it is
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minus VDC by 2. Only thing is the durations
for the top switch and bottom switch of A
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and B is the same, only the sequence is delayed
by 120 degree. It goes like this.
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So, this is the pole B with respect to the
fictitious centre tab VB0 plus VDC by 2 minus
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VDC by 2 VB0. Now VC0, let us take a different
colour for VC0, let us take green. This is
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the X axis, so it should be the top switch
of the starting instant for the C phase should
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be 240 from A or 120 from B. So, 120 from
B this is 120, 120 means 2 pi by 3. 120 means
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again here; this is 120, one more 120 means
it will be from here, this 240. So, C phase
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starts from here.
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Now, this is the pole voltage C that is VC0
shape. So, if you see the shape, the positive
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the positive amplitude, positive duration
for the A phase, B phase, C phase are the
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same. They are same wave from but there is
a time delay or phase shift of 120 degree
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with respect to each other is there. Now,
here also like before, our single phase half
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bridge converter, it has the fundamental V
harmonics if you say, it has the fundamental
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third, fifth, seventh, ninth all harmonics
will be there for each pole voltage waveform.
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Now, as I told, we are more interested in
the fundament and we want to suppress the
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low frequency, high amplitude low frequency
harmonics as much as possible. So, the next,
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after the fundamental, the most a predominant
harmonic is 3. It has the amplitude inversely
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proportional to the harmonics, so it will
go like this. If fundamental is there, third
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will be here, fifth will be here, this way
it goes.
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So, how we can eliminate the third harmonics?
Is it possible? Let us start with third harmonics;
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see third harmonics if you see here, fundamental
moves by 120 degree. When the fundamental
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moves by 120 degree means that is let us say
the fundamental for this one is something
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like this; the third harmonics, how much it
will move? Third harmonics how to find out?
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For fundamental 120 degree for third harmonics,
it will be 360. That means third harmonics,
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if the third harmonics starts from here, 120
degree; see it will be again here, it will
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go like this and come here. So, 120 degree
from here, it is the same thing starts as
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if it starting from here. So, that means third
harmonics is concerned, it will be in same
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phase.
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So, if you draw here, here also if you see
here that the third third harmonics; this
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is the third harmonics for this one, we will
draw with this, the third harmonics and what
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about the third harmonics for this one? Third
harmonics for this one from here to here,
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if it is starting from here, third harmonics
here, 120 degree, it will have the same effect.
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So, third harmonics if you see, so let us
say third harmonics starting from here; so,
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that means
the third harmonics for all the three phase
waveform are in same phase because 120 into
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3 is equal to 360 degree.
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So, this waveform also from here to here,
it is 120 degree phase shifted. So, starting
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from here, it is same; exactly the way it
starts here, it will start here. So, if you
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see here, all the third harmonics are in the
same phase, there is no phase shift. So, this
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will be true for ninth multiple of three,
ninth. All these are with three ninth, then
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what you say multiple of 3 -- 27, all these
harmonics, 15, all odd multiple will have
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all the multiple of 3, 3, 9. These types of
harmonics will have will be in the same phase.
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So, these are called triplen order.
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Any way because of the symmetry triplen order
even multiple 6, 12, all those harmonics will
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not be there because the waveform has odd
symmetry. For 6 and 12, it is even symmetry.
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Even symmetry means if you take the waveform,
the centre point left half and the right half
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will be same. But here the left half and right
half will be the same but it is inverted.
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So, it is called odd symmetry.
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So, all the odd multiple of 3 will be absent
here. That means all the triple n order will
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be absent here sorry triple n order will be
in the same phase. So, all the triple n order
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harmonics are in phase, in phase in all the
pole voltages, in the pole voltages, pole
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voltages. So, they are in same phase means
if you see here, let us say the voltage waveform
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is like this; due to this voltage if the current
exceeds, the current will be if you see here,
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the sum of these three, the sum of these three,
only the triple n order if the current exists
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or the harmonics is V third, V3, VA0 third
plus VB0 third plus VC0 third.
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If you add all these harmonics, it is not
0. For symmetrical three phase, what we used
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to say, VAN plus VBN plus VCN phase voltage
should be 0. But if you see, this is not zero.
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So, for this connection, IA plus IB plus IC
triple n order if you see, the current will
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be coming in same phase. So, if IB is coming
in this direction, the IA triple n current
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will also be in this direction and triplen
order C phase also will be in this direction,
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it will sum here and it is not zero. So, it
will come back to the phase through the DC
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link if the current exists there.
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Now suppose, if I break this one; now this
is for the triple n, now if I draw the fundamental,
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let us draw the fundamental with a different
colour. Fundamental, here it is drawn; this
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fundamental will be approximately like this.
Here the fundamental will be like this; if
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you take the fundamental three fundamentals
if you sum it, the VA0 1 plus VB0 1 plus VC0
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1 will be zero. Triple n will not be 0, the
fundamental will be 0. For fifth and seven,
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fifth and seventh will also will be 0. So,
if all these, if I break this one, the triple
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n current which is coming here this direction,
they will sum here and if you remove this
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connection to the central point, what will
happen? There will not be any return thought
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for the triple n. So, what is meant by this
one? So, let us redraw this one again, once
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again.
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This is A, this is B, this is C; so we found
for the load. We will draw it in a different
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way; this is A, this is B and this is C. Previously,
this point, we have connected to O. That means
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previously, this point was connected to this
side. So, we found that if the current triple
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n that is IA 3, IB 3 and IC 3, 3 means triplen
order; all will come here, it is not zero,
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it will sum up and it will go like this. So,
when this connection is there, the triplen
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order will go here. That is in this current
will be IA 3 plus IB 3 plus IC 3; this current
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will flow through this point. These triplen
voltages are there, they are in the same phase
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across this one.
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Now, if we remove this one that means I will
remove this path, then what happens? The triplen
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order third, ninth, fifteenth, all that order
currents will not flow through even though
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voltage is present. What you mean by voltage
is present? Still our VA0, VA0 is like this;
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we will draw once again. This is our VA0,
this our VA0, even though that centre point,
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fictitious centre point connection is removed
that is this connection, broken line we will
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show, that shows this is removed; this connection
is removed, so still we can according to our
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switching sequence, we can turn on top and
bottom and we can measure the voltage with
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respect to A0, VA0 is like this and VB0 120
degree phase shifted.
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So, let us draw the VB0 waveform after 120
degree phase shifted. So, it will be starting
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from here
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and VC0 waveform will be here. See, VC0 120
degree phase shifted with respect to B or
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240, so it will start from this is 240, this
is our VC0. Now, if you see here, let us mark
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this 60 degree points; these are the 60 degree
points, here also
here the 60 degree points are there. So, what
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we did? We removed the triple n order harmonics
current by removing this connection. So, without
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using filter, a simple connection removed
all the triple n order currents. But triplen
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order voltages are there in the pole voltages.
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That means by removing this connection, now
let us mark this point as N. So, this is our
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N. So, the connection, by removing the connection
NO, all the
triple n order currents are zero through the
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load in IA IB IC; so, by removing this one.
But still the pole voltages are same that
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means in our pole voltages VA0 VB0 VC0, all
the triple n order voltages will be there.
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So, what we are more interested is in the
current.
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So now, in this phase current IA IB IC 3,
we have only fundamental, fifth, seventh,
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eleventh, 13, all these harmonics only will
be there. Now, the question is previously
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with the NO connection, our load voltages
and the pole voltages were same. Now, the
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load voltages and VAN, now the load voltages
are VAN VBN and VCN; so we have to find out
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what is VAN VBN VCN load voltage. How to find
out? Let us find out. VA0 is known, VA0 is
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equal to if you measure, it is VAN plus VNO.
Previously, because of the connection VNO
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was 0 because it was shorted; now it is open,
so there is a voltage VN.
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Now, what is VB0? VB0 will also be equal to
VAN plus VNO, VC0 also is equal to VAN plus
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VNO. But what we want is VAN sorry this is
VB0 is equal to VBN, VC0 is equal to VCN.
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Let me correct it; this is VCN, VNO is the
same. What we want? VBN VAN VBN and VCN; how
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to find out? See, I said, in current which
is passing through IA IB IC are one, fifth,
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seven, eleven, thirteen harmonics and we know
for a three phase system, the triple n order
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will not be 0, all the fundamental will be
0 and we can see all the harmonics fifth,
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seventh, eleventh, thirteenth if you take
the fifth, seventh, eleven, thirteen harmonics
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individually from VAN VBN VCN, this will also
sum to 0; but VA0 contain triple n but this
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waveform VAN VBN will not contain triple n
order.
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So, according to Kirchhoff's law, where
the triple n order will be? The triple n order
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will be always dropped across NO. So, all
the triple n order will be dropped across
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VNO. So, before coming to naught one; see
we have marked the switches but we had not
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numbered it. So, let us take starting from
omega T is equal to 0, starting from here,
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we are turning on the switch, top switch of
phase A. So, this is one.
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After that, after 120 degree, top switch of
B is turned on here, the next turn. So, here
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it will be this is one; so at this point omega
T is equal to zero, we are turning on phase
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A, top switch. So, one is marked top switch.
Then, after this one, the transition happens
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here that means VC0 going from positive to
negative that means at this instant, the negative
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C phase C pole switch is turned on. So, this
will be number two; negative means this is
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one, this is number two.
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Then after this, again we are turning on B
phase top switch. So, this is top switch is
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number three. So, three is here. So, here
we have one, here you have two, here we have
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three. Then after this, here we are turning
on the bottom switch of A phase. So, this
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will be four. So, this we can say one, this
we can say two full this duration, this will
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be three, this full duration will be four,
so this is 4.
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So, here we are turning on four. After four,
we are turning on this side that is top switch
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of C phase; this is five, so this is five.
So, let me correct this one. So, five is here.
40:13.869 --> 40:20.869
Now, after this one, we are turning on the
bottom of B phase. So, this is 6, so this
40:24.460 --> 40:31.460
is 6. So, the sequence is same like thyristor
135, 462. So, top switches are marked 135,
40:32.460 --> 40:39.460
462 and you can see, every 60 degree there
is a change. That is true also for symmetric
40:39.550 --> 40:45.490
operation. There are switch 6 switches, so
in a symmetric if you have to turn on the
40:45.490 --> 40:52.490
switches, every 60 degree we have to turn
on a switch and turn off a switch. So, this
40:55.200 --> 40:57.390
sequence is 135, 462.
40:57.390 --> 41:04.390
Now, as I told before, this pole voltage waveform
VA0 VB0, it will contain fundamental triple
41:06.980 --> 41:13.980
n all the harmonics. But the moment NO is
removed, all the triplen harmonics are harmonic
41:15.010 --> 41:22.010
currents are removed. Now, the new load voltage
waveform VAN is equal to the relation is VA0
41:22.220 --> 41:29.220
is equal to VAN plus VNO. So, if you do the
Kirchhoff's voltage law for the loop, VA0
41:29.339 --> 41:36.339
content all harmonics. So, this harmonics
should be dropped either in VAN or VNO.
41:36.430 --> 41:43.430
Now, VAN since the NO is removed, VAN will
not have triplen order harmonics. So, VAN
41:45.760 --> 41:52.760
should not contain VAN triple order voltages
also. If voltage is there, there should be
41:52.809 --> 41:59.390
a current and drop to take care of the Kirchhoff's
law. So here, what happens? VAN will not contain
41:59.390 --> 42:06.390
all the triple n order. So, where the VAN
will go? VAN will all go to VNO. VNO contain
42:06.559 --> 42:13.559
all the triplen order. Now, we want what is
V; we want to find out VAN VBN VCN. Let us
42:18.270 --> 42:19.950
find out.
42:19.950 --> 42:26.950
So, VAN VBN VCN, we have got the pole voltages
VA0 is equal to VAN plus VNO and VB0 is equal
42:38.940 --> 42:45.940
to VBN plus VNO and VC0 is equal to VCN plus
VNO. Now, we know that all the triplen order
42:56.369 --> 43:03.369
will be dropped across VNO. So if you sum
it, VA0 plus VB0 plus VC0 is equal to, we
43:13.050 --> 43:18.260
know VAN VBN VCN will not contain the triple
n order, so all the harmonics when you sum
43:18.260 --> 43:24.550
it with the 120 degree phase shift including
the fundamental, it will be VAN plus VBN plus
43:24.550 --> 43:31.550
VCN will be zero and triple n order is not
zero, they are equal; so it is VNO. So, this
43:33.430 --> 43:40.430
shows the value VNO, VNO is equal to 1 by
3 VA0 plus VB0 plus VC0.
43:48.270 --> 43:55.270
Now, let us substitute this one to this equation
one; let us, see this is 2, this is 3, this
43:59.290 --> 44:06.290
is 4. So, substituting equation 4 in 1; now
our new VA0 is equal to VAN plus VNO is equal
44:16.790 --> 44:23.790
to 1 by 3 into VA0 plus VB0 plus VC0. That
means VAN is equal to VA0 minus 1 by 3 VA0
44:38.440 --> 44:45.440
because this has to go to other side and minus
of 1 by 3 into VB0 plus VC0, we can write
44:51.609 --> 44:58.609
it like this. This is equal to VA0 minus 1
by 3 VC, 2 by 3 VA0 minus 1 by 3 VB0 plus
45:05.180 --> 45:12.180
VC0, this is equal to our VAN value. So, in
a similar way, we can write down the VBN and
45:17.210 --> 45:24.210
VCN also, using the symmetry. So, what will
be VBN? VBN is equal to again 2 by 3 VB0 minus
45:32.530 --> 45:39.530
1 by 3 instead of VB0, VA0 plus VC0. Similar
way, VCN also we can do it.
45:47.069 --> 45:54.069
Now, let us draw what is our VAN waveform.
VAN is equal to 2 by 3 VA0 minus 1 by 3 VB0
45:54.800 --> 46:00.849
plus VC0. Let us go to go back to our waveform,
we will try to draw that. See, let us draw
46:00.849 --> 46:07.849
the waveform. See, VAN if we know, we can
find out VBN VCN also because 120 degree phase
46:09.210 --> 46:16.210
shifted. So, let us take VAN first 60, these
are the 60 degree points; so, first 60 degree.
46:24.220 --> 46:31.220
That means this is equal to 2 by 3 VA0. Let
us take VAN is equal to VB0 plus VC0. So,
46:35.869 --> 46:42.869
first 60 degree VB0 and VC0 are equal and
opposite and it will cancel. So, VAN will
46:43.270 --> 46:50.270
be 2 by 3 VA0; 2 by 3 VA0 means this is VDC
by 2, VDC by 2 into 2 by 3 is equal to VDC
46:52.040 --> 46:59.040
by 3. So, let us take VDC by 3 is this much
because that is equal to 2 by 3 VDC, 2 by
47:06.829 --> 47:13.829
3 VA0; 2 by 3 VA0 is equal to VDC by 2 and
2, 2 cancel and VDC by 3.
47:16.510 --> 47:23.510
Now, during the next 60 degree that is from
here to here; so, from here. So, from the
47:24.569 --> 47:31.569
next sixty degree if you see, VB0 and VC0
are is same phase. So, VA0 plus VB0, VA0 plus
47:37.099 --> 47:44.099
VB0 will be minus of because both are minus;
VDC by minus VDC by 2, minus VDC by 2, so
47:50.690 --> 47:57.690
minus VDC. So, the VA0 plus VB0 will be minus
of minus 1 by 3 into VDC, minus VDC. So, it
48:02.430 --> 48:09.400
will be equal to 1 by 3 VDC and here again
it will be 1 by 3 VDC by 2.
48:09.400 --> 48:16.400
So, here if you see here, this will go to
2 VDC here. So, this will be 2 times into
48:25.520 --> 48:32.520
VDC by 3. Again, again if you see, here it
will be like this. So, the waveform will be
48:35.059 --> 48:42.059
if this is the X axis, the waveform will have
a shape like this; every 60 degree there is
48:42.579 --> 48:49.579
a change and it goes like this
and this duration is VDC by 3 amplitude and
this amplitude is equal to 2 VDC by 3. So,
49:06.470 --> 49:12.359
this is also called as six step waveform.
Why? For one cycle that is from here to here,
49:12.359 --> 49:19.359
there are six steps are there; this is 1,
2, 3 then 4, then 5, then 6, it comes here,
49:24.819 --> 49:29.150
starting point. Again, it starts here, one
level. So six, these are called, this are
49:29.150 --> 49:36.150
all called as six step waveform.
49:38.309 --> 49:45.309
This waveform will contain fundamental sorry
fundamental, fifth, seventh, eleventh, all
49:53.710 --> 50:00.710
harmonics, all the triplen will not be there.
This is our VAN waveform, VAN. VBN VCN will
50:02.030 --> 50:09.030
be 120, 120 degree phase shifted. So, if you
see here, to get this six step waveform, we
50:11.829 --> 50:16.250
have not done anything, we had done only a
very simple thing.
50:16.250 --> 50:23.250
The VNO connection, the neutral connection
we removed. That is we only removed this neutral
50:27.900 --> 50:30.829
connection; from here to what is the connected
centre point.
50:30.829 --> 50:37.829
Once you remove that one, all the triple n
order harmonics are removed because triple
50:38.109 --> 50:44.760
n order harmonics are in same phase. So, it
will sum up. So, unless there is a return
50:44.760 --> 50:49.460
path, it will not be able to flow. So, by
removing the return path that is connecting
50:49.460 --> 50:55.839
this point to this O point, we are removing
the triple n order. But in the pole voltage
50:55.839 --> 50:57.809
waveform, the triple n harmonics are there.
50:57.809 --> 51:04.809
Now, the question is what is our phase voltage
waveform? So, the phase voltage waveform that
51:06.220 --> 51:13.220
is VAN, VAN plus the triplen order will be
VA0. So, all this NO, VAN will be dropped
51:14.680 --> 51:21.680
across the load neutral point and our DC link
midpoint, we can measure. That will give the
51:23.819 --> 51:30.250
triple n order voltages and the phase voltage,
we got like this. So here, all the triplen
51:30.250 --> 51:35.740
order harmonics we have eliminated. But in
many of the motor drive applications, we have
51:35.740 --> 51:41.270
to eliminate other harmonics also, low order
fifth, seventh, 11, 13. At the same time,
51:41.270 --> 51:46.380
we have to control the fundamental. So, how
we can do this one? What is the technique
51:46.380 --> 51:47.650
we have to use it?
51:47.650 --> 51:54.049
See, here also because the pole voltage, we
can do independal switching and we are using
51:54.049 --> 52:00.869
equal degree on equal duration off but how
to control the amplitude? To control, now
52:00.869 --> 52:05.740
from this waveform if you want to control
the amplitude, we have to change the DC link.
52:05.740 --> 52:10.510
But DC link changing is very difficult. Suppose,
a battery operated vehicle for high dynamics,
52:10.510 --> 52:17.319
we cannot keep on changing the battery. So,
DC link we call it a stiff DC link but for
52:17.319 --> 52:23.410
many motor drive applications along with frequency,
the fundamental amplitude also we have to
52:23.410 --> 52:27.319
vary. At the same time, the harmonics also
we have to suppress.
52:27.319 --> 52:31.670
Now, the triple n order we have eliminated
but still the other harmonics are fifth, seventh;
52:31.670 --> 52:38.280
how to suppress? There again we use the PWM
technique, sin triangle PWM technique or other
52:38.280 --> 52:45.280
PWM techniques are there. We will study various
PWM techniques in the coming classes.