WEBVTT
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So, last class we were trying to find out
the volt second average that is the average
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variation of the pole voltage waveform for
our space vector PWM. That means for switching
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between inner sectors the vectors forming
the sector that is two active vectors and
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zero vectors.
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So, for sector one, so we found the T0 is
divided into two parts that is T01 and T02
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and traced at the start and end of the sampling
period and the next sampling period, the sequence
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is reversed that means the first sampling
period is T01 T1 T2 T02, next sampling period
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is T02 T1 T2 T01 and T1 and T2 are the periods
during which the active vectors forming the
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sectors or the sectors are switched and T1
is for the vector which is at the start of
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the sector. That means if the rotation is
from the anticlockwise, it is start of the
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sector and T2 is the n vector.
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So, for the general expression, we got for
sector one VA0 average VB0 or VC0 average.
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If you see here, these equations are irrespective
of the sector. So, if you know the T1 T2 in
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a sector, we can find out the average value
in that sector. So, let us see what is the
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average value in sector one. This is easily
we can get by substituting the value of substituting
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values of T1 T2; T1 T2 we have already found
out. So, let us go to the next page.
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Let us again write for our clarity; VA0 average
is equal to in sector one, this is in sector
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one, VA0 is equal to VDC by 2 divided by TS,
TS is the sampling period into T1 plus T2.
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Now, VB0 average that is average variation
that is why we are dividing by TS, this will
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be equal to VDC by 2 divided by TS into minus
T1 plus T2. Since the T0 periods are equally
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divided at the start and end and for the start
if it is 000 is used, end it is 111 and equal
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duration the volt second, they will cancel.
So, C T0 will not come into picture for the
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average variation.
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Now, V0 zero average equal to VDC by 2 divided
by TS into minus, minus T1 minus T2 that means
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this also equal to minus of VS average in
sector one. Now, we have already found out
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the values of T1 T2. Now, three phase channel.
So, for our reference, let us note it here,
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T1. T1 is equal to TS into VS, mode of our
reference space vector divided by VDC that
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is our radii of the hexagonal structure, VDC
into root 2, 2 by root 3 that is sin 60 is
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there so that means sin 60 root 3 by 2; so
this will be 2 by root 3 into sin 60 minus
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alpha. So, alpha is measured from the start
of a sector. So, alpha can vary from 0 to
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60 degree. So, we are finding out the average
variation in a sector. The sector is an equilateral
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triangle formed by the active vectors and
the zero vectors. T2 is equal to again TS
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into VS mode divided by VDC into 2 by root
3 into sin alpha, this is our ...
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Now, let us put these values T1 T2 in equation
1 and 2. So, if A0 and VB0 average if we can
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find out, VC0 average we can find out; it
is only equal to minus VA0 average. So, let
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us say VA0 average, this is equal to VDC by
2 divided by TS into T1 we have put, VS mode
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divided by VDC 2 by root 3 into sin 60 minus
alpha plus TS into VS mode divided by VDC
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into 2 by root 3 into sin alpha. So, this
we can, this is of the form sin A plus sin
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B, we can use the standard trigonometric relations
and we can reduce this one to finally it will
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come to VS by root 3, VS mode divided by root
3 into sin 60 plus alpha.
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Now, let us say what is VB0 average? So, this
is the one we require. So, in a sector with
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alpha vary from 0 to 60 degree, the average
variation of VA0 average varies like this
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that is VS by root 3 sin 60 plus alpha. Now,
let us take VB0 average. Now, let us see the
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VB0 average is equal to again VDC by 2 divided
by TS into TS into VS mode divided by VDC
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into 2 by root 3 sin alpha minus TS VS divided
by we are substituting only the T1 T2 values
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into 2 by root 3 sin 60 minus alpha; so, this
is that. So, if you see here, this will reduce
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to VB0 average will reduce to VB0 average
will be VS sin alpha minus 30; this way it
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will happen this is VB0 average. So now, let
us plot, let us find out this only 60 degree,
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so let us find other sectors how it works.
So, let us go to the next page.
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We know or let us draw our hexagonal voltage
space vector structure; this will be like
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this. So, sector one, they are all equilateral
triangle, my figure is not that clear but
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it is hexagon with equal radii, here. So,
this is our sector one. Then sector two, sector
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three, sector four, sector five, sector six
and switching sequence is 100 110 010 011
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001 then 101. Now, so VA0 average in sector
one, VA0 average is equal to the sampled VS
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that is this is our VS is somewhere here,
so the sampled VS magnitude into by root 3
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sin 60 plus alpha and VB0 average is equal
to VS sin alpha minus 30 degree and VC0 average
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is equal to minus VA0 average.
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See, we want the let us see the average variation
for pole A. If you know the pole A average
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variation; pole B, pole C will be 120 120
degree phase shifted. So, in this phase, see
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our ABC phase will be so far for the conversion,
if AB if this is A, B is in this direction,
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C is in this direction, C A B that means A
is along this axis in this direction. Now
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A phase, so let us take the average variation
along pole A, VA0 average for a cycle that
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means we are trying to plot VA0 average for
a cycle of operation.
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So, A phase is here; so here we are using
space vector, average variation of the voltage
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space vector. So, when it is here, A phase
will have maximum. So, let us take here A
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phase will be minimum. So, omega t is equal
to 0 is here. So, the voltage space vector
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rotates from here; when it comes here, A phase
will have maximum; here it will when it comes
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here, it will be again zero; here it will
come, it will be negative and opposite. So,
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omega t C T0 starts from here. So omega t,
first let us see omega T0 is equal to 0 to
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30 that means in sector 5, what is the average
pole voltage variation for phase A from here
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to here or that means omega t C omega t C
omega t is equal to 0 to omega t is equal
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to 30 degree. How to find out?
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If you see here, the average pole voltage
variation is that switching is from 0 to 1,
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0 to 1 and we know for the inner sector, the
inner sector if alpha is taken from the start
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of the sector for here it is start of sector
is here, for sector one it is here; the T1
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T2 are the same in any sector, only the switching
vectors are different. Now, let us see find
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out the average variation VA0 average variation.
It is varying, the switch vector 0 to 1 here.
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But we have derived the equations for sector
one, so sector one it is 0 to 1 is B phase.
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So, in sector 5 for omega t less than is equal
to 0 to less than equal to 30 degree; see
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now we are talking about the omega t general
variable, VA is or we will say the VA0 average
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VA0 average is the
same as the VB0 average in sector one, VB0
average in sector one. So, what is the VB0
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average? The VB0 average is equal to VS sin
alpha minus 30 degree.
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Now alpha, we have to replace with omega t,
alpha to be replaced with our general variable
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omega t with omega t. So, what is the relation
between omega t and alpha? So, in sector five
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if you take when alpha is equal to 30; see,
alpha we are measuring from the start of the
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sector, so from here to here it is 30 degree.
When alpha is equal to 30, omega t is equal
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to 0. So, what is the relation between alpha
and omega t? So, this indicates that alpha
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is equal to omega t plus 30 degree. So, when
omega t is equal to 0, alpha is equal to 30;
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then omega t is equal to minus 30 alpha is
equal to 0. So, that means in this equation,
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VA0 average in sector 5 from omega t is equal
to 0 to 30, VB0 average for omega t less than
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equal to 0 greater than is equal to 0 less
than equal to 30 degree that means from here
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to here, end of the sector; middle of the
sector 5 to end of the sector 5 that means
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middle of S5 to end of S5, sector 5.
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What is the equation? We have to simply replace
alpha with alpha in terms of omega t that
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is equal to VS is equal to sin of omega t
plus 30 minus 30 degree. So, that is equal
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to VS sin omega t. This shows this is only
valid for omega t less than equal to 30 degree
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less than equal to 0 degree. Now, the next,
what we have to find out? What is the VA0
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average from omega t is equal to 30 to 90
degree? So now, if you know that relation
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0 to 90 degree we have found out, so if the
sinusoidal variation if the average variation
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is sinusoidal variation, then automatically
we can draw the other waveform from the symmetry.
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So, let us find out what is the average variation
for VA0 from omega t is equal to 30 to 90.
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Again, let us go to the next page.
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These are hexagon with radii equal to VDC;
this is the radii, this is 100, 110, 010,
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011, 001, 101. Now, we have found out VA0
variation from omega t that is omega t is
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equal to 0 to omega t is equal to 30 degree,
we know it, VA0 average. Now, we have to find
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out omega t less than equal to 90 degree greater
than or equal to 30 degree that means this
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interval. So, this is if this is sector 1,
sector 2, sector 3, sector 4, sector 5 and
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sector 6 and omega t less than or equal to
30, less than equal to 0 degree, the VA0 average,
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VA0 average we got it from the previous slide,
it is equal to VS amplitude of the sampled
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reference phase vector into sin omega t. So,
this you do not confuse with the continuous
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sin omega t for full cycle; this sin m, this
equation is valid only for our omega t that
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is what do you mean by omega t? Omega t is
the speed with which our reference space vector
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rotates, 0 to 30.
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Now, we have to find out for omega t 30 to
90 degree that is in S6. S6 if you see, the
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variation of the variation of the switches
is from one to one. This is same as in sector
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one also. So, that means here the variation
with respect to alpha, the alpha measure from
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the start of the sector, the VA0 average is
same as in sector 1. Only thing we have to
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change the origin. VA0 average is same as
the sector 1 that means VS sin alpha minus
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30 degree. So, if you see here, for alpha
varies from alpha varies from in sector 6
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is the same as alpha varies from sector 1
because the vector change is one to one here
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and here also one to one.
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So, alpha varies from less than or equal to
0 degree, less than equal to 60 degree but
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omega t varies from less than equal to 90
degree less than equal to 30 degree. This
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shows omega t minus 30 degree is equal to
alpha 0. That means when omega t is equal
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to 30, alpha is equal to 0 here that is here.
Omega t is equal to zero means alpha is equal
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to 0 and the omega t is equal to 90, alpha
equal to 60. Here, omega t is equal to 90
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degree; alpha is equal to 60 degree.
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So here, VS by or here one mistake is there,
VS by root 3 is also there that I forgot to
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write; this is VS by root 3 from the previous
equation. So now, the VA0 average is equal
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to VS mode divided by root 3 into sin alpha
is equal to omega t minus 30. So, this will
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be equal to sin omega t minus 60 degree. So,
VA0 average in sector 1 is VS by root 3 sin
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60 plus alpha, sin 60 plus alpha. Now, alpha
is varying from 0 to 60 degree that is from
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here to here and omega t is varying from 30
to 90 degree; so what is the relation between
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omega t and alpha? Omega t 60 plus alpha that
will be equal to omega t minus 30 is equal
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to alpha that means when omega t is equal
to 30, alpha is equal to 0; when omega t is
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equal to 90, alpha will be equal to 60 degree
that is from here to here. So, when you substitute
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this one, omega t minus 30 here, alpha; equation
will be sin omega t plus 30 degree.
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So, VA0 average for omega t varying 30 to
90 degree is equal to sin omega t plus 30.
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So, what will be the type of waveform this
will generate? Let us try to plot the waveform,
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write down the waveform.
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The required wave from for omega t is equal
to omega t less than 30 degree greater than
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0 degree, this is equal to VA0 average is
equal to VS the sampled VS amplitude into
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sin omega t. Now, omega t less than equal
to 90 degree greater than equal to 30 degree,
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VA0 average equal to VS by root 3 into sin
omega t plus 30 degree. So, these are the
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two equations, 30 degree. So, let us plot
this waveform. So, let us draw the angle;
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this is our y axis and this is our X axis.
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So, alpha is equal to 0 to 30 degree that
means here, this is VS sampled value of VS
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plus sin omega t that means part of a sin
omega t with amplitude is equal to sampled
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value of VS and from 0 to 30 degree. So, this
will be of this form, this one. So, this is
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our 30 degree. After this, from here to 90
degree, this is part of VS by root 3, now
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the amplitude is reduced; VS by root 3 sin
omega t sin omega t plus 30 and omega t varies
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from 30 to 90 degree. If you plot it, it will
be hope something like this. If you fix Vs,
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then VS by root 3 is this amplitude. Now,
if 90 degree; so this again repeats, so if
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you see, if you repeat this one for symmetry,
it will go like this, 0.
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So, the variation of our, see we expected
a sinusoidal variation and we got something
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like this; this is for the A phase. B and
C, it will be 120 degree phase shifted like
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this will be there. If you draw it, we can
get it. Let us say the B phase; B phase will
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be like this, 120 degree phase shifted, VB0
average variation, then VC0 average will
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be like this. We were trying to expect or
we are expecting a sinusoidal variation and
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we got a waveform like this. So, what all
this contains? Let us see.
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So, if you plot the line to line voltage,
let us take the line to line voltage for if
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you use, see we know VA VB VC; if you can
plot the line to line voltage, typical line
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to line voltage will be something like this
it will happen. See, line to line voltage
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is sinusoidal; line to line voltage is the
subtraction of two pole voltages. So, something
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common to this both the waveforms; that waveforms
get subtracted and we got the line to line
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to voltage. Now, what is that one?
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Let us sum all the three phases; if you sum
all the three phases, what will you get? If
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it has contains a triple n harmonics, you
will get a waveform like this. So, that means
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the average variation contains a triple n
order. So, triple end order even if it is
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there, it will not produce an average voltage
variation, average voltage variation only
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due to the fundamental. The fundamental component
of this wave form will be like this; let us
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the typical fundamental component if you see
that we can get from the line to line to voltage.
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The typical fundamental component for the
A phase will be this one. So, that is sinusoidal.
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So, how to find out? See, we got the VA0 VB0
pole voltages; VA0 average variation, VA0
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VB0 VC0. Now VA0 VB0, add VA0 VB0; if it contains
only positive and negative sequence, then
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VA0 VB0 VC0 will be 0. But here it is not
0, we got a triple n order here, all the multiplication
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of the triple n, third ninth all that harmonics
are there. So, you sum it and divide by 3
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that is you get this waveform that is this
waveform. This waveform let us call V3; this
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you subtract from VA0 average that means VA0
VA0 average minus V3, the triple end order
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you remove it, then that will be our the sinusoidal
waveform, this one. So, what is the advantage,
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the sinusoidal waveform?
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Let us say, even though triple n is there,
the fundamental only will generate power.
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So, the moment we add the triple n, the peak
of this one will be slightly suppressed. See,
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this is the level, this level that is this
level. The sin was there, it would have here.
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So, that means and for sin triangle PWM if
you use if you use the same way with the triangle,
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we get extra modulation that means some more
we can increase. So, the space vector computed
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sin triangle, we get a boost in the voltage.
So, how much that boost in the voltage? Let
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us find out.
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This is our V alpha, this is our V beta and
maximum voltage space vector, we can get the
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maximum inscribed circle. This amplitude let
us say this OA, OA is equal to our DC link
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voltage VDC. So, what is the radii radius
of the maximum inscribed circle, this one?
34:04.950 --> 34:11.950
That is inscribed circle touch the hexagonal
peripheral here. So, this will be equal to
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30 degree, this will be equal to that means
OC, OC is equal to VDC into cos 30 degree.
34:23.550 --> 34:30.550
That means cos 30 equal to root 3 by 2, VDC
into root 3 by 2. So, as the voltage space
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vector rotates, that is our VS rotates when
the VS is along V alpha, as it rotates; we
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will have the V alpha, the magnitude of the
alpha component will be maximum. So, V alpha
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maximum is equal to VDC into root 3 by 2.
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So, V alpha component we got. Now, we can
find out what is the VA maximum. That will
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give the maximum, maximum force will from
our space vector PWM. So, let us go back to
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our alpha beta transformation. So, previously
we found that V alpha is equal to VA plus
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VB cos 120 degree plus VC cos 240 degree.
So, what is cos 120? This is equal to VA,
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cos 120 is minus 1 by 2, VA minus 1 by 2 VB
cos 240 also minus 1 by 2 that is minus VC
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1 by 2. This is equal to VA minus 1 by 2 into
VB plus VC.
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Now, we are talking about the sinusoidal components.
For sinusoidal components, VA plus VB plus
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VC equal to 0. So, this will be VB plus VC
equal to minus VA. So, minus of minus, this
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will become plus. So, this will be equal to
3 by 2 VA; so, V alpha maximum. What is V
36:48.550 --> 36:55.550
alpha maximum? V alpha maximum is equal to
VDC into root 3 by 2. So, what is VA maximum?
37:04.420 --> 37:11.420
VDC into root 3 by 2 into from this equation
V alpha is equal to, so if V alpha is not,
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VI is equal to 2 by 3 V alpha. So, V multiplied
by 2 by 3 here. So, VA maximum is equal to
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VDC by VDC by root 3; this is the maximum,
VDC by root that is VA maximum using space
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vector PWM is equal to VDC by root 3 is equal
to 0.577 VDC.
37:47.740 --> 37:54.740
So, you get, in space vector PWM if you read
our sin triangle PWM, we get extra boost in
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this. How this extra boost has come?
38:09.050 --> 38:16.050
For sin triangle PWM, when we use sin triangle
PWM, you have the sin wave and you have the
38:22.770 --> 38:29.770
triangle wave and if you take the basic inverter
configuration, one pole is 0, VA0 maximum;
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the sinusoidal average variation maximum is
equal to 0.5 VDC, sin triangle. Now, space
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vector instead of the triangle, what is the
wave form we are getting? Fundamental is a
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triangle waveform but actual variation is
a sin plus a triple end content that means
39:01.619 --> 39:08.619
something like this; this is the one, this
is sin plus a triple n content that is multiple
39:17.450 --> 39:20.140
of 3, triple n content.
39:20.140 --> 39:27.140
Then the VA0 maximum is equal to 0.577 VDC
that means instead of using this sin wave,
39:37.220 --> 39:43.100
if you use this sin plus triple n; what we
can or we have some extra modulation. If you
39:43.100 --> 39:49.230
see here sin triangle PWM, extra range is
possible here because this is flattened here.
39:49.230 --> 39:54.580
So, that is why we are getting maximum modulation
of 0.577 VDC.
39:54.580 --> 40:01.580
So, let us see, what is or let us study what
is the similarity between sin triangle and
40:05.440 --> 40:11.109
the space vector PWM. Let us go back to our
sine triangle PWM.
40:11.109 --> 40:18.109
I will extend this one for your clarity. This
is A phase, this is our B phase; let us say
40:42.670 --> 40:49.670
this is our C phase. Now, let us say sin triangle
PWM, let us say
triangle PWM something like this it goes.
41:03.490 --> 41:10.490
See, we are using a high frequency triangle
waveform; just for the analysis, we expanded
41:16.980 --> 41:23.980
a waveform like this. Now, our triangle waveform
is going from here to here, it goes like this;
41:28.150 --> 41:35.150
this is our A B C till it reaches this point,
till it reaches this point, all the phases
41:43.920 --> 41:50.920
are let us mark this point is some P, till
it reaches P point, triangle reaching, triangle
41:58.369 --> 42:05.369
reaching the point P. Here, all the ABC phases
are above, the magnitude of till that one,
42:12.590 --> 42:17.430
all the sin greater than the triangle.
42:17.430 --> 42:24.430
So, what happens? All the top switches are
on; all the top switches are on, we are marking
42:26.330 --> 42:33.330
as 111. Now, from P to Q, P to Q let us for
my analysis P to Q, here both are equal; P
42:48.200 --> 42:55.200
to Q what happens? Till that one, A and B
are above triangle but C is below. So, upto
42:59.220 --> 43:06.220
here, this is the sequence, the inverter sequence
will be upto here, upto PQ, the inverter switching
43:09.250 --> 43:16.250
state is 110. Here, triangle upto the P, it
will be 111.
43:27.910 --> 43:34.910
Now, above that one, see here both are equal,
so above that one, the Q to S, Q to S triangle
43:38.700 --> 43:45.700
is greater than the sin wave. So, all the
switches will be it will be 000. So, Q to
43:52.290 --> 43:59.290
S if you see here, Q to S, these are zero
period; this is zero period, this is also
44:01.640 --> 44:08.640
zero period. So, if you see, in sin triangle,
the zero periods which are see vector if you
44:08.820 --> 44:13.980
say here, this vector switching let us take
another switching sequence, let us take another
44:13.980 --> 44:20.980
sequence here. Let the triangle is going like
this, so this I will mark that some triangle
44:25.440 --> 44:32.440
for hit us let us say P Q then R, then S.
44:41.359 --> 44:48.359
So P Q region, P Q region, the inverter states
is all 111 because sin is greater than that
44:59.250 --> 45:06.250
triangle. Then Q to R, Q to R, B is below
triangle; so this will be 101. Then R to S,
45:15.420 --> 45:22.420
then let us see this is T, R to S, R to S
C has also gone the below the triangle. So,
45:25.560 --> 45:32.560
that means 100. Then from S to T, it is 000.
So, if you see here, same like space vector
45:36.320 --> 45:41.700
PWM, it is going from zero state to active
vector, then the next active vector, then
45:41.700 --> 45:48.700
here. So, it is switching between 101 to 100
that means 101 to 100, it is switching in
45:50.080 --> 45:57.080
S6, S6 same like sin triangle space vector
PWM but the zero periods are not equally distributed
46:01.530 --> 46:02.570
in a sampling period.
46:02.570 --> 46:07.490
What is the sampling period? Now, the triangle
which going from P to T that is this period,
46:07.490 --> 46:14.490
half of the triangle is our sampling period;
so during the sampling period, during a sampling
46:23.359 --> 46:30.359
period TS, TS in sin triangle period, the
zero vectors are zero vector periods are not
46:35.119 --> 46:42.119
equal, zero vector periods are not equal in
sin triangle, in sin triangle PWM. But in
46:55.280 --> 47:00.520
space vector, they are zero. But otherwise
if you see when it goes from zero to active
47:00.520 --> 47:05.080
vectors, only one inverter leg is switched.
So, all other factors are taken here, only
47:05.080 --> 47:12.020
the zero vector periods are not equally distributed.
When the zero vectors are equally distributed,
47:12.020 --> 47:19.020
the average variation in the space vector
becomes something like this. So, that shows
47:20.030 --> 47:21.900
this get flattened here.
47:21.900 --> 47:28.900
So, we can extra boost sorry for the triangle
waveform, it will be something like this,
47:30.210 --> 47:36.490
so here. So, we can get extra boost is possible
but triangle only this much only possible.
47:36.490 --> 47:42.200
So, this extra boost is possible that will
make the modulation index slightly increased
47:42.200 --> 47:48.960
in modulation in sin triangle PWM. So, we
have talked about the space vector PWM. In
47:48.960 --> 47:55.960
space vector PWM, we require the instantaneous
amplitude of VS and the alpha that is the
47:56.480 --> 48:03.339
sector in which it states in it in which it
is at the sampling period it is there and
48:03.339 --> 48:08.710
the alpha angle is required to find out the
switching intervals that is T1 T2 T0.
48:08.710 --> 48:14.420
Once you do the switching intervals in various
sectors, we will get average variation sinusoidal
48:14.420 --> 48:21.420
with triplen addition. So, we can get the
extra boost in the fundamental voltage because
48:21.820 --> 48:26.890
trip triplen VA VB VA plus VB plus VC is not
equal to 0. So, that will not contribute to
48:26.890 --> 48:33.890
the space vector generation. So, we get an
extra boost in the modulation index and for
48:35.520 --> 48:42.099
the space vector PWM, we will geT0 0.577 VDC.
For the same with the sin triangle, it is
48:42.099 --> 48:42.780
0.5 VDC.
48:42.780 --> 48:49.550
Now, next class we will study how to generate
the PWM based only on or there are various
48:49.550 --> 48:55.330
schemes available in literature, here we will
proposes scheme based only on sampled reference
48:55.330 --> 49:01.310
phase amplitudes not based on VS. We know
that VS is generated from VA VB VC; so instead
49:01.310 --> 49:07.339
of sampling VS, we will sample at their instant
the VA VB VC amplitude and from that one VA
49:07.339 --> 49:13.500
VB will be rotating sinusoidally. So, from
the VA VB VC, we will find out the timings
49:13.500 --> 49:18.250
T1 T2 a simple and fast algorithm for in digit
implementation; we will be studying in the
49:18.250 --> 49:21.440
next class.
49:21.440 --> 49:27.400
Thank you.