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00:01:06,000 --> 00:01:12,300
Last class, we concluded with three phase
mid point configuration and the firing angle
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00:01:12,750 --> 00:01:19,750
range for inductive load. Now, let us take
the discontinuous operation of this three
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00:01:21,470 --> 00:01:26,729
phase mid-point configuration. This will happen
when you have a resistive load; this is for
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00:01:26,729 --> 00:01:33,729
understanding the firing scheme of how the
converter works.
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00:01:34,450 --> 00:01:41,450
So, operation with resistive load with resistive
load; we are talking about 3 phase midpoint
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00:01:54,280 --> 00:02:01,280
configuration. So, the three phase midpoint
configuration that last time we have drawn,
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00:02:01,890 --> 00:02:08,890
it will be e1, e2, e3, this is our neutral,
e1, e2 and e3. So, we have the small leakage
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00:02:22,740 --> 00:02:29,740
inductors. Take care of the finite on and
off duration so that we can avoid the dead
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00:02:30,150 --> 00:02:37,150
short circuits between the phases. So, these
are highly inductive load.
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00:02:43,300 --> 00:02:50,300
Now, in this case, suppose the load is resistive
that means resisting, load is resisting; in
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00:02:54,240 --> 00:03:01,240
that case, what will happen? Let us draw the
three phase waveform. This, with practice
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00:03:10,710 --> 00:03:17,710
you can do it; A phase, this is the B phase
and this is the C phase. So, from this, let
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00:03:45,440 --> 00:03:52,440
us this is okay, this is our e1 this is our
e2 or e3 it can be.
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00:03:59,610 --> 00:04:06,610
Now, last time we told that the firing angle
starts from here alpha is equal to 0 here
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00:04:09,470 --> 00:04:16,470
and there are three devices. So, each device
will be conducting for 180 degree no sorry
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00:04:17,959 --> 00:04:24,959
120 degree. So, alpha is equal to 0, this
is the envelope that is the e1 envelope coming
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00:04:29,440 --> 00:04:34,740
across the node when alpha is equal to 0.
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00:04:34,740 --> 00:04:41,740
Now, let us take when alpha is equal to 30
that means alpha is equal to 30, suppose next
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00:04:42,600 --> 00:04:48,360
phase we are going to switch to alpha is equal
to 0 or efi is going to switch on at alpha
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00:04:48,360 --> 00:04:53,380
is equal to 30 that is here; alpha is equal
to 0 here, this alpha - alpha is equal to
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00:04:53,380 --> 00:05:00,380
0 here, alpha is thirty here. Then what will
happen? Immediately it will go here, it will
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come here; like this.
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00:05:05,390 --> 00:05:12,390
Again, the next phase starts here. So, for
resistive load, if you see here the voltage
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00:05:13,290 --> 00:05:20,290
wave current waveform will be in phase with
the voltage waveform. So, if slightly voltage
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00:05:20,530 --> 00:05:26,030
goes negative here, it is not possible here.
There is no inductive load, so the inductive
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00:05:26,030 --> 00:05:31,400
current - di dt cannot change, the slot current
change and you cannot force the thyristor
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00:05:31,400 --> 00:05:35,370
in the on state. So here immediately, the
moment voltage come to 0, resistively current
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00:05:35,370 --> 00:05:41,680
will also come to 0 because voltage and current
are in phase. The thyristor will switch on
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00:05:41,680 --> 00:05:46,630
immediately. But for voltage control, alpha
as I said told before, alpha can go from 0
30
00:05:46,630 --> 00:05:48,180
to 180 degree.
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00:05:48,180 --> 00:05:54,780
Now, let us take alpha more than 30 degree,
alpha more than 30 degree. So, I will again
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00:05:54,780 --> 00:06:01,780
re draw for clarity. This is our A phase,
this our B phase, this is our alpha more than
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00:06:14,550 --> 00:06:21,550
30 degree, alpha 30 here more than let us
take at this portion. So, the envelope will
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00:06:22,669 --> 00:06:29,669
be like this. At this point, the commutation
stops, the current thyristor switched off.
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00:06:30,889 --> 00:06:37,889
But the next thyristor again will be turned
on here only. So, the voltage ripple coming
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00:06:39,010 --> 00:06:46,010
across the load is different from that of
an inductive load. So, this is that portion.
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00:06:48,570 --> 00:06:55,570
What it shows? With resistive load, there
is a discontinuous conduction depending on
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00:07:02,300 --> 00:07:08,370
the firing angle. That means alpha more than
30 degree, there is a discontinuous conduction.
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00:07:08,370 --> 00:07:15,370
Now, what will happen as alpha increases?
The repel part, the input mains; what are
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00:07:17,120 --> 00:07:21,840
the mains coming across the load will slowly
get reduced at alpha is equal to 180 degree,
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00:07:21,840 --> 00:07:26,040
the output repel will be coming across load
is 0 that means output voltage is equal to
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00:07:26,040 --> 00:07:33,040
0. So, this way also a resistive load, we
can have a 0 to positive voltage variation
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00:07:37,870 --> 00:07:38,449
possible.
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00:07:38,449 --> 00:07:45,400
Now, see is it totally possible with a resistive
load? Can we have a discontinuation with discontinuity
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00:07:45,400 --> 00:07:52,400
hope with an inductive load? It is possible
that means when the voltage waveform goes
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00:07:53,570 --> 00:08:00,570
negative, we should give alternate path for
the inductors so that instead of forwarding
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00:08:02,419 --> 00:08:07,260
the thyristor and the current flowing through
the thyristor, it can free feel through another
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00:08:07,260 --> 00:08:08,389
path. So, it is possible.
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00:08:08,389 --> 00:08:15,389
Suppose, the load is inductive or see, if
you give a diode like this; this is inductive,
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00:08:19,500 --> 00:08:26,500
load is inductive and with diode. The moment
the voltage waveform, this is positive, it
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00:08:28,190 --> 00:08:35,190
can only positive, any time it is negative;
what will happen? Current will free feel through
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00:08:35,380 --> 00:08:42,380
this way. This is possible. So, this way we
can ensure. Suppose, output voltage we want
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00:08:43,180 --> 00:08:48,399
to limit it, we do not want output dc voltage
or instantaneous voltage, we need not want
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00:08:48,399 --> 00:08:55,399
to go to negative, we can use a free willing
diode and with the firing angle variation,
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00:08:56,129 --> 00:09:02,319
we can have a output voltage variation from
0 to positive.
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00:09:02,319 --> 00:09:09,319
Now here, what will be the output dc voltage
equation with respect to the firing angle?
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00:09:12,050 --> 00:09:19,050
That we can find out here that is E0 alpha.
Again for a general m phase, it will be 1
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00:09:22,809 --> 00:09:29,809
by 2 phi m. So, m by 2 phi integral; so integral
- lower limit and upper limit depending on
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00:09:32,259 --> 00:09:36,660
whether we are sine omega t or cos omega t.
So, let us take Em - maximum value of the
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00:09:36,660 --> 00:09:43,660
phase voltage and cos omega t d omega t. That
means we are finding the area in the period
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00:09:45,300 --> 00:09:49,740
and divide by the period. That will give the
average value that will give the dc value.
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00:09:49,740 --> 00:09:56,059
Now, what will be the lower limit? Sine omega
t means, cos omega t means here. So, firing
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00:09:56,059 --> 00:10:01,740
angle minimum is here. So, it will be this
distance, from here to this distance. This
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00:10:01,740 --> 00:10:08,740
will be for three phase, it is phi by 3; for
m phase, it is phi by m. So, lower limit is
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00:10:11,249 --> 00:10:18,249
if omega t is equal to 0 here, omega t is
equal to 0 here, for cos omega t waveform.
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00:10:23,499 --> 00:10:30,499
So, when you go to the lower limit, it will
be minus phi by m then plus alpha. Alpha can
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00:10:32,290 --> 00:10:39,290
vary from 0 to vary from here to here for
symmetry operation that is that we have studied
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00:10:43,259 --> 00:10:46,800
in the last class for example inductive load.
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00:10:46,800 --> 00:10:53,800
But for a resistive load, the upper limit
is for a cos omega upto here. You know that
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00:10:55,100 --> 00:10:58,809
output repel is not coming across the load,
this can be used as conductor. So, upper limit
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00:10:58,809 --> 00:11:05,809
is phi by 2. So, if you do the integration
here, finally you will get m by 2 phi into
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00:11:08,220 --> 00:11:15,220
Em into sine omega t minus phi by m plus alpha
upper limit phi by 2. So, final equation in
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00:11:22,709 --> 00:11:29,709
terms of rms value - root 2 E, E here is the
rms value m by 2 phi into 1 minus sine alpha
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00:11:37,449 --> 00:11:43,350
minus phi by m. This is the general equation.
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00:11:43,350 --> 00:11:50,350
So, for a three phase waveform, for the three
phase case, three phase midpoint configuration
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00:11:52,629 --> 00:11:58,459
with the resistive load or inductive load
with the free willing diode; why free willing
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00:11:58,459 --> 00:12:05,459
diode? We require the thing to go through
the diode only. The output repel waveform
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00:12:07,610 --> 00:12:14,610
will be 0 to positive. Negative, it will not
be going, we can plot it. This will be for
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00:12:18,749 --> 00:12:20,980
a three phase case.
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00:12:20,980 --> 00:12:27,980
For a three phase case, from the previous
equation that is E0 alpha, it will be equal
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00:12:32,889 --> 00:12:39,889
to root 2 E divide by 2 phi into m into 1
minus sine alpha by phi by m.
For a three phase case, we can put all m is
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00:12:58,259 --> 00:13:05,259
equal to 3. So, approximately the curve, approximately
will be something like this; at omega t is
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00:13:13,660 --> 00:13:18,389
equal to 150 degree we will have to start
this continues conduction. Maximum will happen
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00:13:18,389 --> 00:13:25,389
alpha is equal to 0 degree. This is E0 alpha
by E rms, E0 alpha by the maximum value that
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00:13:30,519 --> 00:13:37,519
is what is the maximum value? Ed0 maximum
here; so when you do it, it will go like this.
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00:13:40,639 --> 00:13:47,639
Now, we have studied about 3 phase midpoint
configuration. Now, again for high power applications,
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00:13:51,980 --> 00:13:58,980
also to increase the output dc voltage and
also to reduce the repel content that means
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00:14:01,040 --> 00:14:07,100
increase the repel voltage, we have to go
for three phase fully controlled converter.
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00:14:07,100 --> 00:14:11,670
That means same like single phase fully controlled
converter, we have to go for the three phase
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00:14:11,670 --> 00:14:13,449
fully controlled converter.
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00:14:13,449 --> 00:14:20,449
So now, we will start studying about the three
phase fully controlled converter that is three
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phase or three phase bridge, three phase bridge.
So, how the configuration will be for the
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00:14:34,970 --> 00:14:41,970
three phase bridge? Let us go to let us start
studying about the three phase bridge. Three
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00:14:47,079 --> 00:14:54,079
phase bridge here; same like three phase half
controlled bridge, we will be rectifying the
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00:14:56,139 --> 00:15:01,739
waveform during the negative portion also.
So, this can be returned like this to start
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00:15:01,739 --> 00:15:08,739
with. This is our previous positive side;
we will go for making the figure clarity better,
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00:15:22,329 --> 00:15:29,329
so we are removing the inductance now. We
have the mains here - e1, e2 then e3, this
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00:15:45,489 --> 00:15:52,489
will go here. As before, cathodes are connected
to wave.
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00:15:55,959 --> 00:16:01,170
Now, we want the negative portion also. So,
what will have? We will connect diodes like
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00:16:01,170 --> 00:16:08,170
this no sorry thyristor like this, 3 thyristors.
So here, the cathodes here are shorted together.
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So, for negative, here we will anodes are
shorted too. So, the load will be across this
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one and it will be connected here. Now this
side, this would again, it will be connected
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to e1 side. What will happen? This will go
to e2 side, then this e3 will be connected
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here, this in the negative side also.
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00:17:05,530 --> 00:17:11,189
So, from the analysis from positive side,
we are exchanged to negative side. Negative
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00:17:11,189 --> 00:17:15,360
side also we want to connect, the thyristor
to conduct and the current conduction that
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00:17:15,360 --> 00:17:20,459
is current will conduct in this direction
that will be even though that will be done
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00:17:20,459 --> 00:17:24,270
through the inductive load, it will force
through this one.
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00:17:24,270 --> 00:17:31,270
Now, in a simple way, we can again but this
is the way it extends from positive side to
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00:17:32,769 --> 00:17:39,769
negative side. Now, if you see, the actual
rectifier in the text book, it will be like
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00:17:41,090 --> 00:17:45,500
this. That means this the same connection,
in a simple way we can represent like this
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00:17:45,500 --> 00:17:52,500
where this is for the positive part and this
is for the negative part and here is our load.
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00:18:11,570 --> 00:18:18,570
e1 and e2 are here; e1, e2 is this way, e3
is this way as before. This is the neutral
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00:18:38,049 --> 00:18:45,049
point that is this point. So, this is an exception
of this one. The concept, negative side also
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00:18:45,730 --> 00:18:52,100
we can use it; the same way then opposite
to way the thyristor has to be connected and
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00:18:52,100 --> 00:18:57,700
this three phase fully controlled configuration
can be derived this way.
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00:18:57,700 --> 00:19:04,700
Now, let us see what is the firing angle range?
And, where the alpha is equal to 0? What is
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the output waveform, output dc, output repel
and also the output dc with respect to firing
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00:19:13,100 --> 00:19:20,100
angle? We will we will study now. So, let
us go to the next page. So, we will redraw
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the three phase wave again here for clarity.
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00:19:32,019 --> 00:19:39,019
This is the positive side, this is the negative
side, so we are trying to make use of the
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00:19:47,710 --> 00:19:53,289
input waveform for power transfer positive
as well as negative side. So, maximum utilize
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00:19:53,289 --> 00:20:00,289
says now the three phase mains. This is e1,
e2, e3, load is here. e1 e2 e3 this e1 e2
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00:20:14,360 --> 00:20:20,510
e3 can be represent as a b c or r y b.
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00:20:20,510 --> 00:20:27,510
Now, how the commutation starts? What are
the sequences of switching? If you see, there
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00:20:36,480 --> 00:20:43,480
are 6 thyristor and in the cycle of operation
that means ts period, the period fundamental
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00:20:46,909 --> 00:20:53,909
period is equal to T, it is a 360 duration.
So, each thyristor, the conduction duration
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00:20:55,070 --> 00:21:02,070
for symmetric operation, the conduction duration
for symmetric operation, the conduction duration
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00:21:10,799 --> 00:21:17,370
for each thyrsistor will be 6 means how much
it will come? It will be equal to 60 degree.
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00:21:17,370 --> 00:21:24,370
See, each thyristor will be conducting, totally
only 60 degree duration in a 360 pm interval.
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00:21:26,470 --> 00:21:33,470
Now previously, when we discussed about the
top and bottom, each thyristor for the half
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00:21:36,940 --> 00:21:42,340
bridge, each thyristor will be conducting
for 120 degree and every 120 degree, there
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00:21:42,340 --> 00:21:48,250
is a commutation process initiated. Now here,
how it will happen? Let us first draw the
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00:21:48,250 --> 00:21:53,120
three phase mains waveform. All these phases
control converter using thyristor, these are
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00:21:53,120 --> 00:21:58,460
naturally computational that means the incoming
thyristor will force the previous conduction
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00:21:58,460 --> 00:22:04,960
thyristor by applying the reverse voltage
across it from the incoming phase.
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00:22:04,960 --> 00:22:11,960
So, let us draw the three phase waveform.
This is our A phase, 120 degree, we have the
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00:22:33,049 --> 00:22:40,049
B phase, then 360 degree, we will have the
C phase or e1 e2 e3 . So, this is e1, this
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00:22:49,120 --> 00:22:56,120
is e2, this is e3. Now, alpha is equal to
0. Let us take e1 when the e1 phase is, at
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00:23:04,860 --> 00:23:11,860
this instant, if you see here, if you take
omega t is equal to 0 from here, omega sine
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00:23:14,630 --> 00:23:21,630
omega t, we are staring from here. Let us
take for e1 to turn on, the positive e1 is
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00:23:22,330 --> 00:23:29,330
positive here. So, from this point to this
point, for the positive side, if you see here
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00:23:34,240 --> 00:23:41,240
e1 will be conductive sorry e1 is positive
and T 1 will be conducting. So, this is T
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00:23:45,360 --> 00:23:52,360
1. So, this is positive, this is T 1. So,
e1 will be conducting.
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00:23:53,889 --> 00:24:00,889
Now, e3 is here; if you see here, e2 is more
positive from this point to this point sorry
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00:24:06,659 --> 00:24:13,659
e2 is more negative. So, the thyristor which
is maximum, negative e2 negative, so that
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00:24:15,889 --> 00:24:22,129
will be conducting now. So now, we have initiated
the firing angle from here that is initiated
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00:24:22,129 --> 00:24:28,090
the firing angle process of alpha is equal
to 0 from here. Now, from here to here, e1
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00:24:28,090 --> 00:24:35,090
will be conducting but at this point, after
that this point, e3 will become more negative.
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00:24:36,669 --> 00:24:43,669
e3 is more negative means the next thyristor
to be turned on. First 1 is turned down, then
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00:24:45,879 --> 00:24:52,879
next is this is T 2. So, this is this direction
is for T 2. So, e3 negative, so this is T
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00:24:56,360 --> 00:25:03,360
2, next is T 2. See, as I told before, see
the positive sides and negative sides are
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00:25:07,309 --> 00:25:10,840
independent of each other from the previous
diagram.
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00:25:10,840 --> 00:25:17,539
So, as before, the conduction duration for
the upper limit as well as the lower limit
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00:25:17,539 --> 00:25:24,539
is 120 degree not 60 degree. So, this is 120
degree not 60 degree. But if you see here,
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00:25:27,029 --> 00:25:34,029
every 60 degree, there is a commutation. That
means there is a new thyristor turned on.
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00:25:48,820 --> 00:25:55,820
This is T 2. Now, at this point; so we are
turning on, alpha is equal to 0, we are turning
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00:25:58,429 --> 00:26:05,429
on T 1 here. Then at this point, we have turned
on T 2. Now, upper limit; e2 is going more
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00:26:05,840 --> 00:26:11,600
positive, so T 1 should be switched off and
T 3 should be turned on, next is T 3. So,
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00:26:11,600 --> 00:26:18,600
this will be T 3, this is T 3, this is T 3
so but at the lower side, here e3 will be
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00:26:35,330 --> 00:26:37,909
e1 more negative.
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00:26:37,909 --> 00:26:44,909
So here, it is T 4 e1 more negative means
T 4 means here, this is T 4. Now again, e3
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00:26:49,419 --> 00:26:56,419
will become more positive here at this point.
At this point T 4, T 1, T 2, this is T 4,
164
00:27:01,259 --> 00:27:08,259
this is T 3, e3 become more positive, e3 will
become more positive here sorry at this point,
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00:27:10,860 --> 00:27:16,940
e3 become more positive; so which has to be
turned on? Now, the T 4 is on. The next thyristor
166
00:27:16,940 --> 00:27:23,940
will take as T 5 here. Then here, e2 become
more negative after 60 degree that is T 6,
167
00:27:26,990 --> 00:27:33,990
this is T 6. So, we will again draw the waveform
and the conduction duration again we will
168
00:27:35,720 --> 00:27:42,720
again write for clarity. So, let us go to
the next page now.
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00:27:50,009 --> 00:27:57,009
So, this is one thyristor, we will make it
bigger; then each phase will have a positive
170
00:28:16,360 --> 00:28:23,360
and negative part and this can be independently
triggered also as I told before, these are
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00:28:25,190 --> 00:28:29,460
independent because this is the positive half
bridge, this is negative half bridge. Now,
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00:28:29,460 --> 00:28:36,460
this is the e1, e1 here, next one e2, next
is e3, this is our neutral, e3 here.
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00:28:52,519 --> 00:28:59,519
Now, let us draw the phase waveform again.
This is e1, 120 degree you can have approximate,
174
00:29:19,210 --> 00:29:26,210
e2. Now, you can have e1 e2 e3. So, our alpha
is so we will try to understand again for
175
00:29:36,049 --> 00:29:43,049
alpha is equal to 0, so e1, this is e1, the
first one is e1 that is this one e1, this
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00:29:45,169 --> 00:29:52,169
is e2 and this is e3. Now, alpha is equal
to 0, we are firing here. At alpha is equal
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00:30:00,840 --> 00:30:07,840
to 0, e1 is more positive. So, thyristor one
is fired. So, e1 side positive, this is e1,
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00:30:08,919 --> 00:30:15,919
this is e1, the positive side thyristor will
be this one that is T 1. So, T 1 conduction
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00:30:16,179 --> 00:30:23,179
duration
will be this period, upto this period, T 1
will be conducting. Then only the next positive
180
00:30:39,230 --> 00:30:46,210
side T 2 will conduct. So, T 2 conduction
were this is the T 2 conduction period. At
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00:30:46,210 --> 00:30:52,299
this point, we will be initiating T 2.
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00:30:52,299 --> 00:30:58,629
But if you see the negative side that is this
portion, if you see here e3 is more negative,
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00:30:58,629 --> 00:31:05,240
after e1 is more positive, e3 will become,
so each phase whether positive or negative
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00:31:05,240 --> 00:31:10,960
limping be at the same firing angle alpha,
alpha is equal to 0, we have to use for positive
185
00:31:10,960 --> 00:31:15,460
also negative. So, positive alpha is equal
to firing angle here; for negative, the firing
186
00:31:15,460 --> 00:31:22,340
angle alpha starts from here. The firing angle
alpha that from there the phase becomes maximum
187
00:31:22,340 --> 00:31:28,889
negative, here the maximum negative will be
e3. e3 will be maximum negative. So, here
188
00:31:28,889 --> 00:31:35,889
the T 2 thyristor will be conducting; this
is T 2, T 1, T 2.
189
00:31:38,549 --> 00:31:45,549
Now, e1 is still conducting. So, at this point,
e1 at sorry e1 phase is not positive, T 1
190
00:31:48,639 --> 00:31:54,960
will be conducting and T 2 will be conducting.
Now, at this point e2, the phase e2 become
191
00:31:54,960 --> 00:32:00,619
more positive. So, T 1 has to be switched
off and next positive one thyristor at the
192
00:32:00,619 --> 00:32:07,619
positive side has be turned on that will be
T 3. So, T 3 will be e2 positive will be this
193
00:32:08,649 --> 00:32:15,649
is T 3. So, T 3 conduction period is from
here. So, we can mark it now with a different
194
00:32:16,159 --> 00:32:23,159
color that will give clarity. So, let us take
green. So, this is the
conduction duration for T 3.
195
00:32:35,669 --> 00:32:42,669
Now, after some time, after T 3 is turned
on, after 60 degree that is this point e3
196
00:32:43,639 --> 00:32:50,639
e1 phase become more negative. So now, the
T 2 has to be switched off and next thyristor
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00:32:55,990 --> 00:33:02,990
has to be turned on. So, next is T 4 at the
e1 phase maximum. So, this is T 4. So, T 4
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00:33:04,230 --> 00:33:11,230
duration will be let us mark it with another
color, let us take this color. So, this is
199
00:33:15,720 --> 00:33:22,720
the duration for T 4. Now again, positive
is after 60 degree, there is a conductor initiation
200
00:33:27,019 --> 00:33:33,710
is required at e3. e3 because more positive
than e2, so T 4 is our next positive side
201
00:33:33,710 --> 00:33:40,710
will be next side will be T 5 at the positive
side of this one. That is T 5, then, last
202
00:33:42,360 --> 00:33:49,149
e2 becomes negative that is from this portion
onwards that is T 6.
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00:33:49,149 --> 00:33:56,149
So, if you see the sequence, 1, 3, 5, in top
and 4, 6, 2 that is T 6, T 6 is this one sorry
204
00:34:00,940 --> 00:34:07,940
so the next phase will be so I will just clear
this part. So, next side part will be here
205
00:34:15,980 --> 00:34:22,980
it is T 6, this is T 6. So, the sequence if
you see, T 1, T 3 and then T 5, this is T
206
00:34:27,500 --> 00:34:34,500
5; T 1, T 3, T 5, T 4, T 6, T 2, this way
we have to turn on, this is the sequence with
207
00:34:37,910 --> 00:34:43,110
which we have to turn on the devices, alpha
is equal to 0. Alpha is equal to 0 for positive
208
00:34:43,110 --> 00:34:48,610
side, negative side at the same point that
is whenever the two adjacent phases just cross
209
00:34:48,610 --> 00:34:53,110
equal that is the point that is for positive
side, this is alpha is equal to 0 starts here;
210
00:34:53,110 --> 00:34:56,820
for negative side, this is the way and these
2 are independent.
211
00:34:56,820 --> 00:35:03,820
But for symmetry of operation if you see,
every 60 degree a thyristor; every 60 degree
212
00:35:05,870 --> 00:35:12,870
duration, a thyristor is turned on is turned
on. Each thyristor
will be conducting for 120 degree, 120 degree
213
00:35:43,640 --> 00:35:50,640
duration. So individually, we have written
the repel content for alpha is equal to 0
214
00:35:51,040 --> 00:35:58,040
for positive side and negative side. So, what
will be the net output repel? Net output repel
215
00:35:59,250 --> 00:36:04,930
will be the positive minus negative. If you
see, instantaneously will you give the next
216
00:36:04,930 --> 00:36:06,870
output repel, that you can do it.
217
00:36:06,870 --> 00:36:13,870
So, if you see, the output repel will have
a 6 pulse content. So, in a 360 degree, the
218
00:36:15,590 --> 00:36:22,590
duration will be that means every 60 degree
you are initiating. So, it is like a 6 pulse
219
00:36:23,190 --> 00:36:27,740
waveform. The repel have a frequency will
be 6 times the fundamental. How do you do
220
00:36:27,740 --> 00:36:34,740
it? The positive side and negative side; positive
side, assume the previous as I told you, positive
221
00:36:35,010 --> 00:36:42,010
side if you see here that is this portion
that is this portion, this voltage or this
222
00:36:45,050 --> 00:36:52,050
voltage with respect to the neutral here.
Let us take this is VA - upper side. V, this
223
00:36:56,760 --> 00:37:03,760
is N, VN. So, total voltage across the repel
that is instantaneous voltage repel, this.
224
00:37:04,970 --> 00:37:11,970
Let us see, this is B. So, Van that is half
bridge, positive half bridge repel minus the
225
00:37:16,810 --> 00:37:22,090
negative half bridge repel. AN minus BN will
be, it is already negative, so when you subtract,
226
00:37:22,090 --> 00:37:28,320
it will become again positive. So, AN minus
VN is a repel content. You can, this we can
227
00:37:28,320 --> 00:37:35,320
find out easily. So, output repel will be
the positive repel minus negative repel. So,
228
00:37:35,570 --> 00:37:41,130
as the alpha is increased for the positive
side, the corresponding increase for the negative
229
00:37:41,130 --> 00:37:45,180
side also should be increased or symmetry
operation.
230
00:37:45,180 --> 00:37:52,180
Now, let us take this is for highly inductive
load. What will happen if it will be a resistive
231
00:37:58,300 --> 00:38:05,300
load? Or here also as before, if we do not
want the output voltage, we do not want the
232
00:38:05,590 --> 00:38:12,590
output voltage, negative portion is not required.
Three phase, we are using just before some
233
00:38:13,130 --> 00:38:17,410
power converter application where we require
large power. So, high voltage is required
234
00:38:17,410 --> 00:38:21,470
but at the same time, we do not want the negative
voltage that is firing angle control. We do
235
00:38:21,470 --> 00:38:28,470
not have to control in such a way so that
output should not got negative. Then what
236
00:38:28,650 --> 00:38:35,650
will, then how this can be achieved? See,
here, we can, lets go to next page.
237
00:38:41,020 --> 00:38:48,020
As before, for the single phase and three
phase mid point configuration, we can conduct
238
00:38:55,120 --> 00:39:02,120
we can connect a diode. This is our one way
to e1 e2 e3. See here, if you do not want
239
00:39:19,030 --> 00:39:25,540
a negative, negative repel if you do not want
to appear on the load side, you can put a
240
00:39:25,540 --> 00:39:32,540
diode. Then, same like midpoint, you will
get a discontinuous conduction. So before
241
00:39:34,090 --> 00:39:38,400
coming to that one, let us find for a three
phase fully controlled converter, what will
242
00:39:38,400 --> 00:39:43,790
be the output voltage repel, output dc with
respect firing angle? For the three phase
243
00:39:43,790 --> 00:39:50,790
half bridge converter, we found it is e0 alpha
is equal to 1.17 E rms into cos alpha - This
244
00:39:56,200 --> 00:39:57,870
is for three half phase.
245
00:39:57,870 --> 00:40:04,870
So, we can assume for the fully controlled
converter with inductive load, not with freewheeling
246
00:40:06,120 --> 00:40:10,510
diode, freewheeling diode is not there that
means we are trying to get more both positive
247
00:40:10,510 --> 00:40:15,360
as well as negative voltage. It will be for
the three phase, e0 alpha three phase fully
248
00:40:15,360 --> 00:40:21,450
controlled converter. It will be two times
with symmetrical firing angle. For the both
249
00:40:21,450 --> 00:40:28,450
positive as well as negative, it will be 1.17
into e cos alpha. So, this we can easily derive
250
00:40:32,650 --> 00:40:38,830
or we can draw the repel and we can find the
integration also, we can find out that one.
251
00:40:38,830 --> 00:40:45,830
Now, if we put an output diode, whenever the
output voltage repels goes negative, the load
252
00:40:46,310 --> 00:40:53,310
will freewheel through the diode. So, that
means the negative portions will be clamped
253
00:40:54,530 --> 00:41:01,530
to zero level across the load. So, this means
with firing angle, we can have 0 to positive
254
00:41:02,060 --> 00:41:09,060
voltage here. But before coming to but we
require one extra diode here. So, in a similar
255
00:41:12,060 --> 00:41:19,060
in the same way, what we have single phase
case; can we get away with diode? Then we
256
00:41:22,100 --> 00:41:29,100
have to use a three phase semi converter.
Before coming to semi converter, let us find
257
00:41:29,180 --> 00:41:35,300
out what is the type of repel for a three
phase converter without diode. What is the
258
00:41:35,300 --> 00:41:41,690
type of nature of output voltage repel that
coming across the load?
259
00:41:41,690 --> 00:41:47,190
Let us take for a firing angle alpha is equal
to 30 for this converter, for the three phase
260
00:41:47,190 --> 00:41:54,190
converter. So, we will remove this diode for
the time being and we will connect the load
261
00:42:03,760 --> 00:42:10,760
here. That is load is connected here. So,
let us see how the output repels comes and
262
00:42:13,870 --> 00:42:20,870
again we will mark the thyristors. This we
know, this is T 1, this is T 3 and this is
263
00:42:26,840 --> 00:42:33,840
T 5; T 5, 135 then this is T 4, T 4 then this
is T 6 and this is T 2.
264
00:42:47,560 --> 00:42:54,560
Now, let us draw the output repel for the
three phase fully controlled converter. Let
265
00:43:04,420 --> 00:43:11,420
us draw the three phase voltage waveform V1,
V2, V3. This is e1, this is our X axis e1
266
00:43:34,030 --> 00:43:41,030
that is e1. Now e2, this is e2. Now e3, e3
will be 240 lagging with respect to e1 or
267
00:44:00,880 --> 00:44:07,880
120 lagging with respect to e2 . We got the
waveform. Now, we can think about the positive
268
00:44:15,520 --> 00:44:22,230
half as well as the negative half. The positive
half, the firing angle alpha is equal to 30
269
00:44:22,230 --> 00:44:29,230
that means we are firing alpha is equal to
30 here, 30.
270
00:44:29,760 --> 00:44:36,760
Now, for the T1, alpha is equal to 30 means
starts from here. Probably, we will use this
271
00:44:42,110 --> 00:44:49,110
is e1, we will use a different color so that
the clarity of the repel will be here, we
272
00:44:54,320 --> 00:45:01,320
will use red. So, this is the voltage repel
coming across the load when T1 is turned on,
273
00:45:08,170 --> 00:45:15,170
T1, this is T1. Now, at this point that means
we are taking the positive half bridge, three
274
00:45:18,560 --> 00:45:25,560
will be fired. So, three will be means it
is here, here three. That is thyrsistor three.
275
00:45:35,160 --> 00:45:42,160
Then here, the thyristor five, it will be
turned on this is 5.
276
00:45:45,330 --> 00:45:52,220
Similarly, for the negative, when the phase
voltage goes negative, the bottom devices
277
00:45:52,220 --> 00:45:59,220
T 4, T 6, T 2 has to be switched on. So, let
us say when Vy and Ve2 is negative that is
278
00:46:00,640 --> 00:46:07,640
this is when e2 is negative that is this portion,
the thyristor T 6 will be turned on. So, T
279
00:46:14,830 --> 00:46:21,830
6 will be V 6 T 6 portion. Then, after T 6,
this waveform; this waveform is the e3 waveform,
280
00:46:33,370 --> 00:46:40,370
e3 going negative means T 2 will be turned
on. So, from here, T 2 will be turned on,
281
00:46:43,570 --> 00:46:50,570
this way. At this point, e1V 1 negative maximum
negative; here T 4 will be turned on. So,
282
00:46:53,990 --> 00:47:00,990
it repeats like this; this is 4, this is 2
and this is 6. So, we have repel voltage for
283
00:47:02,830 --> 00:47:06,400
the positive and repel voltage for the negative.
284
00:47:06,400 --> 00:47:13,400
So, if you see here with respect to our mains
negative, mains neutral, the positive part,
285
00:47:14,900 --> 00:47:21,900
we can call as positive part we can call as
A, the negative part we can call as; instantaneous
286
00:47:30,020 --> 00:47:34,980
values of the repel due to the positive, we
can call as AN, AN is the neutral point BN.
287
00:47:34,980 --> 00:47:41,860
So, the net voltage output voltage is positive
instantaneous subtraction of positive minus
288
00:47:41,860 --> 00:47:48,860
negative that is VAN minus VBN. So, VAN minus
VBN will be there.
289
00:47:57,310 --> 00:48:04,310
So, the positive say that the positive repel
is here, the positive repel is here and the
290
00:48:05,800 --> 00:48:12,800
negative repel is here. So, what will be the
net? What will be the net repel coming across
291
00:48:17,970 --> 00:48:24,970
the output? VAN minus VBN that means instantaneous
values; the positive as well negative if you
292
00:48:33,460 --> 00:48:40,460
subtract, the repel waveform will be like
this. So, this will go, the cycle. So, this
293
00:48:47,230 --> 00:48:54,230
is 60 degree duration. So, we have the repel
will be 6 times the fundamental, this is 0.
294
00:48:56,060 --> 00:49:03,060
Now, as the firing angle varies; so this will
have an average dc voltage when alpha is equal
295
00:49:04,510 --> 00:49:11,430
to 90 degree, here also it is output voltage
is equal to output voltage is equal to e0,
296
00:49:11,430 --> 00:49:18,430
alpha is equal to 2 into 1.17 into E rms into
cos alpha. So, alpha can vary from 0 to 180
297
00:49:23,590 --> 00:49:30,590
degree. When alpha is equal to 90 degree,
the repel will have equal positive and equal
298
00:49:32,040 --> 00:49:36,110
negative output dc will be 0.
299
00:49:36,110 --> 00:49:43,110
Now, we want the output voltage fully from
0 to positive and we do not want the negative
300
00:49:45,360 --> 00:49:52,360
excursion. Then I told you that we can put
a diode across the load so that load can freewheel
301
00:49:54,770 --> 00:50:01,670
but one more extra diode is required. So,
here also three phase fully controlled converter
302
00:50:01,670 --> 00:50:06,340
can work as in a semi converter mode that
is three thyristor can be controlled; other
303
00:50:06,340 --> 00:50:13,340
three can be used diode so that the voltage
can go from 0 to positive all the time. How
304
00:50:15,050 --> 00:50:17,720
it will be? Let us see.
305
00:50:17,720 --> 00:50:24,720
We want only positive voltage 0 to positive
output dc voltage, so the bottom converter,
306
00:50:33,310 --> 00:50:40,310
the semi the converter we will use diodes.
The bottom switches, we will use diodes. So,
307
00:50:43,590 --> 00:50:50,590
this is the e1 e2 e3, here is the load.
308
00:51:02,200 --> 00:51:09,200
Now, we have diodes here. The diodes one advantage
of diodes; the moment it is forward biased,
309
00:51:15,640 --> 00:51:22,340
it does not require a gate pulse to turn on.
The moment it is forward biased that is anode
310
00:51:22,340 --> 00:51:29,340
is positive with respect to cathode, it will
conduct immediately and as I told before sorry
311
00:51:34,070 --> 00:51:41,070
we
will just correct it. We will redraw once
again.
312
00:51:50,670 --> 00:51:57,670
So, this is diode, so here diode, so that
the diode will; so here, the moment the diode
313
00:52:03,280 --> 00:52:08,760
is forward biased, output will be quickly,
diode will be forward biased, immediately
314
00:52:08,760 --> 00:52:09,820
it will conduct.
315
00:52:09,820 --> 00:52:16,670
Now, as I told before, these two can also
independently we can control as the previous
316
00:52:16,670 --> 00:52:23,210
case of midpoint conversion during the positive,
converted during the positive half cycle of
317
00:52:23,210 --> 00:52:27,820
the waveform converted during the negative
half cycle of the convertor. So, that means
318
00:52:27,820 --> 00:52:34,820
we have the e1 here, e2 here, e3 here. So,
the positive side, it will be like this. This
319
00:52:49,900 --> 00:52:56,900
is our A side that is A side. In the A side,
we will connect the load.
320
00:53:03,440 --> 00:53:10,440
So similarly, the diodes are like this; here
this point will go here, this is called B,
321
00:53:32,370 --> 00:53:39,370
this is also B, e1 e2 e3, this our neutral
point, load is connected like this. e2 we
322
00:53:58,030 --> 00:54:05,030
will redraw this one that is e1 and here,
this is our neutral three phase neutral side.
323
00:54:24,620 --> 00:54:31,620
Now, output repel, total output repel is output
repel due to the top thyristor converter that
324
00:54:35,290 --> 00:54:42,290
is VAN minus VBN is equal to VAB that is a
instantaneous voltage repel across the load.
325
00:54:52,500 --> 00:54:59,500
So, we can consider these two independently,
then we can find out what is the repel. Independently
326
00:55:00,980 --> 00:55:07,940
means the top thyrsistor with firing angle
alpha, the bottom diodes with firing angle
327
00:55:07,940 --> 00:55:14,940
alpha is equal to 0. So, the net voltage VAB,
we know for that VAN will be equal to 1.17
328
00:55:20,230 --> 00:55:27,230
E rms into cos alpha.
329
00:55:29,080 --> 00:55:36,080
What about VBN? VBN is due to diode with firing
angle alpha is equal to 0. So, it is already
330
00:55:39,550 --> 00:55:46,550
negative. So, minus of that one will be plus
positive. So, this will be equal to 1.17 E
331
00:55:48,620 --> 00:55:55,620
rms because diode cos alpha is equal to alpha
is equal to 0 cos alpha is equal to 1. So,
332
00:55:58,910 --> 00:56:05,910
the net voltage for the semi converter is
equal to 1.17 E into rms that into 1 plus
333
00:56:08,900 --> 00:56:14,720
cos alpha.
334
00:56:14,720 --> 00:56:21,720
Now, in the previous slide, we showed that
for a three phase fully controlled converter
335
00:56:23,500 --> 00:56:29,570
output repel is 6 times the fundamental frequency.
But let us see what will be the type of repel
336
00:56:29,570 --> 00:56:35,330
for this diode, thyristor and diode configuration?
Here, we have introduced diode configuration
337
00:56:35,330 --> 00:56:42,330
so that output voltage we do not want to go
negative. If we does not want to go negative
338
00:56:43,370 --> 00:56:48,530
means let us see whether this will happen
that means we can only with the firing angle,
339
00:56:48,530 --> 00:56:55,530
we can have only it can have only 0 to positive
value. Let us see a typical output voltage
340
00:56:56,740 --> 00:57:02,370
repel for firing angle alpha is equal to 60
and alpha is equal to 90.
341
00:57:02,370 --> 00:57:09,370
So, let us draw the waveform for alpha is
equal to that is upper bridge alpha will be
342
00:57:11,000 --> 00:57:18,000
turning will be firing with respect to angle
alpha is equal to 60 degree. So, let us draw
343
00:57:18,010 --> 00:57:25,010
the VAB e1 e2 e3 waveform. This is our e1
waveform. Let us draw e2. e2 will be 120 degree
344
00:57:40,580 --> 00:57:47,580
lagging with respect to e1. Let us draw the
e3 waveform.
345
00:57:59,040 --> 00:58:06,040
Now, let us take the voltage repel due to
the diode. Diode will conduct at when firing
346
00:58:12,580 --> 00:58:18,400
angle alpha is equal to 0. So, alpha is equal
to 0 any phase starts from here; here and
347
00:58:18,400 --> 00:58:25,400
here. So, the conductance, so the repel due
to the diode, the negative side is like this.
348
00:58:28,470 --> 00:58:35,470
This is the diode. Diode will act as negative
side. That always for act as a three phase
349
00:58:42,380 --> 00:58:49,380
half bridge configuration, three phase half
bridge, three phase half bridge rectifier.
350
00:58:49,400 --> 00:58:56,400
So, the repel is like this.
351
00:58:57,960 --> 00:59:04,960
Now, depending on the firing angle, the upper
limp repel will vary. So, now here, in the
352
00:59:10,310 --> 00:59:16,300
present case, we are studying for alpha is
equal to 60 degree. So, 60 degree means it
353
00:59:16,300 --> 00:59:23,300
will be like this. See, this is the from here
to here and then from here it will immediately
354
00:59:33,550 --> 00:59:40,550
go here, this way. So, this is so every 60
degree, it starts from here. So, this is 60
355
00:59:47,280 --> 00:59:54,280
degree. Similarly, this is 60 degree. It will
come here. So, this is the repel, this is
356
00:59:55,800 --> 01:00:02,800
the AN repel, this is the VAN repel and this
is the diode that is the VBN repel, VBN. So,
357
01:00:13,810 --> 01:00:17,370
VAN minus VBN.
358
01:00:17,370 --> 01:00:24,370
Instantaneous value if you plot it, it will
be like this. Here if you see, this is the
359
01:00:37,930 --> 01:00:44,930
0 line, alpha is equal to 0. The repel exactly
comes to zero line and then switch. But if
360
01:00:51,700 --> 01:00:55,930
you see, compared to the fully controlled
converter, the semi converter repel is three
361
01:00:55,930 --> 01:01:02,420
times the mains. But the fully controlled
converter it is 6 times. So, the repel content
362
01:01:02,420 --> 01:01:06,570
the 0 repel current the current the repel
current repel content of the current will
363
01:01:06,570 --> 01:01:12,490
be much less in a three phase fully controlled
converter compared to three phase semi controlled
364
01:01:12,490 --> 01:01:17,860
converter that means with thyrsistors and
diode. So, this is 60 degree.
365
01:01:17,860 --> 01:01:23,870
Now, let us say alpha is equal to 90 degree;
alpha is 90 degree, we can draw it. Then the
366
01:01:23,870 --> 01:01:30,870
repel will be like this; alpha is equal to
90 degree, this portion goes negative. The
367
01:01:38,150 --> 01:01:45,150
moment goes negative, the thyrsistor and diode
will act as a short circuit. So, they are
368
01:01:45,250 --> 01:01:49,830
discontinues that means output repel will
suddenly, output is shorted equivalent to
369
01:01:49,830 --> 01:01:56,830
putting a freewheeling diode. It will come
here, this is the one.
370
01:01:56,990 --> 01:02:03,990
So, as the firing angle increases, slowly
this discontinuation period will increase
371
01:02:06,700 --> 01:02:13,700
and output voltage will come to 0. So, the
equation for the output diode using thyristor
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01:02:17,730 --> 01:02:24,730
and diode as I told you, Ed0 into 1 plus cos
alpha for the semi controlled converter; so,
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01:02:29,200 --> 01:02:31,880
it will come like this.
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01:02:31,880 --> 01:02:38,880
Now, why we have come to the semi converter?
So, many drive application, we may not require
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01:02:43,670 --> 01:02:49,230
the output voltage negative. We want only
a voltage control from 0 to positive. Then
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01:02:49,230 --> 01:02:55,750
we need not use thyrsistors because thyrsistor
if we use, with using thyrsistor and with
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01:02:55,750 --> 01:03:01,020
alpha control, we can have always from 0 to
positive voltage. But the problem is with
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01:03:01,020 --> 01:03:06,130
thyrsistors again we require extra gate drive
circuits. So, it is expensive. So, instead
379
01:03:06,130 --> 01:03:10,380
of that one, we can do away with diode that
is the purpose.
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01:03:10,380 --> 01:03:17,380
Now, so far, we have assumed that the conductions
are instantaneous; conductions are instantaneous
381
01:03:18,420 --> 01:03:25,420
and that there is no delay between the incoming
switch incoming thyrsistor and the commutating
382
01:03:25,800 --> 01:03:30,910
thyristor. The moment reverse bias is applied;
we assume that the conducting thyrsistor will
383
01:03:30,910 --> 01:03:36,320
immediately switch off. But that is not the
practical case. In many practical cases, there
384
01:03:36,320 --> 01:03:42,470
is a finite time for turn on and turn off
process. So, let us see what will happen due
385
01:03:42,470 --> 01:03:49,470
to finite time required for turn on and turn
off process? Is there, we will introduce any
386
01:03:50,140 --> 01:03:56,370
problem in the output voltage? Let us study
that one.
387
01:03:56,370 --> 01:04:03,370
Now, so we will go to
the effect of finite on time. So, this is
called commutation process. So far, the formulas
388
01:04:16,910 --> 01:04:22,630
derived in the previous sections are based
on the assumption and commutation process
389
01:04:22,630 --> 01:04:29,020
is instantaneous. There is a current switch
on and off are instantaneous. But due to the
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01:04:29,020 --> 01:04:36,020
presence of line inductance AC supply, the
transformer and the SCR circuitry will not
391
01:04:36,500 --> 01:04:43,500
permit instantaneous current change. See,
even though the thyrsistor can be easily turned
392
01:04:44,290 --> 01:04:51,290
on and turned off but there will be if finite
line inductance, transformer inductance. So,
393
01:04:52,020 --> 01:04:59,020
we are conducting limp cannot switch on switch
on or switch off the current instantaneously
394
01:04:59,900 --> 01:05:02,160
due to the finite inductance.
395
01:05:02,160 --> 01:05:09,110
So, apart from the finite turn off and turn
on time, the circuit inductance because of
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01:05:09,110 --> 01:05:15,560
the circuit inductance, the leakage inductance,
due to wiring or power circuit bus structure
397
01:05:15,560 --> 01:05:21,140
or the due to the transformer, there is a
delay in the turn on and turn off process.
398
01:05:21,140 --> 01:05:28,080
So, this can create a decrease, even though
the decrease in the output dc voltage due
399
01:05:28,080 --> 01:05:34,170
to this commutation overlaps. What I mean,
turning on and turn off time, they are not
400
01:05:34,170 --> 01:05:39,070
instantaneous. There is a period, the commutation
process goes through. That is called the commutation
401
01:05:39,070 --> 01:05:45,150
overlap. Due to that, there can be a reduction
in the output voltage. So, we will study about
402
01:05:45,150 --> 01:05:46,420
that one in the next class.