1
00:01:06,000 --> 00:01:17,700
Last class we concluded with single phase
half controlled converter. That means we have
2
00:01:17,960 --> 00:01:23,810
only even though the positive and the negative
cycle we are making use of but the control,
3
00:01:23,810 --> 00:01:30,810
we are doing only with 2 devices and the other
two are diodes. Then we found the voltage,
4
00:01:31,070 --> 00:01:36,100
the voltage vary, output DC voltage, it can
go from 0 to positive value.
5
00:01:36,100 --> 00:01:43,100
But many motor drive applications, we want
0 to positive as well as 0 to negative also
6
00:01:43,310 --> 00:01:50,310
that means the DC voltage should be variable
from minus value to 0 to the positive value.
7
00:01:52,159 --> 00:01:59,159
Then what will happen to the quadrant region
if you see in this figure with the voltage
8
00:02:02,179 --> 00:02:02,979
here?
9
00:02:02,979 --> 00:02:09,979
That is this part voltage and this part we
have Ia, voltage can change from positive
10
00:02:10,289 --> 00:02:17,289
to negative but the current will be only in
one direction that is this direction that
11
00:02:17,760 --> 00:02:23,310
is this direction. So, current is only from
the positive that is from the source type
12
00:02:23,310 --> 00:02:28,800
to the load. But if you see here, two quadrant
operations is possible that is one is this
13
00:02:28,800 --> 00:02:35,800
quadrant, other one is this quadrant; the
reverse regeneration is not reverse regeneration,
14
00:02:38,310 --> 00:02:43,540
forward motoring, this part is possible. The
current is one direction, this part is possible,
15
00:02:43,540 --> 00:02:47,470
not sorry not the other part, this part sorry
this part is possible.
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00:02:47,470 --> 00:02:54,470
So, we will change it. See, when the current
is voltage is negative and the current is
17
00:02:57,690 --> 00:03:04,690
positive; we can have only this direction
only possible. So here, the four quadrant
18
00:03:07,400 --> 00:03:14,400
operation, so here the current can conduct
only in one direction because of the thyristor
19
00:03:15,200 --> 00:03:20,860
or the diodes, for the silicon wire. So what
will happen? Current will go only in the positive
20
00:03:20,860 --> 00:03:27,130
direction but the voltage can go in the negative
direction. Voltage cannot go in the negative
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00:03:27,130 --> 00:03:30,000
direction for the previous converter.
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00:03:30,000 --> 00:03:37,000
Now, if we use all thyristor instead of diodes;
all 4 thyristors are shown here in this figure.
23
00:03:38,670 --> 00:03:45,670
That is thyristor 1 that is Th 1, then we
have Th 2, Th 2 is this one, then we have
24
00:03:50,570 --> 00:03:57,570
Th 3 here, then we have Th 4. So, if we use
this one; we have all 4 thyristors, so we
25
00:04:05,860 --> 00:04:10,880
will soon we will see that we can have voltage
in the negative direction also. So, we can
26
00:04:10,880 --> 00:04:15,810
have two quadrant operation is possible. So,
we will come to that one now. So, we will
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00:04:15,810 --> 00:04:22,810
go, we will study now, single phase fully
controlled converter. So, let us go to the
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00:04:24,940 --> 00:04:31,940
next page now. So, we will talk about single
phase fully controlled converter. That is
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00:04:37,659 --> 00:04:44,659
single phase fully sorry that is a fully controlled
converter.
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00:05:00,909 --> 00:05:07,909
So, let us again draw our power circuit schematic.
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00:05:10,310 --> 00:05:17,310
Now, thyristor 1 that I will mark it as Th
1, then you have Th 2, here you will have
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00:05:27,199 --> 00:05:34,199
bottom also we will have controlled device
that is Th 1, Th 2, Th 4 and Th 3. This is
33
00:05:45,029 --> 00:05:52,029
our mains, so here we will represent as a
and b here, phase voltage V4. Now, see here,
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00:05:59,710 --> 00:06:06,710
all the 4 devices we can control. So, how
we can get here positive as well as negative
35
00:06:06,789 --> 00:06:08,590
DC voltage? We will see now.
36
00:06:08,590 --> 00:06:15,590
Let us draw our mains voltage V4 here, this
is Vab and the inverse also we will put here
37
00:06:21,770 --> 00:06:28,770
like this.
Now here, this is our y axis that is amplitude,
38
00:06:30,710 --> 00:06:37,710
this is our omega t axis. We are firing with
respect to a firing angle alpha so that means
39
00:06:38,419 --> 00:06:45,419
we are firing here. So, if you see here, alpha
starts from here. When we fire this one, which
40
00:06:47,340 --> 00:06:52,490
are the thyristors we have to fire? That is
Vab is positive at the positive side. So,
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00:06:52,490 --> 00:06:58,550
whenever we fire a thyristor, the most forward,
the only forward bias devices will only be
42
00:06:58,550 --> 00:07:03,279
conducting. So, we have to fire the forward
bias thyristors.
43
00:07:03,279 --> 00:07:09,839
If you see, a is positive here, so during
the portion Th 1 will be forward biased; this
44
00:07:09,839 --> 00:07:16,839
point is negative that is this point is negative,
so Th 3 will be Th 3 will be also forward
45
00:07:17,919 --> 00:07:24,919
biased. So, at this point, we will be turning
out Th 1 and Th 3. So, till that what point,
46
00:07:26,349 --> 00:07:32,599
it will happen? So, next thyristor, so for
symmetry operation, if we are firing Th 1
47
00:07:32,599 --> 00:07:39,599
and Th 3 here that is Th 1 and Th 3 here,
the next thyristors for symmetrical operation
48
00:07:43,699 --> 00:07:49,629
during the negative portion; we should also
same firing angle alpha, we should fire next
49
00:07:49,629 --> 00:07:50,749
thyristors.
50
00:07:50,749 --> 00:07:57,749
That is at this point, you will be firing
Th 2 and Th 4. So, if you see here, till this
51
00:08:00,319 --> 00:08:07,319
point, this part of the input mains will be
coming across the load. So, the load we have
52
00:08:08,529 --> 00:08:14,020
marked, we can assume there is a load here,
across this one that is V0, here, across this
53
00:08:14,020 --> 00:08:21,020
portion that is our V0 alpha. So, Th 2 Th
3 will be fired. So, what will happen when
54
00:08:24,930 --> 00:08:31,930
the voltage goes negative here? We are assuming
the load is highly inductive, so current will
55
00:08:32,010 --> 00:08:39,010
be flowing through this path to the load and
returns through this path. So, the moment
56
00:08:41,339 --> 00:08:48,339
voltage goes negative, current cannot die
instantaneously. So, the di by dt changes,
57
00:08:48,670 --> 00:08:54,089
the voltage negative; so the current will
slowly slow down. So, load will force the
58
00:08:54,089 --> 00:09:00,279
thyristor to conduct till the next 2 thyristors
that is Th 2 and Th 2 are turned on.
59
00:09:00,279 --> 00:09:07,279
So, the load will keep on conducting; Th 1
and Th 3 will keep on conducting during this
60
00:09:07,779 --> 00:09:14,779
portion. So, during this portion, Th 1 and
Th 3 will be conducting. Now, when again we
61
00:09:18,139 --> 00:09:25,139
are turning on Th 2 Th 4, this full portion
that is upto here, Th 2 and Th 4 will be conducting.
62
00:09:39,810 --> 00:09:46,740
So, what happened to this portion? This portion
also, if you have continuous operation, this
63
00:09:46,740 --> 00:09:53,740
portion also will be that is this portion
also will be Th 2 and Th 4 are conducting.
64
00:09:57,540 --> 00:10:03,240
So, if you see here; whenever a thyristor,
two groups of thyristors that is 1 top and
65
00:10:03,240 --> 00:10:09,500
1 bottom are turned on only, the previous
conducting thyristor will be switched off.
66
00:10:09,500 --> 00:10:15,980
This is natural commutation because from the
mains voltage without any extra circuit, we
67
00:10:15,980 --> 00:10:20,399
are trying to turn off the devices this is
called natural commutation.
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00:10:20,399 --> 00:10:27,399
So, depending on the firing angle, the negative
excursion can go still further. So, what will
69
00:10:30,339 --> 00:10:36,740
happen here? There is a positive side positive
part of the voltage wave form coming across
70
00:10:36,740 --> 00:10:40,819
the load and the negative portion also coming
across the load that is this portion that
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00:10:40,819 --> 00:10:47,720
is here also. So, average value, if you take
like the previous case, if you find out the
72
00:10:47,720 --> 00:10:52,019
average value, integrate the waveform that
means we are trying to find out the area,
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00:10:52,019 --> 00:10:56,610
the positive or negative, the total area will
slowly get reduced, positive area will slowly
74
00:10:56,610 --> 00:11:03,610
get reduced because of the negative conduction.
That way the average value will get reduced.
75
00:11:05,550 --> 00:11:12,550
Let us take a case where positive or negative
are equal that is the case like this. We will
76
00:11:17,110 --> 00:11:19,860
go to the next figure.
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00:11:19,860 --> 00:11:26,860
So, I will draw the waveform; this is our
Vab, this is our the Vab negative. Let us
78
00:11:35,690 --> 00:11:42,509
take a case where positive or negative are
equal that is this portion. That means from
79
00:11:42,509 --> 00:11:49,509
the symmetry of the figure Vab, we know alpha
is equal to 90 degree is equal to omega t
80
00:11:49,509 --> 00:11:56,509
is equal to pi by 2. Then what happens? There
is equal positive area and equal negative
81
00:12:04,130 --> 00:12:11,130
area that is here. This is positive, this
is negative and this is positive.
82
00:12:13,360 --> 00:12:20,360
So, as before from the previous circuit, this
period Th 1 and Th 3 and this period Th 2
83
00:12:33,889 --> 00:12:40,889
and Th 4; again this period, Th 2 and Th 4.
So, if you see
the 2 pi s that is Th 1 and Th 3 and Th 2
84
00:12:56,930 --> 00:13:01,269
Th 4, they will be conducting for irrespective
of the foreign angle, they will conducting
85
00:13:01,269 --> 00:13:07,360
for 180 degree duration but here depending
on the firing angle, the duration of the 180
86
00:13:07,360 --> 00:13:14,329
degree period that is 180 period during that
period, the two pairs either Th 1 or Th 3
87
00:13:14,329 --> 00:13:20,509
or Th 2 and Th 4 are conducting, slowly get
shift and the positive and negative area becomes
88
00:13:20,509 --> 00:13:21,220
equal.
89
00:13:21,220 --> 00:13:28,220
So, if we find out for this equation the average
value; how, do we find out the average value?
90
00:13:29,600 --> 00:13:36,600
We are integrating the Em sin omega t d omega
t. We will be integrating between sin omega
91
00:13:45,690 --> 00:13:52,690
t sin omega t omega t series omega t 0 will
be here, omega t is equal to 0 sorry omega
92
00:13:53,910 --> 00:13:59,509
t is will be for sin omega t, it will be omega
t will be at this point. Here is the omega
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00:13:59,509 --> 00:14:06,509
t equal to 0 sorry so omega t will be 0 here;
omega t will be 0 here.
94
00:14:17,629 --> 00:14:24,629
So, what is the limit of integration? Lower
limit is equal to alpha, upper limit is equal
95
00:14:26,600 --> 00:14:33,480
to pi plus alpha. So, the next firing next
thyristor is fired here, alpha starts from
96
00:14:33,480 --> 00:14:40,480
here. So, for the sin omega t, this is pi
plus alpha, here it will come. So, pi plus
97
00:14:40,970 --> 00:14:47,720
alpha. Now, if you see for present dashed
area integration, the positive and negative
98
00:14:47,720 --> 00:14:52,220
will be equal. So, that average if you want,
it will be 0. So, during this portion, during
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00:14:52,220 --> 00:14:59,220
this firing, for this firing angle when alpha
is equal to pi by 2, the output Vdc will be
100
00:15:00,029 --> 00:15:02,230
0.
101
00:15:02,230 --> 00:15:09,230
Now, let us say the load is highly inductive
or with back emf load whether it can send
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00:15:12,620 --> 00:15:19,620
power by from load to the source; then what
will happen? The thyristor can be forced to
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00:15:20,329 --> 00:15:26,079
conduct that is Th 1 and Th 2 can be forced
beyond this region also. Let us take a typical
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00:15:26,079 --> 00:15:33,079
example. Again, let us go to the next page
again also. Let us draw the waveform here.
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00:15:48,560 --> 00:15:55,560
So Vab, this is the negative part, this is
Vab; let us take the firing angle starts from
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00:16:05,259 --> 00:16:12,259
here, let us take we are taking the firing
angle here, alpha. So, at this point, here,
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00:16:19,470 --> 00:16:26,470
we will be turning on Th 1 and Th 3. So here,
the moment it turn on assuming its continuous
108
00:16:34,970 --> 00:16:41,970
operation, this much positives area will be
shifted to the load side. Again, same equal
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00:16:45,180 --> 00:16:50,220
distance starting from this point, the next
thyristor is fired that is this is somewhere
110
00:16:50,220 --> 00:16:57,220
here. At this point, we will be turning on
Th 2 and Th 4.
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00:17:01,600 --> 00:17:08,600
So, till this portion, negative portion if
the load is highly inductive, it will force
112
00:17:09,949 --> 00:17:16,949
Th 1 and Th 3 to conduct. So, at this point
if you see, during this portion, the negative
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00:17:17,640 --> 00:17:23,270
side is more than the positive side. Again
at this point, the voltage polarity or the
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00:17:23,270 --> 00:17:28,380
part of the voltage waveform coming across
the phase will be positive.
115
00:17:28,380 --> 00:17:35,380
So, this will keep on continuing. So, till
the firing angle is concerned, this portion
116
00:17:35,700 --> 00:17:42,020
also if you can see for study state operation,
this also will be during this portion also
117
00:17:42,020 --> 00:17:49,020
Th 2 and Th 4 will be conducting Th 2 and
Th 4 and this portion, Th 1 and Th 3 will
118
00:17:51,100 --> 00:17:58,100
be conducting. Now, if you see the previous
integration, if we can see, we can see that
119
00:17:58,260 --> 00:18:05,260
what will happen? The net average voltage
is negative. So, the load is capable of forcing
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00:18:06,620 --> 00:18:13,620
the current in the negative portion also.
That means that is road is capable of regenerating;
121
00:18:14,650 --> 00:18:19,419
then we can have output negative also negative
voltage also output DC.
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00:18:19,419 --> 00:18:26,419
So, let us find out the general equation for
this semi for this single phase fully controlled
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00:18:28,010 --> 00:18:35,010
converter with firing angle alpha varying
from 0 to pi, theoretically 0 to pi. We will
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00:18:35,549 --> 00:18:39,929
come to the practical limitation later. So,
what happens? When the voltage is negative,
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00:18:39,929 --> 00:18:44,559
current is always in one direction. So, power
can be positive as well as negative. So, we
126
00:18:44,559 --> 00:18:49,179
can have two quadrant operations. What you
mean by two quadrant operation?
127
00:18:49,179 --> 00:18:56,179
If you have axis here, this is our E0 alpha,
the output voltage with respect to firing
128
00:18:56,400 --> 00:19:03,400
angle, this is our current I0, then we can
have operation in these two regions, two quadrant
129
00:19:05,240 --> 00:19:11,360
operation is possible. Now, let us find out
what do the general output DC voltage equation
130
00:19:11,360 --> 00:19:18,360
where we are worried about the output DC voltage
with respect to firing angle alpha. So, let
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00:19:19,580 --> 00:19:26,580
us V0 alpha is equal to alpha to pi plus alpha
Em sin omega t d omega t.
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00:19:47,740 --> 00:19:54,740
Now, you know integration cc, this is equal
to Em; Em is the peak value of the sinusoidal
133
00:19:56,179 --> 00:20:02,200
voltage, Em by when you integrate, we have
to find out the value, we have to find out
134
00:20:02,200 --> 00:20:08,640
the average value also that means it has to
be divided by 1 by pi. Previously, it was
135
00:20:08,640 --> 00:20:15,640
2 pi, why 1 by pi? The periodicity, every
pi it changes. So, the period is pi not 2
136
00:20:15,820 --> 00:20:22,820
pi as before. So, you will have Em by pi into
minus cos omega t integrating limits alpha
137
00:20:31,600 --> 00:20:38,600
to pi plus alpha. No, this easily we can integrate
it symbol.
138
00:20:40,090 --> 00:20:47,090
So, this will equal to Em by pi minus cos
pi plus alpha plus cos alpha. So, this will
139
00:21:00,700 --> 00:21:07,700
be equal to Em by pi cos alpha minus cos pi
plus alpha, we can cause a plus b form. That
140
00:21:18,620 --> 00:21:25,620
is cos pi cos alpha minus sin pi sin alpha.
This will finely come to Em by pi, you can
141
00:21:40,110 --> 00:21:47,110
derive this one by cos alpha. So, here also
we found previously, it was 1 plus cos alpha.
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00:21:48,549 --> 00:21:55,549
Here, the output DC voltage is proportional
to cos alpha. This is for a single phase fully
143
00:21:57,159 --> 00:22:00,510
controlled thyristor converter.
144
00:22:00,510 --> 00:22:07,510
Now, let us see
let us see if alpha is equal to 0 that means
firing angle starting at the zero crossing
145
00:22:24,750 --> 00:22:31,580
of the waveform. Then what happens? It is
equivalent like a diode. So, you will have
146
00:22:31,580 --> 00:22:38,580
single phase rectifier fully rectified waveform
like all the thyristors are replaced with
147
00:22:38,820 --> 00:22:45,820
diodes. So, final, for a diode, for a diode
case, single phase converter, AC to DC converter
148
00:22:48,260 --> 00:22:55,260
with diode; the output maximum voltage is
will be equal to 2 Em 2 Em by pi. This is
149
00:23:02,840 --> 00:23:06,240
the one you will get.
150
00:23:06,240 --> 00:23:11,529
So, what we have studied from the single phase
case? We have gone through a half controlled
151
00:23:11,529 --> 00:23:16,840
converter where you want only the voltage
0 to positive. You can appropriately choose
152
00:23:16,840 --> 00:23:23,840
the diodes, two diodes and we can have only
the control function using only two thyristors.
153
00:23:24,179 --> 00:23:31,179
Now, if you want to have the output voltage
both positive and negative, then we should
154
00:23:31,580 --> 00:23:37,210
have gone for a full three phase fully controlled
converter where the maximum output voltage
155
00:23:37,210 --> 00:23:44,210
will be 2 Em by pi for a firing angle alpha
is equal to pi or for the zero voltage, output
156
00:23:47,590 --> 00:23:51,710
DC voltage zero, you should have firing angle
alpha is equal to from the equation you can
157
00:23:51,710 --> 00:23:57,450
alpha is equal to pi by 2 that is 90 degree,
then cos pi by 2 will be 0.
158
00:23:57,450 --> 00:24:04,230
Then the next when for negative maximum, you
have to have firing angle alpha is equal to
159
00:24:04,230 --> 00:24:09,630
pi assuming the load is able to send back
that is force the thyristor to conduct in
160
00:24:09,630 --> 00:24:16,120
the forward direction that is load is able
to regenerate back to the source. So now,
161
00:24:16,120 --> 00:24:22,630
that way, we can have two quadrant operation
is possible. But we have used only single
162
00:24:22,630 --> 00:24:27,860
phase rectification. Some many applications
for high power where more power is required,
163
00:24:27,860 --> 00:24:34,860
we cannot draw all the power from the single
phase. Then, you may have to go for a three
164
00:24:36,519 --> 00:24:43,230
phase rectification so that output voltage
will be more, also the output repel that is
165
00:24:43,230 --> 00:24:44,480
also very important.
166
00:24:44,480 --> 00:24:49,269
So far, we have not talked about the repel
content; we are assuming that load is highly
167
00:24:49,269 --> 00:24:55,010
inductive, the repel frequency will be high
for the repel frequency, then impedance l
168
00:24:55,010 --> 00:25:01,570
omega will be higher and current will be repel
current will X plus. But for the single phase
169
00:25:01,570 --> 00:25:06,600
if you know the symmetry of the waveform,
we can see the repel frequency will be two
170
00:25:06,600 --> 00:25:09,730
times the input frequency.
171
00:25:09,730 --> 00:25:15,640
So, the harmonics, the repel frequency, the
harmonic and frequency also shifted to the
172
00:25:15,640 --> 00:25:22,640
higher to the frequency say compared to the
input frequency. Now, if you go for three
173
00:25:23,080 --> 00:25:28,970
phase rectification, we can have increase
DC voltage. At the same time, the repel frequency
174
00:25:28,970 --> 00:25:34,140
can be either three times or six times depending
on whether it is half bridged converter or
175
00:25:34,140 --> 00:25:35,350
fully controlled converter.
176
00:25:35,350 --> 00:25:40,659
So, let us start from the half bridged converter.
So, in many of the text books, it is called
177
00:25:40,659 --> 00:25:44,840
three phase midpoint configuring, half bridged.
That means we are trying to do only with the
178
00:25:44,840 --> 00:25:51,840
positive side that is one side only. Let us
see the configuration, let us go to the next
179
00:25:52,289 --> 00:25:59,289
page. So here, three phase midpoint configuration,
midpoint configuration.
180
00:26:11,320 --> 00:26:18,320
So, our three phases or AC source; let us
represent it by source e1, it can be a¬1
181
00:26:42,870 --> 00:26:49,870
b1c1 also but I will put e1, then e2, then
you have e3. So, this is our neutral, this
182
00:26:57,039 --> 00:27:04,039
point we will say as N, our three phase mains
neutral. Then we will have I have put some
183
00:27:10,169 --> 00:27:16,140
inductance here, I will come I will talk about
that one later, some small leakage inductance.
184
00:27:16,140 --> 00:27:23,140
Then you have the
thyristors
according to our source, source e1 e2 e3.
We can also name this one as Th 1, Th 2, Th
185
00:27:41,460 --> 00:27:42,139
3.
186
00:27:42,139 --> 00:27:46,870
So, we will more talking about the diodes
and you know from the single phase technique,
187
00:27:46,870 --> 00:27:51,570
for hiring angle alpha is equal to 0, we will
get the equation for a diode. So, thyristor
188
00:27:51,570 --> 00:27:56,570
will give the general equation whether diode
or thyristor control rectification. So, we
189
00:27:56,570 --> 00:28:03,570
will start with thyristors. Then where the
load is connected? Load will be connected
190
00:28:05,750 --> 00:28:12,210
here at this point assuming we are using for
high power operation. Let us a take a DC motor
191
00:28:12,210 --> 00:28:17,880
load, this is the inductance and the armature
inductance and the resistance here, it is
192
00:28:17,880 --> 00:28:19,950
like this.
193
00:28:19,950 --> 00:28:26,460
Now, let us see how, what we have to know
about when we are going to fire the thyristors?
194
00:28:26,460 --> 00:28:33,460
We know for the thyristor to be turned on,
it should be forward biased or we are using
195
00:28:34,639 --> 00:28:41,639
natural commutation. Natural commutation in
the sense, the incoming thyristor; there is
196
00:28:41,750 --> 00:28:48,750
a thyristor which is turned on now turned
on and the corresponding voltage phase voltage
197
00:28:50,380 --> 00:28:56,090
at that thyristor when it is turned on, we
will assume the thyristor voltage again on
198
00:28:56,090 --> 00:29:00,909
thyristor is 0, thyristor is turned on, the
top is negligible and that phase voltage should
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00:29:00,909 --> 00:29:06,659
be able to reverse bias the previously conducting
thyristor and should be able to switch off
200
00:29:06,659 --> 00:29:08,039
immediately.
201
00:29:08,039 --> 00:29:13,950
So, this point, we have to take care of when
we decide about the firing angle range. So,
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00:29:13,950 --> 00:29:19,159
for that one, let us draw the three phase.
Previously, we are drawing about the single
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00:29:19,159 --> 00:29:22,870
phase wave form; now let us draw the three
phase wave form.
204
00:29:22,870 --> 00:29:29,870
So, I will try to draw a symmetric three phase
waveform. This is the 120 degree one, then
205
00:29:48,480 --> 00:29:55,480
you have this is our e1 e2 and e3. Now, let
us take the acceleration with respect to thyristor
206
00:30:05,289 --> 00:30:10,950
one. So, when the thyristor one is turned
on, if you say e1 e2 e3 is taken, the previous
207
00:30:10,950 --> 00:30:17,950
conducting was Th 3. So, Th 3 means this is
the phase voltage waveform that where in that
208
00:30:22,730 --> 00:30:29,730
limb only the Th 3 is connected from the figure;
so, this one. So, when the thyristor connected
209
00:30:31,730 --> 00:30:38,630
to the e1 phase is turned on; as I told before,
the e1 should be able to turn off Th 3. So,
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00:30:38,630 --> 00:30:42,740
we are talking about 1 2 3 sequence.
211
00:30:42,740 --> 00:30:49,740
So, at this instant or instant from that onwards
e1 should be positive than e3, then only e1
212
00:30:51,950 --> 00:30:58,950
can reverse bias e3. If you see here, when
T 1 is turned on at this point, full e1 will
213
00:30:59,700 --> 00:31:06,659
come across Th 3. Previously, T across conducting,
it will come here and Th 3 will be e1 is more
214
00:31:06,659 --> 00:31:11,720
positive than e3; so thyristor will be reverse
bias and we can assume that thyristor will
215
00:31:11,720 --> 00:31:13,080
be immediately switched on.
216
00:31:13,080 --> 00:31:18,630
So, starting point of e1, assuming we will
assume the thyristor conduction is instantaneous,
217
00:31:18,630 --> 00:31:25,630
the starting point of turn on is this point.
So previously, for single phase, alpha 0 was
218
00:31:27,389 --> 00:31:32,649
when the zero crossing. Now, if you see for
a three phase, alpha 0 is equal to for a sign
219
00:31:32,649 --> 00:31:37,940
wave it starts from this degree is equal to
30 degree that is pi by pi by 6. After that
220
00:31:37,940 --> 00:31:43,169
only alpha 0 starts, the firing angle starts.
We can ensure the firing angle so that label
221
00:31:43,169 --> 00:31:48,389
commutation will happen. So, this is the starting
point of thyristor one Th 1.
222
00:31:48,389 --> 00:31:55,389
So, we can turn on Th 1 here, at this point
also we can turn on, at this point also, at
223
00:31:56,789 --> 00:32:03,789
this point also we can turn on, at all these
points the voltage given will be greater than
224
00:32:05,460 --> 00:32:12,460
e3. So, if you take the region, the duration
of the region along which e1 is always more
225
00:32:15,049 --> 00:32:22,049
positive than e3 is this region. That is from
here to here, this region. So, what do the
226
00:32:26,880 --> 00:32:33,880
advantage here? If you see here sorry we can
fire thyristor Th 1 at any of this point with
227
00:32:36,940 --> 00:32:43,909
respect to e1 that is from here to here. This
is the firing range.
228
00:32:43,909 --> 00:32:50,909
Alpha, firing that is gate pulse you can give
it to firing range of Th 1 so that Th 3 will
229
00:32:59,840 --> 00:33:06,840
be, the previous conducting Th 3 will be successfully
switched off. Now, what do the range? Firing
230
00:33:09,460 --> 00:33:15,100
range is a full one, there is a 180 degree
range is there. Alpha can be zero, alpha can
231
00:33:15,100 --> 00:33:21,679
be 180 degree here, alpha also firing angle
for thyristor one as alpha 1; alpha 1 is equal
232
00:33:21,679 --> 00:33:22,679
to pi.
233
00:33:22,679 --> 00:33:29,279
So, anywhere we can turn on thyristor assuming
the load is highly inductive; load will force
234
00:33:29,279 --> 00:33:36,279
the thyristor when the voltage also goes negative.
So, at this portion, the e1 voltage, at this
235
00:33:36,559 --> 00:33:43,549
portion if you see here, the e1 voltage is
going negative here, from here to this point.
236
00:33:43,549 --> 00:33:50,549
So, this is a firing range. So, also we should
know for symmetry operation, we should also
237
00:33:53,570 --> 00:33:59,940
at equal duration, for both the output DC
watts, so constant DC; every thyristor should
238
00:33:59,940 --> 00:34:03,700
forward the same firing angle with respect
to its starting point.
239
00:34:03,700 --> 00:34:10,700
And three phase, for three phase operation,
so here, it is if you see here, there are
240
00:34:11,060 --> 00:34:17,320
3 thyristor. Each thyristor will be conducting
for 120 degrees. 123 into 3 is 360 degree
241
00:34:17,320 --> 00:34:24,320
and firing angle range is alpha to alpha 0
to alpha is equal to pi. This is true with
242
00:34:26,089 --> 00:34:31,929
the previous single phase also but alpha 0
was equal to omega t0 for our sin omega t.
243
00:34:31,929 --> 00:34:38,929
Here, alpha is equal to 0 means for sin waveform,
omega t is equal to pi by 630 degree.
244
00:34:39,730 --> 00:34:46,730
Now, why I put this inductance here? These
are some of the particle limitation. We assume
245
00:34:49,169 --> 00:34:56,169
that previously the conduction is instantaneous
but there will be for any thyristor to turn
246
00:35:00,819 --> 00:35:06,109
on and turn off, there will be some finite
time required that is turn off minimum. So,
247
00:35:06,109 --> 00:35:13,109
even though Th 1 is turned on, at any of the
point as marked here; Th 3 will not be immediately
248
00:35:13,119 --> 00:35:17,609
turned on. it will take some time. So, momentarily
there is a short circuit Th 1 and Th 3 is
249
00:35:17,609 --> 00:35:24,609
conducting. It can be a converter grind all
times, it can be microsecond range 20 to 30
250
00:35:24,990 --> 00:35:29,720
microsecond range, it can be both Th 1 and
Th 3 is on.
251
00:35:29,720 --> 00:35:36,240
So, during this portion, the mains are e1
e2 and e3 are short circuited. So, heavy circulating
252
00:35:36,240 --> 00:35:43,240
current can flow. To take care of the difference
in voltage and to avoid the various heavy
253
00:35:45,069 --> 00:35:52,069
short circuit current, these inductance are
provided in series with the device, small
254
00:35:52,069 --> 00:35:57,530
inductance. How to come design this inductance?
We will come to that one later. That is very
255
00:35:57,530 --> 00:35:57,780
easy.
256
00:35:57,660 --> 00:36:04,000
You know, if for a thyristor is a taken, what
is the T of minimum? So, during the T of minimum,
257
00:36:04,000 --> 00:36:08,170
these both inductance voltages across the
inductance should be the difference between
258
00:36:08,170 --> 00:36:14,780
e1 and e3. So, based on that one, we can decide
the inductance and for a highly output DC
259
00:36:14,780 --> 00:36:19,640
voltage with much repel current, for DC current,
we can assume the drop across the inductance
260
00:36:19,640 --> 00:36:25,240
is negligible. These are theoretical assumptions
to start with but there are practical polo's
261
00:36:25,240 --> 00:36:26,780
associated with it.
262
00:36:26,780 --> 00:36:32,690
We come to the practical problems because
of the non-ideality of the devices and the
263
00:36:32,690 --> 00:36:38,310
other components in the power circuits later.
Let us see, here also, our aim is to find
264
00:36:38,310 --> 00:36:42,890
out the output voltage with respect to the
firing angle. The firing angle also, it can
265
00:36:42,890 --> 00:36:49,890
vary from alpha to pi by alpha to alpha 0
to alpha pi is equal to pi. Let us find out
266
00:36:50,280 --> 00:36:57,280
the output e0 alpha for this one. Let us go
to the next page. So, expressions for the
267
00:36:58,319 --> 00:37:04,280
output volt with inductive load, here, for
a three phase midpoint configuration; so,
268
00:37:04,280 --> 00:37:05,290
let us find.
269
00:37:05,290 --> 00:37:12,290
This is our e0 alpha that is equal to... as
I told, every device will be conducting for
270
00:37:16,230 --> 00:37:22,869
120 degree that means period is equal to 120
degree. So, for a three phase, what is the
271
00:37:22,869 --> 00:37:29,869
period? Period will be 2 pi by 3, so for a
three phase. Then, what is that for a general
272
00:37:30,990 --> 00:37:36,420
m phase? It will be, now we will be deriving
with for a general m phase to more generalize
273
00:37:36,420 --> 00:37:43,420
it. So, it will be 2 pi by m. So, the average
value that we divide by the periods; so for
274
00:37:43,660 --> 00:37:48,740
three phase, it should be 120 degree that
is 1 divided by 2 pi by 3 for a general m
275
00:37:48,740 --> 00:37:55,740
phase, one divided by 2 pi by m that will
be equal to m divided by 2 pi, then integrate.
276
00:37:58,270 --> 00:38:05,270
Now, integral limit, we take either we can
use it sin omega t or cos omega t. So here,
277
00:38:06,450 --> 00:38:13,450
I am taking Em cos omega t, Em cos omega t
d omega t, we are integrating. What do the
278
00:38:17,859 --> 00:38:24,859
lower limit with respect to the, let us go
to the previous slide.
279
00:38:27,990 --> 00:38:33,859
This is our, if you see the lower limit if
a cos omega t is this one. This is the point
280
00:38:33,859 --> 00:38:39,790
where we have the omega t is equal to 0.
281
00:38:39,790 --> 00:38:46,650
Now, for understanding, again we will draw
the waveform. If you see here, this is the
282
00:38:46,650 --> 00:38:50,310
sin wave form.
283
00:38:50,310 --> 00:38:56,920
Let us take the e1 waveform. Because of the
symmetry, we can only we have to integrate
284
00:38:56,920 --> 00:39:03,920
only the any part of e1 e2 or e3 for 2 pi
by m say 3 phase 2 pi by 3. So, Em sin omega
285
00:39:06,750 --> 00:39:12,790
t, we are integrating between I am taking
cos omega t, so cos omega t means omega t
286
00:39:12,790 --> 00:39:19,790
is equal to 0 here. But our lower limit integration
can be for a three phase, it is pi by 6 as
287
00:39:24,069 --> 00:39:29,910
I told below, this is 30 degree pi by 6, this
is the point where e3 will be touching here.
288
00:39:29,910 --> 00:39:36,910
So, this is the point, pi by 6 or for a three
phase, this is pi by 6.
289
00:39:39,500 --> 00:39:45,900
Now, when we want to generalize this one;
what is this one? Lower limit omega t0, we
290
00:39:45,900 --> 00:39:52,900
are going this direction that is negative.
How much it is going? If it is pi by 6 here,
291
00:39:54,030 --> 00:40:01,030
for a general m phase, this period is equal
to pi by 3 for a three phase and for a general
292
00:40:01,730 --> 00:40:04,819
m phase, it is pi by m.
293
00:40:04,819 --> 00:40:11,819
So, omega t0 is equal to 0, we are going to
negative side that is the lower limit of integration
294
00:40:12,440 --> 00:40:16,540
that is the firing angle at which only we
can start firing. So, for a general m phase,
295
00:40:16,540 --> 00:40:23,540
it will be minus pi by m that is the starting
point. From there, alpha we can increase;
296
00:40:25,730 --> 00:40:31,240
alpha from 0 to whatever values. So, minus
pi by m, general alpha if you take this way,
297
00:40:31,240 --> 00:40:38,240
then what is upper limit? Same thing; other
phase also, we should start where for alpha
298
00:40:40,589 --> 00:40:47,589
is equal to 0, for other phase also, it will
start from here that is this is e3, this is
299
00:40:47,599 --> 00:40:54,599
e1, this is e2. Alpha is equal to 0 for e2
starts from here. So, with respect to e1,
300
00:40:55,780 --> 00:41:00,920
this point is equal to if it is pi by m, this
is upper limit also pi by m.
301
00:41:00,920 --> 00:41:07,920
Now, alpha increases. So, the upper limit
will be pi by m plus alpha. So, that means
302
00:41:15,790 --> 00:41:19,740
if you see, from this waveform, lower limit
for a general m phase, we are starting from
303
00:41:19,740 --> 00:41:25,900
this point omega t is equal to 0 comes to
this one that is this point then from here
304
00:41:25,900 --> 00:41:30,430
lower limit. Then the alpha whatever starts
from here alpha starts from anywhere. So,
305
00:41:30,430 --> 00:41:36,069
minus pi by m plus alpha and the upper limit
pi by m for the other phase again alpha, alpha
306
00:41:36,069 --> 00:41:39,410
has to be equal for every phase.
307
00:41:39,410 --> 00:41:46,410
So, if you do here, what is the output when
you integrate this one and find out it will
308
00:41:47,059 --> 00:41:54,059
be e0 alpha is equal to m Em by pi sin pi
by m and cos alpha. So, this is the general
309
00:42:11,920 --> 00:42:18,920
equation for a general m phase with respect
to firing angle alpha. Now, for three phase,
310
00:42:20,390 --> 00:42:26,700
you put n is equal to 3, then it will be for
three phase, let us say for the three phase,
311
00:42:26,700 --> 00:42:33,700
three phase midpoint configuration, this will
be equal to 3 Em by pi sin pi by 3 into cos
312
00:42:40,940 --> 00:42:44,290
alpha, this is the one.
313
00:42:44,290 --> 00:42:49,010
Now, for the three phase diode rectifier not
the thyristor that we have divide the general
314
00:42:49,010 --> 00:42:56,010
phase, three phase diode rectifier that is
semi controlled say half bridge; we can say
315
00:42:58,250 --> 00:43:05,250
from the single phase terminology, three phase
midpoint configuration with diode, three phase
316
00:43:05,380 --> 00:43:10,809
diode rectifier diode. Not fully controlled,
semi controlled; it will be equal to alpha
317
00:43:10,809 --> 00:43:14,640
will be 0, firing angle will be 0.
318
00:43:14,640 --> 00:43:21,640
So, 3 Em by pi into sin pi by 3; so we can
easily derive this one. Now, with firing angle
319
00:43:31,559 --> 00:43:38,559
alpha is equal to 0, the general equation
for a general m phase for a general m phase,
320
00:43:49,280 --> 00:43:54,720
the maximum output voltage that is Ed0 alpha
maximum. We are more interested in maximum
321
00:43:54,720 --> 00:44:01,720
value because we have to choose our devices
based on that one. It will be equal to m Em
322
00:44:08,000 --> 00:44:15,000
by pi into sin pi by m, this is the maximum
equation.
323
00:44:22,200 --> 00:44:29,200
Now usually, our mains voltage we are electrical
engineers, we represent it as a rms value
324
00:44:30,599 --> 00:44:37,599
not the maximum values. So, how do we represent
this one if it is rms? It will be root 2 m
325
00:44:37,770 --> 00:44:44,770
by pi into Em ef note e, rms value, so we
will represent as E; E into sin pi by m. This
326
00:44:58,530 --> 00:45:04,369
is the maximum value we get from this one.
327
00:45:04,369 --> 00:45:11,369
Now, firing angle as a tool, powering firing
angle range is for 180 degree, alpha 0 to
328
00:45:12,829 --> 00:45:19,829
180 degree; now output voltage is sin pi by
m into cos alpha. So, alpha is equal to pi
329
00:45:20,579 --> 00:45:27,579
by 2 E 0 alpha will be 0. Where is the pi
by 2 ranges? Where is the pi by 2 point? Let
330
00:45:30,990 --> 00:45:37,990
us see. We will go to the next figure now.
Let us again draw the three phase waveform
331
00:45:38,180 --> 00:45:43,790
from this point for us to understand.
332
00:45:43,790 --> 00:45:50,790
This is our a phase and e1 voltage, this is
the e2 voltage. From experience, we can also
333
00:45:55,859 --> 00:46:02,859
equally approximately draw it. Alpha is equal
to 0 starts from here for e1; this is e1,
334
00:46:06,130 --> 00:46:13,130
this is e2, this is e3. This is alpha is equal
to 0 starts from here, alpha is equal to pi
335
00:46:19,390 --> 00:46:26,390
is this range, this is the range. So, the
half distance is the pi by 2 periods. So,
336
00:46:26,780 --> 00:46:31,300
at the pi by 2 period, what will happen?
337
00:46:31,300 --> 00:46:36,640
Suppose for the output voltage to be 0, the
repel voltage is still coming across the load,
338
00:46:36,640 --> 00:46:43,640
there should be equal positive and equal negative
region will be there. So, the firing angle
339
00:46:44,280 --> 00:46:51,280
if you take at the pi by 2 period alone; so
there we will have the approximately equal
340
00:46:56,010 --> 00:47:02,329
voltage duration somewhere here. So, alpha
is equal to 0 here and at this point, this
341
00:47:02,329 --> 00:47:07,770
is from here to here pi by 2 will be, for
a sin wave from here to here, it is pi by
342
00:47:07,770 --> 00:47:14,770
2. Already pi by 3 is gone here, so next pi
by 3, we have to add here. So, somewhere here,
343
00:47:15,079 --> 00:47:18,540
it will come. Also, the next waveform, it
will start from here.
344
00:47:18,540 --> 00:47:23,530
So, if you see here, for pi by 2 something
like this will happen; equal positive and
345
00:47:23,530 --> 00:47:30,530
equal negative, positive and equal negative.
So, when you draw neat waveform, you can say
346
00:47:32,200 --> 00:47:37,480
with respect to alpha is equal to 0, alpha
is equal to pi by 2 that is the starting point
347
00:47:37,480 --> 00:47:42,640
of the firing angle, we can have equal positive
and equal negative repel and output voltage
348
00:47:42,640 --> 00:47:43,940
will be 0.
349
00:47:43,940 --> 00:47:49,930
So, let us find out, see our converter, we
are controlling the firing angle only depending
350
00:47:49,930 --> 00:47:56,930
upon the voltage what we require. So, let
us have the graphics, the output Ed0 alpha
351
00:47:57,859 --> 00:48:02,230
with respect to our firing angle. So, we will
draw here.
352
00:48:02,230 --> 00:48:09,230
If you see, draw here, let us take Ed e0 alpha
here, e0 alpha divided by the e0 maximum e0
353
00:48:13,299 --> 00:48:20,299
maximum. So, maximum will be at point one
that is one. Then the output DC from the equation
354
00:48:28,960 --> 00:48:35,960
we know it that is e0 alpha is equal to root
2 m E by pi sin pi by m into cos alpha. So,
355
00:48:51,720 --> 00:48:58,720
the output e0 is a function of alpha, not
alpha cos alpha. So, the output with respect
356
00:48:59,170 --> 00:49:06,170
to firing angle, it will as a cosine function;
so this is our omega t, this is pi that is
357
00:49:14,440 --> 00:49:21,440
alpha, alpha, this is not omega t, this is
alpha in radians per second; when you represent,
358
00:49:22,079 --> 00:49:26,530
this is pi, this is 0, this is pi by 2.
359
00:49:26,530 --> 00:49:33,530
So, if you see when alpha is equal to 0 that
is at this point, at this point, we will have
360
00:49:34,359 --> 00:49:41,099
the maximum voltage, when alpha is equal to
pi by 2, the output voltage is 0; then alpha
361
00:49:41,099 --> 00:49:47,420
is equal to pi, it will be negative maximum.
So, it will go like this.
362
00:49:47,420 --> 00:49:54,420
Now, let us go for a general m phase. We have
calculated this one for a general m phase.
363
00:49:59,170 --> 00:50:06,170
If you see here, let us mark m is equal to
2, 3 then 6, it can go back to a. Then, let
364
00:50:14,470 --> 00:50:21,470
us also mark it here that is e0 alpha by the
e0 maximum. If you put it, for the two phase
365
00:50:29,369 --> 00:50:36,369
operation, it will be 0.9; for the three phase,
our midpoint it is 1.17 that is the ratio
366
00:50:45,440 --> 00:50:52,440
- the output DC with respect to the maximum
value that is what is happening here 1.17;
367
00:50:53,220 --> 00:51:00,220
then for six phase, it is 1.35 sorry I have
made a mistake here, the ratio it is maximum,
368
00:51:04,700 --> 00:51:11,700
it cannot be 1.17, it is with respect to Ed
E0 maximum given by E rms.
369
00:51:14,309 --> 00:51:21,309
So here, what we are going to write? It is
not rms, it is not the output voltage, the
370
00:51:22,849 --> 00:51:29,849
ratio it is with respect to the maximum value
divided by the E rms value that is with respect
371
00:51:31,859 --> 00:51:38,859
to maximum; so it will be e0 maximum output
maximum divided by E rms. So here, if you
372
00:51:46,510 --> 00:51:53,510
see here, it will be 0.9 that is here; for
three phase, 1.17; then for six phase, it
373
00:51:54,859 --> 00:52:01,119
is 1.35; we can have a twelve phase that is
extend this one and write twelve phase here,
374
00:52:01,119 --> 00:52:08,119
here it will be 1.40; again we can have eighteen
phase, here it will be approximately again
375
00:52:10,290 --> 00:52:10,720
1.40.
376
00:52:10,720 --> 00:52:17,720
So, as the number of phases if it can increase;
so why I am saying this one, from two phase
377
00:52:17,960 --> 00:52:23,940
to three phase we have gone then what we found
is the repel, the output repel will be three
378
00:52:23,940 --> 00:52:30,619
times the fundamental frequency that is also
clear from this figure from the output repel
379
00:52:30,619 --> 00:52:34,089
frequency, three times the fundamental frequency.
380
00:52:34,089 --> 00:52:40,299
So, as the number of phases increases, the
output repel can also increase but we are
381
00:52:40,299 --> 00:52:45,559
not going to get a considerable margin in
the output DC voltage. So, what we are mostly
382
00:52:45,559 --> 00:52:50,339
interested in the three phase side and the
single phase. So, three phase half bridge;
383
00:52:50,339 --> 00:52:57,339
the maximum value with respect to E rms is
equal to 1.17. What is meant by this one?
384
00:53:00,530 --> 00:53:07,530
For a three phase, the output e0 alpha is
equal to 1.17 E rms value into cos alpha.
385
00:53:15,799 --> 00:53:19,030
This is for a three phase midpoint configuration.
386
00:53:19,030 --> 00:53:25,470
We will assume a symmetrical operation like
the single phase; if we can go for the full
387
00:53:25,470 --> 00:53:29,040
phase control three phase fully controlled
convert, then we have the positive side equal
388
00:53:29,040 --> 00:53:34,530
way controlling, negative side also. Then
what should be for any firing angle; if you
389
00:53:34,530 --> 00:53:39,059
generally think what would be the maximum
output voltage? Multiplied by 2, it has to
390
00:53:39,059 --> 00:53:46,059
be 1.17 into 2 that means 2.34 four into E
rms into cos alpha.
391
00:53:49,140 --> 00:53:56,140
So, now we will go for the three phase configuration.
How it is possible? Where is the firing range?
392
00:53:59,480 --> 00:54:05,720
How the repel comes? Most of the high power
applications or drive applications, three
393
00:54:05,720 --> 00:54:08,990
phase will move or drive application, we will
be drawing power from the three phase and
394
00:54:08,990 --> 00:54:14,260
the how the output voltage can be controlled
with respect to firing angle for m inductive
395
00:54:14,260 --> 00:54:14,690
load?
396
00:54:14,690 --> 00:54:21,170
Now, we have talked about we have not talked
about the resistive load for the three phase
397
00:54:21,170 --> 00:54:28,170
as well as three phase case. That also we
will cover in the next class. So, next class
398
00:54:28,260 --> 00:54:33,609
we will go for the three phase and first we
will start with resistive load, then inductive
399
00:54:33,609 --> 00:54:38,869
load with firing angle; how the output repel
will happen? We will conclude on the three
400
00:54:38,869 --> 00:54:41,260
phase controlled converters.
401
00:54:41,260 --> 00:54:45,480
Then we will talk about the non-idealities.
As I told, we will put the small leakage inductance
402
00:54:45,480 --> 00:54:52,480
small inductance across each thyristors. Why
it is turned? To take care of the commutation,
403
00:54:52,660 --> 00:54:56,359
during the commutation period that means when
thyristor is one thyristor is turned off and
404
00:54:56,359 --> 00:55:01,410
another thyristor is turned on during the
delay period, final turn on and turn off time.
405
00:55:01,410 --> 00:55:06,160
So, the two phases can be short circuited.
So, to avoid short circuit: why this is getting
406
00:55:06,160 --> 00:55:11,589
short circuited? The incoming phase and outgoing
phase, the voltages are not equal. So, there
407
00:55:11,589 --> 00:55:17,220
is a difference in voltage. Because of the
difference in voltage, heavy and the diodes
408
00:55:17,220 --> 00:55:20,740
are conducting simultaneously conducting that
can be a short circuit.
409
00:55:20,740 --> 00:55:25,750
So, to avoid the short circuit, avoid the
heavy current; so the difference in voltage
410
00:55:25,750 --> 00:55:30,460
has to be for a momentarily for microsecond
duration during the turn on and turn off period,
411
00:55:30,460 --> 00:55:36,089
it has to be across some impedance we have
to provide. That is why we have to provide
412
00:55:36,089 --> 00:55:39,780
the leakage inductance and how to design?
What are the effects of this one? Whether
413
00:55:39,780 --> 00:55:44,609
it is going to affect our output DC voltage
or extra component we have put; we will study
414
00:55:44,609 --> 00:55:45,900
in the next classes.