1
00:01:06,000 --> 00:01:11,400
Last class, we talked about PWM converter
using ac to dc converter that is using thyristors
2
00:01:11,530 --> 00:01:18,530
and phase control converters. There we required,
for turn on and turn off of the thyristor,
3
00:01:19,170 --> 00:01:25,509
especially turn off; turn on, we can give
gate pulse, to turn off, we have to give forced
4
00:01:25,509 --> 00:01:32,509
commutation. So, we will study a very popular
forced commutation of thyristor and this forced
5
00:01:33,780 --> 00:01:40,780
commutation, we will study in the context
of a dc to dc converter that is a chopper.
6
00:01:41,100 --> 00:01:47,179
So, let us talk about dc to dc converter.
It is a very popular scheme used before by
7
00:01:47,179 --> 00:01:50,990
thyristors, for thyristors for commutation.
8
00:01:50,990 --> 00:01:57,990
So, dc to dc conversion: here, the circuit,
let us say this is our input, let us say this
9
00:02:12,069 --> 00:02:19,069
is our battery. Then you have the thyristor,
main thyristor to be commutated, this a basic
10
00:02:27,590 --> 00:02:34,590
buck converter configuration, this is your
freewheeling diode DFW freewheeling, then
11
00:02:36,370 --> 00:02:43,370
you have the motor load here. For forced commutation,
for turn on this device, we can give a gate
12
00:02:50,920 --> 00:02:57,920
pulse here. Thyristor will be turned on. So,
this we will talk about switch 1 S1.
13
00:03:02,739 --> 00:03:09,739
Now, to turn off this thyristor, we will give
a reverse bias across the thyristor from a
14
00:03:11,090 --> 00:03:18,090
charged capacitor. So, this is the capacitor
to give. Initially, this capacitor should
15
00:03:19,540 --> 00:03:26,540
be charged, it will be charged through a resistance.
This is the switch, then a resistance. So,
16
00:03:29,099 --> 00:03:35,790
during turn on of the chopper circuit or dc
to dc converter, first thing is you have to
17
00:03:35,790 --> 00:03:42,790
turn on this switch S1 SW such that the capacitor
will be charged to input voltage E. Let us
18
00:03:44,849 --> 00:03:51,849
say, this is E. So, knowing the R and C, we
know the approximate time constant. So, through
19
00:03:53,950 --> 00:03:59,359
a monoshot, we can turn on this switch and
after that we can switch off it. By the time,
20
00:03:59,359 --> 00:04:05,349
capacitor will be charged to this voltage
here.
21
00:04:05,349 --> 00:04:12,349
Now, but this is not the voltage we have to
apply across the thyristor. For thyristor
22
00:04:13,659 --> 00:04:18,070
to turn off, you have to apply or we have
to reverse bias that means anode should be
23
00:04:18,070 --> 00:04:23,960
negative compared to cathode. That means we
have to reverse the capacitor voltage before
24
00:04:23,960 --> 00:04:28,500
the thyristor can be turned off. So, there
is an auxiliary resonance circuit we will
25
00:04:28,500 --> 00:04:35,500
be using here. This is the resonance circuit,
it constitute one diode, then L, there is
26
00:04:40,300 --> 00:04:47,300
a resonance circuit comes across this one.
This diode is given, this LC will act as the
27
00:04:47,820 --> 00:04:52,680
resonance circuit such that the diode is given
such that once the capacitor charge in the
28
00:04:52,680 --> 00:04:58,850
opposite direction due to LC, it should not
discharge. So, there is no reverse conduction.
29
00:04:58,850 --> 00:05:05,690
To block the reverse conduction, we have given
this diode D. So, this capacitor C will be
30
00:05:05,690 --> 00:05:12,340
charged. So, L and C will act as the resonance
circuit.
31
00:05:12,340 --> 00:05:19,340
Now, there is another auxiliary thyristor
is there. This is this one, this we will call
32
00:05:25,020 --> 00:05:32,020
as S2; this is S1, this is S2, S2 we will
mark with a different colour so that it is
33
00:05:34,200 --> 00:05:41,200
visible, this is S2. Now, how the chopper
works? This is a buck converter that means
34
00:05:44,950 --> 00:05:51,780
we can have voltage less than, dc voltage
less than the input E voltage, variable voltage.
35
00:05:51,780 --> 00:05:58,780
So, this is a dc to converter, buck converter.
So, this whole thing will act as the commutation
36
00:06:00,110 --> 00:06:05,490
circuit including the switch S1.
37
00:06:05,490 --> 00:06:12,490
So, as I told, the first thing is one to start
on; turn on SW for a small period. This period
38
00:06:37,050 --> 00:06:44,050
depends on the time constant RC of the system
so that capacitor will be such that the capacitor
39
00:06:44,860 --> 00:06:51,860
is charged to E such that initially, the capacitor
is charged as shown in figure. Capacitor is
40
00:07:04,250 --> 00:07:11,250
charged to E as shown in figure 1. So, let
us talk about this is figure 1. Now, capacitor
41
00:07:26,720 --> 00:07:32,150
is charged. Now, we are ready for turning
on this chopper. So now, you are turning on
42
00:07:32,150 --> 00:07:36,970
the chopper. First, this is only the initial
charging process.
43
00:07:36,970 --> 00:07:43,970
Now, once the capacitor is charged to E here;
the chopper is ready for action. Now, S1 is
44
00:07:46,870 --> 00:07:53,870
turned on, S1 is turned on. S1 is turned on;
what will happen? Immediately, this E will
45
00:08:07,160 --> 00:08:14,160
come across the DF. So, DF is reverse biased
and E will come across the load side. This
46
00:08:16,450 --> 00:08:21,950
E which is coming across the load depends
on when we are the duration, depends on when
47
00:08:21,950 --> 00:08:28,950
we are turning off S1. So, first let us say,
we are turning on S1. So, how the circuit
48
00:08:30,680 --> 00:08:34,940
will be? See, the circuit, I will draw here.
49
00:08:34,940 --> 00:08:41,940
We have E here that is E here, thyristor is
turned on. So, this we will say, this is a
50
00:08:42,300 --> 00:08:49,300
turned on switch. This is off S1 is on, diode
is reversed biased so that will not come into
51
00:08:51,620 --> 00:08:58,620
picture, we have this load. Now, there is
a capacitor here, charged capacitor in this
52
00:09:05,870 --> 00:09:12,870
direction and there is this diode and L. So,
moment it is turned on, the load with, load
53
00:09:20,770 --> 00:09:26,720
is, we want the load, as a dc circuit, so
load because of the high inductance, the load
54
00:09:26,720 --> 00:09:33,720
will flow here, IL dc. But because of this
L and C, there is a resonant discharge of
55
00:09:38,420 --> 00:09:43,790
C or resonant charge of C in the opposite
direction through which path the L and C.
56
00:09:43,790 --> 00:09:49,920
So, we will mark that one with a different
colour so that it will be visible. So, apart
57
00:09:49,920 --> 00:09:56,920
from the load, there is a resonant current
here. This resonant current, we will mark
58
00:10:00,760 --> 00:10:04,470
it as IR.
59
00:10:04,470 --> 00:10:11,470
So, how long it will... because of the LC
resonance, this capacitor will charge in the
60
00:10:16,510 --> 00:10:23,510
opposite direction. So, how the waveform will
be? So, the resonant charging, the capacitor
61
00:10:24,670 --> 00:10:31,670
with the whole voltage and camp waveform will
look like this. So, the capacitor, initially
62
00:10:33,660 --> 00:10:40,660
charged to E here; with resonant charging,
it will go like this and come to minus E here,
63
00:10:42,480 --> 00:10:49,440
it is a part of a sinusoidal that is it will
charge in the opposite direction. So, what
64
00:10:49,440 --> 00:10:56,440
will happen to the current? Current if you
see here that is IR current. It will slowly
65
00:10:57,430 --> 00:11:04,430
increase, at this point, current is maximum,
so di by dt is minimum. Then it will go like
66
00:11:05,279 --> 00:11:10,490
this, this is our IR. The moment the current
is 0; assuming ideal LC circuit, the capacitor
67
00:11:10,490 --> 00:11:17,490
will charge to the minus V here. So, this
is our IR. Once the capacitor is charged,
68
00:11:23,090 --> 00:11:29,940
the other direction; this capacitor is ready
for commutation of S1 at any time.
69
00:11:29,940 --> 00:11:36,940
Now, this capacitor can whenever the turn
on this S2 here, whenever turn on this S2;
70
00:11:37,690 --> 00:11:42,940
now the capacitor is charged in the opposite
direction, it will apply a reverse voltage
71
00:11:42,940 --> 00:11:48,560
across the thyristor and thyristor will switch
off immediately. Now, this period, we should
72
00:11:48,560 --> 00:11:53,850
note this one, this is the resonant half period,
this resonant half period if you see what
73
00:11:53,850 --> 00:12:00,850
the period; the omega r, the resonant frequency
omega r is equal to 1 by this standard circuit,
74
00:12:05,060 --> 00:12:11,800
1 by root LC, 1 by root LC.
75
00:12:11,800 --> 00:12:18,800
Now, this is the omega r; so what is a period,
half resonant period? Omega r is equal to
76
00:12:24,250 --> 00:12:31,250
omega r is equal to if Tr is the half resonant
period, it will be 2 pi by Tr, this is the
77
00:12:36,190 --> 00:12:43,190
half resonant period. So, 2 pi by omega r,
so Tr will be equal to we can find out pi
78
00:12:48,870 --> 00:12:55,870
into root LC because we know it is omega r
is equal to omega r is equal to 2 pi into
79
00:13:03,779 --> 00:13:10,779
Fr and Fr is equal to 1 by Tr. Now, why this
period is very important? So, during the turn
80
00:13:19,230 --> 00:13:26,230
on process, this period half Tr, we are using
for resonant charging of the capacitor in
81
00:13:26,830 --> 00:13:33,830
the opposite direction. So, what it means?
This means sorry this is Tr is equal to 2
82
00:13:34,790 --> 00:13:41,790
pi, Tr is equal to sorry I will again mark
this one; Tr is equal to here, Tr is equal
83
00:13:45,110 --> 00:13:52,110
to Tr is equal to 2 pi into root LC, this
is the one.
84
00:13:58,480 --> 00:14:05,480
Now, this is very important so this much period,
we require for capacitor to resonant charging.
85
00:14:07,350 --> 00:14:14,210
So, once the capacitor is ready for commutation
that means capacitor is charging the opposite
86
00:14:14,210 --> 00:14:18,589
direction, then only it will be ready for
commutation, commutating the main thyristor
87
00:14:18,589 --> 00:14:25,589
S1that is this S1 here, here is the S1, this
S1. Now, so this period, resonant period is
88
00:14:30,830 --> 00:14:36,660
as small as possible. How much small? That
we will come to that one later for designing
89
00:14:36,660 --> 00:14:41,230
L and C; so, assuming the capacitor is charged
now, it is ready for commutation.
90
00:14:41,230 --> 00:14:48,230
Now, let us take the condition when we turn
on S2 that means we are ready for the commutation
91
00:14:49,750 --> 00:14:56,750
and now we are turning on S2 to turn off S1.
So, let us go to the next page. Here, what
92
00:14:57,000 --> 00:15:04,000
will happen? S1 is turned on immediately.
So, let us draw the circuit during commutation.
93
00:15:07,070 --> 00:15:14,070
Now, turn off period that is turn off of S1
that is a main thyristor. Now, how the circuit
94
00:15:22,740 --> 00:15:29,740
looks? We have E here, S1 is turned on when
S2 is turned on. So, the capacitor, the condition
95
00:15:34,990 --> 00:15:41,399
immediately S1 is turned on, it will look
like this. Capacitor is in this direction
96
00:15:41,399 --> 00:15:48,399
charged; now S2 is turned on. So, S2 turned
on, we will make it like this, the circuit
97
00:15:49,279 --> 00:15:56,279
is like this. We have the diode, see diode
is not yet turned on. Why? when S1 is turned
98
00:15:59,010 --> 00:16:05,510
on, the capacitor voltage C that is also equal
to E, the capacitor voltage, the capacitor
99
00:16:05,510 --> 00:16:12,510
C, the voltage across the capacitor is E.
Then E, this 2 E will come here at this point,
100
00:16:13,680 --> 00:16:20,490
let us say this is A, at this A point and
diode is reverse biased, diode will not conduct
101
00:16:20,490 --> 00:16:21,860
now.
102
00:16:21,860 --> 00:16:28,860
So, still load is highly inductive and load
should get the full dc current. So, current
103
00:16:31,680 --> 00:16:38,680
cannot changes in this inductance. So still,
this dc IL will flow that means this IL will
104
00:16:39,660 --> 00:16:46,490
come through C1 S2 and through L and go to
the load. So, the current path, we will mark
105
00:16:46,490 --> 00:16:53,490
now. The current path is like this and returns;
this is our IL dc current. So, what it means
106
00:17:07,779 --> 00:17:14,779
is the capacitor will slowly discharged, capacitor
will be charged by a constant current IL or
107
00:17:20,370 --> 00:17:25,770
the capacitor will discharge, discharge and
then charge in the opposite direction.
108
00:17:25,770 --> 00:17:32,770
Now, the capacitor will be
discharged with a constant current equal to
IL, discharged with a constant current, current
109
00:17:54,539 --> 00:18:01,539
that depends on the load that is IL. So, capacitor
will slowly discharge and again see, till
110
00:18:06,360 --> 00:18:11,940
the diode is turned on, load cannot free freewheel.
Once the load freewheels, the commutation
111
00:18:11,940 --> 00:18:18,940
circuit is totally isolated. So, for the diode
to be this diode to be forward biased, this
112
00:18:22,619 --> 00:18:29,619
capacitor should charge in the opposite direction
that is in this direction to E through using
113
00:18:30,539 --> 00:18:31,489
IL.
114
00:18:31,489 --> 00:18:38,190
So, when it becomes E, it will oppose the
input voltage E and any slight overcharge,
115
00:18:38,190 --> 00:18:44,419
diode will be forward biased due to the capacitor.
Then the diode will discharge. So, what will
116
00:18:44,419 --> 00:18:51,419
happen? Here, now the capacitor will be discharged
with constant current IL, then after that
117
00:18:51,590 --> 00:18:58,590
capacitor is discharged, again capacitor is
charged to capacitor will be charged to E;
118
00:19:08,019 --> 00:19:15,019
capacitor will be charged to E with the constant
load current. So, then what happens? Once
119
00:19:29,970 --> 00:19:36,970
the capacitor charged to this point that is
this, this way, it will oppose E and the diode
120
00:19:39,100 --> 00:19:45,299
assuming very small difference voltage, diode
forward bios very small voltage only required;
121
00:19:45,299 --> 00:19:49,779
so diode will turn on and the load will freewheel,
load will freewheel through this path. Now,
122
00:19:49,779 --> 00:19:50,759
the commutation is completed.
123
00:19:50,759 --> 00:19:57,759
What happens to
the capacitor voltage at this point? Capacitor
voltage again, it charges in the opposite
124
00:20:05,659 --> 00:20:11,499
direction. Now, you do not require the initial
original that turn on circuit to charge the
125
00:20:11,499 --> 00:20:18,499
capacitor that is this switch SW and R is
not required to turn on the capacitor. Now,
126
00:20:18,970 --> 00:20:25,529
the capacitor itself will take care of the
thing. How? If you see here, because of this
127
00:20:25,529 --> 00:20:32,529
load current; now, how the capacitor condition
will be? So, diode is
this is switched off, this is S1 here, capacitor
128
00:20:39,720 --> 00:20:46,720
is charged to in the opposite direction E,
then you have the E here, this is E, this
129
00:20:51,989 --> 00:20:58,989
is also E. So, this is the circuit, it is
ready. Capacitor is charged and again as when
130
00:21:00,350 --> 00:21:05,580
S1 is turned on, it will through resonance
charging, it will charge in the opposite direction
131
00:21:05,580 --> 00:21:06,940
and is ready for commutation.
132
00:21:06,940 --> 00:21:13,940
Now, what is more important thing is how to
find out this L and C so that the resonant
133
00:21:15,749 --> 00:21:22,259
charging period is as small as possible compared
to the Ton time of the S1. Why? The output
134
00:21:22,259 --> 00:21:29,259
voltage depends on V0; from the buck converter,
we know. V0 is equal to input E into Ton divide
135
00:21:38,529 --> 00:21:45,529
by the total period T. So, this is the way.
So, this Ton for a variable voltage, the output
136
00:21:46,320 --> 00:21:52,289
voltage depends on the Ton. So, the minimum
duty cycle, so our resonant, half resonant
137
00:21:52,289 --> 00:21:59,289
period Tr should be much much less than the
minimum Ton period; so one condition.
138
00:22:02,159 --> 00:22:09,159
So, one condition only there, now we have
two values to find out L and C. So, two conditions
139
00:22:09,909 --> 00:22:16,909
are required. So, how to find out the C? Let
us start with, the C, we have introduced the
140
00:22:17,779 --> 00:22:24,259
C in the circuit so that we can give a voltage,
a force. We can force a voltage across the
141
00:22:24,259 --> 00:22:30,850
thyristor to switch off that is forced commutation
and we assume during the previous analysis
142
00:22:30,850 --> 00:22:37,739
that the S1, switch S1 will commutate, switch
out immediately. But in all, on all practical
143
00:22:37,739 --> 00:22:44,289
circuit, in all practical circuit, the thyristor
will take some finite turn on time turn off
144
00:22:44,289 --> 00:22:50,239
time. So, this we can find out from the datasheet.
So, that means any thyristor will have a minimum
145
00:22:50,239 --> 00:22:56,919
toff time. So, during the full toff period,
till the thyristor is turned off that is during
146
00:22:56,919 --> 00:23:03,919
the full toff period, we have to apply the
reverse voltage across the thyristor C. So,
147
00:23:05,330 --> 00:23:12,330
from this circuit when T1 is turned on, the
voltage across the C was this one, plus and
148
00:23:12,840 --> 00:23:14,450
minus here.
149
00:23:14,450 --> 00:23:21,450
Now, due to the resonant charging, it will
come to the negative voltage here. With resonant
150
00:23:26,739 --> 00:23:33,739
charging, this negative voltage will come
here. Now, when S2 is turned on when S2 is
151
00:23:40,700 --> 00:23:47,700
turned on, this negative voltage which is
marked with blue colour will come across the
152
00:23:47,869 --> 00:23:54,869
thyristor and the thyristor will discharge
due to the load current. So, this diode S1
153
00:23:55,109 --> 00:24:02,109
will be reverse biased till this diode, till
this thyristor discharge from, it will come
154
00:24:03,809 --> 00:24:10,809
from E to 0 through the constant dc current.
So, this much time, we will be reverse biasing
155
00:24:11,119 --> 00:24:16,769
the thyristor. So, for a circuit, the load
current can vary. So, if the load current
156
00:24:16,769 --> 00:24:20,609
is maximum, high; the capacitor discharge
will be fast.
157
00:24:20,609 --> 00:24:27,609
So, for a maximum load current also, the discharge
time should be more than the Toff minimum
158
00:24:29,909 --> 00:24:36,909
of the thyristor. That means during discharging
of the capacitor, during discharging of the
159
00:24:57,309 --> 00:25:04,309
capacitor, capacitor with constant load current;
the discharge period that means it is completely
160
00:25:25,049 --> 00:25:32,049
discharged to zero period that should be more
than the Toff minimum of the thyristor. So,
161
00:25:37,929 --> 00:25:42,039
it will be capacitor is assuming, this is
the discharge period; let us take the discharge
162
00:25:42,039 --> 00:25:46,479
period is equal to tq.
163
00:25:46,479 --> 00:25:53,479
So, tq into I load maximum that is for any
circuit; what is the maximum I load current?
164
00:26:08,350 --> 00:26:15,350
This is the total charge; this will be equal
to C into E. So, I will say this is the I
165
00:26:18,499 --> 00:26:25,499
load maximum current that depends on the load.
So, this is the equation. From this one, the
166
00:26:27,249 --> 00:26:34,249
tq, the time required for the capacitor discharge
fully from E to zero voltage is equal to tq
167
00:26:38,460 --> 00:26:45,460
is equal to CE, CE divide by IL, CE divide
by IL. So, from this one and the tq, tq should
168
00:27:08,489 --> 00:27:15,489
be more than the Toff minimum; so from that
one, let us go to the next page so that we
169
00:27:16,029 --> 00:27:18,479
can make it clear.
170
00:27:18,479 --> 00:27:25,479
So, from the CE, CE is equal to tq, the discharge
period tq into IL maximum, maximum load current.
171
00:27:32,850 --> 00:27:39,759
Load current can vary; we know the circuit,
we know the load, maximum, we know load, current
172
00:27:39,759 --> 00:27:46,759
should be known. So, for the maximum current,
this tq, tq should be; what is tq? tq is equal
173
00:27:50,940 --> 00:27:57,940
to CE divided by IL. This tq should be greater
than or equal? So, greater: how much greater?
174
00:28:03,769 --> 00:28:10,470
That depends on your design. You have to go
back and forth and decide; 10% more of the
175
00:28:10,470 --> 00:28:12,929
tq minimum, so depending on your design.
176
00:28:12,929 --> 00:28:18,210
So, there is not; here, we cannot say, exactly
this one. This depends on the designer's
177
00:28:18,210 --> 00:28:23,849
capability how to give a cause effective solution
but this is the approach. So, it should be
178
00:28:23,849 --> 00:28:30,849
greater than Toff minimum, Toff minimum that
will be specified in the datasheet; Toff minimum
179
00:28:33,279 --> 00:28:40,279
for the main thyristor, for the main thyristor
S1. From this equation, we can find out tq,
180
00:29:02,590 --> 00:29:09,590
this is C. So, from this equation, we can
find out, from this, E is known, IL is known,
181
00:29:14,379 --> 00:29:21,379
Toff minimum is known; from this equation,
we can find out C. How? From this inequality,
182
00:29:21,570 --> 00:29:28,570
C will be greater than or equal to Toff greater
than or equal to Toff minimum, there is a
183
00:29:32,049 --> 00:29:39,049
thyristor turn off time divided by
E into IL load, I load maximum.
184
00:29:43,229 --> 00:29:50,229
So, first you compute this value and the value
C chosen should be greater than this value,
185
00:29:51,739 --> 00:29:58,119
this will ensure your commutation. Now, there
is another thing. Capacitor we have found
186
00:29:58,119 --> 00:30:05,119
out, there is one more component, L we have
to find out. So, how we will find out L? Let
187
00:30:06,080 --> 00:30:12,710
us go back to our original circuit again.
Let us go to the next page. See, during turn
188
00:30:12,710 --> 00:30:19,349
on, the moment S1 is turned on, the load current
will flow, also the resonant current will
189
00:30:19,349 --> 00:30:23,460
be flowing through the main switch S1.
190
00:30:23,460 --> 00:30:30,460
That is this is our E, main switch is turned
on that is this is our main switch, capacitor
191
00:30:36,509 --> 00:30:43,509
initially charged to this value, then you
have the diode L. Here, you have the load
192
00:30:53,599 --> 00:31:00,599
circuit. So, the resonant current which flows
through C and L, so we will mark it like this
193
00:31:09,440 --> 00:31:16,440
that is this current IR; what is the equation
for the resonant current for the LC circuit?
194
00:31:20,950 --> 00:31:27,950
This IR, we know it, IR is equal to voltage
across the capacitor is E. So, E into root
195
00:31:32,359 --> 00:31:39,359
C by L into sin omega resonant frequency that
is 1 by root LC into t.
196
00:31:44,190 --> 00:31:51,190
So, this resonant peak current that which
we have drawn before, the current which go
197
00:31:52,940 --> 00:31:59,940
and the capacitor and charge the capacitor
in the opposite direction
that is this one - plus E and minus E; this
198
00:32:12,039 --> 00:32:19,039
resonant current that is this peak value is
equal to E into root C by L, this value will
199
00:32:26,570 --> 00:32:30,619
go through the S1.
200
00:32:30,619 --> 00:32:37,619
So, during S1 on, the current flowing through
the S1 are IL plus the peak current E into
201
00:32:46,099 --> 00:32:51,609
root C by L, this is the peak current. This
much peak current, every turn on; this will
202
00:32:51,609 --> 00:32:58,609
happen for every turn on, every turn on, so
load current that is a maximum load current
203
00:33:04,090 --> 00:33:11,090
plus E into root C L will go through the thyristor.
So, this current, it is a, this E into root
204
00:33:12,729 --> 00:33:19,729
C by L is a small duration. So, any thyristor
will have a repetitive peak current that is
205
00:33:19,729 --> 00:33:26,729
repetitive peak current. Let us see that we
will make it as Ip Ip or repetitive peak current.
206
00:33:30,129 --> 00:33:37,129
Any thyristor datasheet will give this one.
207
00:33:38,840 --> 00:33:45,840
So, this data, so we should ensure that any
time this repetitive peak current should be
208
00:33:47,389 --> 00:33:54,389
our IL plus is IL plus E into root C by L
that is IL plus E into root C by L should
209
00:34:01,830 --> 00:34:08,830
be always less than or equal to the repetitive
peak current or the resonant current E into
210
00:34:22,659 --> 00:34:29,659
root C by L should be less than or equal to
Ip minus IL maximum, load maximum current.
211
00:34:40,450 --> 00:34:44,270
Let us represent this as I0.
212
00:34:44,270 --> 00:34:51,270
So, this condition shows that the condition
what you have written here; E into root C
213
00:34:57,520 --> 00:35:04,520
by L should be less than or equal to I0. What
is I0? I0 is equal to repetitive maximum Ip
214
00:35:07,700 --> 00:35:14,700
minus I load maximum, I load maximum. This
is equal to E into E square into C by L should
215
00:35:24,579 --> 00:35:31,579
be less than or equal to I0 square. That means
L should be less than or equal to E square
216
00:35:42,690 --> 00:35:49,690
C by I0 square; one condition. So, from here,
we found L should be or E0 square E0 square
217
00:35:59,319 --> 00:36:05,950
C by I0 square should be less than or equal
or else should be greater than or equal to
218
00:36:05,950 --> 00:36:09,240
this value.
219
00:36:09,240 --> 00:36:16,240
But there is not sufficient, there is one
more condition; we want to do it. From the
220
00:36:16,630 --> 00:36:23,630
resonance circuit, we found that omega r is
equal to 1 by root LC. This is equal to 2
221
00:36:29,140 --> 00:36:36,140
pi into fr or 2 pi by Tr where Tr is the Tr
is the resonant period. So, Tr is equal to
222
00:36:46,809 --> 00:36:53,809
2 pi into root LC. But what we are talking
about the half resonant period because we
223
00:37:01,240 --> 00:37:05,880
are not allowing the current to because of
the diode, we are not allowing the current
224
00:37:05,880 --> 00:37:12,880
to flow in the negative direction. The half
resonant period, so the half resonant period,
225
00:37:14,720 --> 00:37:21,720
half period is equal to Tr by 2; we will mark
as Tr dash is equal to pi root LC and we want
226
00:37:38,369 --> 00:37:45,369
this half resonant period pi root LC should
be smaller than the Toff minimum much smaller
227
00:37:48,960 --> 00:37:52,680
than or the condition Toff minimum required.
228
00:37:52,680 --> 00:37:59,680
So, to ensure we will say 10%, so these are
all design criteria, 10% of the duty cycle,
229
00:38:01,520 --> 00:38:08,520
minimum duty cycle; so 10% 0.1 of Ton minimum.
Let us square this one that means pi square
230
00:38:23,410 --> 00:38:30,410
into L into C less than or equal to zero point
squaring 0.01 into Ton square minimum. This
231
00:38:41,200 --> 00:38:48,200
implies, from this inequality; what we can
find out? L should be less than or equal to
232
00:39:06,099 --> 00:39:13,099
0.01 Ton square minimum divided by pi square
C.
233
00:39:20,260 --> 00:39:27,260
So, let us go back to the equality that means
L, this should be 0.01 Ton square divide by
234
00:39:38,410 --> 00:39:45,410
pi square C, this is the condition. So, we
can compute this value; E is known, C is known,
235
00:39:49,770 --> 00:39:56,770
I0 known, I0 is equal to Ip minus IL, I0 is
equal to this value. Then the Ton minimum
236
00:40:01,319 --> 00:40:06,109
is known that is this one depending on the
variable duty cycle, minimum chopper. Capacitor
237
00:40:06,109 --> 00:40:11,579
value we have initially found out from the
Toff minimum required to switch off the thyristor.
238
00:40:11,579 --> 00:40:17,859
So, from this value, we can find out the maximum
range of L and the minimum range of L; we
239
00:40:17,859 --> 00:40:24,859
have to choose L in between. So, commutation
will be pakka.
240
00:40:25,200 --> 00:40:31,309
So, using simple rules, using engineering
approximation; one we said, the load is highly
241
00:40:31,309 --> 00:40:36,559
inductive, so load is linear. So, from the
linear charging of discharging period, we
242
00:40:36,559 --> 00:40:43,559
have found out the Toff maximum or Toff minimum
required for the capacitor to discharge fully
243
00:40:44,670 --> 00:40:50,170
so that the thyristor will be reverse biased,
we found out the C. Then from the repetitive
244
00:40:50,170 --> 00:40:55,670
peak, there is known the resonant peak and
the load current; one condition for L, one
245
00:40:55,670 --> 00:41:00,359
range of L we have found, the lower range.
Then the upper range, we found out the minimum
246
00:41:00,359 --> 00:41:07,319
resonant period, resonant period. That should
be much less than Ton duty cycle minimum of
247
00:41:07,319 --> 00:41:13,990
the circuit. From this one, we can find out
the L and C and the computation is complete.
248
00:41:13,990 --> 00:41:20,390
So, with this, that force commutation; we
are only talking about the dc to dc converter,
249
00:41:20,390 --> 00:41:27,059
just to introduce the dc to dc converter we
introduce this commutation, very widely used
250
00:41:27,059 --> 00:41:34,059
commutation circuit, popular commutation circuit.
Now, what we want to do? The purpose of this
251
00:41:35,029 --> 00:41:42,029
introducing this force commutation; what we
want? We want a PWM operation of the thyristor
252
00:41:45,970 --> 00:41:52,970
so that we can have unity power factor from
the mains. So, unity power factor, at the
253
00:41:54,619 --> 00:42:01,269
same time, the converter should be able to
work in the four quadrant region so that power
254
00:42:01,269 --> 00:42:04,760
flow can be from the input to the load from
load to the mains.
255
00:42:04,760 --> 00:42:11,400
So, how and what is the basic configuration
for a four quadrant converter? Let us start.
256
00:42:11,400 --> 00:42:18,400
Let us talk about the converter; a front end
ac to dc converter with unity power factor,
257
00:42:19,289 --> 00:42:24,849
at the same time four quadrant operations
is also possible. Let us talk about how we
258
00:42:24,849 --> 00:42:31,260
can have a converter with four quadrant operation.
259
00:42:31,260 --> 00:42:38,260
Converter; it can be used that is converter
ac to dc converter; so for we are talking
260
00:42:47,609 --> 00:42:54,609
about ac to dc converter with four quadrant
operation. We talked about our basic dc to
261
00:43:13,019 --> 00:43:20,019
converter previously that is the buck converter,
the configuration is like this; we have this
262
00:43:23,299 --> 00:43:30,299
one, then the inductance, we have the load,
freewheeling diode d here. This is our E.
263
00:43:36,380 --> 00:43:43,380
Now, we are not showing the commutation circuit;
commutation circuit is inside it is there.
264
00:43:47,579 --> 00:43:54,579
Here if you see here because of the thyristor
main thyristor S1, we will do with a different
265
00:43:56,039 --> 00:44:03,039
colour, because of this thyristor S1, the
load current can conduct only in one direction.
266
00:44:08,559 --> 00:44:15,559
That means this output voltage can vary from
zero to positive only and current only one
267
00:44:16,369 --> 00:44:16,880
direction.
268
00:44:16,880 --> 00:44:23,880
So, if you see the power diagram that is E0
versus I0 let us say load current I0, E0 versus
269
00:44:26,049 --> 00:44:33,049
I0; only one quadrant is only possible, single
quadrant operation. This is E0 and this is
270
00:44:39,170 --> 00:44:46,170
I0. Now, we want a power flow from load side
to the, load side to the source side. So,
271
00:44:53,240 --> 00:45:00,240
from the buck converter, we know that E0 will
be less that E always, E0 will be less than
272
00:45:03,089 --> 00:45:10,089
E. Now, to want a power flow from E0 side
to E that is power flow from E0 to E; what
273
00:45:27,079 --> 00:45:34,079
is the configuration? We know the boost configuration;
this is buck, this is buck, you know the boost
274
00:45:38,440 --> 00:45:44,500
configuration. So, how do you draw the boost
configuration? Now, the source is our E0,
275
00:45:44,500 --> 00:45:47,109
the power flow is from the load side to the
E.
276
00:45:47,109 --> 00:45:54,109
So, let us draw our load side like this. So,
how do you boost it? First, any boost circuit,
277
00:46:04,519 --> 00:46:11,519
we know that inductor is charged first. So,
here is our thyristor. So, we are using thyristor
278
00:46:13,740 --> 00:46:18,319
here or IGBT or MOSFET, whatever you can use
it. So, since so far we are talking about
279
00:46:18,319 --> 00:46:19,910
thyristor, we will talk about the thyristor
now.
280
00:46:19,910 --> 00:46:26,910
So, this one; so you initially, this is the
main switch S1 is turned on. When S1 is turned
281
00:46:29,250 --> 00:46:36,250
on, full E0 will come across L then S1 is
turned off; what will happen? We will put
282
00:46:41,390 --> 00:46:48,390
a diode here. This is our E. When S1 is turned
off, the current will flow through L and D
283
00:46:56,799 --> 00:47:03,799
that means the current flow when S1 is turned
on, S1 is turned on, the current flow will
284
00:47:05,730 --> 00:47:12,029
be it will flow from because of that stored
energy in the inductor, it will show from
285
00:47:12,029 --> 00:47:19,029
this side to this way. So, what you mean by
that one? Here, we are not changing the polarity
286
00:47:19,410 --> 00:47:25,910
of the voltage, current direction we have
changed. So, we can have two quadrant operations.
287
00:47:25,910 --> 00:47:31,369
Voltage is still positive, only positive.
288
00:47:31,369 --> 00:47:38,369
So, we can have two quadrant operations. Now,
so for one quadrant, we use buck converter;
289
00:47:45,410 --> 00:47:51,650
for the two quadrants, we use other converter.
Can we combine these two so that the circuit
290
00:47:51,650 --> 00:47:58,039
can be buck as well as boost? That mean or
from the power flow point, we can have power
291
00:47:58,039 --> 00:48:05,039
can flow in both direction. So, let us draw
that one. See here, so now, we are going to
292
00:48:09,539 --> 00:48:16,539
combine these two. So, what we want is the
numbers we will change it, this one we will
293
00:48:18,880 --> 00:48:22,920
make it S2. See, our mains, another one is
S0.
294
00:48:22,920 --> 00:48:29,920
So, how the circuit will be? So, E is still
E here, this is our E, this is our old S1,
295
00:48:40,769 --> 00:48:47,769
S1 comes here, this is our load, the
296
00:48:57,990 --> 00:49:04,990
freewheeling diode. This freewheeling diode
is this one. Now, for boosting, S2 is connected
297
00:49:11,230 --> 00:49:18,230
between inductant this side and the load,
so this is here. Same way, we can use it here,
298
00:49:18,890 --> 00:49:25,890
S2 and the diode, this diode will come here,
this will make it D2, this make D1. So, if
299
00:49:32,930 --> 00:49:39,930
you see here, to make the buck and boost independently,
we will colour it here; this is S1 sorry we
300
00:49:46,789 --> 00:49:53,789
will colour the different colour, this is
S1, this is the D2, these two are for boosting
301
00:49:57,839 --> 00:49:59,400
action.
302
00:49:59,400 --> 00:50:06,400
So, we can have two quadrant operations. So,
if you repeat the same thing other side, we
303
00:50:11,670 --> 00:50:16,740
can have four quadrant operations. So, the
circuit here will be final circuit will be
304
00:50:16,740 --> 00:50:23,740
see, I will connect the diode like this; this
side also it is there, now this point, we
305
00:50:51,480 --> 00:50:58,480
will remove and connect it here. So, the load
will be like this. So, what it shows the quadrant
306
00:50:59,789 --> 00:51:06,789
operation here? We can have any operation
from this side, we can have negative operation
307
00:51:08,930 --> 00:51:14,650
also. So, we can all four quadrant operation
is possible that means both the voltage and
308
00:51:14,650 --> 00:51:20,240
current as far as the load is concerned, both
the voltage and current can be positive as
309
00:51:20,240 --> 00:51:27,240
well as negative. So, this can be used, this
configuration is called a four quadrant chopper.
310
00:51:32,160 --> 00:51:39,160
So, if you know this is a single phase inverter
configuration; see, as now the load, we want
311
00:51:39,789 --> 00:51:44,309
both positive and negative, so load will have
alternating current and voltage.
312
00:51:44,309 --> 00:51:51,309
So, if you say positive and negative means
PWM operation is possible. So, if you go to
313
00:51:55,089 --> 00:52:02,089
the next page here sorry we want the load
positive and negative; so positive waveform
314
00:52:07,609 --> 00:52:14,609
and negative waveform, so I put sinusoidal
voltage. Is it possible? See, with duty cycle
315
00:52:16,250 --> 00:52:20,680
control, I can have variable voltage. At many
instant, I can have different voltage so that
316
00:52:20,680 --> 00:52:27,680
I can have, I can have a sinusoidal voltage.
At the same time I can have a sinusoidal current
317
00:52:28,190 --> 00:52:34,930
also that means both positive and negative
is possible; a current something like this
318
00:52:34,930 --> 00:52:37,849
is possible.
319
00:52:37,849 --> 00:52:44,849
Now, for unity power factor, for unity power
factor, the current here should be like this;
320
00:52:50,869 --> 00:52:57,869
unity power factor. So, ac to dc converter,
we are talking about ac to dc converter; so
321
00:52:58,349 --> 00:53:05,349
this is the dc, this is the ac part, ac to
dc with unity power factor. Is it possible?
322
00:53:06,769 --> 00:53:13,769
We will try to study that part in the next
class.