1
00:00:25,070 --> 00:00:32,070
We continue our discussion on the frequency
response of the amplifier. We have seen that
2
00:00:34,550 --> 00:00:41,550
there are three factors, which make result
in three different values of lower cut off.
3
00:00:44,080 --> 00:00:51,080
But in the graph when we plot frequency versus
gain, then we get actually just one lower
4
00:00:57,260 --> 00:01:04,260
cut off which is f 1. Now, three factors may
contribute to the lower cut off, one arises
5
00:01:07,680 --> 00:01:14,680
from the first coupling capacitor at the input.
The other cut off may come from the second
6
00:01:14,710 --> 00:01:21,710
cut off at the from the second capacitor at
the output. And similarly, bypass capacitor
7
00:01:22,399 --> 00:01:29,399
which we use with R E to enhance the gain,
that may also give the third cut off. But
8
00:01:31,060 --> 00:01:38,060
as in the plot we get only one value f 1.
So, which of these three f 1 f 1 prime f del
9
00:01:41,560 --> 00:01:48,560
f 1 double prime we have to consider, the
one which is out of these three, which is
10
00:01:49,220 --> 00:01:56,220
having highest value of cut off, higher value
of frequency at that will be the one with
11
00:01:56,600 --> 00:02:03,600
that we call f 1. Similarly, we discussed
that for the upper cut off for the upper cut
12
00:02:08,119 --> 00:02:15,119
off the coupling capacitors, and bypass capacitors
are not responsible, they have no role to
13
00:02:16,940 --> 00:02:23,940
play beyond these frequencies they behave
like a short. So, why the gain falls at higher
14
00:02:25,360 --> 00:02:32,360
frequencies? And as said in the beginning
that this fall occurs, because of the contributions
15
00:02:35,470 --> 00:02:42,470
of a junction capacitances.
Now, junction capacitances we have seen that
16
00:02:43,940 --> 00:02:50,940
the analysis will require the use of Miller’s
theorem. Miller’s theorem is the one which
17
00:02:51,000 --> 00:02:58,000
we use to to reduce the the impendence. Whose
one end is connected at input and other end
18
00:02:59,769 --> 00:03:06,340
connected at the output to its equivalent
value at the input terminals and the equivalent
19
00:03:06,340 --> 00:03:12,620
value at the output terminals, these are given
by Miller’s theorem. And we have seen that
20
00:03:12,620 --> 00:03:19,620
for example; the junction capacitance between
collector and base.
21
00:03:21,910 --> 00:03:28,910
That is the situation suitable for the application
of the Miller’s theorem and so, at the input
22
00:03:30,000 --> 00:03:37,000
the Miller equivalent of this capacitance
is multiplied by the gain and gain is quite
23
00:03:38,040 --> 00:03:45,040
high. We are talking about for example; for
common emitter amplifier in which the gain
24
00:03:46,720 --> 00:03:53,720
is 100, 80, 200. So, a small capacitance gets
magnified to 200 times or 100 times, 150 times
25
00:03:56,250 --> 00:04:01,079
and it is responsible to give you the upper
cut off.
26
00:04:01,079 --> 00:04:08,079
So, this is one capacitance. Similarly, at
the output there is the effective equivalent
27
00:04:10,329 --> 00:04:17,329
capacitance from by using Miller’s theorem
we get that may give another cut off. And,
28
00:04:17,449 --> 00:04:24,449
third cut off we were talking that comes because
current gain beta which is the ratio of I
29
00:04:27,509 --> 00:04:33,840
C the collector current output current to
input current, that is base current. And this
30
00:04:33,840 --> 00:04:40,840
falls at very high frequencies. And this fall
we can express in terms of the value f t which
31
00:04:43,320 --> 00:04:50,320
is the gain at which the frequency at which
the gain falls to unity which is provided
32
00:04:50,370 --> 00:04:56,900
by the manufacturer.
And out of these three whichever has the lowest
33
00:04:56,900 --> 00:05:03,900
of the upper cut off lowest of the upper cut
off that is the frequency f two here that
34
00:05:04,200 --> 00:05:11,200
we get in the plot. So, having said this now,
we should see how multi stage amplifiers will
35
00:05:17,230 --> 00:05:17,850
behave.
36
00:05:17,850 --> 00:05:24,850
So, we take next the multi stage multistage
amplifiers. We have been considering so far
37
00:05:42,100 --> 00:05:48,120
single stage amplifiers. That means, one transistor
with its biasing network and coupling networks
38
00:05:48,120 --> 00:05:55,120
that forms a single stage amplifier. The gain
of the single stage amplifier may not be sufficient
39
00:05:57,640 --> 00:06:04,640
for many applications. Whether the application
includes, running of a motor or some other
40
00:06:05,070 --> 00:06:12,070
equipment that may not be sufficient. In practice
actually, single stage amplifiers are having
41
00:06:12,990 --> 00:06:19,990
less use. multi stage amplifiers are used.
What is multi stage amplifier?
42
00:06:20,190 --> 00:06:27,190
A amplifier in which more than one stage it
may be two, three, four they may be identical
43
00:06:28,000 --> 00:06:35,000
or different type of a stages they are joined
together, they are connected normally in series
44
00:06:37,450 --> 00:06:44,450
and so, this is also known as Cascade amplifier.
So, multi stage amplifier or Cascade amplifier
45
00:06:53,840 --> 00:07:00,840
we have when more than one amplifying circuits
are connected. Like, we take for simplicity
46
00:07:04,510 --> 00:07:11,510
a two stage amplifier which is a multi stage
amplifier. And we talk about the salient features
47
00:07:11,990 --> 00:07:18,990
of this.
So, here there are two amplifiers. This is
48
00:07:40,860 --> 00:07:47,860
amplifier 1 having voltage gain A V 1, this
is amplifier 2 having gain A V 2. This is
49
00:08:04,550 --> 00:08:11,550
the input and this is the net output V out.
Now, 1 and 2 i am writing this is the input
50
00:08:19,860 --> 00:08:26,860
to the first amplifier and this is the output
from the compound amplifier. Now, here this
51
00:08:30,230 --> 00:08:37,230
output here is, V out 1 from the first amplifier
whatever is the output that is here.
52
00:08:44,820 --> 00:08:51,820
And this is the, here we feed what is known
V in 2. And obviously, the output because
53
00:09:00,170 --> 00:09:07,170
the output of the first amplifier is fed to
the input to the next amplifier. So obviously,
54
00:09:08,990 --> 00:09:15,990
V out 1 is equal to V in 2. The output of
first stage is equal to the input to the next
55
00:09:23,680 --> 00:09:30,680
stage. Now, in this two cascaded stages, what
is most important? There are few things which
56
00:09:34,400 --> 00:09:41,400
are most important; one is what is the overall
gain? If the individual stages, individually
57
00:09:43,230 --> 00:09:50,230
amplifier 1 has a voltage gain A V 1 and
amplifier 2 has a voltage gain A V 2. Then,
what is the net voltage gain in this amplifier?
58
00:10:12,880 --> 00:10:19,880
This is A V voltage gain of 1 and 2 combined
that is why, i have written the gain as 1
59
00:10:24,980 --> 00:10:31,980
2. And this is obviously, look at this circuit
the gain the the input here is V in 1 and
60
00:10:36,630 --> 00:10:43,630
output is V out 2.
So obviously, the gain is output voltage to
61
00:10:43,649 --> 00:10:50,649
the input voltage. So, here it is V out 2
V in 1 and this can be written in the form
62
00:10:58,330 --> 00:11:05,330
V out 1 V in 1 into V out 2 and V in 2. But
this is equal to we have seen that V in 2
63
00:11:38,980 --> 00:11:45,980
is equal to V out 1. So, this is further V
out 1 V in 1 and V out 2 V out 1. Because
64
00:12:10,529 --> 00:12:17,529
V here V out 1 is equal to V in 2.
So, V in 2 i have written as V out 1. So,
65
00:12:20,709 --> 00:12:27,709
this cancels out and we are left with what
we have written V out 2 by V in 1. And this
66
00:12:30,360 --> 00:12:37,360
is therefore, for the compound amplifier the
gain is this is A V 1 and this is A V 2 very
67
00:12:45,350 --> 00:12:52,350
important relation. The what if individually
the two amplifiers are coupled or cascaded
68
00:12:57,370 --> 00:13:04,370
having the individual values of the gain A
V 1 and A V 2? Then what is the value of the
69
00:13:06,279 --> 00:13:12,940
net gain? That is the product of the two gain.
Remember product of the two gains. Let us,
70
00:13:12,940 --> 00:13:19,940
take an example; that in a two stage two stage
amplifier, let A V 1 be equal to 20 and A
71
00:13:32,170 --> 00:13:39,170
V 2 be equal to 30 they may be identical or
they may be different. So, i have taken 20
72
00:13:40,190 --> 00:13:41,730
and 30.
73
00:13:41,730 --> 00:13:48,730
Then gain of the two stage amplifier gain
of 2 stage that is multi stage amplifier.
74
00:13:57,180 --> 00:14:04,180
That is the product of the two so, 20 into
30 this is 600 it is the product. If we feed
75
00:14:11,060 --> 00:14:18,060
here 1 milli volt this will appear at the
output of 600 milli volts. If the gains of
76
00:14:19,420 --> 00:14:26,420
individual stages are 20 and 30 so, it is
a big gain.
77
00:14:29,100 --> 00:14:36,100
Now, we have talked about decibel scale d
B scale. When gains are expressed in d B(s)
78
00:14:39,339 --> 00:14:46,339
then for multi stage multistage amplifier
the d B(s) are additive. Because d B is a
79
00:14:48,970 --> 00:14:55,970
log scale so, the product becomes a addition
in in a logarithmic ah algebra. So therefore,
80
00:14:57,920 --> 00:15:04,920
we can write when we talk of voltage gains
then in d B(s) we can express 20 log 10 V
81
00:15:13,970 --> 00:15:20,970
out 2 V in 1 and this is equal to 20 log base
10.
82
00:15:25,480 --> 00:15:32,480
And here V out, out i am writings in brief
V o 1 V in i 1 into V out 2 and V in 2 and
83
00:15:51,720 --> 00:15:58,720
this can be written as, 20 log base 10 A V
1 into A V 2 which is A 1 2. That is gain
84
00:16:09,420 --> 00:16:16,420
of the multi stage amplifier in d B(s) this
is A V 1 in d B(s) plus A V 2 in d B(s) additive.
85
00:16:28,000 --> 00:16:35,000
And, we are for simplicity considering only
2 stages. Actually, there can be many stages.
86
00:16:37,779 --> 00:16:44,779
So, for example; if 3 amplifiers having gain
of 20 d B each, then for 3 stages of the multi
87
00:16:59,510 --> 00:17:06,510
stage
amplifier gain will be
88
00:17:13,240 --> 00:17:20,240
Gain for 3 stages will be 20 d B plus 20 d
B plus 20 d B for 3 stages gain will be 60
89
00:17:27,839 --> 00:17:34,839
d B. So, while the gains when not express
in d B(s) they are product but once they are
90
00:17:40,789 --> 00:17:47,230
expressed in d B(s) they are additive. So,
this is about gains.
91
00:17:47,230 --> 00:17:54,230
I repeat that gains are the product of individual
gains and when products when the gains are
92
00:17:54,749 --> 00:18:01,749
expressed in d B(s) then they are additive.
This is about the gain. Gain may be voltage
93
00:18:03,519 --> 00:18:10,519
gain, current gain anything or power gain.
Now, and we can extend it to any number. Two
94
00:18:12,999 --> 00:18:18,720
stages we are considering three, four, five
n number of stages the same rules are applicable.
95
00:18:18,720 --> 00:18:25,720
Now what about the phases? These are algebraically
additive.
96
00:18:29,850 --> 00:18:36,850
These are by simple intuition we can see that
they have to be these are additive. So, if
97
00:18:41,480 --> 00:18:48,480
we couple two amplifiers for example; both
have a phase difference of pi. Then if there
98
00:18:51,799 --> 00:18:58,230
are only two stages, then pi plus pi so the
phase difference will be 2 pi. That means,
99
00:18:58,230 --> 00:19:05,230
output will be in phase with the input. So,
phases are additive. And, if one has a phase
100
00:19:07,259 --> 00:19:14,259
of pi other has a phase of minus pi then 0
phase will appear and so on. So, phases are
101
00:19:16,779 --> 00:19:21,289
additive.
Now so, these are two important considerations
102
00:19:21,289 --> 00:19:28,289
about multi stage amplifiers, how the the
voltage gain or current gain so, in general
103
00:19:30,210 --> 00:19:37,210
how gains are related with the individual
values. So, they are normally products and
104
00:19:37,330 --> 00:19:44,220
when expressed in d B(s) they are additive.
And phases they are additive algebraically.
105
00:19:44,220 --> 00:19:51,220
Now, frequency response frequency response
of multi stage amplifier frequency response
106
00:20:10,230 --> 00:20:17,230
of multi stage amplifier. It is now very clear
that, by adding a stages we amplify the signals
107
00:20:22,489 --> 00:20:29,489
much more and gains go very high.
If you remember we have said that gain and
108
00:20:30,919 --> 00:20:37,919
bandwidth product is constant. This is constant
for a individual circuit this is constant
109
00:20:37,999 --> 00:20:44,999
for a system. System means multi stage amplifier.
So, when the gain goes high, the bandwidth
110
00:20:49,899 --> 00:20:56,899
has to fall. So, remember that in multi stage
amplifiers the gain fall at the gain increases
111
00:20:59,269 --> 00:21:05,269
bandwidth falls.
Now, numerically we have to calculate this
112
00:21:05,269 --> 00:21:12,269
a fall in bandwidth. The calculation of bandwidth
of the multi stage amplifier is not difficult
113
00:21:19,419 --> 00:21:26,419
in two cases and the two cases are; case 1,
case 1 when stages which have identical lower
114
00:21:35,559 --> 00:21:42,559
cut off and upper cut off, when they are coupled
when they form the cascading cascade amplifier,
115
00:21:44,850 --> 00:21:51,850
the multi stage amplifier, then the calculation
is simple. And, I repeat what am saying, suppose
116
00:21:54,519 --> 00:22:01,519
we connect two amplifiers, they have identical
lower cut off and or identical out of the
117
00:22:05,200 --> 00:22:12,200
the the upper cut off then it is very easy
to calculate the cut off lower as well as
118
00:22:14,100 --> 00:22:21,100
upper for the multi stage amplifier.
And the expressions are, so the case 1 is:
119
00:22:22,249 --> 00:22:29,249
stages having same closely equal lower and
or upper cut off frequencies. If for an example;
120
00:22:53,749 --> 00:23:00,229
lower cut off is the same uppers they are
different then this calculation in case 1
121
00:23:00,229 --> 00:23:07,229
what we are talking will be valid only for
the upper, for the lower cut off which are
122
00:23:07,899 --> 00:23:13,499
identical.
And if only upper is is identical, then the
123
00:23:13,499 --> 00:23:19,450
rule which I am just putting here that will
be applicable to the upper cut off. And if
124
00:23:19,450 --> 00:23:25,409
both are identical, then these both expressions
can be used.
125
00:23:25,409 --> 00:23:32,409
So, if n stages are cascaded, then the lower
cut off for the system that means, for the
126
00:23:43,729 --> 00:23:50,729
multi stage amplifier this is f 1 is the cut
off for the individual stage. Then this is
127
00:23:53,539 --> 00:24:00,539
2 to the power 1 by n minus 1 and f 2 of the
system is equal to f 2 by 2 to the power 1
128
00:24:15,009 --> 00:24:22,009
by n minus 1. This is the expression for lower
cut off, upper cut off of a multi stage amplifier,
129
00:24:27,869 --> 00:24:34,869
if they have same lower cut off(s) and higher
cut off(s), if only 1 is identical then only
130
00:24:38,229 --> 00:24:45,229
corresponding expression is to be used.
We can see that how the lower cut off is increased
131
00:24:46,320 --> 00:24:53,320
and upper cut off falls. Here the gain if
this is f 1 this is f 2 if this falls then
132
00:25:00,419 --> 00:25:07,419
we are talking about something of this kind.
It has fallen and this increases. So, earlier
133
00:25:28,929 --> 00:25:35,929
bandwidth was this one. And now, the bandwidth
will be reduced to 1. We take an example;
134
00:25:46,149 --> 00:25:53,149
let f 1 that means, the lower cut off for
for the individual amplifier is 20 hertz and
135
00:26:01,450 --> 00:26:08,450
upper cut off is 20 kilo hertz the bandwidth
is f 2 minus f 1which is 20 kilo hertz minus
136
00:26:23,690 --> 00:26:30,690
20 hertz. 20 hertz is negligible in comparison
to 20 kilo hertz so, this is simply 20 kilohertz.
137
00:26:36,799 --> 00:26:43,799
This is the bandwidth of the individual amplifier.
Now, suppose 2 stages are cascaded, then f
138
00:26:50,559 --> 00:26:57,559
1 for the two stage amplifier from this expression
1 by n becomes equal to 1 by 2 because n is
139
00:27:03,259 --> 00:27:10,259
equal to n is number of stages n is number
of a stages. So, that gives 1.55 f 1 that
140
00:27:14,580 --> 00:27:21,580
is 1.55 into 20 hertz and it comes out to
be 31 hertz. Lower cut off increases from
141
00:27:27,919 --> 00:27:34,919
20 hertz to 31 hertz when 2 stages are joined
together to form a multi stage.
142
00:27:38,210 --> 00:27:45,210
And the upper cut off f 2 prime two stage
amplifier this is 0.64 f 2 from this expression
143
00:27:51,940 --> 00:27:58,940
2 to the power half minus 1 when you solve
it it comes 0.64. So that, this is reduced
144
00:28:02,330 --> 00:28:09,330
from 20 kilo hertz to 12.8 kilo hertz. And
because 31 is very small so, bandwidth for
145
00:28:19,219 --> 00:28:26,219
this two stage amplifier actually is reduced
from 20 to 12.8 kilo hertz.
146
00:28:31,649 --> 00:28:37,909
So, if more stages identically even this is
stage if you solve for four this will be much
147
00:28:37,909 --> 00:28:44,909
reduced, bandwidth falls and the the physical
reasoning we can get from the fact that, gain
148
00:28:47,450 --> 00:28:54,450
bandwidth product is constant. So, we have
seen numerically that while the gains are
149
00:28:55,519 --> 00:29:02,519
enhanced drastically but the bandwidths fall.
So, in a multi stage amplifier to maintain
150
00:29:05,109 --> 00:29:11,889
a reasonable bandwidth even for the multi
stage amplifier we have to start with individually
151
00:29:11,889 --> 00:29:18,889
stages which should have much higher bandwidth
than required in the application.
152
00:29:22,369 --> 00:29:29,369
How multi stage amplifiers are obtained? What
are the methods of coupling? Either a stages
153
00:29:31,999 --> 00:29:38,999
or even I mean how we couple a source source
signal source to the amplifier? There are
154
00:29:53,440 --> 00:30:00,440
three methods methods of coupling there are
3 methods; one is R C coupling, other is direct
155
00:30:14,440 --> 00:30:21,440
coupling, third is transformer coupling these
are the 3 methods, which can be used to couple
156
00:30:37,859 --> 00:30:44,859
either various stages of amplifiers or at
the input the source can be coupled to the
157
00:30:47,529 --> 00:30:54,529
amplifier by any of these three methods.
And similarly at the final stage the load
158
00:30:57,210 --> 00:31:04,210
can be a loud speaker, a printing machine
or some other ah gadget, that can be the power
159
00:31:07,299 --> 00:31:14,299
can be coupled by any of these 3 methods we
discuss one by one R C coupled, this is what
160
00:31:15,669 --> 00:31:21,919
we have been discussing the lower cut off,
upper cut off we have said and that was the
161
00:31:21,919 --> 00:31:28,919
R C coupled amplifier. When we take a source
here and we source resistance r s and this
162
00:31:30,489 --> 00:31:37,489
was the capacitor and we have derived the
expressions also for it, that was R C coupling.
163
00:31:39,589 --> 00:31:46,589
R c coupling is very widely used because it
is most convenient it is integrable but only
164
00:31:48,450 --> 00:31:55,450
difficulty is as we have seen in our studies
that it puts a restriction on the lowest frequencies
165
00:31:58,019 --> 00:32:05,019
which can be amplified, the graph for a R
C coupled amplifier gain versus frequency
166
00:32:07,070 --> 00:32:14,070
goes like this so, this is the lower cut off.
So, for example; 0 frequency d c signals cannot
167
00:32:18,149 --> 00:32:25,049
be coupled by R C coupling, the signal will
not go to the next stage.
168
00:32:25,049 --> 00:32:32,049
So, anyway we have discussed enough the all
analysis of the R C coupled circuits that
169
00:32:32,119 --> 00:32:34,269
we have discussed.
170
00:32:34,269 --> 00:32:41,269
And now, we take another method of coupling
which is Direct coupling. In Direct coupling
171
00:32:52,609 --> 00:32:59,609
the signal is coupled just it is connected
by a conductor. For example; in direct coupling
172
00:33:00,669 --> 00:33:07,669
if this is the signal source or the previous
stage it is just coupled to the base of the
173
00:33:08,499 --> 00:33:15,499
next. Direct coupling there is no capacitor
involved in it. This is also quite in recent
174
00:33:18,999 --> 00:33:25,999
years say last 20 or 25 years, this coupling
has found big applications and for example;
175
00:33:30,139 --> 00:33:37,139
in operational amplifiers op amps. Which we
are going to study in details in later modules.
176
00:33:38,749 --> 00:33:45,749
So, op amps most of the op amps make use of
direct coupling. Their I said if you remember
177
00:33:47,289 --> 00:33:54,289
that, op amp is a multi stage amplifier and
there are several stages which are directly
178
00:33:55,450 --> 00:34:02,450
coupled. So, there there will not be any restriction
on the lower cut off. When this is connected
179
00:34:05,119 --> 00:34:12,119
directly even if it a d c signal, it will
be able to propagate to couple.
180
00:34:12,540 --> 00:34:19,540
So, the response for direct coupled will have
no cut off lower cut off no but upper cut
181
00:34:23,169 --> 00:34:30,169
off for which this is gain this is frequency
there is no lower cut off we start from zero
182
00:34:32,970 --> 00:34:39,970
frequency is still the gain will be there
and the upper cut off falls if you remember
183
00:34:42,090 --> 00:34:48,750
because of the junction capacitances. And
whether we are using a bipolar transistor
184
00:34:48,750 --> 00:34:55,750
or a F E T this is always there so, this cut
off will be there this will be there.
185
00:34:57,170 --> 00:35:04,170
So, in direct coupling in I C(s) and in operational
amplifier, operational amplifier is also in
186
00:35:06,150 --> 00:35:13,150
a I C form. So, integrated circuits and op
amps etcetera they make use of direct coupling.
187
00:35:15,140 --> 00:35:21,220
So, while the biggest advantage, there are
two advantages. There is no need of capacitor.
188
00:35:21,220 --> 00:35:27,970
So, capacitor formation in I C will not be
required, other thing is there is no limit
189
00:35:27,970 --> 00:35:34,970
on the lower frequency right from d c 2 higher
frequencies you can couple. Of course, the
190
00:35:36,120 --> 00:35:43,120
higher fall at very high frequencies beyond
mid band the cut off will be there and that
191
00:35:43,610 --> 00:35:50,610
is because of the junction capacitances.
Now, there is one disadvantage with the direct
192
00:35:54,050 --> 00:36:01,050
coupling. That the disadvantage is rather
one is not very serious, though it is a disadvantage.
193
00:36:06,850 --> 00:36:13,850
When we couple these stages directly, then
even d c are coupled. If we put R C coupling
194
00:36:15,580 --> 00:36:22,580
then we have seen that the capacitor isolates
one stage with the other. Capacitor acts as
195
00:36:22,690 --> 00:36:29,690
a block for d c. So, we can design the amplifiers
more accurately individually and then we can
196
00:36:33,030 --> 00:36:39,950
use R C coupling and because for d c purposes
the capacitors will keep them isolated.
197
00:36:39,950 --> 00:36:44,900
So, the biasing etcetera are not affected
the queue points the operating points are
198
00:36:44,900 --> 00:36:51,900
not affected. But when we use direct coupling
then we are connecting them for direct voltages
199
00:36:54,970 --> 00:37:01,970
also d c voltages also and hence, the design
has to be done more carefully. But more serious
200
00:37:04,310 --> 00:37:09,630
than this we will take a design example to
illustrate it what i have just said.
201
00:37:09,630 --> 00:37:16,630
And more serious than this is that low frequency
noise also gets amplified. Because, that capacitor
202
00:37:23,870 --> 00:37:30,270
will block it capacitor with block this low
frequency noise. Here in direct coupling there
203
00:37:30,270 --> 00:37:37,270
is no blockage, there is no restriction on
the frequency. So, even noises will propagate
204
00:37:38,100 --> 00:37:45,100
through the amplifier and they will be amplified.
Now, this problem can be taken care of by
205
00:37:48,770 --> 00:37:55,770
taking more advance circuits like differential
amplifiers. Where we can filter out these
206
00:37:55,850 --> 00:38:02,850
noises by a different mechanics that we will
see then we talk of differential and operational
207
00:38:02,860 --> 00:38:09,570
amplifiers.
Let us, take so direct coupling is nothing,
208
00:38:09,570 --> 00:38:16,570
simply a connector connects the sample the
signal or the output from the last stage to
209
00:38:19,490 --> 00:38:26,290
the load directly by a connecting wire. If
it is a individually I mean, if it is discrete
210
00:38:26,290 --> 00:38:33,290
amplifier then a connecting wire otherwise
a conducting path on the I C will do the direct
211
00:38:33,920 --> 00:38:39,260
coupling, we take a design example;
212
00:38:39,260 --> 00:38:46,260
Let us, consider this circuit. This is a direct
coupled two stage amplifier I take just one
213
00:39:31,660 --> 00:39:38,660
design example; this is plus 12 volts, and
there are two transistors which are directly
214
00:39:42,510 --> 00:39:49,510
coupled. And let us say that base of transistor
Q 1 initially is at plus 2.1 volts, that is
215
00:39:54,760 --> 00:40:01,760
2.1 volt is available here. And, these resistance
are given this is 700 ohms, and this is 3.3
216
00:40:07,050 --> 00:40:14,050
kilo ohms, this is 4 k and what is given is
that for both these transistor Q 1 and Q 2
217
00:40:20,460 --> 00:40:27,460
the current gains are same, beta is 100 each.
And the voltage drop this drop here, this
218
00:40:29,540 --> 00:40:36,540
is 0.7 volts here, and this is also 0.7 volts
here. So, V BE as we call it this is 0.7 volt.
219
00:40:41,430 --> 00:40:48,430
So, the problem now is that what should be
the value of R C to put the collector to put
220
00:40:52,610 --> 00:40:59,610
the collector at plus five volts, we have
to find out the value of this resistor.
221
00:41:02,550 --> 00:41:09,550
Now, the solution I give. The emitter current
in Q 1, so I E 1 for Q 1 this you know now,
222
00:41:20,790 --> 00:41:27,790
what we have done earlier in our earlier module.
That how to calculate this voltage minus this
223
00:41:29,360 --> 00:41:36,360
divided by this will give you the I E. So,
I E 1 of Q 1 is V B B minus V B E divided
224
00:41:40,390 --> 00:41:47,390
by R E 1 and this is 2.1 volts minus 0.7 volts
by 700 ohms, and this gives is 2 milli ampere.
225
00:41:53,170 --> 00:42:00,170
And, so these two milli ampere I C and I E,
they are the same I C always in transistor,
226
00:42:02,340 --> 00:42:09,340
bipolar transistor analysis is equal to I
E. So, 2 milli meters flows through here,
227
00:42:09,830 --> 00:42:16,830
then we know how much will be that at what
voltage base of Q 2 will be.
228
00:42:17,200 --> 00:42:24,200
Let us, put this point A here. What is point
A? That is the same the potential here will
229
00:42:25,520 --> 00:42:29,930
be the same as the collector of Q 1 or base
of Q 2.
230
00:42:29,930 --> 00:42:36,930
So, here V A will be simply Kirchhoff’s
voltage law you apply 12 volts here. This
231
00:42:41,000 --> 00:42:48,000
is connected to 12 volts and 2 millli ampere
current is flowing. So obviously, 4 K into
232
00:42:49,000 --> 00:42:56,000
2 milli ampere current is flowing through
that so, this is 8 and so, this is 12 minus
233
00:42:57,870 --> 00:43:04,870
8 is equal to 4 volts. So, the potential here
is 4 volts 4 volts now we proceed in the same
234
00:43:07,250 --> 00:43:12,590
way here out of 4 volts 0.7 volt drops across
this junction.
235
00:43:12,590 --> 00:43:19,590
Which i write the i 2 i e 2 this is for q
2 this will be equal to 4 volts minus 0.7
236
00:43:28,850 --> 00:43:35,850
volts divided by R E so 4 volts minus 0.7
divided by R E is 3.3 K and this is also 3.3
237
00:43:40,110 --> 00:43:47,110
volts. So, 3.3 volts divided by 3.3 K gives
1 milli ampere 1 milli ampere current, this
238
00:43:51,330 --> 00:43:58,330
is same as I C 2 here. This current is flowing
through this resistor, with this we have to
239
00:44:00,760 --> 00:44:05,460
find out and again we apply remembering that
this is to be maintained at 5 volts
240
00:44:05,460 --> 00:44:12,460
Then, R C we can find out like this, that
12 volts out of that 5 volt is the requirement
241
00:44:19,760 --> 00:44:26,760
that the collector should be collector should
be at 5 volts. And what is the drop here?
242
00:44:28,090 --> 00:44:35,090
I C into R C. So, plus R C into I C and I
C we have found out to be 1 milli ampere and
243
00:44:43,730 --> 00:44:50,730
from here, we can find R C and that comes
out to be twelve minus 5 volts divided by
244
00:44:52,370 --> 00:44:59,370
I E, which is 7 volts 1 milli ampere so, this
7 kilo ohms.
245
00:45:00,790 --> 00:45:07,790
So, we have to use a resistance of 7 kilo
ohms to get the voltage at 5 volts. So, this
246
00:45:09,680 --> 00:45:15,070
is the design principal i will have illustrated
through this example.
247
00:45:15,070 --> 00:45:22,070
Now, one of the very popular direct coupled
multi stage circuit is what is known as Darlington
248
00:45:23,980 --> 00:45:30,980
pair, Darlington pair is a 2 stage amplifier
where two transistors are actually directly
249
00:45:39,000 --> 00:45:46,000
coupled, the circuit is this. This is sold
in the market as a single device. This is
250
00:46:02,760 --> 00:46:09,760
all enclose these are two transistors Q 1
and Q 2; this is Q 1 this is Q 2 and this
251
00:46:10,270 --> 00:46:17,270
is base, this is collector and this is emitter
and here the current i b 1 flows and this
252
00:46:24,320 --> 00:46:31,320
is i C 1 this is i e which is same as i c
1 and the same current becomes i b 2 and here
253
00:46:37,980 --> 00:46:44,980
it is i c 2, as i said this is enclosed as
in a single packaging and the three leads
254
00:46:49,270 --> 00:46:56,270
like a transistor they are available.
And what is the use of why we are doing this?
255
00:46:56,680 --> 00:47:01,800
Voltage gain we have seen now, this is the
example of current gain. That, if the current
256
00:47:01,800 --> 00:47:08,800
gain for q 1 is beta 1 and current gain for
2 is beta 2, then for this Darlington pair
257
00:47:12,540 --> 00:47:19,540
the current gain beta will be the product
of the 2 and this we can very simply see that
258
00:47:22,080 --> 00:47:29,080
what will be if the input current is i b 1
what will be this current? i c 1 you know
259
00:47:30,140 --> 00:47:37,140
it is beta 1 times i b 1 and i b 1 this is
i b 2 is same as i c 1, this current is the
260
00:47:46,960 --> 00:47:53,960
same as this current and this current is going
on, so i b two is equal to i c 1.
261
00:47:54,770 --> 00:48:01,770
And then i c 2 is beta 2 times i b 2 and i
b 2 is simply i b 2 is same as here, this
262
00:48:20,780 --> 00:48:27,780
is same as i b 2. This we substitute in this
so, i c 2 becomes beta 1 beta 2 into i b 1
263
00:48:34,560 --> 00:48:41,560
and what is the current gain? The current
over all current gain beta which is the the
264
00:48:41,700 --> 00:48:47,310
ratio of output current to input current.
Now, for the collective compose it direct
265
00:48:47,310 --> 00:48:54,310
coupled amplifier output current is i c 2
input current is i b 1 and from here we find
266
00:48:57,020 --> 00:49:04,020
that, this is the beta 1 into beta 2. This
is a very useful connection. Because, from
267
00:49:05,580 --> 00:49:12,580
a single transistor many times the current
gain is a not very high. It is 100 200 maximum
268
00:49:13,910 --> 00:49:20,910
three 400. But this is the product, if we
couple this way we form a Darlington pair
269
00:49:21,480 --> 00:49:25,120
out of the two transistors then given will
be very high.
270
00:49:25,120 --> 00:49:32,120
As an example, let beta 1 be 50 and beta 2
be 60, then for the compound this Darlington
271
00:49:39,610 --> 00:49:46,610
pair the current gain will be equal to 50
into 60, which is 3000. Darlington pair which
272
00:49:52,780 --> 00:49:59,780
has been formed with two transistors individually
having current gain of 50 and 60. When they
273
00:50:01,530 --> 00:50:08,530
form a Darlington pair then, the current gain
available will be 3000. This has wide applications
274
00:50:11,060 --> 00:50:18,060
it is formed even in integrated circuits and
otherwise also in discrete circuits Darlington
275
00:50:18,910 --> 00:50:23,980
pair is widely used.
Darlington pair is actually in reality these
276
00:50:23,980 --> 00:50:30,980
are two emitter followers two emitter followers
with infinite emitter resistance, emitter
277
00:50:43,350 --> 00:50:50,350
resistance with first with first transistor
with first transistor. How we can show? That,
278
00:50:59,540 --> 00:51:06,540
this is this. For a c purposes we ground this
otherwise it will be a biasing we will have
279
00:51:29,660 --> 00:51:36,660
to provide and this is what we said, this
r e 1 is infinitely high and this the load
280
00:51:37,800 --> 00:51:43,180
this is emitter follower we are connecting
the output from emitter to the next stage
281
00:51:43,180 --> 00:51:50,180
and at the next stage this is q 1, this is
q 2 we are taking the output here and we are
282
00:51:50,790 --> 00:51:57,180
giving input here with respect to ground of
course, here also with respect to ground the
283
00:51:57,180 --> 00:52:04,180
output is taken and this is r l. Then since,
beta is beta 1 by beta 2 now, see here there
284
00:52:09,330 --> 00:52:14,180
is another advantage gain, current gain of
the Darlington pair is very high.
285
00:52:14,180 --> 00:52:21,180
So, is the input impendence what is the impendence
here? That is z i at the second stage you
286
00:52:25,570 --> 00:52:32,570
remember the fundamental thumb rules, that
for a circuit in which the input is given
287
00:52:34,550 --> 00:52:41,550
at base whether it is common emitter or this
emitter follower. This is equal to the beta
288
00:52:43,950 --> 00:52:50,020
into load resistance. Load is here. Now, r
l at the emitter because it is emitter follower.
289
00:52:50,020 --> 00:52:57,020
We always connect load at the emitter.
So, this will be z i 2 is equal to the current
290
00:53:00,810 --> 00:53:06,820
gain into the load resistance r l, this we
have done earlier and i am sure you understand
291
00:53:06,820 --> 00:53:13,820
that, this is the beta times the load so,
it is this. And, this will act z 1 will act
292
00:53:15,670 --> 00:53:21,750
this is infinity this is in parallel with
infinity. So, the result will be simply z
293
00:53:21,750 --> 00:53:28,750
1 2 z i 2 z i 2 is the load for the first
stage because this is infinity resistance
294
00:53:29,830 --> 00:53:36,770
and this in parallel.
So, the input impendence at the first stage
295
00:53:36,770 --> 00:53:43,770
here this is equal to the beta of it beta
times the the input impendence seen here which
296
00:53:47,320 --> 00:53:54,320
is beta 2 r l very high input impendence.
Beta 1 into beta 2 into r l in the above example,
297
00:54:01,870 --> 00:54:08,870
if r l is 1 kilo hertz and this collective
beta which is the product of the two is 3000
298
00:54:09,270 --> 00:54:16,270
so, from 1 kilo hertz it will show a input
impendence of 3 mega hertz very high impendence.
299
00:54:21,280 --> 00:54:28,220
So, sometimes Darlington pairs are used to
achieve very high degree very high magnitudes
300
00:54:28,220 --> 00:54:31,030
of input impendence.
301
00:54:31,030 --> 00:54:38,030
So, we talked about the R C coupling in details
and a the response of a R C coupled amplifier
302
00:54:46,020 --> 00:54:53,020
was like this. This is gain versus frequency
and these two cut off frequencies we defined.
303
00:54:56,510 --> 00:55:03,450
In direct coupling there is no limit on this
lower cut off. And hence, the frequency response
304
00:55:03,450 --> 00:55:10,450
of that direct coupled amplifier is like this.
Where that upper cut off of course, will be
305
00:55:12,119 --> 00:55:19,119
here, but lower cut off is 0 this is gain.
And now so, this is the direct coupling and
306
00:55:21,430 --> 00:55:28,430
the a special case of direct coupling we talked
Darlington pair which is the direct coupled
307
00:55:28,480 --> 00:55:35,480
two stage transistor and which is sold as
a unit. Now, finally, very briefly because
308
00:55:36,310 --> 00:55:43,310
ah we have to talk about transformer, the
third coupling transformer coupling, transformer
309
00:55:47,800 --> 00:55:54,090
coupling is very rarely used and there are
few reasons for it.
310
00:55:54,090 --> 00:56:01,090
One reason is transformers are bulky they
are bulky, they are they cannot be integrated
311
00:56:04,410 --> 00:56:11,410
not integrable and their frequency response
in general is poor and to have a high frequency
312
00:56:15,950 --> 00:56:22,950
response the transformer becomes very expensive.
So for frequency response poor for these three
313
00:56:24,820 --> 00:56:31,670
reasons transformer coupling is not used.
But few applications is still reserved for
314
00:56:31,670 --> 00:56:38,670
example, often now, how the transformer coupling
for example; this is the final stage. Where
315
00:56:41,020 --> 00:56:48,020
this is the primary is connected to the collector.
And here directly a load like a speaker is
316
00:56:50,150 --> 00:56:56,030
connected this is transformer coupling, this
is transformer.
317
00:56:56,030 --> 00:57:03,030
And it is used if you open your radio sets,
transistor sets then the a speaker you may
318
00:57:03,060 --> 00:57:10,060
find connected to a transformer. So this is
transformer coupling, in which a it can take
319
00:57:11,950 --> 00:57:18,950
a large currents and hence, a large signals
can be can be passed through the transformer
320
00:57:20,240 --> 00:57:26,510
and without and d c will be checked, no d
c will propagate in the transformer so, that
321
00:57:26,510 --> 00:57:31,270
is the advantage.
But, these three disadvantages they restrict
322
00:57:31,270 --> 00:57:38,270
the use of a transformer coupling and it is
very bulky they are having much weight as
323
00:57:38,750 --> 00:57:45,300
compared to other parts and not integrable
transformers cannot be integrated like capacitor
324
00:57:45,300 --> 00:57:52,010
can be integrated transistors are of course,
integrable and frequency response in general
325
00:57:52,010 --> 00:57:59,010
is poor. So, that finishes our module four.