1
00:00:24,900 --> 00:00:39,900
We were discussing the inverting amplifier,
and two things which we did; one was that
2
00:00:40,170 --> 00:00:47,170
the practically used expression for voltage
gain with feedback amplifier. This was minus
3
00:00:49,760 --> 00:00:56,760
R F by R 1; that is the ratio of these two
resistances R F and R 1 that gives the voltage
4
00:01:03,260 --> 00:01:10,260
gain. We can choose these resistances in the
clom range to get a particular ratio, and
5
00:01:13,360 --> 00:01:19,750
that will be the gain. This was the point
which we discuss. Another important point
6
00:01:19,750 --> 00:01:26,750
was that v 2 was at zero potential; that means,
it was at ground potential, and this is called
7
00:01:30,369 --> 00:01:37,369
virtual ground. Ground, because which is at
ground potential and virtual ground, because
8
00:01:38,259 --> 00:01:45,259
actual ground can observe almost any amount
of current, but this is a small signal device.
9
00:01:47,479 --> 00:01:54,479
So, hardly fraction of ampere will be the
current, which this will be able to handle.
10
00:01:55,549 --> 00:02:02,549
So, these two points we discussed. Another
feature as it was pointed out, that inverting
11
00:02:04,090 --> 00:02:11,090
non inverting amplifiers. These are both very
widely used amplifiers. So, let us look for
12
00:02:12,640 --> 00:02:19,640
other important features of this amplifier.
Next parameter is the input impedance, input
13
00:02:23,790 --> 00:02:30,790
impedance
of inverting amplifier. For this we first
draw a simple equivalent circuit for op amp.
14
00:03:06,430 --> 00:03:13,430
Simple it is this, this is the voltage source,
where this is v i d, and this resistance is
15
00:03:55,010 --> 00:04:02,010
R O the output resistance. And this is the
inverting input and this is the non inverting
16
00:04:05,540 --> 00:04:12,540
input, and here we give this signal v i 2,
v i 1 and we take v out here. So, this is
17
00:04:18,729 --> 00:04:25,729
the equivalent circuit, and amplifier we remember
that the basic amplifier is this. This R F
18
00:04:45,710 --> 00:04:52,710
this is R 1 and here is the input for the
inverting amplifier. Now this resistance r
19
00:04:56,629 --> 00:05:03,629
i is quite high. Here this resistance we have
been saying, that input impedance first stage
20
00:05:06,830 --> 00:05:13,830
of op amp is a differential amplifier, and
differential amplifier has very high input
21
00:05:16,310 --> 00:05:23,310
impedance. So, this is r i and it is more
than one mu ohms in general. So, we can apply
22
00:05:27,620 --> 00:05:34,620
Miller's theorem which we have used earlier.
Miller's theorem is applicable to find out
23
00:05:41,509 --> 00:05:48,509
the equivalent resistance of this R F at input
and output, because this is in Miller's
24
00:05:50,939 --> 00:05:57,939
configuration, what is Miller's configuration.
One end of the resister is connected at the
25
00:05:58,310 --> 00:06:05,310
input and the other end is connected at output.
So, this is in Miller's configuration, and
26
00:06:07,889 --> 00:06:10,120
we can apply Miller's theorem.
27
00:06:10,120 --> 00:06:17,120
So, in that case the impedance which we will
be getting is this. This is R 1, and this
28
00:06:40,779 --> 00:06:47,779
resistance this is the Miller's equivalent
at the input, and this is equal to R F by
29
00:06:55,550 --> 00:07:02,550
1 plus a where a is the open loop gain. And
this is that r i the input impedance here.
30
00:07:07,969 --> 00:07:14,969
This r i and this is in Miller's form, so
we transform it to its equivalent at the input.
31
00:07:15,909 --> 00:07:22,909
And this is R 1 and of course here, this is
v i. So at the input, the net resistance of
32
00:07:30,389 --> 00:07:37,389
feedback amplifier; this one. The input impedance
of the feedback amplifier becomes z i with
33
00:07:52,689 --> 00:07:59,689
feedback, so feedback amplifier. This is R
1 plus this combination in parallel. So, R
34
00:08:02,319 --> 00:08:09,319
1 plus R F 1 plus a in parallel with r i.
Now, here a is very large. So, this R F will
35
00:08:20,490 --> 00:08:27,490
be very close to zero. R F by 1 plus a is
closed to zero, because a is very large, of
36
00:08:36,589 --> 00:08:43,589
the order of ten to power five. So, this is
almost zero and any smaller resistance in
37
00:08:46,520 --> 00:08:53,520
parallel with high resistance, the result
in case, the smaller one. So, R F 1 plus a
38
00:08:56,240 --> 00:09:03,240
in parallel with r i is almost zero. Therefore,
input resistance is only r i.
39
00:09:07,070 --> 00:09:14,070
Input resistance or input impedance, input
impedance of amplifier is
R 1, this is. Whatever value we choose in
40
00:09:36,509 --> 00:09:43,509
the inverting amplifier, the value of r one
that will be the value for input impedance
41
00:09:44,220 --> 00:09:51,220
how; simple, it is if you want to have a inverting
amplifier with input impedance of say five
42
00:09:55,220 --> 00:10:02,220
kilo ohms and gain twenty and tape this R
1 as five k and because gain is twenty. So,
43
00:10:03,019 --> 00:10:10,019
twenty times of this; that means, hundred
k hundred kilo ohms R F and five k r one will
44
00:10:12,769 --> 00:10:19,769
give input impedance of five k and gain of
hundred. Very simple design and that is one
45
00:10:20,670 --> 00:10:27,670
reason of the popularity of op amp. And from
a similar consideration also, the value comes
46
00:10:30,810 --> 00:10:37,810
out to be R 1. The similar consideration is,
that this point is at ground potential, v
47
00:10:38,050 --> 00:10:44,680
2 is virtual ground. So, what is the impedance
which you expect between these two points.
48
00:10:44,680 --> 00:10:51,680
The virtual ground and this point R 1. So,
R 1 ends from a more rigorous simple analysis,
49
00:10:55,660 --> 00:11:02,660
though we have found that input impedance
is R 1. Now the output impedance, output impedance
50
00:11:10,269 --> 00:11:17,269
can also be derived from similar considerations,
but we are using another concept. Since it
51
00:11:18,740 --> 00:11:25,740
is a voltage shunt feedback, which we are
using with the inverting amplifier, then you
52
00:11:34,300 --> 00:11:41,279
recall from what has been said earlier when
we were discussing feedback in amplifier,
53
00:11:41,279 --> 00:11:48,279
that voltage shunt feedback brings the output
voltage down be a vector of 1 plus a b, that
54
00:11:49,569 --> 00:11:56,569
is the output impedance as shown in this equivalent
resistance is r o. This r o, because of that
55
00:12:00,399 --> 00:12:07,399
feedback it will be reduced, and this is R
O with feedback. This is sorry Z O with feedback
56
00:12:12,920 --> 00:12:19,920
is R O, 1 plus A B.
And so this is drastically reduced here, R
57
00:12:27,779 --> 00:12:34,779
O is the output resistance or impedance same
thing, at a small frequencies. So, output
58
00:12:38,720 --> 00:12:45,720
resistance without feedback, and A is the
open loop gain, and B is equal to R 1 by R
59
00:12:58,639 --> 00:13:05,639
1 plus R F. So, this way this is drastically
reduced, which is again which goes in fever
60
00:13:14,889 --> 00:13:21,889
of a voltage amplifier. Inverting amplifier
is a voltage amplifier, which has a high input
61
00:13:23,720 --> 00:13:30,720
impedance. We can choose the value of R 1,
and gain is high we can choose the ratio R
62
00:13:31,540 --> 00:13:38,540
F by R 1, and the output impedance is low
as given by this equation. So, this is about
63
00:13:41,060 --> 00:13:48,060
the output impedance. Now we have done non
inverting amplifier, we have done inverting
64
00:13:52,569 --> 00:13:59,569
amplifier, and we have evaluated the basic
parameters of the amplifier. And the next
65
00:14:02,220 --> 00:14:09,220
is the differential amplifier.
66
00:14:13,370 --> 00:14:20,370
We have a studied differential amplifier using
a two transistors q 1 and q 2 in connecting
67
00:14:32,339 --> 00:14:39,339
any particular way, and when we collect the
outputs we take the outputs from the two collectors,
68
00:14:40,959 --> 00:14:47,959
then a differential amplifier can be constructed.
That is the basic design and that was investigated
69
00:14:53,399 --> 00:15:00,399
for various basic understanding. Actually,
instead of using, the transistors discreet
70
00:15:05,949 --> 00:15:12,949
transistor units, and then providing a constant
current and all that. Actually the differential
71
00:15:13,279 --> 00:15:20,279
amplifiers are realized, by using a operational
amplifier and using it in a differential mode.
72
00:15:21,439 --> 00:15:28,439
So, this is the basic circuit, gain open loop
gain is A. This is inverting and this is non
73
00:15:36,620 --> 00:15:43,620
inverting input, the batteries, and now, here
we connect R 3 R 2 and this input is the voltage
74
00:16:06,129 --> 00:16:13,129
is v 1, and here we apply the signal source
v b. Similarly, here R 1, this is v 2 and
75
00:16:42,910 --> 00:16:49,910
this R F, and here is the other input signal
v a. v a v b and this is one and two, so this
76
00:17:02,110 --> 00:17:09,110
is the differential
amplifier circuit.
And v out, as we have studied in the differential
77
00:17:30,000 --> 00:17:37,000
mode gain into v a minus v b. This is the,
whatever is the signals which will be available
78
00:17:44,309 --> 00:17:51,309
here. We actually I should have written v
1 minus v 2 for this, but anyway the difference
79
00:17:58,130 --> 00:18:05,130
here that will be amplified; whatever is the
voltage at a v 1 and v 2 the difference will
80
00:18:05,710 --> 00:18:12,710
be amplified. Now to study these two signals,
again we take help of the super position theorem.
81
00:18:16,340 --> 00:18:23,340
In such cases, it also phase to apply the
super position theorem, and we will take one
82
00:18:25,059 --> 00:18:32,059
input signal active at a time, the other will
be grounded. We will see what will be the
83
00:18:32,179 --> 00:18:39,179
output of that, and then the other will be
activated and the first one will be in grounded.
84
00:18:40,010 --> 00:18:46,070
Again we will take the output and net output
will be the. This is super position theorem,
85
00:18:46,070 --> 00:18:53,070
that algebraic of the two will be the net
output of the differential amplifier. So,
86
00:18:56,179 --> 00:19:03,179
remembering this design this of course, we
will be needing in our discussion further,
87
00:19:03,240 --> 00:19:09,600
that the relationship between input and output
voltages can be obtained.
88
00:19:09,600 --> 00:19:16,600
Let us take case one, and case one is, let
v b be grounded. This is grounded on we are
89
00:19:26,260 --> 00:19:33,260
investigating the output, because of v a,
v a as you will see if this is a grounded.
90
00:19:36,679 --> 00:19:43,679
This is a inverting amplifier. So, we are
then having
inverting amplifier, so the output v o, because
91
00:20:00,529 --> 00:20:07,529
of a, will be minus R F by R 1 into v a. This
is the gain, gain is v o a by v a, which is
92
00:20:18,340 --> 00:20:25,340
equal to the ratio of that two resistances.
These resistance R F by R 1, this is what
93
00:20:27,220 --> 00:20:34,220
for inverting amplifier, this is the gain.
So, this is because of the first signal source
94
00:20:41,210 --> 00:20:48,210
v a active and v b is grounded. We take case
two; case two is let v a be grounded and v
95
00:20:58,010 --> 00:21:05,010
b is connected to non inverting input, with
resistance R 2 and R 3 this is implied. So,
96
00:21:16,270 --> 00:21:23,270
then we are dealing with
97
00:21:32,399 --> 00:21:39,399
a non inverting amplifier. And here what is
this voltage v 1, this will be, this signal
98
00:21:45,730 --> 00:21:52,730
source will send the current, and that current
will be v b by R 2 plus R 3 and when this
99
00:21:54,490 --> 00:22:01,490
current passes through this resistor R 3,
whatever is the voltage developed here that
100
00:22:01,970 --> 00:22:08,970
will be v 1. So, v 1 obviously is, v 1 is
R 3 R 2 plus R 3 into v b, voltage divider.
101
00:22:19,200 --> 00:22:25,450
This network is acting as a voltage divider
for this voltage. So, what is the voltage
102
00:22:25,450 --> 00:22:32,450
which will be available at R 3 is this, and
output due to v b, because of this input signal.
103
00:22:38,570 --> 00:22:45,570
Because of this input signal, the output then
will be v o b this is equal to 1 plus R F
104
00:22:53,700 --> 00:23:00,700
by R 1, this is the gain which we have still,
which we have arrived at earlier for non inverting
105
00:23:03,020 --> 00:23:10,020
amplifier this is the gain, and into the input
signal. This v o b by v is the gain of the
106
00:23:13,070 --> 00:23:19,860
non inverting amplifier, which is equal to
this and hence v o b is equal to this. Now
107
00:23:19,860 --> 00:23:26,860
we substitute for sorry v 1. This v 1 and
what will, it appear at output this is this,
108
00:23:35,809 --> 00:23:42,809
and now we substitute for v 1, as a we have
got here this is the value of v 1, which we
109
00:23:47,450 --> 00:23:54,450
substitute. And then v o b is equal to R 1
plus R F by R 1 into R 3, R 2 plus R 3 into
110
00:24:08,000 --> 00:24:15,000
v b. This is the output and. Now, let resistance
R 1 is equal to R 2, R 1 R 2 we take identical
111
00:24:30,600 --> 00:24:37,600
and R 3 and R F we take identical. So, let
R 1 is equal to R 2 and R F equal to R 3,
112
00:24:43,529 --> 00:24:50,529
then when we substitute this, this will cancel
and we are left with v o b equal to R F R
113
00:24:54,860 --> 00:25:01,860
1 v b. This is the contribution under this
condition, from the second input source v
114
00:25:05,490 --> 00:25:12,490
b.
Now we apply the differential or the effective
115
00:25:13,710 --> 00:25:20,710
super position theorem, and then it will be
A D differential gain, which is equal to v
116
00:25:26,510 --> 00:25:33,510
0, v a minus v b, and this is v o a plus v
o b, v a minus v b. This is the general expression
117
00:25:53,820 --> 00:26:00,820
for the differential amplifier, and when we
substitute these values of the two, by super
118
00:26:04,940 --> 00:26:11,940
position theorem then it comes out to be this,
and which obviously, is using one and two
119
00:26:14,929 --> 00:26:21,929
it comes out to be R F by R 1; that means,
the gain A D is equal to minus R F by R 1,
120
00:26:35,519 --> 00:26:42,340
this is under the condition. When in this
design, we have taken R 1 equal to R 2 and
121
00:26:42,340 --> 00:26:49,340
R F equal to R 3, which is a symmetrical design
which is normally used, and then this gain
122
00:26:50,330 --> 00:26:51,750
is this much.
123
00:26:51,750 --> 00:26:58,750
And the output will be with this gain, and
two inputs v a minus v b. It is the difference
124
00:27:07,750 --> 00:27:14,750
of the two signals will be amplified, and
whatever is common will be filtered out. Another
125
00:27:17,380 --> 00:27:24,380
important point is, that because the input
impedance is different for inverting and non
126
00:27:25,950 --> 00:27:32,950
inverting amplifier. So, let us briefly talked
about the input impedances of differential
127
00:27:42,350 --> 00:27:49,350
amplifier. The inverting part of it, that
is R i a input impedance, because of the inverting
128
00:28:02,730 --> 00:28:09,730
amplifier, this we know, that this is equal
to R 1. And for non inverting amplifier, what
129
00:28:12,350 --> 00:28:19,350
is the impedance which will be seen here,
between this and ground these two in series.
130
00:28:20,450 --> 00:28:27,450
So, this is one, and two is R i b, because
of this v what the b source will see as the
131
00:28:31,230 --> 00:28:38,230
impedance, that will be R 2 plus R 3, and
because we have taken it R 1 equal to r 2.
132
00:28:42,860 --> 00:28:49,860
So, this is R 1 plus R F.
The inverting amplifier will see R 1 as the
133
00:28:54,769 --> 00:29:01,769
impedance, while the non inverting will see
the impedance R 2 plus R 3, which is equal
134
00:29:01,769 --> 00:29:08,769
to R 1 plus r f. So, this is a important conclusion,
that besides the magnitude the fact is, that
135
00:29:14,070 --> 00:29:21,000
the impedance is seen by two input sources,
they will be different for a differential
136
00:29:21,000 --> 00:29:28,000
amplifier now. So, the major three amplifiers,
which can be easily accomplished with the
137
00:29:32,159 --> 00:29:39,159
help of a operational amplifier we have completed;
that is a non inverting amplifier, a inverting
138
00:29:41,370 --> 00:29:48,370
amplifier, and a differential amplifier. Now,
there are some other forms of the amplifiers,
139
00:29:49,200 --> 00:29:55,480
which are a small modifications over this
inverting and non inverting once, which are
140
00:29:55,480 --> 00:29:56,700
also widely used.
141
00:29:56,700 --> 00:30:03,700
So, let us take one or two cases. The case
of a voltage follower for example, voltage
142
00:30:10,639 --> 00:30:17,639
follower, this is one and another one which
we can study, is current to voltage converter.
143
00:30:29,889 --> 00:30:36,889
These are both these amplifiers; voltage follower,
and current to voltage convertor. Both these
144
00:30:37,149 --> 00:30:44,149
amplifiers have a good number of applications
and they are used. So, let us see how the
145
00:30:46,049 --> 00:30:52,250
inverting and non inverting amplifiers, with
little modification can give these functions,
146
00:30:52,250 --> 00:30:59,250
which find wide acceptability in systems.
Let us first take voltage follower; voltage
147
00:31:08,720 --> 00:31:15,720
follower is synonym to what we have done,
in emitter follower, and discreet circuits
148
00:31:16,299 --> 00:31:23,299
with b j t, we had emitter follower.
Now with op amp the circuit is voltage follower,
149
00:31:28,600 --> 00:31:35,600
and this is you remember this is the modification
of a non inverting, modified non inverting
150
00:31:43,779 --> 00:31:50,779
amplifier. So, the basic amplifier was this.
This is R 1, R F, v i, v o and this is the
151
00:32:25,600 --> 00:32:32,600
batteries, the d c sources, and the open loop
gain is A. This is a non inverting amplifier,
152
00:32:48,149 --> 00:32:55,149
and this is modified to this form, and how
this modification comes I will just talk,
153
00:33:08,559 --> 00:33:15,559
v i and here
154
00:33:25,259 --> 00:33:32,259
this is voltage follower, and how it has been
obtained. Here we have made R F equal to zero
155
00:33:40,870 --> 00:33:47,679
in the basic non inverting amplifier. This
R F has been made zero, so it is just a connector
156
00:33:47,679 --> 00:33:54,679
here. And R 1 is taken as infinitely large,
very high value, so not there. A gap is infinite
157
00:34:01,090 --> 00:34:08,090
a resistance, so R 1 has been taken infinite,
and R F has been taken zero. The circuit is
158
00:34:10,320 --> 00:34:12,280
reduced to this.
159
00:34:12,280 --> 00:34:19,280
And the gain of the non inverting amplifier
is 1 plus R F by R 1, when R F is a zero then
160
00:34:31,200 --> 00:34:38,200
this is reduced to one. When gain is one,
and then v out is gain into v in. And since
161
00:34:45,169 --> 00:34:52,169
this gain is one, v out is equal to v i. This
is what in emitter follower, why from where
162
00:34:55,829 --> 00:35:02,829
the name comes in emitter follower. The output
voltage follows the ammeter. Here we are seeing
163
00:35:03,280 --> 00:35:10,280
that this output which will appear here, this
will be of exactly equal magnitude as input,
164
00:35:13,800 --> 00:35:20,800
and the face will be in face; that means face
is also same, so that is why the name voltage
165
00:35:22,560 --> 00:35:29,560
follower. Now you may question that if the
same voltage appears and same a face appears,
166
00:35:31,060 --> 00:35:37,440
what is the function of a voltage follower.
The application of voltage followers is the
167
00:35:37,440 --> 00:35:44,440
same as for emitter follower; that is as a
buffer amplifier. The input impedance is very
168
00:35:45,490 --> 00:35:52,490
high, output impedance is very low. So, this
can be used to for matching purposes, matching
169
00:35:53,730 --> 00:36:00,730
impedance. So, the application is as a buffer
amplifier. But it is preferred over emitter
170
00:36:13,849 --> 00:36:20,849
follower for two reasons that the input impedance
is much higher in this case, in the case of
171
00:36:23,400 --> 00:36:30,400
voltage follower as compare to emitter follower.
And here output is exactly equal to input.
172
00:36:33,940 --> 00:36:40,940
They are more equal than in the previous case.
So, these are the two points, because of that
173
00:36:42,089 --> 00:36:49,089
a voltage follower, which is a modified non
inverting amplifier, that can be realized
174
00:36:52,060 --> 00:36:57,000
and it is used as a buffer amplifier.
175
00:36:57,000 --> 00:37:04,000
Another example let us take and that is of
current to voltage convertor. This is a modification
176
00:37:23,200 --> 00:37:30,200
of the inverting amplifier; modified inverting
amplifier, which can work as a current to
177
00:37:39,930 --> 00:37:46,930
voltage converter. There are many situations
where such kind of application will be required.
178
00:37:47,660 --> 00:37:54,660
For example, in a photo sensitive devices
photo divots. The currents are available which
179
00:37:57,839 --> 00:38:04,180
will be varying within ten stay or whatever
it is. So, those currents have to be converted
180
00:38:04,180 --> 00:38:11,180
into voltage format for further processing.
And so current to voltage convertor will be
181
00:38:12,250 --> 00:38:19,250
a requirement. Another other situations may
be in a digital to analog convertors. There
182
00:38:21,810 --> 00:38:28,400
are several sensors which may give it different,
which may give you the current, and we have
183
00:38:28,400 --> 00:38:35,400
to get the corresponding voltage. Now as I
said it is a modified inverting amplifier.
184
00:38:37,510 --> 00:38:44,510
For a inverting amplifier the gain with feedback
your, which is the ratio of output v to input,
185
00:38:48,040 --> 00:38:55,040
this is equal to minus R F by R 1, this we
have been saying on through that for inverting
186
00:38:58,560 --> 00:39:05,510
amplifier, gain is the ratio of the two resistance;
the feedback resistance and the input resistance.
187
00:39:05,510 --> 00:39:12,510
And from here v out is simply minus R F, R
1 into v 1 or this is equal to minus, same
188
00:39:19,960 --> 00:39:26,960
v 1, R 1, v i R 1 into R F. Now v 2 is 0 here.
This was the total amplifier, and now what
189
00:40:03,030 --> 00:40:10,030
we are seeing, this is the virtual ground
v 2, v 2 was 0, because it was because it
190
00:40:13,440 --> 00:40:20,440
is virtual ground. This is v 2 is virtual
ground, in that case the what will be this
191
00:40:28,319 --> 00:40:35,319
current, v i minus v 2 is divided by R 1 is
i 1, but because this is zero. So, because
192
00:40:51,140 --> 00:40:58,140
v 2 is zero, as it is virtual ground. The
v i R 1 this is equal to i 1, and this we
193
00:41:07,140 --> 00:41:09,579
can substitute, so this is the currents.
194
00:41:09,579 --> 00:41:16,579
So, what we are getting. We are getting v
out equal to minus i i into R F, that is output
195
00:41:25,040 --> 00:41:32,040
voltage is propositional to the current. This
is what is expected out of current to voltage
196
00:41:37,480 --> 00:41:44,480
convertor, and what we have done, this is
R F this is current i 1 and this is same as
197
00:42:04,609 --> 00:42:11,609
i i here, because this current i b is very
close to zero. And
we have replaced
the voltage source v i and resistance R 1
198
00:42:35,050 --> 00:42:42,050
by its current equivalent and we get this
circuit, and here obviously, the output varies
199
00:42:51,359 --> 00:42:58,359
according to the input, and hence this is
the current to voltage convertor. And here
200
00:43:05,809 --> 00:43:12,809
since the output as we have seen, output resistance
of the inverting amplifier is very low. So,
201
00:43:15,329 --> 00:43:22,329
this current, this voltage is developed, because
of this current variations; this will be independent
202
00:43:24,579 --> 00:43:31,579
of the load resistance. Because again the
same thing, low and high impedances in parallel,
203
00:43:35,579 --> 00:43:41,569
the lower one dominates. So, because the input
impedance is very low, this value of r l does
204
00:43:41,569 --> 00:43:48,569
not affect that and it works in a prefect
manner. And applications I said photo detector
205
00:43:51,150 --> 00:43:58,150
that photo currents are converted into voltages
and so on, so this is another application.
206
00:44:01,400 --> 00:44:08,400
Now moving further towards the application,
now we go for amplifiers; that means operational
207
00:44:10,650 --> 00:44:17,650
amplifiers for mathematical operations.
208
00:44:26,270 --> 00:44:33,270
Applications for mathematical operations;
we will
209
00:44:47,770 --> 00:44:54,640
take will good number of mathematical operations
and the circuits for them, that how we can
210
00:44:54,640 --> 00:45:01,640
realize those operations, how those operations
can be obtained electronically. Let us start
211
00:45:04,559 --> 00:45:11,559
with the simplest case, the
sign changer; that means electronics understand
the quantities in the form of voltage. So,
212
00:45:20,760 --> 00:45:27,760
if we can change the sign of voltage, then
that is the sign changer, and this can be
213
00:45:29,059 --> 00:45:36,059
easily obtained by inverting amplifier, using
inverting amplifier. This is inverting amplifier,
214
00:46:19,730 --> 00:46:26,730
in this the gain, you remember the gain is
minus R F by R 1 and which is equal to v 0
215
00:46:34,410 --> 00:46:41,410
output by input voltage. Now, if we choose
here the two resistances is as of same magnitude.
216
00:46:42,170 --> 00:46:49,170
So, R 1 is equal to R F; then this term will
be unit, and gain in this case will be one,
217
00:46:55,050 --> 00:47:02,050
and v zero will be minus v i. And these two
will be of identical magnitude, because gain
218
00:47:10,490 --> 00:47:17,490
is one. So, the sign has been changed. If
you want to change the sign, let us use a
219
00:47:19,940 --> 00:47:26,940
inverting amplifier with unit gain, then it
can amputate sign changer.
220
00:47:30,829 --> 00:47:37,829
Now more important application, for applications
of op amp, so we go for summing, scaling and
221
00:47:42,819 --> 00:47:49,819
averaging amplifiers. Summing
222
00:48:03,920 --> 00:48:10,920
amplifiers
223
00:48:15,359 --> 00:48:22,359
is the one, where there are more than one
inputs, and the output will be the sum of
224
00:48:24,460 --> 00:48:31,460
all the inputs. If that we can achieve that
will be is summing amplifier. And similarly
225
00:48:33,180 --> 00:48:40,180
is scaling and averaging amplifiers, we will
take one by one. So, let us first take. Now,
226
00:48:40,880 --> 00:48:47,880
these amplifiers can be realized either in
inverting mode and using inverting amplifier,
227
00:48:55,670 --> 00:49:02,670
or by using non inverting amplifier. To save
time we will go for the case inverting amplifier.
228
00:49:14,260 --> 00:49:21,260
So, let us take first a summing amplifier
using an inverting amplifier, what is the
229
00:49:23,220 --> 00:49:30,220
basic circuit. There will be more than one
input two three four five six any number.
230
00:49:31,170 --> 00:49:38,170
So, for simplicity we take three inputs. This
is v 1 v 2 v 3 and R F, and this is R 1 R
231
00:50:18,210 --> 00:50:25,210
2 R 3 and this is the inverting input, and
this is non inverting input. So, far we have
232
00:50:30,220 --> 00:50:37,220
being connecting it to ground, but now we
introduce another concept.
233
00:50:37,500 --> 00:50:44,500
Instead of directly connecting to ground,
it has been shown that the, if we connect
234
00:50:49,579 --> 00:50:56,579
the resistance here R x, which is roughly
equal to the parallel combination of all these
235
00:51:02,000 --> 00:51:09,000
resistance of R 1 R 2 R 3, it works batter.
Here the zero errors for example, when there
236
00:51:17,130 --> 00:51:24,000
is no input signal the output should be zero,
such conditions they are slightly battery
237
00:51:24,000 --> 00:51:31,000
achieved, and hence it is a common practice
to use here, a resistance of appropriate value
238
00:51:31,280 --> 00:51:38,280
which is given by the parallel combination
of these resistances, and that can be connected.
239
00:51:38,760 --> 00:51:45,760
So, this is the basic summing amplifier. And
as I said we have taken for simplicity three
240
00:51:58,839 --> 00:52:05,839
inputs; voltages v one v two v three, output
v zero and this is a inverting amplifier.
241
00:52:08,640 --> 00:52:15,640
Now, first I take the general expression,
and then one by one we will go for summing
242
00:52:17,150 --> 00:52:24,150
amplifier, scaling amplifier, small modifications
in the three now, but major thing is the same.
243
00:52:26,160 --> 00:52:33,160
So, this voltage v b, remember v b zero virtual
ground. Now, we
will take this and we can write, that the
244
00:52:55,849 --> 00:53:02,849
currents here; this is I 1, this is I 2, this
is I 3. Now, because this voltage is zero,
245
00:53:09,069 --> 00:53:16,069
what will be this current I 1, I 1 will be
v 1 minus v zero by R 1. Similarly these currents,
246
00:53:19,069 --> 00:53:26,069
but this is zero. So, the current is actually
v 1 R 1, this is equal to I 1, v 2, R 2 is
247
00:53:28,480 --> 00:53:35,480
I 2, so all these currents. And the sum up
of the currents will be I F. We will continue
248
00:53:41,540 --> 00:53:48,540
that this simple circuit will give always,
the output that will be the sum up input voltages,
249
00:53:50,530 --> 00:53:56,720
and if these voltages represent numbers two
three five whatever, then the output will
250
00:53:56,720 --> 00:54:03,720
be the sum of these numbers. So, we will continue
this analysis, that form the main analysis,
251
00:54:06,280 --> 00:54:13,280
we can arrive at scaling, averaging, summing
amplifiers and that all operations, mathematical
252
00:54:14,980 --> 00:54:21,980
operations, then we will go for differentiator
and integrator circuits and so on, we will
253
00:54:22,780 --> 00:54:23,180
continue .