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 .