1 00:00:28,019 --> 00:00:35,019 We were discussing the method, mathematical operations through amplifiers, and op amps 2 00:00:37,539 --> 00:00:44,100 are used in this process, and in fact that is the reason, that is why these are called 3 00:00:44,100 --> 00:00:51,100 operation amplifiers. So, the mathematical operation, which was currently under discussion, 4 00:00:56,860 --> 00:01:02,479 is summing amplifier is scaling amplifier and an averaging amplifier. These are three 5 00:01:02,479 --> 00:01:09,479 different amplifiers, but because they all the three, can be drawn from the basic amplifier. 6 00:01:09,600 --> 00:01:16,600 So, that is why I am taking them all under single heading. Now, these amplifiers can 7 00:01:19,850 --> 00:01:26,850 be realized by using a inverting amplifier, inverting op amp or by non inverting amplifier 8 00:01:30,700 --> 00:01:37,700 equally well. We are taking just one case that is of inverting amplifier, so how to 9 00:01:42,380 --> 00:01:49,380 realize a summing amplifier with inverting configuration. The summing amplifier is the 10 00:01:57,200 --> 00:02:04,200 one, in which the output is the sum of all the inputs. I repeat, that summing amplifier 11 00:02:08,970 --> 00:02:15,970 in is the one which gives a output which is the sum of all the inputs. This is summing 12 00:02:16,390 --> 00:02:23,390 amplifier. Now, the inputs can be more than one, so may be two, three, four, five, six 13 00:02:26,080 --> 00:02:33,080 or it can be number eight or ten, but we just to make the diagram simpler. We take three 14 00:02:36,060 --> 00:02:43,060 inputs and when we use a inverting in configuration; then this circuit is this. v 1 v 2 v 3 are 15 00:03:19,780 --> 00:03:26,780 the three voltages, which are to be summed up and output is here, and this non inverting 16 00:03:34,220 --> 00:03:38,900 input. Normally, so far we have been putting at a 17 00:03:38,900 --> 00:03:45,900 ground potential, by directly grounding it, but for a better arrangement is, when this 18 00:03:50,020 --> 00:03:57,020 resistance R x equivalent to the parallel combination of these resistance is attached. 19 00:03:58,100 --> 00:04:05,100 It is a still at grown potential, but not directly, but through a resistance and this 20 00:04:07,260 --> 00:04:14,260 reduces. It reduces the offset effects; we have not yet talked about the offset effects. 21 00:04:25,350 --> 00:04:32,350 We will talk little later about these things, but they arise, because of the a symmetry 22 00:04:36,240 --> 00:04:43,240 in the design of the circuit. And even when the inputs are absent some currents is still 23 00:04:45,229 --> 00:04:52,229 flow. They are of course, very mild, very small currents, but still, so those off set 24 00:04:52,439 --> 00:04:59,439 effects can be reduced drastically, if instead of directly connecting the non inverting input 25 00:05:00,460 --> 00:05:07,460 to ground. It is connected through a resistance equal to the parallel combination of all the 26 00:05:07,560 --> 00:05:14,560 input resistances. Now here this is at a potential v b, and the three currents i 1 i 2 and i 27 00:05:26,449 --> 00:05:33,449 3 they flow here, and from here will flow i f. 28 00:05:33,809 --> 00:05:40,809 Now, remembering that v b is at zero potential, remember we have said that the inverting input 29 00:05:49,270 --> 00:05:56,270 is virtual ground. So, this is at ground potential. Hence v b is zero, if this is so, this voltage 30 00:05:59,069 --> 00:06:06,069 is zero. Here it is v 1, then we can simply write the expression for the current i 1, 31 00:06:10,199 --> 00:06:17,199 this is actually speaking v 1 minus v b by R 1, but because v zero. So, i 1 is simply 32 00:06:21,159 --> 00:06:28,159 v 1 by R 1. Similarly, we can write for i 2 and i 3. So, i 2 is v 2 R 2, i 3 is v 3 33 00:06:42,409 --> 00:06:49,409 R 3. These are the currents and now we are seeing that this is simplified. And the whole 34 00:06:53,210 --> 00:07:00,210 expression gets simplified by this concept of virtual ground, and that is why emphasis 35 00:07:01,339 --> 00:07:08,339 was given to this point, that the inverted input is a virtual ground. Now, these three 36 00:07:14,889 --> 00:07:21,889 currents are flowing, there will be a current here I B, but I B as we are seeing that, because 37 00:07:24,439 --> 00:07:31,439 the input impedance of the amplifier is extremely high. So, this I B can be taken as zero. 38 00:07:35,389 --> 00:07:42,389 I B is equal to zero, we neglect this current. It is extremely small current it is extremely 39 00:07:49,059 --> 00:07:56,059 is small current, if the neglect this current I B is equal to zero, then; obviously, the 40 00:07:58,300 --> 00:08:05,300 summation of these currents is equal to I F. Therefore, I 1 plus I 2 plus I 3 this is 41 00:08:14,099 --> 00:08:21,099 equal to I F. And we can write for I F, all the three currents we have written. We can also write 42 00:08:28,270 --> 00:08:32,750 for I F the current we can write here. 43 00:08:32,750 --> 00:08:39,750 This three currents we write here. Now, what will be I F. This is v B minus this terminal 44 00:08:44,520 --> 00:08:51,520 is at potential v b, which is zero, we will write, finally zero minus v 0 by R F and this 45 00:08:53,839 --> 00:09:00,839 is equal to minus v 0 by R F. So, substituting the values of all the currents in this expression, 46 00:09:05,790 --> 00:09:12,790 we get v 1 by R 1, which is equal to I 1 plus v 2 by R 2 plus v 3 by R 3 and this is equal 47 00:09:21,320 --> 00:09:28,320 to minus v 0 R F, or the output we can write from here, or v 0 is equal to minus R F R 48 00:09:44,560 --> 00:09:51,560 1 by v 1 plus R F R 2 v 2 plus R F R 3 v 3. This is the relation which is we may, say 49 00:10:06,410 --> 00:10:13,410 the basic relation for this circuit, and then by choosing. You recall that R F by R 1 is 50 00:10:16,570 --> 00:10:23,570 the magnitude of the gain by which v 1 will be amplified. Similarly, R F by R 2 is the 51 00:10:26,690 --> 00:10:33,690 gain by which v 2 will be amplified, and f appears at the output and so on. So for now, 52 00:10:36,570 --> 00:10:43,570 from this basic equation which we call as first equation, we can now go for summing 53 00:10:45,270 --> 00:10:52,270 amplifier. Summing amplifier in this equation one, if we choose R 2 equal to R 2 equal to 54 00:11:10,520 --> 00:11:17,410 R 3, say this is R. 55 00:11:17,410 --> 00:11:24,410 Then what we have is this, v 0 is equal to minus R F by R and v 1 plus v 2 plus v 3. 56 00:11:38,920 --> 00:11:45,920 If further we choose R F this gain as one. So, R F further, if R F is taken equal to 57 00:11:57,370 --> 00:12:04,370 R, then magnitude of 58 00:12:10,720 --> 00:12:17,720 gain is equal to one, and in that case this term will be one, and v 0 is simply minus 59 00:12:20,120 --> 00:12:27,120 v 1 plus v 2 plus v 3; that means, the output voltage for this summing amplifier under these 60 00:12:32,600 --> 00:12:39,600 conditions of choosing R 1 equal to R 2 equal to R 3, and let us say that is R. And R F 61 00:12:41,470 --> 00:12:48,470 also equal to have that is alright this resistance is in this circuit are taken as single value 62 00:12:48,830 --> 00:12:55,830 resistance. So, all these resistance is R F r 1 r 2 r 3 are taken as a single value 63 00:12:56,540 --> 00:13:03,540 resisters say 10 k, 20 k and so on. Then the output will be, the sum of all the input voltages 64 00:13:05,990 --> 00:13:12,990 with a reverse sign, with a negative sign; that is sign is not really material, if you 65 00:13:14,020 --> 00:13:21,020 want positive we can use the sign changer circuit, after that which we have already 66 00:13:21,110 --> 00:13:28,110 discussed, and because this minus sign is coming, because we have taken a inverting 67 00:13:28,180 --> 00:13:35,180 amplifier. If we take non inverting this sign automatically will disappear, but as I said 68 00:13:35,600 --> 00:13:42,600 it is not is significant point. So, this is the summing circuit. 69 00:13:43,880 --> 00:13:50,880 Then we can go for is scaling, scaling or weighed amplifier, in the 70 00:14:14,230 --> 00:14:21,230 simply summing amplifier, we have seen that all the inputs are amplified by the same amount, 71 00:14:22,560 --> 00:14:29,560 choosing all the resistance is as single value are in R F also of R. Then simple summing 72 00:14:31,710 --> 00:14:38,710 amplifier results is scaling are weighted amplifier is the one, in which the inputs, 73 00:14:43,430 --> 00:14:50,430 each input voltage is amplified differently. I repeat that in a scaling, a scaling amplifier 74 00:14:55,370 --> 00:15:02,370 is also called weighted amplifier, in which the amplification in factor for individual, 75 00:15:03,120 --> 00:15:10,120 input voltage is weighted, is not same, but they are different, they are weighted. And 76 00:15:12,050 --> 00:15:19,050 so then the amplifier is called a scaling are weighted amplifier. Such amplifiers are 77 00:15:20,190 --> 00:15:27,190 needed. For example, if we want that the output should be when v 1 v 2 v 3 are the input voltages, 78 00:15:36,370 --> 00:15:43,370 then v the output should meet this requirement; that means, this equation should be satisfied, 79 00:15:51,540 --> 00:15:56,280 this is equation. The output should we according to this equation, 80 00:15:56,280 --> 00:16:03,280 where v 1 v 2 v 3 are the inputs. Now here same circuit and same fundamental equation, 81 00:16:07,550 --> 00:16:14,550 that circuit is the same, but here the values of R 1 R 2 R 3 and R F we have choose differently, 82 00:16:18,380 --> 00:16:25,380 and we will take help of this basic equation, which we have written for that amplifier. 83 00:16:26,130 --> 00:16:33,130 Now, in this case if we choose a particular value of R F. For example, let R F the feedback 84 00:16:40,760 --> 00:16:47,760 resister is of the value of twelve kilo ohms. We can choose 24, but just as an example I 85 00:16:50,670 --> 00:16:57,670 am taking 12. Then we should choose R 1 is 3 k, 3 kilo ohms. So, that this will be amplified 86 00:17:07,240 --> 00:17:14,240 by R F by R 1, look here R F by R 1 into v 1. Here we want affecter of four, according 87 00:17:18,299 --> 00:17:25,299 to the requirement. And if we choose R F at 12 kilo ohms, then R 1 necessarily we will 88 00:17:26,919 --> 00:17:32,770 have to take at 3, so that if factor of 4 comes. 89 00:17:32,770 --> 00:17:39,770 And R 2, we are suppose to take as 24 kilo ohms, because R F by R 2 here, we need is 90 00:17:53,470 --> 00:18:00,470 point five. So, this will v point five, when 12 kilo ohms divided by 24 kilo ohms. So that 91 00:18:01,750 --> 00:18:08,750 means, 12 by 24, which is point five. So, we will get again of point five. And similarly 92 00:18:11,210 --> 00:18:18,210 R 3 we take as at 8 kilo ohms. These resistance are whole different R 1 R 2 R 3, they are 93 00:18:23,590 --> 00:18:30,590 not same they are all different, and they are different, then R F as well. So, this 94 00:18:31,809 --> 00:18:38,210 scaling can be done according to this equation, by choosing this resistances, and then keeping 95 00:18:38,210 --> 00:18:45,210 that we will get, which will be 4 v 1 point 4 v 2 plus 1 point 5 v 3 with a reverse sign, 96 00:18:53,260 --> 00:19:00,260 with a minus sign, if this minus sign we already talked, it is not really much important, it 97 00:19:00,910 --> 00:19:07,910 can be taken care of without much problem. So, this is the scaling are weighted amplifiers. 98 00:19:09,870 --> 00:19:16,870 Then the third one was the averaging amplifier. Averaging amplifier, what say averaging amplifier. 99 00:19:37,780 --> 00:19:44,780 Averaging amplifier is the one in off the inputs. Average of all the inputs put together, 100 00:19:48,300 --> 00:19:55,300 SO that is averaging amplifier. Now averaging amplifier again by the same fundamental equation 101 00:19:56,910 --> 00:20:03,910 we can get. If R 1 is equal to R 2 is equal to R 3, say R. This is the condition and R 102 00:20:19,100 --> 00:20:26,100 F by R is taken as 1 by n, this is additional point, first point is this. This is the second 103 00:20:30,390 --> 00:20:37,390 point in R F by R should be taken as 1 by n, where n is the number of inputs, in the present case 104 00:20:51,500 --> 00:20:58,500 which is under discussion. There are three inputs. So, this factor R F by R should be 105 00:20:59,440 --> 00:21:06,440 equal to, in the present case under discussion, we should have R F by R equal to 1 by 3. Now, if we substitute 106 00:21:27,650 --> 00:21:34,650 R F by when we take R 1 equal to R 2 equal to R 3 equal to R. So, here this is common term 107 00:21:43,130 --> 00:21:50,130 which we can take out, R F by R, and R F by R in this case, is taken as 3. So, obviously, 108 00:21:52,280 --> 00:21:59,280 v zero is minus R F by R into v 1, plus v 2, plus v 3. And since this is taken as one, 109 00:22:08,290 --> 00:22:15,290 so v 0 will be minus v 1 plus v 2 plus v 3 by 3, because R F by R is 1 by 3. So, this 110 00:22:27,179 --> 00:22:34,179 is the result for averaging amplifier. The output is the average of all the inputs with 111 00:22:42,960 --> 00:22:49,960 this sign reversed. So, this is the averaging amplifier, we continue with these amplifiers, 112 00:22:53,760 --> 00:23:00,760 we will take other circuits also and we continue, let us talk of a logarithmic amplifier. 113 00:23:35,120 --> 00:23:42,120 Logarithmic amplifier is the one, in which the output is the log of the input. I repeat 114 00:23:46,110 --> 00:23:53,110 what is logarithmic amplifier. Logarithmic amplifier is the one in which output is proportional, 115 00:23:55,890 --> 00:24:02,890 to the log of input, let us say, that is v i, this is the logarithmic amplifier. Now 116 00:24:06,240 --> 00:24:13,240 if in the simple inverting amplifier, if we replace the resistance R F by a diode, what 117 00:24:17,970 --> 00:24:24,970 will the circuit. This will be the circuit, v out and this is a D C voltage v s, d c source. 118 00:25:21,110 --> 00:25:28,110 This is the circuit of a logarithmic amplifier. Here the I B characteristic of diode, we remember 119 00:25:32,690 --> 00:25:39,690 that I equal to I 0, exponential v Q by K T minus 1. This is the equation of diode, 120 00:25:52,490 --> 00:25:59,490 which describes the current voltage relationship. This is applied voltage in this the resulting 121 00:26:00,179 --> 00:26:07,179 current, and here I 0 is the saturation current; the saturation current. This we are going 122 00:26:17,750 --> 00:26:24,750 to use and here this K T by q, this is Boltzmann’s constant, this is temperature in Kelvin and 123 00:26:30,480 --> 00:26:37,480 this is q we charge electronic charge when. So, the thermal equivalent of that room temperature, 124 00:26:38,080 --> 00:26:44,429 the voltage equivalent this is that, the voltage equivalent of thermal energy at room temperature. 125 00:26:44,429 --> 00:26:51,429 This comes how to be roughly 26 milli volts at temperature T equal to 300 degrees Kelvin. 126 00:26:58,549 --> 00:27:05,549 So, as long as this v applied field, is say of the order of 500 milli volts or one volt. 127 00:27:10,690 --> 00:27:17,690 This one will be neglected, because this is exponential term which will rise very fast. 128 00:27:19,370 --> 00:27:26,370 And so neglecting that one, we can use the equation I equal to I 0 exponential v q by 129 00:27:33,870 --> 00:27:40,870 K T from here, or exponential v q by K T is equal to I by I 0, or we take log of both 130 00:27:55,419 --> 00:28:02,419 sides; v equal to K T by q into log I minus log I zero, and this current I here in this 131 00:28:24,799 --> 00:28:31,799 circuit. This is I S from the source, this current and this current is I and this voltage 132 00:28:37,820 --> 00:28:44,820 across this which will are applying that is v. And this current the current here, this 133 00:28:47,730 --> 00:28:54,730 is actually zero, because of very high impedance of the operational amplifier. The current 134 00:28:59,679 --> 00:29:06,679 going in the inverting input is zero. So, taking I as I s equal to v s by R. And further 135 00:29:18,880 --> 00:29:25,880 neglecting small contribution, this contribution of this term is negligibly small. So, we neglect 136 00:29:41,059 --> 00:29:48,059 that, in that case this v will be equal to minus. It will v 0, because see here. Here 137 00:30:03,750 --> 00:30:10,750 this is the voltage developed across this, and because it is inverting. So, with the 138 00:30:13,419 --> 00:30:20,419 minus sign it will be K T by q into log v s by R. Thus this is a constant. So, the output 139 00:30:35,620 --> 00:30:42,620 voltage is proportional to the natural log of the input v s. Output voltage is proportional, 140 00:30:53,309 --> 00:31:00,309 to the natural logarithmic of the input voltage. This is the logarithmic amplifier. Now we 141 00:31:19,850 --> 00:31:26,850 will go for other mathematical operations, using these op amp circuits, we go for the 142 00:31:32,610 --> 00:31:38,330 integrator. 143 00:31:38,330 --> 00:31:45,330 The integrator, this is also called integration amplifier. Here the output voltage is integral 144 00:32:16,679 --> 00:32:23,679 of the input voltage, output voltage is the integral of the input voltage then the amplifier 145 00:32:27,000 --> 00:32:34,000 is called, the circuit is called the integrator. Integrator can be easily realized, by using 146 00:32:37,030 --> 00:32:44,030 the op amp in the inverting mode, and by replacing the feedback resistance R F by a capacitor. 147 00:32:47,750 --> 00:32:54,750 So, the circuit of the integrator, the basic circuit is this. This is R, this is that cap 148 00:33:12,870 --> 00:33:19,870 stance, this is the point x. And here we attach the input signal, and output is taken here 149 00:33:28,400 --> 00:33:35,400 v out. This is the circuit, R F has been replaced by the capacitor c. And here this is the current 150 00:33:46,870 --> 00:33:53,870 i which would be flowing, and here also, because this current we taking as zero. So, the same 151 00:33:55,460 --> 00:34:02,460 current will flow here, in the capacitor. Now, from the simple basic capacitor theory; 152 00:34:08,200 --> 00:34:15,200 from the basic capacitor theory or the theory of the capacitor, the voltage developed across 153 00:34:24,419 --> 00:34:31,419 the capacitor and the charging current have a relationship; that i is equal to c d v by 154 00:34:35,179 --> 00:34:42,179 d t. If this is the relationship between the charging current and the voltage, developed 155 00:34:46,840 --> 00:34:53,840 across the capacitor, this is simple relationship. The charging current gives a rate of change 156 00:34:55,780 --> 00:35:02,780 of voltage, and the capacitor is charged. Now, a applying this equation in the present 157 00:35:04,240 --> 00:35:11,240 case, then it will be. Let us say which of course, this voltage is v 2, which is zero 158 00:35:17,090 --> 00:35:24,090 of course, but for the sake of the argument, this is c into d v 2 minus v 0. The two sides 159 00:35:28,720 --> 00:35:35,720 of the capacitor, one is at a voltage of v 2, and the other and is at v 0 and this is 160 00:35:40,670 --> 00:35:47,670 d t, but v 2 is 0, as it is virtual ground. 161 00:35:54,770 --> 00:36:01,770 So, i is equal to minus c d v 0 by d t, and i this is the same as this i. So, i is; obviously, 162 00:36:18,869 --> 00:36:25,869 here v i minus v 2 by R, i is equal to v 1 minus v 2 by R, v to v in 0, is equal to v 163 00:36:37,270 --> 00:36:44,270 i by R. This is i. So, this i we substitute in this expression, therefore we get v i by 164 00:37:01,290 --> 00:37:08,290 R, the current is equal to minus c into d v 0 by d t. or from here we can write d v 165 00:37:20,210 --> 00:37:27,210 0 by d t is equal to minus 1 by R c into v i by d t. integrating both sides, we get v 166 00:37:54,609 --> 00:38:01,609 0. So, this will not be there, when we take this, this is simply this will be in the case. 167 00:38:09,220 --> 00:38:16,220 So, v 0 is equal to minus 1 by R c, integral 0 to t, v i d t plus c 1, where c 1 is a integration 168 00:38:27,510 --> 00:38:34,510 constant, c 1is a integration constant and represents initial capacitor voltage at t 169 00:38:46,310 --> 00:38:53,310 equal to 0. And it represents initial voltage at capacitor at t, the time equal to zero. 170 00:39:05,400 --> 00:39:12,400 And this we can set c 1 can be set to zero. In that case v 0 the output voltage will be 171 00:39:21,910 --> 00:39:28,910 1 by R c to integral v i over d t. This is the expression for the output voltage, and 172 00:39:42,349 --> 00:39:49,349 the amplifier, therefore provides an output voltage, which is proportional to the integral 173 00:39:50,849 --> 00:39:57,849 of input voltage. Output is proportional this is a constant R c, time constant of the circuit. 174 00:39:59,990 --> 00:40:06,990 So, v 0, the output voltage is proportional to the integral of the input voltage. This 175 00:40:10,760 --> 00:40:17,760 is the integration and the integrator or a integration amplifier. Now let us discuss 176 00:40:22,520 --> 00:40:29,520 further more, and what are the problems in this basic integrator, and how we take care 177 00:40:31,869 --> 00:40:38,869 of those things. Now here if the input voltage is constant, that is v i is a constant, say 178 00:40:54,369 --> 00:41:01,369 let us write with capital V, then by substituting that in this equation, we will get v 0 will 179 00:41:06,170 --> 00:41:13,170 be V t y R c. So, from time to zero to time t, this will be keeping on changing. So, it 180 00:41:17,859 --> 00:41:24,859 will be a ramp, the output will be a ramp, because of this negative sign, it will be 181 00:41:26,210 --> 00:41:33,210 going this way. This is time and this is the voltage v zero. So, a constant input voltage 182 00:41:33,820 --> 00:41:40,820 if it is positive, the ramp will be like this. If v i is a negative voltage then the ramp 183 00:41:43,089 --> 00:41:50,089 will go in the other direction, this is t. So, this will be the ramp. This is v zero 184 00:41:52,440 --> 00:41:59,440 verses t, and for a square wave input, we will get a triangular shape output. 185 00:42:08,420 --> 00:42:15,420 If the input is a square wave, the output will be triangular; that means these once. This 186 00:42:40,400 --> 00:42:47,400 is input in square view, this is time, then the output, this is constant, so it will be 187 00:42:58,410 --> 00:43:03,599 a ramp, and then this is also constant, it will be ramp and that these two ramps will 188 00:43:03,599 --> 00:43:10,599 be going in a opposite direction. This is t, so this is the output, a triangular. So, 189 00:43:21,609 --> 00:43:28,609 if the input is a square wave, output is triangular. In square wave input will give a triangular 190 00:43:44,000 --> 00:43:45,510 move. 191 00:43:45,510 --> 00:43:52,510 And if the input is a sinusoidal signal, let input voltage be sinusoidal, say v i is a 192 00:44:09,560 --> 00:44:16,560 zero sin omega t. We put this in the fundamental equation, in this equation. Then v 0 will 193 00:44:28,339 --> 00:44:35,339 be minus 1 by R c into a 0 sin omega t d t, and this v 0 comes out to be a 0, omega R 194 00:44:51,460 --> 00:44:58,460 c cos omega t. Here this is important that the maximum amplitude of the output, will 195 00:45:05,430 --> 00:45:12,430 be is frequency dependent. This is frequency dependent. Therefore, whatever is the maximum 196 00:45:13,140 --> 00:45:20,140 output for a signal of a frequency, say one kilo hertz at two kilo hertz, it will be half. 197 00:45:22,030 --> 00:45:29,030 Maximum output will be half, so it will be frequency dependent. Frequency dependent maximum output voltage. Now let us talk, we 198 00:45:49,050 --> 00:45:56,050 continue with the integrated amplifier, that what are the problems with the basic circuit, 199 00:45:57,410 --> 00:46:04,410 which we have drawn simply by replacing R f by a capacitor c. 200 00:46:05,160 --> 00:46:12,160 Now, there are two problems; problems with basic integrator, problems are that, for v 201 00:46:30,810 --> 00:46:37,810 1 when it is zero, the integrator x as in open circuit for d c. Look at this circuit, 202 00:46:51,520 --> 00:46:58,520 here when v i is zero, then in this circuit, the integrator x is a op amp circuit, because 203 00:47:07,040 --> 00:47:14,040 this capacitor will act as a blocking capacitor, and it will not permit any d c currents to 204 00:47:16,410 --> 00:47:23,410 be fed back to the input. I repeat that when v i is zero capacitor c blocks d c current, 205 00:47:34,849 --> 00:47:41,849 and from where this current will come. These are drift currents, drift and off set current. 206 00:47:46,220 --> 00:47:53,220 I said about offset currents, we will be talking later, but at least for the time being, we 207 00:47:56,200 --> 00:48:03,200 can say offset currents are there, because of a symmetric design of this circuits, which 208 00:48:05,849 --> 00:48:12,849 are almost impossible to get them in perfect mast conditions. So, this small current, there 209 00:48:13,990 --> 00:48:20,990 will be blocked by the capacitor, and these currents will not be fed back to the inverter, 210 00:48:24,220 --> 00:48:31,220 but they will continuously charge the capacitor. And hence the voltage gets developed, and 211 00:48:35,240 --> 00:48:42,240 then it is amplified and the output voltage appears, while the input is zero. So, this 212 00:48:44,609 --> 00:48:51,190 presents of this output voltage, it is a error voltage and it should not b when input voltage 213 00:48:51,190 --> 00:48:58,190 is zero, the output voltage should be zero, but it will be finite, because of the charging 214 00:48:59,800 --> 00:49:06,800 of the capacitor, and then that voltage is amplified by the circuit and there will be 215 00:49:08,250 --> 00:49:15,250 a error voltage, gives error voltage at the output. Another problem; that in this circuit 216 00:49:31,640 --> 00:49:38,640 at very low frequencies close to, for example zero. The gain of this inverting amplifier, 217 00:49:39,980 --> 00:49:46,980 is impedance of the capacitor divided by R F by R. R F is now replaced by X c, that is 218 00:49:49,339 --> 00:49:56,339 the capacitive reactants of the capacitor. So, the gain is this, and this is close to, 219 00:50:04,099 --> 00:50:11,099 this is this is very high reactance, because the capacitor 1 by omega c is the reactance. 220 00:50:13,329 --> 00:50:20,329 So, omega approach is zero, this reactance X c approach is infinity, as omega approach 221 00:50:23,900 --> 00:50:30,900 is zero. So that means, the gain will be very high for is small frequencies. These problems 222 00:50:33,510 --> 00:50:40,510 can be taken care, by attaching a register R F with the capacitor, and that is the practical 223 00:50:48,560 --> 00:50:52,240 circuit. 224 00:50:52,240 --> 00:50:59,240 Practical integrator; the circuit is this. C F is that capacitor, and 225 00:51:50,060 --> 00:51:57,060 v zero we take here. Now, this R F we will take care of the problems, which we have discussed, 226 00:52:02,240 --> 00:52:09,240 that it minimizes. R F minimize is the output error voltage, and it will also take care 227 00:52:25,530 --> 00:52:32,530 at very low frequencies, the gain will not be infinity, which will be at very low frequencies; 228 00:52:39,880 --> 00:52:46,880 that means omega going to zero. Now, this is infinity we can neglect, because R F is 229 00:52:48,320 --> 00:52:55,320 in parallel, and gain will be R F by R which is finite. And we will see that normally this 230 00:52:56,700 --> 00:53:03,700 is kept around ten by design, we keep it around ten. So, this takes care, but it puts a limitation. 231 00:53:12,960 --> 00:53:19,960 True integration will be there, when the combination R F and C F produce a capacitive impedance. 232 00:53:29,170 --> 00:53:36,170 So, there is a restriction that at very low frequency, this circuit will not be a true 233 00:53:42,280 --> 00:53:49,280 integrator, and what is that condition we will just see, but this is the practical circuit 234 00:53:50,050 --> 00:53:57,050 and inclusion of R F that gives the modification and the performance. 235 00:53:58,710 --> 00:54:05,710 Now, we discussed the gain magnitude and frequency response of this circuit. Frequency response of the integrator; 236 00:54:27,770 --> 00:54:34,770 if we take v i, we have done it above as sin omega t and v 0 is minus 1 by R C F into, 237 00:54:53,510 --> 00:55:00,510 this is omega a zero, integral of this a zero into cos omega t. This is input. This is output 238 00:55:05,950 --> 00:55:12,950 after integration. And if we take the ratio of the peak voltages, v zero peak and v i 239 00:55:19,240 --> 00:55:26,240 peak, peak voltages, ratio of peak voltages, then this is a zero, a zero omega R C F and 240 00:55:34,180 --> 00:55:41,180 this is 1 by omega R C F. So the gain, this is gain, gain falls. This is v 0 by v i by 241 00:55:59,109 --> 00:56:06,109 taking the peaks. This is gain and this is in d B. this is frequency f, and this is f 242 00:56:12,060 --> 00:56:19,060 1, and we will see what it is. This will come out from here. f 1 comes out to be 1 by 2 243 00:56:22,310 --> 00:56:29,310 pi R 1 R, let we have taking only R C F. And this is a slop minus 20 d B fall per decade 244 00:56:33,150 --> 00:56:40,150 of change of frequency. So, this is the frequency response, and this is for the basic amplifier, 245 00:56:42,730 --> 00:56:49,730 basic integrator and how this is modified for the case of the practical integrator, we will take further 246 00:56:56,839 --> 00:57:03,839 and discuss other circuits also.