1 00:00:00,590 --> 00:00:05,930 Now let us consider nonlinear circuits; a circuit with a two terminal nonlinear element 2 00:00:05,930 --> 00:00:33,510 such as diode, and see how to analyze it. And let me take this circuit with a voltage 3 00:00:33,510 --> 00:00:48,450 source, a resistor – R, and a diode. And of course this could be any other nonlinear 4 00:00:48,450 --> 00:01:00,500 element as well. I will call the voltage across this V D, and the current through it as I 5 00:01:00,500 --> 00:01:12,259 D. Let us see how to analyze this. Now a linear circuit with this same structure is to have 6 00:01:12,259 --> 00:01:21,630 R and R 1. Now, of course this one after you have done your basic electrical circuits or 7 00:01:21,630 --> 00:01:25,240 even maybe before, it does not even require a moment thoughts right, you know what the 8 00:01:25,240 --> 00:01:30,450 solution is, it is just the resistive divider. Now we will see how to solve this and you 9 00:01:30,450 --> 00:01:35,719 will understand that the analysis of nonlinear circuits is considerably harder than linear 10 00:01:35,719 --> 00:01:39,349 circuits. 11 00:01:39,349 --> 00:01:47,619 So, now really this is an extremely simple circuit. So, this is an independent voltage 12 00:01:47,619 --> 00:01:53,499 source and all way have to do is write k c l at this node that will give us the solution. 13 00:01:53,499 --> 00:02:01,340 So k c l at that node says that the current flowing through R in this direction equals 14 00:02:01,340 --> 00:02:10,079 I D which is current flowing through a diode in the downward direction. Now, the diode 15 00:02:10,079 --> 00:02:18,340 itself tells you that I D is f of V D, this is the general case for nonlinear element; 16 00:02:18,340 --> 00:02:37,349 and in case of the diode, it is I s exponential V D by the thermal voltage minus 1. And the 17 00:02:37,349 --> 00:02:45,510 current in the resistor R, this equals V S minus V D which is the voltage across the 18 00:02:45,510 --> 00:02:54,100 resistor divided by the resistance value. So, our k c l equation simply says that V 19 00:02:54,100 --> 00:03:04,519 S minus V D by R equals f of V D in case of general nonlinearity or in the specific case 20 00:03:04,519 --> 00:03:16,760 of the diode, it is I s exponential V D by V t minus 1. 21 00:03:16,760 --> 00:03:23,489 Now there is only one variable here; V s is the source, we know that. The only one variable 22 00:03:23,489 --> 00:03:30,750 is V D; so all others are parameters of the components. R is the parameter of the resistor, 23 00:03:30,750 --> 00:03:36,980 and I s is the parameter of the diode, and V t is the fundamental constant. So, you may 24 00:03:36,980 --> 00:03:44,110 be familiar with solving nonlinear equations, you may be familiar with methods like bisection 25 00:03:44,110 --> 00:03:51,209 method or Newton-Rhapson iteration and so on. So, if you do then you know how to solve 26 00:03:51,209 --> 00:03:57,170 this numerically and you know that unlike linear equations, there is general inverse 27 00:03:57,170 --> 00:04:02,499 in formula or general solution formula for nonlinear equations. So, if you have a set 28 00:04:02,499 --> 00:04:08,010 of linear equations by a matrix inversion you can find the solutions; but for nonlinear 29 00:04:08,010 --> 00:04:12,989 equations, there is no such general method and depending on the type of nonlinearity, 30 00:04:12,989 --> 00:04:18,039 you may have to choose the different methods. Usually the solution is numerical, and there 31 00:04:18,039 --> 00:04:21,459 may be multiple solutions, there may be no guarantee that you will reach the solution 32 00:04:21,459 --> 00:04:30,250 and so on. But usually the ways to solve nonlinear equations are numerical and iterative. 33 00:04:30,250 --> 00:04:33,860 Of course, in this course we would not go through that; we just assume that the solution 34 00:04:33,860 --> 00:04:41,510 can be found somehow. What we will do is to bypass explicitly solving the nonlinear equation, 35 00:04:41,510 --> 00:04:58,300 at least for hand calculations. So, this can be solved numerically. Now what is also useful 36 00:04:58,300 --> 00:05:05,720 for engineers is to find the solution pictorially or graphically, this is for the sake of visualization. 37 00:05:05,720 --> 00:05:11,490 Now usually you do not do graphical solutions when you want accuracy in the solution, but 38 00:05:11,490 --> 00:05:13,849 you do it when you want to visualize the behavior. 39 00:05:13,849 --> 00:05:21,539 So let us do that now for this particular circuit. Now that technique is also limited, 40 00:05:21,539 --> 00:05:26,500 so when you have too many variables, you cannot draw multi-dimensional graphs. In this particular 41 00:05:26,500 --> 00:05:32,099 case, it turns out you can do it, because you have only two variables – one voltage 42 00:05:32,099 --> 00:05:48,030 and one current. So the equation to be solved is V S minus V D by R which equal to I s exponential 43 00:05:48,030 --> 00:06:08,830 V D by V t minus 1. So, what I can do is I can essentially plot the right hand side and 44 00:06:08,830 --> 00:06:14,720 left hand side separately. So, there is some function on the left hand side, there is some 45 00:06:14,720 --> 00:06:19,289 other function on the right hand side, both are functions of V D which my independent 46 00:06:19,289 --> 00:06:28,780 variable. And I can find the point of equality or the point of intersection that will be 47 00:06:28,780 --> 00:06:34,240 the solution. Now the right side is nothing but the diode current, so that is why I marked 48 00:06:34,240 --> 00:06:40,129 it as I D versus V D, so plotting the right side, it simply means plotting the diode characteristic, 49 00:06:40,129 --> 00:06:55,120 which are like that; it is the exponential. And plotting the left side that is what I 50 00:06:55,120 --> 00:07:06,039 want to plot is V S minus V D by R versus V D. 51 00:07:06,039 --> 00:07:11,600 So, this is just the straight line you can see that this is the linear function of V 52 00:07:11,600 --> 00:07:22,550 D, so when V D equals zero, this function will be V S by R; and when V D equals V S, 53 00:07:22,550 --> 00:07:28,129 this function will be zero. And it obvious that it also has a negative slope and the 54 00:07:28,129 --> 00:07:35,780 slope is minus 1 by R. Remember V D is the independent variable here, and the coefficient 55 00:07:35,780 --> 00:07:45,000 of V D in this function is minus 1 by R. So, very easy to plot also; so when V D is zero, 56 00:07:45,000 --> 00:07:52,830 it is V S by R; when V D equals V S, it will be zero. And it will be a line – straight 57 00:07:52,830 --> 00:07:59,979 line connecting these two points, and the slope of this is obviously minus 1 by R. Now, 58 00:07:59,979 --> 00:08:05,930 clearly in this case, there is only one point of intersection and that is the solution. 59 00:08:05,930 --> 00:08:14,699 Now if you solve this equation numerically, you would find this value of V D, and if you 60 00:08:14,699 --> 00:08:19,180 find the value of the current corresponding to that that is V S minus V D by R or this 61 00:08:19,180 --> 00:08:36,150 side, it will be this value of current. So this is the solution. So, by plotting different 62 00:08:36,150 --> 00:08:41,750 parts of the equation and finding the point of equality that is finding the point of intersection, 63 00:08:41,750 --> 00:08:47,820 you can find the solution. So, say somehow you find the solution either numerically or 64 00:08:47,820 --> 00:08:52,540 graphically. Now I will just give you an illustration; if you want, you can solve this numerically 65 00:08:52,540 --> 00:09:06,850 and find the solution. So, let me take a case, where V S is 5.7 volts, R is 5 kilo ohms and 66 00:09:06,850 --> 00:09:16,920 the diode has a saturation current of 10 to the minus 15 amperes. It turns out that if 67 00:09:16,920 --> 00:09:24,650 you solve this numerically, you will find that V D is approximately 0.72 volts, and 68 00:09:24,650 --> 00:09:44,350 the diode current is approximately 1 milli amp that is the solution; let say we have 69 00:09:44,350 --> 00:09:48,600 solved it for a particular value of V S, which is 5.7 volts. 70 00:09:48,600 --> 00:10:01,421 And now the value of V S changes, so let say it becomes 6.7 volts. So, what you have to 71 00:10:01,421 --> 00:10:08,460 do, you have to solve this all over again that is pretty clear right. So, with let say 72 00:10:08,460 --> 00:10:13,910 this particular curve, whatever I have shown here, this is the part the straight line is 73 00:10:13,910 --> 00:10:19,280 what is dependent on V S right in this particular case. This exponential the right hand side 74 00:10:19,280 --> 00:10:23,420 of it it is independent of V S, it happens to be so in this particular case. And the 75 00:10:23,420 --> 00:10:29,010 left hand side is dependent on V S, it is this line. So, let say this particular one 76 00:10:29,010 --> 00:10:34,890 which I have already drawn corresponds to V S equals 5.7 volts. If you change the value 77 00:10:34,890 --> 00:10:40,530 of V S, what you have to do, you have to redo this. So, let say V S changes to 6.7 volts, 78 00:10:40,530 --> 00:10:50,070 you will get a line like that and you will get a different solution. So let say V S is 79 00:10:50,070 --> 00:10:58,740 6.7 volts or alternatively if V S is 4.7 volts, you would get a solution like that and you 80 00:10:58,740 --> 00:11:07,050 will get yet another solution. So this corresponds to V S equals 4.7 volts. So, the point is 81 00:11:07,050 --> 00:11:14,020 that any time the input changes, you have to solve the nonlinear equation all over again. 82 00:11:14,020 --> 00:11:17,690 Now this is the big difference from linear circuits. 83 00:11:17,690 --> 00:11:27,510 So, if I go back to the linear circuit, I showed earlier again this is extremely simple 84 00:11:27,510 --> 00:11:35,670 and if you want to solve it, you can do it from scratch also right. So, in this case, 85 00:11:35,670 --> 00:11:42,620 let say you knew the solution for V S equals 5.7 volts then if V S changes to 6.7, you 86 00:11:42,620 --> 00:11:48,470 know that the new solution is simply proportional to the new value of V S, so linearity holds. 87 00:11:48,470 --> 00:11:55,760 Even when you have multiple sources, the solution will be linear combination of all the sources. 88 00:11:55,760 --> 00:12:03,010 So, if you find solutions, if you find the solution to the circuit for one particular 89 00:12:03,010 --> 00:12:09,320 value of V S, you can find it for any other value simply by scaling. Now in case of nonlinear 90 00:12:09,320 --> 00:12:13,300 circuits, first of all finding the solution is the very complicated; and top of it, if 91 00:12:13,300 --> 00:12:18,620 you change the value of V S, you have to go through the nonlinear calculations all over 92 00:12:18,620 --> 00:12:21,450 again. Now this case where the input is changing 93 00:12:21,450 --> 00:12:30,370 is a very relevant case for us, because we want to amplify some signals in our amplifier. 94 00:12:30,370 --> 00:12:36,220 And signal usually is a time varying signal that is when I am speaking into the microphone, 95 00:12:36,220 --> 00:12:40,350 it is giving a time varying voltage to the circuit. So we have to be able to analyze 96 00:12:40,350 --> 00:12:42,720 our circuit varying inputs. 97 00:12:42,720 --> 00:12:56,930 So, I will show the same example with let say two sources in series, 98 00:12:56,930 --> 00:13:03,860 so this could be let say 5.7 volt source, and this could be a sinusoidal source, this 99 00:13:03,860 --> 00:13:18,130 is what I call a signal. This could be let say some 100 milli volt sin omega t or maybe 100 00:13:18,130 --> 00:13:27,990 one volt sin omega t. Now if you want to find V D in this particular case, what will you 101 00:13:27,990 --> 00:13:36,710 have to do. So, the way we set it up, the circuit is exactly the same, so the voltage 102 00:13:36,710 --> 00:13:45,820 source that is applied to the circuit, it is now time varying. So this is the V S right, 103 00:13:45,820 --> 00:13:57,150 this is the value of V S, and V S in this case is 5.7 volts plus 1 volt sin omega t. 104 00:13:57,150 --> 00:14:12,220 And if I plot V S versus time, it will look like that. The maximum value will be 6.7 during 105 00:14:12,220 --> 00:14:19,470 positive peak of the sin wave and minimum value will be 4.7 volts. So, how would you 106 00:14:19,470 --> 00:14:25,200 find the solution or the response V D to an input like this, you would have to find it 107 00:14:25,200 --> 00:14:34,500 for 5.7, there is some value of V D that you get that is see from the previous graph. So 108 00:14:34,500 --> 00:14:41,630 this is the value for 5.7. You have to do it for 6.7 and 4.7 and so on, and also many 109 00:14:41,630 --> 00:14:49,740 points in between. So, essentially this is V S, you have to keep 110 00:14:49,740 --> 00:14:57,410 varying V S, and keep drawing these straight lines, find the solution and plot the solution 111 00:14:57,410 --> 00:15:02,950 V D from that one. So, there will be some solution here, something else for 5.7, something 112 00:15:02,950 --> 00:15:08,860 else for 4.7, some other solution for intermediate value and so on. It is extremely cumbersome. 113 00:15:08,860 --> 00:15:15,340 And you can feel that to get a reasonable plot, you have to take many points in this 114 00:15:15,340 --> 00:15:20,270 cycle. So, first of all, finding the solution for each point is very tedious and doing it 115 00:15:20,270 --> 00:15:29,810 for so many points this is just next to impossible. So, you can find 116 00:15:29,810 --> 00:15:35,920 V D like this in principle , we know that value of V D, when the input is 5.7 that is 117 00:15:35,920 --> 00:15:47,170 about point seven volts. So, the 0.72 volts, so that is what I said and when it is 6.7, 118 00:15:47,170 --> 00:15:55,500 you can calculated numerically, it turns out to be something slightly higher. And for 4.7, 119 00:15:55,500 --> 00:15:59,400 it will be something slightly lower. And you can calculate the intermediate points and 120 00:15:59,400 --> 00:16:05,200 find that the solution looks somewhat like that. 121 00:16:05,200 --> 00:16:13,620 Now, this while ay in principle, it is simply not practical. So, what we will do is to find 122 00:16:13,620 --> 00:16:20,100 an alternative way of analyzing nonlinear circuits which is approximate but at least 123 00:16:20,100 --> 00:16:25,340 something that we can carry out by hand and that we need to able to do in order to get 124 00:16:25,340 --> 00:16:27,800 any insight into our circuits.