1 00:00:00,500 --> 00:00:04,680 We have seen how cumbersome it is to analyze circuits with nonlinear elements. 2 00:00:04,680 --> 00:00:15,210 What we will do now is to look at approximate ways of solving nonlinear circuits. 3 00:00:15,210 --> 00:00:36,160 We have previously considered an example with a two terminal nonlinear element. 4 00:00:36,160 --> 00:00:51,059 And we can solve this numerically or graphically, but the real difficult 5 00:00:51,059 --> 00:00:56,350 is that for every change in the input, we have to redo this. 6 00:00:56,350 --> 00:01:01,609 This by itself is cumbersome and then for every value of input we have to redo this. 7 00:01:01,609 --> 00:01:05,740 And of course, when you have time varying signals in our circuit, the input will be 8 00:01:05,740 --> 00:01:09,960 changing; so for every point of the time varying signal, you have to redo the numerical or 9 00:01:09,960 --> 00:01:15,610 graphical calculation, and this is very difficult. 10 00:01:15,610 --> 00:01:20,159 At least to get some insight into the circuit and for hand analysis, we would like to avoid 11 00:01:20,159 --> 00:01:21,159 these. 12 00:01:21,159 --> 00:01:22,909 So let us see what we can do. 13 00:01:22,909 --> 00:01:37,140 I will repeat the graphical solution which we had got. 14 00:01:37,140 --> 00:01:45,030 Essentially, we are solving for this equation, the current in the resistor being equal to 15 00:01:45,030 --> 00:01:46,500 the current in the diode. 16 00:01:46,500 --> 00:01:51,850 The current in the resistor is V s minus V d by R and that is equal to the current in 17 00:01:51,850 --> 00:02:02,720 the diode, which is the diode saturation current times exponential V d by V t minus 1. 18 00:02:02,720 --> 00:02:09,250 So, what we do is we draw a graph for the right side and the left side and see where 19 00:02:09,250 --> 00:02:16,200 they meet. 20 00:02:16,200 --> 00:02:29,670 So, this is the plot of the right hand side, this is the plot of the left hand side, and 21 00:02:29,670 --> 00:02:33,580 they meet here, which is the solution. 22 00:02:33,580 --> 00:02:42,819 Now, as V s changes, we have to repeat this. 23 00:02:42,819 --> 00:02:52,760 So, for a larger value of V s, in this particular case the right hand side does not change, 24 00:02:52,760 --> 00:03:00,420 but the left hand side does, and we have a new solution, new point of intersection with 25 00:03:00,420 --> 00:03:04,069 the nonlinear curve. 26 00:03:04,069 --> 00:03:07,050 The simplification that we will make is the following. 27 00:03:07,050 --> 00:03:11,520 You see that the original solution is here, and the new one is there. 28 00:03:11,520 --> 00:03:17,120 We will assume that as V s changes, it will move to a new point on the nonlinear curve, 29 00:03:17,120 --> 00:03:22,950 but it will not move too far away, that is this point is close to this point. 30 00:03:22,950 --> 00:03:30,180 Now, we will see later what close means, the point is that if you do not move too far away 31 00:03:30,180 --> 00:03:38,230 on the nonlinear curve, then around this point, if you do not move too far away on nonlinear 32 00:03:38,230 --> 00:03:43,280 curve, we can approximate the nonlinear curve by straight line that is you take any curve, 33 00:03:43,280 --> 00:03:47,900 you zoom into very small part of it, it will look like a straight line, so that is the 34 00:03:47,900 --> 00:03:49,150 principle we use. 35 00:03:49,150 --> 00:03:56,400 So the idea here is that if we do not move too far away, let me call this as the original 36 00:03:56,400 --> 00:04:01,459 solution, this is really just the first solution that you calculate. 37 00:04:01,459 --> 00:04:06,270 And this has to be got from nonlinear numerical or graphical solutions, there is no shortcut 38 00:04:06,270 --> 00:04:07,270 to this. 39 00:04:07,270 --> 00:04:13,680 What we are trying to do is at least to save some trouble while calculating the solution 40 00:04:13,680 --> 00:04:16,200 for subsequent points when V s changes. 41 00:04:16,200 --> 00:04:23,030 So, for one value of V s, we calculated exactly and the other values we approximate this nonlinearity 42 00:04:23,030 --> 00:04:28,050 by a straight line, assuming that we do not go too far on the nonlinear curve. 43 00:04:28,050 --> 00:04:40,521 Now, what is the straight line that we can use, so it is the tangent of this curve at 44 00:04:40,521 --> 00:04:50,630 the original point, and the original solution is known as the operating point of the circuit. 45 00:04:50,630 --> 00:04:56,650 So, I first show this graphically, because it is very easy to understand, it can also 46 00:04:56,650 --> 00:04:59,660 be proved formally from the equations. 47 00:04:59,660 --> 00:05:05,970 So, we are approximating this curve, which is the blue curve - the nonlinear one by its 48 00:05:05,970 --> 00:05:08,840 tangent which is of course this straight line at the operating point. 49 00:05:08,840 --> 00:05:13,789 And operating point for now you can consider it as the very first value that you calculate. 50 00:05:13,789 --> 00:05:18,110 For some value of V s, you calculated the exact solution that is the operating point. 51 00:05:18,110 --> 00:05:25,270 So, now, how does this help, because you know this is the straight line. 52 00:05:25,270 --> 00:05:30,740 For all other points, we are computing intersection of two straight lines and this is very easy. 53 00:05:30,740 --> 00:05:34,340 In fact, we do not even have to calculate the intersection explicitly, I will show the 54 00:05:34,340 --> 00:05:44,311 method by which we will do that, because when V s changes to this corresponding to this 55 00:05:44,311 --> 00:05:50,900 red curve the actual solution is that one, but if you consider the intersection with 56 00:05:50,900 --> 00:05:56,720 this approximate straight line the approximation solution is there, and they are sort of close 57 00:05:56,720 --> 00:05:57,720 to each other. 58 00:05:57,720 --> 00:06:01,669 And if you consider a point that is closer to the original operating point, they will 59 00:06:01,669 --> 00:06:04,319 be even closer to each other. 60 00:06:04,319 --> 00:06:12,870 So for now we will state the criteria vaguely; if 61 00:06:12,870 --> 00:06:32,850 the change is small then approximating the nonlinear 62 00:06:32,850 --> 00:06:54,970 curve by its tangent yields good enough solutions. 63 00:06:54,970 --> 00:06:57,479 So this is what we will do. 64 00:06:57,479 --> 00:07:03,520 Now this is approximate, but it turns out this approximation is quite good and lot of 65 00:07:03,520 --> 00:07:05,889 practical context, so we will use this widely. 66 00:07:05,889 --> 00:07:11,229 But of course, you must understand the limitations of this approximation; you must not go too 67 00:07:11,229 --> 00:07:12,750 far from the operating point. 68 00:07:12,750 --> 00:07:23,120 And how far is too far very much depends on the context that you are working in. 69 00:07:23,120 --> 00:07:26,479 And it also depends on how much accuracy you want in the first place. 70 00:07:26,479 --> 00:07:31,639 If you want the solution to be very accurate then you cannot go too far from the operating 71 00:07:31,639 --> 00:07:37,990 point, but if crude solutions are ok in some cases; many cases you just want to estimates, 72 00:07:37,990 --> 00:07:41,460 so in those cases, you could go further from the operating point.