1 00:00:00,700 --> 00:00:05,629 So, what are we really doing when we approximate the curve based straight line, so let us see. 2 00:00:05,629 --> 00:00:13,559 So, in general terms, the equation for this circuit were V S minus V D by R equals f of 3 00:00:13,559 --> 00:00:22,350 V D, and for the particular case of the diode, this f of V D was I s exponential V D by V 4 00:00:22,350 --> 00:00:44,020 t minus 1. Now, let me say that the first solution or the operating point that I calculate 5 00:00:44,020 --> 00:00:54,579 is for V S equals V S 0. So, that means that this blue stuff here calculated when V S was 6 00:00:54,579 --> 00:00:57,990 equal to some value V S 0, it does not matter what it is, it is basically the first thing 7 00:00:57,990 --> 00:01:04,659 I calculated. So this is V S 0 and this is V S 0 by R, and this is the operating point 8 00:01:04,659 --> 00:01:15,970 solution or the original solution. Let me call those values as V D 0 and I D 0. So, 9 00:01:15,970 --> 00:01:29,479 V D 0 and I D 0 are the operating point of the diode and this has to be got by nonlinear 10 00:01:29,479 --> 00:01:43,950 analysis 11 00:01:43,950 --> 00:01:53,060 and this is of course usually numerical. And let me say that at the operating point 12 00:01:53,060 --> 00:02:03,740 the diode voltage is V D 0, I denoted like this; and the diode current is I D 0. We have 13 00:02:03,740 --> 00:02:12,890 taken an example earlier where if I s is ten to the minus 15 amperes, and V s is 5.7 volts 14 00:02:12,890 --> 00:02:27,410 and R is 5 kilo ohms, so V D 0 turns out to be about 0.716 volts, and I D 0 is approximately 15 00:02:27,410 --> 00:02:34,450 1 milli amp. So, this I got from numerical analysis and that is the operating point. 16 00:02:34,450 --> 00:02:42,680 Now, let say that we have a different value of V S, and I will represent the new value 17 00:02:42,680 --> 00:03:04,400 of V S as V S 0 plus some lower case v s so that means that new V S represented as old 18 00:03:04,400 --> 00:03:17,290 value plus an increment lower case v s. I can always do this, I mean my original value 19 00:03:17,290 --> 00:03:23,260 of V S is 5.7 volts, if V S changes to 6.7 volts, I will say that it is one volt increment 20 00:03:23,260 --> 00:03:41,969 over 5.7 volts. I can do this for any V S right. And similarly I represent every quantity 21 00:03:41,969 --> 00:03:48,810 in this circuit, that is the actual diode voltage as the original diode voltage, the 22 00:03:48,810 --> 00:03:58,560 operating point diode voltage plus an increment in the diode voltage and so on. 23 00:03:58,560 --> 00:04:12,680 So all quantities are represented as the original 24 00:04:12,680 --> 00:04:28,280 value which of course, it is called the operating point plus increments over the operating point. 25 00:04:28,280 --> 00:04:38,540 So now I can write my equations in terms of this definition. So I had V S minus V D by 26 00:04:38,540 --> 00:04:50,130 R equals I S exponential V D by V t minus 1. The operating point of course satisfies 27 00:04:50,130 --> 00:05:07,810 the equation, so V S 0 minus V D 0 by R equals I S exponential V D 0 by V t minus 1. And 28 00:05:07,810 --> 00:05:14,570 the new value also satisfies this that is V S 0 plus v s, I am writing this V S new 29 00:05:14,570 --> 00:05:27,230 value as V S 0 plus an increment minus V D 0 minus v d by R equals I s exponential V 30 00:05:27,230 --> 00:05:33,240 D 0 plus v d by V t minus 1. 31 00:05:33,240 --> 00:05:54,180 Now let me write it in general terms as well not just for the diode. So this would be in 32 00:05:54,180 --> 00:06:03,610 general some function of V D, and this will be function of V D 0 – the operating point. 33 00:06:03,610 --> 00:06:10,880 So this is the general equation. And in the third case, I represent the general case as 34 00:06:10,880 --> 00:06:25,290 the operating point plus an increment, so I have f of V D 0 plus v d. Now what I do 35 00:06:25,290 --> 00:06:39,510 is expand this in a Taylor series about the operating point, this is the significance 36 00:06:39,510 --> 00:06:44,340 of the operating point. I choose some operating point which is the original case, for which 37 00:06:44,340 --> 00:06:48,860 I calculate the nonlinear solution exactly. And above that operating point, I expand the 38 00:06:48,860 --> 00:06:55,860 new case in a Taylor series. And what do I get, I will get the function at the operating 39 00:06:55,860 --> 00:07:07,960 point plus the first derivative calculated at the operating point times this increment 40 00:07:07,960 --> 00:07:15,360 V D plus higher order terms, because I am going to neglect this, I am not including 41 00:07:15,360 --> 00:07:23,760 them. Now what is this mean, we have already seen 42 00:07:23,760 --> 00:07:40,750 this in the graphical solution, so let say this is the operating point V D 0. So the 43 00:07:40,750 --> 00:07:51,690 very first term here, f of V D 0 is nothing but this part, this is f of V D 0. And the 44 00:07:51,690 --> 00:08:00,370 second term means a linear dependents on V D or increment over V D 0, so that corresponds 45 00:08:00,370 --> 00:08:08,310 to this straight line that is the sum of these terms corresponds to this straight line. And 46 00:08:08,310 --> 00:08:12,050 then you can add higher order terms, the next one will be parabolic and the next one will 47 00:08:12,050 --> 00:08:15,910 be cubic and so on. And if you add all of that you will get the exact nonlinearity, 48 00:08:15,910 --> 00:08:21,550 but we will stop here and use the straight line. So whatever I said earlier if you are 49 00:08:21,550 --> 00:08:26,430 not move too far from the operating point, you can use the straight line approximation, 50 00:08:26,430 --> 00:08:29,710 and the straight line is the tangent to the curve at the operating point, this is what 51 00:08:29,710 --> 00:08:35,110 I meant. I expand the nonlinearity in a Taylor series above the operating point, and neglect 52 00:08:35,110 --> 00:08:44,060 all of the higher order terms that is second order and higher than second order terms. 53 00:08:44,060 --> 00:08:50,810 The Taylor series expansion goes like this right f of V D which is f of V D 0 plus the 54 00:08:50,810 --> 00:09:02,800 increment will be f of V D 0 plus f prime calculated V D 0 times V D plus the second 55 00:09:02,800 --> 00:09:20,700 derivative by two factorial times V D square, of course everything is calculated V D 0 and 56 00:09:20,700 --> 00:09:34,570 so on. Now, you can see that these terms the higher order terms reduce more rapidly with 57 00:09:34,570 --> 00:09:40,330 decreasing value of V D compared to the first one. So, you can always find a sufficiently 58 00:09:40,330 --> 00:09:47,750 small value of lower case v d – the increment where these terms are negligible, as long 59 00:09:47,750 --> 00:10:13,750 as you have a smooth continuous function. So, this approximation is always valid, the 60 00:10:13,750 --> 00:10:22,000 only thing is you have to restrict the range of V D over which it is valid, that is all. 61 00:10:22,000 --> 00:10:36,170 Now, what does it mean if I neglect these, I will be left with this f of V D 0 and the 62 00:10:36,170 --> 00:10:43,680 first order dependence on this lower case v d or the incremental v d. Then, what I do 63 00:10:43,680 --> 00:10:54,310 is I subtract the operating point from this case 64 00:10:54,310 --> 00:10:57,520 that is I simply take this equation and that equation and subtract. 65 00:10:57,520 --> 00:11:10,800 What do I get in easily verify that we will get V S minus 66 00:11:10,800 --> 00:11:28,050 V D by R to be equal to f prime calculated at the operating point times V D. Now what 67 00:11:28,050 --> 00:11:45,529 is the significance of this, this is linear in the increment V D. Now there is this constant 68 00:11:45,529 --> 00:11:52,100 that dependents on the operating point, but as far as the equation is concern that is 69 00:11:52,100 --> 00:11:59,470 just the constant; V D is the variable and this equation is linear in V D, so that means 70 00:11:59,470 --> 00:12:05,880 that it is much easier to solve than the nonlinear equations. So for a changed value of V S, 71 00:12:05,880 --> 00:12:11,130 we have to solve the nonlinear equation all over again. Now we do not do that, we find 72 00:12:11,130 --> 00:12:19,730 the approximate solution using a linear equation, now that is a great simplification. Also because 73 00:12:19,730 --> 00:12:24,270 there are lots of nice techniques for handling linear equations and we can use all of those 74 00:12:24,270 --> 00:12:51,130 things as far as the incremental quantities are concerned. 75 00:12:51,130 --> 00:12:58,260 So this is the great simplification; so in summary, we first calculate the exact solution 76 00:12:58,260 --> 00:13:02,520 that is the solution from the nonlinear equation for the operating point, and that we will 77 00:13:02,520 --> 00:13:08,010 do numerically or graphically later we will see how to do it by hand. Then for any other 78 00:13:08,010 --> 00:13:13,140 value of the input, there can be more than one input it does not matter. You express 79 00:13:13,140 --> 00:13:17,960 every quantity in the circuit both inputs and internal variables as the original the 80 00:13:17,960 --> 00:13:26,700 operating point cases plus increment. And then you find that wherever you have a nonlinearity 81 00:13:26,700 --> 00:13:30,830 you can expanded in a Taylor series about the operating point and neglect higher order 82 00:13:30,830 --> 00:13:35,130 terms that is second and higher order terms then you will be left with a linear equation 83 00:13:35,130 --> 00:13:39,890 relating the increments. So now you can use everything that you know about linear circuit 84 00:13:39,890 --> 00:13:45,370 analysis whatever you learned in basic electrical circuits or some equivalent course to analyze 85 00:13:45,370 --> 00:13:48,920 nonlinear circuits as well in a approximate way.