1 00:00:05,360 --> 00:00:07,720 PROFESSOR: Last time we talked about op-amps and the fact 2 00:00:07,720 --> 00:00:10,120 that they allowed us to abstract away certain 3 00:00:10,120 --> 00:00:13,070 components of a circuit and sample particular voltages 4 00:00:13,070 --> 00:00:15,820 from a circuit and then modify those voltages as we would 5 00:00:15,820 --> 00:00:18,720 like in an LTI fashion. 6 00:00:18,720 --> 00:00:22,030 Today I'd like to talk to you about some other interesting 7 00:00:22,030 --> 00:00:24,470 things that fall out of the fact that we're only dealing 8 00:00:24,470 --> 00:00:25,760 with LTI systems. 9 00:00:25,760 --> 00:00:29,840 In particular, Thevenin/Norton equivalence and superposition. 10 00:00:32,460 --> 00:00:35,700 Thevenin/Norton equivalence is the idea that if you have a 11 00:00:35,700 --> 00:00:39,100 very complex circuit, but it's still a LTI circuit, you can 12 00:00:39,100 --> 00:00:43,540 express it using a linear curve. 13 00:00:43,540 --> 00:00:46,060 And superposition is the idea that because you're also 14 00:00:46,060 --> 00:00:48,900 dealing with LTI components, if you're trying to solve a 15 00:00:48,900 --> 00:00:54,260 circuit, you can take the individual contributions of 16 00:00:54,260 --> 00:00:58,820 independent sources to that circuit and sum them in order 17 00:00:58,820 --> 00:01:01,845 to find out either the total current flowing through a 18 00:01:01,845 --> 00:01:04,170 particular component or the total voltage drop across a 19 00:01:04,170 --> 00:01:07,220 particular component. 20 00:01:07,220 --> 00:01:08,665 At this point, I'll walk you through both. 21 00:01:16,270 --> 00:01:18,970 Thevenin/Norton equivalence is an important concept in that 22 00:01:18,970 --> 00:01:20,880 you may have a very complicated circuit. 23 00:01:20,880 --> 00:01:22,700 And you don't really want to talk about the entire 24 00:01:22,700 --> 00:01:23,490 complicated circuit. 25 00:01:23,490 --> 00:01:28,220 You just want to sample the voltage drop or current in a 26 00:01:28,220 --> 00:01:29,470 very particular location. 27 00:01:32,250 --> 00:01:35,090 Because we're dealing with LTI systems we can actually 28 00:01:35,090 --> 00:01:40,115 express that particular sample as its relationship between I 29 00:01:40,115 --> 00:01:43,880 and V and possibly whatever resistive component is 30 00:01:43,880 --> 00:01:50,170 associated with the voltage drop across that sample. 31 00:01:50,170 --> 00:01:53,530 We can solve for this curve by looking at the location that 32 00:01:53,530 --> 00:01:56,800 were sampling and finding the open circuit voltage 33 00:01:56,800 --> 00:01:58,690 associated with that position. 34 00:01:58,690 --> 00:02:02,170 Or if we were to leave the two terminals from where we are 35 00:02:02,170 --> 00:02:06,250 sampling open, what is the voltage drop across that 36 00:02:06,250 --> 00:02:10,419 section of our circuit? 37 00:02:10,419 --> 00:02:13,540 That's the point right here, where the current flowing 38 00:02:13,540 --> 00:02:16,090 through the system is 0. 39 00:02:16,090 --> 00:02:19,550 And there's a given voltage associated with it. 40 00:02:19,550 --> 00:02:26,180 Likewise if we want to find the other intercept, we can 41 00:02:26,180 --> 00:02:29,430 close those two terminals by running a wire across them and 42 00:02:29,430 --> 00:02:32,900 then look at the current that flows across that wire. 43 00:02:32,900 --> 00:02:34,260 That's the close circuit current. 44 00:02:41,950 --> 00:02:44,620 The slope is going to tell us about our resistance, if any. 45 00:02:44,620 --> 00:02:46,960 If we're only dealing with a voltage source or a current 46 00:02:46,960 --> 00:02:48,990 source, then we're only going to be dealing 47 00:02:48,990 --> 00:02:50,780 with a straight line. 48 00:02:54,460 --> 00:02:57,300 Once we've solved for these values, we tend to express 49 00:02:57,300 --> 00:03:02,410 them either as a Thevenin equivalent circuit or a Norton 50 00:03:02,410 --> 00:03:03,680 equivalent circuit. 51 00:03:03,680 --> 00:03:07,240 And you can convert between the two. 52 00:03:07,240 --> 00:03:08,490 Let's walk through an example. 53 00:03:13,620 --> 00:03:14,870 Here I've got a simple circuit. 54 00:03:17,550 --> 00:03:19,750 And I would like to find the Thevenin equivalent of that 55 00:03:19,750 --> 00:03:23,750 circuit, given that I'm sampling from above this 56 00:03:23,750 --> 00:03:26,200 resistor and below this resistor. 57 00:03:28,790 --> 00:03:30,900 The first thing that I'm going to do is find the 58 00:03:30,900 --> 00:03:32,730 open circuit voltage. 59 00:03:32,730 --> 00:03:38,330 Or if I were sampling from this point to this point, what 60 00:03:38,330 --> 00:03:42,658 is the voltage drop across this section of the circuit? 61 00:03:42,658 --> 00:03:43,990 Well, I've got 36 Volts. 62 00:03:46,510 --> 00:03:48,530 There's 12 Ohms of resistance in the 63 00:03:48,530 --> 00:03:49,885 section that I'm sampling. 64 00:03:49,885 --> 00:03:52,060 And there's 6 Ohms of resistance outside the section 65 00:03:52,060 --> 00:03:53,380 that I'm sampling. 66 00:03:53,380 --> 00:03:57,200 So 2/3 of my total voltage drop across this voltage 67 00:03:57,200 --> 00:04:00,570 divider is going to be accounted for inside my 68 00:04:00,570 --> 00:04:03,910 Thevenin equivalent circuit. 69 00:04:03,910 --> 00:04:07,200 So my Thevenin equivalent voltage is 24 Volts. 70 00:04:12,290 --> 00:04:15,410 Now I want to go after the short circuit current. 71 00:04:15,410 --> 00:04:17,589 If I were to draw a wire from this terminal to this 72 00:04:17,589 --> 00:04:22,480 terminal, I'm curious what this value would be. 73 00:04:22,480 --> 00:04:26,230 As a consequence of connecting these two terminals, these 74 00:04:26,230 --> 00:04:28,700 resistors are going to be completely bypassed. 75 00:04:28,700 --> 00:04:32,880 We don't actually have to drop any voltage or put any current 76 00:04:32,880 --> 00:04:35,420 through these resistors because there's actually an 77 00:04:35,420 --> 00:04:39,500 infinitely easier way for the current to go. 78 00:04:39,500 --> 00:04:41,310 In this case, we're just going to follow Ohm's law. 79 00:04:44,800 --> 00:04:49,850 36 divided by 6 is 6 Amperes. 80 00:04:49,850 --> 00:04:52,380 But because of the directionality on this short 81 00:04:52,380 --> 00:04:56,160 circuit current, current in this system is going to flow 82 00:04:56,160 --> 00:04:57,580 in this direction. 83 00:04:57,580 --> 00:04:58,960 So are short circuit current is 84 00:04:58,960 --> 00:05:00,970 actually negative 6 Amperes. 85 00:05:10,130 --> 00:05:13,160 That just leaves Rth. 86 00:05:13,160 --> 00:05:15,460 Once again we're going to rely on Ohm's law. 87 00:05:20,760 --> 00:05:26,160 If we divide our voltage by our current-- 88 00:05:26,160 --> 00:05:27,670 you can't have a negative resistance. 89 00:05:27,670 --> 00:05:29,860 We're actually going to divide by the negative of this. 90 00:05:33,070 --> 00:05:34,320 We'll get out 4 Ohms. 91 00:05:41,230 --> 00:05:43,750 Let's do it again, this time with a slightly different 92 00:05:43,750 --> 00:05:45,810 sample point and find the Norton 93 00:05:45,810 --> 00:05:47,060 equivalent of our circuit. 94 00:05:51,200 --> 00:05:54,345 So if I'm only sampling across the 9 Ohm resistor-- 95 00:05:57,120 --> 00:05:59,970 I personally still solve for the Thevenin equivalent 96 00:05:59,970 --> 00:06:01,900 voltage first because I think it's easiest. 97 00:06:06,350 --> 00:06:09,780 I'll just make note of it over here. 98 00:06:09,780 --> 00:06:12,670 The voltage drop across this resistor is determined by the 99 00:06:12,670 --> 00:06:14,060 fact that this is a simple voltage divider. 100 00:06:17,080 --> 00:06:18,440 The other resistors in this circuit-- 101 00:06:18,440 --> 00:06:20,910 7, 2, 9 Ohms. 102 00:06:25,110 --> 00:06:27,200 So the voltage drop across this resistor is going to 103 00:06:27,200 --> 00:06:30,100 represent half of the total voltage in our circuit. 104 00:06:33,040 --> 00:06:34,290 That's 18 Volts. 105 00:06:41,150 --> 00:06:43,190 The short circuit current associated with connecting 106 00:06:43,190 --> 00:06:50,570 these two terminals is 36 Volts divided by the sum of 107 00:06:50,570 --> 00:06:52,160 these two resistors. 108 00:06:52,160 --> 00:06:54,300 Once again we're going to end up bypassing this resistor 109 00:06:54,300 --> 00:06:58,610 entirely because, think about it this way, if there was a 110 00:06:58,610 --> 00:07:04,850 resistor across here with 0 resistance, all the current 111 00:07:04,850 --> 00:07:08,480 would sink through it, or it would represent the location 112 00:07:08,480 --> 00:07:09,900 for which infinite current would flow. 113 00:07:19,840 --> 00:07:24,210 36 divided by 9 is 4. 114 00:07:24,210 --> 00:07:26,520 And because we're following in the opposite direction we're 115 00:07:26,520 --> 00:07:29,110 going to say this is negative 4 Amperes. 116 00:07:34,000 --> 00:07:41,560 In this orientation, the Norton equivalent amperage is 117 00:07:41,560 --> 00:07:43,220 actually the negative of this value. 118 00:07:43,220 --> 00:07:44,875 You sometimes see the arrow pointing downwards. 119 00:07:54,340 --> 00:07:57,050 This preserves our property of short circuit current. 120 00:07:57,050 --> 00:07:59,960 If this was short circuited, Isc would still 121 00:07:59,960 --> 00:08:03,090 be negative 4 Amperes. 122 00:08:03,090 --> 00:08:04,340 And our Vth-- 123 00:08:06,610 --> 00:08:10,155 if our Vth was the voltage drop across these two points-- 124 00:08:16,040 --> 00:08:19,550 ha, I haven't solved for Rth yet. 125 00:08:22,380 --> 00:08:23,470 There are a few ways to do this. 126 00:08:23,470 --> 00:08:26,870 But I'm going to use Vth and In. 127 00:08:48,640 --> 00:08:51,890 18 over 4, we can reduce it to 9 over 2. 128 00:09:00,980 --> 00:09:03,320 And if I were to solve back out for the voltage drop 129 00:09:03,320 --> 00:09:07,305 across this resistor just as a consequence of I Norton, I 130 00:09:07,305 --> 00:09:10,520 would get 18 Volts, which is also the Thevenin equivalent 131 00:09:10,520 --> 00:09:12,000 voltage for this circuit. 132 00:09:16,320 --> 00:09:19,450 That covers Thevenin/Norton equivalence. 133 00:09:19,450 --> 00:09:22,050 And we're going to talk about superposition really quickly. 134 00:09:27,050 --> 00:09:30,350 Superposition is yet another circuit-solving strategy. 135 00:09:30,350 --> 00:09:32,440 It falls out as a consequence of only working with LTI 136 00:09:32,440 --> 00:09:33,070 components. 137 00:09:33,070 --> 00:09:37,100 And it means that in order to solve for a given component 138 00:09:37,100 --> 00:09:40,880 current or component voltage, you can actually find it by 139 00:09:40,880 --> 00:09:43,790 taking the linear combination of every contribution as a 140 00:09:43,790 --> 00:09:46,290 consequence of independent sources. 141 00:09:46,290 --> 00:09:47,800 What do I mean when I say that? 142 00:09:47,800 --> 00:09:50,050 Well, if you have a circuit like this-- 143 00:09:50,050 --> 00:09:51,410 and you might recognize it from the 144 00:09:51,410 --> 00:09:52,930 NVCC lecture earlier-- 145 00:09:56,670 --> 00:10:01,170 you can actually express it as a linear combination of just 146 00:10:01,170 --> 00:10:05,150 the voltage source and just the current source. 147 00:10:05,150 --> 00:10:08,120 Note what when I remove the current source, I end up with 148 00:10:08,120 --> 00:10:11,650 no connection because if you have 0 current flowing through 149 00:10:11,650 --> 00:10:14,480 a system, it's the same as if the 150 00:10:14,480 --> 00:10:17,270 connection has been removed. 151 00:10:17,270 --> 00:10:20,760 Note that when I remove the voltage source, I no longer 152 00:10:20,760 --> 00:10:23,000 have a voltage drop across that connection, but I 153 00:10:23,000 --> 00:10:24,250 maintain the connection. 154 00:10:29,180 --> 00:10:32,330 If in this example I was just looking for I1. 155 00:10:32,330 --> 00:10:35,180 I1 would be the linear combination of the 156 00:10:35,180 --> 00:10:40,570 contribution of I1 in this particular circuit plus I1 in 157 00:10:40,570 --> 00:10:43,410 this particular circuit. 158 00:10:43,410 --> 00:10:44,920 So this is the expression that I'm looking for. 159 00:10:51,150 --> 00:10:55,620 In the first circuit, I'm just going to use V over IR -- 160 00:10:55,620 --> 00:10:56,870 or V equals IR -- 161 00:11:00,680 --> 00:11:03,060 and I'm going to have 3 Amperes in this direction. 162 00:11:18,860 --> 00:11:26,600 In this circuit I'm actually going to have to 163 00:11:26,600 --> 00:11:27,850 use a current divider. 164 00:11:29,910 --> 00:11:32,150 Identify that I have five parts total to distribute 165 00:11:32,150 --> 00:11:34,060 among these two branches. 166 00:11:34,060 --> 00:11:37,000 Identify that this side is going to get the inverse of 167 00:11:37,000 --> 00:11:40,370 this side's contribution to the total parts, or this side 168 00:11:40,370 --> 00:11:45,180 is going to get the proportion of parts equivalent to this 169 00:11:45,180 --> 00:11:46,430 side's contribution. 170 00:11:48,850 --> 00:11:50,630 And vice versa. 171 00:11:50,630 --> 00:11:53,800 That this side is the side that I'm interested in. 172 00:11:53,800 --> 00:11:57,150 And so 2/5 of the total current flowing through this 173 00:11:57,150 --> 00:11:59,650 current divider is going to end up in this branch. 174 00:12:03,060 --> 00:12:06,050 2/5 of 10 is 4. 175 00:12:06,050 --> 00:12:10,810 It's in the opposite direction of this current flow. 176 00:12:10,810 --> 00:12:17,550 So I'm actually going to contribute negative 4 Amperes 177 00:12:17,550 --> 00:12:18,800 from this subcircuit. 178 00:12:27,790 --> 00:12:31,440 This should match the results from the NVCC lecture. 179 00:12:31,440 --> 00:12:34,690 That's my super fast coverage of superposition and also 180 00:12:34,690 --> 00:12:36,550 Thevenin/Norton equivalence. 181 00:12:36,550 --> 00:12:38,370 Next time we're going to start a whole new module.