1 00:00:01,620 --> 00:00:05,340 In this video we're gonna demonstrate the technique for determining the Thevenin 2 00:00:05,340 --> 00:00:08,560 equivalent circuit by looking at a relatively simple circuit and 3 00:00:08,560 --> 00:00:12,440 going through the different steps you will be using to do this. 4 00:00:13,570 --> 00:00:16,270 The first thing we need to determine is the open circuit voltage. 5 00:00:16,270 --> 00:00:20,682 The Thevenin voltage or the voltage in our Thevenin model. 6 00:00:23,708 --> 00:00:28,222 Consists of a Thevenin voltage which we pointed out is just the open 7 00:00:28,222 --> 00:00:30,210 circuit voltage. 8 00:00:30,210 --> 00:00:33,170 So in this case we need to know what the voltage between a and 9 00:00:33,170 --> 00:00:36,740 b is with nothing connecting a and b. 10 00:00:38,090 --> 00:00:39,490 As you look at this, you can see then, 11 00:00:39,490 --> 00:00:44,110 that the open circuit voltage is the voltage across R2. 12 00:00:44,110 --> 00:00:48,320 Now, we can use any method we want to determine the voltage across R2. 13 00:00:48,320 --> 00:00:51,950 We could calculate the current running through these. 14 00:00:51,950 --> 00:00:54,740 Which, because this open circuit, 15 00:00:54,740 --> 00:01:00,414 we should point out there is no current- [SOUND] Going through that branch. 16 00:01:00,414 --> 00:01:05,080 That means that these two resistors are in series. 17 00:01:05,080 --> 00:01:07,430 And by definition the current is [INAUDIBLE] the same through both of them. 18 00:01:07,430 --> 00:01:12,460 So we could calculate the current through those two resistors and 19 00:01:12,460 --> 00:01:15,160 we multiply R2 by that current and 20 00:01:15,160 --> 00:01:18,180 that would give us the Open circuit volts through the volts across R2, 21 00:01:18,180 --> 00:01:23,810 we're gonna also use our voltage divider which was developed, we derive a voltage 22 00:01:23,810 --> 00:01:27,910 formula by taking advantage of the fact that those two cells are in series. 23 00:01:27,910 --> 00:01:28,710 So let's just do that. 24 00:01:28,710 --> 00:01:34,264 V open circuit, is equal to then, V sub s, times. 25 00:01:34,264 --> 00:01:39,159 R2 over R1 + R2. 26 00:01:39,159 --> 00:01:42,741 For this circuit, that is our V Thevenin. 27 00:01:45,714 --> 00:01:48,717 So to get the open circuit voltage using any of the voltage, 28 00:01:48,717 --> 00:01:52,622 any of the certain analysis techniques that you have to determine the voltage 29 00:01:52,622 --> 00:01:54,080 across the open terminals. 30 00:01:55,550 --> 00:01:58,390 Next, we need to determine the thevenin resistance. 31 00:02:00,521 --> 00:02:05,650 And at that point, we will have our thevenin equivalent circuit. 32 00:02:05,650 --> 00:02:08,650 Now we're gonna see there are at least three different methods for 33 00:02:08,650 --> 00:02:11,650 determining the thevenin resistance. 34 00:02:11,650 --> 00:02:16,195 And we're going to Demonstrate each of those models on the simple circuit. 35 00:02:18,008 --> 00:02:22,532 The first method takes advantage of the definition of R Thevenin, 36 00:02:22,532 --> 00:02:27,541 you'll recall from our previous video we said that R Thevenin was defined 37 00:02:27,541 --> 00:02:32,330 as the ratio of the open circuit voltage to the short circuit current. 38 00:02:33,410 --> 00:02:36,780 Now in the previous slide we determine that the open circuit voltage 39 00:02:38,360 --> 00:02:42,100 which is the voltage across our to in this case which was vs so 40 00:02:42,100 --> 00:02:47,130 times R two over R one plus R two.Now we need 41 00:02:47,130 --> 00:02:52,480 to determine what the short circuit current is by short circuit currently mean 42 00:02:52,480 --> 00:02:57,330 if we short The wire between the two terminals A and 43 00:02:57,330 --> 00:03:03,180 B and then determine what that current is that flows through that short circuit. 44 00:03:03,180 --> 00:03:05,310 That's the short circuit current. 45 00:03:05,310 --> 00:03:09,640 Well, in this case the short circuit pulls the voltage here to zero. 46 00:03:09,640 --> 00:03:11,660 So, there would be no current going through R2. 47 00:03:13,790 --> 00:03:18,390 And the short circuit current then is simply the current that would be flowing 48 00:03:18,390 --> 00:03:24,760 through R 1 or I short circuit is equal to V sub S divided by R 1. 49 00:03:24,760 --> 00:03:29,690 So you've been using this first method method 1 we say 50 00:03:29,690 --> 00:03:33,440 then that open the R thevenin is equal to the ratio of the open circuit voltage 51 00:03:33,440 --> 00:03:38,490 short circuit current, we form that And say that R 7 is equal to V 52 00:03:38,490 --> 00:03:43,740 open circuit divided by I short circuit which is equal to V sub 53 00:03:43,740 --> 00:03:48,020 S times R 2 over R 1 plus R 2. 54 00:03:48,020 --> 00:03:51,550 That's our open circuit voltage 55 00:03:51,550 --> 00:03:55,248 divided by the short circuit current which is V sub S. 56 00:03:55,248 --> 00:03:58,650 Over R one. 57 00:03:58,650 --> 00:04:00,550 Well, the v of s is cancelled. 58 00:04:00,550 --> 00:04:04,560 We take this one over our one in the denominator, invert and multiply, and 59 00:04:04,560 --> 00:04:10,784 we get then that our Is equal 60 00:04:10,784 --> 00:04:17,420 to R 1 R 2 over R 1 + R 2 At least it works 61 00:04:17,420 --> 00:04:21,750 if you've got an independent source in the circuit that you're attempting to model. 62 00:04:23,110 --> 00:04:25,850 Method one then is determine the open circuit voltage and 63 00:04:25,850 --> 00:04:31,080 the short circuit current using any circuit analysis method you'd like. 64 00:04:31,080 --> 00:04:33,850 And then forming the ratio of the open circuit voltage to the short 65 00:04:33,850 --> 00:04:34,680 circuit current. 66 00:04:36,690 --> 00:04:41,250 The second method works when all you have or 67 00:04:41,250 --> 00:04:45,680 the only types of sources you have are independent sources. 68 00:04:45,680 --> 00:04:50,357 It works only with independent 69 00:04:51,762 --> 00:04:58,950 Sources and in this method, or to apply this method, 70 00:04:58,950 --> 00:05:04,720 we deactivate any of the sources present. 71 00:05:04,720 --> 00:05:06,510 And by deactivate, 72 00:05:06,510 --> 00:05:09,730 we mean the same thing that we did back when we were talking about super position. 73 00:05:09,730 --> 00:05:13,930 When you deactivate a voltage source, you turn the voltage source to zero. 74 00:05:15,320 --> 00:05:17,780 Or in this case, or 75 00:05:17,780 --> 00:05:20,660 a voltage source turned to 0 is effectively a short circuit. 76 00:05:21,870 --> 00:05:27,010 So having deactivated the independent sources present, we then calculate 77 00:05:28,170 --> 00:05:33,050 the resistance seen looking back into the circuit. 78 00:05:34,070 --> 00:05:38,450 Well, looking back into it with this point pulled to ground by deactivating 79 00:05:38,450 --> 00:05:44,270 the voltage source, we see then that R1 and R2 are in parallel with each other. 80 00:05:44,270 --> 00:05:48,845 And the resistance that we see going back into the ab terminals is 81 00:05:48,845 --> 00:05:54,300 R1R2 over R1 + R2. 82 00:05:54,300 --> 00:05:57,420 Which is the same result we got on the previews. 83 00:05:57,420 --> 00:05:58,500 Using the previews method. 84 00:06:00,990 --> 00:06:08,470 The third method, involves applying a test voltage at the terminals, and 85 00:06:08,470 --> 00:06:13,210 pushing the current back into the circuit, pushing the current back into the circuit. 86 00:06:13,210 --> 00:06:15,120 After deactivating the sources. 87 00:06:16,780 --> 00:06:20,510 This is somewhat like going back to our power training analogy, or 88 00:06:20,510 --> 00:06:22,550 an automobile analogy. 89 00:06:22,550 --> 00:06:26,410 It'd be like sort of turning off the engine, and 90 00:06:26,410 --> 00:06:31,140 then blowing back into the exhaust pipe, and measuring 91 00:06:31,140 --> 00:06:36,680 the resistance to the movement of air going back into the Into the exhaust pipe. 92 00:06:36,680 --> 00:06:42,640 So, Method 3, deactivate the independent sources. 93 00:06:43,680 --> 00:06:47,660 If you have dependent sources present, don't deactivate them. 94 00:06:47,660 --> 00:06:52,480 Deactivate just the independent sources, and then 95 00:06:54,000 --> 00:06:59,540 apply a test voltage We'll call it V Test, VT for test. 96 00:06:59,540 --> 00:07:04,170 That's not the Thevenin volt, it's just V Test which will 97 00:07:04,170 --> 00:07:10,100 force a test current to flow back into the circuit. 98 00:07:10,100 --> 00:07:13,470 Again, after we have deactivated the source. 99 00:07:17,280 --> 00:07:22,000 The resistance that we feel going back into it is simply the ratio 100 00:07:22,000 --> 00:07:26,360 of the test voltage divided by the test current. 101 00:07:28,330 --> 00:07:33,140 So our approach now requires us to 102 00:07:33,140 --> 00:07:35,740 come up with an equation or more than one equation. 103 00:07:35,740 --> 00:07:40,570 It allows us then to by just algebraic manipulation 104 00:07:40,570 --> 00:07:43,880 form the ratio of Vtest over Itest. 105 00:07:43,880 --> 00:07:46,120 Let's just do it. 106 00:07:46,120 --> 00:07:48,060 It's easier to show that it is to tell about it. 107 00:07:49,220 --> 00:07:50,450 When we apply this, 108 00:07:50,450 --> 00:07:55,750 after deactivating the voltage source, we see that we have only one node here. 109 00:07:55,750 --> 00:08:01,320 And that node is in fact, at least in this circuit, The voltage, 110 00:08:01,320 --> 00:08:03,710 our test voltage being Test. 111 00:08:03,710 --> 00:08:07,920 So, let's write a node equation here at this node in terms of the node voltage 112 00:08:07,920 --> 00:08:09,480 V Test. 113 00:08:09,480 --> 00:08:14,250 The current leaving this node going through R1 is going to be V Test divided 114 00:08:14,250 --> 00:08:19,180 by R1, is the current leaving going this way. 115 00:08:19,180 --> 00:08:23,740 The current coming down here through R2 Is gonna be added to it 116 00:08:23,740 --> 00:08:26,350 plus V test divided by R2. 117 00:08:26,350 --> 00:08:31,140 Now you'll notice that I test is directed into the nodes, so 118 00:08:31,140 --> 00:08:35,840 that would be minus I test equals 0. 119 00:08:35,840 --> 00:08:42,460 Now let's factor out the common V test here Times 1 over R1 120 00:08:42,460 --> 00:08:48,070 plus 1 over R2 and take I test to the other side as a positive I test. 121 00:08:49,900 --> 00:08:54,800 Getting a common denominator we have then V test time 122 00:08:54,800 --> 00:08:58,690 R1 plus R2 over R1 times R2. 123 00:09:00,640 --> 00:09:03,070 Is equal to I test. 124 00:09:04,180 --> 00:09:08,600 Now that we've got a factor of the test on this side, 125 00:09:08,600 --> 00:09:13,870 I test on that side divide both sides by I test, divide both sides by this term 126 00:09:13,870 --> 00:09:19,269 here and we get that then v test over I test 127 00:09:21,330 --> 00:09:26,041 Which is our R thevenin is = R1, 128 00:09:26,041 --> 00:09:32,029 R2 over R1 + R2. 129 00:09:33,815 --> 00:09:37,405 Which is the same resistance we determined through the other two methods. 130 00:09:37,405 --> 00:09:41,975 So, again, this approach drives home the idea that what this Thevenin resistance 131 00:09:41,975 --> 00:09:46,255 represents is the restriction or the resistance seen 132 00:09:46,255 --> 00:09:50,635 looking back into the terminals of our circuit. 133 00:09:52,990 --> 00:09:57,590 Let's review then, determining we find the open circuit voltage 134 00:09:57,590 --> 00:10:00,790 using any method of circuit analysis we want 135 00:10:01,930 --> 00:10:06,470 we then use one of three methods to determine our thevenin. 136 00:10:06,470 --> 00:10:13,700 The first method works When you have at least one independent source present. 137 00:10:15,070 --> 00:10:20,100 Method one then is too short between the output terminals and 138 00:10:20,100 --> 00:10:21,640 determine the short circuit current and 139 00:10:23,530 --> 00:10:28,290 then form the ratio V open circuit divided by high short circuit. 140 00:10:28,290 --> 00:10:29,230 That's method one. 141 00:10:31,220 --> 00:10:35,970 Method two works only if you have independent sources 142 00:10:37,460 --> 00:10:43,060 and with only independent sources present, And deactivate the independent sources. 143 00:10:43,060 --> 00:10:45,740 In this case, we had a voltage source that we shorted out. 144 00:10:45,740 --> 00:10:49,780 Had it been a current source, we would have opened circuited that current source, 145 00:10:49,780 --> 00:10:51,930 and then determined the resistance, 146 00:10:51,930 --> 00:10:56,190 the equivalent resistance seen looking back in from a to b. 147 00:10:58,410 --> 00:11:00,900 Method two works only if independent sources are present. 148 00:11:02,540 --> 00:11:10,710 Method three works when you have both independent and dependent sources. 149 00:11:12,210 --> 00:11:17,990 If you have dependent sources, if you have independent sources you deactivate Any 150 00:11:17,990 --> 00:11:22,480 of the independent sources by replacing voltage sources with short circuits, and 151 00:11:22,480 --> 00:11:25,190 current sources with open circuits, and 152 00:11:25,190 --> 00:11:28,430 then apply a test voltage to the terminals. 153 00:11:29,440 --> 00:11:32,420 And then using algebraic manipulation techniques, 154 00:11:32,420 --> 00:11:36,790 derive the ratio of V test over I test. 155 00:11:38,430 --> 00:11:40,230 Three different options. 156 00:11:40,230 --> 00:11:42,240 Some options work under all circumstances, 157 00:11:42,240 --> 00:11:45,520 other options work only at given certain circumstances. 158 00:11:45,520 --> 00:11:49,890 And in the next two or three videos, we'll go through and give examples demonstrating 159 00:11:49,890 --> 00:11:53,390 each of these three techniques in somewhat more complicated circuits.