WEBVTT 00:00:01.962 --> 00:00:04.939 Welcome to an Introduction to Electrical & Computer Engineering. 00:00:04.939 --> 00:00:06.300 My name is Lee Brinton, 00:00:06.300 --> 00:00:11.150 I'm an electrical engineering instructor at Salt Lake Community College. 00:00:11.150 --> 00:00:14.730 In this video we'll be talking about ways to analyze circuits using the node 00:00:14.730 --> 00:00:16.219 voltage method. 00:00:16.219 --> 00:00:20.810 We're gonna start by making a, looking at the similarities between elevation and 00:00:20.810 --> 00:00:21.940 voltage. 00:00:21.940 --> 00:00:26.100 We'll then discuss the difference between branch voltages and node voltages. 00:00:26.100 --> 00:00:30.920 We'll introduce the concept of nodes and critical or extraordinary nodes. 00:00:30.920 --> 00:00:35.854 We'll then look at the actual process of analyzing circuits using node voltages, 00:00:35.854 --> 00:00:40.011 and we'll look at how that applies when they are dependent sources and 00:00:40.011 --> 00:00:42.210 supernodes present in the circuit. 00:00:43.690 --> 00:00:46.710 First of all, the similarities between elevation. 00:00:46.710 --> 00:00:49.420 If we wanna talk about the elevation of, say Mount Olympus, 00:00:49.420 --> 00:00:53.180 we've got to define first of all what we're measuring relative to. 00:00:53.180 --> 00:00:55.970 In other words, we establish our reference. 00:00:55.970 --> 00:01:00.410 Typically sea level is our reference at zero feet. 00:01:00.410 --> 00:01:03.660 Mount Olympus then is about 9,500 feet above sea level, 00:01:03.660 --> 00:01:08.030 and the Salt Lake Valley Floor is around 4,100 feet above sea level. 00:01:09.550 --> 00:01:12.630 On the other hand, if we're standing on the Valley Floor, and 00:01:12.630 --> 00:01:16.290 looking up to the mountains to the east, we'll see that the elevation, 00:01:16.290 --> 00:01:20.055 the amount that they rise, can be calculated by taking the higher elevation, 00:01:20.055 --> 00:01:27.194 9,500 feet less the lower elevation of 4,100 feet, and we've got a 5,400 00:01:27.194 --> 00:01:31.400 foot change in elevation going from the Valley Floor to the top of Mount Olympus. 00:01:32.550 --> 00:01:35.950 On the other hand, if instead of calling sea level our reference, 00:01:35.950 --> 00:01:44.250 we made the Valley Floor our reference and said then the elevation here = 0 feet, 00:01:44.250 --> 00:01:49.240 then Mount Olympus would be 5,400 feet, 00:01:49.240 --> 00:01:54.200 and sea level would be -4,100 feet. 00:01:54.200 --> 00:01:57.780 Elevations are all relative to a reference, and 00:01:57.780 --> 00:02:01.090 similarly, that's true with voltages. 00:02:02.360 --> 00:02:06.770 To understand that, let's make sure we understand the difference between a branch 00:02:06.770 --> 00:02:08.970 voltage, and a no voltage. 00:02:08.970 --> 00:02:13.650 A branch voltage is the voltage across a branch within a circuit. 00:02:13.650 --> 00:02:17.629 In this case here, we've got a 3V drop across this resistor. 00:02:18.790 --> 00:02:22.695 Over here, we've got a 10V drop across that resistor, similarly a 2V, 00:02:22.695 --> 00:02:24.840 and a 12V drop across those resistors. 00:02:26.010 --> 00:02:29.980 On the other hand, if we wanted to talk about the voltage, 00:02:29.980 --> 00:02:36.570 at a point in the circuit, we would need to specify what we were referring it to. 00:02:36.570 --> 00:02:43.540 So let's create, we'll call this node here 0V, 00:02:43.540 --> 00:02:49.170 and then as we come along this branch we go up from 0, up 15V, 00:02:49.170 --> 00:02:56.020 so at this point, the voltage there is 15V above 0, or above our reference. 00:02:56.020 --> 00:03:01.108 Now as we traverse this branch here, we drop 3V, 00:03:01.108 --> 00:03:08.030 getting to a voltage here of 15- 3 = 12V above our reference. 00:03:08.030 --> 00:03:10.715 Continuing on along here, 00:03:10.715 --> 00:03:15.968 we drop another 10V to 2V above our reference, and 00:03:15.968 --> 00:03:23.790 then continuing on down across these 2V to back to our zero reference. 00:03:23.790 --> 00:03:28.389 Thus the distinction a node voltage where the voltage add a node is a voltage at 00:03:28.389 --> 00:03:33.061 the node relative to some reference, whereas the voltage across the branch is 00:03:33.061 --> 00:03:36.498 just a drop across the single element within the branch. 00:03:39.737 --> 00:03:44.664 Up until now, as we've been analyzing circuits, we've identified branch 00:03:44.664 --> 00:03:49.020 currents and voltages, and worked with those as our variables. 00:03:49.020 --> 00:03:53.830 For example, we have a branch current here, call it i1, we have another branch 00:03:53.830 --> 00:03:59.580 current here, call it i2, another here, i3, i4, 00:04:01.330 --> 00:04:06.538 and perhaps referencing like that, i5, and of course we know that i5 in this 00:04:06.538 --> 00:04:11.860 case equals i0, but in order to analyse this circuit using branch currence, 00:04:11.860 --> 00:04:16.350 we would have five different variables. 00:04:16.350 --> 00:04:20.406 And when using those variables we could then write KDL and KCL equations, and 00:04:20.406 --> 00:04:24.540 solve for any branch voltage or branch current in that circuit we wanted to do. 00:04:27.049 --> 00:04:32.599 The node voltage method involves, rather than branch currents, 00:04:32.599 --> 00:04:36.079 it involves our defining node voltages. 00:04:37.490 --> 00:04:41.850 In order to do that, we need to specify or to make the distinction between nodes and 00:04:41.850 --> 00:04:43.470 extraordinary nodes. 00:04:43.470 --> 00:04:47.516 A node is a point where two or more branches are joined. 00:04:47.516 --> 00:04:52.155 Here we've got a node, there's another node here, there's another node there, and 00:04:52.155 --> 00:04:56.353 here's a node, here's a node, and then all the way along the bottom here is yet 00:04:56.353 --> 00:04:57.180 another node. 00:04:57.180 --> 00:05:00.972 So we have one, two, three, four. 00:05:00.972 --> 00:05:02.907 I said that was, I made a mistake there. 00:05:02.907 --> 00:05:06.030 This is all one node, five nodes. 00:05:06.030 --> 00:05:11.680 So one, two, three, four, five regular nodes, 00:05:11.680 --> 00:05:17.270 and now we have this node here where we have more than two, 00:05:17.270 --> 00:05:21.610 we have three or more branches coming to these. 00:05:21.610 --> 00:05:26.340 Those types of nodes are referred to as extraordinary nodes, or critical nodes. 00:05:26.340 --> 00:05:29.610 In other words, a node is a place where two or 00:05:29.610 --> 00:05:32.880 more branches come together, and extraordinary or 00:05:32.880 --> 00:05:37.110 critical node is a place where three or more branches come together. 00:05:37.110 --> 00:05:45.220 In this case we have one, two, three extraordinary or critical nodes. 00:05:47.510 --> 00:05:52.710 Our approach is going to be then to identify the critical nodes, 00:05:52.710 --> 00:05:56.438 choose one of them as our reference, and 00:05:56.438 --> 00:06:01.480 define variables at the other 00:06:01.480 --> 00:06:05.870 two nodes, and 00:06:05.870 --> 00:06:11.160 then with those voltages equation 00:06:11.160 --> 00:06:17.780 at each of those critical nodes, in terms of V1 and V2. 00:06:17.780 --> 00:06:23.070 For example, let's start by summing the currents leaving this node right here. 00:06:24.460 --> 00:06:29.340 In terms of V1 and this, our voltage source, 00:06:29.340 --> 00:06:33.730 we can first of all note that, the voltage at that node 00:06:33.730 --> 00:06:38.850 is V0 volts above our reference. 00:06:39.880 --> 00:06:41.680 Now, we can write an expression for 00:06:41.680 --> 00:06:46.710 the current leaving this node by taking the voltage of this side 00:06:46.710 --> 00:06:52.120 minus the voltage at this side of those resistors and dividing by the resistence. 00:06:52.120 --> 00:06:54.900 In other words, we're specifying the branch voltage across those two 00:06:54.900 --> 00:06:57.770 resistors in terms of our node voltages, or 00:06:57.770 --> 00:07:05.226 thus we would write (V1-V0)/(R2+R3) 00:07:05.226 --> 00:07:10.600 will be the current leaving our first node going to the left. 00:07:12.270 --> 00:07:15.487 Now let's do similarly for the other two branches, and 00:07:15.487 --> 00:07:19.799 sum those three currents together and acknowledge that the sum of the three 00:07:19.799 --> 00:07:24.460 currents leaving that node must equal zero, just cuz current law. 00:07:24.460 --> 00:07:29.689 So the current now leaving that node coming downs through our one would be, 00:07:29.689 --> 00:07:34.005 V1, the voltage at the top, minus the voltage at this side, 00:07:34.005 --> 00:07:37.260 which in this case is just 0, divided by R1. 00:07:37.260 --> 00:07:43.020 And, finally, the current leaving node 1 going to the ride would be, 00:07:43.020 --> 00:07:48.108 the voltage of the left hand side of that resistor would be V1, 00:07:48.108 --> 00:07:52.140 the voltage on the right hand side would be V2, so 00:07:52.140 --> 00:07:58.573 (V1-V2)/R4 represent the sum of the three currents leaving that node, 00:07:58.573 --> 00:08:02.428 and the sum of those three things must equal 0. 00:08:02.428 --> 00:08:05.420 Similarly, we write another KCl at node 2. 00:08:05.420 --> 00:08:10.144 So this is node 1, and then at node 2 we have, 00:08:10.144 --> 00:08:14.628 the current leaving node 2 going this way is 00:08:14.628 --> 00:08:19.596 going to be the voltage at the right-hand side, 00:08:19.596 --> 00:08:26.170 V2 minus the voltage of the left-hand side V1 divided by R4. 00:08:26.170 --> 00:08:30.809 Note right now, that the current leaving node 2 going to the left 00:08:32.140 --> 00:08:36.940 is equal, but of opposite sign to the current leaving node 1, and 00:08:36.940 --> 00:08:41.539 going to the right, and you'll notice those two terms are the same in each of 00:08:41.539 --> 00:08:43.740 these equations other than the different bias sign. 00:08:44.790 --> 00:08:49.526 This term in the first equation, (V1-V2)/R4, 00:08:49.526 --> 00:08:54.474 and that term in the second equation (V2-V1)/R4. 00:08:54.474 --> 00:09:01.320 All right continuing on, we now add the current leaving node 2 going down. 00:09:02.350 --> 00:09:07.373 that current will be (V2/R5) + the current 00:09:07.373 --> 00:09:11.783 leaving the node going in this direction, 00:09:11.783 --> 00:09:18.887 well actually the current is going in so we'll subtract -I0 = 0, 00:09:18.887 --> 00:09:23.188 the sum of those three currents equal 0. 00:09:23.188 --> 00:09:27.768 So here we have two equations with two unknowns, 00:09:27.768 --> 00:09:33.573 it becomes simply a matter of algebra at this point to solve for 00:09:33.573 --> 00:09:36.043 those two node voltages.