WEBVTT 00:00:00.000 --> 00:00:05.865 Hello, this is Doctor Cynthia, first at the University of Utah, and today we're going to talk about 00:00:05.865 --> 00:00:09.662 Thevenin and Norton equivalent circuits. Thevenin and Norton are used when 00:00:09.662 --> 00:00:15.393 we analyze systems to allow us to simplify a complex system into simpler block diagrams that we can 00:00:15.393 --> 00:00:19.649 analyze or design individually. The concept of input and output resistance 00:00:19.649 --> 00:00:24.200 becomes very important in this case and can or not lead to voltage loading. 00:00:24.200 --> 00:00:27.662 We always consider designing our circuit as simple feminine equivalent 00:00:27.662 --> 00:00:32.740 to a Norton equivalent shown here, and they can be converted back and forth between each other using 00:00:32.740 --> 00:00:38.075 source transformation. We'll also talk about maximum power transfer and the updated ECE toolbox 00:00:38.075 --> 00:00:42.460 using these two powerful methods. We actually do this all the time we if 00:00:42.460 --> 00:00:47.264 we want to design a simple lamp or a fan, we don't actually consider the entire 00:00:47.264 --> 00:00:52.570 Power Distribution shown in this box. We convert it instead to a feminine equivalent circuit. 00:00:52.570 --> 00:00:58.666 Here is the 110 Volt source in your house and an equivalent source resistance, say that's 10 ohms. 00:00:58.666 --> 00:01:03.942 What we're saying is that the voltage and the current seen at these two 00:01:03.942 --> 00:01:08.330 points are the same as the voltage and current that they're seen at these two points. 00:01:08.330 --> 00:01:12.048 So we've done an equivalent circuit. We can then analyze this circuit 00:01:12.048 --> 00:01:18.020 as a set of blocks, as shown here. Let's consider the total resistance and find the current, 00:01:18.020 --> 00:01:21.940 and let's see what voltage we're measuring across our fan in our lamp. 00:01:21.940 --> 00:01:26.700 We can see that these two are in parallel, so 100 in parallel with 100 is 50. 00:01:26.700 --> 00:01:32.700 Plus the 10 in series means that our total resistance is 60 ohms. If we consider the current here, 00:01:32.700 --> 00:01:38.620 that's 110 volts divided by the total resistance, or 1.83 lamps, 3 amps. 00:01:38.620 --> 00:01:43.270 Now let's consider what's the voltage across the fan in the lamp using a voltage divider. 00:01:43.270 --> 00:01:50.127 That's going to be the 110 volts divided by their fifty ohms divided by sorry times their fifty ohms 00:01:50.127 --> 00:01:55.486 divided by 10 + 50 or 60 ohms, which gives us 92 volts. Now wait a minute. 00:01:55.486 --> 00:01:59.934 That is significantly less than the voltage that we started with. That's called voltage loading. 00:01:59.934 --> 00:02:06.078 Where did the voltage go? It went across our S right here. What could we have done to reduce 00:02:06.078 --> 00:02:10.561 this voltage loading? We could have reduced our source resistance to a much smaller value 00:02:10.561 --> 00:02:15.173 so that we got a higher value here. Take a minute, stop the video and see if you 00:02:15.173 --> 00:02:18.350 can calculate what the voltage is across the fan in the lamp. 00:02:18.350 --> 00:02:24.790 Is it closer to 110 volts? Now let's talk about the input and output resistance. 00:02:24.790 --> 00:02:30.613 The input resistance is the resistance looking into the circuit, and the output resistance is the is 00:02:30.613 --> 00:02:33.705 the resistance that's looking into the output part of the circuit. 00:02:33.704 --> 00:02:39.337 So here is an output resistance and here is another input resistance. Let's number these. 00:02:39.338 --> 00:02:42.343 This is R out for the first block, R in and R out for 00:02:42.342 --> 00:02:45.350 the second block, and R in for the third block. 00:02:45.350 --> 00:02:48.950 Here's how you can decide if you need to consider voltage loading or not. 00:02:48.950 --> 00:02:54.668 If R out from our first block is much less than, not less than, 00:02:54.668 --> 00:03:01.115 or equal to much less than R in for our second block, then no voltage loading occurs and the 00:03:01.115 --> 00:03:04.754 circuits can be considered decoupled. So in this case, 00:03:04.754 --> 00:03:09.770 is 10 ohms much less than 100 ohms? No, In fact, probably not. 00:03:09.770 --> 00:03:16.650 How about one ohms? Yes. So in this case they would still be considered coupled. 00:03:16.650 --> 00:03:22.770 In this case they would be considered decoupled. Let's consider the next set. 00:03:22.770 --> 00:03:26.330 Let's say R out two. 00:03:26.330 --> 00:03:31.490 Is that much less than R in 2R in three? Absolutely not. 00:03:31.490 --> 00:03:38.334 In this case we have 100 ohms. Is that much less than the 100 ohms input to R3? No. 00:03:38.334 --> 00:03:45.230 So these circuits must always be considered coupled because they are not much less than. 00:03:45.230 --> 00:03:49.447 This tells us that we need to analyze these two together, and in fact we did because we 00:03:49.447 --> 00:03:53.310 considered them in parallel. But if we were using our one ohm source, 00:03:53.310 --> 00:03:58.056 if this was a simple one ohm source, we could at least consider the source 00:03:58.056 --> 00:04:04.174 block independent from the load blocks. OK, now let's go back to look at 00:04:04.174 --> 00:04:08.420 our Thevenin equivalent circuit. What we have done is considered two points, 00:04:08.420 --> 00:04:15.357 A and B in our circuit, and we have analyzed those as a very We've taken our actual circuit and 00:04:15.357 --> 00:04:19.860 converted it into a much simpler Febinine equivalent circuit. In order to do this, 00:04:19.860 --> 00:04:25.900 we need to find two things, the V Thevenin and the R Thevenin. Let's see how to do that. 00:04:25.900 --> 00:04:30.100 In order to find the V thevan and the first thing we do is what's called the open circuit method. 00:04:30.100 --> 00:04:35.204 We remove the loads and we open circuit the actual circuit and we measure, calculate or simulate the 00:04:35.204 --> 00:04:41.414 voltage between points A and B. That's called the open circuit voltage. So similarly, 00:04:41.414 --> 00:04:48.140 as soon as we measure that open circuit voltage, we can see that Vth is equal to Voc. 00:04:48.140 --> 00:04:54.005 Again, we can measure, calculate, or simulate this. There are three ways to 00:04:54.005 --> 00:04:58.398 find the Fevodin resistance, and I'm going to show you all three. The first one can only be used if 00:04:58.398 --> 00:05:01.910 there are no dependent sources, and it's probably the easiest of the methods. 00:05:01.910 --> 00:05:07.150 What you do is you deactivate all the independent sources, you convert a voltage to a short circuit, 00:05:07.150 --> 00:05:11.390 a current to an open circuit, and of course you can't have dependent sources. 00:05:11.390 --> 00:05:16.590 Then you simply look into points A and B and you see what the equivalent resistance is there. 00:05:16.590 --> 00:05:22.100 And that's equal to the Fevodin resistance. Let's do this. Here's an example. 00:05:22.100 --> 00:05:28.300 We want to do a feminine equivalent circuit of this more complicated circuit between points A and B. 00:05:28.300 --> 00:05:31.700 The first thing that we do is deactivate the voltage source by shorting it, 00:05:31.700 --> 00:05:34.754 and deactivate the current source by opening it 00:05:34.754 --> 00:05:38.084 up as we have here. Then we look to see what does the 00:05:38.084 --> 00:05:42.812 resistance look like when we look inside. Well, we can see that we have 250 00:05:42.812 --> 00:05:49.266 ohm resistors in parallel, so that gives us a 25 ohm resistance. That block of two would be 00:05:49.266 --> 00:05:55.230 in series with this 35 ohm. So combine the 25 and 35 and that's going to give you 60 ohms, 00:05:55.230 --> 00:06:01.430 which when you put it in parallel with this 30 ohm is going to be the Rth is equal to 20 ohms. 00:06:01.430 --> 00:06:07.390 We could have measured, simulated or calculated this. In this case we calculated it. 00:06:07.390 --> 00:06:12.440 Now here's the 2nd way to do this. Let's use the open circuit short circuit method. 00:06:12.440 --> 00:06:18.647 We have already opened the circuit in order to find Voc. We can and so that's our. That's the Voc. 00:06:18.647 --> 00:06:23.824 Then we can short circuit the load and measure the current. In that case, 00:06:23.824 --> 00:06:29.280 R Thevenin is equal to Voc divided by ISC. Again, we can measure, 00:06:29.280 --> 00:06:34.130 calculate or simulate this. Let's calculate it. Here's another example. 00:06:34.130 --> 00:06:39.384 Let's find the Thevenin equivalent between points A and B. Now, the first thing that we're going to 00:06:39.384 --> 00:06:45.627 do is measure the Voc across here. So let's do that using node voltage. 00:06:45.627 --> 00:06:48.643 In this case, we're going to have VC 00:06:48.643 --> 00:06:54.120 -24 / 6 plus VC -0 / 12 plus A7 amp outgoing. 00:06:54.120 --> 00:06:57.734 And then we don't need to consider this because any resistor that is connected to 00:06:57.734 --> 00:07:01.440 an open circuit does not have a current going through it and can be neglected. 00:07:01.440 --> 00:07:07.080 So we can use this to calculate that V thevenant is -12 volts. That shows up right here. 00:07:07.080 --> 00:07:13.110 And notice we simply turned the V thevenant upside down. OK, now let's find the R thevenin. 00:07:13.110 --> 00:07:18.270 We need the short circuit current, so let's take VC prime because this is a different circuit. 00:07:18.270 --> 00:07:22.540 VC prime -24 / 6 plus VC 00:07:22.540 --> 00:07:27.510 prime -0 / 12 + 7 amps. And this time the two ohms does show up. 00:07:27.510 --> 00:07:31.726 So it's VC prime -0 / 2 ohms that I'll 00:07:31.726 --> 00:07:36.910 write here says VC prime is -4 volts. Then we consider the I short circuit current. 00:07:36.910 --> 00:07:41.432 We know what VC prime is, so we can find ISC. That's -2 amps. 00:07:41.432 --> 00:07:44.649 In this case R Thevenin is V Thevenin 00:07:44.649 --> 00:07:50.310 which was -12 / I short circuit -2. That's six ohms. A common mistake. 00:07:50.310 --> 00:07:54.670 Please don't make this is do not take the VC prime and put it in here. 00:07:54.670 --> 00:08:00.270 It's supposed to be the V Thevenin divided by the I short circuit current. 00:08:00.270 --> 00:08:03.407 So that gives us V Thevenin is 6 ohms.