WEBVTT 00:00:00.000 --> 00:00:04.755 >> Now, let's do an example with our non-inverting operational amplifier. 00:00:04.755 --> 00:00:06.960 So, here comes an example. 00:00:06.960 --> 00:00:09.750 Let's start with our non-ideal model. 00:00:09.750 --> 00:00:12.120 This is the non-ideal model and let's connect it 00:00:12.120 --> 00:00:15.195 up in the non-inverting amplifier case that I showed you. 00:00:15.195 --> 00:00:18.120 So, we want to define our gain. 00:00:18.120 --> 00:00:20.340 This way we want to find the circuit gain of 00:00:20.340 --> 00:00:25.305 this amplifier circuit and the gain is defined as the output divided by the input. 00:00:25.305 --> 00:00:28.020 So, let's use some, 00:00:28.020 --> 00:00:29.340 the node-voltage method. 00:00:29.340 --> 00:00:32.100 Let's first consider this node A. 00:00:32.100 --> 00:00:35.505 So, let's consider our currents that are going out this way. 00:00:35.505 --> 00:00:38.520 We know sum of those must be equal to zero. 00:00:38.520 --> 00:00:40.950 So, what voltage do we have here? 00:00:40.950 --> 00:00:43.745 V out minus what voltages over here. 00:00:43.745 --> 00:00:51.560 That is A times Vp minus Vn and divide that by the resistor R0. 00:00:51.560 --> 00:00:54.545 Now, let's consider this value right here, 00:00:54.545 --> 00:00:59.530 V out minus this voltage is Vn. 00:00:59.530 --> 00:01:04.250 We're going to divide that by R1 and that is going to be equal to zero. 00:01:04.250 --> 00:01:08.165 Now, let's do our node at this point. 00:01:08.165 --> 00:01:18.244 We can say Vn minus V out divided by R1 plus Vn minus zero divided by R2, 00:01:18.244 --> 00:01:25.520 plus right here, Vn minus going all the way, 00:01:25.520 --> 00:01:28.440 keep going right there. 00:01:28.440 --> 00:01:34.315 Vn minus Vp divided by Ri and this all equals zero. 00:01:34.315 --> 00:01:36.005 Those are the two values. 00:01:36.005 --> 00:01:38.900 These are the two equations and we'd call them the A equation and 00:01:38.900 --> 00:01:42.320 the B equation that define the two nodes that we have. 00:01:42.320 --> 00:01:44.060 Let's also take a look at the circuit, 00:01:44.060 --> 00:01:45.680 see if there's anything else that we know. 00:01:45.680 --> 00:01:48.875 Because at the moment, we have three unknowns and only two equations. 00:01:48.875 --> 00:01:51.560 We don't know V out, Vn or Vp. 00:01:51.560 --> 00:01:52.925 But if you look right here, 00:01:52.925 --> 00:01:57.695 you can see that Vp is equal to Vs. Because of the way the circuit is hooked up. 00:01:57.695 --> 00:02:01.290 So, let's just substitute that back in for 00:02:01.290 --> 00:02:06.140 Vp in each of those cases right there to make these equations a little bit simple. 00:02:06.140 --> 00:02:07.970 Now we have two unknowns, 00:02:07.970 --> 00:02:12.870 Vn and V out and we have two equations to solve them. 00:02:13.150 --> 00:02:20.255 So, in order to write those as a matrix equation and apply Cramer's rule, 00:02:20.255 --> 00:02:22.310 I've given you a little example here. 00:02:22.310 --> 00:02:24.470 You might want to just stop the video for a minute and 00:02:24.470 --> 00:02:26.720 take a look at the [inaudible] this example. 00:02:26.720 --> 00:02:29.630 I took the two equations right here and I converted 00:02:29.630 --> 00:02:32.630 them into matrix form for V out and Vn. 00:02:32.630 --> 00:02:34.205 Then I applied Cramer's rule. 00:02:34.205 --> 00:02:37.100 You may have to look that up online if you don't remember it from math. 00:02:37.100 --> 00:02:41.120 Let's stop the video right now and do a little bit of practice on 00:02:41.120 --> 00:02:46.910 this particular problem and see that you can get the answer for the gain of this Op Amp. 00:02:47.630 --> 00:02:50.660 Okay, I'm hoping that you spent a little bit of time 00:02:50.660 --> 00:02:53.120 with that example and did the non-ideal case. 00:02:53.120 --> 00:02:59.180 That's the last time you're going to have to do the non-ideal amplifier in this class. 00:02:59.180 --> 00:03:04.790 Now, let's rewrite our node a equation V out minus A times V 00:03:04.790 --> 00:03:12.980 n minus V p divided by R out plus V out minus Vn divided by R1 equals zero. 00:03:12.980 --> 00:03:16.510 Here's the KCL node that I had at node B. 00:03:16.510 --> 00:03:20.855 Now, let's apply our ideal Op Amp equations. 00:03:20.855 --> 00:03:23.585 Remember we know that Vs is equal to Vp. 00:03:23.585 --> 00:03:27.305 But we also know because its ideal that Vp equals Vn. 00:03:27.305 --> 00:03:29.949 So, if Vn equals Vp, 00:03:29.949 --> 00:03:31.440 then this is zero. 00:03:31.440 --> 00:03:35.100 Vn is Vs, Vn is Vs, 00:03:35.100 --> 00:03:40.230 Vn is Vs, and Vn minus Vp is equal to zero. 00:03:40.230 --> 00:03:43.200 Great, our A equation and our B equation, 00:03:43.200 --> 00:03:44.945 we've applied that simplification. 00:03:44.945 --> 00:03:46.885 Do we need to apply anything for the currents? 00:03:46.885 --> 00:03:49.095 No, so we don't need to use that information. 00:03:49.095 --> 00:03:50.600 But keep your eye on this one. 00:03:50.600 --> 00:03:53.995 Remember that the output resistance is zero. We're going to need that. 00:03:53.995 --> 00:03:58.220 Right here this equation gives us the idea that we could solve for V out 00:03:58.220 --> 00:04:02.510 very easily because the only values we have are V out and Vs. 00:04:02.510 --> 00:04:05.270 But remember that this value is zero. 00:04:05.270 --> 00:04:07.220 So, it's like we're dividing by zero. 00:04:07.220 --> 00:04:09.650 This is not a good equation for us to use. 00:04:09.650 --> 00:04:11.930 Let's use the B equation instead. 00:04:11.930 --> 00:04:13.370 So, for the V equation, 00:04:13.370 --> 00:04:18.709 I want to solve for V out and I can say right here there's my V out. 00:04:18.709 --> 00:04:25.995 So, I can say that V out is equal to Vs times one divided by R2, 00:04:25.995 --> 00:04:31.415 plus one divided by R1 and then I have a V out over R1 on the other side. 00:04:31.415 --> 00:04:36.200 So, if I solve this, I can say that V out over Vs and happens to be 00:04:36.200 --> 00:04:43.760 my gain is going to multiply this thing out R1 plus R2 over R1 R2. 00:04:43.760 --> 00:04:46.340 Then I'm going to bring this R1 over to the other side, 00:04:46.340 --> 00:04:48.100 see this R1 came over here. 00:04:48.100 --> 00:04:52.160 So, now my R1 cancel out and my gain which is my output 00:04:52.160 --> 00:04:56.210 divided by my input is R1 plus R2 over R2. 00:04:56.210 --> 00:05:00.320 Awesome. If we actually calculate that for the numbers that we have here, 00:05:00.320 --> 00:05:01.520 the value is five. 00:05:01.520 --> 00:05:06.860 If we had calculated it using the non-ideal case, it was 4.99975. 00:05:06.860 --> 00:05:11.330 So, this is a very good approximation for an ideal Op Amp. 00:05:11.330 --> 00:05:15.240 This is how we are going to do our problems in the future. 00:05:15.740 --> 00:05:19.775 So remember, that the circuit gain depends on a circuit. 00:05:19.775 --> 00:05:21.665 Here's the gain we just derived it. 00:05:21.665 --> 00:05:23.915 Remember saturation is going to take over too. 00:05:23.915 --> 00:05:26.015 We can't make this gain just as large as we want. 00:05:26.015 --> 00:05:30.940 It's controlled by the power supplies that we put on this Op Amp. 00:05:30.940 --> 00:05:34.450 Again, back to our Wild Horse Corral.