WEBVTT 00:00:01.670 --> 00:00:05.280 >> The first of the op-amp configurations that we're going 00:00:05.280 --> 00:00:08.250 to consider is known as the non-inverting amplifier. 00:00:08.250 --> 00:00:12.270 It gets its name from the fact that the source voltage to 00:00:12.270 --> 00:00:16.560 be amplified is connected to the non-inverting terminal. 00:00:16.560 --> 00:00:21.315 Because of that, it turns out that the output voltage will be of the same sign, 00:00:21.315 --> 00:00:24.075 S-I-G-N, as the input voltage. 00:00:24.075 --> 00:00:26.495 So, for example, if V_s is a positive voltage, 00:00:26.495 --> 00:00:28.755 the output voltage will also be positive. 00:00:28.755 --> 00:00:31.095 On the other hand, if the input voltage is negative, 00:00:31.095 --> 00:00:33.390 then the output voltage will be negative also. 00:00:33.390 --> 00:00:35.515 The sign is not inverted. 00:00:35.515 --> 00:00:38.030 I have here two different schematics. 00:00:38.030 --> 00:00:41.900 I'd encourage you to stop the video for just a second and convince 00:00:41.900 --> 00:00:46.625 yourself that these two schematics are identical as far as function is concerned. 00:00:46.625 --> 00:00:49.160 The things that you're going to be looking at are that 00:00:49.160 --> 00:00:52.190 the V_s is connected to the non-inverting terminal. 00:00:52.190 --> 00:00:55.355 Of course, the non-inverting terminal is the one that's got the positive sign on it. 00:00:55.355 --> 00:00:57.530 So, if this one is connected to the non-inverting, 00:00:57.530 --> 00:01:00.440 over here, V_s is also connected to the non-inverting. 00:01:00.440 --> 00:01:05.425 So, this amplifier over here is upside down from this op-amp here. 00:01:05.425 --> 00:01:09.305 The other thing that you look at is where does the feedback go. 00:01:09.305 --> 00:01:10.700 We'll talk more about feedback later, 00:01:10.700 --> 00:01:15.065 but suffice for now to say that feedback is the process or the act of 00:01:15.065 --> 00:01:21.560 taking the voltage at the output and running it back to, 00:01:21.560 --> 00:01:25.835 in this case, the negative or the inverting terminal. 00:01:25.835 --> 00:01:27.950 Again, we'll talk more about feedback later, 00:01:27.950 --> 00:01:30.865 but this is referred to as negative feedback. 00:01:30.865 --> 00:01:33.350 You can remember that because the feedback loop comes back 00:01:33.350 --> 00:01:37.385 to the terminal of the negative sign which is the inverting sign. 00:01:37.385 --> 00:01:39.800 So, in both of these feedback circuits, 00:01:39.800 --> 00:01:42.820 we're going from V_out through R_1. 00:01:42.820 --> 00:01:45.470 The node beat between R_1 and R_2 is 00:01:45.470 --> 00:01:48.440 then tabbed and brought back to the inverting terminal. 00:01:48.440 --> 00:01:50.915 Notice, over here, we have the same situation, 00:01:50.915 --> 00:01:52.205 coming from the output, 00:01:52.205 --> 00:01:55.580 going through R_1, connecting to R_2, 00:01:55.580 --> 00:01:59.970 and the node where R_1 and R_2 are connected is connected 00:01:59.970 --> 00:02:04.940 then to the inverting terminal and then brought back the ground. 00:02:04.940 --> 00:02:11.390 Now, let's analyze these op-amps at both of these circuits and see what it is, 00:02:11.390 --> 00:02:13.910 why we might draw it this way under certain circumstances 00:02:13.910 --> 00:02:16.790 and why we might draw it this way under other circumstances. 00:02:16.790 --> 00:02:18.950 Let's start here with the one on the right. 00:02:18.950 --> 00:02:21.320 By drawing it in this configuration, 00:02:21.320 --> 00:02:27.830 it makes it obvious that the output voltage goes through a voltage divider circuit, 00:02:27.830 --> 00:02:34.260 and only a portion of the output voltage is fed back to the inverting terminal. 00:02:34.700 --> 00:02:36.720 In this case here, 00:02:36.720 --> 00:02:38.920 V_n then, 00:02:39.010 --> 00:02:41.630 using our voltage divider formula, 00:02:41.630 --> 00:02:45.250 V_n is equal to the voltage across R_2, 00:02:45.250 --> 00:02:53.430 which is V_out times R_2 over R_1 plus R_2. 00:02:53.430 --> 00:02:57.435 Now, in order to get V_out as a function of V, 00:02:57.435 --> 00:03:01.130 our input, we're going to reverse the roles 00:03:01.130 --> 00:03:04.730 here and multiply both sides by the inverse of this. 00:03:04.730 --> 00:03:06.590 In solving or solving for V_out, 00:03:06.590 --> 00:03:16.225 we get that V_out is equal to V_n times R_1 plus R_2 over R_2. 00:03:16.225 --> 00:03:18.325 Now, we don't want it as a function of V_n. 00:03:18.325 --> 00:03:21.800 We want to know what the output is as a function of the input voltage 00:03:21.800 --> 00:03:26.825 V_s. We now apply one of our ideal op-amp approximations. 00:03:26.825 --> 00:03:29.950 That was the one that referred to as the virtual short. 00:03:29.950 --> 00:03:35.000 That V_p and V_n are going to be so close to each 00:03:35.000 --> 00:03:40.835 other that the difference V_p minus V_n is zero, 00:03:40.835 --> 00:03:45.800 or what we can say then is that V_n is approximately equal to V_p. 00:03:45.800 --> 00:03:49.175 Now, what is V_p? 00:03:49.175 --> 00:03:51.545 Using another op-amp approximation, 00:03:51.545 --> 00:03:55.585 the current going into the input terminals is zero. 00:03:55.585 --> 00:03:58.645 Therefore, I_p is zero. 00:03:58.645 --> 00:04:00.710 There will be no voltage drop across 00:04:00.710 --> 00:04:03.275 V_s because the current going through it is zero. 00:04:03.275 --> 00:04:07.910 So, V_p is in fact just our source voltage V_s. 00:04:07.910 --> 00:04:15.060 So, V_p equals V_s. V_n equals V_p due to the virtual short. 00:04:15.060 --> 00:04:21.820 We have then that V_out again replacing V_n with V_s. 00:04:21.820 --> 00:04:31.225 V_out is equal to V_s times R_1 plus R_2 over R_2. 00:04:31.225 --> 00:04:33.960 Generally speaking, we'll take this and say, well, 00:04:33.960 --> 00:04:38.270 note that R_2 is a denominator that's common to both of those two terms. 00:04:38.270 --> 00:04:46.110 So, we can then have V_out is equal to V_s times R_2 over R_2, 00:04:46.110 --> 00:04:53.350 that's one, plus R_1 over R_2. 00:04:53.350 --> 00:04:57.570 We say that the gain, G, the closed loop gain, 00:04:57.570 --> 00:05:01.640 the gain that we get because of this feedback circuit is equal 00:05:01.640 --> 00:05:08.615 to one plus R_1 plus R_2. 00:05:08.615 --> 00:05:13.940 We're going to refer to that as the gain of the non-inverting amplifier. 00:05:13.940 --> 00:05:16.690 We'll see when we get to the inverting amplifier configuration that 00:05:16.690 --> 00:05:20.540 its gain is just slightly different than this. 00:05:20.540 --> 00:05:24.910 But let's just point out now that V_out is in fact going to be 00:05:24.910 --> 00:05:29.440 the same sign as V_s. R_1 and R_2 are both positive quantities. 00:05:29.440 --> 00:05:32.970 So, the ratio of a positive quantities is positive, 00:05:32.970 --> 00:05:34.670 plus one is a positive number, 00:05:34.670 --> 00:05:38.525 times whatever the sign is on V_s gives us a V_out, 00:05:38.525 --> 00:05:41.785 which is the same sign as V_s. Now, 00:05:41.785 --> 00:05:44.375 let's look at this circuit over here. 00:05:44.375 --> 00:05:50.190 To analyze this, we're going to use a technique that use a node analysis, 00:05:50.190 --> 00:05:56.655 which is an analysis technique that we'll frequently use on op-amp circuits, 00:05:56.655 --> 00:06:02.005 especially as the op-amp circuit gets to be a little bit more complex. 00:06:02.005 --> 00:06:04.450 So, here's the deal. 00:06:04.450 --> 00:06:13.770 This is now V_n. We're going to write a node equation at V_n. But first, 00:06:13.770 --> 00:06:18.630 we're going to note that this voltage here, V_p, 00:06:18.630 --> 00:06:21.640 is going to equal our source voltage. 00:06:21.640 --> 00:06:25.575 As we saw over here, the R_s had no influence on it. 00:06:25.575 --> 00:06:28.315 So, I've left that out of this circuit here. 00:06:28.315 --> 00:06:36.470 So, V_p equals V_s. V_n equals V_p because of the virtual short. 00:06:36.470 --> 00:06:40.200 So, V_n then is going to equal V_s. 00:06:40.200 --> 00:06:50.115 Now, let's write a node equation summing the currents leaving this node. 00:06:50.115 --> 00:06:57.570 The current leaving this node going through R_2 to ground is V_s minus 00:06:57.570 --> 00:07:04.690 zero divided by R_2 plus the current going from this node in this direction. 00:07:04.690 --> 00:07:07.730 It's going to be the voltage across R_1, which is V_s, 00:07:07.730 --> 00:07:16.270 minus V_out divided by R_1, 00:07:16.270 --> 00:07:23.405 and then plus the current going into or leaving this node and going into the op-amp. 00:07:23.405 --> 00:07:26.600 But our ideal op-amp approximation tells us 00:07:26.600 --> 00:07:29.300 that the current going into the input terminal is zero. 00:07:29.300 --> 00:07:31.955 So, we could write a plus zero there. 00:07:31.955 --> 00:07:37.855 But let's that off and simply say that the sum of those two currents must equal zero. 00:07:37.855 --> 00:07:40.215 So, what that's saying is that in fact, 00:07:40.215 --> 00:07:42.055 just as we saw over here, 00:07:42.055 --> 00:07:45.920 the current is going from the output through these two resistors back. 00:07:45.920 --> 00:07:52.520 It's going from the output through these two resistors to ground. 00:07:52.520 --> 00:07:58.355 These two resistors are in series because the current going in here is zero. 00:07:58.355 --> 00:08:03.395 Now, let's solve this equation for V_out in terms of these paths. 00:08:03.395 --> 00:08:11.910 We have V_s factoring out V_s times one over R_2 plus one over 00:08:11.910 --> 00:08:20.950 R_1 minus V_out over R_1 is equal to zero. 00:08:22.490 --> 00:08:26.520 That bring this term over to the other side, and solving for V_0, 00:08:26.520 --> 00:08:36.794 We get then that V_0 is equal to V_s times R_1 times 00:08:36.794 --> 00:08:44.810 one over R_2 plus one over R_1 or distributing that 00:08:44.810 --> 00:08:47.315 through the R_1 over the R_1 gives you the one, 00:08:47.315 --> 00:08:49.190 the R_1 over R_2 gives you the other term, 00:08:49.190 --> 00:08:50.975 and we get the V_out is equal to 00:08:50.975 --> 00:08:59.995 V_s times one plus R_1 over R_2. 00:08:59.995 --> 00:09:02.885 So, the gain terms are the same. 00:09:02.885 --> 00:09:05.195 These two circuits are equivalent 00:09:05.195 --> 00:09:09.870 and that gives you an idea of what the non-inverting amplifier does.