0:00:00.800,0:00:04.110 Okay, now we're going to work on our first op-amp circuit. 0:00:04.110,0:00:08.610 here's what the circuit is going to look like watch where I put the plus sign is 0:00:08.610,0:00:22.890 on the top on this one and we're going to have a voltage source over here this 0:00:22.890,0:00:31.529 will be plus or minus Vin that's our input signal and over on the output will 0:00:31.529,0:00:40.800 have Vout and it's hooked up this way there's a resistor another resistor to 0:00:40.800,0:00:48.300 ground and this goes back to the inverting input and we're going to look 0:00:48.300,0:00:52.170 at this circuit and see what it does now we know that connected up here the 0:00:52.170,0:00:58.760 power supplies hooked up to these points here and the ground symbol is zero volts 0:00:58.760,0:01:02.940 and we want to analyze this circuit and what do we know about this 0:01:02.940,0:01:10.200 we know that Vout equals some gain. We'll write the gain right there, a big big 0:01:10.200,0:01:18.900 number times v-, sorry v+ minus v-. And let's label that v+ is this 0:01:18.900,0:01:25.170 point right here and v- is this point right here. And we also know that 0:01:25.170,0:01:34.770 the currents, let's call them i+ and i- equals zero and that's the 0:01:34.770,0:01:39.299 currents going in here this is i- here and that's i+. And we know those 0:01:39.299,0:01:47.579 are both zero. So now what I want to do is I want to describe our amplifier with 0:01:47.579,0:01:52.079 some sort of a circuit model with these properties, alright? And the circuit model 0:01:52.079,0:01:59.820 for an amplifier looks like this: We have v- here v+ here and over on 0:01:59.820,0:02:04.290 this side we haven't here's a new symbol that you haven't seen before it's 0:02:04.290,0:02:09.649 usually drawn as a diamond shape and this is a voltage source 0:02:10.649,0:02:18.950 but is this it's a special kind of voltage source it's called a voltage 0:02:21.920,0:02:30.170 dependent voltage source 0:02:32.849,0:02:36.959 and it's the same as a regular ideal voltage source except for one thing, it 0:02:36.959,0:02:45.920 says that the V in this case Vout equals gain times v+ minus v- 0:02:45.920,0:02:51.689 so the voltage here depends on the voltage somewhere else and that's what 0:02:51.689,0:02:54.030 makes it a voltage dependent that's what that means 0:02:54.030,0:02:59.549 so we've just taken our gain expression here and drawn a circuit diagram that 0:02:59.549,0:03:06.150 represents our voltage expression for our circuit now specifically over here 0:03:06.150,0:03:11.549 we draw on an open circuit on v+ + v - so we know that those currents are zero 0:03:11.549,0:03:18.449 so this model this circuit sketch represents our two properties of our of 0:03:18.449,0:03:25.349 our op-amp. So I'm going to take a second here and I'm going to draw the rest 0:03:25.349,0:03:33.510 of our circuit surrounding this model but I need a little bit more space. 0:03:33.510,0:03:39.290 So let's put in the rest of our circuit here we had our voltage source 0:03:40.069,0:03:50.069 connected to v+ and that's Vin, and over here we had Vout let's check V out 0:03:50.069,0:04:01.979 was connected to two resistors and the bottom is connected to ground and this 0:04:01.979,0:04:10.229 was connected there. So what our goal is right now we want to find Vout as a 0:04:10.229,0:04:19.139 function of Vin. That's what we're shooting for so let's see if we can do 0:04:19.139,0:04:24.630 that. Let's give our resistor some names, let's call this R1 and R2, our favorite 0:04:24.630,0:04:30.690 names always. And now everything is labeled. Oh, and we can label this point 0:04:30.690,0:04:36.740 here and this point we can call v-, v- 0:04:38.520,0:04:44.830 So that's our two unknowns. Our unknowns are Vo (Vout), and V-. So let's 0:04:44.830,0:04:49.240 see if we can find them. So what I'm going to do is just start writing some 0:04:49.240,0:04:54.520 expressions for things that I know are true. For example, I know that I know that 0:04:54.520,0:05:06.460 Vout equals A times v+ - v-. Alright, that's what 0:05:06.460,0:05:11.380 this op-amp is telling us is true. Now what else do I know? Let's look at this 0:05:11.380,0:05:16.390 resistor chain here. This resistor chain actually looks a lot like a voltage 0:05:16.390,0:05:20.320 divider. And it's actually a very good voltage divider. Remember we said this 0:05:20.320,0:05:26.850 current here, what is this current here? It's zero. I can use the voltage divider 0:05:26.850,0:05:31.600 expression that I know. In that case I know that v-, this is the voltage 0:05:31.600,0:05:42.960 divider equation, equals Vout times what? times the bottom resistor. Remember this? 0:05:42.960,0:05:52.810 R2 over R1 plus R2. So the voltage divider expression says that when you 0:05:52.810,0:05:56.860 have a stack of resistors like this with a voltage on the top and ground on the 0:05:56.860,0:06:06.190 bottom, this is the expression for the voltage at the mid point. Okay, so what 0:06:06.190,0:06:11.140 I'm going to do next is I'm going to take this expression and stuff it right 0:06:11.140,0:06:19.840 in there. Let's do that see if we get enough room okay let's go over here now 0:06:19.840,0:06:36.430 I can say that Vout equals a times v+ minus Vout times R2 over R1 plus 0:06:36.430,0:06:38.760 R2. 0:06:39.950,0:06:45.300 Right, so far so good. Let's keep going, let's keep working on this Vo 0:06:45.300,0:07:00.500 equals a times v+ minus A Vo R2 over R1 plus R2. 0:07:01.430,0:07:06.660 All right, so now I'm going to gather all the V naught terms over on the left hand 0:07:06.660,0:07:21.300 side. Let's try that so that gives me V naught plus A V naught times R2 over R1 0:07:21.300,0:07:30.900 plus R2. And that equals A times v+, and actually I can change that now, 0:07:30.900,0:07:43.730 v+ is what? v+ is Vin. Okay, let's keep going. I can factor out the V naught. 0:07:43.730,0:08:01.470 V naught is 1 plus A R2 over R1 plus R2, and that equals A Vin. Alright, so 0:08:01.470,0:08:06.990 we're getting close and our original goal, we want to find Vout in terms of 0:08:06.990,0:08:12.120 Vin. So I'm going to take this whole expression here and divide it over to 0:08:12.120,0:08:16.050 the other side, so then I have just V naught on this side and Vin on the 0:08:16.050,0:08:27.800 other side. Let's make some more room. I can do that. I can say V naught equals A 0:08:28.250,0:08:43.250 Vin divided by this big ol' expression 1 plus A R2 over R1 plus R2. 0:08:43.250,0:08:50.790 Alright, so that's our answer. That's the answer, that's Vout equals some 0:08:50.790,0:08:54.800 function of Vin. 0:08:54.949,0:08:58.980 Now I want to make a really important observation here. We're going to, this is 0:08:58.980,0:09:03.899 going to be a real cool simplification. Okay so this is the point where op-amp 0:09:03.899,0:09:08.730 theory gets really cool. Watch what happens here. We know that A is a giant 0:09:08.730,0:09:12.899 number. A is something like 10 to the fifth or 10 to the sixth, and it's 0:09:12.899,0:09:17.160 whatever we have here, if our resistors are sort of normal sized resistors, we 0:09:17.160,0:09:21.389 know that a giant number times a normal number is still going to be a very big 0:09:21.389,0:09:27.029 number compared to 1. So this 1 is almost insignificant in this expression 0:09:27.029,0:09:31.889 down here. So what I'm going to do, bear with me, I'm going to cross it out, I'm going to say 0:09:31.889,0:09:36.509 no, I don't need that anymore. So if this if this number here if A is a 0:09:36.509,0:09:40.970 million, 10 to the sixth, and this expression here is something like 1/2 0:09:40.970,0:09:47.100 then this total thing is 1/2 of 10 to the 6 or a half a million, and that's 0:09:47.100,0:09:52.250 huge compared to 1, so I can pretty safely ignore the 1, it's very very small. 0:09:52.250,0:09:57.839 Now when I do that, well look what happens next. Now I have A top and bottom 0:09:57.839,0:10:03.480 in the expression and I can cancel that, too. So the A goes away. Now this is 0:10:03.480,0:10:07.649 pretty astonishing. We have this amplifier circuit, and all of a sudden I 0:10:07.649,0:10:12.029 have an expression here where A doesn't appear the gain does not appear. And what 0:10:12.029,0:10:20.910 does this turn into? This is called V naught equals Vin times what? times R1 0:10:20.910,0:10:31.740 plus R2 divided by R2. So our amplifier, our feedback circuit, came down to Vout 0:10:31.740,0:10:37.490 is Vin multiplied by the ratio of the resistors that we added to the circuit. 0:10:37.490,0:10:41.790 This is one of the really cool properties of using op amps in circuits, 0:10:41.790,0:10:46.379 really high gain amplifiers. What we've done is we have chosen the gain of our 0:10:46.379,0:10:51.029 circuit based on the components that we pick to add to the amplifier. It's not 0:10:51.029,0:10:55.379 determined by the gain of the amplifier as long as the amplifier gain is really 0:10:55.379,0:11:01.380 really big. And for op amps that's a good assumption it is really big. So this 0:11:01.380,0:11:06.779 expression came out with a positive sign, right, all the R's are positive values, so 0:11:06.779,0:11:12.950 this is referred to as a non-inverting op-amp 0:11:15.019,0:11:25.829 circuit, amplifier. So just to do a quick example, if R1 and R2 are the same then 0:11:25.829,0:11:30.779 we end up with an expression that looks like this Vout equals R1 plus R2, R 0:11:30.779,0:11:42.779 plus R over R is equal to 2, so the gain is 2 times Vin. So just to do a quick 0:11:42.779,0:11:51.420 sketch, just to remind ourselves what this looks like, this was Vin and we had 0:11:51.420,0:12:05.399 what out here? We had a resistor and a resistor to ground, and this is Vout. 0:12:05.399,0:12:09.329 So this is the configuration of a non-inverting amplifier built with an 0:12:09.329,0:12:15.240 op-amp. The two resistors in this voltage divider string connected to the negative 0:12:15.240,0:12:19.110 input, and it's going to be one of the familiar patterns that you'll see over 0:12:19.110,0:12:24.649 and over again as you read schematics and you design your own circuits.