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Current time:0:00Total duration:12:50

Video transcript

- [Voiceover] Now I come to another configuration for an op-amp. And it's partially drawn here. And I'm gonna talk about this as I draw the rest of this circuit in. So this is gonna be made from a resistor configuration that looks like this. We'll have a resistor on the top. And this will be v-out, as we did before. And now we have a connection like this and the connection here to ground. This terminal is the minus terminal; and this terminal is the positive terminal. This is upside down from what we've done so far. But now pay particular attention here, this one has the minus on top. So now we have R1 and R2. And this is v-in. In particular, what we wanna do is find an expression for v-out as a function of v-in. All right. And in this video, I'm gonna do it the hard way. And what the hard way means is we're gonna do all the algebra to do this. And then in the next video I'm gonna show you the easy way. The easy way is really fun to learn, and it really helps to see it the hard way one time just so you appreciate the easy way. The other thing we get to do in this video is we'll do the algebra and we'll see how this gain, we take advantage of this gain to make some assumptions. OK, so let's go after this. Let's develop an expression for v-out in terms of v-in. First, let's write some things we know about v-out, OK? We know that v-out equals A times... Now, it's usually v-plus minus v-minus. This is v-plus. This is v-minus. Usually the expression here is v-plus minus v-minus. And since v-plus is zero, we're just gonna put in minus v-minus. This is equivalent to saying that v-minus equals minus v-out over A. So what else can we write for this circuit? OK, let's look at these resistors. Let's call this plus and minus vR1; and we'll call this one plus minus vR2. So there's a current flowing here, and that we'll call I. So I'm just gonna use Ohm's Law on R1 here and write an expression for I. I equals vR1 over R1. Right? Another way I can write that. What's this voltage here? This is v-minus. So I can write this in terms of v-minus, and that equals v-in minus v-minus over R1. That's the current going through this guy here. Now I'm gonna use something special. I'm gonna use something special that I know about this amplifier. What I know about an op-amp is that this current here is equal to zero. There's no current that flows into an ideal op-amp. So I could take advantage of that. What that means is that I flows in R2. So let me write and expression for I based on what I find over here, based on R2. I can write I equals, let's do it, it's vR2 over R2. And I can write vR2 as: v-minus minus v-out over R2. All right, so I took advantage of the zero current flowing in here to write an expression for current going all the way through. All right. So now we're gonna set these two equal to each other. Now we're gonna make these two equal to each other. Let me go over here and do that. V-in minus v-minus over R1. That equals this term here, which is v-minus minus v-not, v-out rather, over R2. How many variables do we have here? We have v-out, we have v-in, and we have v-minus. And what I want is just v-out and v-in, so I'm gonna try to eliminate v-minus; and the way I'm gonna do that is this expression over here. We're gonna take advantage of this statement right here to replace minus v-out over A. So I'll do that right here. So let me rewrite this. It's gonna be v-in minus minus v-out over A, so I get to make this a plus, and this becomes v-out over A all divided by R1. And that equals... V-minus is minus v-out over A minus v-out over R2. All right, let's roll down a little bit, get some room, and we'll keep going. What am I gonna do next? Next, I'm gonna multiply both sides by A, just to get A out of the bottom there. So now I get A times v-in plus v-out over R1 equals minus v-out minus Av-out over R2. There's a lot of algebra here, but trust me, it's gonna simplify down here in just a minute. All right, so now I'm gonna break this up into separate terms so I can handle them separately. Av-in over R1 plus v-out over R1 equals minus v-out over R2 minus Av-out over R2. Let's change colors so we don't get bored. Next, what I'm gonna do is start to gather the v-out terms on one side and the v-in terms on the other side. So that means that this v-out term here is gonna go to the other side. Av-in over R1 equals, let's do minus v-out over R2 minus Av-out over R2. And this term comes over as minus v-out over R1. Haven't made any sign errors yet. Now let me clear the R1. We'll multiply both sides by R1. Av-in equals R1 over R2 minus v-not minus R1 over R2 Av-not minus v-not. Yeah, the R1s cancel on that last term. All right. Av-in equals. Out of here I can factor this term here. Minus R1 over R2 times v-not, I can factor that out of here and here. So I can do minus R1 over R2 v-not times one plus A minus v-not. So let's take a look at this expression and use our judgement to decide what to do next. Now, because A is so huge, that means that this first term is gonna be gigantic compared to this v-not term here. V-not is some value like five volts or minus five volts or something like that. And A times this is something like 100,000 or 200,000, something like that. It dwarfs this v-not, so I'm gonna ignore this for now. I'm just gonna cross that out, and we'll move forward without that little v-not on the end of the expression. This is after we've left that out. Now we have v-in on this side. And I'm gonna take A over to the other side. One plus A over A. And this is a point where we get to use our judgement again. Again, A is a huge number, like a million; and so A plus one is a million and one. That fraction is really really close to one, so I'm gonna ignore it; I'm gonna just say it's one. So we'll send this one to one. And let me roll up a little bit more, just to have a little bit more room. Now what we have is what? V-in equals minus R1 over R2 times v-out. And I want the expression just in terms of v-out, so I'm gonna spin this around, and we'll get v-out equals minus R2 over R1 times v-in. So this is what our op-amp is doing for us. It basically says v-out is the ratio of two resistors times v-in. The gain of the overall circuit is determined by the ratio of those two resistors. And very importantly, there's a minus sign in front of it. The minus sign came all the way through. Again, let me sketch the circuit real quick. We had the minus sign on top. There's v-out. And there was R1 and R2. And the positive input, the non-inverting input, was connected to ground. OK? So this pattern with the resistor going over the top to the minus, this is called a inverting op-amp. And this is a really familiar pattern in op-amp circuits. You can see these on schematics, and you'll be designing these on your own. So this was quite a bit of algebra it took to get down to this point, and in the next video I'm gonna show you a really easy way to shortcircuit all this and be able to do this analysis really quickly; and that's called the virtual ground.