Inverting op-amp circuit

Now I come to another configuration for an op-amp and it's partially drawn here. I'm going to talk about this as I draw the rest of this circuit in. So this is going to 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. 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.

Now, 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 Vn. In particular, what we want to do is find an expression for V out as a function of V in. All right, and in this video, I'm going to do it the hard way.

What the hard way means is we're going to do all the algebra to do this. Then, in the next video, I'm going to 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.

Okay, 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. Okay, we know that V out equals A times... now, it's usually V+ minus V-. This is V+; this is V-. Usually, the expression here is V+ minus V-. Since V+ is zero, we're just going to put in minus V-. This is equivalent to saying that V- equals minus V out over A.

So what else can we write for this circuit? Okay, let's look at these resistors we have. Let's call this plus or 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 going to use Ohm's law on R1 here and write an expression for I, and I equals VR1 over R1.

Right? Another way I can write that—what's this voltage here? This is V-. So I can write this in terms of V- and that equals Vn minus V- over R1. All right, so that's the current going through this guy here. Now I'm going to use something special. I'm going to 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 can take advantage of that. What that means is that I flows in R2. So let me write an expression for I based on what I find over here based on R2, and I can write I equals... let's do it's VR2 over R2. I can write VR2 as V- 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 going to set these two equal to each other. Now we're going to make these two equal to each other. Let me go over here and do that. Vn minus V- over R1 equals this term here, which is V- minus V out over R2.

So what I'm going for here is I have, how many variables do we have here? We have V out, we have V in, and we have V-. What I want is just V out and V in, so I'm going to try to eliminate V-. The way I'm going to do that is this expression over here. We're going to take advantage of this statement right here to replace V- with minus V out over A, so I'll do that right here.

Let me rewrite this. It's going to be Vn 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. That equals V- 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 going to do next? Next, I'm going to multiply both sides by A just to get A out of the bottom there. So now I get A times (Vn plus V out over R1) equals minus V out minus A times V out over R2. There's a lot of algebra here, but trust me, it's going to simplify down here in just a minute.

All right, so now I'm going to break this up into separate terms so I can handle them separately: A times Vn over R1 plus V out over R1 equals minus V out over R2 minus A times V out over R2. Let's change color so we don't get bored.

Okay, next, what I'm going to 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 going to go to the other side. So we got A times Vn over R1 equals minus V out over R2 minus A times V out over R2, and this term comes over as minus V out over R1.

Okay, haven't made any sign errors yet, and now let me clear the R1. We'll multiply both sides by R1. A times Vn equals R1 over R2 minus A times V out over R2 minus R1 over R2 times V out. Yeah, the R1s cancel on that last term.

All right, A times Vn equals... out of here I can factor this term here, minus R1 over R2 times V out. I can factor that out of here and here. So I can do minus R1 over R2 times V out times (1 + A).

Minus V out. Let's take a look at this expression and use our judgment to decide what to do next. Now, because A is so huge, that means that this first term is going to be gigantic compared to this V out term here. You know V out is some value like 5 volts or minus 5 volts or something like that, and A times this is something like a 100,000 or 200,000, something like that—it dwarfs this V out.

So I'm going to ignore this for now. I'm just going to cross that out, and we'll move forward without that little V out on the end of the expression. Now, this is after we've left that out. Now we have Vn on this side, and I'm going to take A over to the other side: 1 + A over A. This is a point where we get to use our judgment again. Again, A is a huge number, you know, like a million.

So A + 1 is a million and one. That fraction is really, really close to one. So I'm going to ignore it. I'm just going to 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 that Vn equals minus R1 over R2 times V out, and I want the expression just in terms of V out.

So I'm going to spin this around and we get V out equals minus R2 over R1 times Vn. 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. Okay, so this pattern with the resistor going over the top to the minus—this is called an 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 going to show you a really easy way to short-circuit all this and be able to do this analysis really quickly, and that's called the virtual ground.