Non-inverting op-amp circuit

Okay, now we're going to work on our first op-amp circuit. Here's what the circuit's going to look like. Watch where it puts the plus sign; it is on the top on this one. We’re going to have a voltage source over here; this will be plus or minus Vn. That’s our input signal, and over on the output, we’ll have Vout. It’s hooked up this way: there's a resistor, another resistor to ground, and this goes back to the inverting input.

We’re going to look at this circuit and see what it does. Now, we know that connected up here, the power supply is hooked up to these points here, and the ground symbol is zero volts. We want to analyze this circuit, and what do we know about this? We know that Vout equals some gain; we'll write the gain right there, a big, big number times V plus minus V minus.

Let’s label that V plus is this point right here, and V minus is this point right here. We also know that the currents—let's call them I plus and I minus—equal zero, and that’s the currents going in here. This is I minus here, and that’s I plus, and we know those are both zero.

So now, what I want to do is describe what's going on inside this triangle symbol in more detail by building a circuit model. Alright, and the circuit model for an amplifier looks like this: we have V minus here, V plus here, so this is Vin, and over on this side we haven’t... here’s a new symbol that you haven’t seen before; it’s usually drawn as a diamond shape, and this is a voltage source. But is this... it’s a special kind of voltage source. It’s called a voltage-dependent voltage source, and it's the same as a regular ideal voltage source except for one thing: it says that the V, in this case Vout, equals gain times V plus minus V minus.

So the voltage here depends on the voltage somewhere else, and that’s what makes it voltage dependent. That’s what that means. So we’ve just taken our gain expression here and drawn a circuit diagram that represents our voltage expression for our circuit. Now, specifically over here, we’ve drawn an open circuit on V plus and V minus, so we know that those currents are zero.

So this model, this circuit sketch, represents our two properties of our op-amp. So I’m going to take a second here, and I’m going to draw the rest of our circuit surrounding this model, but I need a little bit more space. So let’s put in the rest of our circuit here. We had our voltage source connected to V plus, and that’s Vin, and over here we had Vout. Let’s check; Vout was connected to two resistors, and the bottom is connected to ground, and this was connected there.

What our goal is right now, we want to find Vout as a function of Vn. That’s what we’re shooting for. So let’s see if we can do that. Let's give our resistors some names; let’s call this R1 and R2—our favorite names always—and now everything is labeled. Oh, and we can label this point here and this point; we can call V minus. V minus. So that’s our two unknowns: our unknowns are Vout and V minus.

So let’s see if we can find them. So what I’m going to do is just start writing some expressions for things that I know are true. For example, I know that Vout equals A times V plus minus V minus. Alright, that’s what this op-amp is telling us is true. Now, what else do I know? Let’s look at this resistor chain here. This resistor chain actually looks a lot like a voltage divider, and it’s actually a very good voltage divider. Remember, we said this current here—what is this current here? It’s zero.

I can use the voltage divider expression that I know in that case. I know that V minus—this is the voltage divider equation—equals Vout times what? Times the bottom resistor, remember? This R2 over R1 plus R2. So the voltage divider expression says that when you have a stack of resistors like this with a voltage on the top and ground on the bottom, this is the expression for the voltage at the midpoint.

Okay, so what I’m going to do next is I’m going to take this expression and stuff it right in there. Let’s do that; see if we get enough room. Okay, let’s go over here. Now I can say that Vout equals A times V plus minus Vout times R2 over R1 plus R2. Right? So far, so good. Let’s keep going; let’s keep working on this. Vout equals A times V plus minus A Vout R2 over R1 plus R2.

Alright, so now I’m going to gather all the Vout terms over on the left-hand side. Let’s try that. So that gives me Vout plus A Vout times R2 over R1 plus R2, and that equals A times V plus. And actually, I can change that now. V plus is what? V plus is Vin.

Okay, let’s keep going. I can factor out the Vout; Vout is one plus A R2 over R1 plus R2, and that equals A Vin. Alright, so we’re getting close, and our original goal—we want to find Vout in terms of Vin. So I’m going to take this whole expression here and divide it over to the other side. So then I have just Vout on this side and Vin on the other side.

Let’s make some more room; I can do that. I can say Vout equals A Vin divided by this big old expression, one plus A R2 over R1 plus R2. Alright, so that’s our answer. That’s the answer; that’s Vout equals some function of Vin. Now I want to make a really important observation here.

We’re going to... this is going to be a real cool simplification. Okay, so this is the point where op-amp theory gets really cool. Watch what happens here. We know that A is a giant number; A is something like 10 to the fifth or 10 to the sixth. Whatever we have here—if our resistors are sort of normal size resistors—we know that a giant number times a normal number is still going to be a very big number compared to one.

So this one is almost insignificant in this expression down here. So what I’m going to do—bear with me—I’m going to cross it out. I’m going to say, "Now I don’t need that anymore." So if this number here—if A is a million, 10 to the sixth—and this expression here is something like one-half, then this total thing is one-half of 10 to the sixth, or a half a million, and that’s huge compared to one.

So I can pretty safely ignore the one; it’s very, very small. Now, when I do that, well, look what happens next. Now I have a top and bottom in the expression, and I can cancel that too. So the A goes away. Now, this is pretty astonishing. We have this amplifier circuit, and all of a sudden, I have an expression here where A doesn’t appear. The gain does not appear.

And what does this turn into? This is called Vout equals Vin times what? Times R1 plus R2 divided by R2. So our amplifier, our feedback circuit, came down to Vout is Vin multiplied by the ratio of the resistors that we added to the circuit. This is one of the really cool properties of using op-amps in circuits. Really high gain amplifiers—what we’ve done is we could choose the gain of our circuit based on the components that we picked to add to the amplifier.

It’s not determined by the gain of the amplifier, as long as the amplifier gain is really, really big, and for op-amps, that’s a good assumption—it is really big. So this expression came out with a positive sign; right? All the R's are positive values, so this is referred to as a non-inverting op-amp circuit amplifier.

So just to do a quick example, if R1 and R2 are the same, then we end up with an expression that looks like this: Vout equals R1 plus R2, or R over R equals two. So the gain is two times Vin. So just to do a quick sketch, just to remind ourselves what this looks like, this was Vin, and we had—what out here? We had a resistor and a resistor to ground, and this is Vout.

So this is the configuration of a non-inverting amplifier built with an op-amp—the two resistors in this voltage divider string connected to the negative input. So that’s what a non-inverting op-amp circuit looks like, and it's going to be one of the familiar patterns that you'll see over and over again as you read schematics and you design your own circuits.