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>> Now, let's do an example with our non-inverting operational amplifier.

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So, here comes an example.

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Let's start with our non-ideal model.

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This is the non-ideal model and let's connect it

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up in the non-inverting amplifier case that I showed you.

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So, we want to define our gain.

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This way we want to find the circuit gain of

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this amplifier circuit and the gain is defined as the output divided by the input.

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So, let's use some,

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the node-voltage method.

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Let's first consider this node A.

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So, let's consider our currents that are going out this way.

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We know sum of those must be equal to zero.

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So, what voltage do we have here?

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V out minus what voltages over here.

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That is A times Vp minus Vn and divide that by the resistor R0.

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Now, let's consider this value right here,

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V out minus this voltage is Vn.

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We're going to divide that by R1 and that is going to be equal to zero.

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Now, let's do our node at this point.

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We can say Vn minus V out divided by R1 plus Vn minus zero divided by R2,

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plus right here, Vn minus going all the way,

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keep going right there.

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Vn minus Vp divided by Ri and this all equals zero.

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Those are the two values.

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These are the two equations and we'd call them the A equation and

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the B equation that define the two nodes that we have.

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Let's also take a look at the circuit,

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see if there's anything else that we know.

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Because at the moment, we have three unknowns and only two equations.

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We don't know V out, Vn or Vp.

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But if you look right here,

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you can see that Vp is equal to Vs. Because of the way the circuit is hooked up.

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So, let's just substitute that back in for

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Vp in each of those cases right there to make these equations a little bit simple.

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Now we have two unknowns,

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Vn and V out and we have two equations to solve them.

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So, in order to write those as a matrix equation and apply Cramer's rule,

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I've given you a little example here.

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You might want to just stop the video for a minute and

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take a look at the [inaudible] this example.

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I took the two equations right here and I converted

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them into matrix form for V out and Vn.

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Then I applied Cramer's rule.

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You may have to look that up online if you don't remember it from math.

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Let's stop the video right now and do a little bit of practice on

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this particular problem and see that you can get the answer for the gain of this Op Amp.

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Okay, I'm hoping that you spent a little bit of time

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with that example and did the non-ideal case.

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That's the last time you're going to have to do the non-ideal amplifier in this class.

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Now, let's rewrite our node a equation V out minus A times V

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n minus V p divided by R out plus V out minus Vn divided by R1 equals zero.

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Here's the KCL node that I had at node B.

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Now, let's apply our ideal Op Amp equations.

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Remember we know that Vs is equal to Vp.

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But we also know because its ideal that Vp equals Vn.

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So, if Vn equals Vp,

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then this is zero.

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Vn is Vs, Vn is Vs,

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Vn is Vs, and Vn minus Vp is equal to zero.

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Great, our A equation and our B equation,

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we've applied that simplification.

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Do we need to apply anything for the currents?

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No, so we don't need to use that information.

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But keep your eye on this one.

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Remember that the output resistance is zero. We're going to need that.

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Right here this equation gives us the idea that we could solve for V out

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very easily because the only values we have are V out and Vs.

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But remember that this value is zero.

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So, it's like we're dividing by zero.

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This is not a good equation for us to use.

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Let's use the B equation instead.

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So, for the V equation,

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I want to solve for V out and I can say right here there's my V out.

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So, I can say that V out is equal to Vs times one divided by R2,

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plus one divided by R1 and then I have a V out over R1 on the other side.

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So, if I solve this, I can say that V out over Vs and happens to be

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my gain is going to multiply this thing out R1 plus R2 over R1 R2.

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Then I'm going to bring this R1 over to the other side,

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see this R1 came over here.

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So, now my R1 cancel out and my gain which is my output

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divided by my input is R1 plus R2 over R2.

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Awesome. If we actually calculate that for the numbers that we have here,

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the value is five.

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If we had calculated it using the non-ideal case, it was 4.99975.

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So, this is a very good approximation for an ideal Op Amp.

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This is how we are going to do our problems in the future.

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So remember, that the circuit gain depends on a circuit.

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Here's the gain we just derived it.

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Remember saturation is going to take over too.

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We can't make this gain just as large as we want.

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It's controlled by the power supplies that we put on this Op Amp.

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Again, back to our Wild Horse Corral.