Lecture 4 1 Intro to Op Amps part 2 of 2

ECE1250 from the University of Utah. This video introduces the Op Amp, which allows us to build a whole new set of operations (division, multiplication, add, subtract, compare, and more). This video covers what's inside the op amp, Op Amp Gain A, Circuit Gain G, Ideal/Non-Ideal Op Amps, and how to use the Op amp as a switch.

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

So, here comes an example.

Let's start with our non-ideal model.

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

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

So, we want to define our gain.

This way we want to find the circuit gain of

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

So, let's use some,

the node-voltage method.

Let's first consider this node A.

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

We know sum of those must be equal to zero.

So, what voltage do we have here?

V out minus what voltages over here.

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

Now, let's consider this value right here,

V out minus this voltage is Vn.

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

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

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

plus right here, Vn minus going all the way,

keep going right there.

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

Those are the two values.

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

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

Let's also take a look at the circuit,

see if there's anything else that we know.

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

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

But if you look right here,

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

So, let's just substitute that back in for

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

Now we have two unknowns,

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

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

I've given you a little example here.

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

take a look at the [inaudible] this example.

I took the two equations right here and I converted

them into matrix form for V out and Vn.

Then I applied Cramer's rule.

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

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

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

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

with that example and did the non-ideal case.

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

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

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

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

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

Remember we know that Vs is equal to Vp.

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

So, if Vn equals Vp,

then this is zero.

Vn is Vs, Vn is Vs,

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

Great, our A equation and our B equation,

we've applied that simplification.

Do we need to apply anything for the currents?

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

But keep your eye on this one.

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

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

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

But remember that this value is zero.

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

This is not a good equation for us to use.

Let's use the B equation instead.

So, for the V equation,

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

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

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

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

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

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

see this R1 came over here.

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

divided by my input is R1 plus R2 over R2.

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

the value is five.

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

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

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

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

Here's the gain we just derived it.

Remember saturation is going to take over too.

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

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

Again, back to our Wild Horse Corral.