This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

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来自 Georgia Institute of Technology 的课程

电子学基础

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This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

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Op Amps Part 1

Learning Objectives: 1. Develop an understanding of the operational amplifier and its applications. 2. Develop an ability to analyze op amp circuits.

- Dr. Bonnie H. FerriProfessor

Electrical and Computer Engineering - Dr. Robert Allen Robinson, Jr.Academic Professional

School of Electrical and Computer Engineering

Welcome back to Electronics.

This is Dr. Ferri.

In this lesson we'll go over some basic Op Amp amplifier configurations.

In a previous lesson we looked at Ideal Op Amp behavior and Buffer Circuits.

Now we want to introduce some specific types of amplifier circuits.

The Inverting and

Non-Inverting configurations, as well as the difference in summing configurations.

We're also going to introduce the gain of a circuit.

Let's look first at the non-inverting amplifier.

We, in order to analyze this,

we want to be able to use our ideal amplifier assumptions.

And that is, that we've got, zero amps going in to each of these terminals.

And a zero, voltage drop across those terminals.

And I show the result of this being right here.

This shows the output as a function of the input, now let's derive it.

As part of their, our derivation, we are going to be doing some KVLs.

The first one, will be up, around this loop right here.

So, going up around here and back down to, the ground.

Going up this way, I get minus Vn, plus the voltage drop here.

Well the current is zero so that means the voltage drop is zero.

And then plus a voltage drop here, but that's also a zero.

So, we're left with the only other voltage drop which would be across this resistor.

And I'll define this current as being i sub 3.

So, it's plus i sub 3, R3 equal 0.

And solving for i3.

We get that is equal to Vn, over R3.

Now my second KVL, I will, look across this loop.

Actually this node voltage right here, let me go from minus to plus right here.

Going from minus to plus we gain V0.

And also looking in this direction,

we can add up this voltage across these two resistors.

That should equal to V0.

So we have V0, is equal to R2 plus R3 times i3.

And notice that,

I'm using the fact that there's zero current going in this direction.

So i3 going down through this resistor, has to be the same i3

as going through this resistor, since no current is going in this direction.

So, if I were to plug in this expression, N for i3.

I gain this expression out here.

So V0 is equal to this quantity times VN.

I'm going to define a quantity like that as being G.

In other words the gain.

The gain here is equal to R2 plus R3 over R3.

So it's whatever multiplies the input gives me the output.

And because my circuit has only resistors in it, my gain is a static gain.

Let's look at a particular example here.

Now in this example, I've given numerical values of 200 to R2 and R3.

Well, remember our gain was equal to R2 plus R3 over R3.

In this case if I plugged in the values I would get 400 over 200,

would be a gain of two.

Since the gain is equal to one the input is amplified.

And remember what our gain equation is, the output is equal to G times the input.

So in this case the gain is two.

So for example, if, V in is equal

to 5 volts, then V out is equal to 10 volts.

So the gain is amp, the output is amplified.

The input is amplified to give me the output.

Now let's look at a different configuration, the inverting amplifier.

We're going to use the same ideal op-amp assumptions,

which again zero amps, zero amps into those two terminals,

and the voltage drop is zero volts.

So, I'm going to do the same type of derivation as I did before,

which is to look at KVLs around different loops.

So, I'm going to start with this loop right here.

Again, we're going to be using KVLs.

So my first loop, I've got, minus V in plus R1 times,

and I'll call this current i1,

because the current goes through here and up through this way.

So I've got zero current here, so

all the current through V1 are, through R1 has to go through RF.

So I've got R1 times i1.

In other words, i1 is

equal to V in, over R1.

Now I want to do a second KVL,

and I'm going to do that KVL up and around this way.

And remember that this being a no voltage, it's as equivalent to having that

voltage here, with it being connected to ground this way.

So then I can actually do my KVL completely around this way.

So going around this leap I'm going to get a minus

V in plus R1 i1 plus Rf i1 plus V0 equal to 0.

We've already established that these first two terms are going to cancel out.

So these two terms cancel out.

And what I'm left is this part here.

So I can solve for i1.

Well actually, what I want to do is plug in for i1 right here.

So I have Rf over R1 times V in is equal to V0,

and actually there's a minus sign here.

So this gives me the equation,

the input output equation, for an inverting amplifier.

And I highlighted it right here.

So this term right here, minus Rf over R1 is equal to the gain.

Now notice in this case the gain could be greater than one or

less than one depending on this ratio.

How I select those resistors, if I want to amplify, or

if I want to make the output smaller, or what we say attenuate.

Lets look at a specific example with numbers in here.

Suppose R1 is equal to 1,000, Rf is equal to 2,000.

So our gain is equal to minus Rf over R1, and

in this particular case that would be equal to minus two.

So V out is equal to minus 2 times V in.

That's my equation.

That's how the input relates to the output.

And in this case, the input is if the magnitude let's say,

if the magnitude of G is greater than one the input is amplified.

If the magnitude is less than one the input is attenuated.

So the magnitude of V out is made larger than that of V in, so

we call it an, an amplified, a, an amplified signal.

So in this case if V in is equal to ten volts,

V out is equal to minus 20 volts.

Now, the term inverting corresponds to

the fact that we've got a negative sign right here.

We negate the voltage.

So that's when we say it's an inverting amplifier.

Now let's look at a circuit that's a little bit more complicated,

because I've got two inputs to it.

And in this particular circuit we're going to be

finding that the output is a difference between the two inputs.

We show the result right here, so we see we're, we're taking the difference between

V2 and V1 and multiplying it by a gain to give me the output.

Let's show the derivation here.

Now this derivation is a little bit easier if we use super position.

Super position is a method that we use for linear circuits.

And this amplifier, we're looking at the linear range so we can use super position.

In super position we look at the output due to

the individual inputs applied separately.

So in one case I would zero out one input, and calculate the output due to,

the other input.

And then I would reverse it.

I would say well,

okay let's, zero out this input, and look at the output due to, that input.

And then I'd sum them together.

So let's try super position.

So in the first case, we'll let V2 equal to 0 and we are going to solve for V0.

So, let me actually redraw this circuit here.

So, I have still got V1, and R1.

I am redrawing this circuit because we can make an assumption.

We can simplify it a bit.

Now this part right here, if V2 is going to be zeroed out.

So, that's actually the same as setting that to ground.

Then that means that there's zero current that flows through R1 or Rf.

And with zero current there, then this right here is equivalent to ground.

I can just connect that to ground, because it's got the same potential as ground.

So in that case, this looks exactly like an inverting amplifier.

And with the inverting amplifier, G is equal to minus Rf over R1.

So V out, due to just V1, is equal

to minus Rf, over R1 times V1.

So now we're going to, to finish super position.

We're going to have to do the opposite which is set V1 equal to 0 and

solve for the output due to V2.

So, taking a look at this, again, I want to set V1 equal to 0, so

let me zero it out.

Set that equal to zero.

If I zero it out, then that's equivalent to setting that to ground.

Now in this particular case, it's a little bit easier if I define a voltage here,

call it V sub b, and a voltage here call it V sub a.

Since this is V sub b right here, then, and

this is ground, this node voltage actually is V sub b.

And this node voltage right here, since this is ground, is V sub a.

Going through the derivation, where I let V sub 1 equal to 0.

We show that V sub a has to equal to V sub b,

because there's zero potential, difference between this.

It's an ideal amp op amp.

Now, if I want to find what Va is, I just have to look at this right here.

And note that this is a complete loop.

So I can use the voltage divider law.

[NOISE] And the voltage divider law says that V sub

a is equal to the ratio of Rf over the sum of the two

resistors in the loop, R1 plus RF times V2.

I can do the same thing across this loop right here, so going from V0,

being treated as if it was a source here, but this potential right

here is divided across these two resistors depending on their ratio.

So, Vb is equal to R1, that resistor,

over the sum of the two resistors, R1 plus RF times V0.

Now, I can equate V2 or V0 and Vb and

if do that I notice they have got the same denominators.

So, just simplifying that and, after I equate them,

I get Rf V2 is equal to R1 V0.

So solving for V0, I get Rf over R1, times V2.

Now, putting them together, the last step.

Maybe I'll put this as three.

The last step in super position is the total value of V0, and

that's the result due to V1 by itself, plus the result due to V2 by itself.

So in that case we got V0 is equal to minus Rf,

R1, V1 that was our first thing that we solve for.

Plus what we just solve for, Rf over R1 times V2.

And that gives us this relationship here,

where this now is the gain of the different circuit.

The last configuration I wanted to look at is a summing amplifier.

Now I'm not going to go through the derivation of that.

We use the same methods that we've used before.

We use the ideal op amp assumptions, and we can also use super position here.

In this particular case, if R1 is not equal to R2 then, we get this result here.

We define a gain, G1, and a gain, G2, like this.

So, V0 is equal to G1 times V1, plus G2 times V2.

This is what we might call a weighted sum, where we weight them differently.

Now if R1 is equal to R2, then this result would actually be equal to

V0 is equal to minus Rf over

R1 times V1 plus V2.

So in that particular case we have one gain G, times the sum of the two things.

If I wanted, I can actually add in another resistor here, and

I can actually keep going like this adding in lots of resistors, and

different voltages, and call this V3.

And if I did that, I would have V0 is equal to

minus Rf over R1, if it's the same thing.

If R1, R2, and R3 are the same, then I'd have V1 plus V2 plus V3.

That's if R1 equal to R2 equal to R3, and

this case to get this equation we had made the assumption that R1 is equal to R2.

So it just sums the, the different voltages together.

So in summary, one of the most important things we've talked about was the gain.

The gain is what relates the input voltage, to the output voltage.

We've been calling it G.

In all these cases, G is the constant, and

it's dependent on the values of the resistors that we've chosen.

We've defined four different amplifier circuit configurations.

The not inverting, inverting, difference and summing amplifiers.

In the next lesson, we will cover differentiators and

integrators, thank you.