L4 2 1 NoninvertingChap 4

Non-Inverting Operational Amplifier Configuration

>> The first of the op-amp configurations that we're going

to consider is known as the non-inverting amplifier.

It gets its name from the fact that the source voltage to

be amplified is connected to the non-inverting terminal.

Because of that, it turns out that the output voltage will be of the same sign,

S-I-G-N, as the input voltage.

So, for example, if V_s is a positive voltage,

the output voltage will also be positive.

On the other hand, if the input voltage is negative,

then the output voltage will be negative also.

The sign is not inverted.

I have here two different schematics.

I'd encourage you to stop the video for just a second and convince

yourself that these two schematics are identical as far as function is concerned.

The things that you're going to be looking at are that

the V_s is connected to the non-inverting terminal.

Of course, the non-inverting terminal is the one that's got the positive sign on it.

So, if this one is connected to the non-inverting,

over here, V_s is also connected to the non-inverting.

So, this amplifier over here is upside down from this op-amp here.

The other thing that you look at is where does the feedback go.

We'll talk more about feedback later,

but suffice for now to say that feedback is the process or the act of

taking the voltage at the output and running it back to,

in this case, the negative or the inverting terminal.

Again, we'll talk more about feedback later,

but this is referred to as negative feedback.

You can remember that because the feedback loop comes back

to the terminal of the negative sign which is the inverting sign.

So, in both of these feedback circuits,

we're going from V_out through R_1.

The node beat between R_1 and R_2 is

then tabbed and brought back to the inverting terminal.

Notice, over here, we have the same situation,

coming from the output,

going through R_1, connecting to R_2,

and the node where R_1 and R_2 are connected is connected

then to the inverting terminal and then brought back the ground.

Now, let's analyze these op-amps at both of these circuits and see what it is,

why we might draw it this way under certain circumstances

and why we might draw it this way under other circumstances.

Let's start here with the one on the right.

By drawing it in this configuration,

it makes it obvious that the output voltage goes through a voltage divider circuit,

and only a portion of the output voltage is fed back to the inverting terminal.

In this case here,

V_n then,

using our voltage divider formula,

V_n is equal to the voltage across R_2,

which is V_out times R_2 over R_1 plus R_2.

Now, in order to get V_out as a function of V,

our input, we're going to reverse the roles

here and multiply both sides by the inverse of this.

In solving or solving for V_out,

we get that V_out is equal to V_n times R_1 plus R_2 over R_2.

Now, we don't want it as a function of V_n.

We want to know what the output is as a function of the input voltage

V_s. We now apply one of our ideal op-amp approximations.

That was the one that referred to as the virtual short.

That V_p and V_n are going to be so close to each

other that the difference V_p minus V_n is zero,

or what we can say then is that V_n is approximately equal to V_p.

Now, what is V_p?

Using another op-amp approximation,

the current going into the input terminals is zero.

Therefore, I_p is zero.

There will be no voltage drop across

V_s because the current going through it is zero.

So, V_p is in fact just our source voltage V_s.

So, V_p equals V_s. V_n equals V_p due to the virtual short.

We have then that V_out again replacing V_n with V_s.

V_out is equal to V_s times R_1 plus R_2 over R_2.

Generally speaking, we'll take this and say, well,

note that R_2 is a denominator that's common to both of those two terms.

So, we can then have V_out is equal to V_s times R_2 over R_2,

that's one, plus R_1 over R_2.

We say that the gain, G, the closed loop gain,

the gain that we get because of this feedback circuit is equal

to one plus R_1 plus R_2.

We're going to refer to that as the gain of the non-inverting amplifier.

We'll see when we get to the inverting amplifier configuration that

its gain is just slightly different than this.

But let's just point out now that V_out is in fact going to be

the same sign as V_s. R_1 and R_2 are both positive quantities.

So, the ratio of a positive quantities is positive,

plus one is a positive number,

times whatever the sign is on V_s gives us a V_out,

which is the same sign as V_s. Now,

let's look at this circuit over here.

To analyze this, we're going to use a technique that use a node analysis,

which is an analysis technique that we'll frequently use on op-amp circuits,

especially as the op-amp circuit gets to be a little bit more complex.

So, here's the deal.

This is now V_n. We're going to write a node equation at V_n. But first,

we're going to note that this voltage here, V_p,

is going to equal our source voltage.

As we saw over here, the R_s had no influence on it.

So, I've left that out of this circuit here.

So, V_p equals V_s. V_n equals V_p because of the virtual short.

So, V_n then is going to equal V_s.

Now, let's write a node equation summing the currents leaving this node.

The current leaving this node going through R_2 to ground is V_s minus

zero divided by R_2 plus the current going from this node in this direction.

It's going to be the voltage across R_1, which is V_s,

minus V_out divided by R_1,

and then plus the current going into or leaving this node and going into the op-amp.

But our ideal op-amp approximation tells us

that the current going into the input terminal is zero.

So, we could write a plus zero there.

But let's that off and simply say that the sum of those two currents must equal zero.

So, what that's saying is that in fact,

just as we saw over here,

the current is going from the output through these two resistors back.

It's going from the output through these two resistors to ground.

These two resistors are in series because the current going in here is zero.

Now, let's solve this equation for V_out in terms of these paths.

We have V_s factoring out V_s times one over R_2 plus one over

R_1 minus V_out over R_1 is equal to zero.

That bring this term over to the other side, and solving for V_0,

We get then that V_0 is equal to V_s times R_1 times

one over R_2 plus one over R_1 or distributing that

through the R_1 over the R_1 gives you the one,

the R_1 over R_2 gives you the other term,

and we get the V_out is equal to

V_s times one plus R_1 over R_2.

So, the gain terms are the same.

These two circuits are equivalent

and that gives you an idea of what the non-inverting amplifier does.