L4 2 7Negative FeedbackChap 4

Explanation of Negative Feedback and how the feedback loop keeps the Op Amp in its linear region

>> To wrap up our discussion on Op-amps,

let's see if we can understand how this negative feedback loop here

constraints the Op-amp output,

so that our ideal Op-amp approximations work.

We've pointed out that an Op-amp is designed as

a high-gain device such that V_out is equal to some A or A,

some really large number,

times the difference between V_p and V_n. A is on the order of 10_5,

say 50,000, 100,000, or something like that.

But our source, our power supply voltages,

are going to be significantly smaller than 10_5.

Which means that, say,

let's just put some numbers on here,

let's say that this is a plus or minus 15-volt source.

That means that the output voltage is constrained to

being within the range of plus or minus 15 volts.

If that's the case,

then this difference V sub p minus V sub n is going to have to be very small,

or V_p minus V_n is going to

equal V_out divided by A. V_out can't be any larger than 15 volts.

So, 15 volts divided by 10 to the fifth means that V_p minus V_n is relatively small.

That's our approximation, that's our assumption

that we used in deriving our ideal Op-amp approximations,

and the gains for these different amplifiers.

So, what is it about this feedback, this R_1,

R_2 voltage divider, that keeps V_p minus V_n so small?

Let's see if we can understand that.

To do that, let's just pick a few more numbers.

Let's say that R_1 is equal to four kiloohms,

and R_2 is equal to one kiloohm.

So, with that then V sub n is just a voltage divided portion of V_out.

So, V sub n is equal to V_out times 1 divided by 1 plus 4, that's 5.

So, V sub n is equal to one-fifth of V_out.

When analyzing loops, you make some assumptions and

then see what this circuit does to the assumptions that you've made.

So, let's assume that V sub p has been 0,

which means V_out would be 0,

and V sub n would be 0.

So, V sub p has been 0,

and it is transitioning to V sub p equals 2.

So, let's just say that it just switched,

and V sub p now equals 2 volts. All right.

An Op-amp doesn't change the output voltage instantaneously.

It's limited by something called the slew rate,

and it doesn't really matter what that is,

you'll learn more about that in later.

In later classes, we'll learn more about Op-amps.

The supply is to say that at an instantaneous change here, the input,

will have some delay before its output is singing or the changes in to the output.

In fact, what happens is V_out starts to grow,

and under these circumstances,

it would be heading towards A,

this big number, times V sub p which is 2,

minus V sub n. So, initially,

this V sub p minus V sub n is going to be about 2, it's right down here.

V_out would be heading towards,

is not going to get there.

This feedback to work it's going to constrain it,

but if it was just open circuit,

V_out would be heading towards A,

which in our case is 10 to the fifth,

times V sub p which is 2,

minus V sub n. All right.

Again, it can't get anywhere close to that because our power supplies are limited,

but it would saturate at plus or minus 15,

in this case it's saturated at 15 volts.

But look what happens, as V_out starts to grow, V sub n,

which is equal to one-fifth of V_out,

let's write that here one-fifth V_out,

V sub n now starts to grow also.

As V sub n starts to grow,

this difference here V_p minus V sub n starts to get smaller.

Let's write that down here.

As V_out starts to grow,

V sub n which is equal to one-fifth V_out starts to increase also,

and thus the difference,

V_p minus V sub n,

starts to get smaller.

V_out continues to grow,

it's still a number times a relatively small number times a big number.

So, V_out is going to continue to grow,

and it will continue to grow until it stabilizes at a voltage where in fact V_out

is equal to A times V sub p minus V_n,

but V_n is minus one-fifth V_out.

Let's say that again. V_out will continue to grow,

until it gets to the point where V_out is equal to

A times the difference V_p minus V sub n at V sub n is one-fifth V_out.

This point then becomes the stabilized value of V_out.

Okay. Let's solve this for V_out.

While distributing A, we've got V_out is equal

to AV sub p minus one-fifth AV_out.

Bring the one-fifth AV_out to the other side.

We've got a V_out plus one-fifth AV_out is equal to AV sub p,

factor out the V_out and you're going to get V_out times 1 plus one-fifth A,

that's a V_out there,

one-fifth A is equal to A times V sub p. So, V_out then stabilizes,

or the final value of V_out is equal to AV_p divided by one plus one-fifth of A.

Let's go ahead and plug in our numbers and see what V_out then would be.

So, V_out would equal A,

which is 10 to the fifth,

times V sub p which we just chose to be 2 divided

by 1 plus one-fifth of 10 to the fifth.

Plug that into a calculator and you get that V_out is equal to 9.9995 volts.

V_n, the sub-divided version of that would just be one-fifth of 9.9995 volts,

which is equal to 1.9999,

and thus V sub p minus V sub n is equal to two minus 1.9999,

or V sub p minus V sub n is equal to 0.0001 volts,

or one-tenth of a millivolt.

When you compare that difference, that 0.0001 volts,

the voltage here to here,

we see that's one-ten thousandth of a fold.

Our approximation again was that this voltage wasn't exactly 0,

but it was so small compared to V sub p which is 2,

or V sub n which turns out to be 1.9999 volts,

or V_out which turned out to be 9.9995 volts.

That this approximate or that

this voltage across here was so small that we can neglect it.

Thus, our approximation V sub p being approximately equal to

V sub n is true under linear conditions,

and this feedback, this voltage divider circuit right here,

constrains the value on V sub n,

thus constrain the value on V_out,

so that this approximation holds as long as we don't allow V sub s to get too big.

Under those circumstances this non-inverting amplifier then would have a gain of

1 plus 4 over 1 or a gain of 5. Pretty close.