0:00:00.000,0:00:02.505
>> To wrap up our discussion on Op-amps,

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

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constraints the Op-amp output,

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so that our ideal Op-amp approximations work.

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

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

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some really large number,

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

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

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But our source, our power supply voltages,

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are going to be significantly smaller than 10_5.

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Which means that, say,

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let's just put some numbers on here,

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

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That means that the output voltage is constrained to

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being within the range of plus or minus 15 volts.

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If that's the case,

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

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or V_p minus V_n is going to

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

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

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That's our approximation, that's our assumption

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

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and the gains for these different amplifiers.

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

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

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Let's see if we can understand that.

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

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

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and R_2 is equal to one kiloohm.

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

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

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

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When analyzing loops, you make some assumptions and

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

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

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which means V_out would be 0,

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and V sub n would be 0.

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So, V sub p has been 0,

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and it is transitioning to V sub p equals 2.

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

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

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

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It's limited by something called the slew rate,

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and it doesn't really matter what that is,

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you'll learn more about that in later.

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

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

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

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In fact, what happens is V_out starts to grow,

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and under these circumstances,

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it would be heading towards A,

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

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minus V sub n. So, initially,

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

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V_out would be heading towards,

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is not going to get there.

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

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but if it was just open circuit,

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V_out would be heading towards A,

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which in our case is 10 to the fifth,

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times V sub p which is 2,

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minus V sub n. All right.

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

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but it would saturate at plus or minus 15,

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in this case it's saturated at 15 volts.

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

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which is equal to one-fifth of V_out,

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let's write that here one-fifth V_out,

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V sub n now starts to grow also.

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As V sub n starts to grow,

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

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Let's write that down here.

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As V_out starts to grow,

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

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and thus the difference,

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V_p minus V sub n,

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starts to get smaller.

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V_out continues to grow,

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

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So, V_out is going to continue to grow,

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

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is equal to A times V sub p minus V_n,

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but V_n is minus one-fifth V_out.

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

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until it gets to the point where V_out is equal to

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

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This point then becomes the stabilized value of V_out.

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Okay. Let's solve this for V_out.

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While distributing A, we've got V_out is equal

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to AV sub p minus one-fifth AV_out.

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Bring the one-fifth AV_out to the other side.

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

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

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that's a V_out there,

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

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

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

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So, V_out would equal A,

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which is 10 to the fifth,

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

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

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

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

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which is equal to 1.9999,

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

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

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or one-tenth of a millivolt.

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When you compare that difference, that 0.0001 volts,

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the voltage here to here,

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

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

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

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

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or V_out which turned out to be 9.9995 volts.

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That this approximate or that

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

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Thus, our approximation V sub p being approximately equal to

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V sub n is true under linear conditions,

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and this feedback, this voltage divider circuit right here,

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constrains the value on V sub n,

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thus constrain the value on V_out,

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

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

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