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In an op-amp (non-inverting), like this picture

when \$V_s\$ increases by \$\Delta V\$, \$V_n\$ approaches \$V_p\$ until the difference is \$\Delta V/ A\$ (where A is infinity).

Why is there the \$\Delta V/ A\$ difference and where does this difference come from? Why doesn't \$V_n\$ just equal \$V_p\$ (so that the difference is 0)?

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  • \$\begingroup\$ No time to post a full answer but if you want a quick answer take a look at allaboutcircuits.com/textbook/semiconductors/chpt-8/… \$\endgroup\$ – hedgepig Oct 27 '16 at 18:32
  • \$\begingroup\$ For Vp=Vn the difference would be zero - and the amplifier unit would have no input diff. voltage. Only for the idealized case (with Ao approaching infinity) we can have - mathematically -a finite value for Vout=0*infinity. \$\endgroup\$ – LvW Oct 27 '16 at 19:19
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Because the opamp has finite gain. Note that in your equation when A is infinity, Vp and Vn are equal.

Since the opamp has finite gain, it needs some small difference between Vp and Vn to produce a non-zero result. For example, let's say the opamp gain is 100,000 and is producing 7 V out. That means it must see a difference of (7 V)/100,000 = 70 µV to produce that 7 V out.

We can usually consider the two inputs to be the same voltage and ignore a small difference like 70 µV. Usually the input offset voltage swamps that anyway. There isn't much point considering 70 µV difference when the input offset voltage is 2 mV. That means that the actual voltage difference between Vp and Vn can be up to 2 mV off from what it should be ideally. Basically, the input offset voltage is the error inside the opamp when interpreting the difference between its inputs. In this example, the additional offset of 70 µV to make the opamp produce its actual output is only 3.5% of the ambiguity in the difference between the inputs anyway.

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  • \$\begingroup\$ Thanks! I don't understand why in an ideal op-amp with negative feedback, Vn = Vp and Vp - Vn = 0. If Vp - Vn = 0 then output voltage = A * (Vp - Vn) = A * infinity? What does that even mean output voltage is? It makes more sense to me that we would want to have the ideal be Vp - Vn = (Vs / A) so that output voltage is Vs. Is there something I'm misinterpreting? \$\endgroup\$ – laura Oct 27 '16 at 19:14
  • \$\begingroup\$ @laura: You are almost right about A being infinity. It's large, but not infinity. As a result, Vp does not quite equal Vn. Vn = Vp is a approximation that is useful much of the time. You could say that's the same as approximating A to infinity instead of the large number it really is. \$\endgroup\$ – Olin Lathrop Oct 27 '16 at 20:11
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It's pretty much by definition. The output is the differential input voltage times the gain (A). Therefore the differential input voltage is the output voltage divided by the gain.

The feedback drives the negative input to very close to the positive input due to the very large gain, again so Vin = Vout/A. If A is infinity as in an ideal op-amp then there's no difference between the 2 inputs. (Or you could say the input voltage is vanishingly small.)

In a non-ideal amplifier the feedback drives the input voltage to a value just large enough to satisfy Vout = A*Vin.

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  • \$\begingroup\$ In the real world, why can't Vn ever reach Vp? Why does the differential input voltage have to exist? Why can't Vn = Vp and then Vn - Vp = 0? (If this happens though, then doesn't voltage output just 0 become since A * 0 = 0 (since A can't be infinity in the real world) and thus there would be no point to the amplifier since it produces 0 output voltage?) \$\endgroup\$ – laura Oct 27 '16 at 19:28
  • \$\begingroup\$ I think you just answered your own question- If Vn = Vp then then Vin=0 and Vout will always equal zero. In the real world some small delta is required to get a useful output from the amplifier. \$\endgroup\$ – John D Oct 27 '16 at 19:46

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