Operational Amplifiers

SkyEye

Why is is that we can assume V(p) = V(n) for an ideal Op-Amp with negative feedback? I understand it intrinsically, that the feedback cuts down the input voltage until it reaches a reasonable level, but not fundamentally; that is, why does feeding back on the negative end guarantee the voltage there will be on the same level as on the positive side?

ink4n3

I guess I can't give you and exact answer, but I think the reason is op-amps are made up of transistors. Here's a link I found that shows exactly what's in it about half way down. http://www.williamson-labs.com/480_opam.htm I don't really feel like analyzing that circuit too in depth and I never asked this question while I was in school. Does that help at all?

ecco the dolphin

SkyEye wrote: »

Why is is that we can assume V(p) = V(n) for an ideal Op-Amp with negative feedback? I understand it intrinsically, that the feedback cuts down the input voltage until it reaches a reasonable level, but not fundamentally; that is, why does feeding back on the negative end guarantee the voltage there will be on the same level as on the positive side?

Well, look at it this way:

1.) The positive terminal, Vp, initially has a higher voltage than Vn.

In a ideal opamp, with a large gain, the output will therefore become positive.

Feedback is connected from the now positive output to the negative input, Vn.

The output becoming positive has thus now pulled the voltage of Vn more positive through the feedback path.

For an ideal opamp, this feedback mechanism will continue until Vp = Vn.


2.) The positive terminal, Vp, initially has a lower voltage than Vn.

The output becomes negative.

The feedback connecting the output to the negative terminal will thus pull the voltage at Vn lower and lower, until Vp = Vn.

If the feedback pulls the negative terminal voltage too low (so that Vp > Vn), we revert back to case 1, where the feedback will pull Vn back higher until it finally settles where Vp = Vn.*


Practically, you'll have offset errors and finite gain/bandwidth, but the approximation that Vp = Vn holds pretty well as long as you remain within spec.

SkyEye

I understand that part pretty well, but I still don't get how the op amp "knows" when to stop at the specific point Vp = Vn. Why not stop at a point where the voltage will be saturated?

GothicLargo

If you have two equivalent springs underneath a seesaw, you compress one side and then let go... what happens?