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We’re considering an ideal opamp here; so does that mean that Vo is infinity ♾ * 0 (A * (V+ - V-))?

Thanks!

thelamp
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  • The statement that V+ = V- holds when A is infinity. When A is finite and the necessary feedback is present, V+ and V- are very close but not exactly equal. – nanofarad Feb 22 '19 at 21:30
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    @EdgarBrown It appears to be a duplicate of a duplicate! –  Feb 22 '19 at 22:24

1 Answers1

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$$ V_0 = A(V_+ - V_-) $$

That's correct.

$$ V_+ = V_- $$

That's not correct. It's more like

$$ V_+ \text {almost} = V_- $$

The gain of the op-amp is typically so high that the output adjusts so that the negative feedback (which is what your question assumes) brings the inverting input to

$$ V_- = V_+ - \frac {V_O}{A} $$

which is very very close to, but not equal to V+.

Note also that they will only equal zero when the V+ input equals zero.

Transistor
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    A better way to put it would be $V_+ = lim_{A \rightarrow \infty} V_-$. – TimWescott Feb 22 '19 at 21:35
  • Possible duplicate of Apparent contradiction on ideal OpAmp equations :-) – Huisman Feb 22 '19 at 22:08
  • So it is ABSOLUTELY not done (check the down vote and 5 upvotes of @Edgar Brown remark) to post a question which looks a bit similar to a quesion of maybe years ago ... and then you can just answer with an answer which has been used for the very same question. Weird community here – Huisman Feb 22 '19 at 22:35
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    @Huisman: Is that addressed to me? I'm not sure what your point is. – Transistor Feb 22 '19 at 22:41
  • Addressed to all down voters: I think it is weird down voting duplicate questions and allow for duplicate answers which already have been reffered to by the remark "this question is a duplicate of ...". You should also downvote these answers. Or stop downvoting / upvoting the remark a question is duplicate. I encourage the latter. – Huisman Feb 22 '19 at 22:47
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    Don't worry about it. I was writing my answer when the duplicate comment was written. The funny thing is that the link is to a question where my answer was accepted. I didn't remember (but I've written > 2,500 answers) even though it was this year. At least I'm consistent. The downvote could have been for lazy writing, lack of research, etc., but it's a sign that the subject is badly taught if the question keeps getting asked. – Transistor Feb 22 '19 at 23:02
  • Use an example to teach. Assume Avol of 100,000, and opamp Vout of 10 volts. Thus the opamp's virtual_ground needs 100 microVolts. Which is far from zero. – analogsystemsrf Feb 23 '19 at 04:40
  • @analog: I did take that approach in my "duplicate" answer. I don't have anything that will measure 100 µV other than as a possible blink of the least significant digit on my DMM. I earn most of my living on mains power and a little on 10 kV so our perspectives may be different. – Transistor Feb 23 '19 at 09:16