I have a few questions concerning opamps.(adsbygoogle = window.adsbygoogle || []).push({});

First what is the big deal about infinite common mode rejection? It seems to me by looking at the open-loop gain equation A(V2 – V1) = V0 it is obvious that when V2 = V1 the output is zero (common mode rejection). This may be where I am confused … when they are talking about common mode rejection are they saying that ideally due to A being really big and V2 – V1 = V0/A that if V2 and V1 are not equal that they will be brought to a common mode and thus rejected? Is this what brings about the virtual short in the circuit? I was under the impression that the feedback loop caused the virtual short. Also if this is what brings about the virtual short wouldn’t that mean the “short” exists without feedback? And if this is not the case what would be wrong about stating that the virtual short is due to V2 – V1 = V0/A in the case where A blows up to infinity (I know this is a mathematical not a physical explanation but why would it be wrong).

Now here is how I understand the cause of the virtual short in the circuit:

When there is a feedback loop into one of the inputs and the inputs are not common V2 != V1 then there will be a large gain due to A(V2 – V1) which will cause some of the output to flow back into the input where the feedback loop is hooked into causing the input to increase or decrease in voltage until there is an equilibrium in which V2 = V1. If this is the correct explanation how would non idealalities in the opamp bring about finite common mode rejection?

These seem to be two contradictory explanations of the virtual short principle and I have convinced myself that they are both correct. Can someone please tell me where I am getting mixed up and what is wrong with my reasoning in either explanation?

Thank you in advance.

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# Opamp common mode rejection and virtual short

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