Quick confirmation on Acm

Mar 27, 2007 1 Replies

I am trying to put together a paper on common-mode rejection, and I am stuck on the concept of common-mode gain (Acm). The 'circuit' shown when common-mode rejection is introduced is an op- amp with no feedback, and inputs tied to Vcm. The output should (ideally) be zero. Practically, one should expect an additional small error voltage on the output. This is of course, when we ignore any offset multiplied by the open loop gain. Jiri Dostal shows the same circuit, but says that the CMRR is equal to the change in Vo divided by the change in Vcm, and this in turn is also equal to the open loop gain, A, divided by the CMRR (X). He goes on to say that the raitio of A/X is on the order of 1. What does he mean by "order of 1." The value of Acm should be 0 or pretty damn close. I am missing something blantantly obvious.



See attached PDF for reference if I make no sense.



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Thanks in advanced.



he's taking orders of magnitude, powers of 10.

compared to A it is.

when you drive the differential par at an op-amps input with a common-mode signal they behave like emitter followers, hence the Acm "about 1"

Bye. Jasen

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