Quick confirmation on Acm

[Kingcosmos]

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.

http://nothingbutnode.freehostia.com/pdf/Jiri Dostal.pdf

Thanks in advanced.

 

[jasen]

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."

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

The value of Acm should be 0 or pretty
damn close.

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"