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Quick confirmation on Acm

Discussion in 'Electronic Basics' started by Kingcosmos, Mar 27, 2007.

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  1. Kingcosmos

    Kingcosmos Guest

    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. Dostal.pdf

    Thanks in advanced.
  2. jasen

    jasen Guest

    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"
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