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Opamp common mode rejection and virtual short

  1. Jan 19, 2008 #1
    I have a few questions concerning opamps.

    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.
     
  2. jcsd
  3. Jan 20, 2008 #2

    dlgoff

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    A simple mathamatical explanation of Common Mode Rejection Ratio can be found here.
     
  4. Jan 20, 2008 #3
    ahh i should have read that before i posted... but i do have another quetion about what i dont understand about CMR: Is the only reason why we want common mode rejection is because of feedback and gain stabilizaiton? meaning that if the common mode gain [tex] A_{s} [/tex] was high that there would be some unwanted oscilations in the closed loop gain?

    I am asking this because it seems to me that the point of the opamp and why its so special is that it provides a "linear" gain rather than gain that would distort the signal. And if [tex] A_{s} [/tex] was big then it would introduce distortion into the signal because it would have a hard time bringing the opamp into equilibrium.
     
    Last edited: Jan 20, 2008
  5. Jan 21, 2008 #4

    dlgoff

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    yes

    I'm not sure on this being a problem with equilibrium other than the common mode signals would show up on the output. I'm sure Berkeman would have a good answer.
     
  6. Jan 21, 2008 #5

    berkeman

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    CMRR is important for real-world applications, like sensor amplification and communication network integrity. If you have a run of twisted pair wire (either from a sensor or from a communication cable), and there are sources of electrical noise nearby (which is common -- even AM radio transmissions get picked up on cables like these), the noise is coupled into the twisted pair in common-mode. So whatever differential amplifier circuit you are using to process the differential information on the cable, will need to have a good CMRR in order not to add some of the common-mode noise into the differential-mode signal.
     
  7. Jan 21, 2008 #6

    berkeman

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    BTW, one of the qualification tests that network communication products need to pass in order to get a CE mark to be marketed in Europe, is a standard called EN 61000-4-6, "Conducted RF Immunity". At the higher levels of this test, you have common-mode noise injected into your communication cable, at about 51Vpp, swept from 150kHz to 80MHz with 80% AM impressed on it. Your communication must continue (in our case with about a 1Vpp differential data signal), even in the presence of this common-mode interference.
     
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