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Op-amps and CMRR

  1. Oct 4, 2014 #1
    In the given op-amp circuit we have to find CMRR..for the we will need to find voltages at the 2 nodes (1 & 2) , we can fibd voltage at node 2 by voltage divider..how to find voltage at node 1 ..??
    I also have a doubt ..that since no current enters the the op=amp the voltage difference Vd across the 2 nodes will be zero , so wouls'nt the voltage at the 2 nodes will be equal??
     

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  3. Oct 5, 2014 #2

    jim hardy

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    no help yet ?

    Isn't that the basic premise of an operational amplifier? It's surrounded by a feedback network that lets it hold its inputs equal ?
    What happens if you use voltage divider rule to solve for voltage at node 1 as function of V1and Vo , then set your two node equations equal...?
     
  4. Oct 5, 2014 #3
    But if the voltages at node 1 & 2 are equal ..then what about Ad ( gain with different input signals) ..it will be infinity..!!!..since the 2 input voltages at node 1 &2 will be equal Vd will be zero then..!!
     
    Last edited: Oct 5, 2014
  5. Oct 5, 2014 #4

    davenn

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  6. Oct 5, 2014 #5

    jim hardy

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    Ad should not appear in the method i suggested. That method will tell you for any V1 and V2 what is Vo.

    Just try it.
     
  7. Oct 5, 2014 #6
    before solving i out.. my main concern is..Vd ..now see.. Vd = Vnode2 -Vnode1 & Vd is zero as we know ..so both voltages are equal ..again I am saying Vd is zero that means Ad is infinity... & CMRR=Ad/Ac.. then CMRR will be infinity..!!!
     
  8. Oct 5, 2014 #7

    dlgoff

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    For an ideal op-amp, Correct.

    http://en.wikipedia.org/wiki/Operational_amplifier#Ideal_op-amps
     
    Last edited: Oct 5, 2014
  9. Oct 6, 2014 #8
    But we use this assumption (Vd=0 ) while solving for expressions of output voltage in inverting or non-inverting amplifiers..so this is applicable there only..right.>?? For numericals in which nothing as such specified..we'll find the 2 input voltages individually and solve it out..!!
     
  10. Oct 7, 2014 #9

    jim hardy

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    In reality Ad is something like 10^5 to 10^8
    so when you say difference at inputs is zero, well, 0.000001 does round off to zero.

    What is important is to get your mind working these circuits as if the gain actually were infinite and the opamp able to balance its inputs exactly,.
    Then they become intuitive.
    Then , as an astute student of operational amplifiers, you back up and apply corrections for the non-idealness of the amplifier.

    Sometimes it's Ad you have to correct for, sometimes its the input current, sometimes it's the zero shift between inputs...

    So work it as if the amp were ideal, then back up and plug in your real gain of say 10^6 and see how much the result changes.
    Then plug in a couple microvolts of input offset. And then a few nanoamps of input current.
    That'll give you a feel for the effects of opamp's non-idealness.
    I used to do that with Basic on my TI99 because it had a great math pack for basic, TI made their home computer report as many digits as did their pocket calculators, Taught me a lot.

    old jim
     
  11. Oct 8, 2014 #10

    donpacino

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    Hey just a quick note. I think you are supposed to find the CMRR for the circuit as a whole, not the op-amp.

    This means finding the common mode gain (setting V1=V2) and differential gain (keeping them independent).
     
  12. Oct 8, 2014 #11

    donpacino

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