I'm studying on my own for an exam. I bumped into this problem that really got me. My answer just doesn't match.

Exercise 4

Find the CMRR of the operational amplifier above, consider Ad = 1000000 (Answer CMRR= 100000)

  • From the theory I know that CMRR = 20*log(Ad/Acm) where Ad is the gain in differential mode and Acm is the gain in common mode.
  • I notice that the answer doesn't seem to be in dBs so I assume the answer is given by simply CMRR = Ad/Acm
  • Why do they give me Ad? since I can get it from a circuit analysis.
  • I got from the picture that Vo/ΔVi = 1/100, so if Vo/ΔVi = Ad my Ad doesn't match with the given Ad
  • How can I get Acm, I think this data is inherent of the op-amp and have to be given information.

I think there's something wrong with the problem, but since I don't have a way to confirm it I think I am missing something. Any help will be appreciated.

The source pdf (ES) Page 6, number 4.

  • \$\begingroup\$ @spehro-pefhany was right, I found that the Diferenciattor circuit is used as a Simple Common-Mode Rejection Ratio test circuit, where there are only two resistor values, so the 100 ohm resistor have to be 100k as the one in the feedback loop. \$\endgroup\$ – Ivancho Oct 8 '16 at 3:09
  • \$\begingroup\$ Also @transistor was right, I was mistaken the differential open-loop gain (Ad) with common mode gain (Acm). With this considetations I could get the true CMRR as the answer says. Tank you \$\endgroup\$ – Ivancho Oct 8 '16 at 3:09

I think the 100 ohm resistor should be 100K. The differential gain is thus very close to 10. The CMRR of the op-amp is 10^5 and the open-loop gain of the op-amp is 10^6.

You then calculate the CMRR of the differential amplifier, assuming perfectly accurate resistors, and I get the suggested answer.


\$ A_d \$ is the differential open-loop gain of the op-amp and in this case is 1M. (It's not the gain of the circuit with the feedback applied as you seem to think.) So if \$ V_{IN+} > V_{IN-} \$ by 1 µV the output would increase by 1µ x 1M = 1 V.

See if you can solve it with that information. Let us know. Meanwhile I'll have to read up on the matter!


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