# Calculating and plotting CMRR

I'm trying to find and plot the CMRR of the following circuit:

I'm getting zero for common mode gain, that gives an infinite CMRR for the circuit. When I followed normal feedback laws, I get (2/SRC)+1 as the differential gain.

• If Q1 and Q2 are identical and C and C are identical, did you expect to get a non zero common mode gain ? How is the output defined ? V3-V4 ? or just V3 ?
– AJN
Commented May 16, 2021 at 9:07
• With perfect and identical components, you would expect to get perfect common mode cancellation. Define the actual level of imperfection you get in the real world (one opamp may have an open loop gain 10dB more than the other, or a pole at a different frequency, or the circuit layout have more stray capacitance at one of the opamp inputs than the other) and then you'll get some non-zero figures. Commented May 16, 2021 at 9:14
• Always use mismatch of cables and components to measure CMRR Commented May 16, 2021 at 11:08

Welcome to Electrical Engineering.

With such a circuit, you have a differential input ($$\V_1 - V_2\$$) and a differential output ($$\V_3 - V_4\$$). But you also have a common-mode input ($$\\frac{V_1 + V_2}{2}\$$) and a common-mode output ($$\\frac{V_3 + V_4}{2}\$$).

Then you can define the differential gain as $$\\frac{V_3 - V_4}{V_1 - V_2}\$$ and the common mode gain as $$\\frac{V_3 + V_4}{V_1 + V_2}\$$.

Then you get the CMRR of this circuit.

Hint: the common-mode gain is not 0.

• Thanks for the answer. The way we were told to do it is finding the output for V1 and V2 and then rearrange the Vout into G(common)*V(common) + G(dif)*V(dif). When I solved for Vout of this circuit. I got a pure differential gain. Am I doing something wrong? Commented May 16, 2021 at 12:08
• However, with your method I'm getting G(common) = 1 I'm not sure if I'm right Commented May 16, 2021 at 12:10
• I do agree with G(common) = 1, but I fail to understand your “When I solved for Vout of this circuit.”… What $V_{\textrm{out}}$? Commented May 16, 2021 at 12:22
• my bad, I'm using Vout = V4-V3 Commented May 16, 2021 at 12:37
• @AkilaUyanwatta I don’t think you should consider $V_{\textrm{out}} = V_4 - V_3$ and use that to derivate the CMRR. Or you might as well consider $V_{\textrm{out}} = V_2 - V_1$ and you get perfect CMRR with no component at all. Commented May 16, 2021 at 13:27