In the CMRR definition, should the denominator be the A(common-mode input to differential output)? It seems you use the A(common-mode input to common-mode output) to calculate the CMRR. I thought the CMRR should be infinity because you have no mismatch in your components (which means the differential output remains zero no matter what your input common mode is). Thanks!

@MSMTUE

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You have a very valid point! Indeed, you can consider the denominator of CMRR either as common-mode-to-common mode gain or common-mode-to-differential-gain. This difference is only when the output is differential. You can use both to demonstrate different effects. 1) common-mode-to-common mode gain would be used when we want to show the ability of the stage to suppress common mode disturbances; 2) common-mode-to-differential-gain would be used when we want to show the ability of the stage to prevent common-mode disturbances injected into the signal. The latter point 2) is much more valid (your point), since ultimately we want to preserve the quality of the differential signal. Point 1) can be also valid if the stage is loaded with circuits that are sensitive to common-mode variations. So, the definitions should be either explicitly made or interpreted in the given context. If we consider the problem (as an exercise) it makes only sense if we consider common-mode-to-common-mode gain, since common-mode-to-differential gain is always 0 (if there is no mismatch). Since a learning objective of this exercise is to show that the time-constant at the common-node of the input pair transistors (Vs) is only relevant for common-mode signals and creates a pole only for common mode signals (and not for differential), we use here the definition 1) common-mode-to-common mode gain.