CMOS differential pair - common-mode rejection ratio

Question

From: Design of Analog CMOS Integrated Circuits by Behzad Razavi

This is a differential pair with a resistive load. The author derives the CMRR of the differential pair in the prescence of a g_m mismatch between M₁ and M₂ and finite R_SS. The result is Equation (4.58)

This equation suggests that if R_SS were replaced with an ideal current source I_SS that has infinite output resistance, then the CMRR should become infinite.

This does not make sense to me intuitively. If we assume that channel-length modulation and body effect are 0 and that M₁ has a higher g_m than M₂ due to mismatch. Then, when the same V_{in, CM} is applied to the gate of M₁ and M₂, M₁ will demand a higher current from current source I_SS and M₂ will receive a smaller current. Thus, the common-mode level at V_out1 will be lower than at of V_out2, hence a differential gain due to CM input and thus a finite CMRR.

Have I missed something?

Answer 1

Remember the ISS current is constant. With different gm's, the transistors may run different DC currents (which sum to ISS). Since you assume ISS is constant, and there are no body effects or ROUT effects, then when CM signal is applied, it doesn't change any of the currents (the mismatched currents remain mismatched), so the CM gain is 0.