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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 gm mismatch between M1 and M2 and finite RSS. The result is Equation (4.58)

This equation suggests that if RSS were replaced with an ideal current source ISS 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 gm than M2 due to mismatch. Then, when the same Vin, CM is applied to the gate of M1 and M2, M1 will demand a higher current from current source ISS and M2 will receive a smaller current. Thus, the common-mode level at Vout1 will be lower than at of Vout2, hence a differential gain due to CM input and thus a finite CMRR.

Have I missed something?


1 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.

  • \$\begingroup\$ You can't assume Iss is constant in the test for common mode gain with an Rss resistor. \$\endgroup\$
    – Andy aka
    Commented Oct 9, 2022 at 20:05
  • \$\begingroup\$ Oh I see. So you are saying that for a certain Vin,CM value, M1 will have more ISS current than M2 and Vout1 will be slightly lower. However, increasing Vin,CM will just mean that Vs also increases by the same amount, so Vout does not respond to this Vin,CM change \$\endgroup\$ Commented Oct 9, 2022 at 21:00
  • 1
    \$\begingroup\$ exactly -- for a constant ISS. If there is an RSS, then you may have to make some assumptions about what is causing gm to differ between M1 and M2 -- Is it W/L, or VTH -- that could affect a CM calculation. \$\endgroup\$
    – jp314
    Commented Oct 9, 2022 at 22:14
  • \$\begingroup\$ Thanks. Felt a little un intuitive at first. \$\endgroup\$ Commented Oct 9, 2022 at 22:16

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