What minimum CMRR in dB is required to keep common-mode error below the signal when Vcm = 1 V and Vdiff = 10 mV?

What minimum CMRR in dB is required to keep common-mode error below the signal when Vcm = 1 V and Vdiff = 10 mV?

• A. 20 dB
• B. 60 dB
• C. 80 dB
• D. 40 dB

Answer: (D)

CMRR measures how much of a common-mode input sneaks into the output compared with the desired differential signal. To keep the common-mode error from swamping the measurement, the output due to the common-mode voltage must be smaller than the output due to the differential signal. Let Ad be the differential gain and Ac the common-mode gain. CMRR = Ad/Ac. The differential output is Ad × Vdiff, while the unwanted common-mode output is Ac × Vcm. Requiring Ac × Vcm < Ad × Vdiff gives Ad/Ac > Vcm/Vdiff. Here Vcm = 1 V and Vdiff = 10 mV, so Vcm/Vdiff = 100. Thus the CMRR must be greater than 100, which is 20 log10(100) = 40 dB. Therefore, the minimum CMRR is 40 dB.

CMRR measures how much of a common-mode input sneaks into the output compared with the desired differential signal. To keep the common-mode error from swamping the measurement, the output due to the common-mode voltage must be smaller than the output due to the differential signal.

Let Ad be the differential gain and Ac the common-mode gain. CMRR = Ad/Ac. The differential output is Ad × Vdiff, while the unwanted common-mode output is Ac × Vcm. Requiring Ac × Vcm < Ad × Vdiff gives Ad/Ac > Vcm/Vdiff.

Here Vcm = 1 V and Vdiff = 10 mV, so Vcm/Vdiff = 100. Thus the CMRR must be greater than 100, which is 20 log10(100) = 40 dB. Therefore, the minimum CMRR is 40 dB.