# Calculate the max common mode voltage at the differential amplifier inputs

• Engineering

## Homework Statement:

An instrumentation measuring system uses a differential amplifier having a CMRR of 120 dB and differential gain of 10^5 . The maximum
differential input signal is 60 μV. If the amplified noise voltage is specified to be not more than 1% of the maximum output voltage, calculate the maximum common mode voltage that can be present in the input to the amplifier.

## Relevant Equations:

Acm= common-mode gain

Max output voltage=max amplified signal voltage x max amplified noise v
Im unsure if I am on the correct track or have gone off on a tangent. Any help or guidance would be appreciated.

120=20log10(10^5/Acm)

120/20=log10(100,000/Acm)

6=log10(100,000/Acm)

taking antilogs 1,000,000=100,000/Acm

Acm=100,000/1,000,000

Acm=0.1

Max amplified signal v=10^5 x (60x10-6) = 6v

max amplified noise v = 0.01x6=0.06

max common mode gain=max amplified noise v/max noise v

rearrange for max noise v

max amplified noise/max common mode gain
0.06/0.1=0.6V

would 0.6V be my max common mode voltage that can be present in the input to the amplifier?

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