Calculate the max common mode voltage at the differential amplifier inputs

[clh99]

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

CMRR=20log10(Adiff/Acm) decibels
Adiff=differential gain
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.

CMRR=20log10(Adiff/Acm)

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=0.1Max 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?

 

[ammarb32]

Just to confirm, is this a MOSFET problem? Would you be able to show us the diagram for this problem?

 

[clh99]

Just to confirm, is this a MOSFET problem? Would you be able to show us the diagram for this problem?

There is no mention of MOSFETs in my lesson work and no diagram to go with the question, sorry