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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= commonmode 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=0.1
Max amplified signal v=10^5 x (60x106) = 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?
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=0.1
Max amplified signal v=10^5 x (60x106) = 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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