What is the Maximum Common Mode Voltage for an Instrumentation Measuring System?

[Ianlamb]

Hi,
I have this question.

An instrumentation measuring system uses a differential amplifier having a CMRR of 120 dB and differential gain of 105. 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.

Just need to know if this is correct or I am on the right track.

I have the formula

CMRR=20log10(Adiff/Acm)

Then

120=20log10(10^5/Acm)

Then

120/20=log10(10^5/Acm)

Then

6=log10(10^5/Acm)

Then taking anti logs

1000000=10^5/Acm

Transpose this into

Acm=10^5/1000000=0.1volt

Is this correct? as when I plug the 0.1 back into the original equation I do not get the original 120dB answer.

Thanks.

 

[Tom.G]

Acm=10^5/1000000=0.1volt...

...is taking the ratio of two gains. How do you get Volts from a pure ratio?

This tutorial may help:
http://www.analog.com/media/en/training-seminars/tutorials/MT-042.pdf