# Common Mode Noise Attenuation

The question is below:

Your raw sensor signal is 6.5 mV, and you are using an instrumentation amplifier to process it. The amplifier has a CMRR of 80 dB and a differential mode gain of 40 dB. If the RF noise on the leads from the thermocouple sensor to the data logger is 79 mV, what will the noise level be on the amplified signal in mV? (Type in a two-decimal number.)

I tried the following steps:

1. Find common-mode gain from CMRR = 20log(Ad/Acm)
2. Multiply 79 mV with Acm to find the noise at the output

But it turns out that I was wrong. I would appreciate if anyone could help me with this.

• What result did you get? Why do you think you were wrong? Dec 1, 2021 at 17:22
• I got Acm to be 0.004, 79 times 0.004 is 0.32 mV. But the answer seems to be wrong. I want to know if my approach is correct. Dec 1, 2021 at 17:25

If the RF noise on the leads from the thermocouple sensor to the data logger is 79 mV, what will the noise level be on the amplified signal in mV?

Here's where the question isn't tied down enough - does it mean that the RF noise on both leads is simultaneously 79 mV i.e. the common-mode interfering noise is truly 79mV or is it trying to tease out something else? I say this because noise induced on one lead wont be the same as the noise on the other lead; it might be close and, it will be close if the sending end device (raw sensor signal) has a balanced output impedance to ground. But the details in the question don't cover this.

Anyway, to get you out of jail as best i can I've had to assume the 79 mV noise is truly common mode. Given that the CMRR is 80 dB, that means that the 79 mV is reduced by 80 dB into a purely differential signal at the inputs. That differential noise signal is then amplified by 40 dB hence, the 79 mV common mode noise appears at the output of the InAmp amplified by a gain of -40 dB i.e. it is 79 mV divided by 100 or 0.79 mV.

• Please note that if a person deems a question to be unanswerable and then leaves that statement as a formal answer, then what does that tell you? Dec 1, 2021 at 23:22

$$\CMRR = (A_{dm}-A_{cm})[dB]= 80~ dB ~, A_{dm}= 40 ~dB,~ thus ~A_{cm}=-40~dB\$$
$$\ V_{IN_{cm}}= 79 mV ~RF???\$$

This CMRR only applies to DC to < 10 Hz and reduces 20 dB per decade to some small value like < 20dB

Considering the question says RF noise, we can safely assume they're is not much differential or CM gain so the output is likely only suppressed < 20 dB but without a datasheet, this question is unanswerable since RF frequency and part datasheet are BOTH undefined.