Question

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.)

Answer #1

**Answer :**

CMRR, by definition, is

CMRR = 20 log(AD/AC) where AD = differential mode gain and AC = common mode gain

**As AD is given in decibels, therefore we first convert
into the ratio of Vout to differential input voltage
(Vdin).**

40 = 20 log(Vout/Vdin)

Therefore Vout/Vdin = 100

Next, using the definition of CMRR

80 = 20 log(100/AC)

4 = log(100/AC)

100/AC = 10000

or AC = 0.01

As input common-mode noise is 79mV, therefore the noise on
amplifier output is: 79*.01 = **0.79mV**

Hope this helps

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