# Calculate the differential common-mode gain of instrumentation amplifier

I am trying to design an instrumentation amplifier with a CMRR of 50 dB.

I have to target a differential gain of 60 dB.

$$CMRR(dB) = 50 dB = 20log(\frac{A_{vd1}}{A_{cm1}})+ 20log(\frac{A_{vd2}}{A_{cm2}})$$

Where $$A_{cm1} = 1$$

$$CMRR(dB) = 50 dB = 20log(1+2\frac{R_1}{R_g})+ 20log(\frac{R_3}{R_2}) - 20log(A_{cm2})$$.

$$A_{v} = 60 dB = 20log(1+2\frac{R_1}{R_g})+ 20log(\frac{R_3}{R_2})$$

I arbitrarily assigned the gain for each stage, e.g., Av1 and Av2 are 40 and 20 dB respectively. This gave resistors of the following

$$R_1 = 100 k\Omega$$ $$R_g = 2.5 k\Omega$$ $$R_2 = 1 k\Omega$$ $$R_3 = 10 k\Omega$$

Resulting in a $$Acm2 = 0.316$$

If I assume a 5% tolerance for the resistor values, what would be the differential and common-mode gain? And what would be the worst cast CMRR? I am very confused on how to approach this. Thank you for any tips/suggestions.

For ideal matched parts with shared Rgain there is (in theory) no CM gain.

$$\V_{out} = (V_2 – V_1 ) (1 + \dfrac{2R_1}{R_g})\cdot \dfrac{R_3}{R_2}\$$

The method of computing sensitivity of any result is a partial derivative with respect to the variable.

In this case the rise in CMRR is due to mismatch tolerance worst case.

The partial derivative of output w.r.t. R1 $$\= \dfrac{2R_3}{R_2}\$$ The tolerance on a single R1 of 0.2% with R3/R2=1 is 1/1000 or CMRR=60dB

For +/-0.2% on each R1 CMRR reduces by 6dB or CMRR=54dB

• Thank you for the comment. However, I am confused by your description. Why is R3/R2 = 1? What is 0.2%? Are you referring to the resistor tolerance? Commented Nov 24, 2019 at 0:05