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.

enter image description here


1 Answer 1


enter image description here

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

  • \$\begingroup\$ 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? \$\endgroup\$
    – user367640
    Nov 24, 2019 at 0:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.