The common mode gain for a differential amplifier in the general case is:

$${V_o \over V_c }={ R_1R_4-R_2R_3 \over R_1(R_3 + R_4) }\tag{1}$$


simulate this circuit – Schematic created using CircuitLab

Suppose the resistances have relative inaccuracy of \$\,\varepsilon_i(i=1\,\text{to 4})\$ so \$R_i=R_i^*(\varepsilon+1)\$ where \$R_i^*\$ is the nominal value of \$R_i\$.

Also suppose that \$\varepsilon_i\lt\lt1\$ and,

$$\tag{2}R_1^* = R_3^*\,\text{and}\,R_2^*=R_4^*\qquad\text{(condition for differential amplifier)}$$

Then, $$\tag{3}A_c={ R_1R_4-R_2R_3 \over R_1(R_3 + R_4) }\approx{R_2^* \over R_1^*+R_2^*}(\varepsilon_1+\varepsilon_4-\varepsilon_2-\varepsilon_3)$$

But I can't find this value algebraically.

This is important because the worst case CMRR is:

$$\tag{4}CMRR={A_d \over A_c}={1+R_2/R_1\over\mid\varepsilon_1\mid+\mid\varepsilon_2\mid+\mid\varepsilon_3\mid+\mid\varepsilon_4\mid}.$$

  • \$\begingroup\$ maybe I should ask this question in the mathematics \$\endgroup\$ Apr 12, 2015 at 15:42
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    \$\begingroup\$ What are you asking here? The expression for \$A_c\$ with your substitutions (i.e., just a bit of algebra), or the expression for the error in the common mode gain with respect to the errors in the resistor values? The former is just $$A_c\approx{R_2^*(\varepsilon_1-\varepsilon_2-\varepsilon_3+\varepsilon_4)\over(R_3+R_4)(1+\varepsilon_1)}$$ if we assume the error products \$\varepsilon_2\varepsilon_3\$ and \$\varepsilon_1\varepsilon_4\$ are negligible. (Note that those are the real \$R_3\$ and \$R_4\$ in the denominator, not the nominal values.) \$\endgroup\$ Apr 12, 2015 at 16:20
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    \$\begingroup\$ Thank you very much for having done the math too. I want to algebraically come to the given value. At first I didn't see that the error products are indeed negligible. So now my numerator is the same as in the penultimate equation. What's different are the denominators, (R3+R4)(1+ε1) cannot be R∗1+R∗2 right? Only if you neglect all the errors in the denominator. If you think this is correct, can you explain why? Thank you again \$\endgroup\$ Apr 12, 2015 at 17:18
  • \$\begingroup\$ I see. I don't know, sorry. Where does your given expression come from? Are you sure it's correct? \$\endgroup\$ Apr 12, 2015 at 19:40
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    \$\begingroup\$ I'm not sure it's correct. No derivation is shown, and it doesn't seem to follow from the definitions given. Maybe with some additional assumptions it would, but apart from neglecting the error products all other simplifications seem questionable. You can of course calculate the worst-case CMRR for yourself without using any approximations, and then simplify it to the extent that you think justifiable. That's what I'd do. \$\endgroup\$ Apr 13, 2015 at 23:55

1 Answer 1


The worst case CMRR is appropriate to the expression you gave, since the absolute values of the error or tolerance gives the maximum common mode gain. And Differential gain is as usual equal to R2/R1


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