CMRR is defined by the ratio of differential gain to common-mode gain:

$$ CMRR = {A_d \over A_{cm}} $$

And CMRR is measured by the change in output voltage (\$ \Delta V \$) when a common-mode input voltage is applied:

$$ CMRR = {V_{cm} \over \Delta V} $$

That implies the following relation, of which I cannot prove mathematically: $$ {A_d \over A_{cm}} = {V_{cm} \over \Delta V} $$

I tried various algebraic combination and the only way I could reach the above relation is this:

$$ \begin{align} V_{out} = A_d(V_2-V_1) + A_{cm}V_{cm} &= A_d(V_2-V_1±\Delta V) \\ \\ A_{cm}V_{cm} &= ±A_d\Delta V \\\\ {V_{cm} \over {\Delta V}} &= ±{A_d \over A_{cm}} \end{align} $$

But this derivation would imply \$ \Delta V \$ is measured at the input instead of a value measured at the output. \$ \Delta V \$ in the equation is also amplified by the differential gain - so it's not a voltage that you can measure at the output either. I researched online but they just laid out these two equations side by side without mathematical proof of their relation.

How to go from \$ A_d \over A_{cm} \$ to \$ V_{cm} \over {\Delta V} \$?

$$ CMRR \equiv {A_d \over A_{cm}} = {\text{my question here for the missing steps}} = {V_{cm} \over {\Delta V}} $$

  • \$\begingroup\$ Well, you also already used the asumption \$A_{cm}\cdot V_{CM} = A_d \cdot\Delta V\$ in your first equation of your derivation... \$\endgroup\$
    – jusaca
    May 7, 2021 at 8:38
  • \$\begingroup\$ Ha!? But you already proved it mathematically in the first three steps with your assumptions. \$\endgroup\$
    – Mitu Raj
    May 7, 2021 at 10:23
  • \$\begingroup\$ The first equation is the definition. The second equation is how one measures it. But mathematically how you go from the first equation to the second? \$\endgroup\$
    – KMC
    May 7, 2021 at 12:28
  • 1
    \$\begingroup\$ iIs this answer gives you help electronics.stackexchange.com/questions/432511/… \$\endgroup\$
    – G36
    May 10, 2021 at 17:33

2 Answers 2


The second definition (\$CMRR=\frac{V_{cm}}{\Delta V}\$) must be wrong. If \$\Delta V\$ is change in output caused by \$V_{cm}\$, then $$\Delta V = A_{cm}\cdot V_{cm}\implies \frac{V_{cm}}{\Delta V}=\frac{1}{A_{cm}},$$ not \$CMRR\$.

\$CMRR\$ is telling you how much is \$V_d\$ amplified compared to how much \$V_{cm}\$ is amplified, not the amplification of \$V_{cm}\$ alone.

The way to measure it would be $$CMRR = \frac{V_{cm}}{\Delta V} \cdot A_d$$

Which makes sense, since \$CMRR\$ should be a measure of "how good the amplifier is at not amplifying common mode". Higher \$A_{cm}\$ shouldn't matter so much if \$A_d\$ is also very high, but if \$A_d\$ is low, then you also want very low \$A_{cm}\$ to get a good difference between wanted and unwanted signal.


I have always been uncertain as to which parameters are defined and which are being measured, or what variable comes before or after a calculation. Thanks to @Jiří Maier's answer above and @G36's link, but the concept only came clear after I drew up a sample circuit and plugged in some values:


The circuit is a diff-amp with CM error due solely to resistor tolerance as an example. There'll be an output offset caused by resistor mismatches, but you can't algebraically (symbolically) resolve differential voltage, CM voltage, or their corresponding gain \$ A_d \$ or \$ A_{cm} \$ as a function of resistor values or tolerances by itself.

Things make sense after we plug in numeric values. After which the equation can be manipulated into a composition of \$ V_{OUT(d)} \$ and \$ V_{OUT(\text{offset})} \$. While it is easy to see 98.04 as the differential gain, you can't say 0.04 is the CM gain because \$ V_2 \$ may not necessarily equate \$ V_{cm} \$ (or \$ {V_1 + V_2} \over 2 \$). The term \$ V_{OUT(\text{offset})} \$ as a whole indicates there is an offset to \$ V_{OUT(d)} \$, but \$ V_{OUT(\text{offset})} \neq A_{cm} V_{cm} \$.

0.04 can only be CM gain \$ A_{cm} \$ if and only if there is zero differential voltage (or \$ V_2 - V_1 = 0\$ ), which automatically implies \$ V_2 \$ equals to \$ V_{cm} \$. Only then you see the ratio of \$ {98.04 \over 0.04} = {{A_d \cdot V_{cm}} \over V_{OUT(cm)}} \$ and define it as CMRR. Only in this case for zero differential input can \$ V_{OUT(\text{offset})} = A_{cm} V_{cm} = V_{OUT(cm)} \$. To demonstrate:

If \$ V_2 = 3V \$ and \$ V_1 = 2.8V \$, the exact equation will yield

$$ \begin{align} V_{OUT(total)} &= V_{OUT(d)} + V_{OUT{(\text{offset})}} \\ &= 98.04 \cdot (3-2.8) + 0.04 \cdot 2.8 = 19.72V \; ✓✓ \; \end{align} $$

My difficulty then was not realizing \$ V_{OUT} \neq A_d V_d + A_{cm} V_{cm} \$ !!! (I was misled by the terminologies used here into thinking they were equal), which otherwise would be off by 0.004V:

$$ \begin{align} \;\;\;\;\;\;\;\; V_{OUT(total)} &= V_{OUT(d)} + V_{OUT{(cm)}} \\ &= A_d V_d + A_{cm} V_{cm} \\ &= 98.04 \cdot (3-2.8) + 0.04 \cdot {{3+2.8} \over 2} = 19.724V \; ✘✘ \; \end{align} $$

To sum things up:

  • CMRR is defined and measured by \$ \text{CMRR} = {{A_d \cdot V_{cm}} \over V_{OUT(cm)}} \$ straightly for zero differential voltage (\$ V_d = 0 \$). And the relation \$ \text{CMRR} = {A_d \over A_{cm}} = {{A_d \cdot V_{cm}} \over V_{}} \$ again presumes zero differential voltage (\$ V_d = 0 \$).

  • When a signal composes of both differential and common mode voltage (\$ V_d \neq 0 \$), there is no straight forward way of calculating \$ V_{OUT} \$ in the form of an equation. And \$ V_{OUT} \neq A_d V_d + A_{cm} V_{cm} \$. Instead, by knowing your gain, you measure \$ V_{OUT(\text{offset})} \$ by adjusting the input voltages.

  • Often diff amp has unity gain and the gain is set at the buffer stage. \$ \; \text{CMRR} = {V_{cm} \over V_{OUT(cm)}} \$ only implies \$ A_d = 1 \$


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