In this document Vout of a non inverting op amp is given as follows:

enter image description here

But why \$\frac{V_{in}}{CMRR} \$ is also multiplied by the gain?

I would write it as:

\$ V_{out} = \big[1 + \frac{R2}{R1} \big] \big[V_{in} \big] + \big[\frac{V_{in}}{CMRR} \big] \$

Common mode input voltage should produce an output of \$\frac{V_{in}}{CMRR} \$ by definition but yet in the document it is multiplied by the gain.

  • \$\begingroup\$ CMRR is an input referred term; as such it is seen at the output as that offset multiplied by any gain. \$\endgroup\$ Apr 14, 2019 at 14:49
  • \$\begingroup\$ forum.allaboutcircuits.com/threads/… \$\endgroup\$
    – G36
    Apr 14, 2019 at 14:54
  • \$\begingroup\$ @PeterSmith I don’t understand there is CMRR when theres no feedback and theres different effect when theres feedback. Sad part is the text books do not elaborate on these. \$\endgroup\$
    – GNZ
    Apr 14, 2019 at 15:30
  • \$\begingroup\$ All books derive using open loop gain. No one uses opamp without feedback. I want to see the definition of CMRR when there is feedback and gain. \$\endgroup\$
    – GNZ
    Apr 14, 2019 at 15:36

1 Answer 1


The common-mode rejection ratio (CMRR), is defined as follows:

$$CMRR = \frac{A_D}{A_{CM}}$$


\$A_D\$ is a differential mode voltage gain (Open loop gain).

\$A_{CM}\$ common-mode voltage gain.

And we usually express the CMRR in decibels.

Notice that from CMRR definition we have:

$$A_{CM} = \frac{A_D}{CMRR}$$

and the output voltage caused by common-mode input voltage (\$V_{CM}\$) is equal to:

$$V_{OUT} = A_{CM} \cdot V_{CM} =\frac{A_D}{CMRR} \cdot V_{CM} = A_D \frac{V_{CM}}{CMRR} $$

AS you can see we can model CMRR as DC offset voltage at the input equal to \$\frac{V_{CM}}{CMRR}\$

enter image description here

And this is why Vin/CMRR is also multiplied by the gain.

And after you apply the negative feedback the differential and common-mode gain will drop, but the ratio between those two will stay the same and will be equal to CMRR. This is why we can apply the same principle to the amplifier with the feedback.

But this time the differential mode gain is in your example equal to (1 + R2/R1).

  • \$\begingroup\$ Im not sure I understand but afaik Ad is not always open loop gain in the definition of CMRR. I saw examples where they dont use open loop gain but just the differential gain. See here youtube.com/watch?v=p4ZPC8J4Nb8 Why do you insist CMRR always needs the value of “open loop” gain I dont get it. Many examples they dont even use Aol when calculating CMRR. \$\endgroup\$
    – GNZ
    Apr 14, 2019 at 18:21
  • \$\begingroup\$ I was trying to explain the general case here. The CMRR as one of an opamp non-idealities (the same as in the Analog document). But in the video, they have the ideal opamp and examine the circuit CMRR with an ideal opamp. \$\endgroup\$
    – G36
    Apr 14, 2019 at 18:36
  • \$\begingroup\$ I see I was mixed up between the CMRR of the amplifier itself and overall CMRR of the total circuit. \$\endgroup\$
    – GNZ
    Apr 15, 2019 at 14:51

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