The definition of CMRR is 20×log10(Aol/Acm) where Aol is the open loop gain and Acm is the common mode gain.

If we have a buffer with 20dB CMRR opamp and if we apply 1V differential input along with 1V common mode input as shown in my hand drawn example:

enter image description here

In this case we dont know the open loop gain but we know the CMRR. What will be the total output voltage? In other words how to quantify the contribution of the 1V common mode input to the output voltage. We are using feedback here so would that effect CMRR?


Look at a data acquisition input parameters:

enter image description here

CMRR is given as 92dB but Aol never mentioned. Does that mean we cannot quantify Acm?

  • \$\begingroup\$ What is the differential gain than (Aol)? \$\endgroup\$ – G36 Apr 13 '19 at 15:24
  • \$\begingroup\$ Im confused because there is feedback and the rest \$\endgroup\$ – pnatk Apr 13 '19 at 15:25
  • \$\begingroup\$ forum.allaboutcircuits.com/threads/… \$\endgroup\$ – G36 Apr 13 '19 at 15:25
  • \$\begingroup\$ Is my buffer diagram with differential amplifier wrong? \$\endgroup\$ – pnatk Apr 13 '19 at 15:34
  • \$\begingroup\$ Yes, it is wrong. Your buffer has only one input (noninverting). \$\endgroup\$ – G36 Apr 13 '19 at 15:36

CMRR is at its most basic the amount by which a circuit rejects common mode input relative a differential input of the same magnitude.

Consider your DAQ card, it has a differential input which drives some doings (fet input instrumentation opamps and shit) that eventually get measured by an ADC and fed to some software that knows the scale factors so it can produce a reading in volts. You have no knowledge of the internal scale factors, on some settings it is amplifying before it hits the ADC, some settings it is probably attenuating, it don't matter. It has (in some settings) gain, but you don't know what it is, you just know what full scale range is in terms of volts per sample value and maximum input.

Now we know (providing we believe the datasheet) that at 60Hz on the +-10V scale we get a CMRR of 92dB, which means that common mode inputs (Providing they remain within the allowed range) at 60Hz will be attenuated by 92dB before they get summed in with the differential mode signal you are measuring. 92dB is (in voltage terms) a factor of about 2.5*10^-5, so 1V or common mode input will appear as about 2.5*10^-5V of noise added to the differential mode signal which in that mode has a full scale range of +-10V.

One thing about that card is that the input impedance is apparently 100G Ohms, I don't believe it, the capacitive loading will screw that massively at the top of the small signal bandwidth! But lets roll with it, it might be sufficiently true close to DC, this means that reasonably small source impedance imbalances will not seriously degrade CMRR. However source impedance may if large enough convert that couple of hundred pA of bias and offset into an annoying voltage, doubly so if it changes with temperature, metrology is a bitch like that.

| improve this answer | |
  • \$\begingroup\$ But why does your first sentence not apply to the original question? Given a diff amplifier CMRR as 20dB can a 1V common mode voltage’s effect on output calculated? If we dont know open loop gain Aol, G36 says in comments we cannot calculate it. \$\endgroup\$ – pnatk Apr 14 '19 at 0:09
  • \$\begingroup\$ Anyway dont bother I dont think I will ever understand this \$\endgroup\$ – pnatk Apr 14 '19 at 0:13
  • \$\begingroup\$ Yes, obviously, it will add a signal 20dB below 1V to the output! So 100mV of noise with a 1V CM noise input.... With opamps you can just assume Aol is 'Large', and the maths works (Particularly at low frequency). Nobody uses opamps open loop, there is always a feedback network wrapped around them to make them reasonably behaved, and this is usually what dominates the performance. Seriously only undergraduate textbooks sweat open loop gain, everyone else just goes 'meh, talk to me about Gain Bandwidth Product, far more useful for actual design'. \$\endgroup\$ – Dan Mills Apr 14 '19 at 0:25
  • \$\begingroup\$ I see thanks the comments under the question made me confused now more clear. \$\endgroup\$ – pnatk Apr 14 '19 at 1:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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