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So I have spent days reading about specifics of op-amps, instrumentation amps, and ECG circuitry and it just isn't clicking for me.

As I understand it, the common mode signal gets amplified and added to the differential output voltage. I don't see how this will be a problem since all it will do is provide a DC shift.

There seem to be several ways of compensating for DC shifts so why is it so important to have a high CMRR?

For example, in the following ECG circuit we already have the RLD and the output DC feedback loop for DC shift correction, so CMRR shouldn't be that big of a concern.

Despite this ECGs seem to have a minimum CMRR standard of 80dB.

Can someone please help clear up these concepts a bit for me?

enter image description here

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    \$\begingroup\$ the common mode signal gets amplified and added to the differential output voltage That doesn't make sense when also to have a minimum CMMR standard of 80dB. So what is is it, amplification of common mode or rejection of common mode? The answer is that you want to reject the common mode. You want to amplify the differential mode signal as that's the signal you're interested in. \$\endgroup\$
    – Bimpelrekkie
    Commented Jun 9, 2021 at 20:53
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    \$\begingroup\$ Suppose you have a large commonmode signal that is not rejected. The signal you want is differential and small. Both are appear at the output of this circuit. The output is single ended so there's no way to treat commonmode signals any differently from differential mode signals. So both are summed and appear at the output. So you'd have a large signal (from the commonmode) that you don't want and also a small signal signal that you do want. In extreme cases the opamp can clip due to the large commonmode signal making it impossible to retrieve the wanted signal. \$\endgroup\$
    – Bimpelrekkie
    Commented Jun 9, 2021 at 20:57
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    \$\begingroup\$ in the following ECG circuit we already have the RLD and the output DC feedback loop for DC shift correction, so CMMR shouldn't be that big of a concern. Actually that DC feedback loop is part of the system which gives this circuit a high CMMR. Without that loop the opamp could clip due to it amplifying the (DC) CMM signal. By rejecting DC and CMM it will be possible to always amplify the differential signal despite large CMM signals. \$\endgroup\$
    – Bimpelrekkie
    Commented Jun 9, 2021 at 20:59
  • \$\begingroup\$ Thank you. This did clear up a few things in my head! \$\endgroup\$
    – needbrainscratched
    Commented Jun 10, 2021 at 5:42

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I don't see how this will be a problem since all it will do is provide a DC shift.

The human body forms significant capacitance with other conductors nearby. Mains frequency electrical noise would be of particular concern when measuring the signal generated by nerve conduction. DC can easily be blocked with capacitors. Low frequency AC noise cannot be filtered out so easily. Hence, the technique of attenuating common mode signals is vital for ameliorating this kind of noise.

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  • \$\begingroup\$ Oh I get it I think. You mean to say that a DC shift won't help with AC common mode right. That filtering has to be done by the instrumentation amplifier. \$\endgroup\$
    – needbrainscratched
    Commented Jun 10, 2021 at 5:41
  • \$\begingroup\$ @needbrainscratched Yes, a DC shift won't help with common mode AC noise. But a differential amplifier with a high CMRR will help get rid of AC noise to the extent that it is common mode. \$\endgroup\$
    – Math Keeps Me Busy
    Commented Jun 10, 2021 at 5:52

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