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In the datasheets of differential input amplifiers there is a figure of merit called CMRR. This is normally experimentally measured at a particular frequency like 60Hz and presented in the datasheet along with the input impedance of the amplifier. But if our source is not balanced, i.e if the source has high output impedance the overall CMRR will degrade.

Below first figure below shows just a madeup example where a datasheet provides CMRR of such an amplifier at 60Hz as 80dB and input impdace as 1Meg Ohm:

enter image description here

Regarding the same amplifier in first figure, if we dont have opportunity to experiment, in the presence of a source imbalance where Rs is 1k as in second figure above, how can we very rougly estimate the new degraded overall CMRR? (Again for 60Hz Vcm and neglecting parasitic capacitances)

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Your model is inaccurate.

Because CM noise has a much higher impedance than the wire, this can be modeled as a weak transformer coupling by mutual inductance ( B field) or stray capacitance (E field).

It must be given a value then also the cable impedance and mismatch which tends to dominate the results for CMRR. Also RLC values of every wire and interference source must be estimated.

But if you were to direct connect a CM voltage of 0 Ohms then yes, the induced differential voltage from CM voltage is reduced by 60dB. But that does not apply to 230V.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Im asking only the effect of Rs neglegting any other effects. \$\endgroup\$ – pnatk Jan 11 '19 at 20:02
  • \$\begingroup\$ Yes that would apply to a common emitter source of 1K with 0 ohm ripple on supply at source with ideal ground and separate supplies. \$\endgroup\$ – Tony Stewart EE75 Jan 11 '19 at 20:03
  • \$\begingroup\$ How did you calculate 60dB? Thats what I really ask about? What is the formula you used? \$\endgroup\$ – pnatk Jan 11 '19 at 20:03
  • \$\begingroup\$ 20 log R ratios \$\endgroup\$ – Tony Stewart EE75 Jan 11 '19 at 20:04
  • \$\begingroup\$ Is the new CMRR 20dB? Just the effect of Rs I mean \$\endgroup\$ – pnatk Jan 11 '19 at 20:06
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Some op amp and many instrumentation amp or diff amp data sheets list more than one type of input impedance (I am looking at AD8067). There is a "common mode" input impedance in addition to the "differential" input impedance. The common mode impedance is the impedance from each input to ground, and this is the value you can use to see the effect on the voltage at the input terminal for inputs with different source impedances. Since each input has a different impedance to ground, you can easily estimate the effect of applying the same voltage on both terminals from different source impedances.

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  • \$\begingroup\$ Interesting. If the amplifier has perfect input output isolation does that mean the common mode input impedance is very high hence the CMRR? \$\endgroup\$ – pnatk Jan 11 '19 at 20:36
  • \$\begingroup\$ I'm not sure about this. I think that devices listing common mode impedance are those that are used for what I guess you are trying to do: build an amplifier that is tolerant of varying input impedance. Vanilla op amps don't usually list it. I think it only means that the input stage has a capacitance and resistance term to ground, regardless of the output, so that the actual voltage at the input terminal will vary with the source impedance. \$\endgroup\$ – John Birckhead Jan 11 '19 at 21:07

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