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I am taking the output of the shown differential amplifier to the microcontroller (1, 2 and 3).

Each opamp here is in a different PCB but all are being fed by the same supply (same ground) and the same current source as in the image.

A colleague has advised being cautious about the common mode issue in this design.

I am not able to understand what is the common mode issue from textbooks, tutorials, etc.

Simulation of the circuit shows different outputs for the same current which is confusing but is still not helpful in understanding.

Would anyone please help explain this issue?

enter image description here

enter image description here

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  • \$\begingroup\$ The current though R4, R9 and R14 may create issues as you move up the ladder and have higher voltage with respect to ground. Have you tried to simulate it? \$\endgroup\$
    – winny
    Jul 26, 2022 at 11:29
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    \$\begingroup\$ It would make sense to ask your colleague. Then it would make sense to use a simulator. Then finally, after doing these things and getting nowhere, it would make sense to ask on this site. \$\endgroup\$
    – Andy aka
    Jul 26, 2022 at 11:34
  • \$\begingroup\$ When the grounds are not the same and there is stray coupling from EMI, you have CM noise and also DM noise inside the loop area....export link from Falstad \$\endgroup\$ Jul 26, 2022 at 11:57

3 Answers 3

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Remember that each op-amp will try to keep the inverting and non-inverting inputs at same voltage.

Because the differential inputs have a current path to ground and output via resistors, each differential input draws current.

If each op-amp draws current, then each measurement resistor has a different current flowing through it.

So each op-amp measures differet current and will output diferent voltage.

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As somebody already mentioned the issue is that the "absolute" voltages with reference to ground, on the input pins of the diff-amps, are not equal, with OP1 having more Common-Mode voltage than say OP3. Your colleague was merely pointing out that your diff-amps outputs will contain some level of error that is a function of the common-mode voltage on the input. This may or may not be a concern for you so work out if it will be. Common-Mode-Rejection (CMR) is the measure of the level of influence the common-mode voltage will have on the differential output voltage, and the better (higher values) the less influence and less to worry about. An ideal diff-amp will have infinite common-mode rejection, but since you don't use ideal amps, the higher the CMR the less your outputs will be affected by common-mode input voltages. So work out if it's a concern for you by estimating the maximum possible error in the various outputs, and then fix it by looking at high CMR valued amps, or some other way that would suit your needs, budget etc. good luck ...

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  • \$\begingroup\$ Look the datasheet for not only rejection, but also common-mode-Range. There are often these strange-looking eyeball plots that show you the allowable CMR (common mode range) for various input voltage levels, gain setting, and power supply levels. Study those. \$\endgroup\$ Jul 26, 2022 at 12:51
  • \$\begingroup\$ Yes indeed !! Both rejection and range will be relevant... \$\endgroup\$
    – citizen
    Jul 27, 2022 at 15:13
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Take this one step at a time.

If the current is 20 mA, the total voltage across each 200 ohm resistor will be about 4 volts. The upper resistor will have 12 and 8 volts at each end.The lower end will undergo a 50% reduction due to the 82k resistors, so the + input to the upper op amp will be at 4 volts. It will be the job of the op amp the adjust the output of the op amp to drive the - input to 4 volts as well. This "common" voltage (as in, "shared" voltage) is what is being discussed.

Your circuit diagram did not list your power supply voltages. Let's assume, as a worst case, that you are using 3.3 volt supplies. Then 4 volts at the + input is almost certainly going to give you problems. Even if you are using a 5 volt supply, a lot of older op amps will not work correctly at this level.

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  • \$\begingroup\$ The current here is shown as 4 mA, so the voltage on each 200 ohm resistor will be 800 mV. The highest common mode voltage will be 2.4V. Depending on the specific op-amp, there will be offset voltage and input bias current that might induce errors of several mV, but not as high as shown. It LOOKS like the power supply rails may be reversed, which would explain the errors. \$\endgroup\$
    – PStechPaul
    Jul 26, 2022 at 19:54

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