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What's the best equipment for testing the common mode rejection of an op-amp? Like can I use an ordinary signal generator and just connect the one output to both inputs of the op-amp?

Or is there a special testing procedure or equipment for this?

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4 Answers 4

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Depending upon the op-amp, connecting the inverting and non-inverting inputs together may, or may not, cause the op-amp to saturate. This is because the offset voltage may be high enough so that when multiplied by the open-loop gain, their product exceeds the output voltage limits.

If the op-amp in question does have significantly high offset voltage, then to test common mode rejection ratio, one will need a circuit that enables zeroing out the offset voltage before applying a common mode signal.

One technique is described in this Analog Devices tutorial.

enter image description here

Unfortunately it requires ultra-high precision resistors. :-(

An alternative test jig is shown below

enter image description here

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  • \$\begingroup\$ I have no idea how you have set things up, and you should not change your question because it already has answers. You should ask a new question, and be as specific as possible explaining your set up. \$\endgroup\$ Commented May 20 at 4:14
  • \$\begingroup\$ The comments section is not for follow up questions to your original question. Stack Exchange is not like some other forums where there are long back and forth discussions. If you have a follow up question to your original question, ask it as a new question. \$\endgroup\$ Commented May 20 at 5:00
  • \$\begingroup\$ Ok. On topic. Why did you describe about shorting the In+ and In-. I was asking best equipment for CMR tests. Why didn't you mention about signal generator? Anyway. here is a separate thread for the a 3rd actual test electronics.stackexchange.com/questions/713780/… \$\endgroup\$
    – Jtl
    Commented May 20 at 5:47
  • \$\begingroup\$ @Jtl You wrote "can I use an ordinary signal generator and just connect the one output to both inputs of the op-amp?" Connecting a signal to both inputs together will give both inputs the same voltage. This will only work if the offset voltage is small enough not to saturate the op-amp. \$\endgroup\$ Commented May 20 at 5:51
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Made with microcap v12.

This seems to be ok. CMRR = ~ 90 dB.

enter image description here

Here is an example using circuit n°2 proposed by "Analog Devices" ...

Note that I can't verify what the simulation "says" (?). Seems that CMRR is too "high".

enter image description here

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If you run open-loop then even the offset voltage might be enough to saturate the output. So I think in general the obvious approach might not work.

I think I have seen test circuits in some op amp spec sheets.

It's not clear why you want to do this, unless you are an op amp designer.

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    \$\begingroup\$ Their previous question may be related, electronics.stackexchange.com/questions/713583/… but as this context wasn't included, I have to assume the independent question is intended as-is. \$\endgroup\$ Commented May 20 at 0:03
  • \$\begingroup\$ Yes i want to test both in op amps and the 2 ADC. so for ADC test, it is easier bec i can just use a signal generator and connect the output to both ADC? \$\endgroup\$
    – Jtl
    Commented May 20 at 1:49
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A half-answer -- just to illustrate the error here,

and just connect the one output to both inputs of the op-amp?

Well, let's give it a try:

schematic

simulate this circuit – Schematic created using CircuitLab

V(OUT) is saturated to +V, so gain is zero (actually exactly zero in this case; but CircuitLab makes fairly gross approximations, and it's never quite zero in practice; neither will a real TL081 saturate to the supply).

What is Vos doing there? Obviously if you feed 3mV into a 10^5 gain amplifier, it's going to saturate, that's self-sabotage, right? Well that's the thing. You can't get rid of it, that one is part of the component itself. And in general, you don't know what value it is, + or -, within the rated range. TL081 isn't a precision type, so its input offset varies by a few mV. Inconsequential for the audio applications it's most often used with, but fatal to a circuit like this.

And even if you had a precision type with Vos very close to zero, 10µV say, it still doesn't take much to ruin this measurement -- op-amps have gain in the millions quite regularly. The sheer voltage drop along the wires to V1, due to temperature fluctuations, will show up in such a setup!

There is a perhaps subtle quirk to your question:

If you had asked "..of the amplifier?", that's a very different question! An amplifier, generally speaking, is a fully realized module, with well defined gain from input to output, voltage/current range, power level, CMRR if differential, etc. Basically everything an operational amplifier has, except the gain: an op-amp is specifically designed for maximum gain, so that it can be tuned as desired in any application. It's what makes them "operational". But so is its downfall here: without the feedback path to define that gain, this circuit is impossible to use.

Hence the resistors to define a feedback path in Math Keeps Me Busy's answer. With a differential amplifier circuit, we can set differential and common-mode gains independently; by setting CM to zero with very well-matched resistors, and shorting the inputs together, we measure CM purely of the amp itself. Or, the alternative with a servo amplifier to keep it stable, while varying the supply voltage instead. In either case, the overall gain is well-defined, and the measurement becomes feasible.

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