I'm wondering what possible different methods there are for measuring the input offset voltage of an Opamp.

So far, all I have been able to come up with is the circuit I've attached, measuring the output voltage and then dividing by the gain when both inputs are grounded.
The other method is to ground one input and apply a voltage at the other input and change it in very small increments until the output switches or begins to move to 0.

I understand that neither of these may be very accurate but would like to know some other, possibly better but also simple methods.

 

hi James
Welcome to AAC.
Check this link out.
Eric
https://www.edn.com/electronics-blo...ffset-voltage--PSRR--CMRR--and-open-loop-gain

More info:
https://www.google.co.uk/search?q=+...x-b&gfe_rd=cr&dcr=0&ei=nTOqWszZDemA8QfQ5b_ACg

 

So far, all I have been able to come up with is the circuit I've attached, measuring the output voltage and then dividing by the gain when both inputs are grounded.

You've got the (+) and (-) inputs of the op amp reversed; as is, the feedback is positive instead of negative. Swap the inputs, and your circuit is fine; it's the same circuit topology as this, from the OP177 data sheet:


It should work just fine.

 

Can´t you just connect it with whatever feedback so that it is in linear mode and then measure the voltage between + and - inputs? Of course with high impedance meter.

 

Can´t you just connect it with whatever feedback so that it is in linear mode and then measure the voltage between + and - inputs?

Sure, if you've got a high-quality voltmeter that can accurately measure voltages that small, and if you take great care to avoid thermocouple effects in the meter's input leads. But why bother with that-- just let the op amp amplify its own offset voltage, as proposed, to get a larger voltage that's more easily measured.

That same circuit, BTW, is also useful for evaluating an op amp's input voltage noise.

 

The circuit in #3 is subject to error due to input bias current. It is important to use a low value resistor from the inverting input to ground since bias current will add a voltage equal to the input bias current times [that resistance in parallel with the feedback resistance].

If you are dealing with an amp with high input bias current, both the II and NII should see the same resistance - i.e. use a resistor equal to the parallel combo of the gain setting resistors from the NII to ground. The match need not be precise. Here just using another 50 ohm resistor would be fine. I think you would be hard pressed to find an op amp with bias current so high that the voltage across 50 ohms would be significant.

Don't forget that the amp must have bipolar power supplies of sufficient magnitude to allow the amp to work properly with zero volts input and 4000 times the offset voltage at the output. Don't assume a rail-to-rail input amp can be operated with a unipolar power supply. If the amp has an input offset voltage of 2 mV and you take a gain of 4000, you'll need more than 8 volts for the supply - on the appropriate side. For a low voltage amp you may have to take substantially lower gain.

 

The circuit in #3 is subject to error due to input bias current.

It would be, if the input bias current were a lot larger than it is; for the example given, an OP177, the maximum input bias current is listed as 2.8 nA, which across 50 Ω will cause a voltage drop of just under 150 nV. For all but the very best precision op amps, I doubt that is a significant error. For op amps with very large input bias currents, such as an LT1028, the circuit given in that part's data sheet would be a better choice:

Don't forget that the amp must have bipolar power supplies of sufficient magnitude to allow the amp to work properly with zero volts input and 4000 times the offset voltage at the output.

That's only true for the component values given. The high voltage gain in that circuit was necessitated (I assume) by the OP177's extremely low input offset voltage (60 μV max. for the OP177G). For other op amps, with higher input offset, the 200 kΩ resistor can be reduced correspondingly, thus lowering the gain.

 

Some ref material -

http://www.analog.com/en/analog-dialogue/articles/simple-op-amp-measurements.html

http://www.analog.com/media/en/training-seminars/tutorials/MT-037.pdf

http://www.analog.com/media/en/training-seminars/tutorials/MT-038.pdf


Regards, Dana.