How this measurement circuit works?

An op-amp doesn't have any ground terminal. It s only sense of what is ground is based on its supply pins. Changing the difference between the center point between the supplies and the common mode input voltage is the same whether the inputs or the supply pins are moved. Moving the supply pins makes doing the measurement easier.

Right, but this is yet another of those CMR test circuits that really measures the open loop gain and/or output impedance. In order for the output to stay still, the amp has to slew all its internal nodes to compensate for the change in the output loading. (The load should be returned to the junction of the power supplies.) At least they aren't measuring the output barefoot, which is the usual mistake--they're trying to take out v_icm.

Also, it could be quite exciting making the two copies of V_icm match well enough to measure the CMR of a good op amp. Pease's circuit is much better. (It's from a 1990 Electronic Design column...anybody have it in softcopy?)

Cheers,

Phil Hobbs

Thanks for the help...

Ok, i understand now why Vicm is connected to Vdd and Vss, but i don't understand why there is another Vicm connected to Vout and opamp not under test

Yes, I hope that in explaining the intent I didn't imply that the circuit was a good one.

I didn't mean to suggest that the circuit was a good one. The best circuit is to power the op-amp from some batteries and build the whole op-amp circuit as though it had the common point of the batteries as the ground. You then can drive this common point with a signal generator and opserve the output.

That is so the output of the DUT remains at the centre of the two power supplies, i.e. its output does not change relative to its own power. If that wasn't done the open loop gain would come into the equation as the input would have to change to cause that output change.

kevin

Check the original text: CMOS Analog Circuit Design, P.E.Allen, copyright 2005.

(paste the WHOLE line into your web) browser

Pages 6.6-8, 6.6-9 give an explanation using the drawing given. There's a proof of how the circuit works (CMRR=1000*Vicm/Vos). The same circuit can be used to determine PSRR.

Paul G.