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For simulation purposes, I know that one can setup a difference amplifier configuration and compensate for offsets for DC current by adding a voltage source on the inverting input, and adjusting it's value until output voltage is zero.

I tried the same thing for AC current, but it doesn't work. The offset is reduced a little bit, but it doesn't go below ~20uV RMS.

Basically I want to output a very small AC voltage from the differential amplifier, in the range of microvolts. Unfortunately, at that range the offset dominates the output.

How does one achieve 0V offset for AC current?

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    \$\begingroup\$ In over 20 years in analog electronics I've never ever heard someone mention offset for an AC current. That's because there is no such thing as AC offset. Offset is a DC phenomenon. Maybe what you mean is noise ? \$\endgroup\$ Oct 11, 2015 at 18:31
  • \$\begingroup\$ are you referring to the input offset voltage of an opamp? \$\endgroup\$
    – user16222
    Oct 11, 2015 at 18:51
  • \$\begingroup\$ Yes, basically I want to apply 1uV AC voltage, and get 1uV output, not the 22-23uV that I'm getting (I don't know from where: offset, noise, etc.) \$\endgroup\$
    – Hassaan
    Oct 11, 2015 at 19:11
  • \$\begingroup\$ Did you not understand my answer? It's addressing exactly this. Do you need a schematic? \$\endgroup\$
    – Fizz
    Oct 11, 2015 at 19:21

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Trivial solution: coupling capacitor on output. This assuming the DC offsets aren't so bad as to cause clipping when the AC signal is overimposed.

If you have more sophisticated needs like preserving high CMR (unclear from the question if you do), then read http://www.ti.com/lit/an/sboa003/sboa003.pdf

If you have high-level DC bias on the input, then removing it before amplification is preferable:

enter image description here

This is the simplest idea, using a passive network; the two coupling capacitors allow only AC in, roughly speaking. More accurately they (together with the resistors) form high-pass filters. More sophisticated solutions are in the aforementioned TI app-note.

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  • \$\begingroup\$ If you could add a schematic, it'd be very helpful, as I have not used coupling capacitors before in a simulation. \$\endgroup\$
    – Hassaan
    Oct 11, 2015 at 19:30
  • \$\begingroup\$ Before I do, you said in a comment "I want to apply 1uV AC voltage, and get 1uV output". Do you want to amplify/buffer the current? Otherwise what's the point of an amplifier in this case? Maybe you have typo in there? \$\endgroup\$
    – Fizz
    Oct 11, 2015 at 19:36
  • \$\begingroup\$ Actually, I need to measure the differential voltage as well as buffer the circuit, so I'm using an instrumentation amplifier. \$\endgroup\$
    – Hassaan
    Oct 11, 2015 at 19:39
  • \$\begingroup\$ By that I assume the input has a high-level DC bias, over which you have this 1uV signal. I'll add a schematic of the general idea; this AC-couples the inputs to remove the DC bias before it's amplified. I'll have sign off for the day after this . \$\endgroup\$
    – Fizz
    Oct 11, 2015 at 19:49

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