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While researching different current measurement methods, I came across a particularly interesting circuit which has a very high input impedance and apparently reduces the effects of the input offset voltage caused by semiconductor mismatches in differential amplifiers. I was a bit skeptical so I decided to build one myself which -to my surprise- did work really well as an accurate differential amplifier (unless I did something wrong during the tests)). I didn't manage to find anything online about how exactly the input voltage followers of this circuit could achieve this, and I would appreciate it if someone were to clarify the exact process of offset nullification in this context.

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    \$\begingroup\$ That's not an instrumentation amplifier. That's a buffered differential amplifier. \$\endgroup\$ Commented Oct 2, 2021 at 23:51
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    \$\begingroup\$ yes but an instrumentation amplifier works the same way ( input stage unity gain) in common mode, or am i mistaken? \$\endgroup\$
    – A.H.Z
    Commented Oct 2, 2021 at 23:52
  • \$\begingroup\$ " reduces the effects of the input offset voltage ". Not true. Where did you hear that? What it does do is reduce the effect of bias current when measuring a source with high impedance. So, it fixes problems with the source, not problems within the amplifiers. \$\endgroup\$ Commented Oct 2, 2021 at 23:56
  • \$\begingroup\$ electricalvoice.com/… . here it is clearly stated that it does if you scroll down "1.It has very low DC offset". Although, now that you mention it , it is possible that they were referring to the instrumentation amplifier ICs themselves. \$\endgroup\$
    – A.H.Z
    Commented Oct 3, 2021 at 0:01
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    \$\begingroup\$ It does not reduce dc offset more than a differential amplifier. \$\endgroup\$ Commented Oct 3, 2021 at 0:40

2 Answers 2

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No, it doesn't reduce offset voltage. Because it has a high input impedance, it will reduce (eliminate) any offset effects from loading current.

The offset V in your circuit is the combination of the offsets of each amplifier plus the mismatch in the sets of resistors.

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  • \$\begingroup\$ understood, so input offset is not nullified at all then? \$\endgroup\$
    – A.H.Z
    Commented Oct 3, 2021 at 0:25
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    \$\begingroup\$ correct; it is not nullified. \$\endgroup\$
    – jp314
    Commented Oct 3, 2021 at 0:29
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As others have already mentioned, instrumentation amplifier by itself does not reduce DC offset when compared to standard differential amplifier. In your case you have buffered differential amplifier because the resistor gain network is missing to make it a "full" instrumentation amplifier (refer to Chapter 1 and Chapter 2 of this book for more details on how instrumentation amplifier architecture looks like: https://www.analog.com/media/en/training-seminars/design-handbooks/designers-guide-instrument-amps-complete.pdf ).

However, there are specialized operational amplifiers which serve exactly that purpose, and they usually come in two flavors, one being called the chopper amplifier, and the other is the autozero amplifier, so that are the terms you need to look for. Operational amplifiers with offset voltage correction are dominantly used in measuring DC or low frequency AC signals. For more details on how exactly offset nulling is done, you can refer to these two resources:

https://www.analog.com/media/en/training-seminars/tutorials/MT-055.pdf https://www.analog.com/media/en/analog-dialogue/volume-34/number-1/articles/demystifying-auto-zero-amplifiers-part-1.pdf

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