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What exactly is offset of the amplifier? How do I measure it?

In my measure lecture, lecturer told me to measure the same voltage in right polarity and opposite polarity. It should be enough to compute offset, but I can't find how to do that.

I know the amplifier has gain \$100\$. And it's noninverting. Values were \$138\ mV\$ with right polarity and \$-152\ mV\$ in the opposite.

I've read, that ideal amplifier has zero Offset when input voltages are equal. So is it possible to say, that offset is the value of the amplifier, when inputs are equal? Than the offset in my case could be \$\frac{|138-152|}{2}\$?


If the scheme is somehow important: enter image description here

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  • 1
    \$\begingroup\$ What is right and opposite polarities? \$\endgroup\$ – jippie Mar 25 '14 at 17:45
  • \$\begingroup\$ Does this help Op Amp Input Offset Voltage \$\endgroup\$ – Warren Hill Mar 25 '14 at 17:55
  • \$\begingroup\$ I mean just changing possitive and negative electrode of the input voltage. The right is the one, when amplifier has positive output. Sorry for poor English. \$\endgroup\$ – user50222 Mar 25 '14 at 18:18
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Offset is (almost) always referred to the input. Your calculation is the output offset.

To find the offset voltage referred to the input, divide the output offset by the gain. In this case, -70\$\mu\$V.

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  • \$\begingroup\$ why is there minus? \$\endgroup\$ – user50222 Mar 25 '14 at 20:53
  • \$\begingroup\$ Because the amplifier, you said, has a gain of +100. If the amplifier had a gain of -100 (inverting) it would be +70uV. \$\endgroup\$ – Spehro Pefhany Mar 25 '14 at 21:18
  • \$\begingroup\$ I computed voltages on the input of the amplifier and on the output. I thought the offset is voltage on the input, if output is \$0\$. That would be \$70\ \mu V\$. If the input is \$0\$, output is \$-70\ \mu V\$. So which one it is by definition? \$\endgroup\$ – user50222 Mar 26 '14 at 12:01
  • \$\begingroup\$ The input offset voltage of an amplifier acts as a voltage source in series with your input. If the sign is + then it adds to the input. So if your input is 1mV and Vos is -100uV, it will give you the same output as if you applied 900uV. To get the equivalent of zero input (to a perfect amplifier) you'd have to apply -Vos. \$\endgroup\$ – Spehro Pefhany Mar 26 '14 at 12:06
  • \$\begingroup\$ I see you've added a schematic. Just to make it 100% clear, the input-referred Vos of the op-amp is like a voltage source in series with the non-inverting input of the op-amp. If you had divided the input down (say you put 100K in series and called that the input of the amplifier) then the input offset of the amplifier would differ from the input offset of the op-amp (it would be doubled). \$\endgroup\$ – Spehro Pefhany Mar 26 '14 at 12:36

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