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The definition of "input offset voltage" is the differential voltage which is required to apply between the two terminals of the op-amp such that the output of the op-amp will become zero.

What I want to learn is: Imagine we have an opamp buffer and we apply(at its non-inverting input) a very well known constant precise voltage 1.0000V for instance and we measure 1.0015V meaning that there is 1.5mV offset. What is this offset voltage called? And does it have any relation with "input offset voltage"?

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ The output is also shorted to ground in your picture \$\endgroup\$ – Andy aka Mar 17 at 12:48
  • \$\begingroup\$ I see thanks, I edited that part. \$\endgroup\$ – ty_1917 Mar 17 at 12:57
  • \$\begingroup\$ I don't think this circuit makes sense. You should see exactly 1 V in the output since it is shorted to the 1 V input. The op amp may try to act to change the output, but it can't do so because of the short. \$\endgroup\$ – swineone Mar 17 at 13:01
  • \$\begingroup\$ I removed the confusing parts so now looks more clear hopefully, \$\endgroup\$ – ty_1917 Mar 17 at 13:09
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The offset at the output could be called the 'output offset voltage'.

If the op-amp has non-zero input offset voltage (but is otherwise ideal) then the output offset voltage is equivalent to (and caused by) the input offset voltage for an op-amp configured as a follower, which you show.

However, if the op-amp has other nonidealities such as input bias/offset current, finite open-loop gain, finite CMRR, etc. then the output offset voltage will not be caused entirely by the input offset voltage but by a combination of these effects. However, in a practical case, if an op-amp has millivolts of input offset voltage, then this will probably be the dominant effect.

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  • \$\begingroup\$ Thanks but for example here: analog.com/media/en/technical-documentation/data-sheets/… it says “Guaranteed 25μV Max Offset Voltage” What is meant by offset voltage here? Does it mean if I use it as a unity buffer the difference between the input and output will be less than 25μV? \$\endgroup\$ – ty_1917 Mar 17 at 16:13
  • \$\begingroup\$ It means the contribution to the output offset voltage due to the input offset voltage will be less than 25μV. However, like I said, other things also contribute to the output offset voltage. Think about what would happen if you had 0 input offset voltage but the dc open-loop gain was only 1000. Then the output would be 1000/1001 = 0.999V, so you'd still be off by a millivolt. Granted that a dc gain of 1000 would be ridiculously low for any op-amp but the point still stands that you need to take ALL sources of error into account if you want to minimize output offset. \$\endgroup\$ – pr871 Mar 17 at 16:58
  • \$\begingroup\$ I already gave you an example calculation of an op-amp having a dc gain of 1000. This yields an output of 0.999V for an input of 1V in the follower configuration (considering no other sources of error). If you don't know how I came to that conclusion, then no offence but I think you're lacking fundamental knowledge required to understand the answer to your question. \$\endgroup\$ – pr871 Mar 17 at 17:12
  • \$\begingroup\$ Type "op amp error budget analysis" into your favorite search engine for plenty of examples. \$\endgroup\$ – pr871 Mar 17 at 17:21
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The input offset voltage would be multiplied by the gain in the circuit you show. There is no feedback at all (ideally). In a model of an op-amp with gain A , infinite CMRR and offset voltage Vos, the output would be Vout = A*Vos. A is usually quite large, so the op-amp output will be railed to one side or the other, depending on the sign of the offset voltage.

If you model common mode rejection there's another term that changes the output voltage from the common mode voltage, but that gain could be positive or negative and even if it happens to be negative, it may not be able to balance out the input offset voltage before the output hits the rails, so the output is unpredictable, but likely railed just as above.

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  • \$\begingroup\$ No one uses opamp without feedback so "input offset voltage" without any extra formula makes no use to me at the moment. How about the last paragraph of the question? \$\endgroup\$ – ty_1917 Mar 17 at 13:14
  • \$\begingroup\$ The lack of (predictable) feedback is due to your circuit. \$\endgroup\$ – Spehro Pefhany Mar 17 at 13:18

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