When we model an operational amplifier, we say that the output voltage \$V_{out}\$ is proportional to the difference between the inputs \$V_\pm\$, \$V_{out}=A(V_+-V_-)\$. I've read that this voltage is always with respect to ground.

enter image description here

However, when we add a double supply to the op-amp, how is ground defined? Does the circuit take it to be \$V_-\$? Or maybe it is \$(V_+-V_-)/2\$?

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    \$\begingroup\$ One possible (but unsatisfying) answer is: it doesn't matter. Since op amps are always used with negative feedback (unless you're doing something quite odd), the feedback cancels out any possible dependency on the supply voltages. \$\endgroup\$ – Hearth Feb 14 at 1:44
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    \$\begingroup\$ I like your suggestion that its \$ {(V_+ - V_-)} \over {2} \$. Another thought: output current goes to zero: an "imperfect" opamp might settle at a strange output voltage, but I'd hope for half-way between the supply voltages. \$\endgroup\$ – glen_geek Feb 14 at 2:12
  • \$\begingroup\$ I think this is a great question. Unfortunately, I can't give an answer with any confidence for the open-loop case. \$\endgroup\$ – Elliot Alderson Feb 14 at 19:06

"Ground" in this context is really a misnomer. It's only a reference. For example, take this circuit that you probably know:


simulate this circuit – Schematic created using CircuitLab

and connect the "ground" to another control instead:


simulate this circuit

You can think of it as being the exact same circuit with the exact same behavior, provided that the reference control stays constant. It simply centers on this new reference, and there's now no need for an explicit ground.

The actual rules for opamps are:

  1. The output goes as far as it needs to, to keep the two inputs equal.
  2. If the "+" input is higher than the "-" input, it goes up; if lower, it goes down.
  3. If the output hits one or the other supply rail, it stops there.
    (actually a volt or two short unless it's explicitly a rail-to-rail design)

You'll notice that there's no mention of "ground" in any of that. Only comparing the two inputs to each other and the output to the supply rails. That's it!

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    \$\begingroup\$ For the benefit of newcomers, rule #1 only applies in circuits where there is negative feedback. \$\endgroup\$ – Elliot Alderson Feb 14 at 12:59
  • \$\begingroup\$ I don't think you answered the question. Suppose the inverting and noninverting inputs have precisely the same voltage. What is the voltage at the output, with respect to the two supply voltages? We can forget about "ground" and still answer that question. \$\endgroup\$ – Elliot Alderson Feb 14 at 13:02
  • \$\begingroup\$ @ElliotAlderson: Then it stays where it is. I think that fits within rule #1. If they're already equal, then there's no need for the output to change. \$\endgroup\$ – AaronD Feb 14 at 18:21
  • \$\begingroup\$ I know that this is kind of a "magical black box" sort of explanation, and some people have a problem with that. But the real-world approximation to it is quite good, thanks to an insane amount of internal gain. Enough that it hardly matters where that gain is referenced to. \$\endgroup\$ – AaronD Feb 14 at 18:24
  • \$\begingroup\$ As for answering the question, I'm taking a sideways approach to it by saying, basically, that there is no ground as such. Logically, then, if there is no ground, then what about the OP's theory? It needs to be rethought, and I'm allowing the OP to use that thought to answer his own question. \$\endgroup\$ – AaronD Feb 14 at 18:30

Your question is a good one. Since the differential gain of an opamp is typically very large 100 dB and up, it is often not a detail that is considered.

Take for example the following simulation of a highly idealized opamp, where each opamp has a different output reference potential:

enter image description here

The output is nearly identical for the three cases, and for basic design, the output reference potential can be ignored.

In cases where you would use the transfer function of the opamp \$A(s)\$ for stability and frequency analysis its also a mute point. Good design practices ensure all three reference potentials become one single AC ground due to decoupling/bypass capacitors.

  • \$\begingroup\$ You have added negative feedback to all of your examples, but the question is asking about the open-loop case. I think you missed the point. \$\endgroup\$ – Elliot Alderson Feb 14 at 19:09

Look at how input offset voltage is defined for that particular op-amp. Remember that's the input differential voltage (at a particular common mode voltage) that is required to drive the output to exactly zero.

Use the same ground reference.

In almost all cases, the maximum offset voltage can drive the output quite a bit more than between the supply rails so it doesn't much matter.


The output voltage as you say is typically approximated by A*(Vp-Vn). However a more accurate equation adds Voffset so Vout=A*(Vp-Vn+Voffset). The offset voltage when multiplied by the open loop gain A gives a zero-input output voltage that may exceed the supply rails. This fact makes your question meaningless in real life. What really happens is that a more positive differential input voltage will cause the output to swing harder in a positive direction. More negative input voltages will cause the output to pull harder in the negative direction. If you don't use negative feedback then the output voltage with both inputs tied together will vary randomly from part to part. All practical feedback circuits are setup to operate properly even with this uncertainty. You can analyze this by using your ideal model with a small voltage added in series with one of the inputs. Or alternatively, put a much bigger battery (A times larger) in series with the output. You will find that the error from either source is reduced by a factor equal to the open loop gain divided by the closed loop gain. Since the gain of most opamps is around a million, these non-idealities get reduced to output errors of only a few microvolts. That's why nearly all opamp analysis ignores these effects. For really precise measurements, though, the effect is real and may require extra circuitry to null the offset voltage.


You can use a Capacitor as your load (CL) to "read" your output voltage (Vout). This capacitance is connected to the ground(gnd), so your Vout measurement will be with respect to the gnd.

  • \$\begingroup\$ We generally try to avoid capacitive loads on op-amps to prevent overloads and instability. \$\endgroup\$ – Transistor Feb 15 at 16:57

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