# Operational amplifiers and virtual ground

Could you physically clarify why we use virtual ground in op amps? I have seen the argument that since A(open loop gain) tends to infinity, we need the difference in inputs to tend to zero(to maintain a finite output.) Now, this is confusing because aren't the inputs under our control? What would happen if I gave one very large positive and one very large negative input to the op amp?

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The term "virtual earth" is a reflection of the fact that if one of the inputs of an op amp, say $V_-$ is connected to the zero volt (earth) line then if the op amp is to work in the non-saturation region the potential difference between the two inputs is going to be very much less than the supply voltage.
So the voltage of the other input $V_+$ is going to be very close to zero volts - virtually at earth potential.

Say that an op amp has an open loop gain $A=10^5$ and the supply voltage is $\pm 10\,\rm V$ then the input-output characteristic of the op amp might look something like this.

You will see from that characteristic that as long as you restrict the potential difference between $V_+$ and $V_-$ to less that about $\pm 10^{-4}\,\rm V$ the output voltage will be proportional to the potential difference across the two inputs.

$V_{\rm out} = 10^5(V_+ - V_-)$

If you wanted to you could have a much larger potential difference across the two inputs but then the op amp output potential would be either $\pm10\,\rm V$ and that output voltage would not change if the input voltages were changed unless the potential difference between the two inputs dropped to below $10^{-4}\,\rm V$.

In my example $V_- = 0\,\rm V$ (earth) then $V_+$ cannot differ in potential by more than $10^{-4}\,\rm V$ ie it is virtually at earth potential.

You will note that as the open loop gain of an op amp increases the approximation that the two inputs are at the same potential gets better and better.

Now, this is confusing because aren't the inputs under our control? What would happen if I gave one very large positive and one very large negative input to the op amp?

If both inputs to the amp are, as you say, under your control, then you don't have negative feedback, and you don't have a virtual ground.

The "recipe" for attaining a virtual ground is high open-loop gain and negative feedback.

A "virtual ground" is a feature of an op-amp in an inverting configuration. It results in (almost) zero common mode voltage on the inputs (images from Wikipedia):

The typical non-inverting configuration has both a very small differential voltage and a (potentially) much larger common mode voltage on the inputs.

The changing (with input signal) common mode voltage can cause the distortion of the op-amp to be much higher when used in the non-inverting configuration (especially at low gains where the relative change in the common mode voltage is great), so the inverting with virtual ground is preferred in (for example) high performance audio applications. There are some things it cannot do (have a very high input impedance, for example), so there are plenty of cases where we use a non-inverting configuration.

Note that a fixed reference that is not ground can be just as good (for example, from a 2.500V bandgap reference).

An opamp is a differential amplifier with an infinite gain. This means that the voltage between the inputs will be infinitely amplified. Practically this simply means that the opamp will apply any gain necessary to keep the viltage between the inputs very close to zero.

In the virtual ground circuit or a rail divider, we need the point between two resistors to be at the ground potential. However, we cannot simply ground this point, because this would ground the output as well. So instead we use an opamp that actively watches the potential in this point and, if it ever so slightly changes from the same as the ground, adjusts the output just enough to compensate. In other words, the opamp allows to practically "ground" this point without the current leaking through it to the ground.

To answer your question, if we apply a large positive and large negative potentials to the opamp inputs, the opamp would fully open putting its full power supply to the output. This means that the opamp would be out of its working range, effectively creating a short in the circit. So yes, it is in our control to force the opamp out of the range, but there is no practical reason for doing this, because the opamp would not be doing anything useful this way.

• The op-amp only works as described after 'practically' if it has a negative feedback loop. – user207421 Oct 15 '17 at 20:15