# Operational Amplifier Virtual ground

I know that an op-amp tries to hold the input voltages the same. So, for virtual ground we can connect + node to the ground and we're getting a 0V on the - node. If the op-amp tries to keep the inputs the same, why are we not getting a virtual ground on the + node if we connect - node to ground?

• That's not what an op-amp tries to do. That's what negative feedback forces it to do. It's much harder to achieve negative feedback if you've tied V-. – Oliver Charlesworth Apr 20 '14 at 12:02
• yeah, i meant opamp with posetive or negative feedback – PYPL Apr 20 '14 at 12:09

I know that Opamp trys to hold the input voltages the same

Not quite. The correct statement is

"When negative feedback is present, the voltage across the (ideal) op-amp input terminals is zero"

why we're not getting a virtual ground on the + node if we connect - node to ground?

In the typical configuration, the op-amp output is connected in some way to the inverting input and, for the ideal op-amp, the output voltage will be whatever it needs to be such that the inverting input voltage is the same as the non-inverting input voltage.

If one grounds the inverting input instead and connects the output in some way to the non-inverting input, it is true that mathematically, one can show that there is an output voltage that will make the non-inverting input voltage zero.

However, it is easy to show that this is an unstable situation - positive feedback - and that if the non-inverting input voltage is disturbed, the output voltage 'runs away', amplifying the disturbance rather than attenuating it as is the case with negative feedback.

Because for an opamp to mimimize the voltage over the two inputs, it needs to be in stable operation.

When making a virtual ground, that stability is guaranteed by connecting the - input to the output, guaranteeing that the output voltage equals the - input.

Then, if for some reason the voltage on the + input rises, the output will rise (due to the opamp's amplification). This is fed back to the - input, so the - input will rise as well, minimizing the difference between + and -.

If the inputs were flipped, the difference would not be minimized by amplification, but be increased--leading to instability, where the output of the opamp will either be the positive or the negative supply voltage, and there will be a large difference between the + and - inputs.

Without any doubt, applying negative feedback (non-inv. inpt grounded) the inverting input will NOT be at ground potential. Instead, it will be at a very tiny voltage (µ volt range) that can be calculated very easily:

Vin(-)=Vout/Ao with Ao being the open-loop gain of the opamp.

However, because for most of the applications (and corresponding calculations) we ASSUME an open-loop gain of infinity at the same time we ASSUME that the voltage at the inv. input node is zero (therefore: VIRTUAL ground, not a real ground). As mentioned, in reality this is NOT the case, however, the error due to this simplification is very small and can be accepted in most applications.

• You are not answering the OP's question just giving an example of how a virtual ground isn't really at ground potential. – Andy aka Apr 20 '14 at 15:35
• Andy - you are right. I didn´t read he question carefully enough. My answer: grounding the inv. input and using feedback to the non-inv. input does not allow operation in the linear range of the amplifier (pos. feedback). Thus, the output clamps to one of the supply rails. – LvW Apr 21 '14 at 8:01

When an OA has very high gain {>10e6} with negative feedback, the output forces the (-) input to match the (+) voltage. Thus the virtual null voltage between +,- inputs can be any voltage biased on (+) in the allowed CM range.

When the + side is connect to ground, we call this null voltage difference a virtual ground. This will remain true so long as the output is not saturated and there is negative feedback with no positive feedback. Grounding the (-) side shunts all the negative feedback thus this negative feedback condition is not satisfied.