# 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?

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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 –  Rocker 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.

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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.

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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.

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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.

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