what is meant by virtual ground in the op amp ?

Ideally the output of op-amp should be zero. So for this to happen the 2 inputs must be same. Hence one of the input is considered to be low or at ground potential. But this input is not actually ground hence it is called as virtual ground.

 The op amp connected in a negative feedback configuration, that is the o/p connected to the inverting terminal, tried to produce the same voltage at the inverting node as we applied to the non-inverting terminal, no matter whatever be the o/p.  This happens because the opamp has a very high differential gain. so only way to not to clip or saturate the o/p in either positive or negative side is to make their both i/ps at the same potential. so when you apply the i/p to the inverting terminal and you connect the feedback to the inverting node, and your non-inverting terminal is grounded, then op amp tries to force the inverting node at the ground potential and whatever the difference between these two nodes are amplified.

in op amp the input resistance is very high due to this very small current (practically zero)  flow through the input terminal to op amp so this implies that at both input point which directly entering into the op amp must have same potential.  

In ideal op-amp gain(a) is infinity... a=Vo/Vi, which means Vo/Vi=infinity n then Vi=0 n as we know that V1-V2=0 in an op-amp,so V1=V2(V1 IS GROUNDED)..so,
V1=0(ACTUAL GROUND)
n V2=0(VIRTUAL GROUND)..

when one terminal is grounded the other terminal is assumed to be at ground potential n that is virtual ground concept of op-amp..

Lets keep it really simple. The first stage of an operational amplifier is the difference amplifier. It detects a difference between the inverting (-) and non-inverting (+) inputs of the op-amp. Observe that a closed loop, inverting amplifier's non-inverting input(+) is tied to ground. The potential at the non-inverting (+) input is therefore 0 volts. Because there is NO difference between inverting (-) and non-inverting (+) inputs, the inverting input (the other input) will also be at a ground potential of 0 volts . It is not tied to ground but has the same voltage potential as the non-inverting (+) input, so it is called a VIRTUAL GROUND. The inputs (- ) (+)should always be at the same potential in a closed loop operational amplifier.

An active virtual ground circuit is sometimes called a rail splitter. Such a circuit uses an op-amp or some other circuit element that has gain. Since an operational amplifier has very high open loop gain, the potential difference between its inputs tend to zero when a feedback network is implemented. To achieve a reasonable voltage at the output (and thus equilibrium in the system), the output supplies the inverting input (via the feedback network) with enough voltage to reduce the potential difference between the inputs to microvolts. The non-inverting (+) input of the operational amplifier is grounded; therefore, its inverting (-) input, although not connected to ground, will assume a similar potential, becoming a virtual ground.

because in the op AMP this uses th DIFFERENCE of input applied..

the input impedance is very high so there is no current flow from the terminal

AS ohms law


if theoretically r= infinite then

i=V/R
i~0;
n in ground there is 0A current

so it is called as virtual ground

Voltage at the non-inverting input terminal of an op-amp can be realistically assumed to be equal to the voltage at the inverting input terminal

Ideally the gain of opamp should be infinite. If I ground any one of the terminal of opamp intentionally say V1=OV, then as per ideal opamp having infinite gain.

gain=Vout/Vid

since, Vid=V1-V2
so according to formula of gain, V1-V2=Vout/gain
now gain = infinite
so on putting the value of gain in formula we have
V1-V2=0
means V1=V2
now we have V1=0V
so V2 will also become equal to zero or in a better way V2 is virtually becoming grounded without even any efforts to ground it.

This is known as virtual ground

Thanks...

ideally inputs of op amp are short and if we apply voltage on one terminal then its mean after resistance the input terminal of op amp is also virtually connected to other terminal which is grounded. DUE to its high input resistance.