# What is meant by virtual ground in the op amp ?

#### priya

• Apr 26th, 2007

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.

#### Deepan

• Nov 11th, 2007

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.

#### ravindra sariya Profile Answers by ravindra sariya

• Apr 11th, 2011

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.

#### MONU S

• Aug 25th, 2011

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..

#### DemplerS Profile Answers by DemplerS

• Aug 25th, 2011

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.

#### ARITRA

• Sep 12th, 2011

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.

#### jaikishan Profile Answers by jaikishan

• Feb 16th, 2012

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

#### SONALI

• Feb 20th, 2012

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

#### shweta bhatt

• Dec 11th, 2012

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...

#### ali asif

• Jan 4th, 2013

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.

#### Samarth

• May 1st, 2014

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.

#### krishnakant

• Jul 30th, 2014

In differential amplifier Va-Vb=0 when Vb is grounded the node Va will be on zero potential the condition is called virtual ground.

#### osman

• Dec 23rd, 2014

if i may correct you its the input of the op-amp that should be ideally zero.

#### Amresh Mishra

• Mar 2nd, 2015

Since the ideal op-amp loop gain is infinite and input impedence is high so there is negligible or zero currents through input entered in op-amp so potential at both the terminal should be same ,hence it is called virtual ground or Rail splitter

#### LION

• Mar 16th, 2015

op-amp sensivity electronics component its take two inputs give to the one out put commenly used electronics circuits

#### Alwin

• Mar 17th, 2015

Actually opamp gain=output voltage/input voltage.For open loop amplifier, it is equal to infinity.In this condition,the denominator must be zero.that is voltage input=0, voltage input=v1-v2.

i.e.v1=v2,v1=0,then v2 must be in ground state for getting input voltage as zero and vice versa.if v1=0 then v2 is in virtual ground state and also v2=0,v1 is in virtual ground state.This is actually the concept of virtual ground.

#### Joe

• Oct 2nd, 2015

I found this whole discussion on a virtual ground amusing.
Only a mathematician could derive that the input has to be zero in an open-loop op-amp circuit from the equation for gain. No reference for where the equation came from.

First an op-amp does not have infinite gain, more likely less than one million, but that is also irrelevant.

Second an op-amp input for a bipolar op-amp is the base of a transistor.

So for an op-amp that is open-loop or no feedback the amount of gain is irrelevant since it does not impact the base of the transistor for either inputs.

The concept of virtual ground is based on an inverting op-amp circuit that has negative feedback. When the non-inverting (positive) input is connected to ground (0 volts) then the inverting input will be driven by the feedback summed with the input voltage toward the same voltage as the non-inverting input of the op-amp.

This concept of virtual ground implies a dual power supply configuration. Assuming both power supplies have equal magnitude but opposite polarities one could say this is a rail splitter or the center of the voltages is at zero. What does this have to do with ground, nothing, even if the voltage is the same?
So to continue this discussion further what happens when the same inverting op-amp circuit is powered by only a single power supply? Let’s replace the op-amp negative power supply terminal with ground instead of the negative supply. Now the op-amp is powered by a single supply.

For a practical application the op-amp non-inverting input must be changed to function properly so this input is now connected to the split rail voltage. In this case assuming that V+=5V then the split voltage is now at 2.5V for the single supply configuration versus 0V for the dual supply configuration.

This is the same circuit that has the op-amp powered differently. It is still going to drive the inverting input toward the non-inverting input via the negative feedback. In the single supply case the inverting input is being driven toward the split voltage of 2.5V.

Would you still call this point (non-inverting input) at virtual ground, probably not? It is still the same circuit functionally.
So the phrase “virtual ground” doesn’t mean anything other than when the non-inverting input is connected to ground then the inverting input via feedback will be driven to the same voltage as the ground reference since it is connected to the non-inverting input.

The phrase "virtual ground" probably was made up by a book author so he had more to write about.

Joe,
A Book Author

#### palguna

• Mar 30th, 2016

I am asking a question. @Deepan
of course not only @deepan all said one thing that in order not to saturate output op-amp maintains inverting input and not inverting input same voltage when it is in negative feedback. that is WHY it maintains, but nobody said HOW it forces both inverting and non-inverting voltages to be same.

#### Rohan Gupta

• Dec 22nd, 2016

The concept of the virtual ground rises from the fact that the input voltage at the inverting terminal of Op. Am. is forced to such a small value that, for all practical purposes it may be assumed to be zero. Hence, the summing point A is essentially at ground voltage and is referred to as Virtual Ground.

#### Abiy

• Jan 21st, 2017

Simply, operational amplifier the two input node voltages tend to have same value. If we connect feedback resistance to inverting op amp and connect non-inverting to the ground (Earth Ground) then the node for inverting are virtual ground (zero voltage) without physically connect to any ground. In other word, in Op amp the inputs node V- = V+ (always equal)
If V- = 0 (gnd) then
node @ V+ =0 virtual gnd)

#### soumya gupta

• Jul 11th, 2017

When the input nodes of an opamp are taken as ground in the negative feedback condition, then its known as virtual ground.

#### kato

• Jul 13th, 2017

When we apply zero volt or ground to the +terminal of an opamp the -terminal also has such low voltage because of high input impedance that is called as virtual ground since it is not ground itself  