# opamp virtual ground

Discussion in 'General Electronics Chat' started by cssc, Oct 19, 2014.

1. ### cssc Thread Starter New Member

Oct 19, 2014
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1
hello...can someone help me with this.?

I learnt that op amp has an infinite input impedance...
but it is said in virtual ground concept that both the inverting and non inverting terminals are at the same potentials...
both the above statements are contradictory....
if both terminals are at same potential(which may imply a short circuit of-course not exactly).,then how can the input impedance be infinite..?

2. ### Papabravo Expert

Feb 24, 2006
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An ideal opamp has an infinite input impedance. A real opamp has a large but finite input impedance. The two input terminals are at the same potential in certain configurations where the output is fedback to one of the inputs. In the open loop case the two terminals can be at different potentials. Any more than a couple of millivolts and the output will be at the rails. Show us a configuration, by means of a schematic diagram, and we'll help explain it to you.

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3. ### Lestraveled Well-Known Member

May 19, 2014
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The two statements are not contradictory. The input resistance of modern op-amps is extremely high, not infinite. The way the external components configure a op-amps function, make the inputs operate at the same voltage.

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4. ### cssc Thread Starter New Member

Oct 19, 2014
26
1
okk...
so.,does that mean, the two terminals of op amps with a closed loop are shortcircuited.....?

5. ### Papabravo Expert

Feb 24, 2006
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No, absolutely not. In the inverting amplifier configuration the output moves in such a way as to make the inverting input equal to the non inverting input. So when the input voltage changes and tries to move the non-inverting input the output moves to make it equal to the non-inverting input. Is that clear? Well it would be if you came up with a picture. I don't seem to have one handy.

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6. ### MrChips Moderator

Oct 2, 2009
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The input terminals are definitely not shorted.

An ideal opamp has infinite gain. This means that if the two input pins are not at the same voltage, the opamp's output pin will hit the power rails, i.e. the output will saturate one way or the other.

It is the purpose of the feed back resistors to bring the input pins to the same voltage.

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7. ### crutschow Expert

Mar 14, 2008
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Another explanation:
Any small voltage difference between the two op amp differential inputs causes a large change in the output voltage due to the high op amp gain.
Thus negative feedback from the output to the inverting (-) input will cause the output voltage to change until the inverting input voltage is very near the value of the non-inverting (+) input (within less than a mV for typical op amp gains).
If the non-inverting input is connected to ground (as, for example, in the common inverting op amp amplifier configuration) then the inverting input is very near the ground voltage also (or 0V). That's why the inverting junction is called a virtual ground point.
Thus, with negative feedback, the two input voltages stay very close to each other, even if their respective input impedances are infinite.

Make sense?

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8. ### cssc Thread Starter New Member

Oct 19, 2014
26
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So., the two input terminals are at the same potential in closed loop., even though the impedance between them is very high (ideally infinite)....
Right..???
And, what about the voltages at the two terminals in open loop...? Are they equal...?

9. ### MikeML AAC Fanatic!

Oct 2, 2009
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Lets talk a real opamp instead of an idealized one. The differential open loop gain of a typical ic opamp is about 100,000. When the opamp is operated closed loop, ignoring the effects of input bias current and intrinsic input offset of the opamp itself, the difference between the two inputs will be Vout/OpenLoopGain. To get 10V out, the differential between the inputs would be 10/1e5= 0.1mV.

Here is a sim that confirms:

I show an ideal opamp except that it has a finite gain ranging from 1000 to 10million. I set the closed-loop gain to be -10 (inverting). I plot the voltage at the - input V(-in) and the output voltage as a function of the OpenLoopGain of the amplifier.

With infinite Open-Loop gain, you expect the output voltage to be 10V and the voltage at the inverting input to be zero. Compare what happens to V(out) and V(-in) with finite gains...

Not likely. I any real circuit, it is impossible to make two voltages "equal". The saving grace is that the Open Loop Gain of any real amplifier is less than infinity.

Last edited: Oct 20, 2014
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10. ### MrChips Moderator

Oct 2, 2009
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Why would you want to configure an opamp with a gain of 100,000 in open loop?

11. ### Papabravo Expert

Feb 24, 2006
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So it can be used as a comparator.

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