Why Virtual Short Concept in an Op-Amp is only applicable to negative feedback

Thread Starter

hoyyoth

Joined Mar 21, 2020
309
Dear Team,

I have a question which is haunting me for a long time.Till now I did not get any proper explanation.

If you can explain with mathematical proof it is appreciated.

The virtual short concept is only applied to op-amps in negative feedback configuration, and not in case of positive feedback.

Is there any plausible reason for this?


My expectation is that mathematical proof which explains why it is applicable to negative feedback and not applicable to positive feedback

Regards

HARI
 

BobTPH

Joined Jun 5, 2013
8,998
Because only with negative feedback are the two inputs at the same potential. And it is only a virtual ground if the + input is at ground. Those two conditions mean the - input is at ground potential even though it is not connected to ground.
 

LowQCab

Joined Nov 6, 2012
4,075
Even though the Output-Stage of an Op-Amp has a definite amount of Resistance,
it will be driven as hard as it is capable of toward one Power-Supply-Rail or the other,
until the Feedback-Circuit is balanced with the Input.

This mostly behaves as, and gives the apparency of,
a "zero-Output-Impedance" Amplifier, ( almost ).
.
.
.
 

WBahn

Joined Mar 31, 2012
30,076
Dear Team,

I have a question which is haunting me for a long time.Till now I did not get any proper explanation.

If you can explain with mathematical proof it is appreciated.

The virtual short concept is only applied to op-amps in negative feedback configuration, and not in case of positive feedback.

Is there any plausible reason for this?


My expectation is that mathematical proof which explains why it is applicable to negative feedback and not applicable to positive feedback

Regards

HARI
Consider what it means for there to be a "virtual short" in the first place.

It refers to the condition where the voltage at one input is equal, for practical purposes, to the voltage at the other input. If the two inputs are at significantly different voltages (and, in this case, much more than a few millivolts, or even microvolts, can be considered 'significant'), then the virtual short doesn't apply.

So, under what conditions are the inputs of an opamp driven to always be extremely close to each other?
 

MisterBill2

Joined Jan 23, 2018
18,584
"Virtual Short" is a poor term used to describe a summing junction driven to some voltage by a number of independent currents. Really, the inverting input is driven to be the same voltage as the non-inverting input. So your confusion is caused by a lazy explanation of what is happening. A bit like the word "Swamp" used by some writers unable ti find the correct words to describe "Absorb excess power", or somesimilar function.
 

WBahn

Joined Mar 31, 2012
30,076
Virtual common or virtual ground are often used to describe the summing junction which seems better than virtual short.
Except it implies that the node is forced to the common or ground potential, which is not what is happening. The inputs are being forced to the same potential as each other, independent of what that potential happens to be relative to whatever node happens to be declared the circuit common or ground. I think that virtual short is a much more apropos description than either virtual common or virtual ground, except when the circuit is such that the voltage at the inputs happens to coincide with the circuit common or ground.
 
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