By virtue of the open-loop gain being infinity for an ideal op-amp, the two input terminals are said to be virtually shorted.

Yet, 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?

Some insight in this regard would be really helpful.

  • 2
    \$\begingroup\$ Positive feedback tends to cause the pin voltages to diverge. Negative feedback tends to cause the pins voltages to converge. It's not more complex than that. \$\endgroup\$
    – jonk
    Commented Oct 25, 2018 at 7:42
  • \$\begingroup\$ Well, what do you mean by convergence and divergence of voltages? \$\endgroup\$
    – Curiosity
    Commented Oct 25, 2018 at 9:22
  • \$\begingroup\$ That would require an answer. Something I'm not currently willing to engage. If what I wrote helps, fine. Otherwise, it will have to wait for the time and inclination to write more arrives. \$\endgroup\$
    – jonk
    Commented Oct 25, 2018 at 9:24
  • 3
    \$\begingroup\$ Possible duplicate of Positive feedback and virtual short in Operational Amplifiers \$\endgroup\$
    – DavidG25
    Commented Nov 16, 2018 at 0:40
  • \$\begingroup\$ Simply put, the output of the op amp is A(V+ - V-). Using an infinite input resistance to the input terminals, derive the output and the voltage at the input terminals. Now, crank A up to infinity and look at the limits. Try for positive and negative feedback. \$\endgroup\$ Commented Nov 20, 2018 at 19:29

2 Answers 2


Positive feedback does not result in stable, linear operation. In a positive feedback amplifier the output adds to the input, resulting in a larger input, resulting in a larger output. In other words, the output accelerates quickly in 1 direction until it can't anymore. In a circuit the limitation is the power supply and the output of the amplifier will be saturated at/near the power supply voltage.

When the output of the amplifier is saturated at/near the power supply voltage, it is not operating as a linear amplifier. The output voltage can no longer be described as GAIN*INPUT. No matter what the input is, the output is the power supply rail voltage.

The reason why you get a virtual short with negative feedback is because negative feedback stabilizes the op amp in linear operation, and the output voltage can be described as GAIN*INPUT. Since the gain is very high, the input is very small. This is the condition of the virtual short.

This is a general explanation, and is overlooking details such as closed-loop vs open-loop gain. Maybe some one else can provide a more detailed answer.


There are a lot of metaphors used to talk about op amps, and none of them should be taken literally, and rarely are they meant to be used in wider contexts.

The metaphor of the virtual short between the inverting and non-inverting pins is just that - a metaphor. That's your plausible reason. The pins aren't actually shorted, so this idea does not carry over to positive-feedback situations.

Another metaphor often used, which is in my opinion slightly better (as talking about things in terms of interconnections like shorts could be mistaken to mean there is actually such a connection, and that it is present in other situations), is that an op amp with negative feed back is 'trying to keep the two input terminals the same'.

In fact, I am just going to say that the virtual short metaphor is flat out incorrect. This makes it seem like a passive effect, and one that would carry over to other situations, when neither is the case. Op amps are still just amplifiers, and all they really do is take the difference of potential between the input terminals and amplifies it to a larger one at the output.

With negative feedback, this amplified difference of potential has the effect of reducing the very input signal being amplified, or rather, the difference between the input terminals, so when the gain is very high, this means the op amp is actively reducing the difference between its two inputs to almost nothing. There is no virtual short, we are just using the amplified output to effect the input, and negative feedback always helps reduce the input.

In positive feedback, the amplified output is going to increase the difference of potential at the input, which makes the output even larger, which makes the input even larger, and in a magic ideal op amp, everything diverges to \$\infty\$ .


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