Ideally, the input impedance is infinite.

But, in calculating the input resistance (Rin) of a difference amplifier, the author took the concept that the two input terminals are short circuited, which is also true, since the open loop gain is infinite. (Which in turn demands that the difference between the input terminal voltages is zero. Hence, short circuit.)

MY Question: Why is that we consider zero input current in few cases (owing to infinite input impedance), and sometimes consider finite current taking the short circuit concept? Is there a logic or is it just a convenience?

This is the snipped out circuit diagram from the book:

enter image description here

  • 1
    \$\begingroup\$ No, there is no real short between them. It just makes it much simpler for calculations. In reality the feedback from the output supplies the current to force the other pin to the same value. \$\endgroup\$
    – winny
    Commented Jul 17, 2016 at 21:54
  • 3
    \$\begingroup\$ @winny has it right, note that in your diagram above there is still no current flowing into the + or - terminals there, those inputs (ideally) still have infinite impedance. However, in this configuration there are currents flowing in the input resistors, supplied by the feedback resistor and the source voltage. Just because the amplifier has infinite input impedance doesn't mean that every circuit you build with it will have infinite impedance on their input terminals. \$\endgroup\$
    – John D
    Commented Jul 17, 2016 at 22:09
  • \$\begingroup\$ This "virtual short circuit" is quite misleading, as it suggests that some current goes between the V+ and V- inputs. \$\endgroup\$
    – Grabul
    Commented Jul 17, 2016 at 22:23

4 Answers 4


The terminology can be confusing for a newbie, actually. The term "virtual short circuit" refers to the fact that in an opamp circuit with negative feedback the circuit is arranged in a way that (ideally) makes the voltage across the two opamp inputs zero.

Since one of the properties of a short circuit between two points is that the voltage across those points is zero, the people who invented that terminology considered (I guess) an intuitive thing to call what happens between the input terminals of the opamp a "virtual short". They called it "virtual" because it lacks the other property of a real (ideal) short: to gobble up any amount of current without problems! Alas, that's no small difference! They could have called the thing in a less confusing manner ("the principle of voltage balancing"!?!), but "the virtual short principle" sounds cooler, probably! Who knows?!

So, when we say that between the two inputs there is a virtual short, it's just an easy and conventional way to say that the circuit strives to balance the voltages at the inputs, i.e. it tries to make them and keep them equal.

Note that the existence of the "virtual short" is a property of the circuit, not of the opamp (although it exploits the ideally infinite gain of the opamp), whereas the fact that no current flows into the inputs is a property of the opamp (ideally).

EDIT (prompted by a comment)

I'll try to be clearer about what I said above. The virtual short is exclusively due to two key factors combined together: very high gain + negative feedback.

Let's do some math do convince ourselves. Let's call \$V^+\$ and \$V^-\$ the voltages at the non-inverting and the inverting inputs of the opamp, respectively, and \$V_o\$ the output voltage. A real opamp, in this respect, is a differential amplifier, i.e. \$V_o = A(V^+-V^-)\$, where \$A\$ is the open-loop gain of the opamp.

Inverting that relationship you get \$V^+-V^-=V_o/A\$. Thus, for finite \$V_o\$ and infinite \$A\$, you get that the difference between the inputs becomes zero.

Where did the negative feedback play a role? Nowhere, till now!!! The catch is that a real opamp needs negative feebdback to keep its output from saturating, in which case the simple linear model of the opamp (i.e. that gain formula) would no longer apply, except outside a very small interval of input voltages (assuming a classic non-inverting configuration where \$V^+\$ is the input voltage and \$V^-\$ is a fraction of the output).

Apply negative feedback and you'll get a zero differential voltage at the inputs over a meaningful range of input voltages.

  • \$\begingroup\$ adding on: the action of an operational amplifier as a very-large-gain, very-high-input-impedance differential amplifier coupled with the output-to-inverting-terminal feedback loop is precisely what drives the differential input voltage to zero \$\endgroup\$
    – obataku
    Commented Jul 18, 2016 at 14:01
  • \$\begingroup\$ Same potential, hence no voltage: equipotential nodes maybe? \$\endgroup\$ Commented Jul 18, 2016 at 15:01
  • \$\begingroup\$ @oldrinb actually the the input impedance of the opamp doesn't play a role for achieving the virtual short. The key factors are very high gain + negative feedback. \$\endgroup\$ Commented Jul 18, 2016 at 20:18

Very good question indeed.

I think much of this can be answered by looking at the equivalent circuit of an op amp.

Op Amp Equivalent circuit

For an ideal op amp, the current flowing into V+ and V- is zero, so this means Rin must be infinite.

When an ideal op amp is setup in a feedback arrangement(Vout is connected to V+ or V- in some way), the voltage at V+ will equal V-. The textbook simulates V+ being equal to V- by making a virtual short there. The input impedance of the op amp is infinite still!

In my circuits class, we didn't make a virtual short between the two because this can be confusing. Instead, we just said V+ = V- and used that as a equation to solve other unknowns.

  • 3
    \$\begingroup\$ You don't "make" a virtual short. It is just a useful concept to describe an op-amp in a negative feedback situation. The "short" is created by the op-amp doing whatever it can to make V+ = V-, assuming normal operating conditions. If the op-amp is acting as a compator, with V+ and V- electrically isolated from Vo, then V+ will not normally equal V-. \$\endgroup\$
    – jbord39
    Commented Jul 17, 2016 at 22:23
  • \$\begingroup\$ I did not suggest that the connection between V+ and V- is physical. I said the textbook simulates this by making a virtual short which is correct. By "making a virtual short," that is intended to mean drawing a line from V+ to V- to represent V+ being equal to V-. \$\endgroup\$
    – Addison
    Commented Jul 17, 2016 at 22:43
  • 3
    \$\begingroup\$ "Since there is no current flowing through Rin, the voltage at V+ must equal V-" - This is not true. Connect a 1k b/w both V+ and V- and put V+ @ positive supply and V- at ground. The output will be the positive rail. V+ will not equal V-. The reason V+ normally equals V- is because the op-amp is setup in a feedback arrangement which attempts to minimize the difference b/w V+ and V-. My point is blindly applying the equations w/o understanding their purpose is going to be misleading. \$\endgroup\$
    – jbord39
    Commented Jul 17, 2016 at 22:48
  • \$\begingroup\$ You make a good point. My reasoning behind V+ being equal to V- is only valid in a feedback arrangement. I have corrected my post. \$\endgroup\$
    – Addison
    Commented Jul 17, 2016 at 23:29

In short, there's a difference between the input impedance of the op amp, and the input impedance of the overall amplifier circuit. Even in terms diff amp you show, there is no current actually entering the op amp, which (ideally) has infinite input impedance.

As an aside, note that the difference amp inputs see different input impedances, which is a built in drawback of the configuration.

  1. Just to clear the air. If an op-amp is NOT being used as a comparator, in other words it has a negative feedback resistor, then it will output the difference between the (+) and (-) input times the gain to keep the (+) and (-) inputs at the same voltage. In the real world, the input impedance of an op-amp can never be infinite or zero ohms, but is somewhere inbetween.

  2. If you use resistors values much too low or high then the op-amp can become unstable, and the voltage between (+) and (-) inputs is unknown. Typically you will see designs where the (+) input is referenced to ground through a resistor and the op-amp has bi-polar power supplies. In this case the (-) input will be a virtual ground because the (+) input is at ground potential.

  3. With single-ended power supplies the (+) input is biased with resistors to 1/2 the supply voltage, so the output has an equal amount of positive and negative swing possible. And yes, do to the feedback loop the (-) input will be at 1/2 the supply voltage as well. Any signal is imposed upon this bias voltage and amplified according to the ratio of the gain and feedback resistors.

  4. Input impedance is controlled by the value of the resistors used, but their minimum and maximum values depends on the op-amp used. A CA3140T op-amp has an input impedance of 1.5 Giga ohms, so using resistors in the megohm range for input/feedback is OK. The op-amp is not loading down the resistors enough to matter.

  5. Now take the LM324 op-amp which has an input impedance about 1,000 times lower. Now you will find out that feedback resistors over 100K begin to not have the gain expected, because the op-amp is acting as a load of its own, putting a severe limit on the maximum value of resistors that can be used.

  6. A good compromise is JFET op-amps like the TL061/TL071/TL081 series which are very quiet for audio use and have an input impedance of 100 Meg ohms or so. You can use resistors up to several megohms without much gain error. One minor drawback of JFET op-amps is the need for a bipolar power supply of +/-5 volts to +/- 18 volts, with +/- 12 volts being typical for power.

  7. Op-amps for RF use have low input impedances (25 to 75 ohms) and output impedances and are powered by 5 or 3.3 volts, with many having +/-5 volt supplies. The low impedances are so high frequencies, sometimes up to 1 GHZ, can charge and discharge the tiny capacitance of the inputs and drive 75 ohm or 50 ohm coax cables (or a twisted pair) with ease. The bias currents in the op-amp are high so signals can swing positive and negative rapidly, with no drag.

I could write a book on op-amps but others already have, including articles on this site. Each op-amp manufacture offers PDF's for the different categories they make, so you could spend years just reading about them.


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