How does an ideal op amp amplify a voltage input when the voltage difference is zero

[newbie_hs]

Looking at the op-amps, there was always this one thing that was confusing to me: in an inverting amplifier, the inputs of the op-amp are both zero, since the one at the + I/P is grounded This will make other I/P also at GND(Virtual Short). But when they are both zero, how can there be any amplification? I would think that you can just draw the line of the circuit without writing down the op amp, since no current flows in the inputs of the op amp.

 

[KlausST]

the inputs of the op-amp are both zero,

A misunderstanding.
The voltage of an OPAMP is close to zero. So close, that you may not be able to measure it with a standard voltmeter.
Additionally there are errors: Like V_OS and I_OS and I_B .. that probably make the (erroneous) voltage higher that the "expectable" voltage difference.
Thus - if the error is bigger than the expectable value one can declare the value as "not measurable".

Now back to ideal opamp.
What makes an OPAMP ideal?
(besides, zero V_OS, zero I_OS, zero I_B. ..)
The gain is infinite.
the formula says: V_out = gain * V_in
now if V_in = 0 .. you expect V_out also to be 0. But here we have the contradiction with gain to be infinite.

0 * infinite = undefined. It is neither zero (because the zero input), nor is it infinte (because of the infinite gain).

***
Let´s take some real numbers of a real OPAMP: Let´s take LM324.
It has a DC gain of typ 100,000.
Let´s imagine you want to get an output voltage of +1V .. then the differential input voltage needs to be 1V/100,000 = +10uV
I doubt your voltmeter is able to measure 10uV correctly. (some advanced ones can)

But even if you could ... would you indeed measure 10uV?

No, because of the OPAMP´s errors. Let´s take its input offset voltage. It is specified to be +/-2mV.

2mV mean 2000uV.... this is 200times (or 400 considering bidirectionality) as high as the expected "gain caused" input voltage.

So indeed: when considering the offset voltage the input voltage could be in the range of -1.99mV to +2.01mV

Even most voltmeters have a hard time to measure those 2mV accurately ..

***
Conclusion:
The "zero input voltage" comes (considering errors) very close to the real value ... but it is not zero-zero!
So considering the (much higher) additional errors, it´s valid to talk about zero.
***

This will make other I/P also at GND(Virtual Short).

From internet search: "What is "virtual short" with OPAMP":
A virtual short-circuit (or simply virtual short) refers to a condition of a differential input amplifier such as an op-amp in which its noninverting and inverting inputs have almost the same voltage.

Klaus

 

[crutschow]

in an inverting amplifier, the inputs of the op-amp are both zero, since the one at the + I/P is grounded This will make other I/P also at GND(Virtual Short). But when they are both zero, how can there be any amplification?

Yes, for an ideal op amp, both inputs are indeed zero with negative feedback and the circuit is in a linear state, while still generating gain.
(You don't need to consider the non-idealities of a real op amp.)

Thus with infinite open-loop gain and negative feedback, the ideal op amp output will go to whatever value is required to generate 0V across the inputs (remember we are talking about idealized infinities and zeros).
To then determine the gain we need to see what condition will cause that input zero.

So, for an inverting amp, we look at the input and feedback currents required to generate that ideal 0V.
The only way that happens is, if the output voltage is such that the Rf feedback resistor current exactly equals the Rin input resistor current.
And the only condition for that is if -Vout/Rf exactly equals Vin/Rin, giving a Vout/Vin gain of -Rf/Rin.

So we thus have gain even though the idealized differential input voltage is zero.
Any other relative input-to-output voltage would generate unequal currents, and a non-zero voltage across the inputs, which then causes the output voltage to change until the zero is achieved.

You can analyze other op amp configurations, such as a non-inverting amp or follower the same way to determine the gain.
Make sense?

 

[D.A.(Tony)Stewart]

We say virtually zero or "virtual null" or differential input = null when loop is closed in linear mode but actually Vdiff.in = Vo * Acl/Aol for Aol=open loop gain

Here's another thing. The differential impedance at source below is 20 k while the single ended impedance is 10+ 100k.

Since this is ground referenced, it uses a bipolar supply.

Note how the common mode inputs of this "virtual null" differential input is all due to the negative feedback attenuation providing gain. = +/-2.5V

If the output ever saturates or clips near either supply rail, the open loop gain Aol immediately drops to zero and you no longer have a "virtual null" input.

Since we as EE's define "ground" as 0V as any reference regardless if it is floating or earth-bonded (PE) , note that the Vin- plot to the right is not "zero volts" but relative to Vin+ it is zero hence the confusion with use of "virtual ground" meaning a relative ground

If you wanted a single supply, that gnd to upper 100 k would be changed to Vdd/2 to satisfy the Vcm input range.

Browser SIM

 

[BradtheRad]

It also makes a difference what voltage is coming from the other side of the load. Yet suppose the op amp has to make its own output 0V, then its internals have to do work in some manner. Some amount of gain, selecting current direction, some amount of current manipulating, etc.