I was reading an old copy of Horowitz and Hill's The Art of Electronics and I am trying to wrap my head around operational amplifiers in negative feedback circuits.

As the book explains, as the op-amp sees a positive voltage at its inverting input relative to its non-inverting input, it drives its output strongly negative, which effectively sets up a voltage divider along the resistor between the signal source and the op-amp's inverting input, and the feedback resistor running between the op-amp's output to the inverting input. The gain will be such that the voltage at the inverting input is the same as at the non-inverting input, which is typically ground. Hence the inverting input in a negative feedback circuit is sometimes called a virtual ground.

Now if this is the case, and the inverting and non-inverting inputs are brought to the same potential, won't the op-amp stop amplifying? I mean, there's now no voltage difference, so there's nothing to amplify.

I suspect that this is actually occurring, but as the op-amp stops amplifying, the output voltage moves up toward zero, which raises the voltage at its inverting input, which triggers amplification again until the virtual ground is re-established and the cycle repeats.

Wait, did I just say the cycle repeats? This implies is that the output voltage is actually oscillating at (probably) a very low amplitude and very high frequency, which for all intents and purposes for common circuits is stable. Is my understanding correct?

I bet if I used a capacitor to introduce a delay in the reaction of the op-amp, I could get oscillation at a lower frequency.

  • \$\begingroup\$ For reference you can also look at this question \$\endgroup\$
    – clabacchio
    Jul 29, 2014 at 11:50

2 Answers 2


At first, the principle of "virtual ground" can be applied during DESIGN of opamp-based amplifiers. This simplifies calculations - and the error is in most cases acceptable. Error? Yes - because there is always a differential voltage between both opamp inputs, which is exactly Vdiff=Vout/Aol. (Aol=open-loop gain of the opamp). Because of the large values for Aol (1E4...1E6 for lower frequencies) this diff. voltage Vdiff is in the µV range.

However, because this is not true for larger frequencies, the closed-loop gain will deviate from the calculated value for rising frequencies.

Regarding your last sentence: Yes - introducing additional delay in the feedback path will cause additional phase shift - and this can lead to instability/oscillations.

EDIT: "...until the virtual ground is re-established and the cycle repeats."

I suppose, with the above cited sentence you are asking for something like a "sequence" which leads to the steady-state conditions after applying an input signal, correct? This is, indeed, a question which deserves some explanations.

Example: Inverting opamp-based amplifier with a gain of "-2". Input: +1V step (t=0).

At the very beginning (t>0), the feedback is not yet active and the output will jump to the maximum negative voltage (supply rail). Now the feedback network causes the inverting terminal to become negative - and the output starts to go to positive voltages. However, this will not continue again and again because the opamp has internal delay elements (causing bandwidth limitations and phase shift). That means: The output does not "jump" to other values but it takes some time to reach the upper rail. But, in reality, the output will NOT reach the upper rail because on the way to the maximum positive output the output voltage crosses some finite negative values - and for an output value of app. Vout=-1.999V there will be an equilibrium between input and output. Explanation:

Vout=-1.999V and Vin=+1V cause a very small voltage between both resistors (at the inv. input terminal) which - when multiplied with Aol - is exactly the assumed output voltage (in the example: Vout=-1.999V.) This equilibrium state is stable.

  • \$\begingroup\$ Thanks for this answer. I appreciate you phrasing it in terms of a sequence. My background is in computer science, which is very sequential. \$\endgroup\$ Jul 30, 2014 at 15:41

Now if this is the case, and the inverting and non-inverting inputs are brought to the same potential, won't the op-amp stop amplifying?

The op-amp amplifies the voltage across (difference) the input terminals regardless.

For the ideal op-amp, the voltage gain \$A\$ is 'infinite' (arbitrarily large) thus, the voltage difference can be arbitrarily small and the op-amp will still produce a non-zero output.

Here's a quick analysis. Connect the signal source to the non-inverting input and connect the inverting input to the output through a voltage divider so that

$$v_+ = v_S$$

$$v_- = \alpha \cdot v_O $$

where \$\alpha\$ is between 0 and 1.

Then, the output voltage is given by

$$v_O = A(v_+ - v_-) = A(v_S - \alpha \cdot v_O)$$


$$v_O = \frac{A}{1 + \alpha A}v_S$$

and the inverting input voltage is

$$v_- = \alpha \cdot v_O = \frac{\alpha A}{1 + \alpha A}v_S$$

and the input voltage difference is

$$(v_+ - v_-) = v_S - \frac{\alpha A}{1 + \alpha A}v_S = \frac{v_S}{1 + \alpha A}$$

So, we see that, for finite op-amp gain \$A\$, the input voltage difference is small but not zero when the signal source voltage \$v_S\$ is non-zero

This difference is amplified by the op-amp which produces a non-zero output voltage.

Only in the limit of \$A \rightarrow \infty\$ does the input voltage difference go to zero regardless but then, with 'infinite' gain, the output can still be non-zero.

  • \$\begingroup\$ This was an excellent, informative answer. If I could accept two answers, I would accept yours too. Nice mathematical derivation, BTW. \$\endgroup\$ Jul 30, 2014 at 15:42

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