Does op-amp virtual ground principle only apply in steady-state?

Assuming a negative feedback config of course.


1 Answer 1


Good question.

It only holds when an op-amp has infinite gain, which is of course never.

However, at DC, in the steady state, it has very high gain, 107 is not uncommon, so the error from a virtual ground is very small.

At progressively higher frequencies, the open loop gain falls, and the error from virtual ground behaviour becomes larger.

Op-amps are usually characterized by the Gain.Bandwidth product, more or less constant for voltage feedback opamps. At a frequency of 0.1% of the GBW, the open loop gain will be about 1000, and the error from virtual ground will be of the order of 0.1%

  • \$\begingroup\$ Or another way to put it: the node has an inductive characteristic, impedance rising with frequency over the same range the amp's gain is falling. \$\endgroup\$ Aug 13 at 22:59
  • \$\begingroup\$ Got it. So if I applied a step signal into my negative feedback op-amp circuit, it is likely that the there would be a brief period where the op-amp input terminals will differ a little due to the lower gain at higher frequencies and then once in steady-state, the op-amp input terminals would be closer to each other since the DC gain is higher. Is that right? \$\endgroup\$ Aug 14 at 13:11
  • \$\begingroup\$ @AlfroJang80 Absolutely. What I used in commercial audio equipment using a virtual ground integrator as the first stage of a delta sigma converter, was to place a small capacitor between the virtual ground and ground. This absorbs any step, keeping the deviation small and reasonable. This prevents the amplifier output slewing violently, possibly going non-linear, and introducing distortion into the system. Eventually the virtual ground op-amp catches up and dischrges the capacitor back to zero. \$\endgroup\$
    – Neil_UK
    Aug 14 at 13:40

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