The following circuit is a simple voltage regulator. The Vout is maintained at -1V, however the opamp in the circuit has a non-zero output impedance. This is represented by the resistor R1.


simulate this circuit – Schematic created using CircuitLab

I have been told that the following procedure will allow me to measure the output impedance of this circuit.

  1. Remove r4 and measure the open circuit voltage
  2. Measure the voltage with r4 (which has been chosen to be close to the expected output impedance)
  3. Now the circuit is modelled as a voltage divider and the output impedance is found


simulate this circuit

Where Vout is the open circuit voltage measured above, and the second voltage is across R4.

The expectation is that adding the resistor will cause a slight decrease in the output voltage. Now, assuming the circuit was accurately represented by the above circuit, this would all make perfect sense. However, as far as I can tell the feedback loop ensures that the output voltage is at -1V, unless the resistance is so low as to cause the output to be clipped. (The person describing the measurement ensures that clipping doesn't happen by varying V1 and making sure the output changes linearly).

My problem is that if the output voltage shouldn't change at all (it is a voltage regulator after all), so that the result will always be close to 0 Ohms. Again, we are assuming the R4 is not so low that that clipping occurs. The person describing the measurement however expects to find around 70 Ohms, and blames the much lower recorded values (2 Ohms) on the variation of the opamps.

So is there any reason that this measurement would get an answer other than 0 Ohms?

  • Consider the inverting op-amp circuit as driving to bring the inverting-input to the same potential as the non-inverting input. i.e. V- = 0 V.
  • To do this I2 = I3 (the feedback current = input current).
  • Since R2 and R3 are both 1 kΩ then the voltages across them are identical. Therefore, since Vin = 1 V then Vout = -1 V.
  • There is no error on the output voltage due to the 70 Ω output impedance of the op-amp provided the op-amp is running within specifications.

I think your advisor is incorrect.


A few pointers: -

  • A perfect op-amp would give zero ohms output impedance
  • An imperfect op-amp with open-loop gain of ten million would have a DC output impedance of micro ohms
  • Respect the maximum current that the op-amp can supply and note that on some op-amps the open-loop gain falls dramatically with load current
  • For AC output impedance this is a different story as the frequency rises because open-loop gain falls (usually) at 6 dB per octave.

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