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I have a question about the load impedance seen by an ideal op-amp output terminal in inverting configuration.

When calculating the load impedance, we set Vin = 0;

In one way I could say that the output terminal sees Rf + Rin as a load since if I place a V_test at the output, the current drawn from the test source would be V_test/(Rf+Rin).

If I think in another way, I could say that the inverting input is a virtual ground since op-amp is ideal, and so the op-amp only sees Rf as a load, and the load impedance is Rf regardless of Rin.

I wanted to see which way of thinking about it is true.

If anyone has any idea about this paradox I'd be glad to hear it.

inverting amplifier

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3 Answers 3

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Feed a voltage at Vin and measure the opamp output voltage and current through Rf. From this, you can calculate the load seen by the opamp output.

Assuming an ideal opamp and linear operation, the negative input is at the same potential as the positive input (virtual ground). Thus, the load seen by the opamp is just Rf.

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  • \$\begingroup\$ The current through Rf contributes only very little to the total output resistance. So - you can forget this portion. Did you overlook the low output resistance of the opamp itself? Your answer applies to an OTA only but not to an opamp! \$\endgroup\$
    – LvW
    Mar 14, 2022 at 8:12
  • \$\begingroup\$ @LvW The OP's question asks about the output load due to the feedback network (Rf+Rin, or, Rf), not the output resistance of the opamp. \$\endgroup\$
    – qrk
    Mar 14, 2022 at 16:12
  • \$\begingroup\$ Yes - it seems that I have misunderstood the question. I think, the the reason was the test signal at the opamps output . \$\endgroup\$
    – LvW
    Mar 14, 2022 at 16:40
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You can't put a test voltage on ideal op-amp output, that itself is a paradox.

That's because the ideal op-amp output has zero output impedance and can use infinite voltages and currents to set the output voltage based on op-amp inputs.

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  • \$\begingroup\$ So where should I put the test voltage source in order to measure it's current? \$\endgroup\$
    – Mahdi
    Mar 13, 2022 at 18:52
  • \$\begingroup\$ There is no need to put any source anywhere else than on the input. And you already showed the right answer with that method. \$\endgroup\$
    – Justme
    Mar 13, 2022 at 19:00
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Mahdi - are you aware that you totally have forgotten the opamps output node and the corresponding very low impedance at that point? So - of course, the feedback network (Rf,Rin) contributes to the total ouput impedance at a negligible amount only.

For an IDEAL opamp the output impedance is zero (without calculation). For a real opamp you have to take into account the influence of negative feedback which drastically reduces the total output impedance.

The resulting expression is R,out=Ro/(1+loop gain) with Ro: output impedance without feedback (data sheet)

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