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Take a standard inverting amplifier: I'm unclear on what the input impedance is. At first I thought it was the equivalent resistance from the inverting input to the ground, which would be Rin || Rf, because there's a ground on the other side of Vin and Vout as well as inside the output of the opamp. However, most sources on the internet seem to claim that the input impedance is Vi/Ii, thus making it Rin. This appears to ignore the fact that the feedback wire connects to other grounds. Then there are answers like this that mention that input impedance is infinity. I'm hoping to figure out a clear definition of what the input impedance represents and a [brief] general approach to calculating it in a more complicated circuit. Thank you!

I read tons of pages like this answer and others, but at my beginner level it was hard to extract a clear answer.

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When we talk about the input resistance of a circuit, we're describing how it affect the other circuit that is providing the input signal.

Specifically, we want to know, in order to change the input current by i amps, how much do we need to change the input voltage? That's why the input resistance is, by definition, \$ \dfrac{\mathrm{d}v_i}{\mathrm{d}i_i}\$.

So what's the input resistance of this circuit?

The key point is that in this configuration, as long as we avoid saturating the op-amp output, the inverting input of the op-amp is a virtual ground. The feedback in the circuit operates to keep that node at 0 V. So whatever input current we want, by Ohm's Law, the required input voltage is \$\mathrm{R_{in}}\times{}i_{i}\$. Therefore the input resistance is Rin.

Then there are answers like this that mention that input impedance is infinity.

That answer was talking about the input resistance of the op-amp, which was assumed to be ideal, not the input resistance of the whole circuit.

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The Photon's answer is absolutely right: ideal op-amp, etc, input impedance is Rin.

In a more general circuit, even one with non-linear components like transistors, input impedance is a small-signal (linearized), frequency-dependent quantity. It's important because it can tell the designer about loading effects between the output impedance of the prior stage and the input impedance of the next one. The general approach to calculate input impedance (or output impedance) is to inject a small current into the input node (di) and look at the resulting change in voltage of the input node (dv). Or, equivalently, to apply a small voltage (dv) and look at the resulting current (di) from your test voltage source. Then compute (dv/di). To make it completely clear that this is what's going on, take a look at my answer to How to compute input impedance, where I demonstrated how to use a circuit simulation program to compute and plot the input impedance by literally adding a test voltage source at the input and plotting a custom expression. Hopefully, seeing the voltage source V1 (or in your case Vin) drawn literally as a voltage source will make it clear how to go about setting up the calculations for doing it by hand!

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The definition of Ri is Vi/Ii. Rf is connected between the output node and virtual ground, and is basically a circuit on its own. It is not involved with the input. What you said is correct; Ri is the equivalent resistance of the input, but that is equal to Rin, not Rin || Rf.

As for “other grounds”, there is only one ground. Ground just means V=0. You can treat “real” ground and “virtual” ground as one and the same when it comes to circuit analysis.

I would like to add an intuition on what the input resistance actually is and why it matters. Input resistance Ri is the equivalent resistance between vi and ground. In other words, if you apply vi to the amplifier, or apply vi to Ri connected to ground, you get the same current flowing through vi in both cases. Hence Ri=vi/ii

As for why it’s important, real signal sources have not only vi, but a signal resistance Rs attached to it. Once you attach the signal source to the inverting amplifier, the input voltage vi would be the node voltage between Rs and Rin. Generally, if you look at an equivalent circuit, the input resistance is the total equivalent resistance between vi and ground. So if you look at the voltage divider rule, Vi=Vs•Ri/(Ri+Rs) Which means the higher the input resistance, the more signal you get at the input.

Keep in mind Ri=vi/ii, NOT Vs/ii. In the second case, Ri would depend on the signal source resistance, which is not the case.

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