Input impedance of inverting op-amp

The input impedance of the inverting amplifier is the resistance seen by the signal source, Rin = vin/iin. Take a practical op-amp with open-loop gain A and very large input resistance Ri.

                Rf
       +------/\/\/-----+
       |                 |
 vin --R1--+--(-)        |
  -> i1    |        >----+---- vo
       GND-+--(+)

The input current i1 flows through R1 into node v− (the op-amp input current is negligible since Ri is very large).

vo = -A·v-          (v+ = 0)
Current through Rf:
  iF = (v- - vo)/Rf = v-(1 + A)/Rf
So, seen from node v-, Rf looks like
  Rf/(1 + A)        (Miller effect)
Node v- to ground: Ri ‖ Rf/(1 + A) ≈ Rf/(1 + A)
Input resistance:
  Rin = vin/i1 = R1 + Rf/(1 + A)
Since A is very large:
  Rin ≈ R1

So the input resistance of the inverting amplifier is set by R1, because the inverting input is a virtual ground. Example: R1 = 10 kΩ, Rf = 100 kΩ, A = 2 × 10⁵: Rin = 10 kΩ + 0.5 Ω ≈ 10 kΩ.

Answer: Rin = R1 + Rf/(1 + A) ≈ R1.

This is low compared with the non-inverting amplifier (Rif = Ri(1 + Aβ)), so the source must be able to drive R1; R1 is chosen large enough (e.g. 10 kΩ or more) not to load the source.