# Input impedence of Op-amp circuit

This is an inverting op-amp with closed loop gain $A$. The question is theoretical, so op-amp is ideal.

simulate this circuit – Schematic created using CircuitLab
I was trying to find the input impedance $Z_i$ of this Op-amp. Let $i$ be the current coming out of $V_i$. Then $Z_i = \frac {V_i}{i}$
In the first case, assuming inverting terminal is at virtual ground $V_i -0 = R_i \times i$ $$Z_i = \frac{V_i}{i}=R_i$$ But when I apply KVL along $V_i - R_i- R_f- V_o$ assuming no current is entering the OP-amp. $$V_i - i.R_i - i.R_f -V_o=0$$substituting $V_o = A.V_i$ $$Z_i = \frac{V_i}{i} = \frac{R_i +R_f}{1-A}$$ Which $Z_i$ is correct

The first one is correct.

Zi = Ri

The input impedance is looking into the opamp with respect to ground. So any resistance between the input signal and ground, and since the inverting terminal is held at ground, the opamp will do everything it can to ensure that it is always held at ground.

This question has also been asked before, so its worthwhile to look through the site before posting.

Input impedance of inverting amplifier clarification

As the amplifier will slew its output to maintain the difference between the inputs at zero (provided it is capable of doing so and is not in saturation), then the first statement must be correct.

As the inverting input is at 0V, then the input resistance (referred to zero volts) is Rin

As a hint for KVL, remember this is an inverting configuration; look carefully at the Vo term.

Following a different approach you can apply the Miller theorem. That means: Zi=Ri+Rf/(1+A) which simplifies to Zi=Ri for ideal opamps (A infinite).