# Input impedance of a non-inverting op-amp

From what I've read (and simulated), the feedback loop of an op-amp modifies the input impedance of the non-inverting input. The value specified in the datasheet is the open-loop input impedance, and the actual closed-loop input impedance will be some much larger number? Why does this happen, and how do you calculate the new input impedance? Does this also reduce effective input capacitance?

Radio-Electronics.com "Op Amp Input Impedance" says it's the differential input impedance of the op-amp plus the impedance to ground seen by the inverting input, with no mention of open-loop gain. So for a voltage follower with no feedback resistor, the impedance seen by the inverting input is zero and the input impedance is unchanged? That doesn't seem right.

HyperPhysics "Practical Benefits:Negative Feedback" says it's $$(1 + A_0 B)\cdot Z_\mathrm{ino}$$ where

• $A_0$ is the gain without feedback (the open loop gain)
• $B$ is the fraction of the output which feeds back as a negative voltage at the input
• $Z_\mathrm{ino}$ is input impedance without negative feedback

So for a voltage follower, B = 1 and it's $\approx A_0 Z_\mathrm{ino}$? • and $Z_\mathrm{ino}$ is the differential input impedance from the datasheet? – endolith Jul 17 '15 at 16:33