2
\$\begingroup\$

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}\$?

\$\endgroup\$
1
\$\begingroup\$

Your hyperphysics link is correct, and so is your conclusion about the input impedance of the voltage follower. The math follows from the basic control system diagram. You can see a nice presentation on it here:

http://slideplayer.com/slide/1496732/

enter image description here

\$\endgroup\$
2
\$\begingroup\$

Feedback always influences the input impedance of an amplifier. However, it depends on the kind of feedback. When the feedback signal adds with the input CURRENT (example: inverting opamp) the input impedance is reduced by the factor (1+loop gain). However, when the feedback signal is superimposed on the input VOLTAGE (example: non-inv. opamp) the input impedance is enlarged by the same factor (1+loop gain).

This can be explained simply as follows: The opamp reacts upon the voltage difference between both inputs. The input resistance can be modelled as a resistance between these inputs. Now - when due to the feedback action the voltage at the negative input node follows the voltage change at the pos. input node, the voltage difference across the input resistance is much smaller (if compared with the case of a grounded inv. node). Hence, the resulting input current drawn from the signal input is much smaller. This explains the increase of the input resistance due to feedback.

\$\endgroup\$
  • \$\begingroup\$ and \$Z_\mathrm{ino}\$ is the differential input impedance from the datasheet? \$\endgroup\$ – endolith Jul 17 '15 at 16:33
  • \$\begingroup\$ Yes - that is correct. \$\endgroup\$ – LvW Jul 17 '15 at 18:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.