I've heard of differential voltages, but the concept of a differential impedance is new to me.

Suppose we are trying to calculate/derive the differential input impedance of the following difference amplifier:

Difference Amplifier

Then what would that mean and how would we go about that?

Pages such as this give information about the impedance seen at the inverting and non inverting inputs to the op-amp but they make no mention of the differential impedance, which leaves me stuck looking for answers. This post is very similar and gives us the answer R1 + R2 (which in our case is really R1 + R3), but there is no explanation of how we arise at that result, and this is where 'im stuck.

Thank you for any help and have a nice day.


The differential input impedance is R1 + R3.

That is because the op-amp actively makes sure via the feedback R2 that both V+ and V- nodes at the op-amp have identical voltages.

Therefore, as V1 terminates via R1 into voltage V-, and as V2 terminates via R3 into voltage V+, and as V+ and V- are identical voltages, there will be V1-V2 over the sum of resistors R1 and R3, and that is the differential impedance.

As an example, let's use the circuit to measure say a 3V battery. One battery terminal, say the positive, is connected to V1, and the other battery terminal, the negative, is connected to V2.

As the battery is a completely floating voltage supply, i.e. it shares no common reference with the supplies of the op-amp or the ground symbol, the measured battery voltage is completely differential.

So, V1-V2 is the battery voltage, 3V. Again, op-amp keeps V+ and V- equal, no matter what V+ and V- are. So there is definitely a 3V drop over resistors R1 + R3 only. Thus, what voltage there is at V+ is, and thus what is the current through R4, has no role in the differential impedance at all.

  • \$\begingroup\$ Ok that's making more sense. My question now is that why isnt the resistor R4 coming into play at all? It divides the voltage supplied by V2 so it would indicate the presence of R4 has some impact on the voltage seen at V+, so why isnt it part of the differential input impedance? \$\endgroup\$ – user53203 Apr 6 at 19:05
  • 1
    \$\begingroup\$ Good question. I will edit it into my answer. \$\endgroup\$ – Justme Apr 6 at 19:07
  • \$\begingroup\$ Thank you for the update, makes sense now. \$\endgroup\$ – user53203 Apr 6 at 19:33

For common mode impedance balance reasons R1 and R3 should be the same value (call it R): -

enter image description here

And, given that both inputs are resistively terminated at the same voltage (Vx), due to op-amp action and negative feedback, the differential impedance is 2R.

But, if you ignored the CM impedance balance requirement the input impedance is R1 + R3.

  • \$\begingroup\$ Ok, thank you. Am I right in saying that R4 is not in the final result because R4 directly controls the value of Vx, and since both V+ and V- have the same voltage (Vx) the 'effect' of R4 is present in both inputs, and so when we take the difference it gets lost? \$\endgroup\$ – user53203 Apr 6 at 19:08
  • 1
    \$\begingroup\$ You can regard both inputs as being connected to Vx via their respective resistors R1 and R3. Nothing else matters. Don’t forget, that a proper diff amp has both those resistors as the same value to cancel common mode noise. \$\endgroup\$ – Andy aka Apr 6 at 19:27
  • \$\begingroup\$ And R4 becomes part of the common mode impedance equation rather than anything related to differential input impedance. \$\endgroup\$ – Andy aka Apr 6 at 19:33
  • \$\begingroup\$ Thanks for the follow up, makes sense now. \$\endgroup\$ – user53203 Apr 6 at 19:34

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.