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For the above circuit, I would like to find the input impedance for the input \$v_1\$.

Assuming that \$R_1 = R_2\$ and \$ R_3 = R_4\$, then: $$v_0 = \frac{R_3}{R_1} (v_2 - v_1)$$

As far as I know, the input impedance will be defined separately for each input (i.e. \$v_1\$ will have an input impedance, and \$v_2\$ will have a separate input impedance, since they are different inputs).

I believe that the input impedance of \$v_2\$ is simply \$R_2\$ + \$R_4\$, because the current flowing in through the input labelled \$v_2\$ passes through those two resistors in series on its way to ground.

However, I don't know how to find the input impedance for \$v_1\$? If there was a virtual earth at the inverting input, then the input impedance would simply be \$R_1\$, but that isn't the case here.

Could someone tell me how to solve this problem?

  • 1
    \$\begingroup\$ If V1 and V2 are uncorrelated (unrelated), then V1 sees an impedance of R1 and V2 sees R2+R4. But if this is a differential amplifier where V2 = -V1 then things get more complicated because the voltage on the right side of R1 is now a function of V1. \$\endgroup\$
    – td127
    Mar 22, 2022 at 1:34

4 Answers 4


An assumption made for an "ideal" op-amp is that the inputs do not source or sink any current. Thus, the impedance seen at the non-inverting input of the differential amp is just the sum of R2 and R4.

Another assumption made for an "ideal" op-amp configured with negative feedback, and not in saturation, is that the voltage on the inverting input is equal to that of the non-inverting input. Because that voltage is common to both inputs, let's call it \$v_c\$. Note that \$v_c\$ is determined by \$v_2\$ and the voltage divider consisting of R2 and R4. Changes in \$v_1\$ do not cause changes in \$v_c\$.

The current through \$R1\$ is


If \$v_1\$ is increased to \$v_1 + \Delta v\$, the current will increase to

$$I_1 + \Delta I = \frac{v_1 + \Delta v -v_c}{R1}$$


$$\Delta I = \frac{v_1 + \Delta v - v_c}{R1}-\frac{v_1 - v_c}{R1} = \frac{\Delta v}{R1}$$

So, the impedance at the inverting input of the differential amp is just

$$Z_1 = \frac {\Delta v}{\Delta I} = R1$$

What is seen at the inverting input is equivalent to the following circuit.


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ I'm pretty sure they the impedance in the inverting input is just R1. \$\endgroup\$
    – Mike
    Mar 22, 2022 at 0:11
  • \$\begingroup\$ @mooshoomatt thank you for fixing typo. \$\endgroup\$ Mar 22, 2022 at 0:45
  • 1
    \$\begingroup\$ Downvoted because answer is incorrect (Mike is correct, impedance is R1) \$\endgroup\$
    – BeB00
    Mar 22, 2022 at 1:00
  • \$\begingroup\$ @Mike You are correct. I have fixed the answer. \$\endgroup\$ Mar 22, 2022 at 1:18
  • \$\begingroup\$ Sorry - but I must say tha the derivation of your answer is not logical. In the second line you have already assumed that the current incease is determined by R1 only. So - the result is no surprise - and no real proof. \$\endgroup\$
    – LvW
    Mar 22, 2022 at 7:32

Simple calculation:

I1=(V1-Vn)/R1 with Vn=V2*R4/(R2+R4)

I1=V1/R1 - V2 * (R4/R1)/(R2+R4).

As we can see, the input resistance for V1 is Rin_1=V1/I1=R1 for V2=0 only.

Special case: R1=R2 and R3=R4 and V1=V2:


  • \$\begingroup\$ The definition of input impedance assumes all independent sources except at the node in consideration are nulled. V2 should be assumed zero and the answer is R1. Did not downvote you, FYI \$\endgroup\$
    – sarthak
    Mar 22, 2022 at 11:04
  • \$\begingroup\$ Thank you for the down-vote. However, I am interested to learn where I was wrong. Would you please so kind and give me a hint? \$\endgroup\$
    – LvW
    Mar 22, 2022 at 11:06
  • \$\begingroup\$ I did NOT downvote... As I said in the previous comment, you have to assume V2 = 0 and V1 = V2 is an incorrect assumption \$\endgroup\$
    – sarthak
    Mar 22, 2022 at 11:08
  • \$\begingroup\$ sarthak - from where did you get this rule? Did you never hear about the classical diff. amplifier (long tailed pair) which has THREE standard input resistances (Differential symm and unsymm. , common mode). Here, we have also a diff. amplifier and - OF COURSE - it is important that the input impedances depend on both input voltages. By the way: What do you mean with "uncorrect assumption"? V1=V2 is a classical application!! \$\endgroup\$
    – LvW
    Mar 22, 2022 at 11:10
  • \$\begingroup\$ Yes for the unsymmetric case which OP is asking the other node should be nulled. For the differential case, the source is connected between the nodes and for common mode the two nodes are connected with a common source. In all cases, only one independent source is present \$\endgroup\$
    – sarthak
    Mar 22, 2022 at 11:14

Assuming that R1 = R2, then the input resistance to ground will be R2 + R4. This is true for V2, and since the two opamp inputs are held at essentially the same voltage, The resistance to ground for V1 will be equal to R1 + R4, which is the same as V2, since R1 = R2.

The differential input impedance will be R1 + R2, since the two inputs to the op amp are essentially at the same voltage, and can be replaced by a short circuit for the purposes of this question.

  • \$\begingroup\$ I have written a comment to your answer at the end of my answer. \$\endgroup\$ Mar 23, 2022 at 12:35

It has always been a good idea to find the answer to a specific question with the help of a more general idea…

Here the answer can easily be found if we see Miller's powerful idea of virtually ​​modifying impedance. The resistor R1 supplied at both ends with the two single-ended voltages V1 and V- is a typical Miller arrangement. Note that both voltage sources are perfect. V1 by definition has low internal resistance. V- is inherently an imperfect voltage source because it consists of a perfect voltage source (the op-amp output) and the resistor R3 in series… but due to the negative feedback it behaves as a perfect voltage source. Thus the op-amp acts as a voltage follower that copies the voltage V+ of its non-inverting input as a voltage V- at its inverting input (the disturbing resistance R3 is eliminated). The op-amp does it by sinking/sourcing a current through R1-R3 network from/to the input voltage source V1.

Let's now consider the four typical cases:

1. V1 varies, V2 is constant (inverting single-ended input). V- = V+ is a part of V2 and is constant. So there is nothing interesting - the input resistance is (the input source V1 "sees") the resistance of R1. If there was no negative feedback (e.g., the op-amp was saturated), V1 would "see" a total resistance of R1 + R3.

2. Both V1 and V2 vary in opposite directions (differential input signal). V- variations are added to V1's variations. As a result, the current through R1 increases and the resistance "seen" by V1 is less than R1. This is a typical Miller arrangement where the resistance is virtually decreased by adding an additional voltage.

3. Both V1 and V2 vary but now in the same directions (common-mode input signals). V- variations are subtracted from V1's variations. As a result, the current through R1 decreases and the resistance "seen" by V1 is more than R1. This is another Miller arrangement (known as "bootstrapping") where the resistance is virtually increased by subtracting an additional voltage. Of course, here it is not presented in its perfect form where V- = V1 and the resistance would be "infinite".

4. V2 varies, V1 is constant (non-inverting single-ended input). As other answers said, V2 "sees" a constant resistance of R2 + R4. But what does V1 "see" now? Maybe it "sees" varying resistance? Or this question does not make sense... I wonder...

I was impressed by WhatRoughBeast's explanations that sound temptingly simple... but I can't agree with most of them. Here is why…

Really, two identical voltage dividers (R2-R4 and R1-R2) are driven, accordingly, by V2 and V1 input voltage sources… but they are grounded differently. While R4 is connected to a fixed ground, R3 is connected to an "oppositely moving ground" (the op-amp output). During the differential mode, this "ground" can be considered as a negative resistor with resistance -R3 that destroys the positive resistance R3. As a result, the total resistance of R1-R3 network is only R1, while R2-R4 total resistance is R2 + R4. So V1 will "see" only R1, while V2 will "see" R2 + R4.

And these two input resistances represent what is called "differential resistance". This is not the resistance (R1 + R2) between the two inputs. This is the same resistance as above between the inputs and ground during a differential mode. Usually, in symmetric differential configurations (long-tailed pair, instrumentation amplifier), they are equivalent but here they differ (R1 for V1 and R2 + R4 for V2).

  • 2
    \$\begingroup\$ Thank you for this analysis. I think, it is very important to realize that the input resistance at the node where v1 is connected is DIFFERENT for the two cases: (a) DC input resistance (depends on the DC voltage V2) and (b) dynamic ac resistance for changes of v1 (does NOT depend on the DC voltage V2). \$\endgroup\$
    – LvW
    Mar 23, 2022 at 10:48
  • 1
    \$\begingroup\$ @LvW, Interesting observations... In my opinion, the static and dynamic resistances will be equal only when V1 variations are symmetric in regards to V- ("virtual ground"). This means that either V2 should be zero (in the case of a bipolar V1) or V2 should be equal to DC V1 (common signal). I am not sure if I have explained it in the best way but I think it will do the job. \$\endgroup\$ Mar 23, 2022 at 13:07

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