Input impedance for a difference amplifier OpAmp circuit

This is my answer to SE EE question Input impedance for a difference amplifier OpAmp circuit.

My answer

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...

My comment to WhatRoughBeast

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).

My comments

1. 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.