What is the input impedance of a differential amplifier?

I should have said for the general case, like

the non inverting input just sees 2 100k resistors in sreries asuming the op amp impedance can be neglected so is just purly 200k.

the inverting input just sees the first 100k as the op amp tries to keep the end of that resistor at half the potential of the non inverting input, however a change in voltage on just the non inverting input cuases a change in curent on the inverting input.

Colin =^.^=

Differential input impedance is 200k on the +in and 66.66k on the -in common mode input impedance is equal and 200k on both inputs. Let us have +Vin= V1 and -Vin= V2 You can find that out by first looking at the +in, where the level is

1/2Vin. The opamp steers its output in such a way that there is the same voltage on the -in. This means the voltage across R1 is 1.5Vin. since the resistor is 100k this corresponds to R= 100k/1.5 =66.66666k Now do the same with V1 and V2 both +Vin to obtain the common mode input impedance. This is taking the opamp as ideal, which is pretty much what you can also measure. I think you should measure the current from a fixed voltage source. The ohms range it is pretty meaningless and might damage the meter depending on the supply voltage of ther opamp. The additional current can lead to unexpected output voltages.

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ciao Ban
Bordighera, Italy

That is what I was referring to. Are you able to follow my explanation? I tried to solve the different possibilities.

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ciao Ban
Bordighera, Italy