Overlooking the obvious: the input impedance of a difference amplifier

RE: Overlooking the obvious: the input impedance of a difference amplifier

Steve O'Connor — Sat, 04 Sep 2021 03:08:55 GMT

Hi, I like this post, however doesn't Equation 7 mean that R_in(N)could be negative.

RE: Overlooking the obvious: the input impedance of a difference amplifier

tim hughes — Thu, 07 Nov 2019 18:06:17 GMT

John, the incremental impedance DeltaV/DeltaI when looking into the positive input is R3+R4 the same as you state, but the corresponding incremental impedance is just R1 when looking into the diff amplifier's negative input because the R1- R2 node is at virtual earth . Direct DC V/I values rather than impedance= DeltaV/DeltaI is what you are discussing above.

For example, the case you indirectly mention where you want to make an AC coupled differential amp and you need to know the impedance when putting Capacitors in series with the inputs you need to to use the above mentioned incremental impedances and make the time constants R1C1 = (R2+R3)*C2 to achieve reasonable common mode rejection through balanced input time constants. (C1 is in series with R1,C2 in series with R3).

( In practice because you cant match capacitors much better than about 1%, CMRR at low frequencies wont be better than 40dB if you are lucky. )

For a concrete example for the case of all equal resistors R1 to R4 then C1 must be twice as large as C2 and the -3dB cutoff will be 1/2*pi*R1*C1 (or equivalently = 1/2*pi*C2*(R3+R4) .

RE: Overlooking the obvious: the input impedance of a difference amplifier

John Caldwell — Tue, 08 Sep 2015 22:20:13 GMT

Hi Ken,

Thanks for bringing this to our attention. To me, the INA133 datasheet is correct and the INA134 datasheet may need to be updated. We can calculate the common-mode input impedance pretty quickly using the equations in the above post. First, if we have a common-mode voltage source the input impedance it "sees" looking into the INA134 is RIN(N) in parallel with RIN(P) (You can imagine tying the two inputs together and then applying a voltage source to them). As usual RIN(P) is 25k + 25k = 50k. RIN(N) in this case is a bit different, since VIN(P) = VIN(N), then VIN(P) / VIN(N) = 1, and equation 7 simplifies to RIN(N) = R1 / 0.5 = 50k. Since RIN(P) is in parallel with RIN(N) for common mode voltages, 50k || 50k = 25k. The differential input impedance can be done "by inspection" by remembering that there is a "vitual short" between the two op amp inputs, and therefore R1 and R3 are in series for differential input voltages, 25k + 25k = 50k.

RE: Overlooking the obvious: the input impedance of a difference amplifier

Ken Dillinger — Tue, 08 Sep 2015 20:44:14 GMT

Because I like being "that guy", there seems to be a problem with the INA134 datasheet.

The INA133 datasheet which has the same value of resistors for the network (25kohm), but states the the common-mode input resistance as 25kohm and the differential input resistance as 50kohm.

The only clue is that for the INA133 the common-mode voltage is specified at VCM=0. So, I presume, that means the input resistance is simply the resistor on the inverting node. If driven from a transformer or other floating source the differential impedance of 50kohm makes sense.

The INA134 has 25kohm resistors for the network and the datasheet specifies 50kohm for both common-mode and differential input resistances.

I always presume people are smarter than me or I missed a subtle clue, but this looks inconsistent.

-Ken

RE: Overlooking the obvious: the input impedance of a difference amplifier

Haroldo Amaral — Mon, 17 Aug 2015 07:15:30 GMT

Hi, Excellent post.

The same analysis can be used with other configurations.