# Is it possible to make a high-impedance voltage follower using a fully-differential op amp?

With a single-ended op amp, I can create a unity-gain voltage buffer like this:

simulate this circuit – Schematic created using CircuitLab

The input impedance of this circuit is equal to the input impedance of the op amp, which is normally very large.

I'm trying to reuse an existing analog front-end circuit that has a differential input connected to a fully-differential op amp. The op amp is wired to provide gain, similar to this:

I know I can set $$\R_f = R_g\$$ to get unity gain, but if I understand right, the input impedance of the circuit is $$\R_g\$$. If I want a differential input impedance on the order of megaohms, is my only choice to put megaohm resistors in this circuit? Or is there some other way to wire it that can take advantage of the naturally high input impedance of the op amp itself?

• @HenningLarsen It's a THS4531A. There are pin-compatible replacements with higher input impedance. Feb 9 at 15:22

The input impedance of this circuit is equal to the input impedance of the op amp, which is normally very large.

It is even higher because of the following negative feedback (bootstrapping). It can be explained in such a way. The op-amp output voltage is equal and opposite to the input voltage (travelling the loop) and connected in series; so it is subtracted from the input voltage. As a result, the resulting voltage variations across the op-amp differential input are small. So the ratio dVin/Rin is small and the current variations are small. This means very high input differential resistance.

If I want an input impedance on the order of megaohms, is my only choice to put megaohm resistors in this circuit?

The professional solution is to put voltage followers (like the one in the first picture) or non-inverting amplifiers before the fully differential amplifiers. This topology is used in instrumentation amplifiers.

"Putting megaohm resistors" will cause undesired voltage drops across them and other problems.

Try this circuit:

Inspired by datasheet for LTC6362

If you have an existing circuit but with resistors, it is probably easy to test.

• This design can't buffer a differential input, however. This is a buffer that also acts as a single-ended-to-differential converter. Feb 9 at 16:23
• @Heath, you are right. However, It was not clear from the question if a differential input is required or not - thus my answer. Feb 9 at 16:50
• Fair point! I assumed it was as a differential input was shown, but it is ambiguous. Feb 9 at 16:51
• @Hearth I do have a differential input. I'll edit the question. Feb 9 at 23:53