Balanced hi-Z buffer/preamplifier - will this opamp circuit work?

[GaussNewton]

NOTE: Electronics noob here, so have patience...

I'm trying to put together a small preamplifier/buffer circuit for hooking up piezo contact microphones. The basic idea is to use two opamps to make a balanced in/balanced out circuit, as shown in the schematic. Since it will be phantom-powered the opamps need to be run single-supply.

My question is: do I need to explicitly reference the source to Vcc/2 or not? Im guessing yes, since without it I don't know where the input bias current will flow from. Also, do I have to do the same thing on the output side (to make way for the second input bias current)?

And if this idea is idiotic for some reason... please do tell.

[GaussNewton]

On second thought... biasing the signal to Vcc/2 like shown in the schematic will result in either low input impedance (low ohm resistors) or poor noise (high ohm resistors). Hmm...

[GaussNewton]

A: Yes. After a little reading, it's obvious that the the input bias currents NEED a return path (or the inputs will float).

[Nigel Goodwin]

Originally Posted by GaussNewton

On second thought... biasing the signal to Vcc/2 like shown in the schematic will result in either low input impedance (low ohm resistors) or poor noise (high ohm resistors). Hmm...

High input impedance preamps have to have high value resistors, there's nothing intrinsically noisey about them - but any high impedance input is more likely to pickup noise and interference.

[GaussNewton]

Originally Posted by Nigel Goodwin

High input impedance preamps have to have high value resistors, there's nothing intrinsically noisey about them

Really? I was under the the impression that all resistors produce noise, and that the noise voltage grows with the square-root of the resistance (Noise (electronics) - Wikipedia, the free encyclopedia).

Originally Posted by Nigel Goodwin

but any high impedance input is more likely to pickup noise and interference.

Can you give an explaination as to why? This doesn't seem obvious to me as a noob...

[Nigel Goodwin]

Originally Posted by GaussNewton

Really? I was under the the impression that all resistors produce noise, and that the noise voltage grows with the square-root of the resistance (Noise (electronics) - Wikipedia, the free encyclopedia).

But you're talking about TINY values - as I said, all high input impedances use high value resistors, you don't have a choice, and it's not a problem.

Can you give an explaination as to why? This doesn't seem obvious to me as a noob...

Think of it as the bottom part of an attenuator, and incoming interference as the top part - the lower the bottom part, the more the interference is attenuated.