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#### hildy

Hi all. I'm new to this forum, and I got here after way too many late nights looking at circuits and whatnot, trying to find an answer to my problem:

How do I make one of those new fancy headphone distribution amps where each musician can select his/her own mix?

details:

1. I know enough to build a high voltage PSU, and then accidentally kill myself with it.

2. I need to take the (6) returns from my headphone busses, buffer them, split each of them into four different phone-amps, then combine the 6 different signals going into each of those phone amps using a little volume matrix that is controlled by each musician.

I think I have the general idea for the topology, but there are a few things (well, a lot, probably) that I don't know, namely, for a low-power amp, is it okay to put a pot on the output, or must the volume control be front-end? I realize this is probably a dumb question, so taunt me at will, but if this were possible, it might allow me to use passive mixers for the musicians.

Later, I'll post a schematic for you all to laugh at, but til then, any help would be mucho appreciated.

-hildy

#### VoltSecond

Idea

Here's my idea: Use 6 volume pots for 6 signals.

Use three cat 5 cables for the routing. One Cat 5 = four decent quality twisted pairs. Use the CAT 5 color code. Center 4 wires are +12, GND and GND -12. (+12 is twisted with one of the grounds.) Outer two wires are audio signals. No matter how you hook the three plugs up, all that happens is the channels get swapped.
If only one cable gets hooked up, the headphone amp still runs.

Cat 5's tight twist gives decent shielding if using a psuedo-balanced signal.

To each of the active wires run a 50K pot. The wiper on each pot goes to a 493K resistor. All 6 resistors sum together at the - input of a quad op-amp running from +12 to -12V. The feedback resistor for this op-amp would be 499K.

There will be two 499K resistors on the + input of the input op-amp. One goes to signal ground; the other goes to power ground. The three 499K resistors and one 493K resistor make a differential amplifier.

The grounds for all signal wires tie together to form signal ground. Signal ground then runs to power ground through ~510 ohms.

The other 3 op-amp outputs are also run as inverters with a gain of ~2, each op-amp output goes through a 10 ohm resistor for load sharing. All three 10 ohm resistors are tied in parallel on the headphone side to drive the headphone. (3.3 ohm output impedance, good for most headphones.)

At each op-amp, put a 220uF, 20V from +12 to power ground and the same on -12V to power ground.

Headphone common returns to power ground.

On the send end, all grounds are common.

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You could also do it with two cat 5 pairs by twisting the +/- 12V and using signal grounds as the ground power return. If you did that, I’d shunt regulate the headphone op-amp and leave out the 510 ohm resistor between grounds.

A simple shunt regulator would be to use two 1n5908 TVS in series to make 12V and raise the distribution supply voltage to +/- 15 or +/- 18V. A resistor from the +18V to the shunt regulator would be poor man’s CCS for the shunt regulator.

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