quad headphone amp

No. They'll all be independent of each other.

The load that this circuit presents to the source is 100k. That's very high. Four of them in parallel, 25k, still quite high. And the position of the pot does not change the load significantly; so the position of the pot will not change the output voltage from the source, which means that the other amps will not notice a difference.

Jester, I didn't look at your schematics, but if they are IC based there shouldn't be any problems. Normal headphones have an impedance between mostly 32 and 600 Ohms(those legacy HiFi ones) and require maybe maximal 100mW for a decent loudness. Some DJ headphones can go up to 400mW, but small earphones are blown with that level. When we calculate the required voltage Vrms= sqrt(P*R) = sqrt (32*0.1) =1.8Vrms for those newer ones and 7.5Vrms for the legacy ones. The DJ type would need 3.6Vrms. Now we can use a resistor of 120ohms with at least 1W rating in series with each output to drop the unrequired voltage and connect to a 10Vrms output amplifier(25W into 4ohms). Any old stereo amp will do. To adjust the level individually, we put a 1W-250 ohms linear potentiometer(best is wirewound or Cermet) in front of each headphone. Maybe a little box for each person with 2 mono pots (dual ones are hard to find, but try PartsExpress), so you can adjust the balance too. You can wire all boxes in parallel and connect as many as you need. They do not influence each other. To avoid damage to the amp in case someone shorts the incoming wires, put two 10R/10W resistors directly on the red speaker outputs and your outgoing wires. You only need one black gnd wire, since these are in parallel.

see diagram with fixed font.

AMP speaker red ___ terminals L o-|___|--o----o----. mini jack .---| 10R/10W | L +red | .------| | | .-. ___ | .---| L | | 250R| |

Thank-you everyone! Jester