I'm building a footswitch for my 6505 as the Peavey one is too big and I don't use the loop any more.
The question I have is this:
On the peavey footswitch, the connector from the amp is stereo, where tip is FX Loop, ring is channel, and the sleeve completes the circuit. The footswitches are latching, and when latched, connect the tip/ring to the sleeve (back to earth, as it were) and turn the FX on/lead channel on. I'll need to get a stereo jack, but can I just leave the tip unwired? I can't take it to ground since it would enable the loop, which I don't want.
The second question I have relates to the LED's; the internet seems to agree that there is some 30V (!) coming out the 'hot' lines for the footswitch, and the amp has an internal resistor for the LED's. If I change the LED's I'll almost certainly have to use a different resistor, is it really as simple as measuring the resistance of the LED and going from there?
It says Classic 30 but they are the same footswitch!
To answer your questions:
1) Yes you can just leave the tip open and not connect it. You also could wire it to a small dip switch and place it inside the enclosure that will let you turn it on if you want.
2) The resistors needed will be dependent on your LEDs.
You can used the calc here to get the correct value. LED Resistor Calculator