Hi PB.
I'm linking a graphic of how the stock switch is wired. It only shows one switch with a mono jack with the sleeve connection feeding the switch directly and the tip connection feeding the switch through an LED in series. For 2 switches, the connection is through a stereo/TRS cable/jack. The sleeve would be common to both switches with the connection to the second switch being the ring connection, also through an LED in series. I'm not sure what type of circuit it's called but basically the switch just shorts the tip (or ring) to the sleeve which activates a relay in the amp and switches channels. There is about 3V DC present from the amp but in the stock switch, there is no resistor on the LED. I assume the output current is limited inside the amp?
(https://forum.bottlehead.com/proxy.php?request=http%3A%2F%2Fi46.photobucket.com%2Falbums%2Ff103%2FMathamology%2Ffootswitchbefore.jpg&hash=2ff10a003ea687a0cbf10c1c52bda12fb614b5a2)
With the stock switch wiring, if an LED fails, I lose the ability to switch channels with the footswitch. I'm wanting to keep the LED's for indication but not have them in series with the connection. I'm thinking now that I could keep it simpler and use DPST switches. I think that would allow me to wire one pole of a switch from the tip connection
through an LED and the second pole of that switch
directly from the tip connection. Then sleeve would be common. Same for the second switch but from the ring connection rather than the tip. If I'm thinking right, this would work ... clicking the switch on would activate both the channel switching relay in the amp and the LED but if the LED fails the channel switching still works and the two poles aren't connected inside the switch so the LED anode stays isolated from the + voltage on the other switch pole (from the tip or ring).
I'll work on drawing a diagram for the DPST switch. Have to do it by hand though so it wont be pretty.