sjaustin wrote:
OK, help me grasp what's going on there. Why are both output jacks wired to the common row of the 3PDT?
- Okay, I am still pretty much in the noob category, but I assume that its not unlike how most 3PDT switch wiring diagrams for effects pedals have you connect the board input to the ground when the effect is bypassed, to reduce noise, especially on high gain pedals. They are doing the same thing here on whichever output channel is not being used at any given time.
sjaustin wrote:
And why is the input wired to both the top and bottom row? It seems like that would make everything be connected all the time. I thought I understood these switches…
I assume that you are referring to the input tip. If so, then take another look.
When the switch is in the "up" position, the signal goes from the input jack tip to lug 1 on the switch. Its then mechanically connected to lug 2 by the switch and on to output jack A.
Yes, a jumper wire also goes go from lug 1 down to lug 6 but it goes no further, because the switch is in the "up" position, so lugs 5 and 6 are not connected at that time. Instead, lug 5 is receiving the ground from lug 4, to pass it along to output B, which is is the output not being used when the switch is in this position.
When the switch is in the "down" position, the signal still goes from the input jack to Lug 1, but now the switch is not connecting Lug 1 and lug 2. Instead, the jumper wire carries the signal to lug 6 (like before) but now the switch mechanically connects lugs 5 and 6, which sends the signal to output B.
Meanwhile lug 2 is now mechanically connected to lug 3, which is receiving ground from lug 4 via a jumper wire. So the ground goes from lug 4 to lug 3, to lug 2, again feeding a ground signal to the unused output A.
At least that is my understanding, though its entirely possible that I may be wrong!
If so, hopefully one of the gurus will ring in and straighten me out...