One of these, I don't have a schematic for it

Last edited by Mr Songwriter at Sep 18, 2006,
I've just found this diagram on Stewmac for a Push/Pull presumably it's the same for a push/push?

Incidentally this is for coil splitting on a pair of humbuckers connected to 1 volume and one tone pot. Does anyone know how the connections in that diagram relate to the potentiometer?
Quote by forsaknazrael
He's talking about a push/pull. Just doesn't know it.

Oh ta, I bought a push/pull in the end, I'm just going to use my multimeter to figure out how it works.

The pot is completely different to the switch, wire the pots as volumes or tones normally. and use the switches for other things. If you can tell me, what pickup layout you wanna use, the brand of pickups, and what kinda selector switch, I can doodle something up.
These go to eleven...
Last edited by JimPlaysGuitar at Sep 18, 2006,
Nice one Jim! thanks a lot; I'm using a Seymour Duncan Jazz in the neck and an SD JB in the bridge (they've both got four conductors) and this is being fitted into a PRS Santana SE with one tone, one volume (which I would like to use for the switch) and a 3 way pickup switch, I want to be able to switch between normal (series wired?) humbucker operation and using both pickups in single coil mode.

Edit: it's a push/pull switch just like the one in your diagram.
Last edited by Mr Songwriter at Sep 18, 2006,
Ah thanks a lot, so that switch is switching between connecting the coil split pickups and them not being connected, isn't it?
Last edited by Mr Songwriter at Sep 18, 2006,
Well pulling it will make them singles. Making your selector switch then select..

1. Bridge inside coil
2. Bridge and neck inside coils parallel
3. Neck inside coil
These go to eleven...
Ah right, I've just rechecked the instructions that came with the pups and they say:

Green = beginning of adjustable coil (south)
Red = finish of adjustable coil
Black = beginning of stud coil (North)
White = finish of stud coil

Adjustable coil = South magnetic polarity
Stud coil = North magnetic polarity.

...putting that together with what you said makes it a bit clearer, I think I'll just draw myself a circuit diagram that includes the info above to be sure.