Switch between pickup #1 soloed and pickup #1 + #2 parallel?

So all I want to do is to be able to switch between the P pickup of my P/J bass being soloed and then the P pickup + the J pickup both full on combined in parallel connection.

I intend to use a DPDT 6 pin On-On switch for it, as that is what I got at the moment, and I think I know how to do that, but not 100% sure, so could anyone please tell me if I got it right?

I should probably point out that I intend to bypass all the pots and let the pickups be connected directly to the guitars output jack socket, as is how I got my bass is wired currently.

Believe me, this is all I need to be able to do, and I don't want more options or anything.


So this is the switch :



So would this be the correct way to do it, to achieve what I want? :

1 = J pickup hot wire

4 = J pickup ground wire

2 = P pickup hot wire + my hot connection to the output jack socket's tip

5 = P pickup ground wire + my ground connection to the output jack socket's sleeve

3 + 6 = Left empty


If not, then how would I do it correctly?

Wire the jack directly to the pot. You only need 1,2 and 3. 2 would go to the pot hot lug and also shorted to 3. Bridge got to 3. Neck hot to 1 The pickup grounds can go to the top of the grounded pot casing

scottosan":6ykn311a said:

Wire the jack directly to the pot. You only need 1,2 and 3. 2 would go to the pot hot lug and also shorted to 3. Bridge got to 3. Neck hot to 1 The pickup grounds can go to the top of the grounded pot casing

Click to expand...

I'm having a little hard time wrapping my head around the way you formulate your suggestion, but I do understand the part about just leaving out the ground all together, just switching between the hot signals of the two pickups.

However if I understand what you are saying right, and got how an On-On switch works right as well, connecting the hot wire of one pickup to lug 1 and the hot wire of the other pickup to lug 3, as well as shorting lug 2 and 3 together would give me the 2 pickups wired in parallel no matter what position the switch is in, in which case I could as well just have skipped the switch all along and wouldn't achieve what I am looking for.

As I understood it an On-On DPDT switch works so that in one position lug 1+2 on one side would be connected, and lug 4+5 on the other side would be connected, and then in the other position lug 2+3 on one side is connected and 5+6 on the other side is connected, which if I got that right, my original idea would work, although could be made simpler by leaving out the ground connections altogether and just connecting all the ground wires one place, whereas your suggestion would just connect the 2 pickups in parallel in both of the 2 possible positions of the switch.

I might very well have misunderstood this though.

Also as I actually wrote in my OP I am not using any of the pots but simply wiring directly to the jack output socket with this switch in between.

Looking at the diagram you posted imagine the switch position on the other side. If the switch is down, pin two would short to pin one. When you flip the switch up, pin two would short to pin three. If I understand you correctly you want one position to be full bridge and the other position to be bridge and neck in parallel. So, if you connect bridge hot to pin 3, assuming up is bridge position, two is always shorted to 3, so it’s always on. Connecting the neck hot to pin one would only engage when switch is down. And it would be in parallel with the bridge.