Wiring a cab for switchable impedence?

The easiest solution is to buy the Marshall switching plate - it should be available through any Marshall main dealer as a spare part (I think you may need to order the plate and the internal electrical bit separately). This will do exactly what you want - it doesn't matter that the cab is a 2x12" with 8-ohm speakers rather than a 4x12" with 16s.

If you don't want to do that (the Marshall unit uses a PCB, which may put you off), you need two sockets and a DPDT switch.

I'll see if I can find (or draw) a diagram of how to wire it up, and post it. I have it in my head but that's not much use to you right now...

 

Hi Leon. I whipped this up...it seems to jive in all instances:

1) using the SPDT switch in the "up" position (the pole makes contact with the lower terminal in the diagram) means each jack gives 8 ohms on each side, "stereo"--two speakers for each jack. Use either jack, or both.

2) using the switch in the "down" position means the left jack gives 16 ohms total (all four speakers)--or using the right jack gives 4 ohms total (all four speakers). Use only one jack at a time.

You'll need two mono switching jacks for this (the standard "Marshall" type jacks which have switches on both contacts), and a SPDT switch (use one side of a DPDT if you can't find a SPDT).

http://petersamps.com/series_parallel_switching_412.gif

After you wire it up, you'll probably want to check each position/jack with a meter (by inserting a speaker cable into a particular jack, and using the meter on the speaker cable to give a resistance reading). I think I got this right, but I am pretty tired at the moment. I'll be sure to look at it later on anyway, after I catch a nap.

 

Oh yeah, and you can omit the bottom two speakers from that diagram for the purposes of your 2x12. Imagine those upper two speakers are both 8 ohm speakers instead of a pair of 16 ohm speakers on each side in parallel. Same thing.