Simulate Pot Load Using DPDT?

[jakeac5253]

A friend of mine wants to build a guitar with one pickup and no knobs at all and I told him that pickups don't produce current without a load present. This got me thinking though, he could simulate a resistive load using a DPDT switch to activate the load full on/full off couldn't he?

Using this as a diagram:


He could, say wire the pickup hot to the B and pickup cold to the A, then run the hot output from D and cold output from C. And using a 500k resistor, he would put it.... umm, between B and A? And putting the switch to the other side would activate the terminals with nothing attached, effectively turning the guitar off.

OR, he could do something cool like jumper B to F and A to E and put a 250K between E and F to simulate a different pot load with the switch in the opposite position. Or simulate a 500k pot turned halfway down by putting a 250K between F and E AND another 250K between F and D. This is a real brain teaser, but I'm convinced that it could be done. Right?

[pdf64]

'A friend of mine wants to build a guitar with one pickup and no knobs at all and I told him that pickups don't produce current without a load present'
The pickup will work fine without a pot. Just wire it straight to the jack socket. A kill switch would be useful - a kill switch just shorts across the pickup.

[David Schwab]

The amp is the load. But a lot of people wire their pickups right to the jack to get a louder brighter tone.

If you want a load to simulate the pots, just wire a 500K or 250K resistor between hot and ground.

If you want to make it switchable, a wire from hot would go to C, and the resistor would from A (or E) to ground. If you use a on-off-on switch, you can use two different resistors, one wired to A and the other wired to E, and connect the other end to ground.

Personally I use the volume knob to mute the signal when I'm not playing, so you can also wire the one side of the on-off-on switch directly to ground (from E) to act as a kill switch.

[jakeac5253]

I stand corrected. See, I just thought that by separating the inductor and the load like that would result in a lot of treble bleed because of the capacitance of the cable.

[Ronsonic]

A kill switch would definitely be a good idea. Both as an effect and for on-stage logistics.

[David Schwab]

I stand corrected. See, I just thought that by separating the inductor and the load like that would result in a lot of treble bleed because of the capacitance of the cable.

You aren't separating the inductor from the load, you are just removing one of the loads. The cable capacitance is going to effect the tone either way. Having pots in the guitar isn't going to make the cable capacitance go away.