software

Does anyone know what value resistor I could use to cut a pup's output in half?

I believe it would depend on the pickup. Also, to make it more complicated, our ears operate in a logarithmic manner, so a 50% reduction in the output voltage of a pickup, would not sound like half as loud. To make it simple, it'd be easiest to use a potentiometer to get the desired desired output. Then measure it's resistance at that setting. Then use a resistor as close to that value as you can find.

BTW what do you need this resistor for?

if you're looking at lower output, have you considered a coil tap?

otherwise, because of the logarithm situation, it'd be a pretty decent maths question to decide on a value

tom

Thanks Guys. The Pot Idea is a good one. And, unfortunately, the coil tap won't work in this case. But, it's also a fine Idea. I think I was being a little too specific with my question. Let me rephrase:

Has anyone ever used a resistor to cut pickup output? If so, what was the value and how were the results?

I want less output. I don't care if it's half as loud or not.

Thanks again for the help.

in effect the volume knob is a variable resistor.. surely leaving it halfway would cut output?

otherwise, it'd be 250k or whatever you can find closest to 125k, which are half the values of the usual volume pot values?

tom

I (as usual) recommend the Spin-A-Split mod.

Imp,
Thanks. That's exactly what I needed.

Zhangliqun,
The spin a split won't work in this mod but thanks.