I have a clapton midboost and was wondering if there was such thing as a pot that cut out of the circuit at zero to conserve the battery. As an alternative I could switch the pot from the J taper that comes stock with it to another taper and just have a push-push pot to bypass it. Does anyone know how I would wire it?
wire a regular no load pot backwards so t he click is at 0 instead of 10.

that or you can take the pot apart and either scrape the carbon half circle disc or paint over it with something. The easiest approach is to just use a push pull that can disconnect the tone capacitor from the circuit.

with a push pull though remember it's a simple on/on double pull double throw switch with an on/on mode. So focus on the bottom two contacts. The middle on the left and right are commons meaning used unconditionally and the one closest to the contacts does our effect. We are doing a simple effect so really we're only using one row of contacts.. anyways put simple if i was to make a push pull a no load.

the capacitor would be wired to the bottom contact of the selector
a wire would be soldered from common (push pull) to ground
this removes the capacitor from the circuit. there is a different way to achieve this as well. If it's a master tone I'd just go about doing this the david gilmour black strat way.

this is an insightful video, i hope this is one of the videos the guy cut his finger nails properly hahah you'll see why the more videos you watch of his but he's a bright guy.
Last edited by Tallwood13 at Oct 31, 2015,
Do you know the difference between a j taper pot and an audio or linear taper one? I assume by the name it is similar to the J shape of the audio taper. Also if I wired a no load pot backwards, wouldn't it also decrease the midboost as I got closer to ten instead of increase it?
Last edited by 70001407 at Oct 31, 2015,
never heard of a J taper pot until you mentioned it. Usually I see
A500k - audio taper / logarythmic
B500k - linear , a smooth 10 to 0 , on a chart it makes a perfect line
D500k - seen on a few ibanez guitars nothing to really go by on google
and anti-log I know exist , they roll off volume (lets say) backwards, hence the "anti"

seymour duncan forums and another blog talked a bit about J-taper , it has to do more with resistance. Here's an article on it. The seymour duncan page is excellent for super technical questions as there is a few engineers on there.

anyways .. lets assume the no load pot is audio taper or linear which are very predictable. 100% is still the same , 50% is still the same just the click happens at 0%, kind of strange but a simple solution.