Making a 500K pot into a 250K pot?

atomheartmother

Senior Member
Messages
594
Is there anyway to essentially turn a 500K volume pot and 500K tone pot into each a 250K pot for just one pickup (of two)?

I have a Tele with a neck humbucker and single coil in the bridge. The problem is that the neck needs 500K pots (it's too dark right now), but the bridge needs 250K pots (too bright with 500K's).

I could just put 500K's in and turn the tone pot down halfway, but that really messes up the middle position.


So I need someway to make the 500K pots each 250K for just the bridge pickup (for both the bridge and middle positions), and keep the 500K pots at 500K for the neck pickup (for both neck and middle positions).

Is there anyway to do this with resistors or anything?
 

Chiba

Platinum Supporting Member
Messages
7,832
I don't think you can make one pickup 'see' 500k on a pot and make the other pickup 'see' 250k. If I was in your situation, I'd wire up two separate volume pots - a 500 for the bucker and a 250 for the single, of course - and put a little 'fake' tone circuit (mimics a dimed tone pot) on each one. That would work for me because I never use my tone pots anyway, I regularly switch them to volume pots.

--chiba
 

fullerplast

Senior Member
Messages
6,781
Originally posted by Clorenzo
Simply put a 240 / 270 kohm resistor between the hot wire of the bridge pu and ground.
That won't really get him what he wants. First of all, that will put the 240K in parallel with 500K giving you only about 170K...which would likely be too dark. If you only had the bridge pickup to deal with you'd want to use a 470K or 510K to get to about 250K total.

The real problem though, is the middle position. Since both pickups are in parallel, anything you do to reduce the pot resistance on one pickup will also affect the other pickup. He may want to try a few different pot or resistor values on the bridge pickup to see if he can come up with a good compromise between bridge only and the middle position. Maybe even put a trim pot in the control cavity to experiment.
 

Clorenzo

Member
Messages
1,930
Originally posted by fullerplast
That won't really get him what he wants. First of all, that will put the 240K in parallel with 500K giving you only about 170K...which would likely be too dark. If you only had the bridge pickup to deal with you'd want to use a 470K or 510K to get to about 250K total.

The real problem though, is the middle position. Since both pickups are in parallel, anything you do to reduce the pot resistance on one pickup will also affect the other pickup. He may want to try a few different pot or resistor values on the bridge pickup to see if he can come up with a good compromise between bridge only and the middle position. Maybe even put a trim pot in the control cavity to experiment.
If I understood him correctly, he wants the bridge pickup to see the load of TWO 250 k pots, which is 125 k (note that the loading of a fully open tone pot is equivalent to that of a resistor of the same value connected to ground, the cap contribution is negligible unless the pot is turned down). 240 k in parallel with two 500 k pots equals 122.4 k, close enough. This is actually done on the American Telecaster HS, see wiring diagram:
http://www.fender.com/support/diagrams/pdfs/AMERTELEHS/SD0118600APg2.pdf
and parts list:
http://www.fender.com/support/diagrams/pdfs/AMERTELEHS/SD0118600APg3.pdf
I have done it myself on guitars where I combined humbuckers and single coils and it works nicely.

You are right about the middle position though: both pu's will see the equivalent load of two 250 k pots. If that isn't acceptable, there's a trick that will improve things and actually solve another problem at the same time: put a resistor in series with the bridge pickup hot wire, after the parallel resistor and before the switch. When two pickups of different dc resistances are connected together in parallel, counterintuitively, the higher resistance one (i.e. normally the one with higher output) contributes to the overall sound less than the lower dcr one, due to its loading. That's why often guitars with two HBs where the bridge one is much hotter than the neck one sound very similar in the neck and neck + bridge positions. Putting a resistor in series with the low dcr pu decreases the loading (increases the load resistance) and there's usually a sweet spot where you get balanced contributions from both pickups. A temporary pot is the best way of finding the right value, then measure and replace with a fixed resitor. I've done this in a couple of guitars, for example my PRS CE22 with Dragon I pu's (neck about 8 k, bridge about 17 k) and in my case the right value was 68 k.
 

John Phillips

Member
Messages
13,040
You can do exactly what atomheartmother wants, but you have to rewire the switch if it's the standard Fender wiring.

In the Fender scheme, the two switch rotors are connected to the pickups, and four of the static terminals to the volume pot.

You need to reverse this so that the two rotors are connected to the volume pot, and each pickup goes to two of the static tags.

Then, connect the static tag for the bridge switch position (on the opposite side from the pickup) to ground via a 500K (470K is the nearest standard value) resistor. This simulates a 250K volume pot in the bridge position only.

If you need to take a tiny bit more top off, you can add another 470K resistor from there to the tone cap as well, which simulates a 250K tone pot, but this is exactly the same as backing off the tone control very slightly, so it may not be necessary. It's the change in the volume pot loading which is the most important.
 

fullerplast

Senior Member
Messages
6,781
Originally posted by John Phillips
This simulates a 250K volume pot in the bridge position only.

John, I think he wanted the 250K load on the bridge pickup in *both* the bridge position *and* the middle position.
 




Trending Topics

Top