Troubles Finding This Part

[sweatyk]

I have a bad mixing/blend pot on a Ken Smith bass with active pickups. Disassembled the pot completely and found one side of it bad. I have looked on line for the last 2 days and cannot seem to find one of these. I was hoping that someone here might know of a place I could find one. Ken Smith doesn't seem to sell individual components rather an entire replacement PCB. I know someone has to sell these.

[drewl]

Looks similar to ones in Boss pedals maybe?

[g1]

The 2 sections have different tapers, so it will be an uncommon part, possibly custom ordered for Ken Smith.

[madkatb]

If you can get a twin 100k pot, linear, strap a 15k resistor from wiper to ground on one of them to emulate a log pot. Probably easier than trying to find a "mixed" twin pot.

[Enzo]

Ken Smith doesn't seem to sell individual components rather an entire replacement PCB.

I know nothing about Ken Smith, but I have to ask this: he doesn't SEEM to sell parts. Does that mean you asked them directly and they said "no", or are we assuming this because we don't see parts on a web site?

madkatb's idea sounds reasonable. But aside from the Smith people, guitar repair shops often have the works out of guitars left over from repairs or scraps. You might inquire of some larger guitar shops if they have one they would part with. Here in Lansing I would call my friends at Elderly Instruments and see if they could help, for example.

[J M Fahey]

If you can get a twin 100k pot, linear, strap a 15k resistor from wiper to ground on one of them to emulate a log pot. Probably easier than trying to find a "mixed" twin pot.

That's a kludge which only "works", sort of, in some places and not in others.
If a pickup drives a Log pot (common case in a pasive Musical Instrument) and you try to simulate a Log one out of a Lin one, you are in trouble.

Problem is that the compensating resistor , 1/6 to 1/10 of the pot value, ends up being in parallel with the full resistance if set to 10 .

Suppose you have a 500k pot, and you "log" it with a parallel 100k resistor ... set it to 10 and the pickup is now loaded with 500k//100k ... some 80k .

The pickup won't like that at all, it will sound very different.

Back to your Ken Smith pot: is this the pickup balance one?

If so, you can use a 2 x 100k linear one, it's not that critical in that position.

As of the pot itself, yes, it's a custom job, specially because of the small square body and the extra long threaded mounting collar.

Funny thing is, the pot carbon tracks are exactly the same ....... just one is mounted "forwards", the other is mounted "backwards", the wipers are in the middle, between the wafers, so mechanically they turn the same way, electrically the opposite way.

I have had such pots custom made in orders as low as 100 of them, first they barked at the "odd/unusual" order, then understood when I explained that I wanted the exact same wafers, but one *assembled* backwards, no big deal.

PD: what is the actual problem with yours?
Maybe a whiff or 2 of "magical" Deoxit or Stabilant can solve it.

[madkatb]

Not disagreeing but I'm missing something.
At 10 ( max volume) there would be signal going into the pot, directly to the wiper and out of the pot without "seeing" any resistance. Similarly, at 1 ( volume off), the signal would pass through the full resistance of the pot and then directly to ground without "seeing" the wiper.
I've used this kludge successfully but never in this situation so, as I said, I'm not disagreeing, just can't get my head around it.

[Richard]

At 10 the signal goes directly to the output with no resistance but is connected to ground through 100k resistance. At 50% the signal passes through 50k to output and passes to ground though 50k. Generally passive pickups seem to have more high end frequency with a greater resistance to ground. Also highs seem to deteriorate more when rolling off the volume. I believe active pickups are less susceptible to these problems.

How much difference would a different value pot make?

[J M Fahey]

Not disagreeing but I'm missing something.
At 10 ( max volume) there would be signal going into the pot, directly to the wiper and out of the pot without "seeing" any resistance.

You are misusing the word "seeing", that's why it does not make sense to you.
It really means whatn load is in parallel
Suppose you have 2 different pots, one a proper 100k Log/Audio pot, the other a 100K linear pot, with a 10k resistor from wiper to ground, to twist its curve into something approaching log, let's see what does the pickup "see" which means what load does it have to drive.
In this case, what resistance it has connected from hot to ground, in parallel with the pickup, and sucking current from it.

1) pots on 0
In this case both are the same (0 output) and the pickup sees 100k in either case

2) pots on 5
a) true Log:
* signal is 1/10th of what the pickup puts out, what's expected from such a pot. (voltage divider 90k/10k)
* pickup sees 100k load
b) fake log
* signal is 1/6th of what the pickup puts out (voltage divider 50k / 50k in parallel with 10k)
* impedance is 50k + 50k//10k , some 58k
So we see it's close to what a true Log would do, but signal is some 20% too high and load some 40% down.
So far so good, most calculate this and become happy, I've even seen posts stating "I'll never buy a Log pot again".
Really? ...let's see what happens when the Musician sets the pot to 10 :
a) real Log:
* signal is 100% what the pickup puts out
* load impedance is still 100 k , the pickup behaves and sounds as expected.

b) fake Log:
*load impedance now is brutally low: 100k//10k , some 9k , 1/11th what is expected.

The pickup loses highs (it's highly inductive) and even mids and lows, because its DC resistance now is significative trying to drive such low impedance.

Where does this kludge work ? (sort of):
at the output of preamps or after buffers, where it's 11:1 impedance variation can be tolerated.

And forget using it in tone controls, where it messes everything up big way.

In such cases if no other option, I prefer to leave the linear pot uncorrected, most of the adjustment will be cramped towards one end or the other, but all other parameters won't be affected.

In this case, a pickup balancing/mix pot, dual linear works quite well and is what most others do.

[Mick Bailey]

The first thing I look for in active circuits is where the pot is located and what the function is. Are they active pickups (embedded electronics), or regular pickups feeding a preamp?

Sometimes the blend control is located after the pickups but before any electronics (even if the pot is mounted on the PCB). Often this situation is more tolerant of pot changes - sometimes for the better - than a pot that's located after the first preamp stage. I always think an MN taper pot works better than an AC pot for a blend control.

[madkatb]

Ah. Gotcha. I was looking only at the signal path and ignoring the load impedance. Thanks Juan.

[sweatyk]

I'm sorry that i left this thread hanging. In conclusion the owner of the bass called Ken Smith himself. He sold him a replacement pot. I installed it and now his world is a happy place......the end.


Thanks for every ones input as usual!