Hey guys,
This might be very straight forward for one of you.
I'm trying to source a replacement POT for a set of Logitech speakers.

It says B50k on the part that's attached to the volume knob (not shown).

Thing is this one seems to have 8 pins.
Front to back: 3 pins, 3 pins, 2 pins.

Am having trouble finding the same type.
I guess it has more pins cause it had the power incorporated in the volume knob.

I was taking them apart to use as the sound for an arcade cabinet that I'm finishing, but the volume control knob was attached to the case and broke the POT when I opened the speaker.

Please help!

 

That's a linear, 250k pot with a power switch on the back. Try to read who made it. www.mouser.com sells parts.

 

As per my book, B denotes linear. Just 2 hrs ago I mounted one of 10K.

 

I don't need the switching capability.
Can I just jump the two pins at the back and use a regular 6 pin potentiometer in place?

 

That looks like a normal 3 pin potentiometer. Look closer.

Edit: Missed again.

 

That looks like a normal 3 pin potentiometer. Look closer.

The whole component has 8 pins for sure. You can see the 3 rows of pins from the side shot.

 

How about replacing with this?
(it doesnt need to be mounted on the board as I need to have the volume accessible from outside the cabinet.

http://www.radioshack.com/100k-dual...fn1=productType&prefv1=Potentiometers&start=3

 

OK. I see. You ripped the conductive element out of the front of the pot on the board. It really is 6 pins.
Your Radio Shack suggestion is an audio taper with less than half the resistance you need.
You might have to order from the big boys, mouser, digikey, Jameco, etc.

 

OK. I see. You ripped the conductive element out of the front of the pot on the board. It really is 6 pins.
Your Radio Shack suggestion is an audio taper with less than half the resistance you need.
You might have to order from the big boys, mouser, digikey, Jameco, etc.

Apologies if this is stupid but isn't the original a 50k?
Am just going by the numbering.... B = linear, so would assume that b50k = linear 50k
If it was a 250 would it not be B250k?

 

I thought you said 250 k in the first post. I guess I need to read slower.

Third one down. $1.57

http://www.mouser.com/Search/Refine.aspx?Keyword=ganged+potentiometer

and there are others, better suited to the shape you need.

 

No worries!
Thanks for the help. I'll try to source a 50k one.
There's a Frys quite near me so probably a better chance there than Radioshack.

* I jumped the two back pins and it turns on. So far, so good.

 

Found the right one at Frys.. but it's just 3 pin.
Is it possible to wire a 6 pin to 3 pin?

I'm hoping, yes....

 

You need a "ganged" potentiometer. That's how they stick 2 of them together and get 6 pins.

 

Any idea what the effect is of using a different rating?
Either a 10 or a 100 ?

 

10k or 100k? Depends on what is driving the signal. 10k might load down the signal. Each 100k pot can be lowered to 50k by attaching a 100k resistor in parallel with the 2 outside pins.

 

Great info. Thanks a lot!!

 

10k or 100k? Depends on what is driving the signal. 10k might load down the signal. Each 100k pot can be lowered to 50k by attaching a 100k resistor in parallel with the 2 outside pins.

I've done that before and the result was a nonlinear output. It wasn't nonlinear like a taper pot, it was nonlinear like a molested bell curve. Think of a car's horsepower chart coming off a dynamometer; it had a peak like 2/3 of the way from 0 to max, and from there it started to go back down. IIRC the final (max) value was the expected 1/2 resistance value.