Reducing a pot’s value

[frets]

Hi guys,
I can’t find the formula to reduce a pot’s resistance. I’m trying to reduce a 10k to a 1k. Thanks😺

[JohnH]

Hi frets

There can be resistors in parallel either across outer lugs, or from outer to inner. It depends on the application,and its not generally possible to match the sweep and consistency of a 'native' 1k. What clever scheme have you hatched up where you might need this?

[frets]

Hi John😺😺😺

Nothing too novel, I’m working on a clone of the Jext Telez White Pedal and don’t have a 1k pot. So I was going to take a 10k and try and reduce its resistance. I’m not too worried about the taper. I can’t find any kind of formula on the internet.

I was thinking putting a 15k in parallel on the 10k pot might be good enough.

I never know how exact has to be on matching the pot resistance called for on pedals. For example, I have a 5k pot, could I just use that instead of the 1k?

[JohnH]

It all 'depends' on the circuit.

If you are just using one outer lug and middle wiper, to make a variable resistor, a parallel 1.2k with a 10k pot will look like (1.2 x 10) / (1.2 + 10) = 1.07k. If you are using all three lugs, then maybe 1.2k from each outer to the middle, but it'll be a bit off at mid turn (itll look like about a 2k pot). Also with any of these, the pot action is all squashed to the end of the pot turn.

[col]

Try this, frets.

www.geofex.com/article_folders/potsecrets/potscret.htm

[mikecg]

Hello frets,

Is this the circuit that you are cloning?



If so, I have 'run the numbers' with the 1 k Ohm treble cut replaced with a 5 k Ohm, and I am getting a pretty good match with wiper positions at 0, 50, and 100%, with 1 k Ohm resistors wired between the wiper lug and each of the track end lugs - that's a 1k resistor on each side of the wiper to the track end lug. I've just simulated the PTB part of the circuit - complete with one of your favourite components - an inductor! I think you will find that it will work pretty much as normal with the 5 k Ohm pot and the 1 k Ohm resistors. Of course the pot 'law' will be altered by the padding resistors, but you probably won't notice that too much. If you are thinking of making a few of these, I would 'splash out' and buy some 1 k pots!

[frets]

Thanks John and Mike and Col. I really appreciate it.

Mike,
Yes, that is the circuit I’ve used. I’ve built one before and it worked great. I’m now building one for myself. It’s really a nice sounding pedal. I highly recommend it. It simulates the sound of George Harrison’s Vox Conqueror amp circa 1968. The real pedal is no longer manufactured.

[mikecg]

Hello frets,

I have to admit that I have very limited experience with pedals, but just recently I've been looking in more detail at them, and today I made a successful bid on fleabay for a second hand Boss ME-80 multi-effects pedal - lots of lovely knobs to twiddle!

[pasqualino63]

Let's talk about that 500 mH inductor.  That's a pricey part and how high does the Q need to be?  
[edit] Moreover, shouldn't there be diode protection to deal with the magnetic field when power is removed or interrupted?

Gowanda electronics mas these toroidal ones that are like $34ea.



The 500mH ones have a .7" dia toroidal core and about .5" high, so it seems like they would fit.  I wonder if it's worth winding your own?

[edit] For a ferrite toroid core 1.2"ODx0.7"IDx0.6"H, you would need to wind (on the order of) 1,000 turns.

I'm basing this on a kinda half-assed formula from this site: hyperphysics.phy-astr.gsu.edu/hbase/magnetic/indtor.html
it has to be incomplete because the gauge of the wire used has to factor into the equation somewhere. [/ edit]