Layout calls for 25k rev. log pot. Can't find one. Suitable replacement?

[skiraly017]

Small Bear has 100k, 150k, 500k and 1meg in reverse log. The only 25k value to be found is a linear. Any ideas? Thanks.

[Toney]

Yep.
Make one using a higher value linear pot and two resistors.
Read GEOFEX "secret life of pots".

[jrem]

Yep.
Make one using a higher value linear pot and two resistors.
Read GEOFEX "secret life of pots".

I=E/R. 

if you put a resistor at each end of a smaller value pot, then you will have a bigger value in total but a pot range that is somewhere between, i.e., 33k - 50k pot -33k = 116k total with a 50k sweep from 33k to 83k. 

If you put some resistor combinations in parallel, then you will have a total value with some sweep, i.e., 100k pot in parallel with a 100k resistor, you will have (100k + 100k0) / 2, or 50k when the pot is all the way up and 0 when the pot is all the way down.

So if the circuit calls for 25k rev log and you need the 25k end-to-end, then use a 100k pots with three 100k resistors in parallel (100/4=25).

Put an ohm meter on it, or someone correct me as I've edited this three times already.

[twabelljr]

     Yes, you could add three 100k resistors in parallel or just one 33k. One 33k will give a total resistance of 24.812k which given the tolerance of pots is pretty close.

[zachary vex]

use an audio or log taper and wire it backwards, and just remember to turn it the wrong way to make it work.

[Toney]

Thats right.
The rat solution.
As ZV says, wire an audio pot backward and just get used to turning it the"wrong" way.... if its for tone
you can call it "filter" "high cut" etc