Potentionmeter question.

[Dylan]

So, I have a 250K pot and I need a 100K; is there anyway I could throw a few resistors on that to make it 100K? If so, what values?

[theshame]

Yes there is. Adding a resistor in parallel will lower the resistance of the potentiometer circuit.

Using: R-total = (R1*R2)/(R1+R2) you get the following:

100K = (250K*R2)/(250K+R2)

R2 ~ 167K

Of course you will probably need to use a few more common value resistors that add up to 167K in series.

Feeling extra bored so I made you a diagram.

[Gordonjcp]

That will *kind of* work, but will make the control law pretty non-linear.

If the 250k pot is wired as a potential divider rather than just a variable resistor, its value might not matter much.  If it *is* wired as a variable resistor, it might just give you a useful range that's all up one end, or it might make things even better.  Without seeing the circuit it's hard to tell.