Circuitbenders Forum

Circuitbenders Forum => Parts, Components, Unbent machines & Junk => Topic started by: Dylan on August 01, 2009, 02:45:45 AM

Title: Potentionmeter question.
Post by: Dylan on August 01, 2009, 02:45:45 AM

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?

Title: Re: Potentionmeter question.
Post by: theshame on August 01, 2009, 05:35:05 AM

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.

(http://imgur.com/aPgov.png)

Title: Re: Potentionmeter question.
Post by: Gordonjcp on August 01, 2009, 01:37:18 PM

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.