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Changing the value of a potentiometer

Started by greyscales, March 09, 2012, 02:32:26 PM

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Wasn't quite sure if this was a beginner question, but I am still enough of a noob that it seems beginner.

How do you change the value of a pot (say 50K) to a higher value (say 100K)?

Is it as simple as adding a resistor between lugs 1 and 3 of the pot (in this case a 50K resistor)?


You can't unfortunately. You can only make it lower.

Witness the math for resistors in parallel: R(x)  = 1/(1/R1 + 1/R2 + ... + 1/Rx)

In the case of two resistors, it can be stated this way:
1/Rx = (R1+R2)/R1*R2

Say R1 = 50 and R2 = 100

1/Rx  = 150/5000 = .03, therefore Rx = 33.3k

In fact, as R2 gets exponentially large, you will only approach the value of R1 by fractional amounts. You would have to have an infinitely large resistor just to get back to your original 50k (only in pure mathematical terms, not practical ones)!

So, you cannot increase the value of a pot with a parallel resistor, but on the other hand, you can reduce the value of a pot that way which sometimes comes in very handy.


Well, glad I asked before I made any assumptions.

Thanks for the info, Bean!


Hey Grey, this is sort of related if you will. I didn't know this for several months when I started and could have used the info.

Resistors in series are additive. As such, you can solder 2 (or more) together to get at some weird value that you dont' have and "teepee" them. While I've got quite a stash now, it took awhile, but I still need to do this for some different values you don't see often.

You basically twists the ends of two together, solder, snip off the excess and there you go. It stands up higher but no more than a cap.