500k pot + 500k resistor = 250 k pot

[getonemaker]

A question:

If I have an audio taper 500k pot and I solder to pin 1 and 3 a 500k resistor would I end up a with a 250k audio taper pot exactly as a stock 250k audio taper pot?

Thanks!

[pdf64]

What application?
If a guitar vol pot, then when set to max they would be equivilant. But turn it down and at any setting, the pot with the 500k track will be more muffled than the real 250k pot, because the impedances are twice as big. Pete.

[defaced]

No, they are not exactly the same. Google" the secret life of pots".

[Enzo]

Or go to the Geofex.com web site and find it there.

[Pedro Vecino]

Why not? There is an element added to the alteration of the original curve?

[bkahuna]

Take a linear taper pot, strap the same size resistor across it, and take some measurements. Measure with wiper at either end and you'll get what you expect, but measure at the middle position and you'll see that you've created a pot that responds more in a log manner than linear. It just isn't as simple as you might think.

[Pedro Vecino]

Thanks! I usually use this procedure to accommodate the tone potentiometer value (connected as a rheostat), so I've never seen any difference added to the alteration of the curve.
But if being used as volume is given a particular effect: in the middle position the sum of the resistances (wiper/pin1 and wiper/pin3) is greater than between the ends. This effect decreases progressively as the wiper is close to the ends.
Regards