Hi, witch elements value I must change to use 50k pot instead the original 20k pot without any sound differences?
Thanks :)
I am kind of new at this but. Could you just solder a 25 resistor on the the 50 K and make it a 25K. I was curious about this as well since I am looking to make a Tube screamer but am having a hard time finding 20K pots
I used the more common 22k value and they worked fine.
Quote from: rosssurf on March 18, 2007, 02:58:00 PM
I am kind of new at this but. Could you just solder a 25 resistor on the the 50 K and make it a 25K. I was curious about this as well since I am looking to make a Tube screamer but am having a hard time finding 20K pots
rosssurf this is not a good idea. This will make linear pot to S-curve pot. The pot will work only at the beginning and at the end. At the center area will not change the sound.
I think that if I must multiple pot valu by 2.5 (from 20k to 50k), I must divine 220ohm resistor to 550 and divide 220n capacitor to 2.5 = 88n. But I'm not sure...
How 'bout this? http://www.smallbearelec.com/Detail.bok?no=463 (http://www.smallbearelec.com/Detail.bok?no=463)
Mark I don't like to use 20k pot. I like to use pots from the local electronic shop. :icon_rolleyes:
You can scale all the parts my multiplying all resistors by (50/20) and all caps by (20/50); that includes the 1k and 220nF at the input.
Did you know the original pot is a 20kG the G means it's a "funny" graphic equalizer taper? What that does is smooth out the control reducing crowding at the end of the control - it doesn't change the tone over a 20kB linear pot.
Another thing you can do is use a 50k and two 18k resistors. You connect one 18k between pot terminals 1 & 2 and another between terminals 2 & 3. With this method there is virtually *no difference* in the response compared to a 20k pot, however the control at the ends becomes about twice as crowded as a 20kB linear pot - ie. worse crowding.