Replace A500k with B500k with in a tube screamer?

[RealLyfeGangsta]

Hey guys,

I want to build a tube screamer and I have all the parts I need on hand except for the A500K pot. I have a B500K and I also have an A100K pot to spare. Could I replace it with the B500K like in the drawing, or could I alternatively make some changes in the circuit to use an A100K instead? Where I live, shipping is very expensive, so that's more or less out of the question for just one part.

 

[allsmoke]

You can replace it with the B500k, but you're going to lose some of the sweep. A linear pot (A500) will have a steady consistent change as you turn from 1-10. A logarithmic pot (B500) will have the same start and end point for your gain, but much of the turn, from say 1 to 7 (out of 10) will not have much of a noticeable effect, and most of the grit will come from the 7-10 sweep position. I will work, but you won't have as much variance on the grit/drive, which might be fine depending on how you intend to use the pedal

 

[Feral Feline]

You can replace it with the B500k, but you're going to lose some of the sweep. A linear pot (A500) will have a steady consistent change as you turn from 1-10. A logarithmic pot (B500) will have the same start and end point for your gain, but much of the turn, from say 1 to 7 (out of 10) will not have much of a noticeable effect, and most of the grit will come from the 7-10 sweep position. I will work, but you won't have as much variance on the grit/drive, which might be fine depending on how you intend to use the pedal

You described the tapers well, but you got your pots wackbards... A500 would be logarithmic (audio), B500 would be linear.

 

[HamishR]

I would use the B500K over the A100K. You'll end up with the same amounts of gain but it will appear in different places on the pot. Using the A100K will give you less gain. You could make it up elsewhere with resistors but you really want the range of gain with your pot.