Blackfire with Recto Tonestack

[YouAre]

I'm using a rectifier tonestack for a blackfire, and don't want six pots. So i opted to use cocentric pots for the tonestack. having bass and mid on one and treb and presence on the other.

Here are the pots i require
Bass 1m
Mid 25k
Treb 250k
Pres 25k
Gain 1m
Vol 100k

Now if i wire up a resistor in parallel with the lugs of a larger value pot will it drop the value? The rule for resistors in parallel, is 1/(1/r1 + 1/r2) right? so i did the math with a 1000k(1m) pot and 25k resistor. The answer i got was about 24k. So Theoretically could i still use A one meg pot, and put a 25k resistor in parallel w/ one set of lugs on the pot, it should be a 24k pot right?
Thanks,
murad

[brett]

Hi.
If I understand what you're trying to do, you are planning on putting that resistor across the outside lugs of the pot, right?  Unfortunately, because the wipe won't wipe the resistor, just the pot, you'll get very non-linear effects (e.g. maximum effect in the centre).  

What you need to do is put the resistor from lug 1 to lug 2 (the wiper).  Do the math on what the parallel resistances give when the pot is 1/4, 1/2, 3/4 and fully turned, and you'll get an idea whether it's useful or not.  In general, you don't want to add a resistor that is less than about 1/3 of the value of the pot, or the effect will be very non-linear.  In a tone stack, it might be really screwing things up.

Anyway, you can always give it a go and switch to a big muff or other stack if it doesn't work out.   Good luck

[YouAre]

I don't know actually which lugs to put the resistors on to make it parallel, i just know that resistors in parallel add down with. But thanks for the tip.