Okay, time for "explain to Hal" again.
How do volume pots work? I'm assuming...by means of common sence...that when resistance is low, "10", the guitar's volume (or effects volume or whatever) is full. when it is turned up, the resistance is enough to in effect kill the signal...
so why do people say that 500k pots "suck tone?" wouldn't any pot at 10 resistance conduct fully?
finally, why 25k for active, 250k for sing and 500k for hum? I mean...if you use 250k for hum, some signal still goes through at "1," I got that one. But wouldn't that mean that active should have even larger than hum, because the signal is even larger?
Please answer the first question first, becuase I think I kinda know the answer to the last one and it lies just beyond my understanding, with concepts of AC current and voltage...
and please don't think any less of me for this post :-D
A volume pot works like a voltage divider (as your AC guitar signal's volume is actually voltage - more voltage is louder guitar). So the pot is 'simulating' 2 fixed resistances that shunts part of the voltage to ground, and part to the output/gainstage/etc. Therefore decreasing the volume. ;)
edit: no-one thinks any less of you, it's a good question if you ask me.
To illustrate Ben's description:
Signal in-> out-< Ground-/
With the vol pot all the way up:
><______________/
The signal needs to cross little or no resistance before leaving the pot.
With the vol pot all the way down:
>______________</
The signal is as far away as ground, and needs to cross the entire resistor before leaving the circuit.
Tone sucking:
One, if the pot is of too small a value, too much of the signal will be shunted to ground. The pot acts as a fixed resistor (first to third lug) connected to ground.
Two, if the signal needs to traverse too much of the pot before reaching the wiper, you can lose high end. This happens when you have the vol pot rolled back a bit.
Some people use a small value cap to send some of the lost high end back to the signal out. Personally, I like to lose a little bit of that sparkle.
Per your 25k active electronics vol pot question, it has to do with the output impedance of the circuitry. Passive p/u output impedance is very high, Mega ohms; while active electronics usually have a relatively low output impedance, tens of k ohms.
Should be right, but I'm pretty tired.
That's right, passive pickups have a huge output Z. :) Impedance is the first thing to make things complicated ;)
Rule of thumb is high input impedance and low output impedance :)