(probably) easy noob question on volume pots/pedals

[rockhorst]

Hey all, I'm a total noob at DIY effects and at the moment not planning on becoming too big an expert on it. However, a friend of mine has a Dunlop Wah that has broken down and we would like to mod it to function as a volume pedal. Now I've put some thought into this and there's some questions popping up. I'd appreciate any help!

What would seem the most obvious way to do this is to take out the switch, power supply and wah-circuit and just use the potmeter (keeping the ground). This would then function as a normal volumeknob on a guitar. My guess is that a totally open volume knob is basically a bypass of the pot-resistance, letting all electric signal pass. Turning the pot puts a load on any current and establishes a voltage drop -> less signal (is this correct?).

I have come to understand that volume pots are typically 250KOhm (single coil pick ups) or 500KOhm (humbuckers). Now here's the part that I really don't get: turning down the volume can completely mute my sound. To my understanding I would be passing my signal through a 500K resistor. But my guitar is also connected to an amp of typically 1MOhm! Those two resistances are in series, meaning that only a max of 1/3 of the total voltage drop is over the volume pot and the rest would go to the amp, yet this is not what happens (cuz in practise my signal is muted). What am I overlooking?

In addition, the power 'consumed' by the pot would manifest itself as heat. Common practise shows this is no problem, but I'm just curious how much heat is produced (in other words: how powerfull is a guitar signal?)

Also, would there be any difference between using the pedal with active or passive pick ups?
Any other things I should consider?

[boogietube]

You could check the Jen wah Volume conversion on the bottom of this page:
http://homepage.ntlworld.com/s.castledine/greenfuz/wah.html
Also some great info on wah mods on this page.

Also some cool info here:
http://www.geofex.com/Article_Folders/wahpedl/wahped.htm

Some cool info here:
http://www.fulltone.com/wah_talk.asp

You could fix it and convert it to a wah volume?

I hope this helps.
Sean

[rockhorst]

Well, problem with that mod is that the Wah is already not functioning. So there might be a circuit break somewhere, that isn't bypassed then, resulting in a nonfunctioning volume pedal I'll look up some schematics I've found to see wich one is the 4.7 uF capacitor and try it anyway...But my guess is that if I can make it work with a minimum amount of circuitry (i.e. just as in a guitar), there the least chance of flaws and breakdowns...

[boogietube]

Yah I gathered that it was kaput. You may be able to fix it with that info.
www.generalguitargadgets.com has a replacement board project that's cool.
I'm going to post in a few minutes. I may have a spare GCB-95 board that works laying around.
BRB
Sean

[boogietube]

I have a functioning GCB-95 board. You would have to replace one of the jacks with the one from your board. If you want it,let me know. We'll work out the postage.
Sean

[Alex C]

Hello, and welcome to the forum.

You may or may not need these, but here are some links about the basics of potentiometers:
http://sound.westhost.com/pots.htm
http://geofex.com/Article_Folders/potsecrets/potscret.htm

Here is a diagram of the pot's schematic symbol and the lug numbering on the actual component:


Between terminals 1 and 3 is a resistive track whose total resistance is the value of the pot (for example 500k). The terminal labeled "2" is the wiper.  This is what moves when you turn the shaft of the pot, and it can contact the resistive track at any point between 1 and 3.  When it is turned fully clockwise, there is no resistance between terminals 2 and 3.
Likewise, when it is turned fully counterclockwise, there is no resistance between terminals 2 and 1.

In the case of a volume control:


Terminal 1 is connected to ground, terminal 3 is connected to the input jack, and terminal 2 (the wiper) is connected to the output jack.  When the shaft is rotated  fully clockwise, there is no resistance between terminal 2 (which in this case is also the output jack) and terminal 3 (which in this case is the input jack), and this is the "maximum volume" position.  Note that there is still a 500k resistance to ground (or whatever the value of the pot happens to be).
In the same way, when the shaft is rotated fully counter-clockwise,  there is no resistance between terminal 2 (the output jack) and ground, so this is the "minimum volume" position.  You won't get any output because the "signal" being sent to the output jack is 0 volts.

What would seem the most obvious way to do this is to take out the switch, power supply and wah-circuit and just use the potmeter (keeping the ground). This would then function as a normal volumeknob on a guitar.

Connect as per the schematic above, and this should work fine.  Keep in mind that the rocker-pinion mechanism in the wah may not be utilizing the entire rotation of the pot.
Here's a diagram with the pot and jack connections:
http://www.generalguitargadgets.com/diagrams/dearm_vol.gif

My guess is that a totally open volume knob is basically a bypass of the pot-resistance, letting all electric signal pass. Turning the pot puts a load on any current and establishes a voltage drop -> less signal (is this correct?).

As noted above, when fully clockwise, it is like a direct connection between input and output, with the addition of a 500k (or the value of the pot) resistor to ground.

[rockhorst]

Thanks Alex...I looked closely to your diagram and then it finally hit me. I had been overlooking the function of the ground! I turned to full (so using the max resistance) all current after the resistance will flow to the ground and no signal will continue down the line to the amp.

@Boogie: If you'd happen to live in Holland, it might be interesting and I suggest you drop me a line, otherwise I reckon it's a bit of a hassle...Besides, I'm gonna be experimenting with volume first .

Other question for if I ever decide to build some stompbox effect:
Is there any real danger of damaging either guitar or amp? I've read some on input and output impedance, but how this actually affects the whole performance is a bit unclear (and yes, I do google).

Tx again!