Jim Dunlop Bass Wah 105Q

[coffee-sipper]

I have read about changing a single resistor on the dunlop bass wah to reduce the delay of of effect off feature.  The harmony-central article was a little vague.  Is there a more detailed (exact) discription of what gets swapped (value and postion of the resistor) out there somewhere?

[Mike Burgundy]

effect off delay?
Is this one of those auto-switching thingies?
You might not want to get rid of it altogether (since your wah will then disengage as soon as you have the treadly fully up - this is annoying) but decreasing it is possible.
It probably uses some kind of scheme with a cap that's either charged or discharged, and uses that charge to switch FETs with. There should be a resistor that regulates rate of charge - but without a schem I couldn't say which.
If it works like Morley (optical detection) there should be a resistor to ground somewhere right after the LDR, or very close to it.
Changing that one should work.

[coffee-sipper]

effect off delay?
Is this one of those auto-switching thingies?
You might not want to get rid of it altogether (since your wah will then disengage as soon as you have the treadly fully up - this is annoying) but decreasing it is possible.
It probably uses some kind of scheme with a cap that's either charged or discharged, and uses that charge to switch FETs with. There should be a resistor that regulates rate of charge - but without a schem I couldn't say which.

You are rigth.  I edited my post.  I meant to say reduce the delay.  I should have been more clear.

[Mike Burgundy]

did an edit for the Morley system above, you just beat me to it
actually Morley even have a trimpot in there.

[coffee-sipper]

Here is the Harmony-Central review that (sort of) details the mod:

There is a delay circuit connected to the switch. As soon as the switch is pressed in, current flows through it, through a resistor, and begins to charge up a capacitor. Once that capacitor is charged (about half a second), the wah switches off and the pedal is by-passed.
Note that when the switch is open (the pedal is presses, you start using the wah) there is no delay. As some of you may know (or remember from physics class), if you reduce the value of the resistor, more current will go through it. And the more current, the faster the capacitor charges up (and the sooner by-pass kicks in).

So what we want to do is reduce the resistance of the resistor. (We could change the capacitor to a smaller one, but for this pedal, the resistor change is easier.)

Take off the panel on the bottom of the wah. Also unscrew the nuts which hold on the input and output jacks. Take out the 2 big screws holding the circuit board, and unplug the cables (just pull) which connect the circuit board to the potentiometer (the round thing with gears and with all the grease slimed over it). You might want to notice which way it was connectedâ€"the wires come out of the plug pointing toward the bottom of the pedal…I think (I’ve since put it back together).

Now that you’re holding the circuit board, look at the switch (on the opposite side as everything else). Press that, you’re in bypass mode. Turn it over again. To the left of the switch (looking from the side with all the components except the switch) is a diode. To the right, you will see two resistors (blue things with colored rings on them.) The top one (the one closest to the edge, for you idiots holding it upside down) is the one we’re concerned with. If you count up the value of the rings, you get 2 megaohms (I think, I’m lazy, so I used a multimeter.) Either chop this resistor off and solder a new one in it’s place, or just go ahead and stick another one on top of it (run them in parallel). Just make sure the new value of the resistor is a lot smaller. I cut the old on off and put in a 10K-ohm resistor…there is no delay at all anymore.

[Mike Burgundy]

That's not that vague. It actually gives a rather detailed description of what resistor to target.
It's not necessary to remove it though - if there's enough exposed lead on it, you can solder a resistor *across* it and not cut anything out, or desolder.
Total resistance of the combo will be slightly less than the value of the smaller resistor you've just soldered across the 2M one.
Look at the component side. Turn the board so you have the diode to the left of the switch connections, and two resistors to the right. The top one should be marked red/black/green/gold or (if there's 5 bands) red/black/black/yellow/red for a 1% resistor.
If you're unsure which one to tackle from this description I suggest you find someone with a little experience. It's quite a simple mod, but if you take on the wrong resistor, the wah might stop working entirely.

[coffee-sipper]

Cool.  Thanks for the extra help.  I will try and tackle this simple one tonight and post some pictures along the way.  I thnk I will shoot for the extra resistor (lower resistance than 2M) to reduce the overall resistance approach.  Any suggestion on what reistor I should use?  Looking at what I have around combined with the 2M already there I will go with a 10K resistor.  From my calculator that will give me 9950ohm resistance.