Wah Wah DC jack problems

[triskadecaepyon]

I am currently having serious problems with my VOX V847.  I tried wiring a standard DC jack to it (i.e. had two pins coming out only), and all I get is silence.  However, when I wiring the battery snap back, the wah works again.  Does anyone have a clue what is wrong? Or did I forget something?

[sir_modulus]

hi trisk,

is it wired like this?
http://www.generalguitargadgets.com/diagrams/opcomp280_sc.gif

if not, post back,

Nish

[triskadecaepyon]

it's not wired like that.  where should the negative of the DC jack go? the chassis?

[sir_modulus]

aight....I'll explain this to you.


First off, you gotta understand how the switching system works. How most wah pedals and other effects work is that, When you plug a cable into the "input" jack, it connects the battery to the circuit. To do this, the output jack is a mono jack (two connectors, signal, and ground(ground being shorted to chassis even when there is no plug, and is put to ground by the plug, when it's inserted)) On the input side, you have a stereo jack. A stereo jack has 3 connections (lugs), the tip, ring, and sleeve. The tip connects to the signal...at the tip of the cable. The ground is grounded through the chassis, and when the cable is plugged in, it pulls it to ground (this so far, is exactly like the mono jack). Now the third lug is what makes this able to switch the power on and off to the effect. You have a "sleeve" lug. What this does, is it connects to the "sleeve" on the cable (ground on a mono cable, second signal on a stereo cable). This is only grounded when a plug is inserted.

How all this is used is that, The 9V line (from effect)goes to the +ve of the battery terminal. All the other ground connections are done to the circuit, which has it's ground connect to either the ring connections (the always grounded one...on both jacks). The sleeve connection of the input jack will be unused so far, so what happens is that the -ve of the battery goes to that. Now what this does is, normally, the -ve of the battery is not connected to the circuit....meaning no current flow. When a plug is inserted into the input jack, the -ve is now grounded, and it lets the current flow, turning on the effect.

There is also a way to wire a DC jack into all this, to have both. If you want to hear that, post back.

Good luck,

Nish

[triskadecaepyon]

I understand what the stereo jack does, but I still am clueless as to where the DC jack is wired into all of this.  Please tell me how to get both on there!

[sir_modulus]

oh aight....sorry didn't mean to underestimate your knowledge.

In terms of the DC jack, you want a mono switching jack. If you look in the schem I showed you, the jack has 3 lugs. One is ground, the little bump is the power from the adaptor. The arrow is the switch. To prevent the DC power from charging the battery, only one must be present at a time in the circuit. The switch does this. When there is no DC jack in, the DC shorts the +ve of the battery to the 9v rail in the circuit. When the adaptor is is, it isolates the 9v from the battery, and lets the 9v from the adaptor go to the circuit.

For that "ground" connection from the battery, you can have it so that, when the DC Power supply is connected, the stompbox is live by putting it to ground...as shown in the schematic. If you put it to the sleeve connection of the stereo jack, like the -ve of the battery, the effect is only on (whether it be battery of DC jack) when the input cable is plugged in.

Hope that helps (post here if you have any questions),

Nish

[triskadecaepyon]

so you never use the neg. of the DC jack?  You only use the + of it?

[sir_modulus]

well, no....

when you look at the schematic that I gave you, follow along.

For battery operation: (let's say the input has a jack in it)
the -ve of the battery is connected to ground via the sleeve
the +ve is connected to the 9v rail, through the switch on the DC jack (it shorts a wire to the +ve when nothing is in the jack...it's represented by and arrow on the schematic)

For DC jack operation: (let's say the input has a jack in it)
the -ve of the power supply is connected to groudn through the outside of the jack (shown by a black box, going to ground on the schematic).
The +ve of the power supply is hooked up to the circuit via the connection from the tip of the DC jack. The switch is open, meaning the +ve of the battery is not connected to anything.

Understand? (If you don't, I'll just draw a quick set of pictures (and it's alright, everyone has to learn sometime...I remember when I asked questions like this....except mine were worse "why does my stompbox blow up! I put the +ve to ground like the schematic said!!!....whoops"))

Nish

[triskadecaepyon]

hmm.  That's weird.  I think mine IS wired like you are saying.   However, is there anything wrong with wiring the DC jack to the same connections as the battery snap?  Or am I totally misinterperating this?

[sir_modulus]

yes, you wire the ground connection of the DC jack to the "sleeve" connection on the input jack (like the -ve of the battery).

here ya go:
http://www.generalguitargadgets.com/diagrams/bluesbreaker_lo2.gif

Nish