Quick question about not using a battery clip and how the "ring" gets wired...

[PedalBuilder]

I am deciding not to use battery clips in any of the pedals I'll be building because I have them all on DC power all the time.  I am using this schematic for my offboard wiring:

http://www.generalguitargadgets.com/diagrams/switch_lo_3pdt_ig_dcjack.gif

If I get rid of the battery clip in that diagram, the red wire from the clip to the DC jack goes away and that's easy enough.  But what happens with the ring of the input jack in this case?  If I wire the negative of the DC jack to the board grounds and then also to the ring, that wouldn't shut it off when unplugged.  It's getting late and I've been staring too closely at perfboard and soldering for way too long.  I appreciate any help.  A lot.

Steve

[railhead]

Just use a mono input jack.

[mth5044]

As long as you don't want it to shut off when you unplug it, just take out the battery clip all together. If you want it to shut off when you unplug, I guess I don't get it. The point for that is to save battery, right?

Maybe you can just run a wire from the ring to where the battery clip connects to the power jack? I'm not totally sure.

[Ripthorn]

The connection to the ring on a stereo jack is primarily to prevent a battery from dying when a cable is not plugged in.  If you want, you can just wire the ground of the dc jack to the ring as usual or you can go straight to the board's ground. It doesn't matter much, but if you don't want the circuit sucking juice while the dc power is connected but the effect is not in use, then run the ground to the ring lug and the sleeve lug to ground.

[PedalBuilder]

if you don't want the circuit sucking juice while the dc power is connected but the effect is not in use, then run the ground to the ring lug and the sleeve lug to ground.

Yeah that's what I was going for.  And that's what I thought, I just wanted to verify.  Thanks guys.