Line 6 DL-4 double preset mod issue - internal jack wiring

[beedoola]

I added a 3PDT switch (had one laying around) to my DL-4, added an led - diode coming on the power jacks positive, the negative going to the middle lug on one of the outer rows, below that goes to the LED - that all works fine.

What doesn't work is the actual preset mode. I had modded a DL-4 a long time ago using the method I'm going to describe, and it worked fine, but that was years ago and I kinda forgot exactly how to do it.

I also did a search here, and though I've seen the other threads regarding the double preset mods, they haven't helped me figure this out.

So. I didn't want to drill a hole on the rear of the pedal and use a external jack, what I had done before was just attach the wires from the switch to the solder lugs of the expression jack on the pcb....

The way I currently have it wired is: outer lug of one of the 3dpt rows>length of wire to a 10k to the TIP

middle lug of switch on same row as above goes to the ring AND ground of the expression jack - when you plug in an external plug from say an expression pedal, there are three tabs on the expression jack: ground, sleeve and tip. When the jack is inserted the ground and sleeve touch, so thats why I have the two soldered together.

What happens is, I can get the switch to change a preset once - I'll hear/see the delay time change, but it wont change back.

What am I doing wrong? This is where I forgot what I did last time. I think I may have had another wire from the switch, not sure where it would go though.....

Here is a pic of the setup. Thanks!!!!

[beedoola]

NM!!! just found this on the interwebs!!

[Processaurus]

There's a way to do it without "lifting the fingers"  , that also doesn't mess with the 10k resistance when the expression pedal is plugged in.  It's been a while, but I remember soldering the 10K (the dummy load, to fool the pedal into thinking there's an expression pedal plugged in when there isn't) resistor onto the jack, from the normaled tip connection (the terminal on the jack that connects to the real tip connection when there is no plug in the jack) to ground.   Then use the footswitch to short the tip to ground, just one wire to the tip and one to ground, and use one pole on the switch to connect them.  It doesn't matter that the 10K is always in the circuit, because you are shorting it out.  The person who made the diagram felt they had to "true bypass" the resistor... 

This way, when there is an expression pedal plugged in the jack, the resistor is out of the circuit, but you can still use your footswitch as  a shortcut to go to the heel back position (or the heel forward, however it works, I forget).

Hook up and power the LED however you want, on another pole.  Looks like you figured that part out.  You only need 2 poles of the 3pdt this way.

[beefyco]

Hey I know this is a few weeks old, but I'm trying this right now, and I got the switch working correctly, but the LED won't work... Can somebody help me out with the wiring for the LED? I have the ground (from the 3pdt switch) wired to the - on the power supply input and positive (LED) wired to + on the power supply.
Is that wrong? Is there somewhere else I should be wiring it? Hopefully somebody's still watching this board
Thanks for the help!

[Barcode80]

Hey I know this is a few weeks old, but I'm trying this right now, and I got the switch working correctly, but the LED won't work... Can somebody help me out with the wiring for the LED? I have the ground (from the 3pdt switch) wired to the - on the power supply input and positive (LED) wired to + on the power supply.
Is that wrong? Is there somewhere else I should be wiring it? Hopefully somebody's still watching this board
Thanks for the help!

Resurrecting this in order to answer this in case anyone comes across this problem. Make sure your LED will run off very low current (such as a super bright LED). Otherwise the LED won't have enough current to light up.