4:1 Compressor LED

[mattlee0037]

So this seems like it should be obvious but for some reason my 4:1 compressor LED is always on. Everything else works fine. Right now I have negative on PSU to neg and pos to pos.
If you number the 3pdt switch as
1 2 3
4 5 6
7 8 9
I have the G pad going to output jack sleeve and output sleeve going to lug 5 then I have the L pad going to 2 with effect in at 1 effect out at 3 input ring to 4 output ring to 6 and 7 and 9 jumpered. Its obviously being grounded in both switch positions somehow but I can't figure out why. The schematic shows the ground pad as general ground which goes to output sleeve and then middle lug, but somehow in both position LED neg has continuity with ground? Those are all the connections I have on this board. Thoughts?

[somnif]

Is it shorting on the enclosure somehow?

[mattlee0037]

Its possible I guess. I can try and check tomorrow. It's soldered to the board and going straight into a fresnel lens so that should be obvious but I have tired eyes from working all day on some pedals.

[somnif]

Should be easy to test. Pull out the jacks (or insulate them somehow) and test for continuity between the negative lead and the enclosure.

[Stomptown]

I would start by looking for a solder bridge between the G and L pads on the PCB since they are located right next to each other and would cause this exact issue.  If it's not that or some other obvious short I would suggest posting pics...

[jimilee]

3pdt is numbered like this.
1 4 7
2 5 8
3 6 9

The madbean wiring diagram has 1 to the LED (-), 2, 3 and 6 are all wired together and wired to input sleeve and to the power input (-),  4 and 9 are wired to input tip, 5 is to the board in, 7 is to the board out, 8 is to the output tip.


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[mattlee0037]

Thanks for all the help with this guys! I tried all the solutions here but it wasn't working so I went to square 1 took the 3pdt switch out and lugs 2, 5, and 8 were always in continuity regardless of switch position. Really odd because I'm almost positive I didn't fry it. I have pretty solid solder work nowadays. Also just so I don't open a new thread what is the benefits of madbeans true bypass vs the one I was using?

[Stomptown]

There are (at least) two benefits:

1. The input of the effect is grounded when the pedal is in bypass.  IIRC this is helpful for high gain circuits that otherwise cause noise/oscillation issues. You can accomplish this with the wiring scheme you are currently using by running a jumper from 1 to 8 (per your numbering scheme in original post).

2. lug 2 & 3 are tied together which is a fail safe for mechanical failure.

You can read more here:  http://www.madbeanpedals.com/tutorials/downloads/MBP_FootswitchWiring.pdf

[mattlee0037]

Awesome thanks so much!

[Stomptown]

Awesome thanks so much!

No problem.  I should clarify that I was I was referring to lug 2 & 3 per Jimilee's numbering scheme when I mentioned them being tied together as a fail safe.

[mattlee0037]

Dont want to start another thread for this but I can't seem to get LED wiring to work how I feel it should with 2 in 1 pedals.
I've built a few and first I normally try running the pedals 9v tap and a single CLR to the 9v pos of the power supply and then 2 wires coming from the other end of the resistor each going to a separate LED, but for some reason when I do this I can't have both LEDs lit at once. To do this I always have to take 2 separate resistors to the + power supply.

1. What are the layman physics behind this phenomenon and why doesn't my way work?
2. Is there a better way than trying to get 3 separate wires going into the +9v (2 CLRs and the board +9v tap)?

[Jebus]

Dont want to start another thread for this but I can't seem to get LED wiring to work how I feel it should with 2 in 1 pedals.
I've built a few and first I normally try running the pedals 9v tap and a single CLR to the 9v pos of the power supply and then 2 wires coming from the other end of the resistor each going to a separate LED, but for some reason when I do this I can't have both LEDs lit at once. To do this I always have to take 2 separate resistors to the + power supply.

1. What are the layman physics behind this phenomenon and why doesn't my way work?
2. Is there a better way than trying to get 3 separate wires going into the +9v (2 CLRs and the board +9v tap)?

1. Ohm's law. When only one of the LEDs are on, the whole amount of current (9V/CLRs value) goes through the LED. When both are on the current is split equally. Always recommended to use separate CLRs for each led. Also helps to get the LEDs to same brightness.
2. Depends on the board you are using. For example most (if not all) Madbean boards have another 9V pad for LED. Also most of the 3PDT breakout boards have multiple 9V pads so you can chain them together. Or if you don't feel comfortable soldering multiple wires to the power connector you can use a piece of veroboard or something similar to split the wire.

[mattlee0037]

Ahh makes sense. Thanks! Vero is a great trick I might have to try. The holes are just so dang tiny on these power jacks

[mattlee0037]

Also I just watched a demo of the 4:1 and mine seems to have much more output. Unity is around 9 o clock but madbeans demo has much less output. Anything specifically I should check for?