SOLVED Illuminated 3PDT Breakout Board (Single Color) Confusion

[BQFS]

This is my first PedalPCB project. I am a retired electronics tech, but brand new to pedal building. I recently bought a bunch of components from a guy who gave up pedal building and this particular board was included. I looked up the info on the PedalPCB site, but there really was no info on how to get the LED on the lighted switch to light up. I bought the Acid Rain fuzz as what I thought would be a simple first effort.

Building the board itself presented no difficulty, but once I had all the holes drilled for mounting, I immediately ran into the problem listed above.

I saw that other people had similar difficulties, but I couldn't understand how they solved the issue. It had to do with actually installing an led on the effects board. This makes no sense to me as the led would be inside the box, serving no useful purpose.

On the breakout board there's two little holes at the bottom into which the wires from the illuminated switch fit perfectly. BUT I don't understand how the power gets to the switch. Unlike other similar breakout boards, there is no place to connect the 9V supply, or a place to put a CLR either.

I built the Acid Rain fuzz board, PCB195. It has a place for an LED right on the board, and I am sure if I study the schematic, I will find a CLR for that LED on the board somewhere. I see 1.8K commonly used, but I typically change to 4.7K to reduce the brightness. Regardless, for now, I am leaving the board as is, built following the BOM exactly.

Clearly, I have to run the 9V supply to the effects board, and not the breakout board, but what do I have to do to get the voltage (presumably reduced by the effects board's CLR) to the LED that is built into the switch itself so that it will light up when the stomp switch is depressed?

It would be very helpful to new people such as me if there were some specific instructions on how to do this.

Thanks very much (in advance) for any assistance anyone can offer me.

Bud

 

[swelchy]

if your using a stomp switch that has the illuminated ring and the illuminated 3dpt breakout board, the only thing you need to do is install a jumper across the cathode and anode in the led spot on the effects pcb main board..... doing that will power the led on your breakout board..

 

[BQFS]

if your using a stomp switch that has the illuminated ring and the illuminated 3dpt breakout board, the only thing you need to do is install a jumper across the cathode and anode in the led spot on the effects pcb main board..... doing that will power the led on your breakout board..

Thank you very much for the quick reply. Okay, I understand what you said, but I don't understand why that will work.

If I apply 9V to the effect board, how does it get to the switch's daughter board? There's only four wire connections between the two boards - In, Ground, SW and Out. I have no idea what SW does or what it means. Intuitively, it might stand for Switch, but that doesn't tell me what the connection does. Perhaps that's the power connection?

 

[BQFS]

Ok, It's official. I am a moron. The schematic clearly shows the path from the 4.7K CLR to the SW connector at the bottom of the board. Sheesh.

Thank you very much for heading me in the right direction. Problem solved!

 

[Robert]

SW is the cathode end of the LED.
By jumpering the LED on the main effect PCB you are shifting the open connection for the LED down to the breakout board.

 

[BQFS]

Thanks, Robert. By the time you responded, I had already seen the connection on the schematic, but I do appreciate your help regardless.

Bud