ilovefuzz.com

I just picked up a Boss PH-1r (sounds so good) but the LED is kinda shot. It comes on when I engage the pedal, but you can't tell unless your holding it right up to your face. Even then, it's baaaarely visible. Anybody wanna help a brother out and fix this for me? I suck at soldering and I'd never let it go if I fucked up this pedal. Of course, I'd never let it go if you fucked it up, either.

Got this sorted out by running it at 12v!

You could cause a lot more harm than good by using a 12v adapter to power a circuit meant to run at 9v. A dim LED is a lot better and easier to fix than a fried circuit. You probably just have a bad connection from the power source to the LED which is a cheap and easy fix.


Besides, if you need an LED to tell you that a Phaser is on, you may want to rethink the whole music playing thing.

MiddleEarthCrisis wrote:
You could cause a lot more harm than good by using a 12v adapter to power a circuit meant to run at 9v. A dim LED is a lot better and easier to fix than a fried circuit. You probably just have a bad connection from the power source to the LED which is a cheap and easy fix.


Besides, if you need an LED to tell you that a Phaser is on, you may want to rethink the whole music playing thing.

I did my homework first. A lot of the old Boss stuff (like the PH-1 and CE-2) is designed to run at 12v. I wasn't about to pump extra electricity through a vintage pedal that I had just bought two hours ago without checking first. And the way I use my pedalboard, in a bypass loop setupd, it helps to know if the pedals are on before I engage the loop they're in.

Wurd! Problemo Solvlemode then.

If you want it to use it at 9V, and you want to increase the brightness of the LED, then you could reduce the resistance of the resistor in series with the LED. A really easy way to do this would be to piggyback another resistor onto the one that's there, so that they're in parallel. Although, you'd want to look at what a good resistance for that LED would be, and what you have, then calculate what parallel resistance would get you there. If whatever's in there is larger than 1k, reducing it to 1k would probably work.

Rparallel = (R1)(R2)/(R1+R2)

Rparallel is the resistance you get by wiring the resistors R1 and R2 in parallel.

Looking at a schematic, it looks like the resistor inside there is 3.9k, so sticking a 1.5k resistor in parallel with that one would give you about 1k.
Or you could just replace the resistor, or the LED.
Or you could just leave it alone like you already said you were going to.

A dim LED is not really a big deal, IMO.