I've just put together a small LED tester on vero, with the possibility to use two in series and a B10k pot to find the right CLR value. Nothing groundbreaking, but it should save me time on future builds (I have a bunch of circuits I still need to box).
(http://monimag.eu/upload/1246/p7284816.jpg)
Now I'd like to make it as foolproof as possible and add a resistor to avoid burning LEDs when I turn the pot to the left, but I don't really know which value to choose, and where to add the resistor. Any idea? I've done a quick search but I don't think I found what I wanted.
By the looks of things the grey jumper is in series with the pot, if so you could just replace it with a ~500Ω resistor.
dave
Quote from: davent on July 28, 2013, 05:48:53 PMBy the looks of things the grey jumper is in series with the pot
Yes it is! (it's where an optional second LED should go) Thanks for your answer, that's exactly what I needed, so just adding a resistor in series works. Actually even a 100R resistor is enough to avoid burning the few LEDs I tried.
I'd put a 470R in there, i used to run LED's on that, and it's fine, some are a bit too bright! But, the brighter they are the less time they last.
George
Let's go at it backwards. Max continuous rated current for most LEDs is 20mA. Figure forward drop at 2v.
(9 - 2) / 0.020 = 350 ohms for min clr
George's 470 is pretty close to this, so use that in series with whatever pot you've selected to determine desired clr value.
Yeah. I wouldn't go any lower than 470ohm, you could put one In, and measure where it goes pop, but there's probably not much point, just put a 470 in there and forget about it :)
Thanks for the additional info guys, the formula is especially welcome.