Electronics, Lighting, & Gadgets - junkbox diving: luxeon star

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View Full Version : junkbox diving: luxeon star


unterhausen
09-30-08, 03:46 PM
I dug out some old Luxeon Star led's I've had for years. I'm thinking about running them at 3.2 volts unregulated off of an A123 1.1ah battery. Am I going to hate myself for this? At 3.2v, two leds in parallel were drawing about .2 amps. Seems like they are rated for .350 A.

It seems like the aluminum plate on the back is isolated from the LED, can anyone verify this?


unterhausen
10-01-08, 01:26 AM
well, it doesn't really light up the all-black clad joggers, I guess I'll have to crank the amps a little. The simplicity is pretty nice though.

brokenknee
10-01-08, 06:18 AM
i have read a lot of threads on modifying led lights. what i don't understand is how they "push" up the amps. is this done through a circuit board? if it is "unregulated" to begin with, does that mean no circuit board to begin with? if so what determines the amount of amps going to the leds then?


unterhausen
10-01-08, 09:43 AM
yes, you have to regulate the current. If you could somehow arrange to have a battery with just the right voltage to make the led pull the current you want, you could go unregulated. In my case, 2 AA is a little two low, and 3 AAA is just a little too high. Basically, you need a regulated current source. Deal extreme seems to be the vendor of choice for that, there are lots of other people that will fix you up.

The simplest way of doing the job is with a resistor in series, but that wastes energy.

brokenknee
10-02-08, 11:37 PM
yes, you have to regulate the current. If you could somehow arrange to have a battery with just the right voltage to make the led pull the current you want, you could go unregulated. In my case, 2 AA is a little two low, and 3 AAA is just a little too high. Basically, you need a regulated current source. Deal extreme seems to be the vendor of choice for that, there are lots of other people that will fix you up.

The simplest way of doing the job is with a resistor in series, but that wastes energy.

thank you