4x Luxeons in series on 6v 2.1amps?

I was digging around and found a 6v 2.1amp ac=>dc adapter, if i ran 4x luxeon 1 watts in series, how fast would i kill my Luxeon on 6v around 500ma each? would it live? Its going on a fairly heavy(1 lb+) copper/aluminum heatsink. I was thinking that it would make a great reading light =). Since the ac adapter has a max output of 2.1a and thats going to be divided 4 ways, i don't think i'll have problems with thermal runway?

Your problem would be that you need to connect them in parallel for them to have 6 volts each, if you connect them in series, all the 4 will get a total of 6 volts which won't even light them up. 6 volts per LED will kill them in micro, no, make that nano seconds. Doesn't matter what heatsink you have, there's just too much heat. On the other hand, your adapter has a max of 2.1 amps, prolly means that the LS would be amps limited instead of volts limited, I think you can connect them in parallel... Each will get around 500 ma I think. I'm not very sure on it though.

yea i meant parallel, i read that Mr.Bulk ran some on 6v and they worked.

Unless your supply is designed to current *limit* (by design), the 2.1 amps is a 'maximum available' number, and could even involve a fuse. The value may not be well controlled (the maker is happy, and so is the customer) if more current is safely available when it's being used as intended. In series, no light. In parallel, no dice (pun intended).

The simple (sound) way IMO is to put a correctly sized resistor in series with each to set it's current to the maximum allowable value, .35 amps. Why abuse the parts (shortening their lifetime, often severely)? It could also easily result in failure of the supply with attendant fire possibilities. A 'back of the envelope' value (six volts less 3 for the LED divided by .35 amps) says 7 ohms. One of these is already in the star (what controls current in the otherwise uncontrolled 'direct drive' circuits), so we need to add six ohms in series with each star. Power across the resistor will be roughly 3 volts times .35 amps, a Watt. Use two Watt resistors. Or two 12 ohm, one Watters (which would be smaller) in parallel for each LED and you should be fine. Leave the resistors out and all sorts of fun stuff can happen. With little to control current division, you could determine destructively the true strength of your LEDs. Weakest first, followed by the others in sequence until the strongest faces the full supply alone....... Or worse.

If it were me, I'd even check the six volts under load and be prepaired to adjust (we call it "trim" or "factory select") the resistor values (easy to do in very small steps with the two resistor's in parallel since you can change one at a time).

And like Mr. Bulk say's "always measure your current".

Or so I see it.

Cheers.

Doug Owen

hmm, i'm wondering how hot a wirewound resistor should run
i have 5x luxeon star 1 watts in parallel.
6v 2.1 amp powersupply(unregulated)
1ohm 10watt resistor.

Anyone see a problem with what i'm doing? Heatsink is more then adequate.

Each Luxeon will have a little different vf. which would make each light use a slightly different percentage of the total curent. If one is significantly lower than the others it will draw a much larger percentage of the power. The best way to drive LED's is with a dropping resistor acrost each one to average out the current better instead of significantly over powering some and under powering others. It is acceptable to conect all LED's in paralel and use only 1 resistor BUT NOT PURFURED. I have done your idea in the past and it has worked alright for me.