Direct Drive Question

For a mod I have a 5w Luxeon Emiter and 3 x 3V batteries which totals up to 9V. The batteries can maximaly deliver a continues current of about 500mA - so 500 x 9V = 4.5W. My question now is, is it save to run the 5W luxeon with 9V instead of the recommended 7V when the batteries aren't able to send more then 500mA or will I still need a step down converter.

Direct driving will put more than 1.8A to the LED initially and taper down exponentially over time. Not sure where you get the 500mA from /ubbthreads/images/graemlins/icon3.gif. Safe? Yes, as long as you give the LED excellent heatsinking. You'd only need a step down converter if you want a regulated current drive (constant brightness) and longer runtime.

The 500mA are from 3 DL1/3N Lithiums. Does the use of a step down converter make sense at all in this setting (f.e. a downboy 700?. The LED needs about 700mA @ 7V while the the batteries deliver max 500mA @ 7V. Will the converter (DB 700) still be able to regulate a constant output even though the battery doesn't deliver the watts the LED needs?

is the 500mA a 'rating' or a tested value? if you try to simulate the LUX with some diodes in series or even a resistor bank that would drop 7V and measure the current draw.... if.. when you draw the current to run 7v Vf and the current is less than the max current of the LED you should have no problem DD from those batteries because though they say 9V... under load maybe you'll have 7.5 or whatever... however... if that is the case.. you'll have a lot of heat dissipated in the battery.

DL1/3N ? is that an 'N' size battery? would esplain (sic) the low current capability.

Thanks for your answer.

To measure the draw I basically short cirq. the 3V batteries with my amp meter... it does 700mA (3V) for about half a second then drops to 500mA where it stays. So 3 batteries in series should do 500mA @ 9V.

Yes it is a 1/3N Lithium battery - has about 220mAh. By the way the 5w Luxeon is an emitter... do you know can I just solder the wires to it, to me the plastic of the LED seems to be really close to the soldering spots... which makes me wonder whether the emitter (and plastic) can take the heat. Have you ever soldered an Luxeon emitter...any ideas/tips?

Ahh... that explains the 500mA. I thought you were taking about a CR123 cell. Those low amperage batteries will exhaust very quickly driving your LuxV. Worth a try to see for yourself though.

Solder to the 2 tabs that extend from the emitter. That is fine. Make sure the emitter slug (silver thing at the bottom of the emitter) is in good contact with a flat piece of heatsink material, eg. CPU heatsink.

Thanks for the heatsink advice while soldering the emitter.
Yeah I estimate about 20min of LuxV power before 3 of those cells are going to be depleted. But I thought it would be fun to have a 5w solitaire sized light - Heatsinking and space is the main issue though and requires really creative solutions... I am having fun though working out all this. I'll post pics once I am done - in about 2 weeks (if the thing works

heat problems... like they say in the military: 'short controlled bursts'.

I had no problem at all soldering to the emitter tabs... in-fact i broke one off and had to solder on to one of those tiny tabs next to the real tabs.

Looking at the energizer spec sheet on 1/3N lithium.. they have 60mA max current rating... it'd be interesting to see what happens when you drain 8 times the recommended max... there will be a TON of internal resistive power losses... if you heat up Lithium hot enough it will explode (this happened in a chemistry class of mine in high school).... so be careful.

I think i'd might try using a lux 3 with two of the batts... but you'll have similar issues.. i was looking into making the tiniest lux flashlight and was looking into using the A123 12V batt with a buck ckt... only draws 255mA to drive lux 3 to 700mA.

I think the estimate of 20 minutes is way off... i'm building a 3W light planning to run off (4) AAA (total of 6.8WH).. and at 3W i'll get about 40 minutes... the 220mAH rating is at low current drive... the 6.8WH at a reasonable current level quickly turns into 2.18WH... so... 68% of the power is lost heating the battery... food for thought... maybe put a thermistor in series with the thing to have it shut off when the batts heat up before they explode.

-awr

Thanks andrew for the exploding hint - I think I then have to rethink that one. I have never planned into my design how much current the batteries can deliver.