Well I hooked up my 4 luxeon mag head!

I just hoked temporey wires to determine how much current it would pull and its basically the same as having one lux 3 hooked up. It was pulling 1.6 amps with all four lux 3's going. I know you guys like to over power your lights, but I do want to use this for more than 10 minutes so is this a good idea can i keep it at this setting direct drive or use a resistor?

Hi - nice start. I assume you are obtaining 1.6 amps total for all 4 Lux IIIs in parallel - right ? If so, that is only 400ma each - so in general it should be no problem. That is actually a very efficient setup from a Lux III perspective. If I understand it correctly, you are basically limited by the current capacity of the battery setup, so the D cells will not provide that for very long.

There are some high capacity NiMH D cells out there, but be cautious - if you obtain a battery setup which can actually delivery more current - it will potentially overpower your Lux IIIs.

If somehow you are pulling 4 x 1.6amps from your existing setup, then I suggest adding resistance until it comes down to around 1 amp / Lux III.

I retested and am only pulling around 400MA per LUX III and have 10 amps NiMh batteries so that should be about 25 hrs. I'd like to get more Ma out of the stars though so I may try to replace one D cell with 4 AA or something I haven't worked it out yet though.

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elfen1 said:
I may try to replace one D cell with 4 AA or something

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You cannot mix battery types. They all have to be AA or they all have to be D. The only way you're going to get more out of them is to use a step up driver or a whole new power source that is a higher voltage.

Any suggestions perhaps Li-on cells ?

If you are running 4x 3watt luxeon in direct drive in parallel. what I would recommend is using 3x C or D nmh.

nmh has a fairly flat discharge rate, and just as important a natural limit to how much current is delivered.

Li-ion discharge at huge rates and need some resistance in-between to keep from killing your luxeon.

If I use three D cells (currently have 4)NiMh then it would reduce the current by 100ma per star. Thats the wrong direction, I want it to be brighter but still last so 800ma sounds good ( unless I can drive it at 1 amp and it will stil last but I think at the current they tend to lose light after a while from what I read). So perhaps if I replace some D cells with 1/2 size D cells anyone think that would increase the current it would voltage at least.

if you are direct driving 3watt luxeon. vf of lux will limit current. assuming cell is able to deliver current. Kbin will draw aprox. 1.1amp each when nmh cell is freshly charged.

I have L bin Lux's there pulling 3.6 V at my latest setup.

I tested it with 5 NiMh and was getting 3.6 amps at 3.9V each star so 900Ma each. I think thats well within the rating for my L Bins the question is how long do you think I can drive it at this setting before something goes wrong or do you think it can handle without diminishing light quality over time?

what kind of nmh are you using? are you going direct drive? 5x 1.25 = 6.25V or too high

I've never used an Lbin, but 900 milliamps sounds about right.

Too bad I tested it with some spare AA's I had hoping they would put out the same current as the larger cells since there the same voltage, but when I tested just 4 D cells it was 5 amps Holy Cow thats 1250Ma per Lux way too much I'll have to use a resistor on these!

you really don't want to go over three D or C nmh cells. You will kill your luxeons.

Try to match cell's final voltage to lux's vf.

Well even with the high amps the voltage on the anode and cathode was within range of 3.7-3.9 for my lux's I should be fine as long as I use a resistor . 3 D- Cells is just way to low of current only 400ma.

There is an interesting post in the battery / electronics secton on the effect of Vf over time. IIRC, if you run a Lux III at nominally 800 - 900ma, well heat sinked, for 24 hours, the Vf will drop down essentially one level, so your L will become approx a K bin. - please check the original post for the details.

I perceive this as some kind of anneal process going on in the LED.

This will allow your light to work potentially at the desired current without the resistor, although, certainly, a resistor is better. I had some custom 2 ohm, 1206 size, power resistors made for my own use because I wanted to run resistored lights from 2x123, which is similar to your application. If you have more space, there are plenty of larger power resistors out there at low cost. You will need at leat 2 watts rating.

Well I tested it and found it already has a resistance of 1.04 ohms and I decided I'll run my stars at 1 amp so a total of 1.3 ohms would be needed since my batts were pulling 5.2 Volts. I guess the extra 1.04 ohms is in the design of the flashlight. And if it did drop in voltage wouldn't that mean I would need a larger resistor of almost .234 ohms X 4(for each LED) . I order 4 {.3ohm }resistors to bring it to 4 amps total.

OH yeah before I install the resistors who thinks 1250Ma per Led is too much because I tried it with some differant batteries and it still pulls around 1.2 amps per LED with batts that have been powering a baby swing for days! So even with new and old its pulling high current! Oh yeah the batteries are Powerizer high current batts from batteryspace.com. (I was asked earlier) /ubbthreads/images/graemlins/ooo.gif

At 1.2 amps / led, it is a push - not tooo much, but a push.

At that power level, heat sinking is a big must, and slight variations in battery / Vf could be your undoing.

Maybe turn it down just a tad - 1 amp, until the LED Vf settles in.

BTW, don't forget that the resistors need to also deal with some power as well.

Well I ordered 4 (.05ohm) 5 watt Ohmite resistors from mouser.com they are actually current sensing ressistors and it was as close to the right one I needed as I could get after some remeasuring. So when they arrive I'll hook them up and see if my calculations were correct.