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linear1 forums  |  LED discussion  |  LED questions and discussion  |  Topic: Cheap Current Regulated Luxeon Star Driver « previous next »
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Author Topic: Cheap Current Regulated Luxeon Star Driver  (Read 5456 times)
fusionxtreme
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« on: November 09, 2005, 10:36:44 PM »

I have reviewed the schematic, but I would like to alter this to work as a running/tail light.  And being new to electronics, I can't fugure out how to do this.  Any help will be greatly appreciated.  Thanks....
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« Reply #1 on: November 10, 2005, 12:29:07 AM »

it really shouldn't need any alteration.

with most automotive applications, estimate you have about 13.6 volts to play with.  so, take 3 volts off for regulation overhead (I think its closer to 2, check datasheet).  this leaves you with 10.6 volts...

the typical red luxeon needs about 2v ... so you could build a series string of 5 luxeons, if you wanted to use white, figure about 3.3 volts each, so you'd only get away with a string of three.

that would be an eye-popping tail light ... 5 watts of leds!  Shocked

of course, you don't need to run 5, you could run one or two, without any changes to the circuit - that is the true beauty of constant current regulation, it'll delivery as much voltage as needed to push the current, given theres enough supply.  be aware these things get HOT ... you'll need some decent mass for both the led and the regulator... mass meaning a chunky thick piece of steel or aluminum.
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« Reply #2 on: November 10, 2005, 09:49:50 AM »

Agreed--it's already pretty well optimized for what you want to do. In fact the thing came about as the result of a lengthy email thread I had with someone wanting to light their ATV with Luxeons, so it got developed for a vehicle application originally.

The only caution I have is a construction one, rather tahn a theory one--the metal boards on the Luxeons are not connected to "ground" so if you directly attache them to a metal heatsink, you'll have issues with series combinations at best, and short circuit your voltage regulator at worst. Luckily the LM317 has good short circuit protection, so the results shouldn't be disastrous. But the tab on the LM317T case is also at Vout--so don't get tempted to heatsink all these hot components to your vehicle chassis.
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justDIY
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« Reply #3 on: November 10, 2005, 10:02:20 AM »

Rob,

Interesting about the metal and ground and whatnot ... I'd been informed of the opposite, or at least, different information?

Luxeon aka Luxeon Emitter ... metal slug at the bottom of the emitter is not electrically neutral and needs isolation.

Luxeon Star ... metal slug of the emitter is thermally interfaced but electrically isolated from the weird little aluminum pc-board, and the pcboard is then electrically neutral.

So I guess we need to know which flavor of the Luxeon the original poster is asking about?
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« Reply #4 on: November 10, 2005, 10:17:18 AM »

Luxeon Star datasheet, p2, note 3
Quote
Electrical insulation between neighboring Stars is required aluminum board is not electrically neutral


Luxeon Emitter datasheet, p2 note 1
Quote
slug of device is not electrically neutral. Do not electrically connect either the anode or cathode to the slug.

And lest you all think I go around memorizing datasheets  oh yeah, it's just that I helped a forum member with a problem in a Luxeon array where this was the underlying cause--the series connected array was getting short-circuited by the heatsink.
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« Reply #5 on: November 10, 2005, 10:46:28 AM »

Hehe ... I was going by info I'd received from a vendor ... so my assumption didn't hold true for luxeons then

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All high powered leds, including Luxeons, are similiar in design. Yes, the back is electrically neutral, you can mount several of them on a flat aluminum sheet as a heatsink for an array.

Luckily, the pcbs for UFOs I purchased are electrically neutral, and I have three of them sharing a metal heatsink nicely.

So there's something to consider Fusion ... might want to go with UFOs instead of Luxeons to simplify mounting... and I think but not 100% sure, the UFOs are similar in size to the new Luxeon Star/C offering, much smaller than the original star.

As far as mounting real luxeons, how do you attach them to a heatsink... I know a long time ago they used mica to conduct heat but insulate electricity... I imagine today there is some where you can buy those adhesive pads like you see on commerical electronics?
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« Reply #6 on: November 10, 2005, 10:51:07 AM »

Haha, I figured mica shims too.

There's a Luxeon Thermal design guide available, but it is fairly silent on the electrical properties of the thermal compounds it mentions.

Thanks for that UFO trivia, I have a project on the drawing board that will be much simpler if I can directly mount to an aluminum chassis, so I'll go the UFO route for simplicity's sake (and cost).
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fusionxtreme
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« Reply #7 on: November 10, 2005, 02:35:24 PM »

I appreciate all of the help.  I am looking at the luxeon star, but it appears that I can not use the same heat sink for the star and the lm317.  If so, this presents another design issue to resolve.  Also,  I would like to use one luxeon star (or UFO) and operate this at 150ma for the running lights and say 300ma for the brake light or turn signal.  I like the simplicity of this design and would like to see if this can be done using another resistor, etc.....

Thanks
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fusionxtreme
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« Reply #8 on: November 10, 2005, 03:01:29 PM »

Also, the luxeons are the Star III.  I believe that these can be run at 700ma, 1000ma and 1400ma.  Will I have any issues with the LM317 design?

Thanks
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« Reply #9 on: November 10, 2005, 06:32:13 PM »

I don't have the parts to test this, but I believe you can use a PNP transistor to select how much resistance is on the OUT pin.

for example, start with 3.9 ohm on the OUT pin connected to ADJ for 320mA

then you'd wire in parallel with that a PNP transistor and another 3.9 ohm resistor.   when you turn the transistor on, it will put the second 3.9 ohms in parallel with the first, halving your resistance to ~1.95 ohms and doubling the current output to ~640 mA

make sure you select a transistor that can handle at least an amp (anything TIP should work)

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« Reply #10 on: November 11, 2005, 10:30:45 AM »

That's an interesting idea--an SPST switch would do it too, wouldn't it?
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fusionxtreme
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« Reply #11 on: November 15, 2005, 04:28:40 PM »

Any suggestions on the PNP transistor.  Also, this will be used in an ATV that will have a stator to run the lights... and sometimes I won't have a battery.  Will this have a negative effect on the transistor?

Thanks
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« Reply #12 on: November 15, 2005, 06:25:52 PM »

There is some math involved with selecting the right transistor, but since I'm not an engineer and I personally hate math, I just leave all that alone, and find the best solution through educated guessing and trial and error.

I have some TIP105 (beefy) PNP transistors on order for another project, they'll be here in a week, I can breadboard it for you at that time... I'm kinda curious to see how it works myself.

Here is a schematic of my idea:


I was going to suggest some values for R1 and R2, however, I can't find anything good that will work for your 3 watt LED, at 700ma you are looking at 875 mw to be dissipated by the resistor, let alone 1400mAa, that puts you past the recommended safe range of a 1 watt resistor ... so you get into weird beasts like wirewound sand or you have to double up your resistors (that would mean four resistors total)

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Hang on! business major trying to do engineering, I hope Rob checks my math  Grin

Ok, so, for "running light mode" and "brake light mode", you want R1 and R2 to be the same, keeps math easy that way

For 700ma as the running lights:
1.25 / If = R is 1.25 / 0.7 = 1.8 ohms

closest easy match: two one watt 3.9 ohm resistors wired in parallel:

RT = 1 / (1 / R1  + 1 / R2 / 1)
RT = 1 / ( 1 / 3.9 + 1 / 3.9) = 1.95 ohms, and the current sent out is
IF = VADJ / RSET = 1.25 / 1.95 = 0.64 Amps which means the resistors will have to 'burn off'
P = I * V = 0.64 * 1.25 = 0.800 watts = 800 mW

Now for braking mode, which is R2, you want the full 1400 mA:
1.25 / 1.4  = 0.89 ohms
using the same resistor(s) you used for R1 again as R2 you have:
RT = 1 ( 1 / 1.95 + 1 / 1.95) = 0.975 ohms, which yields output current of
IF = VADJ / RSET = 1.25 / 0.98 = 1.28 Amps which means the resistors will have to 'burn off'
P = I * V = 1.28 * 1.25 = 1.6 watts ... but 1600 / 4 = 400 mW each, so four 1 watt resistors should be fine.

As for R3, I'm not sure enough on how to calculate that, it needs to be based on the amount of current needed to be switched and the amplification factor of the given transistor (QFE) ... when you find a PNP transistor that can handle at least 3 amps, let us know what it is and I or someone else can help you figure out the value of R3
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fusionxtreme
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« Reply #13 on: November 15, 2005, 06:54:44 PM »

I was almost with you till you got to the end and have 4 1 watt resistors.
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justDIY
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« Reply #14 on: November 15, 2005, 07:00:21 PM »

the four 1-watt resistors are nessecary due to the current you'll be passing through them with your 3-watt LED

the four resistors work in parallel groups of two as single resistors, one for high brightness, one for low brightness.
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