Luxeon III Heatsink

What sort of heatsinking does a luxeon III need?
right now I've got three white ones on a standard Intel P4 heatsink. W/ the fan, it doesn't get warm. I just noticed that the fan wire came loose, and that heatsink is HOT. Not too hot to touch, but still damn hot.. Im guessing about 130 degrees F. (and room temp is about 60) Im using a 1000ma driver, so Im pushing these to their recommended limit.

Ive got a project planned w/ 6 luxeon III's and I was planning on using the same heatsink.
I really dont want to worry about a fan either, so just how much heatsinking do I need on these things?

thanks

6 times as much as you have now?

you need to factor dumping 9W of power.. a lot of heat sinks will actually say how much power they will dump... often they will have a rating of C/W which i'm pretty sure means how many degrees C the temperature will rise dumping W watts.. for example.. 13C/W.. meaning there will be a 13 Deg. C difference in temperature from the air to the heat source per watt.. that would mean a difference of 117C .. or 'holy smoke' hot.. much too hot for lux3 to survive..

I used a heatsink that is about 4 inch square cut down to a 4 inch circle... and i run 4 lux 3's at almost full power (about 850mA)... it only gets 'quite warm', but never too hot to touch.

with decent ventilation i'd suspect a square p4 heatsink would only be good for a single L3.. the old pentium II bigger square heatsink probably more like 2-3 w/o fan.. I'm building a 3xL3 light right now and it will use a 2" square heatsink with a little blower.. cooler = brighter is a key thing with luxeons aw well.. up to 20% of brightness is lost from heat under normal temperature levels.

In Lumiled's document AB23 (Luxeon V thermal design) two types of heatsinks are mentioned, with and without fan cooling.

You can read exact heatsink sizes and air convection rates in the document.

In their test, without fan (free convection) the heatsink-ambient part have 8.5-9[C/W] thermal resistance. With fan it is 3-4[C/W].

Using 4xLuxIII @1000mA (Vf=3.5V) means that each emitter has 3.5Watts => total of 14 Watts should be absorbed to the air.

Let's say we are using fan. That is 14[W]*4[C/W]=56 degrees C between LuxIII slug and room temp (now ignoring the thermal resistance of the adhesive between heatsink and the emitter).

Between LuxIII die (junction) and the slug there is another 3.5[W]*13[C/W] = 45.5 degrees C (114F).
If you are using stars instead of emitters it would be 59.5 degrees C (139F).

So LuxIII emitter junction temperature is 45.5+56=101.5 degrees C (215F) above room temp in a Lumiled tested fan cooled environment.

Heatsinks seem to be a matter of mass and surface area. The more mass, the more heat it can readily absorb; the greater the surface area, the more heat it can release via radiation and convection. I believe a combination of these two factors leads to the C/W rating.

CPU heatsinks with fan have nowadays 0.3-0.6 C/W rates

well.. thanks for your help. It looks like I need a really big heatsink, or I need to incorporate the fan into the design.

I think that w/ a fan, the P4 heatsink will work. Hell, its keeping my overclocked, overvoltaged hot running p4 cpu cool.

BTW, was on lumileds website to look up those data sheets, and noticed you can purchase the luxeons directly. They are cheaper than the usual surplus sources, too!

i made this simple calc based on AB05. i entered some numbers based on your post, i added 20 degrees to your 60, and according to the datasheets the Vf is 3.9v typical @ 1000mA. the results are for 1 LUXIII. this is only a guide. i use this all the time for my custom projects.

http://www.atlevcah.com/images/alittlehelp.jpg

please see the AB for more information on how the results work. hope this helps you. good luck.

keith /ubbthreads/images/graemlins/smile.gif

thats great.. thanks.

man.. this heat problem is really a big issue. It makes these things really hard to deal w/.

A lot of people must be ignoring the ammount of heatsinking these need. Ive seen three luxIII's all in a little halogen accent light size fixture. there is no way that could have enough heatsinking. I guess a flashlight isn't too big of an issue, because it isn't on for long, but I dont see how even the entire flashlight tube could be enough for some of the three luxIII or lux5 lights that are out.

i got one Q about the luxeon !!
When u say "driver board" what is that ?

I got luxeon 3w diode, and i only use a standard 3volts battery on one luxeon... is that wrong of me ???

thanx for answer about my worry haha /ubbthreads/images/graemlins/wink.gif /ubbthreads/images/graemlins/wink.gif

the lux can actually handle a lot more heat than we'd like to think.. 200F ish is about the limit.. i think one of the waynes told me once that 205 or 210 will make it pop.. the thing is.. that the higher the temp, the greater the temp differential to the ambient so the more power can transfer.

My nano light AAA got the lux up to 185F reading with an IR thermometer.. the lux didn't mind half as much as the arctic silver which fell apart the next time i pulled the pill out of the light.

my 4" round heatsink that is about 3/4 thick with about 20 fins works with just convection to cool 4 Lux3s at 850mA.

liteeglow: a 'driver board' is the electronics that match the battery voltage to the luxeon voltage.. since the optimal range of voltage for a Luxeon is a very tiny range.. like 3.7V to 3.8V for example.. and the battery obviously won't stay within that range.. the 'driver' is the magic that makes luxeon friendly voltage from luxeon non-friendly sources.

3V applied to a Luxeon 3 will run it between 1/3 and 1/2 power, actually even a little less.. my driver circuit runs about 3.3V at the emitter at 1W.

Thanx for the help /ubbthreads/images/graemlins/smile.gif

So i understand correct when i beliwe the luxeon got the 100% output when the battery is 3,8v (example) /ubbthreads/images/graemlins/smile.gif when i not use driver board ...

but the light is prety bright now when i use 4v battery power!
but i belive when the battery drop to 3,4v the light goes badly down...

well ok /ubbthreads/images/graemlins/smile.gif i think i understand !
but these driver board is hard to find !

Caution should be used with blanket statements such as Andrewwynn's "the lux can actually handle a lot more heat than we'd like to think".

For instance, the datasheet ds45 from lumleds quotes 135 degrees C as the "absolute maximum" junction temperature for a lux III. That's 275 F. But at that junction temperature, the heatsink temperature will be much lower (depending on how well it's attached to the luxeon) and the light output will have fallen to as little as 60 percent of it's maximum output. Extended runs at high temperatures will also shorten the life of the luxeon considerably.

But the real reason I'm writing this is to warn about thermal runaway. Once a critical temperature is reached, the luxeon will pull more current from the battery (if it can) and as it pulls more current it will heat up even more. This becomes a loop of escallating heat and current until it exceeds a critical limit and stops working.

If you use a currnt controlled driver the risk of thermal runaway will be fairly small. If you use direct drive or a resistor, the risk is much higher.


Daniel

what gadget said :-D

my point was... say with a 13C/W difference between the junction and slug.. that means 39C difference at 3W between the junction and slug.. than if there is maybe a 10C/W differnce from slug to your hand... means that is 150F... ouch.. that's my point... it will get quite warm before it's really a problem.

I would not build a light that has the possibility of thermal runaway... one of mine 'tries' but the increasing current pulls the battery voltage down and i just get a little 'bump' in power output that makes a DD light act like it has a driver for 5-10 mintues.

The reason i said what i did about 'what we'd like to think'.. is that with a bigger heatsink that has reasonably airflow, the lux will be a lot cooler than you'd usually imagine.. .i have a lux 3 that runs at 5W initially and i've never measured a slug or heatsink temp over 130F.

-awr

[ QUOTE ]
idleprocess said:
Heatsinks seem to be a matter of mass and surface area. The more mass, the more heat it can readily absorb; the greater the surface area, the more heat it can release via radiation and convection. I believe a combination of these two factors leads to the C/W rating.

[/ QUOTE ]

There is also the ability of the heatsink to transport the heat, there is a thermal resistance in the metal itself. It is also the reason why the tips of the fins of a heatsink are typically much cooler than where the heat source is mounted. A good example is copper vs. aluminum. Copper has half the thermal resistance of aluminum, and makes a wonderful heat spreader.

An additional factor is the surface finish, especially with heatsinks that are designed to work more as heat radiators. Freshly machined aluminum is a very poor radiator. But if you Hard Anodize III it, the amount it radiates can go up by over a factor of 10, reaching in the 77% range. Copper, oxidized black hits an emissivity in the 88% range. For example, if really good machining is done, such that it has a mirror finish, aluminum can have an emissivity of only 4%. Part way in between is a alodyne finish (chromate conversion), which typically sits in the 40%-60% range. Black paint is a blessing and curse, depending on it's formulation. Many can act as a thermal blanket over the aluminum. Then there are others, such as black body paint, and certain barbaque black (Krylon is a good one), that can be in the 95%-99% emissivity range, with some special paints, like 3M Black Velvet 9560 can hit 100%. Believe it or not, there are even white paints which are heavily loaded with TiO2 that are in the 94% emissivity range, but also reflect nearby heat sources at a 90% range. Keep in mind that these numbers move around alot, depending on the temperature of the heatsink.

A start as a resource is found here:
http://www.infrared-thermography.com/material.htm

Google makes a fine teacher for the curious.

Two notes.

It is always a good reason using emitters rather than stars. LuxIII thermal resistance is 13C/W (emitter) vs 17C/W (star). You win 16C less heat on junction with emitters at 4Watts.
The LuxV has the difference of 5C/W between emitter and star configuration, so junction is 30C cooler at 6Watts when using emitters.

For more exact calculation we can decrease the power by 10-15% calculated with formula Vf*I_led. This power becomes to light emission, only the remaining 85-90% is dissipated as heat.

holyshite..
I never noticed the 2nd page to the question I asked.

Thanks to all who answered my question..
thats all a lot of great info.