# Heat sink length?



## sortafast (Feb 27, 2006)

Alright, i finally got some time on a lathe and roughed out a decent looking little light. the ID is about 7/8" +/-. About the only thing left, is I need to know how long should i make the heat sink for this bad boy? This first light is going to be using a White Lux V driven by a bad boy 750mA converter. The light is gonna be used as a bike light, so there should be a good amount of air passing over the light most of the time. I know that this sucker might get a bit on the warm side of things, so i just want to make sure that i build it so that the heat does not be come a big issue. I am thinking about adding some additional fins of some sort to the outside of the light to help pull the heat away, but i am not sure if it is warranted at this time. Any help would be much appreciated.

Thanks

dave


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## gadget_lover (Feb 27, 2006)

I'd like the answer too. Even better, I'd like to know how one would verify that it was properly sized.

It seems like if you are burning up 5 watts, and you have a 1/4 diameter slug to pass that heat to your heat sink, the bottle neck is right there. But how do we check it? What formula would give us the proper thickness for an aluminum (or copper) disk that is 7/8 inch diameter to pass as much of the heat as possible?


I don't pretend to understand the theory behind moving heat around.

Daniel


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## MoonRise (Feb 27, 2006)

Heat transfer depends, among other things, on the temperature differential between objects and how close you want the 'hot' side to be in temp to the 'cold' side.

A bike light at 7/8 inch diameter really isn't all that big. Running 5 watts through it will tend to get it warm/hot. Unless you have finned the entire exterior of that 7/8 inch diameter!

That said, I'd make the heat sink at least 1/2 inch wide, and 3/4 would be better. More would be better if you can have the size and weight. Remember to thermal compound and close-fit the sink to the body so that the heat can get from the sink to the body and then out to the air. This based on the usual Luxeon on a circular heatsink inside a tubular body a few inches long.


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## gadget_lover (Feb 27, 2006)

MIke, What formula did you use to get the "1/2 to 3/4" figure?

Thanks.

Daniel


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## MoonRise (Feb 27, 2006)

Formula? I just looked at the width of someone else's heatsink and went from there.

If I have to look up and use formulas, it's too much like work and I'd expect to get paid.  

And I'm not going to do partial differential equations for fun. Nope, not me, not my general idea of fun. Maybe js (who was that masked mathematician?)would. :nana: 

mumble, mumble, ... dT/dt = [ (T2 - T1) ..... bunch of rho's, k's, and maybe some other Greek in there as well ...


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## sortafast (Feb 27, 2006)

MoonRise said:


> Heat transfer depends, among other things, on the temperature differential between objects and how close you want the 'hot' side to be in temp to the 'cold' side.
> 
> A bike light at 7/8 inch diameter really isn't all that big. Running 5 watts through it will tend to get it warm/hot. Unless you have finned the entire exterior of that 7/8 inch diameter!
> 
> That said, I'd make the heat sink at least 1/2 inch wide, and 3/4 would be better. More would be better if you can have the size and weight. Remember to thermal compound and close-fit the sink to the body so that the heat can get from the sink to the body and then out to the air. This based on the usual Luxeon on a circular heatsink inside a tubular body a few inches long.


The OD of the tube is 1" and the id is about .824"ish (kinda fudged it a bit) At this point the OAL of the body is kinda hinging on the length of the heat sink to mount the emitter to. I want to make the whole package as small as possible with out causing the sucker to overheat and damage any of the components. As long as the helmet mount itself doesnt get hot, everything should be ok. I might just wing it and see if things start to catch fire:devil: .

Also, is there much difference in heat output from driving a Lux V at 750mA and like a Lux III at 1000mA? I am also thinking of building another light around a white Lux K2 emitter once they are available (and i acquire more funding).


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## gadget_lover (Feb 27, 2006)

Thanks Mike. Now where do I buy a rho? 

Based on observations, I think the attached gif represents what I see as far as heat transfer when I have a heat sink (in grey) inside a metal body that will absorb and dissipate the heat. The yellow is the hottest. The black is the coolest. The grey is ambient temperature.







I could, of course, be way off, but it seems the hotest spot on the heat sink is where the luxeon slug touches it. The coolest is the furthest from the slug. That's thermal resistance in play???

The heat transfer from the cool end of the sink would appear to be much less effective tnan the transfer at the front. 

Now I'm just guesing, but a quarter inch slug will transfer only so much heat, and a 1 inch diameter 1/4 inch thick heat sink will be radiating the heat in all directions to a body with a lot more mass. The interface to the body will be .25 * 3.12 or about 3/4 of a square inch. That's 12 times the surface area of the slug itself.

I know that with electricty, you can increase the resistance of a circuit by using a thicker wire. I don't know what the thermal equivilence is. I suspect that a 4 ounce heat sink will not perform better (in extended run) than a 1 ounce heat sink if both heat sinks connect to a flashlight with the same surface area.

Now I'm guessing that it's not that simple. I'm wondering if it's close enough?


Daniel


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## MoonRise (Feb 27, 2006)

Daniel,

You pretty much have the general idea right, just a few minor picky tech details slightly off.

For rho, first you need a boat. :nana:

Your animated gif is close, but the thermal profile at equilibrium will still show a hot spot under the heat source (LED) and the outer surface of the heatsink would show cool along the surface with some (slight) temp variation from the top of the heatsink to the bottom (3-d temp profile instead of your more 2-d profile).

For electricity, if you use a thicker wire you -decrease- the resistance of the wire. The thermal equivalent to electrical resistance is thermal resistance. But thermally there is also heat capacity (electrical equiv = capacitance).

Once you reach steady-state conditions, it boils down to thermal resistances in the heat path and surface area to conduct/radiate the heat away. We usually won't have much thermal radiation in flashlights (think IR from a big incan as a prime example of thermal radiation). We have mostly conduction going on, from the LED slug to the heatsink, then the heatsink to the flashlight body, then from the body to the air or your hand or water/etc.

Your question of the 4 oz heatsink vs the 1 oz heatsink in bodies of the same size depends on the thermal resistance between the heatsink and the body, and then on the thermal conductivity of the body material, and then of the thermal resistance/conductivity of the flashlight to the surroundings.

So, it all depends.

If you have a heatsink and the flashlight body is HOT right at the heatsink and cool elsewhere, then your flashlight body isn't getting rid of the heat too well. You could try a wider heatsink to get more of the body involved in the heat transfer, or a different (more thermally conductive) body material.

Oh, and your area calculation above should use 3.14 (pi) not 3.12 (almost pi). 

surface_area_of_a_cylinder = pi x diameter x width

1 inch dia x 0.25 inch thick x pi = 0.79 sq inch (smidge over 3/4 sq inch)


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## gadget_lover (Feb 27, 2006)

Yeah, I knew resistance decreases with the cross-section of the wire. Sloppy edit there. I also know that Pi is not really 3.12... I should have used 22/7 or Pi to be exact. 

Ok, so much for the Mea Culpa's. Thanks for being patient. I've been trying to understand this for a while.

In electrical resistance, the maximum current that is passed is limited by the total resistance. Thicker wires don't make much difference if you have a low amp, high resistance circuit. In heat, is the maximum heat transfer limited by the contact area with the relatively small slug? 

I guess that I should stop worying about the theoretical and set up a test jig with my temperature controlled soldering iron , various pieces of aluminum and an infrared thermometer. I'd just like to understand it a little better.

Daniel


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## Christoph (Mar 3, 2006)

I think DougS did a thread on this a while back. 

C


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## gadget_lover (Mar 3, 2006)

A search for "heatsink AND length" is not finding anything in any of the forums. 

Does anyone remember that thread? Did it have an actual formula for the loptimal ength of a heat sink?


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## sortafast (Mar 4, 2006)

I think i am just gonna wing it. I feel that i have adaquate length with the body I just got built up. It will probably be over 1" long and 7/8" in dia. or so. I think it should be fine.


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## Lost_In_Beijing (Mar 5, 2006)

Maybe this site will help with how to size the heatsink:
http://web.telia.com/~u85920178/begin/heat-0.htm

Also, since you'll need to use some kind of thermal compound, here is one manufacturer's instructions on how to apply it (with pictures):
http://www.arcticsilver.com/arctic_silver_instructions.htm

Weldon


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## HEY HEY ITS HENDO (Mar 18, 2006)

Hey Lost_In_Beijing, interesting links  thanks !!

..... sortafast, i`ve saved a link to this page so keep us updated on the progress of your lights,
was the 1"x7/8" the finished size? WOW !! ....


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## wtraymond (Mar 20, 2006)

This question has been brought up before and I think it warrants a more complete and scientific answer.

Gadget asked the same question in a thread about aluminum vs. copper heat conductivity and diffusivity. I started to build a spreadsheet incorporating LED efficiency, wattage, material characteristics, diameter, area, mass, etc... but it's become rather complicated. Any suggestions for keeping this simple?

Any idea where I can find accurate thermal efficiency of the Luxeon series LEDs?

If we collaborate, maybe we can come up with something that can be used as a guide for planning a custom flashlight body.

This is a great topic that needs to be addressed.


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## n_den (Mar 31, 2006)

hello,

first you need to find the total surface area required. you use the following based on worst case scenario, light-on bike standing still:

Junction Temp=135 C, max. from DS40
Amb Temp=70 C, max. from DS40
Current=750mA, user determined
Voltage=6.84v, avg. from DS40
Luxeon V Portable C/W=11, from DS40

Required minimum=14.132 sq. in.

now just calc & modifiy your tube to meet the minimum.

OD=1", user determined, see thread, so, to answer the question the minimum length will be: 4" this will give you 14.137 sq. in.

that was for worst case. i don't think you'd be in a 70 degree C environment.

hope this helps,
n_den


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## gadget_lover (Mar 31, 2006)

What formula did you use to determine the "Required minimum=14.132 sq. in." ?

I'm just hoping to find the formula for the cross-section needed for conducting a set amount of heat (in watts) from point A to point B. Of course, I also need at least some explanation of how to apply that formula.

All the formulas based on CPU heat sinks are concerned with square inches of surface as opposed to thickness of the cross-sections.

Maybe the answer is that the thickness does not matter as much as the contact area between the flashlight body (the radiator) and the heatsink (the heat conduit). I don't know.

Daniel


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## n_den (Mar 31, 2006)

the formulas and how to use them can be found in application brief AB05, it does not contain cross-section formulas only minimum heat sink requirements.

n_den


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## gadget_lover (Mar 31, 2006)

I think I have an answer from a pretty credible source. He can jump in if he wants....

In essence, (and very roughly) the point at which extra thickness makes no difference in heat conduction is just under 1/2 of distance from the slug to the flashlight wall.

In application: Given a minimag with a 1/2 inch diameter bore and a Luxeon with a 1/4 inch diameter slug. That gives you 1/4 inch between wall and slug. 1/2 of that is 1/8 inch, so a heatsink over 1/8 inch thick would be overkill.

Scale that up to a 2 inch Maglight head, and you have a different story. 1 - 1/4 = .75 so a .375 inch thick heatsink would be as thick as you'd need.

There are still valid reasons for a thicker heatsink. One would be to provide heat storage for instances where the heat source is way too big for the body to disipate. A thicker heatsink has more mass to soak up the heat for short runs. A thicker sink would also provide more contact area between the wall and the heatsink, wich may be needed to overcome the thermal resistance at that junction.

The actual formulas are more complex than that. I simplified it for the purposes of discussion.

Daniel


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