# calculating heat sink

#### degarb

##### Flashlight Enthusiast
I am reading 12 in square per watt of heat? This sounds large

I want to drive an xml2 t6 at 700 ma max. No barrel, possible plastic reflector, possible aluminum. Plastic battery housing. What is smallest, lightest, most diminuative heatsink I can use?

I tried to wrap the plastic 2p18650 case in painted aluminum flashing, but think the heat is not transfering, even with heat paste. So at most I just now have the p60 brasspill and aluminum reflector, thin flashing, and possibly 2 pennies, to sink the heat with this design. Wrist light, that needs to be lightweight.

##### Flashaholic
You are doing an ambitious project. Light weight and high power are not a good match. On top of that, a brass pill is NOT a good heat conductor. My understanding is that heat stops at the first insulator.

The 12 sq inches of surface area is to keep the temperature at the LED junction below a critical point indefinitely. For the xml2 that point is 150C That assumes you are conducting heat to still air. Some lights get away with smaller area by using your hand as a way of removing heat from the light. Others have fins to increase the surface area or are in moving air. 700 ma means TWO watts.

Since both sides of a disk are the surface, a 1/4 inch thick 2 inch disk is 3.14 sq inches per side. Stack two with a 1/4 inch gap and you have 12 sq inches.
You get .785 sq inches per side in a 1 inch disk. Play with the circle calculator at http://www.calculatorsoup.com/calculators/geometry-plane/circle.php for other values.

The 12 sq inch rule assumes that you have good heat conductivity from the source all the way through to the radiating surface.

Daniel

#### degarb

##### Flashlight Enthusiast
You are doing an ambitious project. Light weight and high power are not a good match. On top of that, a brass pill is NOT a good heat conductor. My understanding is that heat stops at the first insulator.

The 12 sq inches of surface area is to keep the temperature at the LED junction below a critical point indefinitely. For the xml2 that point is 150C That assumes you are conducting heat to still air. Some lights get away with smaller area by using your hand as a way of removing heat from the light. Others have fins to increase the surface area or are in moving air. 700 ma means TWO watts.

Since both sides of a disk are the surface, a 1/4 inch thick 2 inch disk is 3.14 sq inches per side. Stack two with a 1/4 inch gap and you have 12 sq inches.
You get .785 sq inches per side in a 1 inch disk. Play with the circle calculator at http://www.calculatorsoup.com/calculators/geometry-plane/circle.php for other values.

The 12 sq inch rule assumes that you have good heat conductivity from the source all the way through to the radiating surface.

Daniel

I artic silvered my star onto a video chip heatsink, like the above spikey one (I guess 3/8 inch high spikes), which measured 30 mm by 40 mm. I did this yesterday, based on advice by a supplier that wrote to me that his rule of thumb is 9 inch square for a 3 watt led and 15 in sq. for a 5 watt led. I am guessing my heat sink has a minimum of 7 in sq, with a chimney effect. (700 ma max). Or am I way off?

I do not quite understand yet the 12 inch rule. Is this 12 in for each led, for each led watt running, or each watt of heat?

Last edited:

#### AnAppleSnail

##### Flashlight Enthusiast
I do not quite understand yet the 12 inch rule. Is this 12 in for each led, for each led watt running, or each watt of heat?

Are you making a hand-held light, or a mounted light? Handheld lights shed most of their heat to the hand, and are not usually run continuously and are turned off when painfully hot. Most handheld lights at the P60 host size, running above 5W continuous power, will get very hot after several minutes of operation. This is a fact of small heatsinks and large heat sources. I have a Nichia 219 triple with the same power rating as my soldering iron. Even if 1/3 of the power escapes as light, that's 10W of heat inside the head.

Mounted lights have to 'fend for themselves' to shed heat, and need big heat-transfer area to do so. 12 square inches per watt of LED input power is reasonably conservative to give long LED life. Using a fan (As is the case inside a computer case) increases heat flow and reduces the heatsink required.

#### degarb

##### Flashlight Enthusiast

I artic silvered my star onto a video chip heatsink, like the above spikey one (I guess 3/8 inch high spikes), which measured 30 mm by 40 mm. I did this yesterday, based on advice by a supplier that wrote to me that his rule of thumb is 9 inch square for a 3 watt led and 15 in sq. for a 5 watt led. I am guessing my heat sink has a minimum of 7 in sq, with a chimney effect. (700 ma max). Or am I way off?

I do not quite understand yet the 12 inch rule. Is this 12 in for each led, for each led watt running, or each watt of heat?

While, I still don't understant the 12 rule, per vagueness above. The supplier will be unnamed that had the "9 inch for a 3 watt LED" rule. This highly-secret, unnamed source has a similar heatsink to the graphic chip one I used yesterday http://www.ledsupply.com/led-heatsinks/heatsink-finned-with-adhesive-tape

He says this one has 30 inches square. It is 35 mm by 35 mm, while mine is 30 by 40. So, obviously, I do not know how to calculate surface area of this spikey design.

I am taking a plastic 18650 holder > switch > buckpuck 700 (ma max) > xml t6 4c 4500k on 25 mm star > arctic epoxied onto heatsink > 30 mm x 30 mm deep reflector mounted on top of battery pack > covered electronics & reflector, not covering heat sink, in foam core for splash and ugliness resistance > using strap to strap onto left wrist and thumb loop. This can be pointed depending on how the strap is velcroed onto rear of battery case. This is a working light, used while both hands are used--left hand primarily holding things and so can hole and point light. Yes, I could do a single 18650, but the runtime and output doesn't work for me of any I see. And the Fenix TK 35 is just too darn heavy to strap on and forget!

Last edited:

##### Flashaholic
The surface area of the spikey thing is the sum of the surface area of each spike. Each spike is height * width * length. Count the spikes and add them up and you are done.

I am trying to imagine how I would use a well focused 280 lumen light when my hands are full. That is way too bright for close up work.

Keep the epoxy layer as thin as possible to maximize heat transfer.

Dan

#### RetroTechie

##### Flashlight Enthusiast
The surface area of the spikey thing is the sum of the surface area of each spike. Each spike is height * width * length. Count the spikes and add them up and you are done.
Around a cooler you have complex interactions of heat flow (through the cooler's material, that material's finish, and in the air itself), radiation and airflow, and the result can not be reduced to a simple calculation of surface area. That calculation is easy, but doesn't help you much.

Which is why coolers come with a Kelvin/Watt rating. Which allows you to calculate how many degrees (Kelvin) your LED would rise above temperature of the surrounding air, as a function of the power that LED is dissipating. For the LED's power dissipation, multiply running Volts x Amps and you have a Watts figure. Note that this is not what goes into the cooler as some of the power is 'lost' in the form of light. But since most of those Watts turn into heat, it's good enough as a ballpark figure.

That K/W figure depends on cooler, how it's mounted, and airflow. So if you find a K/W figure for a cooler, check under what conditions that K/W figure could be expected. Calculation example: let's say you have a small passive cooler, that does ~30 K/W:

Your LED: ~2.85V x 0.7A = 2W. Maybe :thinking: 0.1-0.2W of that is lost as light (1.8-1.9W into the cooler), but 2W is close enough.

XM-L2 datasheet says ~2.5 K/W from die -> solder point. The star it's mounted on -> cooler: who knows, let's say another 2.5 K/W. Add those all up: 2.5 + 2.5 + 30 = 35 K/W from LED die -> ambient air.

2W and 35 K/W -> your LED sits ~70 o​K above ambient air. Air = 25 o​C -> LED die is ~95 o​C, etc.

Note that such thermal calculations aren't exact science: good enough / close enough will do. More important is where the figures come from, under what conditions they're measured, and whether they are realistic.

Have you given any thoughts to how you want the finished design to look? Perhaps it would be easier to go in reverse: start with the look / construction, estimate cooling capacity, and use that to set the LED's power level.

#### degarb

##### Flashlight Enthusiast
Around a cooler you have complex interactions of heat flow (through the cooler's material, that material's finish, and in the air itself), radiation and airflow, and the result can not be reduced to a simple calculation of surface area. That calculation is easy, but doesn't help you much.

Which is why coolers come with a Kelvin/Watt rating. Which allows you to calculate how many degrees (Kelvin) your LED would rise above temperature of the surrounding air, as a function of the power that LED is dissipating. For the LED's power dissipation, multiply running Volts x Amps and you have a Watts figure. Note that this is not what goes into the cooler as some of the power is 'lost' in the form of light. But since most of those Watts turn into heat, it's good enough as a ballpark figure.

That K/W figure depends on cooler, how it's mounted, and airflow. So if you find a K/W figure for a cooler, check under what conditions that K/W figure could be expected. Calculation example: let's say you have a small passive cooler, that does ~30 K/W:

Your LED: ~2.85V x 0.7A = 2W. Maybe :thinking: 0.1-0.2W of that is lost as light (1.8-1.9W into the cooler), but 2W is close enough.

XM-L2 datasheet says ~2.5 K/W from die -> solder point. The star it's mounted on -> cooler: who knows, let's say another 2.5 K/W. Add those all up: 2.5 + 2.5 + 30 = 35 K/W from LED die -> ambient air.

2W and 35 K/W -> your LED sits ~70 o​K above ambient air. Air = 25 o​C -> LED die is ~95 o​C, etc.

Note that such thermal calculations aren't exact science: good enough / close enough will do. More important is where the figures come from, under what conditions they're measured, and whether they are realistic.

Have you given any thoughts to how you want the finished design to look? Perhaps it would be easier to go in reverse: start with the look / construction, estimate cooling capacity, and use that to set the LED's power level.

I really like your post, with example!

So, what do you think the k/w of these are?:

http://www.ebay.com/itm/121237694801

http://www.ebay.com/itm/281236091534

digikey might have the spec, but no part that is apparent: http://www.digikey.com/product-search/en?vendor=0&keywords=finned+heat+sink

Gadget lover, well I am used to working outside where 30k lux is typical on cloudy day. A shady side is around 7k lux. I like to keep my walls 700 to 1400 for fixed lighting, and 7,000 to 14,000 with headlamps and wristlights. It may be because you cannot control angle or glancing at the fixed lights that you cannot go beyond this without glare. 200 candela, just doesn't cut it for demanding work like drywalling or painting or working on computer (tiny writing), for that matter. I was trying to work with a 2k candela wristlight today, but it was just too dim. About 3k is bottom for a wristlight. Pointing is no problem, when you need it. Obviously, no time to constantly click on or off. I was surprised at how useful the 4c is over a colder bin in a wristlight. Fewer out the front lumens, but I have yet to master the art of reflector picking. I am thinking 30 by 30 mm may at least get me 3k candela, though the hotspot is huge (about width of my shoulders if worn on head and looking at the ground.). And thanks for the tip on keeping the epoxy thin!

Also, in my design, the heat sink is on its side and glued to battery case. I was using a thin sheet of aluminum, but so much velcro on it that I don't see any aluminum sheet exposed. So, I am thinking of just JB welding it to plastic battery case. The heat will rise away from battery case anyway. Maybe a little heat will enter plastic case. My reasoning is that a putty epoxy would conduct heat better than a rubbery one. Jb weld or mightyputtypurple is cheap. I googled a laser forum that praised JB weld's thermal conductivity, but see here that people don't like it's thermal properties. Another strategy is to put two dots of silver and rest JB weld.

Last edited:

#### degarb

##### Flashlight Enthusiast
Have you given any thoughts to how you want the finished design to look? Perhaps it would be easier to go in reverse: start with the look / construction, estimate cooling capacity, and use that to set the LED's power level.

Well, the light has to meet spec or not worth it. But then, if the heat sink doesn't work, the led will burn out.

The graphic card heat sink seemed only a little warm, running for several hours today. I may rip off the glued-on reflectorc yet, as unhappy with this 2.0 version of the design's reflectors 2k candela hotspot. I have a ir thermometer designed for surface, room and forehead. Assuming it can go to 100 C, could I just measure the temp of the led when I rip off the reflector? (After a new one ships from china.) (version 1.0, is just a brass pill and money glued on, which gets really hot. That light I really like, though and have used for over two weeks.)

An example I think of heat sink necessity: I had an xml last sunday start to smoke and led quit. Well the only thing I did was to remove optic to look at the led for about 10 seconds (I think...maybe I was dazed by brightness and it could have been longer) while running at 1.6 amps. I think the optic was pushing down on star to contact the flashlight's metal. When I removed optic, she burned up in about 10 seconds to, maybe, 2 minutes. That's my theory.

Last edited:

#### degarb

##### Flashlight Enthusiast
Okay, the led supply wrote back and their 35 mm finned heat sink is said to have a thermal resistance of .5. I believe the inverse of this is 2, which is thermal conductance. However, based on graph, it looks more like 15 k/w.

Also, how big is the difference between arctic silver epoxy and arctic aluminum epoxy?

#### RetroTechie

##### Flashlight Enthusiast
That graphic is a bit hard to decipher... But afaict, the ~45o​ angled line refers to passive cooling (like, in free air?). Looks about 17 K/W to me... 15.3 K/W? Okay, whatever.
That curved/horizontal line refers to forced airflow, with K/W figure as a function of airflow (5.15 K/W at 1 m/s or 200 feet/min).

Not bad figures for a small "PC chipset" sized heatsink I think. For figures of ~1 K/W or less (and passively cooled!), think about bulky heatsinks like you find on the back of audio power amplifiers. So

Okay, the led supply wrote back and their 35 mm finned heat sink is said to have a thermal resistance of .5.
K/W? That's BS for a small heatsink like that.

Also, how big is the difference between arctic silver epoxy and arctic aluminum epoxy?
Small. The important thing is to remove air gaps, especially near where the heat is produced. Remove as in: replace air with a better heat-conducting material. And do it in such a way that those air gaps stay removed as the equipment ages. Since those air gaps are usually very thin, it isn't too important what you put in there as long as it's not air.

Beyond that, you quickly get into "my speaker cable sounds better than yours!" territory. :laughing:

As for the design: why not use a (thick!) aluminium or copper U-shaped plate, with LED attached to the U's bottom? Maybe that's enough, maybe you'd need small (finned) heatsinks attached to either side of the U. Batteries could then go 'inside the U', and driver/switch on top, bottom, or back. U shape dimensions determined by whether batteries sit horizontal or vertical. The U would double as a heat distributor + frame to attach everything to. A metal box with thick enough walls would also work.