# Heat Sink calculations - Serious Engineer Stuff



## Blindasabat (Feb 4, 2009)

So I'm sketching up some new lights and wondering "How big of a heat sink do I need?" I have a given diameter, but how thick will it need to be?

So I do some comparison calculations to see the difference between 5mm thick and 3mm thick heatsinks made from aluminum.

The results were enlightening.

Ideally, the following calculations should be solved simultaneously, but I didn't want to do that if I didn't need to (I've forgotten more math then I ever knew) so I decided to see if the results warranted it first.
I first did the steady state convection equations with the assumption that the body would add some area over the sink area alone to convective heat transfer. My old Heat Transfer book from college said that convection coefficient to air was from 5-30 Watts/square meter-C. Since it used 5 for a coal powder example, I picked 20 for nearly ideal aluminum condition. I also assumed 0.6W total power dissipated into the aluminum sink (divided by two as I calculated it in halves below).

*Convection*
Temp at surface = T(air) + (W/m^2 energy flow)/h

h = convection coefficient = 12W/m^2 * C

Energy flow for HALF the cylindrical sink = 0.3W power / (10mm x 40mm convectin area on the surface of the body) = 0.3W/(400mm^2) = 0.3W/(.0004m^2) = 750W/m^2

Ts = 20C + (750W/m2) / (12W/m2-C) = *82.5 deg C *(181 F)

*Conduction inside Sink*
Flux for half the sink = 0.3W/(5mm x 15mm) = 0.3W/0.000075m^2 = 4000W/m^2

for Auminum k = 167 W/m-C, Q= heat flux, L = length heat travels to outside surface, A= sink cross section area (estimated for my round sink)

T at center = T outside + (QL/kA) = 82.5C + (0.3W * 0.013m) / (167W/m-C * 0.000075m^2) = *82.8 deg C*

The slight temperature rise across the sink means that the sink is doing a spectacular job of conducting the heat. and that I am close enough that I don't need to solve the Convection and Conduction equations simultaneously. The result shows that convection on the outside (this is for a headlight) is extremely important and needs to be minded.

I redid the above calculations for a *3mm sink* and the results were *109.3 deg C* (229 F)outside surface and *109.8 deg C* at center of heat sink. I assumed less casing area for heat dissipation as the sink contacts less case area which made the entire difference as the small temp difference across the sink again shows. But this convection area may not really be the case, and the thinner sink may work just about as well in the same size outer housing. 

Some conclusions :
1) A thin sink will do almost as well as a thick sink - especially if it is still as thick at contact to the outside body.
2) A large, good contact from sink to housing is critical for keeping the sink cool.
2) The insignificant temperature drop across the sink (0.1 deg C) shows that Aluminum is as good as Copper, just lighter.

Commentary:
A hot outside housing may mean you have good heat sinking or that your emitter is very hot from small heat transfer area. A cooler housing may mean your heat is trapped inside, or that the sink is well connected to the body and the body is also highly thermally conductive. A small hot area on the outside means the sink is only conducting in a small area and the body is not highly thermally conductive. Some aluminums are better than others for thermal conductivity.

If anyone sees error in my assumptions or calculations, please let me know! I may be way high on the amount of power dissipated.

<edit> asdalton found an error in my sink diameter, which I corrected above. It was 0.0013m (1.3mm!?) now corrected to 0.013m (13mm). The temperature drop from center to outside (in my non-cylindrical approximation) is now 0.52 deg C. Still small, but pointing to larger overall error in approximation. Of course, my approximate power consumption may be a bigger factor. I think it is high, but if anyone has data, let me know!

<edit #2> I updated power to 0.6W total & 0.3W per side as I calculated it in halves.
I also updated the convection coeff to 12 W/sq meter * deg C


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## asdalton (Feb 4, 2009)

I don't have time to look over this in detail now, but I strongly suspect that your estimate of the convective heat transfer coefficient in air is too high. I did an experiment once to determine the heat loss from an aquarium in still air, and it was around 8 W/m^2/K. The fact that the material is aluminum in your case (a good thermal conductor) will not necessarily make the value of _h_ higher.

Also, heat conduction through in cylindrical geometry requires a more complicated formula than the one you are using. The actual temperature delta across the heatsink will be worse than what you estimated, because the heat must pass through a more constrained cross-sectional area near the center of the heatsink.


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## Blindasabat (Feb 4, 2009)

I did shoot a little high on the convection coeff due to this being a bike headlight and the air will not be still, but seeing that it may be too optimistic, I will redo them with lower coeff.


asdalton said:


> I don't have time to look over this in detail now, but I strongly suspect that your estimate of the convective heat transfer coefficient in air is too high. I did an experiment once to determine the heat loss from an aquarium in still air, and it was around 8 W/m^2/K. The fact that the material is aluminum in your case (a good thermal conductor) will not necessarily make the value of _h_ higher.


 
I know the formula for conduction from the center of a round sink is more complex, but I didn't have the time to find or derive that eq, so i assumed a lower area for conduction (5x15mm) than I actually have. If the temperature drop across the sink had been any larger, I would seek out more precise calculations for the geometry and attempt to solve the convection and conduction equations simultaneously (or iteratively) to get a more accurate answer. I could approximate by adding thermal impedance in small steps starting with a very small cross section at the center and increasing from there. I'll look at that if I get the time.
As a side note, the LED is on a star on top of the sink, so that adds some mass & area.


asdalton said:


> Also, heat conduction through in cylindrical geometry requires a more complicated formula than the one you are using. The actual temperature delta across the heatsink will be worse than what you estimated, because the heat must pass through a more constrained cross-sectional area near the center of the heatsink.


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## Blindasabat (Feb 4, 2009)

OK, if I use 10W/m^2 C, then the outside temperature jumps to 145deg C (293 F) and the internal temp becomes 145.1 deg C.

If I drop the sink area assumption down further, then the temperature rise goes from 0.052 deg C to 0.086 deg C from center to outer surface. Still insignificant. I rounded the initial 0.052 deg up to 0.1 deg in my first post.

If my dissipated power assumption is correct, then I need some cooling fins.


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## asdalton (Feb 4, 2009)

It would help if you included a diagram of the geometry. From your calculations it looks like you are assuming a heatsink 1.3 mm in radius. That's smaller than most power LEDs!


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## Justin Case (Feb 4, 2009)

Your first equation looks like Newton's Law of Cooling. However, convective heat transfer coefficient units are not quite right -- should be W/m^2/K. Not sure where you get 10mm x 40mm for your "energy flow" calculation.

The second equation looks like the steady state solution to the 1-D Laplace equation. In your equation, you state that T1 = T2 + (QL/kA). Based on this, I'm not sure where you get L = 0.0013m = 1.3mm. The way I envision your heat sink geometry, I would have thought that L = 5mm or 3mm (you have your LED thermally attached to the top of a cylindrically-shaped aluminum heat sink of thickness 5mm or 3mm). Based on your calculation of cross sectional area, I assume that the diameter of your heat sink is 15mm? But it seems like you are calculating the area for the wrong x-section. Shouldn't it be pi * (15/2)^2 mm^2? It also appears that you are assuming that the heat source is not a point source, but instead is uniformly applied to the entire face of the heat sink.

A 6P LED drop-in often is only about 2/3 efficient when driven at the expected 6V or so. That translates to dissipating about 1.5 watts as heat.

Carslaw and Jaeger, Conduction of Heat in Solids, has every heat conduction equation and solution known to man, including solutions for cylindrical geometries.


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## Blindasabat (Feb 4, 2009)

You are correct sir! The length was off by tenfold. 
I have corrected that in the first post along with adding text similar to this: 
It was 0.0013m (1.3mm!?) now corrected to 0.013m (13mm) also a low number for approximation purposes. The temperature drop from center to outside (in my non-cylindrical approximation) is now 0.52 deg C. Still small, but pointing to larger overall error in my approximation.

Now I know why the AL seemed to be a superconductor! It made little difference in calculated temperature, but tells me more work is needed for a true model. The conclusions still stand that good conductance to the shell or body is paramount!

Of course, my approximate power consumption may be a bigger factor (of error too) along with my assumption of convection area. I think the power assumption is high since my outside temperatures are now 300 & 400 degrees F!!!! I think that or the convective cooling coefficient are off. If anyone has better power dissipation data, let me know!

I can not post a picture or diagram from where I am now. I will try later.
Thanks asdalton!


asdalton said:


> It would help if you included a diagram of the geometry. From your calculations it looks like you are assuming a heatsink 1.3 mm in radius. That's smaller than most power LEDs!


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## Blindasabat (Feb 4, 2009)

OK, the power dissipated is too high or the convection coefficient is too low because of the extremely high outside temperatures compared to my experience in letting well sinked lights sit running for a long time. 
I made a couple of adjustments to my spreadsheet. I lowered the power dissipated to 0.25W since my geometry is for only half the sink anyway (40mm length versus my actual circumference of 80mm assuming a 1" diameter)
Then I jacked the convective coefficient back up to 12 Watts/sq meter-C to account for cooling outdoors (at night) airflow. Remember, my textbook gave a range of 5-30W/m^2-C in still air, moving air is 20-200!!! so I am in the low range.

My new numbers are *72.1 deg* *C* (162 F) at the outside surface with 5mm sink and *72.3 deg C* at LED in center of sink. These seem more reasonable.

The 3mm sink temperatures are 94.4 C (202 F) outside & 94.8 C center of sink.

I acknowledge the actual numbers are based in large part on assumptions and reasonable sounding results compares to observation, but the conduction (temp at LED in center of sink) versus convection (temp at outer surface) temperatures tell a story.


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## 65535 (Feb 4, 2009)

If I where you I'd save my energy and just focus on building a heatsink and housing. Honestly the variables with LED's and those kinds of things. As a general rule you can't overdo heatsinking. Thermal management has no overkill.


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## LukeA (Feb 4, 2009)

Blindasabat said:


> Of course, my approximate power consumption may be a bigger factor (of error too) along with my assumption of convection area. I think the power assumption is high since my outside temperatures are now 300 & 400 degrees F!!!! I think that or the convective cooling coefficient are off. If anyone has better power dissipation data, let me know!



If a power LED is powered such that it emits 242.5lm (for simplicity's sake), then exactly 1.000W of input power is being radiated away. Let's say it takes 3.700V and 1.000A of input power get that output. The remainder of the 3.700W input power (2.700W) will be dissipated into the heatsink. 

For 0.5W of power dissipated into the heatsink the LED could be powered by 3.200V, 0.2604A if it's 40% efficient (97lm/W) at that low current. 

If the driver is dissipating heat into the heatsink as well then that energy needs to be added to the total power absorbed and (hopefully) dissipated by the heatsink.


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## Blindasabat (Feb 4, 2009)

Hey Justin, missed your post while I was typing my own further down. 
Oh yeah, I left "C" off the coeff when I first listed h. Kelvin, Celsius, all the same here. I use degrees Celsius since more people will be used to it.

You may have seen the correction to the 0.0013m, it is now 0.013m = 13mm for the radius of a 25.5mm round sink in a thin shell casing (tube).
The 10mm is assumed width of the hot area subjected to convection around the circumference of the tube. This is an assumption and it may be way wrong since AL conducts well enough to heat up more area for effective convection, but I am shooting low for safety sake. L = the distance the heat must travel from center of sink to the outside where convection occurs, basicall y the radius of the sink = 13mm. The back of the sink is assumed insulated - in this case by air trapped in the driver & switch housing, so the conduction is assumed linear from center to outside. In reality it will lose some heat to that area, which will only improve the situation. I applied the heat to a section of the heat sink as if the LED was on one edge of a 5 * 15mm cross section bar 13mm long. I traded width of the 'bar' for approximation of a half-cylinder heated on the flat. I know, it's really kludgy, but I can adjust parameters when smart guys like you & asdalton correct me (by sending me one of those cylinder conduction equations from your book... :twothumbs). 

Good point about the wattage. I should have used known driver efficiencies of as low as 65-70% like your example to calculate power dissipated! How hot does your 6P get after sitting in air for a while??? My quality sinked known examples (SureFire E-series, HDS) get hot but not too hot to touch and I know they have less than 5mm thick sinks. I once ran a small light, a Fenix L1P, in my pocket (insulated = worst case = NO convection) overdriven on RCR123 (Nekomane body) for a long time and I could still handle it, so it was less than ~200F.

Thanks for the input!



Justin Case said:


> Your first equation looks like Newton's Law of Cooling. However, convective heat transfer coefficient units are not quite right -- should be W/m^2/K. Not sure where you get 10mm x 40mm for your "energy flow" calculation.
> 
> The second equation looks like the steady state solution to the 1-D Laplace equation. In your equation, you state that T1 = T2 + (QL/kA). Based on this, I'm not sure where you get L = 0.0013m = 1.3mm. The way I envision your heat sink geometry, I would have thought that L = 5mm or 3mm (you have your LED thermally attached to the top of a cylindrically-shaped aluminum heat sink of thickness 5mm or 3mm). Based on your calculation of cross sectional area, I assume that the diameter of your heat sink is 15mm? But it seems like you are calculating the area for the wrong x-section. Shouldn't it be pi * (15/2)^2 mm^2? It also appears that you are assuming that the heat source is not a point source, but instead is uniformly applied to the entire face of the heat sink.
> 
> ...


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## asdalton (Feb 4, 2009)

The first parameter to check is the temperature delta across the heatsink. This does not depend on the air heat transfer coefficient.

In cylindrical geometry, I obtained the following:

Ti = To + Q*ln(Ro/Ri)/(2*pi*z*k)

To = heatsink outer temperature
Ro = heatsink radius
Ti = heatsink inner temperature (near LED slug)
Ri = LED slug radius
z = heatsink thickness
k = heatsink thermal conductivity
Q = thermal power

So if we use the following,

Ro = 13 mm
Ri = 2 mm
z = 2 mm
k = 167 W/m/K
Q = 1 W

then (Ti - To) = 0.9 K.

Most of the temperature difference between the LED and the ambient air will be due to the convective component, because heat transfer through air is poor. Estimating this component is difficult, because the entire outside area of the flashlight is involved. And if you have ever held a warm light like the Surefire L4, you know that the outside surface in nowhere near uniformly warm. 

But we do know that a wide variety of high-power LED lights function at reasonable temperatures without forced air cooling. So there is something wrong with your model (probably in the assumed surface area) if you have to tweak parameters to avoid getting a ridiculously high LED temperature.


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## Blindasabat (Feb 4, 2009)

Trust me, I wouldn't waste my time either, but I am actually brushing up on some of this so I can use it at work.
And it doesn't hurt to understand how things work and if 'overkill' is even enough.


65535 said:


> If I where you I'd save my energy and just focus on building a heatsink and housing. Honestly the variables with LED's and those kinds of things. As a general rule you can't overdo heatsinking. Thermal management has no overkill.


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## Justin Case (Feb 4, 2009)

I tested a Deal Extreme SKU 6090 many months ago, and can't recall some of the specifics anymore. But I believe that I ran the drop-in for something like an hour continuous. After about 10 or 15 min the light got noticeably warm, but it never got too hot to touch. You could press it against your face and not feel discomfort. Certainly not as burning hot as a car steering wheel on a sunny day.

I built an LED tower module for SureFire KT TurboHeads and the tower barely got warm after 15 minutes of continuous operation. That tower uses a Seoul P4 U2-bin LED and an SOB 1000 driver board. I never got around to measuring current input to the LED terminals, but I did measure current draw at the tailcap for 2x17670 Li-ions. I got 0.50A. Resting voltage was 7.93V. I figure at that current load, the voltage sag for the 17670 Li-ions isn't all that much, so the total input wattage is ~4W. If we assume a Vf of 3.5V-3.7V for the SSC P4 and exactly 1000ma current delivered to the LED by the SOB 1000, that gives 3.5W-3.7W, or only ~0.3W-0.5W dissipated as heat. That seems consistent with the fact that the tower was barely warm after 15 minutes of running. That's also consistent with the advertised efficiency for the SOB driver (75% to 85%).


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## mudman cj (Feb 4, 2009)

You are considering the power dissipated by the driver, but the LED is dissipating more. See LukeA's post above for an estimate of the heat generated by a typical 3W LED at a typical level of overdrive (1A). And again, the driver is often times not connected to a heatsink at all.

One other comment I want to make is about the KL4 anecdote above. The KL4 is not a typical setup since it uses a 5W LED and the heatsink portion of the head is essentially thermally isolated from the body. See this post by Energie for the proof.


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## Blindasabat (Feb 4, 2009)

Thanks Andrew. As I assumed, the temperature delta across the sink was close - within a half degree - to the cylindrical equation, but I will update my spreadsheet.



asdalton said:


> In cylindrical geometry, I obtained the following:
> 
> Ti = To + Q*ln(Ro/Ri)/(2*pi*z*k)
> 
> then (Ti - To) = 0.9 K.


I do think my surface area assumption is low, but I didn't use the entire area just because I figured the temperature across the surface of a high power light would vary just as you said.


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## Blindasabat (Feb 4, 2009)

Using this EQ, I now get 0.36 deg C drop across the heat sink. Less than my estimated equation. You used a 2mm thick sink. And more wattage - which is still a good number. 


asdalton said:


> In cylindrical geometry, I obtained the following:
> 
> Ti = To + Q*ln(Ro/Ri)/(2*pi*z*k)
> 
> ...


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## Blindasabat (Feb 4, 2009)

I am using the LED power. I hope to sink the driver (if needed) to the back wall or cap of the housing so it won't heat up the LED.


mudman cj said:


> You are considering the power dissipated by the driver, but the LED is dissipating more. See LukeA's post above for an estimate of the heat generated by a typical 3W LED at a typical level of overdrive (1A). And again, the driver is often times not connected to a heatsink at all.
> 
> One other comment I want to make is about the KL4 anecdote above. The KL4 is not a typical setup since it uses a 5W LED and the heatsink portion of the head is essentially thermally isolated from the body. See this post by Energie for the proof.


That area actually matches my assumption very well, which makes for a conservative calculation.


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## mudman cj (Feb 4, 2009)

Blindasabat said:


> I am using the LED power.



Well, it looks like you started at 0.5W and most recently have used 1W. This is still below the typical amount of heat dissipated by a power LED. It should be more like 2.7W.


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## IMSabbel (Feb 4, 2009)

But basically, his results are right:

A couple of mm thickness are more than enough. A fat block just means a few minutes of buffer, no sinking.
And yes, the contact with the housing is critical (this is where more surface might make sense).
The goal is a thermal grease film as thin as possible. To put it into perspective: Depending on the respective generations, CPUs in computers used pressures up to several Kg/cm^2 to get a good surface contact for heat transfer. 
If you just push a puck into a tube, and it is kinda stuck (with the grease,etc), the interface will cause an order of magnitude more thermal resitance than the puck material


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## Justin Case (Feb 4, 2009)

mudman cj said:


> You are considering the power dissipated by the driver, but the LED is dissipating more. See LukeA's post above for an estimate of the heat generated by a typical 3W LED at a typical level of overdrive (1A). And again, the driver is often times not connected to a heatsink at all.
> 
> One other comment I want to make is about the KL4 anecdote above. The KL4 is not a typical setup since it uses a 5W LED and the heatsink portion of the head is essentially thermally isolated from the body. See this post by Energie for the proof.



You are quite correct that I forgot about LED efficiency in converting power into light. Good catch. Lopez et al, Electro-Thermal Modelling of High Power Light Emitting Diodes Based on Experimental Device Characterisation, estimated 1.97W for a Luxeon K2 at 0.74A drive current. So, the estimate of 2.7W at 1A seems in the ballpark.

Be that as it may, my LED tower for SF THs still was barely warm. thus, ~2W-3W LED waste heat plus ~0.5W-1W driver waste heat was easily conducted away in that case.


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## mudman cj (Feb 4, 2009)

I agree with you guys. The heatsink is really just a way to connect the LED thermally to the body of the light. These connections are more important than the size of the heatsink or whether its made of silver, oxygen free copper, or aluminum. The size and shape of the heatsink does affect how large the heatsink/body interface is though. I should not have been nitpicking over the assumptions that were used if they didn't result in a fundamentally flawed conclusion. My wife loves that about me. :laughing:


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