# Copper vs. Aluminum Location



## jar3ds (Sep 4, 2009)

In the above diagram... If an LED is the heat source on this heatsink sandwich.... which one #1 or #2 would keep an LED (located at the heat source cooler)?

I seriously am a complete moron and I am not a physicist or an engineer.... So don't 'hate' on me if I am completely moronic... I just want to open this up for ideas... My only goal is to make a heatsink that is the best possible design for LEDs in a flashlight! :thumbsup:


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## darkzero (Sep 4, 2009)

I would think #1 would be best. Copper absorbs heat better & aluminum dissapates heat quicker. So I would think it's best to have copper closer to the heat source to absorb the heat quicker then have aluminum surround the copper to help dissapate the heat quicker. The aluminum would also be closer to the outside air (assuming as if used for a flashlight). :shrug::thinking:


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## jar3ds (Sep 4, 2009)

but why should the copper transfer its heat to the aluminum? It has better thermal conductivity... 

others have seem to express this happening from copper sinks.... that the LEDs are almost being bathed in their own heat... 

this is a quote from milkyspit in another thread:



> It has been reported that heatsinking is copper, and that is partially true... we did originally machine copper heatsinks for the lights... but performance tests actually suggested the copper wasn't helping to any significant degree with the heatsinking, and in fact might have been worsening it a bit (my theory here is the copper was getting saturated then holding more heat against the emitter slug, allowing the emitter to stew in its own heat, whereas aluminum was actually holding onto less and moving it more steadily out to the skin of the light)... most of the production lights will incorporate aluminum heatsinks, and believe me, folks, that's the one you probably want. If it matters at all, Leef and I have aluminum heatsinks in our personal lights and have no intention of changing.



https://www.candlepowerforums.com/posts/2095752&postcount=18


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## darkzero (Sep 4, 2009)

jar3ds said:


> but why should the copper transfer its heat to the aluminum? It has better thermal conductivity and it has a higher specific heat...
> 
> others have seem to express this happening from copper sinks.... that the LEDs are almost being bathed in their own heat...
> 
> ...


 
I'm not an engneer either. In a single die flashlight such as a Cree or Seoul P4 the use of a copper heatsink is probably not beneficial whatsoever (although the slug of a P4 & P7 are copper). I would think the use of copper is only beneficial when the size of the heatsink is limited in size and/or the heat source exceeds or comes close to the heat capacity of the aluminum.


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## jar3ds (Sep 4, 2009)

we know that since the LED is a continous heat source... its not as important to suck the heat away from the source but rather get rid of it, so it doesn't build up... .. (hence using cooling fins) etc...

I guess #2 was based off of that idea... that the copper backplate will help keep the majority of the heat AWAY from the LED BASEplate...

but as always... i don't know if my logic is sound....


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## jar3ds (Sep 4, 2009)

crap... i am stupid... copper has a lower specific heat than alumnium.... 

right???? i was incorrect above?


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## wquiles (Sep 4, 2009)

We should leverage decades of state of the art heatsink design coming from the PC industry were they are removing anything from 40-50 watts, up to about 150 watts from the CPU die:
- the absolutely best heatsinks have a copper core, or are solid copper
- if there is a hybrid copper/al heatsink, it is always like in image #1 with the copper facing the heat
- all of the cheap and least performing heatsinks are always pure aluminum

Now, in our handheld flashlights this is not as relevant since we have a limited of about 8-10 watts before things get too toasty for handheld operation (see recent LONG thread on this), but from a theory side of things, I would pick image #1 100% of the time over #2, and I recently did:






Will


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## wquiles (Sep 4, 2009)

jar3ds said:


> others have seem to express this happening from copper sinks.... that the LEDs are almost being bathed in their own heat...



Once the heat from the LED gets to the body of the light, if you don't remove the heat faster than the heat is being generated, the LED will certainly be bathed in its own heat since the heat does not has anywhere to go, and the temps will keep climbing until an equilibrium point is reached (usually much higher that one would like in a hand held light). 

This (the LED getting too hot) is not a problem with the heatsink material, although copper probably makes it more obvious since it transfers the heat to the body quicker and the body gets hotter quicker (which in terms of a heatsink is a good thing!).

Will


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## jar3ds (Sep 4, 2009)

wquiles said:


> ...
> Now, in our handheld flashlights this is not as relevant since we have a limited of about 8-10 watts before things get too toasty for handheld operation (see recent LONG thread on this), ...
> 
> Will



can someone link this thread?


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## mudman cj (Sep 4, 2009)

If you look up specific heat capacities of aluminum and copper in a reference, you will see that aluminum's heat capacity is 2.33 times higher than that of copper. It might be tempting to conclude that aluminum can absorb more heat than copper before rising in temperature, but you have to realize that heat capacity is defined as the energy required to raise the temperature of the metal one degree _per unit mass_. Since the density of aluminum is only 0.3 times that of copper, the amount of energy required to raise a _given volume_ of metal by one degree has to factor in the density. The energy required to raise a _given volume_ of copper by 1 degree is 42% higher than for aluminum. So, in the short run copper is better than aluminum at absorbing heat to bring the heat sink up to a given equilibrium temperature. This means that an LED mounted on an aluminum heatsink will dim more quickly than a comparable copper heatsink due to a rising die temperature.

The other factor here is thermal conductivity through the bulk of the metal. Here again, copper is better than aluminum, this time by a factor of about 2. This factor depends upon the temperature of interest as well as the purity of the materials. Another consideration is corrosion or oxidation of the surfaces, but these layers will typically be very thin and therefore not of much concern. 

In this case surface emissivity is not an issue because the heat is being transferred to the flashlight body in both instances, but if the material for the body of the light were in question, then the best material would be anodized aluminum. The reason is that the anodized surface has a high emissivity and therefore it is better able to lose heat to the air. This is the reason that aluminum is perceived to dissipate heat quicker in some circumstances.

The bottom line here is that copper is better in this application, but the actual difference in LED die temperatures in practice from using copper vs. aluminum are small. Since aluminum is cheaper, more common, and lighter it makes a good choice for most applications. The picture Will posted is a good example where copper is superior to aluminum. An even more effective version of the heatsink shown would use copper all the way out to the body of the light instead of switching over to aluminum, but again, the difference would be very small and not worth the increase in weight.


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## EngrPaul (Sep 4, 2009)

In your diagram, if both are insulated on the top and bottom, they are equal. If not insulated top and bottom, #1 is better.

Nobody would make a flashlight this way. Instead, it would be in a circle

(Al(Cu)Al)

vs

(Cu(Al)Cu)

Copper close to the emitter is better because despite it's small mass moves heat quickly. Once the copper gets to a larger surface area, the need for high thermal conductivity is less. Hence, the copper pill pressed into aluminum.

Doesn't make sense to make a copper flashlight. Copper is expensive, heavy, and soft (easy to damage).


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## wquiles (Sep 4, 2009)

jar3ds said:


> can someone link this thread?



Sorry my bad:
https://www.candlepowerforums.com/threads/240896


Will


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## GLOCK18 (Sep 4, 2009)

I have 2 light both have 3 Q5 emitter running the same driver off an AW C battery. Only difference one has a copper heat sink one has an aluminum heat sink the copper light weighs 480 grams and the aluminum heat sink light weighs 400 grams, I ran both light on high for 10 minutes the copper light measured 91 degrees and the aluminum light measured 98.8 degrees. You can really feel the difference when holding both lights the light with the copper heat sink feel substantially cooler.


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## The Dane (Sep 5, 2009)

The heatcapacity of:
Alu is 0.9 joule/gram
Cup is 0.39 joule/gram

The heattransfer of:
Alu is 237
Cup is 401

So in short:
Alu absorbs heat better/faster than copper but copper transfers heat better than alu.

So for the first seconds solution #2 is better but in the end they're the same!


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## mudman cj (Sep 5, 2009)

Your statement regarding the heat capacities of these materials is incorrect due to the higher density of copper. Refer to my earlier post for a more thorough explanation.


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## EngrPaul (Sep 5, 2009)

jar3ds said:


> . which one #1 or #2 would _*keep*_ an LED (located at the heat source cooler)?


 
Italics mine.

OP is looking for a steady state answer... only thermal conductivity matters.


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## gadget_lover (Sep 5, 2009)

If the aluminum light is 8 dgrees hotter, then that is the most effective at moving heat to the body of the light. That IS the objective. 


Pretty lights, BTW. 

Daniel


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## EngrPaul (Sep 5, 2009)

In order to make an assumption based on external temperature, one would have to assume the flashlight with an emitter that is cooler must more efficiently, thereby producing more photons and less heat.

Or that the copper distributes heat more evenly along the flashlight's body, thereby reducing the localized peak temperature.

This is because two identical bodies producing __ watts of heat must have the same average external temperature when placed into the same environment.


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## mudman cj (Sep 5, 2009)

gadget_lover said:


> If the aluminum light is 8 dgrees hotter, then that is the most effective at moving heat to the body of the light. That IS the objective.
> 
> 
> Pretty lights, BTW.
> ...



If the aluminum is more effective as you suggest, then how do you make sense of his statement that the copper heat sink feels much cooler?


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## 65535 (Sep 5, 2009)

The Dane said:


> The heatcapacity of:
> Alu is 0.9 joule/gram
> Cup is 0.39 joule/gram
> 
> ...



If those figures are right.

Al has a Hc of 2.43 joules/cm^3
Cu has a Hc of 3.4944 joules/cm^3

So copper should be the best heat sinking material all around.

Although I believe Silver is a better conductor and has a higher Hc than copper.

Being around $15 an ounce it isn't cheap, but for custom lights it could be beneficial.


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## gadget_lover (Sep 5, 2009)

mudman cj said:


> If the aluminum is more effective as you suggest, then how do you make sense of his statement that the copper heat sink feels much cooler?




If you think about the purpose of a heat sink, it is to move heat from a source to a place it can radiate away ( the sink) . In this case the destination is the flashlight body. From there it can radiate away (infrared) or be exchanged with the air.

Since both lights are externally the same, it's reasonable to assume that they get rid of heat equally well.

We are assuming that both of the heat sources (LEDs, Batteries and converters) are equal.

Given those assumptions, the more efficient heat sink is the one that raises the temperature of the body to a higher level, since is is conducting a higher level of heat to the body.

If the heat sink magically eats energy (making it disappear ) then the cooler light would be the best one. Nothing actually eats energy without transforming it into something. There's some silly law about that. 

Think of coffee in a steel coffee mug and a thermos mug made of steel. The plain one will be hot to touch outside but the coffee will quickly cool. The thermos one will be cool outside but the coffee will stay hot for hours.

Daniel


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## jar3ds (Sep 5, 2009)

so in regards to the 'copper plug' or 'copper pill' that is then inserted into the center of a piece of aluminum:







what is better?:

1) make the light out of a single piece of solid aluminum (no thermal barriers jump the heat over)

2) or use the copper pill surrounded by aluminum?

how much does a thermal gap or barrier create a negative effect?


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## mudman cj (Sep 5, 2009)

gadget_lover said:


> If you think about the purpose of a heat sink, it is to move heat from a source to a place it can radiate away ( the sink) . In this case the destination is the flashlight body. From there it can radiate away (infrared) or be exchanged with the air.
> 
> Since both lights are externally the same, it's reasonable to assume that they get rid of heat equally well.
> 
> ...



First of all, I must apologize for confusing the situation with a misinterpretation of GLOCK18's post. I thought I read that the copper heat sink was much cooler, but after re-reading it I see that he was saying the light with the copper heat sink was much cooler as the numbers show.

There is no reason why aluminum should perform better than copper in this situation unless either one or more of the metals used to make the heat sinks is not pure or annealed, there are differences in the surface roughness or gap sizes of the interfaces, or there are differences with respect to the use of thermal grease. There may also be differences in the amount of heat being generated in each case due to differences in drive current or Vf of the LEDs. I don't know what the explanation is, but I know that if we apply the Physics of heat transfer to this situation then pure copper must perform better than pure aluminum.

As for jar3ds's question - your intuition is spot on. The addition of the interface between the copper pill and surrounding aluminum heat sink will be detrimental. The larger you make the copper pill, then the greater will be the surface area of the interface with the aluminum heat sink, which will result in better heat transfer. Of course, you also want the two surfaces as smooth as possible and with the smallest possible gap, which should have thermal grease in it. 

In the end though, the differences in performance are probably not worth the extra work and cost. As EngrPaul has pointed out, once steady state is reached each of the lights must dissipate the same amount of heat to the environment. Add to that what Will (wquiles) mentioned about the limiting step being heat transfer to the environment in these higher power builds, and this whole debate of copper vs. aluminum can be seen as rather inconsequential.


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## gadget_lover (Sep 5, 2009)

I was curious, so looked at some of the properties of aluminum and copper.

We (as machinists) almost never work with pure metals. We use alloys because they work better and are more readily available.

Looking at http://www.engineersedge.com/properties_of_metals.htm , we see that pure aluminum is 30% better at conducting heat than 6061-T6.

Pure copper can be more than 1,500% better at conducting heat compared to some of the alloys.

I guess my point is that it's not sufficient to say that copper is better than aluminum, but rather to compare specific alloys (or the pure metal)


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## KowShak (Sep 5, 2009)

GLOCK18 said:


> I have 2 light both have 3 Q5 emitter running the same driver off an AW C battery. Only difference one has a copper heat sink one has an aluminum heat sink the copper light weighs 480 grams and the aluminum heat sink light weighs 400 grams, I ran both light on high for 10 minutes the copper light measured 91 degrees and the aluminum light measured 98.8 degrees. You can really feel the difference when holding both lights the light with the copper heat sink feel substantially cooler.


 
The simple view is that if the two lights need to get rid of the same amount of heat the bodies will get exactly as warm as each other. The LEDs and drivers in the better heatsinked light will run cooler if it has better heatsinking.

The more complex view is that differences in heatsinking in one light may mean that a small part of that first light is getting very hot with the rest of the light cold, where as the second light other may warm up the whole of the body of the light and use the whole body more effectively as a heatsink.

If the numbers you've quoted are the external body temperatures and both lights have reached their peak / operating temperature, they can't both be dissipating the same amount of energy as heat unless there is something different between the heatsinks other than the material.


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## 65535 (Sep 7, 2009)

For all practical purpose, make the LED mount as big as possible with as much cross sectional area contacting the outer wall as possible. The material really won't matter in this case as long as you stick with aluminum.


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## mudman cj (Sep 7, 2009)

65535 said:


> The material really won't matter in this case as long as you stick with aluminum.



 

That reminds me of junior high shop class when the teacher said, "You can paint your birdhouse any color you want as long as it's gray". 

But seriously, I assume you mean the body of the flashlight should be aluminum. Right?


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## Linger (Sep 7, 2009)

GLOCK18 said:


> I ran both light on high for 10 minutes the copper light measured 91 degrees and the aluminum light measured 98.8 degrees.



Fantastic - Excellent use of real data to supplement theory. However it seems another measurement is needed. Outside tempurature, and inside tempurature (at the base of the emitters, the 'start' of the heat-sink.
With only one number it is not possible to tell which light has been giving up more heat to the environment, so we can't tell if the hot light absorbed more from the emitter (is better) or retained more from the environment (is worse).

Please repeated the same test, measure temp at outside and temp at inside when finished.


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## gadget_lover (Sep 7, 2009)

> we can't tell if the hot light absorbed more from the emitter



Heatsinks absorb heat much as a paper towel absorbs water in a puddle. The same amount of water is there, but now you have a soggy paper towel too. When it is saturated it will absorb no more water. When you wring it out over a bucket you remove that water, but it's the work of moving the paper towel out of the puddle that removed the water.

Heat sinks are just a way to move heat from a hot spot to a cooler one. Without a way to exchange the heat with a cooler mass, it will eventually be the same temperature as the heat source.

Some materials conduct heat faster, some conduct it more efficiently ( or so I've been told). I guess that's why there's a 'copper VS aluminum' debate.

Daniel


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## wykeite (Sep 7, 2009)

mudman cj said:


> As for jar3ds's question - your intuition is spot on. The addition of the interface between the copper pill and surrounding aluminum heat sink will be detrimental. The larger you make the copper pill, then the greater will be the surface area of the interface with the aluminum heat sink, which will result in better heat transfer./QUOTE]
> 
> But the larger the copper pill then geometrically the larger the (relatively) insulating interface will be. The interface is the crippler for all the equations given here. Ideally an interference fit between faces is needed but is impractical. It's the weak link in the conductivity chain. The larger copper pill will give better short term performance than the smaller and then will deteriorate once saturated to the point that not enough heat can cross the interface.


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## EngrPaul (Sep 7, 2009)

wykeite said:


> But the larger the copper pill then geometrically the larger the (relatively) insulating interface will be. The interface is the crippler for all the equations given here. Ideally an interference fit between faces is needed but is impractical. It's the weak link in the conductivity chain. The larger copper pill will give better short term performance than the smaller and then will deteriorate once saturated to the point that not enough heat can cross the interface.


 
Nothing detrimental about increasing the surface area of the heat transfer interface. Check the equation... :wave:


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## Linger (Sep 7, 2009)

gadget_lover said:


> but it's the work of moving the paper towel out of the puddle that removed the water.


This is an unknown part. While running two identical lights is very helpful, with-out knowing how much heat has been dissipated to the environment during the run-time test leaves a large variable in the equation.

What we are looking for in this thread is what keeps the emitter cooler. Testing emitter temperature at the end of this comparison is an excellent way of getting to that answer.


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## wykeite (Sep 7, 2009)

EngrPaul said:


> Nothing detrimental about increasing the surface area of the heat transfer interface. Check the equation... :wave:


 
I stand by my statement. Heat transfer requires that the interface be able to conduct the heat. Hence the reference to fit. Air is a pretty good thermal insulator compared to metals.


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## mudman cj (Sep 7, 2009)

OK, if you don't want to check the equation then I will try to explain it in what is hopefully plain language. While it is very true that having no interface is better than an air gap, however small, there does exist a tipping point at which the trade-off is worth the gain due to the copper pill being more effective at 'spreading' the heat than aluminum. This tipping point is reached once the copper pill is large enough to have sufficient contact area with the aluminum around it so that there is a sufficient flow of heat through the interface. 

Let's think of it in terms of a flowing liquid. In this analogy, the heat flow is represented as fluid flow. Areas with a higher temperature are represented by a deeper liquid level. Since the heat has some resistance to flowing, let's use something like oil or syrup as the fluid. Imagine that the copper pill is a cone with a steep slop and the aluminum is a continuation of that, but with a less steep slope. Oil or syrup poured over the center (this is the LED location) will flow relatively quickly out to the aluminum and then slow down as the slope decreases. 

Now, let's add an interface between the copper and aluminum sections, which for this analogy will be a perforated wall. The wall is sufficiently high enough to prevent the liquid from ever going over the top of it. The quality of the interface, such as how well the surfaces fit, how finely polished they are, and the quality of the thermal grease between them, determines the size and number of the holes in the wall that let the liquid through. For the purpose of this analogy, just imagine that the holes are all the same size and with the same spacing between them regardless of the size of the copper pill section.

So, if we make the copper pill smaller, then the wall has to form a smaller ring and then there are fewer holes for the liquid to flow through, so it will flow outward at a slower rate. If we increase the size of the copper pill then the wall has much more area and therefore there are many more holes for the liquid to flow through to get to the aluminum. Once the liquid gets to the body of the flashlight, there will be another perforated wall that represents the last boundary for the heat to escape to the air. The buildup of heat at this interface is analogous to a pool of liquid whose depth is dependent upon how quickly liquid is being added at the center (the LED), how much area of wall there is, and of course how large the holes are and the spacing between them. For small lights with high power, this last wall can result in the liquid backing up all the way to the center at the top of the hill (the LED). I hope this helps.


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## kwkarth (Sep 7, 2009)

mudman cj said:


> OK, if you don't want to check the equation then I will try to explain it in what is hopefully plain language. While it is very true that having no interface is better than an air gap, however small, there does exist a tipping point at which the trade-off is worth the gain due to the copper pill being more effective at 'spreading' the heat than aluminum. This tipping point is reached once the copper pill is large enough to have sufficient contact area with the aluminum around it so that there is a sufficient flow of heat through the interface.
> 
> Let's think of it in terms of a flowing liquid. In this analogy, the heat flow is represented as fluid flow. Areas with a higher temperature are represented by a deeper liquid level. Since the heat has some resistance to flowing, let's use something like oil or syrup as the fluid. Imagine that the copper pill is a cone with a steep slop and the aluminum is a continuation of that, but with a less steep slope. Oil or syrup poured over the center (this is the LED location) will flow relatively quickly out to the aluminum and then slow down as the slope decreases.
> 
> ...



Basically, you're saying that we have to concern ourselves with matching thermal impedances in order to have effective heat transfer from one medium to another. Correct?


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## mudman cj (Sep 8, 2009)

kwkarth said:


> Basically, you're saying that we have to concern ourselves with matching thermal impedances in order to have effective heat transfer from one medium to another. Correct?



If you mean that adjoining materials need to have matching values for thermal conductivity, then no. The implication here is that the interface hurts thermal conductivity, but the use of a copper pill can make up for that loss and then some provided that there is enough contact area between the copper pill and the aluminum. What size it needs to be in order to realize an advantage depends upon the quality of the thermal connection.


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## kwkarth (Sep 8, 2009)

mudman cj said:


> If you mean that adjoining materials need to have matching values for thermal conductivity, then no. The implication here is that the interface hurts thermal conductivity, but the use of a copper pill can make up for that loss and then some provided that there is enough contact area between the copper pill and the aluminum. What size it needs to be in order to realize an advantage depends upon the quality of the thermal connection.



Nope, not what I was thinking. What I had in mind is that the copper slug's thermal mass is greater per unit volume than aluminum, so for an effective interface between the copper and the aluminum, of course the interface must be without any thermal barriers, but beyond that, there must be enough surface area of the copper in contact with the aluminum for the aluminum to be able to receive heat at the rate that it will need to be conducted away from the copper. Does that make sense? It's just like the interface between the aluminum and air. The cooling fins on the flashlight's surface are there to present enough surface area of the aluminum to the air to facilitate removing heat at a rate that meets the need. 

If there were enough surface area of aluminum to air interface, the heat would be removed as fast as it is generated, as a result, the flashlight remains at room temperature, or close to it. Same concept with the copper to aluminum interface. Assuming you have a thermally non lossy connection aluminum to copper, there has to be enough surface area of copper-aluminum contact to move the amount heat the copper is capable of conducting from the pill to the aluminum in order for the copper to be beneficial in the first place. Make sense? I don't know this by training, but it just makes intuitive sense. I'm asking/musing.


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## mudman cj (Sep 8, 2009)

kwkarth said:


> Nope, not what I was thinking. What I had in mind is that the copper slug's thermal mass is greater per unit volume than aluminum, so for an effective interface between the copper and the aluminum, of course the interface must be without any thermal barriers, but beyond that, there must be enough surface area of the copper in contact with the aluminum for the aluminum to be able to receive heat at the rate that it will need to be conducted away from the copper. Does that make sense? It's just like the interface between the aluminum and air. The cooling fins on the flashlight's surface are there to present enough surface area of the aluminum to the air to facilitate removing heat at a rate that meets the need.
> 
> If there were enough surface area of aluminum to air interface, the heat would be removed as fast as it is generated, as a result, the flashlight remains at room temperature, or close to it. Same concept with the copper to aluminum interface. Assuming you have a thermally non lossy connection aluminum to copper, there has to be enough surface area of copper-aluminum contact to move the amount heat the copper is capable of conducting from the pill to the aluminum in order for the copper to be beneficial in the first place. Make sense? I don't know this by training, but it just makes intuitive sense. I'm asking/musing.



Yes, that makes sense now, but i would like to clarify something. What is important here in the steady state condition, ie. when the light has been operating a while and the temperatures have stabilized, is not the thermal mass of the materials but the thermal conductivity of the materials. This point was made earlier by EngrPaul. But your point is essentially correct in that the aluminum needs to have a certain minimum area of contact with the copper in order for the aluminum to be able to receive heat at the rate that it is being produced.


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## EngrPaul (Sep 8, 2009)

mudman cj said:


> the aluminum needs to have a certain minimum area of contact with the copper in order for the aluminum to be able to receive heat at the rate that it is being produced.


 
Actually, at steady state, the aluminum will receive a steady rate of heat regardless of how good the interface is. The difference is that the temperature of the copper will be higher when the connection is less effective. The emitter will be correspondingly hotter.

Another possible benefit of copper is drawing heat down deeper into the flashlight, if it is a long cylinder instead of a disc. This moves the heat down into the flashlight body, increasing the effectiveness of the flashlight body at dissipating heat. In this case, the flashlight body's peak temperature will indeed be less, because the temperature will be spread out more on the surface.


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## kwkarth (Sep 8, 2009)

EngrPaul said:


> Actually, at steady state, the aluminum will receive a steady rate of heat regardless of how good the interface is. The difference is that the temperature of the copper will be higher when the connection is less effective. The emitter will be correspondingly hotter.
> 
> Another possible benefit of copper is drawing heat down deeper into the flashlight, if it is a long cylinder instead of a disc. This moves the heat down into the flashlight body, increasing the effectiveness of the flashlight body at dissipating heat. In this case, the flashlight body's peak temperature will indeed be less, because the temperature will be spread out more on the surface.



This makes sense. Except for the part that you seem to be saying aluminum receiving same amount of heat regardless of the quality/thermal conductivity of the interface. I'm probably interpreting what you said, but that's the way it strikes me.
Thx.


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## mudman cj (Sep 8, 2009)

EngrPaul said:


> Actually, at steady state, the aluminum will receive a steady rate of heat regardless of how good the interface is. The difference is that the temperature of the copper will be higher when the connection is less effective. The emitter will be correspondingly hotter.



Yes, but I don't think that what I said is in conflict with this. Note that I said in order for the rates to be the same. What I meant is the rate heat arrives at the interface can be very nearly equal to the rate of heat leaving the interface if there is enough contact area. If there is not enough contact area, then the rates are not equal and the result is that the copper pill develops a higher temperature, just as you said.


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## kwkarth (Sep 8, 2009)

mudman cj said:


> Yes, but I don't think that what I said is in conflict with this. Note that I said in order for the rates to be the same. What I meant is the rate heat arrives at the interface can be very nearly equal to the rate of heat leaving the interface if there is enough contact area. If there is not enough contact area, then the rates are not equal and the result is that the copper pill develops a higher temperature, just as you said.



This is what I thought I understood as well.


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## tino_ale (Sep 10, 2009)

All this makes sense to me.
I think the lesson to learn here is that thermal design is not obvious and simply throwing copper in a flashlight doesn't necessarily improve the LED thermal relief.
Many on CPF seem to think that it is that simple. Copper is always seen as a superior material than aluminium and I think it is sometimes a plain lack of insight.

One thing also I have noticed is how "pills" are implemented in flashlight heads. They are often inserted in a bored hole in the head, which IMO doesn't provide a very good thermal path, since there is basically no contact pressure between the pill and the host. Sure enough thermal compound will help but again with this implementation it is difficult to make sure the compound is filling the cavities, since inserting the pill in it's bored hole will scrap the compound away from the interface.

For this reason I have never really liked cylindrical heatsinks that are inserted in Mags tubes.


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## AnAppleSnail (Sep 10, 2009)

Heat capacity is often given as a function of mass. For a given volume, copper will be heavier and have a higher specific heat (specific == per mass).
Copper: .385 J/g*K
Aluminum: .9 J/g*K

Density makes the difference, though. For a given _volume_, Copper will be a better conductor.
Copper: 8.96 gm/cm^3
Aluminum: 2.7 g/cm^3

This means that, per cm^3, these are the heat capacities:
Copper = (0.385 J/g*K) * (8.96 g/cm^3) = (3.4496 J/cm^3*K)
Aluminum = (.9 J/g*K) * (2.7 g/mc^3) = (2.43 J/cm^3*K)

In short bursts, a copper heatsink is going to stay cold longer. Copper is more thermally conductive as well - a square meter of copper will carry 400 watts of heat over a 1C temperature difference, while aluminum will get around 235 watts. The math says copper is better for the thermal bits of a heatsink. This requires that you have copper going from the emitter to the outside of the light, as material junctions affect conductivity quite a lot.

Copper heats up a bit slower, and at steady state its surface will radiate away about 2x as much heat as aluminum of equal cross-section. It weighs around 3x as much though, and I can't easily account for aluminum oxide coating aluminum, nor green rust on copper.


It makes an interesting math problem to calculate the steady-state temperature of a heatsink. All you need are the input heat and heat dissipation of the heatsink - which for a simple metal heatsink is mostly a property of surface area. A larger heatsink or one with greater dissipation will have a lower steady-state temperature for a given input heat.


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## 65535 (Sep 10, 2009)

I suppose it boils down to what is most convenient. It's not a whole lot of heat, and chances are the body be it aluminum or other will be less conductive than the point where the LED is attached.

I figure anodized aluminum would be a simple convenient choice, no worry about die-body shorts.


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## jar3ds (Sep 10, 2009)

AnAppleSnail said:


> Heat capacity is often given as a function of mass. For a given volume, copper will be heavier and have a higher specific heat (specific == per mass).
> Copper: .385 J/g*K
> Aluminum: .9 J/g*K
> 
> ...



wow... this thread is really getting some good info... i was forgetting that while aluminum does have a higher specific heat, that copper is more dense and therefore actually has a larger heat capacity... 

i think one thing that is important to keep in mind (which i believe was brought up earlier in this thread)... is to keep in mind that we work with aluminum alloy's not pure aluminum... so that means that if you do have a very pure copper heatsink w/ a 6061 T6 body flashlight... the thermal conductivity gap is even larger!:

Aluminum 6061 T6 = 167 W/m-K
Pure Copper = 401 W/m-K

... thanks guys for all of your input!


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## Ekke (Sep 12, 2009)

jar3ds said:


> the thermal conductivity gap is even larger!:
> 
> Aluminum 6061 T6 = 167 W/m-K
> Pure Copper = 401 W/m-K



And Aluminium oxide 40 W/m-K. Do you think these very thin anodizing layers (<0.1mm, even hard anodizing?) affect cooling?


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## AnAppleSnail (Sep 14, 2009)

Ekke said:


> And Aluminium oxide 40 W/m-K. Do you think these very thin anodizing layers (<0.1mm, even hard anodizing?) affect cooling?



It's almost guaranteed to, but I'm having quite a bit of trouble finding thermal properties for those. In general I don't think you'll encounter limits by the metal's conductivity or heat capacity. That is, like any system you aren't going to reach as far as the best component, but instead you are held back by the weakest step. If the LED is properly heatsinked (Metal all over the most stuff closest to the emitter without shorting out), then the big deal is getting the heat out of the flashlight body. I wonder if they'll make syringes so that the light can more effectively use your body for active cooling? Eew... the  looks strangely appropriate here!

A simple calorimeter can be made with two nested styrofoam cups, one larger than the other. With a known volume of water at a given temperature in the inner cup, and a waterproof light (at that same temperature) sitting in the cup, you can measure the heat increase of the water (and approximate the heat transfer of the flashlight body), and compare that to total power input. It's nearly the same setup used to determine heat of reaction, although running the flashlight in the water makes it impossible to measure the heat emissivity of HAIII or other coatings, which I think was pretty much the point. The same principle would work with air and a thermometer, but would quickly harm the flashlight. A remote switch, activation for a given time, and measure the power put into the light and the heat change in a specific time might work. Any thermodynamics people here who want to take a crack at making a legitimate test? I'm unqualified


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