Emissivity of tin-plated vs. nickel-plated copper for heatsink application

[T3sl4co1l]

Yeah, useless without temperature.

Y'know, so as not to go too long with utter hearsay -- how about some physics:



Bastard unit chosen, just because I for some reason memorized the reference point of 150 C in^2/W for typical convection in still air.

Which crosses at merely 136°C, so radiation doesn't seem too bad after all.

But that assumes emissivity = 1, and that doesn't take into account the synergy of heatsink fins and convection.  Probably a heatsink at the same temperature is doing a lot better than 150 (how much better, you'll have to find some estimate of the active area and see, I guess; I don't have any measurements handy).

Note that fins don't participate radiatively, at least beyond whatever help they provide in creating an effective black body cavity (which can only increase emissivity to a maximum of 1).  Consider the root between fins: the surface there is mostly surrounded by other surfaces, and very little of it sees free space, so contributes little to the total.  The effective radiating surface is more or less the convex hull around the object.  So, there's much less radiation area for a finned heatsink, than its actual (physical) surface.  But for sheer surfaces, flat plates, enclosures, etc., the areas are pretty similar.

Note also that space is radiating back upon the surface, so the ambient must be known.  This has been assumed at 25°C.  Rth goes to infinity as delta T goes to zero, and that asymptote is mostly cut off in these data (starting at 50°C).

Note also that the effective resistance goes down considerably as temperature goes up.  That is, given say a 25°C temp difference, the heat flux will be sixteen times higher by merely doubling the absolute temperature (say 600K vs. 300K).  Not that you gain anything from raising ambient, but it's interesting that the conductivity effectively goes down so sharply even with the high ambient.  Whereas a linear material will always drop say 25°C when 1W is applied (for Rth = 25 °C/W), the radiative case only improves with rising ambient (when measured strictly in terms of temp rise).


So: the contribution for plain, convex objects can be fairly high, though also such objects tend not to be very hot (e.g. enclosures), so it may still not be a dominant fraction.  Certainly not for bare metal.  But painted or plastic, pretty good, maybe on the order of 50%.  (If my reference figure is representative at this temperature, then actually more than 50%, since the 150 C in^2/W figure is effectively the 227... resistance from the table, in parallel with an even higher resistance from convection alone.)

Meanwhile, the contribution for concave shapes like heatsinks, may be significantly less.  If the active-convection-surface-area to convex-hull-surface-area ratio is say 5 or 10, and the heatsink is bare metal, radiation might be merely a few percent.  With good emission, 10 or 20% is very believable.  Shallow fins or poor orientation (effectively reducing the active area by creating pockets of still air), even higher.

Tim

[jonpaul]

Tim excellent, mille mercis,

Wonder if a study or paper exists re the ratio conduction/convection/radiation for typical electronic HS at usual ambient and HS temps?

Jon

[coppercone2]

I wonder if ferric chloride staining improves it as much as black anodization, because you can stain copper or aluminum black IIRC for much less effort

[bdunham7]

Y'know, so as not to go too long with utter hearsay -- how about some physics:

Bastard unit chosen, just because I for some reason memorized the reference point of 150 C in^2/W for typical convection in still air.

Which crosses at merely 136°C, so radiation doesn't seem too bad after all.

Physics is great, thanks for constructing and posting all that.  It looks like radiation is more than a rounding error in at least some circumstances.  However, I'm not so quick to dismiss the 'utter hearsay' because a lot of that is based on practical experience.  There are assumptions, rules-of-thumb and so on that come into play and can change the result quite a bit.  Some experimental data would be helpful.

The 150K/in²/W, for example, seems a very rough rule of thumb.  I found this site--I haven't checked their math, but it seems well researched:

https://www.heatsinkcalculator.com/free-resources/flat-plate-heat-sink-calculator.html

I found that I could rearrange a square inch of flat plate and get very different results just by changing the ratio of length to width--which one might expect from a convective model.

Also, IEC 60950 sets limits as to exterior temperatures and I think most products would not exceed these.  Thus for an external case, the temperatures would need to be on the far left side of your graph.  136C would be very hot for any heat sink, enclosed or not.



So the situation changes quite rapidly as the temperatures go up--a cheap, hot plastic power brick underneath a desk in completely still air might indeed lose a significant portion of its heat by radiation.  I would expect a large, external finned heat sink on a power supply or audio amplifier to rely much more on convection.  If I can dig up some aluminum heat sinks that come in both plain and black anodized, perhaps an experiment is in order.

The heat sink calculator is a pay site, but they do also have this heat sink size calculator.  You can pick some arbitrary sizes and then use different temperature and emissivity settings to see how much the calculated heat sink size varies when change the emissivity from 0.1 to 0.9, leaving all the other variables alone.  I'm not sure it is accurate--it may not be taking into account the reduction in emissivity due to geometry. 

https://www.heatsinkcalculator.com/heat-sink-size-calculator.html

[TimNJ]

First of all...thank you everyone for all the interesting discussion. I must admit, I am still a bit murky about my original question. 

There is a noticeable amount of cooling performance to be gained from a high emissivity coating, so unless the product is being seriously cost optimized for large scale mass production it does not make sense to save a few cents with raw aluminum. In fact the OP is thinking of copper for heatsinking, this makes for a rather significant price increase for the extra performance, so it does not make sense to shave cents here by skipping a high emissivity coating.

Before I started at my present job, my ex-boss ran some experiments on one of our open-frame power supplies which ran pretty hot, given the power density and the year it was designed. He ran some experiments replacing the original brass heatsinks (why brass, I have no idea) to plain aluminum, and then to black anodized aluminum. Apparently, the black anodized aluminum outperformed the plain aluminum in a significant way...on the order of 5C reduction in semiconductor case temperature just by anodization.  Of course, this was for a vented open-frame design, and not enclosed plastic adapter, so it's a different animal in some respects.

But, yes, you understand my original intention for starting this thread. Basically, forgetting about "whether it's worth it or not", what would be the theoretical best coating/surface finish for a copper heatsink to maximize radiation performance?

[bdunham7]

Apparently, the black anodized aluminum outperformed the plain aluminum in a significant way...on the order of 5C reduction in semiconductor case temperature just by anodization.

But, yes, you understand my original intention for starting this thread. Basically, forgetting about "whether it's worth it or not", what would be the theoretical best coating/surface finish for a copper heatsink to maximize radiation performance?

Yay!  A data point!  I was looking to see if I could find any heat sinks that came in both varieties but were identical otherwise, I only found one and the plain one doesn't seem to be available. 

To answer your question, the answer of course is the coating with the best emissivity without any insulating effect.  I suppose if it is thin enough, the insulating effect will be minimal.  There's a paint called Parson's Optical or something like that that is supposedly very high emissivity, barbecue grill paint probably isn't far off the mark.

EDIT: There is also a process called Ebonol C which chemically produces a layer of cupric oxide on the surface.  I believe this is fairly exotic, but would be similar to anodizing in that the insulation effect would be minimal.

[TimNJ]

Yeah, useless without temperature.

Y'know, so as not to go too long with utter hearsay -- how about some physics:



Bastard unit chosen, just because I for some reason memorized the reference point of 150 C in^2/W for typical convection in still air.

Which crosses at merely 136°C, so radiation doesn't seem too bad after all.

But that assumes emissivity = 1, and that doesn't take into account the synergy of heatsink fins and convection.  Probably a heatsink at the same temperature is doing a lot better than 150 (how much better, you'll have to find some estimate of the active area and see, I guess; I don't have any measurements handy).

Note that fins don't participate radiatively, at least beyond whatever help they provide in creating an effective black body cavity (which can only increase emissivity to a maximum of 1).  Consider the root between fins: the surface there is mostly surrounded by other surfaces, and very little of it sees free space, so contributes little to the total.  The effective radiating surface is more or less the convex hull around the object.  So, there's much less radiation area for a finned heatsink, than its actual (physical) surface.  But for sheer surfaces, flat plates, enclosures, etc., the areas are pretty similar.

Note also that space is radiating back upon the surface, so the ambient must be known.  This has been assumed at 25°C.  Rth goes to infinity as delta T goes to zero, and that asymptote is mostly cut off in these data (starting at 50°C).

Note also that the effective resistance goes down considerably as temperature goes up.  That is, given say a 25°C temp difference, the heat flux will be sixteen times higher by merely doubling the absolute temperature (say 600K vs. 300K).  Not that you gain anything from raising ambient, but it's interesting that the conductivity effectively goes down so sharply even with the high ambient.  Whereas a linear material will always drop say 25°C when 1W is applied (for Rth = 25 °C/W), the radiative case only improves with rising ambient (when measured strictly in terms of temp rise).


So: the contribution for plain, convex objects can be fairly high, though also such objects tend not to be very hot (e.g. enclosures), so it may still not be a dominant fraction.  Certainly not for bare metal.  But painted or plastic, pretty good, maybe on the order of 50%.  (If my reference figure is representative at this temperature, then actually more than 50%, since the 150 C in^2/W figure is effectively the 227... resistance from the table, in parallel with an even higher resistance from convection alone.)

Meanwhile, the contribution for concave shapes like heatsinks, may be significantly less.  If the active-convection-surface-area to convex-hull-surface-area ratio is say 5 or 10, and the heatsink is bare metal, radiation might be merely a few percent.  With good emission, 10 or 20% is very believable.  Shallow fins or poor orientation (effectively reducing the active area by creating pockets of still air), even higher.

Tim

I appreciate the calculations and visuals. Thanks. I understand you are showing the drop in thermal resistance, presumably from heatsink to ambient, for radiation. You noted that emissivity = 1, so you are assuming the heatsink material is a perfect black body radiator? Are the calculations you show independent of actual heatsink size, dimensions, and other factors?

Can you explain what a concave and convex object is, in the context of this discussion? Obviously I am familiar with the terms in general. By concave, do you mean a radiating object which radiates back on itself due to its shape? Whereas, for an enclosure, the radiating surfaces (i.e hot surfaces facing "cool" ambient) are all pointing away from each other?

Finally, the numbers you suggest 50% for convex objects, and 10-20% for concave objects...are you saying this is the fraction of the total energy released from the heatsink? (i.e. radiation + convection)

Thanks!

[TimNJ]

Apparently, the black anodized aluminum outperformed the plain aluminum in a significant way...on the order of 5C reduction in semiconductor case temperature just by anodization.

But, yes, you understand my original intention for starting this thread. Basically, forgetting about "whether it's worth it or not", what would be the theoretical best coating/surface finish for a copper heatsink to maximize radiation performance?

Yay!  A data point!  I was looking to see if I could find any heat sinks that came in both varieties but were identical otherwise, I only found one and the plain one doesn't seem to be available. 

To answer your question, the answer of course is the coating with the best emissivity without any insulating effect.  I suppose if it is thin enough, the insulating effect will be minimal.  There's a paint called Parson's Optical or something like that that is supposedly very high emissivity, barbecue grill paint probably isn't far off the mark.

I'd honestly like to run some sort of experiment like this myself, with better documented test setup...a few different scenarios representing typical use cases (i.e. the one's we've talked about, enclosed, open-air, etc.). Not that I don't believe my ex-boss's results but sometimes he had a tendency to want to prove himself right 

Regarding the original question, the stipulation was that it had to be a solderable material, as most likely the whole thing would be plated/coated with it. I could not find clear data on tin vs. nickel, but suspect they are probably comparable. It seems that a dull vs bright plating finish is more important. But there are lots of different plating processes which I am just not familiar with, so there may be an ideal process (i.e. I saw "electroless black nickel" plating as possible option.)

[T3sl4co1l]

Apparently, the black anodized aluminum outperformed the plain aluminum in a significant way...on the order of 5C reduction in semiconductor case temperature just by anodization.

But, yes, you understand my original intention for starting this thread. Basically, forgetting about "whether it's worth it or not", what would be the theoretical best coating/surface finish for a copper heatsink to maximize radiation performance?

Yay!  A data point!

Ahhhh!  A half a data point!

Critial information is: out of what HS and ambient temperatures?

I appreciate the calculations and visuals. Thanks. I understand you are showing the drop in thermal resistance, presumably from heatsink to ambient, for radiation. You noted that emissivity = 1, so you are assuming the heatsink material is a perfect black body radiator? Are the calculations you show independent of actual heatsink size, dimensions, and other factors?

Right.  The way radiation works is, everything is radiating, all the time.  Maybe not 100% in all frequency bands, but always something, somewhere.  An ideal dielectric is the only thing that doesn't (there are ideal conductors but not at AC, so superconductors also radiate; granted, it's not much at their low operating temperatures).  And vacuum is really the only thing of that sort.  Which means the temperature of a vacuum is defined by the temperature of its bounding surfaces, if that's any help.  (The universe itself is bounded by the cosmic horizon, which comes to us as cosmic background radiation ~2.7K.)

Radiation is a power density, so we can compare types of heat transfer in these terms, to the extent they are applicable to the other types anyway.  It makes intuitive sense that bigger things can dissipate more heat.  How much more, is the question; convection is very sensitive to geometry and orientation.

Conductivity, is linear or very nearly so, so we can approach it in the same way.  It does depend on thickness being conducted through, but if we define that -- say looking at a particular thermal pad -- we find an area thermal resistance (r_th = thickness / bulk conductivity).

Can you explain what a concave and convex object is, in the context of this discussion? Obviously I am familiar with the terms in general. By concave, do you mean a radiating object which radiates back on itself due to its shape? Whereas, for an enclosure, the radiating surfaces (i.e hot surfaces facing "cool" ambient) are all pointing away from each other?

Basically this,
https://en.wikipedia.org/wiki/Convex_hull

I'm not going to try and explain the whole thing in detail (as I don't understand radiation rigorously enough myself to be able to prove something like that..), but that's the sort of idea.

More specifically I think, if you look at any point on a surface, take however much solid angle of space is visible.  Radiation is Lambertian, so integrate solid angle over the object.  This should be better than the convex hull, as long spindly projections have excellent visibility, like a, think of whatever the star-shaped polyhedra are.  Or, consider the point cloud in the article itself: if we suppose a small sphere around every point, the visibility of any given sphere (as in, solid angle not obstructed by other spheres) is very good, even in the middle of the cloud (though certainly worse there, and depending on density).  We should expect such a configuration to radiate very effectively.

Though I'm also not sure if such shapes radiate as well as their convex hull, which takes up the same space (if the bounded volume is what's important in your application..) but has the most unobstructed surface area in that shape(?).


In any case, in terms of conventional shapes, heatsink fins, don't count towards radiation, at least very much.  Broad surfaces or overall dimensions, yes.  Pointy shapes, not so sure, but you rarely see, like, a pin array type where the pins are very far from the surface (and from each other), so it's at least not a question of high commercial interest.

Finally, the numbers you suggest 50% for convex objects, and 10-20% for concave objects...are you saying this is the fraction of the total energy released from the heatsink? (i.e. radiation + convection)

Yes, breakdown between radiation and convection.

Again, as mentioned, the comparison depends wholly on a reasonable figure for convection; the 150 figure I pulled out is probably from the low end of the scale, and obviously not the high end where radiation would handily account for that alone.

Tim