I want copper copper copper! (well... maybe not)

Wolfman29 said:

Basically, what I am saying is that, because copper is sufficient to absorb the heat from a heatsink (it certainly if if SS is - it's heat capacity is only slightly lower), then what we want is the highest conductivity to get the heat away from the diode quickly. Of course it will fit up faster, but that's not our concern. We're okay with a filled heatsink, as long as its the diode that fills up last (i.e. there will be a traffic jam near the door BEFORE a traffic jam near the room/hallway entrance). However, if more people are trying to go into the steel hallway than they can exit (i.e. if its not wide enough to hold everyone), then we will have a traffic jam near the fire (diode) and the heatsink won't be doing its job.

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Yes! AND on top of that, even though radiant heating to the air isn't necessarily the most important aspect of our sinking - with copper's very high thermal conductivity it actually gets those "people" to the exit door where they CAN leave the building.

tsteele93 said:

Yes! AND on top of that, even though radiant heating to the air isn't necessarily the most important aspect of our sinking - with copper's very high thermal conductivity it actually gets those "people" to the exit door where they CAN leave the building.

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If your host is steel (which a lot are), there's not benefit to using a metal that can get that heat TO the steel host any faster than STEEL can.

The point isn't getting it to the steel host - it's getting it away from the diode. If the diode generates 5J of heat (5 seconds at 1W, let's say), and your material has perfect heat conductivity (for the purposes of example), then your diode will only be subject to the equilibrium temperature of the heatsink (where the 5J are equally distributed throughout the heatsink). However, with low thermal conductivity, you may have twice as much energy stored in heat near the diode as you would further away from the diode, so the diode would be exposed to higher temperature.

Wolfman29 said:

The point isn't getting it to the steel host - it's getting it away from the diode. If the diode generates 5J of heat (5 seconds at 1W, let's say), and your material has perfect heat conductivity (for the purposes of example), then your diode will only be subject to the equilibrium temperature of the heatsink (where the 5J are equally distributed throughout the heatsink). However, with low thermal conductivity, you may have twice as much energy stored in heat near the diode as you would further away from the diode, so the diode would be exposed to higher temperature.

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Yes, but what do you think happens if that heatsink's specific heat capacity is lower?

Well certainly it will rise in temperature with less heat. However, copper isn't that much lower. That's the key here. Copper has a pretty damn high heat capacity, AND it is the second best thermal conductor.

I'm going to break etiquette and double or triple post here. I think it makes sense to break up the various forks in conversation into different posts. If someone - besides T_J - who needs to hush when grown folks is talkin! :crackup: - has a problem with it then I can come back and combine posts.

Doesn't matter now, several posts have happened since I started writing this...

Anyway, in this post I wanted to look closer at Steel as a sink. I found this page very interesting, especially chart 3.


Thermal by tsteele93, on Flickr

Structural Material Behavior in Fire: Steel: Stainless Steel Thermal Properties

Carbon steel might actually make a decent heat sink at the temps we are working with vs stainless which is pretty low on thermal conductivity.

I bet it is hard to machine though.

There are a lot of different grades and variations of steel... carbon, high carbon, low carbon, hot rolled, cold rolled, annealed, tempered, work hardened, CrMo, and there are a lot of grades and variations of stainless steel... 440, 304, 309, 316, 420, 409... so you can't just say "carbon steel vs. stainless steel" because there is a tremendous amount of variation in each steel.

Both "sides" of the debate have merit and are right about some things. I suspect what needs to happen is for someone to take a non-finned small-ish host and make two heatsinks for it: one copper and one steel. Then determine which one will yield a longer run time based on diode temperature.

By the way, where are many of you getting the information that aluminium dissipates heat faster than copper or steel or [anything else]? I can't find anything other than internet rumor that suggests this in any metalurgical references, and the article linked earlier even said that it is just a myth.

How would we measure diode temperature? I suppose we could just use a thermal probe?

Yeah or carefully thermal epoxy a thermistor to the underside near the unused pin. We wouldn't even need to convert the resistance to a temperature, just see which way the resistance changes the least from being off/cold.

Sigurthr said:

By the way, where are many of you getting the information that aluminium dissipates heat faster than copper or steel or [anything else]? I can't find anything other than internet rumor that suggests this in any metalurgical references, and the article linked earlier even said that it is just a myth.

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I agree with you on this, there is a LOT of belief that this is true but I don't see much actual data that shows it to be true.

Found this here...

Does aluminum dissipate heat better than copper? - Overclockers Forums

"OK, specific heat (Cp) is the measure of how much energy it takes to raise the temperature of 1kg of material 1degree (C or K). The Cp for copper is 385 Joules/kg*K and for aluminum is 903 J/kg*K (at 300K = 23 deg. C). density for both is copper: 8933 kg/m^3 and alum: 2702 kg/m^3

So, if we have two same SIZED blocks of material, say 1m x 1m x 1m for simplicity, we have different masses, namely 2700kg of alum. and 8930 kg Cu. if we put 2.438 x 10^6 Joules of energy into each block, we can calculate the temp raise:

Q = m*Cp*deltaT --> deltT = Q/(m*Cp)

deltaT alum: = 2.4381 x 10^6 J/(2700 kg * 903 J/kg*K) =1 degree

deltaT Cu: = 2.4381 x 10^6 J /(8933 kg * 385 J/kg*K) = 0.71 degree

so for the same SIZE blocks, more heat is energy is required to heat the copper by the same delta T than the aluminum. In other words from the moment you turn your computer on it will take a longer time for a copper heatsink to reach steady state conditions.

Now since these blocks have the same size they have the same surface area (that's what we're comparing, right - same size/shape/etc... heatsinks)

Now, that is really transient heat transfer, where there is a buildup of heat in one of the components of the system (the heatsink). At steady state, BOTH Al. and Cu. heatsinks will stabilize at a certain temperature and the heat per unit time (J/s = Q_dot) will be exactly equal to the heat leaving the heatsink per unit time. It will just take a longer time for the Cu one to reach this point.

Now we're talking steady state:
2 processes going on:
(1) is conduction of heat through the heatsink
(2) is convection of heat from surface of heatsink to air

for conduction: Q_dot = (kAc/L)*(Tdie - Tsurface) <-- equation (1)here Q_dot is joules/second or Watts, k is the thermal conductivity of the material in Watts/meter*K, Ac is cross sectional area through the heatsink in meters^2, L is the lenth between Tdie and Tsurface in meters, Tdie is temperature at the die or temperature at the bottom of the heatsink (here we'll consider them equal) and Tsurface is the temp. at the surface of the heatsink, where the air contacts it. (temps in degrees C or K)

for convection: Q_dot = hA(Tsurface - Tair) <-- equation (2)
where A is the area of the surface of the heatsink (giving heat to the air) and h is the convection coefficient and has units Watts/meters^2*K.

from equation (1): Tdie = Q-dot * L/(k*Ac) + Tsurface
from equation (2): Tsurface = Q_dot/(h*A) + Tair

combining: Tdie = Q_dot*L/(k*Ac) + Q_dot/(h*A) + Tair

Taking the Tair, Across section of the heatsinks, A surface of the heatsinks, L length of the heat must travel between the die and surface, and Q_dot to be constant and the same in both cases (Ac, A, L wil be same because we are comparing the same heatsinks just of different materials, but the physical dimensions are the same, Q_dot is constant because the cpu puts out a fixed amount (ideally) say xxxWatts under load, and Tair is constant for it to be a fair comparison)

k for alum = 237 W/m^2*K
k for Cu = 401 W /m^2*K

--->You can see from the resulting equation that to make Tdie lower the only thing you can change is to make k, the thermal conductivity bigger - which is why COPPER WILL ALWAYS YIELD LOWER TEMPS FOR THE STEADY STATE HEAT TRANSFER. "

Well, that is quite interesting and supports what I had thought - aluminium is not better at "giving off heat" than copper. But again it doesn't directly transfer to our application. In a computer you have a fan using convection to cool the heatsink, essentially all that matters is the steady state temperature. In our fin-less and fan-less handhelds we never reach a useable steady state temperature without reaching unlimited duty cycle, as that is the definition of the requirements for unlimited duty cycle.

With our laser builds we rely on transient heat transfer state to set the duration of ON time for all non unlimited duty builds. The "once device is off" heat transfer paradigm sets the denominator of our duty cycle, or "off time" which is also important for our application, but meaningless to a computer application.

Yeah, there are differences once we get to the convection/radiation stage - but until then we are dealing with pretty much the same issues.

I still say cramming as much copper in as a sink is the easiest, best way to go.

I think RHD has brought up one point that is REALLY important and that is the brass modules we use to mount the diodes.

I had already been buying copper (and some Aluminum 3.8's) modules because I wanted maximum heat sinking and it seemed crazy to me to have a crappy metal surrounding the diode and then get nice copper sinks to wrap around them.

Also, I like the idea of making new "modules" preferably something bigger than 12mm o.d. out of copper.

It would be nice to look at e-machine shop for making a bunch of these.

Absolutely! I couldn't agree more there. Though from my experience for <700mW the brass module will do fine if you have a good heatsink and host. My phobos 12x @ 450mA which is putting out 687mW can run for 30minutes continuously without getting warm to the touch if you're not holding it. If you're holding it the heat of your hand contributes to the thermal load and it gets warm after about 13minutes... but cools off in 3minutes.

Nice read, interesting idea!

I think I disagree though...

The heat sink's first duty is to conduct heat energy away from the diode itself. If you were to use a metal like steel with low thermal conductivity, it could act as an insulator rather than a heat sink. If the amount of heat energy produced in a given amount of time is more than the metal can dissipate, the core of the heatsink will heat up. It would be possible for the metal around the diode to heat up, and your diode would overheat before the entire heatsink was even warm.

Assuming the thermal conductivity is fast enough to keep up with the heat produced by the diode, and it can dissipate the heat away from the diode in time, a higher specific heat could lead to really crappy duty cycles.

Specific heat works both for and against you. Rather than a 60sec on/ 30sec off duty cycle, you might be able to run it for 10 minutes before it got too hot, but then since the metal will have absorbed so much heat and since it dissipates it more slowly, it could take another 10-15 minutes to cool down again. While the times would be extended, the ON/OFF ratio would decrease as thermal conductivity decreases, regardless of specific heat. That's fine if you only plan to run it for 10 minutes, then quit.

Metal with higher thermal conductivity will transfer heat away from the diode and to the surface faster where it can be then transferred to the air. Higher thermal conductivity will increase the ON/OFF ratio of the duty cycle.

A larger heat sink with high thermal conductivity will increase heat capacity while retaining or improving the rate of transfer to the air.

I think the shape of the heatsink is much, much more important than the specific heat. Fins which are open to the air will increase surface area allowing the heat o dissipate faster, but again, metal with higher thermal conductivity will give you better duty cycles.

Sigurthr said:

Well, that is quite interesting and supports what I had thought - aluminium is not better at "giving off heat" than copper. But again it doesn't directly transfer to our application. In a computer you have a fan using convection to cool the heatsink, essentially all that matters is the steady state temperature. In our fin-less and fan-less handhelds we never reach a useable steady state temperature without reaching unlimited duty cycle, as that is the definition of the requirements for unlimited duty cycle.

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I've often thought that the whole aluminum is better at giving off heat vs. copper was a load of wash. However, I think the "truth" in the statement comes from the fact that the copper can store more heat due to its mass and SHC, so it doesn't heat up as quickly, and therefore it has more "heat inertia." The result would be that the mass of copper needs to absorb more heat before it creates a higher temperature gradient at the fins. Aluminum can't store much heat, so it heats up more quickly, which means there is a higher temperature gradient at the fins.

Ultimately, cost aside, I think using copper would be the best material for any heatsinking short of exotic materials, because it can both store heat, and transfer it. Aluminum would still be my choice as it cheap, conducts well, doesn't weigh much, and is easy to work with. It may also be good for having the host heat up more for a higher temperature differential at the surface.