The thread for LED heat sinking questions

[widmer]

So I'm starting this thread because I have a question of my own about heat sinking, but others feel free to ask your heat sinking related questions here as well.

My question: I will be mounting my LEDs on aluminum U channel with 1" base and 2" sides. I have the option of buying this in 1/8" thick (0.125) or 1/16" thick (0.0625). I would like to go with the 1/16" because it is lighter and a little bit cheaper, but I am curious: Being that it is thinner cast, would heat not propagate as well up into the fins?

Another way of thinking about this is, at what point does the aluminum become too thin to be able to pull heat away from the LEDs efficiently? :hmm4:

[RedM3]

If I remember correctly from my heat transfer classes in Engineering school (took them 5 years ago and haven't used HT since!), the thickness doesn't really have much, if any, effect. There are several main factors in heat sinks. Length of the "fins," number of fins, material's heat transfer coefficients, and temperature of the source and ambient temperature. The most important thing you'll have to worry about is the surface area, which is a factor of fin length and number of fins.

Since a U-channel will only have two "fins," air flow becomes more important. If you don't have air flow, the ambient temperature increases, which means less heat will be pulled away from the heat sink. With no practical experience, just theoretical, I'd say you'd be fine assuming you have proper air movement.

[ghellin]

Has anyone put a smaller C channel inside of a bigger C channel? For example place a 3/4" channel inside of a 1" channel to add 2 more fins.

[smithj108]

Yes, thickness does matter. The way we pull heat away from our LEDs is via conduction, and conduction simply "pipes" heat to achieve equilibrium. The reason I use the piping or plumbing analogy is because when we try to keep one side of the heat sink cool (aka the fins) and the other side hot (nearest LED mount) we are effectively transferring heat from one location to another. The heat sink itself is the medium the heat obviously moves through. Given that, things arguably "move" better through larger/wider pipes. More heat sink medium - or in the case thicker - helps move more heat. We can't say it moves heat faster, we can only say it moves more heat at once. For example, if you wanted faster, consider this chart:

Metal: Thermal Conductivity

* Steel: 50
* Brass: 109
* Aluminum: 250
* Gold: 310
* Copper: 401
* Silver: 429

The reason we (the industry) chooses aluminum is due to it's excellent conduction properties combined with it's price. That's not saying the aren't better metals:
But we're talking "best bang for the buck".

So, what we are talking about is getting a quantitative amount of heat way from your LED in a reasonable amount of time or rate. We must consider the two main dynamics: 1) speed (rate of conduction), 2) bandwidth (amount of conduction). Consider having to move 50 people per hour from point A to point B, every hour of the day. When deciding on a vehicle to do this, you will consider not only how fast the vehicle is, but also how many persons it can carry.

Now we come to your original question: use 1/8 or 1/16 stock?? The real answer is we need a degree in thermodynamics, etc... to truly know exactly how many BTU's/sec of heat need to be drawn away from the LED and depend on how fast we'll know how much. A metal rated at "250" in conduction that is 1/16 thick might indeed be enough, then again it might not. If it's not, your LED's core temp would simply run hotter and possibly shorten it's life down the road.

I think there's much more science in this than people give credit. But because we really don't know all this math stuff (me included), it's just simpler to over-kill on heatsink - a better safe than sorry philosophy.

[kcress]

Widdy; Don't you recall my build? That's on 0.062" Al. It's run everyday since. It doesn't get hot.

[LargeAngels]

I will be mounting my LEDs on aluminum U channel with 1" base and 2" sides. I have the option of buying this in 1/8" thick (0.125) or 1/16" thick (0.0625). I would like to go with the 1/16" because it is lighter and a little bit cheaper, but I am curious: Being that it is thinner cast, would heat not propagate as well up into the fins? :

How many LED's and how far apart are you mounting them? Are you using fans or not? Forced conduction versus natural convection makes a big difference in the heat transfer. I've taken quite a few heat transfer classes at North Campus and as stated previously there is quite a bit more science to heat transfer than most people think. If in doubt go thicker and add a fan if possible.

[RedM3]

Yes, thickness does matter. The way we pull heat away from our LEDs is via conduction, and conduction simply "pipes" heat to achieve equilibrium. The reason I use the piping or plumbing analogy is because when we try to keep one side of the heat sink cool (aka the fins) and the other side hot (nearest LED mount) we are effectively transferring heat from one location to another. The heat sink itself is the medium the heat obviously moves through. Given that, things arguably "move" better through larger/wider pipes. More heat sink medium - or in the case thicker - helps move more heat. We can't say it moves heat faster, we can only say it moves more heat at once.


That information isn't completely correct. Heat transfer within the solid only is considered conduction. The transfer of heat to the air is the most important part of the cooling process, and that is done by convection. The surface area of your heat sink is THE single most important part of your sink, and that drives its ability to cool.

I broke out my heat transfer texts to check out the equations. It actually turns out a thicker base reduces the ability of a heat sink to reject heat. A thinner base is actually IDEAL for heat extraction, although you don't always see very thin bases for manufacturing and durability reasons.

The plumbing analogy used actually isn't valid, because all materials and mass provides a RESISTANCE to heat flow. Imagine the mass as a BARRIER, rather than a conduit, to heat flow. A thicker base actually increases the resistance to heat extraction. An ideal heat sink actually doesn't have a base and only has fins. The best analogy would be comparing heat transfer to electricity, although that can still be difficult to conceptualize.




Now, onto the more technical engineering jargon so you know I actually know what I'm talking about. I'll go over the equations to solve for the temperature of the LED for a basic plate as a heat sink. For simplification, we'll assume the LED is somehow direct attached to the heat sink, and there is no adhesive or connecting material used. Air temperature will be the same in all scenarios because we're using a powerful enough fan to constantly replace any air that is heated with room temperature air.

Conduction - Transfer of heat within the aluminum:
Q = k*Ac*dT / L, where Q is the LED Output, k is the thermal conductivity of the material, L is the thickness (or length of conduction), and Ac is the cross sectional area across which the length of conduction occurs. dT is the temperature difference, or (Tled - Tsurface).

Convection - Transfer of heat from aluminum to surrounding air:
Q = h*As*dT, where Q is the output of the LED (remains constant since the LED's output the same heat regardless of heat sink size), As is surface area, and dT is the temperature difference between the surface and air, or (Tsurface - Tair)
Notice in convection, there is no mention of the thickness of the heat sink.

Combine these two equations to solve for Tled (I'm substituting the dT's in both equations), and you get:

Tled = (Q * L) / (K*Ac) + Q / (h*As) + Tair

The first term is from Conduction, the second term is Convection, and the Third is also from convection.

Now let's take two heat sinks made of aluminum - identical dimensions, except one is half the thickness of the other. The parts of the equation related to convection will not be affected. The coefficients will not be affected because it is the same material between both heat sinks. The cross sectional area doesn't change. The LED output (Q) doesn't change. All that changes is L, the thickness. To look at the equation simply, imagine it with everything that doesn't change to have 1 as its value:

Tled = 1 * L / (1*1) + 1 / (1 * 1) + 1

It simplifies down to:
Tled = L + 2

That means the temperature of the LED increases as the thickness of the heat sink increases.

Questions?

[lucidvtec]

^^^^^^^^^^^^^ show off!!!

jk.....thats some good work

[smithj108]

That means the temperature of the LED increases as the thickness of the heat sink increases.

Questions?

Your fancy math isn't taking into account that in the scope of conduction temperature within the metal itself is trying to achieve a balance. This can be referred to heat storage or capacitance. I don't have my math book handy so I stick to layman phraseology and terminology. While it is true that convection is what is ultimately taking the heat away from the heat sink, it is conduction that it piping the heat from the LED towards the area that is shedding the "most" heat (aka da fins). Again point A to point B. It was in the context of conduction where I feel thickness does in-fact count and is applicable. Your formula is valid "if" we assume that the heat source is very very close (aka thin) to the heat dissipation area (aka the fins). To take my point to the extreme, go ahead and mount an LED to aluminum foil - I don't care how many "fins" you put on that foil, it won't be "thick" enough to pull that heat away from the LED at an acceptable rate! In your defense, you are not wrong, math isn't wrong, just talking about a specific (narrow) property - i.e. if the metal is too thick, it will hinder the rate that heat gets to the fins as fast as thinner metal would. I get that! But you must factor in metal-speed and capacitance as well as all come into play at once to deliver proper cooling.

Sorry - but I had to refute the "thicker metal = hotter LED notion".

[addonnis242]

Just wondering if this type U channel will work, heres the specs and a pic.

Outside width is 1.25", side is 0.5"H, 1/16" thick aluminum profile, inside of channel is 1-1/8."

http://i194.photobucket.com/albums/z130/addonnis242/102197a.jpg

[lordofthereef]

Ok, so being that my expertise is in the medical field (aka I understood little of what was being calculated above), what can we walk away with? Thickness matters to a point, after which thicker does NOT equal better? What is the answer to the original question? Also, as has been asked, would adding a smaller U channel inside the bigger one be helpful at all or just a waste of effort/money.

[kcress]

addonnis242; While not the best that will be fine. That lip will reduce convection a little but probably not enough to make a difference.

lordofthereef; A single piece of U-channel has proven to be more that good enough. Adding more would be a dubious improvement. If it didn't sit ABSOLUTELY flat against the first one the result would be considerably less effective than without it.

[marc111]

1) In regards to the channel shown. An ordinary U channel would be preferred. The desired flow of air is upwards past the "fins" that lip will hinder the internal air flow such as it is.

2) This illustrates one of the flaws I see in our approach. With a continuous flat bottom to the U channel there is little chance for efficient air circulation on the inside of the U channel without a fan. I think we should consider periodic holes in the bottom surface to aid convection air flow.

I am using some surplus heatsinks with four fins where a slot was milled across the width every 3 inches. The heatsink is about 1 in wide with 0.5 in fins. With an LED mounted in each 3 in section there is minimal heat rise with a temp of only 40 C measured by thermal couple right next to the LED.

3) I would also concur that for practical heat sinks the 1/8 in bottom would be preferred over the 16th. (To qualify this answer I am an EE not a thermal expert so I haven't done the math to prove this.)

On a related topic, thermal compound use:
4) A Question for the group: How much artic silver thermal compound should be used under the star board?

Should this be just a tiny drop scraped across the heat sink surface to fill micro voids?
OR
Should this be a slightly larger amount that is spread into an even layer across the area where the star bd is mounted?

Regards,
Mark
MSEE

[kcress]

Hi mark.

It should be a tiny amount put in the center of the star. Like about the volume of the tapered end of a toothpick. You have enough if you see the compound barely,(just barely), coming out from under the screwed down star on most sides.

[Steve Sullivan]

The thickness of the base of an extrusion adds mechanical, structural rigidity to the heat sink, it also adds thermal mass. The thermal mass of the base metal actually impedes the transfer of heat to the fins, which impedes the transfer of heat energy to the primary heat transfer mechanism, the convective surfaces (i.e the fins), as was stated before. The thicker the base of the heat sink, the more thermal gradient (i.e temperature drop) there will be across the thickness of the base, resulting in a lower temperature at the fins, the lower the temperature, the less energy is removed from the system via convection.

If the thinner base provides sufficient structural support for your design, then by all means use the thinner base, you will get better heat transfer. The same also applies for the fins by the way, you want the fins to have as much surface area and as little thickness as you can get away with mechanically, to optimize the heat transfer.