You are trying to dissipate 700ma x 3.60v = 2.5W (not accounting for the 
energy that is actually being converted into light).

Assumptions:

(1) you have a perfect thermal connection between the LED and the aluminum 
(which it looks like its pretty good based on the fact the Al gets hot)

(2) the Al is a perfect conductor of heat (close enough in a steady state 
condition after 30minutes or so)

(3) that you are actually using a K2 Luxeon Star package

(4) the entire surface of the Al, other than what is covered by the Star is 
exposed to "still air" and is effectively helping to "heatsink" the K2.

We'll look at this two ways. First we'll do a seat-of-the-pants thing, then 
we'll be a little bit more analytical. But first a little side discussion.

<ASIDE>

By the way: one thing to note: the thickness of your Al plate is nowhere 
near as important as its total area, that's why heatsinks have fins. You 
just need enough thickness to provide a low enough thermal resistance so 
that the temperature gradient across the Al doesn't result in too high a die 
temperature for the K2. Given (a) the K2's thermal resistance of 13 degrees 
C per Watt from junction to case for the Star package, (b) as 2.5W 
dissipation, and (c) a maximum junction temperature of 150 degrees C, you 
have to limit the temperature of the Al at the mounting of the Star to no 
more than:

150 C - ( 2.5W * 13 C / W) = 117 C

You can help this process in two ways:

(1) Spread the heat around a chunk of Al. In the end this is self-defeating 
because eventually the whole piece of Al heats up and you are in trouble.

(2) Get the heat out of the Al so that in the steady state the Star is 
sitting on something (much) cooler than 117 C (the lower the better for 
lumen output, reliability and lumen maintenance).

The only ways to get the heat out of the Al are:

(a) Conduction - connect the Al to some bigger chunk of something to "pull" 
the heat away (but see (1) above)

(b) Convection - air moving across the surface will take some of the heat 
with it, and the hot Al will tend to move the air itself (hot air rises).

(c) Radiation - hot objects emit heat to their environment in manner that 
depends on many factors, including the surface characteristics of the object 
and environment, relative temperatures, etc.

Note that all of the above is "recursive". Each object to which you transfer 
heat must:

(I) be outside your 'system' (radiation to distant objects), or
(II) get away from you system (air moving thru convection or forced-air 
cooling)
(II) pass it onto something else using (a), (b), and/or (c), above

</ASIDE>

<SEAT_OF_PANTS>

A little "thought experiment":

The Star package, at 20.2mm wide covers 811 sq mm.

The total surface area of your Al is:

1.5" x 1.5" x 2 faces = 2.25 sq "
+ 1.5" x 0.25" x 4 faces = 1.5 sq"

For a total of 3.75 sq " or 2419 sq mm

Deducting the area covered by the Star you have 1608 sq mm or 16 sq cm to 
dissipate the heat.

This comes to 2.5W / 16 sq cm = 0.16 W/sq cm

Now at first guess, that doesn't seem very high. But look at is this way:

Think of how big a 0.5W thru hole carbon resistor is. I would guess it's 
about 1.2cm long by about -0.4cm in diameter. This gives a surface area of 
1.2cm x 0.4cm x pi = 1.5 sq cm (ignoring the ends). This corresponds to a 
dissipation of 0.5W/1.5 sq cm = 0.30 W/sq cm, which is twice what you are 
seeing. From this, you might think that your numbers aren't too bad, but 
here are the catches...

(1) 1/2W resistor *really* doesn't want to dissipate a full 1/2W
(2) A *major* part of the resistor's dissipation is typically through the 
leads to the foil of the PCB to which it is soldered.
(3) The resistor is a small object and is typically not placed directly 
adjacent to similar resistors.
(4) The design temperature rise of the resistor (even accounting for (1) 
above) is probably a lot higher than your 25 degrees C.

CONCLUSION:

So, what does that come to: it doesn't surprise me that it gets hot!

</SEAT_OF_PANTS>

<ANALYTICAL>

To be a little more rigorous we can use the formula:

The heat, Q, removed from a surface is simply equal to:

Q =A x h x deltaT(surface to ambient)

Where A is area
"h" is a 'magic' coefficient depending on deltaT and airflow rate
Q is the heat (your 2.5EW)
deltaT is the difference in temperature between the surface and ambient

We can reverse that to get:

deltaT = Q / ( A x h )

Unfortunately this isn't quite that simple because "h" isn't really very 
constant, especially for still-air conditions. You actually have to already 
know a deltaT to determine the corresponding "h"!! But, since you already 
reported a deltaT of about 25 degrees C (50 degrees - 25 degrees ambient), 
we get an "h" of approximately:

12 W/sq m K

K is degrees kelvin (the same size as a degree C)

So....

deltaT = 2.5W / ( 16 sq cm x 12 W/sq m K )

since 16 sq cm = 0.0016 sq m, we get

deltaT = 2.5W / (0.0016 sq m x 12W/sq m K)

cancelling some units:

deltaT = 2.5 / (0.0016 x 12/K )

evaluating we get:

deltaT = 130 K

Now the fact that you don't actually see a rise of 130 degrees is probably 
related to several factors:

(1) you are not really in still air (probably the biggest part). even at an 
air velocity of only 1 m/s (about 2 miles per hour) "h"  is about 31 W/sq m 
k, so your computed deltaT would be 12/31 or 130 K or about 49 K. Higher 
velocities result in diminishing returns: 2.5 m/s (5 miles per hour) gives 
an "h" of 43 W/sq m k, only 38% for a velocity 250% higher.

(2) I misestimated the value of "h" from a graph at 
http://www.electronics-cooling.com/articles/1998/jan/calcorn.php (pretty 
likely)

(3) we didn't take into account the power emitted as light (see "Note" 
below)

(4) the phase of the moon

(5) the number of dead fish waved in Australia (or is that New Zealand)

(6) my maths are seriously compromised somewhere in here (I did only get 3 
hours sleep last night)

Note: Power emitted as light:

Being too lazy to convert the lumen values of the various white K2's to mW, 
I used the values for Royal Blue K2's which are already specified in mW (I 
think they are used in dental curing lights). The LXK2-PR12-M00 outputs 
480mW at 700ma current. If the white devices are similar, that would account 
for about 20% of your total power dissipation.

Momentarily and partially curing my laziness, I see that Wikipedia tells me 
that 1 lumen-meter = 0.0015W (at least at the wavelength to which we are 
most sensitive). So, taking the typical lumen-meter value of about 75 for a 
white K2 at 700ma we get only 112mW, not nearly as efficient as the Royal 
Blue at converting electricity to light. [This last has to be taken with a 
grain of salt because you really cannot directly convert lumen-meters to 
watts. This is because the conversion factor is dependent on the wavelength 
(color) of the light!!

</ANALYTICAL>

Hope this helps------


---- Bob Ammerman
RAm Systems

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