Rebel LED Array Heat Calculations

Ok, I've got a little project started (not automotive) with a few Rebels. I need some confirmation on heat transfer to avoid frying these little pricey babies.

Anyways, the numbers I've come up with so far are:

Rebel @350mA (1W)
Max Ambient Temp: 40 ºC
Junction Transfer: 10.5 ºC/W
Base Transfer: 1.2 ºC/W
Adhesive Transfer: 4.5 ºC/W
Heatsink Transfer: 1.4 ºC/W

Which would give me a junction temperature of approx 57.6 ºC. Nice and safe. The confusion for me starts with adding additional 1W LEDs to the array. As I see it, for each LED I add to the array (approx 1.5 - 2 cm between them), the heatsink should increase in temperature by Ambient + (1.4 ºC * n), where n is the number of LEDs.

Now, what I'm unsure about is how that heat transfers to the other LEDs. Since the heatsink is only increasing by 1.4 ºC, does that mean that the junction temperature of each LED is also only increasing by 1.4? IE: 57.6 + (1.4 * n) ?

Basically I just need to know what my max heatsink temperature can be and how it relates to the junction temperature so that I can set up a thermal switch as a safety feature.

By heatsink transfer I am assuming that this is for transfer of heat from the heatsink to the environment. So, yes, each additional watt into the heatsink will increase its temperature above ambient by 1.4 degC (BTW, how do you make the degree symbol?).

That is quite an effective heatsink you have there.

I did figure out the answer to my questions. Turns out you can treat multiple LEDs on the same heatsink as if they were parallel resistors.

Link to answer:
http://www.sharpleds.eu/resources/MINI-ZENIGATA_THERMAL.PDF

You can make the º symbol by holding the alt key and typing 167 on your keyboard.

The corresponding numbers for my situation are:


So the array junction temperatures are:

Each LED has a total thermal resistance of 10.5(junction to case) + 1.2(case to mounting pad) + 4.5(mounting pad to heat-sink) = 16.2 ºC/W

1/(((1/16.2 ºC/W) * 18 LEDs )) = is an array thermal resistance of 0.9 ºC/W

0.9 ºC/W + 1.4 ºC/W (heatsink) is a total thermal resistance of 2.3 ºC/W for all junctions.

Each junction would be (2.3 ºC/W * 18W) + 40ºC = 81.4 ºC junction temperature @ 40ºC ambient. That looks pretty safe to me.

And my heatsink should be at 40ºC + (1.4ºC/W * 18) = 65.2ºC @ 40ºC ambient.

The answer is correct but you make it needlessly difficult.

You have 18 watts going into a heatsink that has a thermal resistance of 1.4 degC/watt so the heatsink temperature will rise to 18*1.4 degC above ambient = 25.2 degC above ambient.

Each of the 1 watt LEDs has a junction to heatsink thermal resistance of 16.2 degC/watt so each one will have a junction temperature 16.2 degC above the temperature of the heatsink so each LED will have a junction temperature of 16.2 + 25.2 degC above ambient = 41.4 degC above ambient. So at an ambient temperature of 40 degC, each junction will have a temperature of 81.4 degC.

All LEDs will be at the same temperature, so there will be no heat flow from one to another.

Parallel resistors have the same voltage across them, so there is no current flow between them.

I know this is an old-ish thread, but i will ask anyway. Why does no one consider the resistance incured through air. From what i am reading it seems as if you are considering the surface temp of the heat sink to be the same as the ambient air temp. I dont belive this is true. There is still a resistance to be calculated from heat sink surface temp to ambient air (1/hA), where h is the coefficient of convection ( usally 10°C/W for natural convection and approx 50°C/W for forced). This extra resistance may put your design over the edge.

Snopro44o said:

This extra resistance may put your design over the edge.

Click to expand...

These are conveniently put in order:

Junction Transfer: 10.5 ºC/W
Base Transfer: 1.2 ºC/W
Adhesive Transfer: 4.5 ºC/W
Heatsink Transfer: 1.4 ºC/W

LED junction to LED base, Base to adhesive, Adhesive to heatsink, and Heatsink to air. His heatsink to air value is suspiciously low, though.

Also, TDP dissipation ratings are always suspect.

Snopro44o said:

I know this is an old-ish thread, but i will ask anyway. Why does no one consider the resistance incured through air.

Click to expand...

Surely, heatsink ratings, if accurate, will already allow for heat loss to the air?