Electronic – Designing a copper plate heat sink


I’m trying to see if a copper sheet I have with me (25 cm X 30 cm, 0.05 cm thick. There's a 6 cm X 6 cm square hole in the center to place a lamp.) is sufficient to keep 36 luxeon rebel LEDs (mounted on coolbase square) at a junction temperature below maximum. The arrangement of LEDs in the copper sheet is fixed, and is as shown in the attached figure. As shown, the arrangement is very compact. So I did the following calculations:

enter image description here

Maximum Junction temperature: For Red LEDs in the array, this is 125°C, so I chose 115 °C for design.

Power dissipated as heat (~80 % of LED power): 67 W for 36 High power LEDs.

Considering individual junction to case thermal resistances, thermal resistance of coolbase MCPCB and interface material, and applying law of parallel pathof thermal resistances, I get a junction to heatsink thermal resistance of 0.3818 °C/W.
Thus allowable maximum heatsink temperature = 115 – 67 X 0.3818 = 25.6 °C = 89.4 °C

The heat transfer from heatsink to ambient air takes place through convection and radiation. The equation is:
$$P = hA(T_{s}-T_{\infty})+\epsilon \sigma A ((T_{s})^4 – (T_{\infty})^4)$$

h is the convective heat transfer coefficient. For forced air convection, (I'm using a fan on top of the heatsink) h begins from 25.

A is the surface area.
\$\epsilon\$ is the emissivity. For burnished copper, emissivity is 0.07 according to this page.

\$T_{s}\$ is surface temperature or heat sink temperature
\$T_{\infty}\$ is ambient temperature.

Substituting an ambient of 43 °C (316 K) and solving for surface area, I get:
Required minimum surface area: 0.056 \$m^2\$.

Available surface area (copper sheet outer surface, excluding 6 cm X 6 cm hole) is 0.0714 \$m^2\$.

I think I should be safe. Did I miss something, or can I use this copper sheet?:D

Best Answer

As mentioned above, existing theory valid for h/s thickness much bigger than 0.05cm. I hope this is not a PCB copper!

Anyway your calculations are quite the same using anothere approach

enter image description here

This temperature is the hotspot temperature (point)


Just to see the effect of the plate thickness: the spreading thermal resistance Rc for the 0.05cm thickness of your h/s, results an additional 1 oC/W. This resistance you have to add to the average h/s performance (i.e 1.3 oC/W+1 oC/W). Increasing the thickness to 0.5cm, the spreading thermal resistace will be 0.1 oC/W and with 1cm thickness Rc becomes 0.05 oC/W. Now you can calculate the total temperature rise