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: Power, $$P = hA(T_{s}-T_{\infty})+\epsilon \sigma A ((T_{s})^4 - (T_{\infty})^4)$$

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

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    \$\begingroup\$ It's not uncommon to neglect radiation, and consider it as extra margin, unless you are designing something for space, where radiation is all you have. \$\endgroup\$
    – Phil Frost
    Commented Mar 19, 2014 at 1:17
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    \$\begingroup\$ If your copper plate really is only 0.05cm = 1/2 mm thick, you cannot expect the entire surface to have the same temp because of internal heat transfer resistance. Can you verify the thickness? \$\endgroup\$
    – posipiet
    Commented Mar 19, 2014 at 7:44
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    \$\begingroup\$ At 89 °C, how long do you expect the copper to remain burnished rather than coated in oxide? And you mention a hole for a lamp: is there going to be a lamp fitted too? \$\endgroup\$ Commented Mar 19, 2014 at 19:17
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    \$\begingroup\$ @Andrew An oxide layer greatly enhances emissivity. That's a plus to the 4th power, as long as it doesn't raise resistance to heat conduction in such a thin sheet of copper. \$\endgroup\$ Commented Mar 19, 2014 at 20:34
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    \$\begingroup\$ @WayfaringStranger I noticed the enhancement to emissivity after commenting and should have mentioned it. But how much oxide could there be on a 0.5mm thick sheet before it is wholly oxide dust? I'm wondering if aluminium sheet would be better (as long as there is no steam around), or a layer of paint. \$\endgroup\$ Commented Mar 19, 2014 at 20:42

1 Answer 1


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

  • \$\begingroup\$ How is heatsink thermal resistance 1.3oC/W? Its obtained if we divide 87.7oC by 67 W. But, shouldn't we be dividing the 'rise in temperature above ambient due to heatsink', that is, 44.7 oC, by 67 oC? Wouldn't that be the thermal resistance? (0.667 oC/W)? Also, by the effect of spreading resistance for a 0.05 cm thick copper sheet, the thermal resistance becomes 1.667 degrees C/W? So the expected temperature of copper sheet is 155 degrees C??? \$\endgroup\$
    – Analon
    Commented Mar 21, 2014 at 2:16
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    \$\begingroup\$ @Analon Rθ 1.3 oC/W is calculated for both sides natural convection, neglecting radiation (which as around 0.15 oC/W in your case). Using forced fan convection with htotal 25, Rθ will be 0.667 oC/W. So even with perfect thermal transfer from LED case to heatsink, the temperature rise will be higher than maximum junction temperature \$\endgroup\$
    – GR Tech
    Commented Mar 21, 2014 at 6:29
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    \$\begingroup\$ @Analon Another interesting thing is that your sheet of copper will reach maximum temperatute in less than 3min time, so your fan is very critical part of your project. \$\endgroup\$
    – GR Tech
    Commented Mar 21, 2014 at 7:15
  • \$\begingroup\$ If Rθ is just 0.667, then the temperature rise from ambient to junction will be (0.667 oC/W + 0.3818 oC/W) X 67 W + 43 oC = 113.26 oC, right? This is very near junction maximum (125 oC). So should I also account for a spreading resistance of 1 oC/W and expect a junction temperature of 180 oC? (Meaning that the heat sink is insufficient?) \$\endgroup\$
    – Analon
    Commented Mar 21, 2014 at 7:47

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