Cooling HP LEDs using Cu sheet

Discussion in 'General Electronics Chat' started by abhaymv, Mar 18, 2014.

  1. abhaymv

    Thread Starter Active Member

    Aug 6, 2011

    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:

    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
  2. Markd77

    Senior Member

    Sep 7, 2009
    0.5mm sheet doesn't conduct heat very far, the outside edges will be cool so not dissipating anywhere as much heat as you calculated. I expect the edges of the cutout will be very hot.
    abhaymv likes this.
  3. abhaymv

    Thread Starter Active Member

    Aug 6, 2011
    I'm placing a fan above the heat sink. Will convection help in spreading the heat?

    If not, what can I do? Should I look for alternate solutions? (Is this one totally useless?)

    Thank you!
  4. Little Ghostman


    Jan 1, 2014
    I have a suggestion,

    Place 2 in a sandwich cold side to hot side, place the side of the sandwich that feels colder on the plate used for heat sink, on the side that feels warmer place a large cpu aluminum heat sink with a fan, something like this
    Note you can find them easy in old pc's at the dump, wire the fan to the wires from the peltier cells, the cells absorb the heat rapidly and move it to the side with the heat sink and fan on, in doing so they produce electric, note make sure the fan is was of those large ones that run at around 18mA. they are not so easy to find but ive found a few, they work much better than the normal ~200mA cpu fans.
    The cells I linked to move heat very fast to the other side, but as the other side gets hotter the fan kicks is as more electric is produced. To be honest even without a fan they can shift some heat! I use them on my aquarium with high power Leds. Works great, just be sure to have a heavily finned heat sink on the warm side of the sandwich, the quicker you remove this heat the colder the other side gets. I find the fan cycles on and off, but My leds are kept pretty cool.
    This method is also used in the all metal fans you can buy for wood burning stoves, that magically seem to power themselves but cost nearly £100.
    abhaymv likes this.
  5. abhaymv

    Thread Starter Active Member

    Aug 6, 2011
    Thank you for the suggestion. I'm trying to avoid using peltiers as much as possible, since I've already designed the power supply for my system. I've also heard that peltiers are somewhat inefficient and produce heat themselves...
    Last edited: Mar 18, 2014
  6. abhaymv

    Thread Starter Active Member

    Aug 6, 2011
    I did a little math on the temperature difference between the inner edges and the outer ones. Approximating four point heat sources of power 67/4 W each on the corners of the inner cut out, the temperature fall from inner corner to outer corner should be 8.7°C. That isn't too severe, right?

    I used the equation P= -k A dT/dx
    k is 384 W/mK for Copper. dx was found by taking the diagonal distance between inner corner and outer corner on the same side.