I am looking for some help in sizing a heat sink for an LED horticulture light I am building. I will be attaching two Bridgelux BXEB-L1190U-30G3000-C-C3 LED strips side by side, link below to specs, to a 1244mm 840AS sink from heatsinkonline.com. The LED strips will be operating at a maximum drive current of 1.52 amps and a forward voltage of 41.5 for a total wattage of 1.52Amps * 41.5Volts * 2strips = 126 watts. The thermal resistance rating of the sink is 4.903 C/W/3”(75mm).
First step is to take the total length of the heatsink and divide by the thermal resistance length.
1244mm / 75mm = 16.58 thermal resistance sections in the total sink length
Now divide the total wattage as heat to be dissipated which would be a worst case scenario. I think the norm is to assume 2/3 of total watts is the amount of heat required to be dissipated.
126 watts / 16.58 = 7.59 Watts to be dissipated over the 3”(75mm) section.
So now take the number of watts times the thermal resistance to get the increase in heat sink temperature from ambient temp.
7.59 watts * 4.903 C/W/3” = 37.24 degrees Celsius increase
Now calculate the temperature of the case as temperature ambient plus heat sink thermal increase.
Tc = 37.24C + 27C = 64.24 degrees Celsius
TcMax for the Bridgelux strip is Tc=85 degrees Celsius.
Since the calculated Tc < TcMax this suggests to me that the heat sink is sized properly to support the two Bridgelux LED strips at the given wattage. Using total watts and assuming that it is all converted into heat should provide an additional buffer of safety. Please comment if there is something I have missed or not accounted for in the above calculation as I have never done this before. Thanks in advance for the help.