A significant design trend for several years has been the switching from incandescent and even CFL lamps to LED lights. The challenge for LED lamps is getting the heat out effectively and precisely. Here are a couple examples of calculating what size heat sink you need for an LED lamp.
The first example is a demonstration of how to figure what heat-sink capabilities you need. The second is a link to an interactive calculator from Luxeon where you can perform “what if” scenarios. You may find this calculator a handy tool to have.
Example calculation:
Five white XPE LEDs are used in an application
1. Maximum ambient temperature (Ta ) of 50°C
2. Assuming a typical forward voltage (Vf ) of 3.2 V at 350 mA
• The total power dissipated is: Ptotal = 5 x 0.350 A x 3.2 V = 5.6 W
• The maximum LED junction temperature (Tj ) provided in the datasheet is 150°C
Therefore:
Determine system thermal resistance RQJ -A:
RQ J -A = DT J -A / Pd
= (150°C -50°C) / 5.6 W
= 17.18°C /W
• The thermal resistance from junction to slug is listed in the datasheet as 9°C/W. The thermal resistance between the LED slug and board (heat sink), depends on the surface finish, flatness, applied mounting pressure, contact area, and the type of interface material and its thickness. With good design, it can be minimized to less than 1°C/W (this will be on the high end, based on the mounting method)
RQJ -A = RQ J – S + RQ S – B + RQ B -A
17.18 = 9/5 + 1+ RQ B -A
RQ B –A = 14.38°C/W
In order to keep the junction temperature below 150°C in worst-case conditions, a heat sink with thermal resistance from heat sink to air less than 14.38°C/W must be chosen. (Courtesy element14)
Here’s a link to a calculator from Luxeon that lets you play around with different values for Vf , current, and number of LEDs to get the right heat sink: Heat sink calculator
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