Heatsink 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 data sheet is 150°C
Therefore:
Determine system thermal resistance RQJ -A:
RQ J -A = DT J -A / Pd
= (150°C -50°C) / 5.6W
= 17.18°C /W
- The thermal resistance from junction to slug is listed in the data sheet 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.