Question

A power semiconductor device with an area of 5.76 mm x 4.32 mm has a thermal conductivity of 136 W/mK. It is to be attached to a piece of copper substrate having an area of 10.4mm x 15 mm and a thickness of 1.25 mm thick with a thermal conductivity of 393 W/mK using solder (thermal conductivity = 50 W/mK, thickness = 0.25 mm) as the die attach. If the junction-to-case thermal resistance is 0.8 K/W, what would be the thickness of the power semiconductor device? [ 15 points ]

Answer 1

total thermal resistance from junction to substrate = Thermal resistance in power semi conductor + thermal resistance in copper+thermal resistance in solder

Thermal resistance in power semiconductor = T/(136*0.00576*0.00432) = 295.49 T where T is thickness

Thermal resistance in copper = 0.00125/393*0.0104*0.015 =0.0203

Thermal resistance insolder = 0.00025/(50*0.0104*0.015)=0.003205

0.8=0.0203+0.003205 +295.49 T

T=0.7476/295.49 = 0.00253 m

so thickness of junction =2.53 mm Ans