A square silicon chip (k = 152 W/m·K) is of width 7 mm on a side and of thickness 3 mm. The chip is mounted in a substrate such that its side and back surfaces are insulated, while the front surface is exposed to a coolant. If 6 W are being dissipated in circuits mounted to the back surface of the chip, what is the steady-state temperature difference between the back and front surfaces?

Question

Hannah

Engineering

22 April, 01:23

A square silicon chip (k = 152 W/m·K) is of width 7 mm on a side and of thickness 3 mm. The chip is mounted in a substrate such that its side and back surfaces are insulated, while the front surface is exposed to a coolant. If 6 W are being dissipated in circuits mounted to the back surface of the chip, what is the steady-state temperature difference between the back and front surfaces?

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Wheeler · Answer 1

The steady-state temperature difference is 2.42 K

Explanation:

Rate of heat transfer = kA∆T/t

Rate of heat transfer = 6 W

k is the heat transfer coefficient = 152 W/m. K

A is the area of the square silicon = width^2 = (7/1000) ^2 = 4.9*10^-5 m^2

t is the thickness of the silicon = 3 mm = 3/1000 = 0.003 m

6 = 152*4.9*10^-5*∆T/0.003

∆T = 6*0.003/152*4.9*10^-5 = 2.42 K