6 A square silicon chip (k 150 W/m K) is of width w 5 mm on a side and of thickness t 1 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 4 W are being dissipated in circuits mounted to the back surface of the chip, what is the steady-state temperature difference between back and front surfaces

1.1⁰C

Explanation:

Width W = 5mm = 0.005

Thickness t = 1 mm = 0.001

K = thermal conductivity = 150W/m.K

P = q = heat transfer rate = 4W

We are to find the steady state temperature between the back and the front surface

We have to make these assumptions:

1. There is steady state conduction

2. The heat flow is of one dimension

3. The thermal conductivity is constant

4. The heat dissipation is uniform

We have:

∆t = t*P/k*W²

= (0.001m x 4W)/150x(0.005)²

= 0.004/0.00375

= 1.06667

This is approximately,

1.1⁰C

Thank you!

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explanation: