Cooling a CPU
A square silicon chip with a thermal conductivity of 150 W/m∙K is of width w = 7 mm on a side and of thickness t = 2 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 60 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?
Expand Hint
Fourier's Law of Conduction:
$$$P=\dot{Q}=kA\frac{\Delta T}{t}$$$
where
$$P$$
is power,
$$\dot{Q}$$
is the rate of heat transfer,
$$k$$
is the thermal conductivity,
$$A$$
is the surface area perpendicular to the heat transfer direction,
$$\Delta T$$
is the change in temperature, and
$$t$$
is the thickness.
Hint 2
Because the silicon chip is a square profile:
$$$A=w^2$$$
where
$$w$$
is the width.
Assumptions
: (1) Steady state conditions, (2) Constant properties, (3) Uniform heat dissipation, (4) Negligible heat loss from back and sides, (5) One-dimensional conduction chip.
All the electrical power dissipated at the chip's back surface is transferred by conduction. From Fourier's Law of Conduction:
$$$P=\dot{Q}=kA\frac{\Delta T}{t}$$$
where
$$P$$
is power,
$$\dot{Q}$$
is the rate of heat transfer,
$$k$$
is the thermal conductivity,
$$A$$
is the surface area perpendicular to the heat transfer direction,
$$\Delta T$$
is the change in temperature, and
$$t$$
is the thickness.
Rearranging the formula to solve for the temperature difference:
$$$\Delta T=\frac{t\cdot P}{kw^{2}}=\frac{0.002m\cdot 60W}{150W/m\cdot K(0.007m)^{2}}=16.33K$$$
16.33 K
Time Analysis
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