A square isothermal chip is of width w 5 mm on a side and is mounted in a substrate such that its side and back surfaces are well insulated; the front surface is exposed to the flow of a coolant at T 15 C. From reliability considerations, the chip temperature must not exceed T 85 C. If the coolant is air and the corresponding convection coefficient is h 200 W/m2 K, what is the maximum allowable chip power

Solution :

All of the electrical power dissipated in the chip is transferred by the convection to the coolant.

So P = q

and from the Newton's law of cooling,

[tex]$P = hA(T-T_ \infty) = hW^2(T-T_ \infty)$[/tex]

And in air,

[tex]$P_{max} = 200 \ W/m^2 . K (0.005 \ m)^2(85 - 15)^\circ C$[/tex]

         = 0.35 W

In the dielectric liquid,

[tex]$P_{max} = 300 \ W/m^2 . K (0.005 \ m)^2(85 - 15)^\circ C$[/tex]

         = 5.25 W

Relative to the liquids, the air is considered a poor heat transfer agent. So, in the air the chip are able to dissipate far less of energy than in the dielectric liquid.

The maximum allowable chip power is 0.35W.

To solve this question, we need to apply Newton's law of cooling.

This states that the rate of heat loss of a body is directly proportional to the difference in temperature between the body and its environment.

Mathematically; P∝ AΔT

[tex]P=hA(T-T_o)\\P=hw^2(T-T_o)\\[/tex]

Let's substitute the values into the above equation.

[tex]P_m_a_x=(200)(5*10^-^3)^2*(85-15)\\P_m_a_x=0.005*70\\P_m_a_x=0.35W[/tex]

From the calculation above, the maximum allowable chip power is 0.35W

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