Electronic Properties of Materials: Conduction - Problems and Solutions

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Consider a thin insulating disk made of mica to electrically insulate a semiconductor device from a conducting heat sink. Mica has thermal conductivity of 0.75 W (m-K)-1. The disk thickness is 0.1 mm, and the diameter is 10 mm. What is the thermal resistance of the disk? What is the temperature drop across the disk if the heat current through it is 25 W?       


Given a thin insulating disk of mica (0.1 mm) between a semiconductor device and a conducting heat sink. In this problem mica acts as an electrical insulator and prevents any charge flow from the semiconductor to the heat sink. The two most common materials used to make heat sinks are aluminum and copper. Although copper has higher thermal conductivity compared to aluminum, aluminum is still preferred because of its lower density and lower price. Aluminum alloys Al6061 and Al6063 in tempered form (normally T4-T7) are used to manufacture heat sinks for a variety of applications.

We have been asked to determine the thermal resistance of the mica disk. We will have to use the Fourier's law of heat conduction, which states that the rate of heat flow per unit area is proportional to the negative thermal gradient developed across the material (at right angles). The proportionality constant is called the thermal conductivity, which is a material dependent constant and it varies with the temperature.

Q{prime} = - A{kappa}{{{Delta}T}/L}, where Q' is the rate of heat flow, which is also called as the thermal current.
Rearranging gives, Q{prime} = - {{Delta}T}/{[{L/{{kappa}A}}]}   [Equation - 1]
Now we need to compare this equation with the Ohm's law (electricity) to figure out the equation for thermal resistance. Ohm's law: I = {{Delta}V}/{R} = {{Delta}V}/{[{L/{{sigma}A}}]}   [Equation - 2]

Comparing equations 1 and 2 gives the thermal resistance as:
{theta} = {[{L/{{kappa}A}}]} = - {{Delta}T}/{Q{prime}}   [Equation - 3]

L is the thickness of the mica disc (0.1 mm), A is the area which can be calculated using the diameter of the mica disc, and K is the thermal conductivity (given to be 0.75 W (m-K)-1). Plugging the values in proper units gives the thermal resistance to be theta = 1.69765 K/W.
From equation 3 we can determine the temperature drop across the disk, which is the product of the thermal resistance and the thermal current (given to be 25 W). It comes out to be - 42.44 K. The negative sign here shows the direction of heat flow, that is towards the colder side (out of the heat sink).