Question

A power transistor is a solid-state electronic device. Assume that energy entering the device at the rate of 1.00 W by electrical transmission causes the internal energy of the device to increase. The surface area of the transistor is so small that it tends to overheat. To prevent overheating, the transistor is attached to a larger metal heat sink with fins. The temperature of the heat sink remains constant at 35.0°C under steady-state conditions. The transistor is electrically insulated from the heat sink by a rectangular sheet of mica measuring 8.25 mm by 6.25 mm, and 0.0852 mm thick. The thermal conductivity of mica is equal to 0.0753 W/m·°C. What is the operating temperature of the transistor?

Answer 1

Step-by-step explanation:

The rate of heat transfer from the transistor to the heat sink through the mica sheet can be calculated using Fourier’s Law of heat conduction. The formula for this law is Q = kA(T1 - T2)/d where Q is the rate of heat transfer, k is the thermal conductivity of the material, A is the cross-sectional area through which heat is transferred, T1 and T2 are the temperatures at either end of the material and d is its thickness.

In this case, we know that Q = 1.00 W (the rate at which energy enters the device), k = 0.0753 W/m·°C (the thermal conductivity of mica), A = 8.25 mm * 6.25 mm = 5.15625e-5 m² (the cross-sectional area of the mica sheet), T2 = 35°C (the temperature of the heat sink) and d = 0.0852 mm = 8.52e-5 m (the thickness of the mica sheet).

Substituting these values into Fourier’s Law gives us:

1.00 W = (0.0753 W/m·°C)(5.15625e-5 m²)(T1 - 35°C)/(8.52e-5 m)

Solving for T1 gives us:

T1 ≈ 35 + (1 / ((0.0753 * 5.15625e-5) / 8.52e-5)) ≈ 35 + 22 ≈ 57°C

So under steady-state conditions with an energy input rate of 1 W and a heat sink temperature of 35°C, we can expect that the operating temperature of this power transistor would be approximately 57°C.