Question

A transistor with a height of \(0.4 \mathrm{~cm}\) and a diameter of $0.6 \mathrm{~cm}$ is mounted on a circuit board. The transistor is cooled by air flowing over it with an average heat transfer coefficient of $30 \mathrm{~W} / \mathrm{m}^{2}\(. \)\mathrm{K}\(. If the air temperature is \)55^{\circ} \mathrm{C}\( and the transistor case temperature is not to exceed \)70^{\circ} \mathrm{C}$, determine the amount of power this transistor can dissipate safely. Disregard any heat transfer from the transistor base.

Short answer

Answer: The maximum power the transistor can safely dissipate is approximately 3.39 watts.

Step-by-step solution

Calculate the Transistor Surface Area

To find the amount of power that the transistor can dissipate, we must first calculate its surface area. The transistor is a cylinder shape with height 0.4 cm and diameter 0.6 cm (radius 0.3 cm). The surface area of a cylinder (ignoring the top and bottom faces) can be calculated using the formula \(A_\mathrm{surf} = 2 \pi r h\), where \(r\) is the radius and \(h\) is the height. $$A_\mathrm{surf} = 2 \pi(0.3 \times 10^{-2} \mathrm{~m})(0.4 \times 10^{-2} \mathrm{~m}) = 7.54 \times 10^{-4} \mathrm{~m}^2$$

Analyze the Heat Transfer Equation

The heat transfer equation states that the rate of heat transfer is a product of the air's heat transfer coefficient, the surface area of the transistor, and the difference in temperature between the air and the transistor. This relation can be expressed as: $$Q = h A_\mathrm{surf} (T_\mathrm{trans} - T_\mathrm{air})$$

Input Given Values and Solve for Power

Now that we have the surface area of the transistor, we can plug in the given values for the heat transfer coefficient (\(h = 30 \mathrm{~W} / \mathrm{m}^{2} . \mathrm{K}\)), the air temperature (\(T_\mathrm{air} = 55^{\circ}\mathrm{C}\)), and the maximum allowed transistor temperature (\(T_\mathrm{trans}= 70^{\circ}\mathrm{C}\)) into the heat transfer equation and solve for \(Q\), which represents the maximum power the transistor can safely dissipate. $$Q = 30 \mathrm{~W} / \mathrm{m}^{2} . \mathrm{K} \times 7.54 \times 10^{-4} \mathrm{~m}^2 \times (70^{\circ} \mathrm{C} - 55^{\circ} \mathrm{C})$$ $$Q = 30 \mathrm{~W} / \mathrm{m}^{2} . \mathrm{K} \times 7.54 \times 10^{-4} \mathrm{~m}^2 \times 15 \mathrm{K}$$ After performing the calculation, we find the maximum power that the transistor can safely dissipate: $$Q \approx 3.39 \mathrm{W}$$ So, the transistor can safely dissipate a maximum power of approximately 3.39 watts.