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 \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). 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 amount of power that can be safely dissipated by the given transistor is approximately 0.339 W.

Step-by-step solution

Calculate the Surface Area of the Transistor

First, we need to find the surface area of the transistor. We are given its height and diameter, so we can assume it has a cylindrical shape. We can, therefore, use the formula for the surface area of a cylinder without its bottom surface (base) to calculate the transistor's surface area, since we are disregarding heat transfer from the transistor base: Surface area (A) = Lateral area of the cylinder = \(2 \pi r h\), where r is the radius, and h is the height. Since the diameter is given, we can find the radius by dividing the diameter by 2: \(r = \frac{0.6 \, \mathrm{cm}}{2} = 0.3 \, \mathrm{cm}\) Now, we can calculate the surface area (A): \(A = 2 \pi (0.3 \, \mathrm{cm})(0.4 \, \mathrm{cm}) = 0.75398 \, \mathrm{cm}^{2}\) Convert the surface area to square meters: \(A = 0.75398 \, \mathrm{cm}^{2} \cdot \frac{1 \, \mathrm{m^2}}{10^4 \, \mathrm{cm}^{2}} = 7.5398 \times 10^{-5} \, \mathrm{m}^2\)

Apply Newton's Law of Cooling

Next, we will apply Newton's Law of Cooling to relate the heat transfer coefficient (h), the surface area of the transistor (A), the temperature difference between the transistor and air (ΔT), and the power that can be dissipated safely (Q): \(Q = h \cdot A \cdot \Delta T\) We have the heat transfer coefficient (h) and the surface area (A), and we can find the temperature difference (ΔT) by subtracting the air temperature from the maximum transistor case temperature: \(\Delta T = 70^{\circ} \mathrm{C} - 55^{\circ} \mathrm{C} = 15^{\circ} \mathrm{C}\) or \(15 \, \mathrm{K}\) Now, we can find the power (Q) that can be dissipated safely: \(Q = (30 \, \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}}) \cdot (7.5398 \times 10^{-5} \, \mathrm{m}^2) \cdot (15 \, \mathrm{K})\) \(Q = 0.33929 \, \mathrm{W}\)

Report the Result

The amount of power that can be safely dissipated by this transistor is approximately 0.339 W.

Key concepts

Transistor Cooling

Cooling is essential for transistors to ensure they function efficiently and reliably. Heat is a byproduct of transistor operation, and if it is not managed properly, it can lead to overheating and damage.
In electronic devices, transistors are cooled using various methods, with "forced convection" being a popular choice. This involves blowing air over the transistor's surface to dissipate the heat.

• "Natural convection" relies on air moving naturally over the component, whereas "forced convection" uses fans or similar devices to enhance air movement.
• It's crucial to maintain a temperature below the maximum allowable case temperature of the transistor to prevent malfunction or degradation.
• Heat sinks are often used along with convection methods to better disperse heat away from transistors.

Cooling transistors effectively prolongs their life and maintains their performance. It is vital to consider the device's environment and cooling methods to ensure optimum conditions.

Newton's Law of Cooling

Newton's Law of Cooling is a fundamental principle to understand when analyzing heat transfer between objects and their environment. It states that the rate of heat loss of a body is directly proportional to the temperature difference between the body and its surroundings. The formula is given by:
\[ Q = h \cdot A \cdot \Delta T \]

• Where Q is the rate of heat transfer (in watts),
• h is the heat transfer coefficient,
• A is the surface area of the object, and
• \Delta T is the temperature difference between the object and the surroundings.

This law helps in designing cooling systems for electronics by predicting how much power can be dissipated safely. Understanding this interaction ensures electronic components like transistors do not exceed their thermal limits.

Heat Transfer Coefficient

The heat transfer coefficient ( h) is a critical parameter in determining the efficiency of heat dissipation from a surface. It quantifies the heat transfer rate per unit area per degree of temperature difference between the surface and the fluid it interacts with.
Several factors influence the value of the heat transfer coefficient, including:

• Type and speed of the cooling fluid (e.g., still air, moving air, water)
• Surface condition (smooth or rough)
• Thermal properties of the fluid and surface material

A higher heat transfer coefficient indicates a higher efficiency in transferring heat from an object to its surroundings. In the context of transistor cooling, it means how effectively the air can remove heat from the transistor's surface.

Surface Area Calculation

Calculating the surface area is an essential step in understanding heat dissipation in transistors. This calculation helps determine how much of the surface is available for heat exchange. For a cylindrical transistor, the formula to calculate the lateral surface area is:
\[ A = 2 \pi r h \]

• Where r is the radius of the cylinder, calculated by halving the diameter.
• h is the height of the cylinder.

Converting this area from the basic measure (square centimeters) to the desired unit (square meters) is important for coherence, especially when plugging into formulas involving heat transfer coefficients generally given in terms of square meters. Proper area calculations ensure the accuracy of thermal management predictions.