Question

A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit board houses 80 closely spaced logic chips on one side, each dissipating \(0.04 \mathrm{~W}\). The board is impregnated with copper fillings and has an effective thermal conductivity of \(30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to a medium at \(40^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the temperatures on the two sides of the circuit board. (b) Now a \(0.2\)-cm-thick, 12-cm-high, and 18-cmlong aluminum plate \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with 864 2-cm-long aluminum pin fins of diameter \(0.25 \mathrm{~cm}\) is attached to the back side of the circuit board with a \(0.02\)-cm-thick epoxy adhesive \((k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Determine the new temperatures on the two sides of the circuit board.

Short answer

Question: Calculate the initial temperature on the front side of the circuit board and the new temperature on the front side after adding the aluminum plate with epoxy adhesive. Answer: To find the initial temperature on the front side of the circuit board, first calculate the total heat generation, thermal resistance, and temperature difference using the given formulas. Add the temperature difference to the temperature on the back side (40°C). For the new temperature after adding the aluminum plate and epoxy adhesive, calculate the new thermal resistance and temperature difference, then add the new temperature difference to the temperature on the back side (40°C) to find the new temperature on the front side.

Step-by-step solution

Calculate total heat generation

We know the power dissipation of a single chip is \(0.04 \mathrm{~W}\), and there are 80 chips on the board. So, the total heat generation can be calculated as follows: $$Q_{total} = N_{chips} \times P_{chip} = 80 \times 0.04 \mathrm{~W} = 3.2 \mathrm{~W}$$

Calculate thermal resistance and temperature difference

The thermal resistance will be the sum of the conductive resistance across the board and the convective resistance from the board to the medium. The conductive resistance can be calculated using the formula: $$ R_{cond} = \frac{t}{kA} = \frac{0.3 \times 10^{-2}\mathrm{~m}}{30 \left(\frac{0.12\mathrm{~m} \times 0.18\mathrm{~m}\right)} $$ The convective resistance is given by the formula: $$ R_{conv} = \frac{1}{hA} = \frac{1}{40 \left(\frac{0.12\mathrm{~m} \times 0.18\mathrm{~m}\right)} $$ Now, sum both resistances and calculate the temperature difference: $$ \Delta T = Q_{total} (R_{cond} + R_{conv}) $$

Calculate temperature on both sides

The temperature on the back side of the circuit board is given as \(40^{\circ}\mathrm{C}\). Since the temperature difference is calculated above, the temperature on the front side of the board can be found by adding the temperature difference to the temperature on the back side: $$ T_{front} = T_{back} + \Delta T $$ Now we have the temperature on both sides of the circuit board. #Phase 2# We have to consider the additional aluminum plate with heat fins and epoxy adhesive in this case. So, we will calculate the new thermal resistance and temperature difference using the same approach as above by adding the conductive resistance of the adhesive and the aluminum plate.

Calculate new thermal resistance and temperature difference

The new conductive resistance can be calculated using the formula: $$ R_{cond,new} = \frac{t_{epoxy}}{k_{epoxy}A} + \frac{t_{plate}}{k_{plate}A} $$ Then, calculate the new convective resistance: $$ R_{conv,new} = \frac{1}{hA} = \frac{1}{40 \left(\frac{0.12\mathrm{~m} \times 0.18\mathrm{~m}\right)} $$ Now, sum the new conductive and convective resistances and calculate the new temperature difference: $$ \Delta T_{new} = Q_{total} (R_{cond,new} + R_{conv,new}) $$

Calculate new temperature on both sides

Again, the temperature on the back side of the circuit board is given as \(40^{\circ}\mathrm{C}\). With the new temperature difference, the new temperature on the front side of the board can be found by adding the new temperature difference to the temperature on the back side: $$ T_{front,new} = T_{back} + \Delta T_{new} $$ Now we have the new temperature on both sides of the circuit board with the added aluminum plate and epoxy adhesive.

Key concepts

Understanding Heat Transfer

Heat transfer is a fundamental concept in physics that describes how thermal energy moves from a warmer object to a cooler one. There are three primary modes of heat transfer: conduction, convection, and radiation.

In the given exercise, we mainly deal with the first two modes. Conduction occurs when heat is passed through a solid material, such as the circuit board in our example. Convection, on the other hand, involves the transfer of heat through a fluid, which could be a liquid or a gas; in our problem, this fluid is the medium at a specific temperature which cools the back side of the circuit board.

As the logic chips on the circuit board dissipate heat, this energy needs to traverse the board (conduction) and then be transferred to the surrounding medium (convection). The effectiveness of this heat transfer process is essential for maintaining the electronic components at operational temperatures, preventing overheating and potential damage.

Conductive Resistance Analysis

Conductive resistance is a measure of how much a material resists the flow of heat through it. Materials with high thermal conductivities, such as metals, have lower conductive resistances, meaning they are better conductors of heat. The equation to calculate conductive resistance is: \[ R_{cond} = \frac{t}{kA} \]where \( t \) is the thickness of the material, \( k \) is the thermal conductivity, and \( A \) is the cross-sectional area through which heat is transferred.

In our textbook exercise, the students first calculate the conductive resistance of the circuit board, which is impregnated with copper fillings, providing it with a high thermal conductivity. The lower the conductive resistance, the more efficiently heat can be transferred across the circuit board. By calculating the conductive resistance, students learn to evaluate how different materials and thicknesses affect the heat transfer rate, which is crucial for thermal management in various engineering applications.

Convective Resistance and Heat Dissipation

Convective resistance, in contrast to conductive resistance, is related to the heat transfer between a solid surface and a moving fluid. The convective heat transfer coefficient, denoted by \( h \), quantifies how well the fluid can remove heat from the surface. The formula used to calculate convective resistance is: \[ R_{conv} = \frac{1}{hA} \]where \( h \) is the convective heat transfer coefficient and \( A \) is the area of the surface in contact with the fluid.

The challenge, as presented in the textbook problem, is determining how the convective resistance is affecting the overall temperature distribution across the electronic components. Factors that affect convective resistance include fluid properties, fluid velocity, and the nature of the fluid flow (laminar or turbulent). Understanding this concept allows students to appreciate how the choice of cooling medium and conditions can dramatically influence the thermal regulation of devices, as well as to calculate and optimize thermal systems for improved performance.