Question

A \(0.4-\mathrm{cm}\)-thick, 12 -cm-high, and \(18-\mathrm{cm}\)-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 $52 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. (a) Determine the temperatures on the two sides of the circuit board. (b) Now a \(0.2-\mathrm{cm}\)-thick, \(12-\mathrm{cm}\)-high, and \(18-\mathrm{cm}\)-long aluminum plate $(k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( with \)8642 \cdot \mathrm{cm}-$ long aluminum pin fins of diameter \(0.25 \mathrm{~cm}\) is attached to the back side of the circuit board with a \(0.02-\mathrm{cm}\)-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

Answer: In part (a), the temperatures on the two sides of the circuit board are \(T_{chipside} = 40.608 \, ^{\circ}\mathrm{C}\) and \(T_{backside} = 40.48944 \, ^{\circ}\mathrm{C}\). In part (b), the temperatures on the two sides of the circuit board with an aluminum plate are \(T_{new\_chipside} = 100.6208 \, ^{\circ}\mathrm{C}\) and \(T_{new\_backside} = 40.3104 \, ^{\circ}\mathrm{C}\).

Step-by-step solution

Part (a) - Determine the temperatures on the two sides of the circuit board

1. Calculate total heat generation: \(Q_{total} = N \cdot Q_{chip}\) where \(N=80\) is the number of chips and \(Q_{chip} = 0.04 \, \mathrm{W}\) is heat dissipation per chip. \(Q_{total} = 80 \times 0.04 \, \mathrm{W} = 3.2 \, \mathrm{W}\) 2. Calculate the resistance for conduction across the circuit board: \(R_{board} = \frac{L_{board}}{k_{board}A}\) where \(L_{board} = 0.4 \cdot 10^{-2} \, \mathrm{m}\) is the board thickness, \(k_{board} = 30 \, \mathrm{W/m\cdot K}\) is the thermal conductivity, and \(A = 12 \cdot 10^{-2} \, \mathrm{m} \times 18 \cdot 10^{-2} \, \mathrm{m}\) is the area. \(R_{board} = \frac{0.4 \cdot 10^{-2}}{30 \times (12 \cdot 10^{-2} \times 18 \cdot 10^{-2})} \, \mathrm{K/W} = 0.000037 \, \mathrm{K/W}\) 3. Calculate the resistance for convection on the back side of the board: \(R_{conv} = \frac{1}{hA}\) where \(h = 52 \, \mathrm{W/m^2 \cdot K}\) is the heat transfer coefficient, and \(A\) is the same as in step 2. \(R_{conv} = \frac{1}{52 \times (12 \cdot 10^{-2} \times 18 \cdot 10^{-2})} \, \mathrm{K/W} = 0.000153 \, \mathrm{K/W}\) 4. Calculate the total resistance: \(R_{total} = R_{board} + R_{conv}\) \(R_{total} = 0.000037 + 0.000153 \, \mathrm{K/W} = 0.000190 \, \mathrm{K/W}\) 5. Calculate the temperature difference between the chip side and medium: \(\Delta T = Q_{total} \cdot R_{total}\) \(\Delta T = 3.2 \, \mathrm{W} \times 0.000190 \, \mathrm{K/W} = 0.608 \, \mathrm{K}\) 6. Determine the chip side temperature: \(T_{chipside} = T_{medium} + \Delta T\) \(T_{chipside} = 40 + 0.608 \, ^{\circ}\mathrm{C} = 40.608 \, ^{\circ}\mathrm{C}\) 7. Determine the back side temperature: \(T_{backside}\) Since the heat generation in the chips is conducted across the board and then dissipated by convection, we can first find the temperature difference across the board, \(\Delta T_{board} = Q_{total} \cdot R_{board}\) \(\Delta T_{board} = 3.2 \, \mathrm{W} \times 0.000037 \, \mathrm{K/W} = 0.11856 \, \mathrm{K}\) Then, we subtract this temperature difference from the chip side temperature to get the back side temperature. \(T_{backside} = T_{chipside} - \Delta T_{board} = 40.608 - 0.11856 \, ^{\circ}\mathrm{C} = 40.48944 \, ^{\circ}\mathrm{C}\) For part (a), we have \(T_{chipside} = 40.608 \, ^{\circ}\mathrm{C}\) and \(T_{backside} = 40.48944 \, ^{\circ}\mathrm{C}\).

Part (b) - Determine the new temperature on the two sides of the circuit board with an aluminum plate

1. Calculate the resistance for conduction through adhesive: \(R_{adhesive} = \frac{L_{adhesive}}{k_{adhesive}A}\) where \(L_{adhesive} = 0.02 \cdot 10^{-2} \, \mathrm{m}\) is the adhesive thickness and \(k_{adhesive} = 1.8 \, \mathrm{W/m\cdot K}\) is the thermal conductivity. \(R_{adhesive} = \frac{0.02 \cdot 10^{-2}}{1.8 \times (12 \cdot 10^{-2} \times 18 \cdot 10^{-2})} \, \mathrm{K/W} = 0.018519 \, \mathrm{K/W}\) 2. Calculate the resistance for conduction across the aluminum plate: \(R_{plate} = \frac{L_{plate}}{k_{plate}A}\) where \(L_{plate} = 0.2 \cdot 10^{-2} \, \mathrm{m}\) is the plate thickness and \(k_{plate} = 237 \, \mathrm{W/m\cdot K}\) is the thermal conductivity. \(R_{plate} = \frac{0.2 \cdot 10^{-2}}{237 \times (12 \cdot 10^{-2} \times 18 \cdot 10^{-2})} \, \mathrm{K/W} = 0.000235 \, \mathrm{K/W}\) 3. Calculate the new total resistance: \(R_{new\_total} = R_{board} + R_{adhesive} + R_{plate} + R_{conv}\) \(R_{new\_total} = 0.000037 + 0.018519 + 0.000235 + 0.000153 \, \mathrm{K/W} = 0.018944 \, \mathrm{K/W}\) 4. Calculate the new temperature difference between the chip side and medium: \(\Delta T_{new} = Q_{total} \cdot R_{new\_total}\) \(\Delta T_{new} = 3.2 \, \mathrm{W} \times 0.018944 \, \mathrm{K/W} = 60.6208 \, \mathrm{K}\) 5. Determine the new chip side temperature: \(T_{new\_chipside} = T_{medium} + \Delta T_{new}\) \(T_{new\_chipside} = 40 + 60.6208 \, ^{\circ}\mathrm{C} = 100.6208 \, ^{\circ}\mathrm{C}\) 6. Determine the new back side temperature: \(T_{new\_backside}\) Since the heat generation in the chips is conducted across the board through the adhesive, aluminum plate, and then dissipated by convection, we can first find the temperature difference across the board, adhesive, and aluminum plate, \(\Delta T_{new2} = Q_{total} \cdot (R_{board} + R_{adhesive} + R_{plate})\) \(\Delta T_{new2} = 3.2 \, \mathrm{W} \times (0.000037 + 0.018519 + 0.000235) \, \mathrm{K/W} = 60.3104 \, \mathrm{K}\) Then, we subtract this temperature difference from the new chip side temperature to get the new back side temperature. \(T_{new\_backside} = T_{new\_chipside} - \Delta T_{new2} = 100.6208 - 60.3104 \, ^{\circ}\mathrm{C} = 40.3104 \, ^{\circ}\mathrm{C}\) For part (b), we have \(T_{new\_chipside} = 100.6208 \, ^{\circ}\mathrm{C}\) and \(T_{new\_backside} = 40.3104 \, ^{\circ}\mathrm{C}\).