Question

The heat generated in the circuitry on the surface of a silicon chip $(k=130 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$ is conducted to the ceramic substrate to which it is attached. The chip is $6 \mathrm{~mm} \times 6 \mathrm{~mm}\( in size and \)0.5 \mathrm{~mm}\( thick and dissipates \)3 \mathrm{~W}\( of power. Disregarding any heat transfer through the \)0.5$-mm- high side surfaces, determine the temperature difference between the front and back surfaces of the chip in steady operation.

Short answer

Answer: The temperature difference between the front and back surfaces of the silicon chip is approximately 0.32 K.

Step-by-step solution

Convert given values into SI units

First, we need to make sure all dimensions are in SI units. The chip's thickness and dimensions should be expressed in meters, so we need to convert them from millimeters. The dimensions of the chip: - Length: 6 mm = 0.006 m - Width: 6 mm = 0.006 m - Thickness: 0.5 mm = 0.0005 m

Calculate the area through which heat is conducted

Next, compute the surface area of the chip: Area (A) = Length * Width = 0.006 m * 0.006 m = 0.000036 m^2

Use Fourier's law to find the temperature difference

Now use the Fourier's law of heat conduction: \(Q = k * A * \frac{\Delta T}{d}\) Rearrange the equation to solve for \(\Delta T\): \(\Delta T = \frac{Q * d}{k * A}\) Plug in the values: \(\Delta T = \frac{3 W * 0.0005 m}{130 W/m*K * 0.000036 m^2} = \frac{0.0015}{0.00468} = 0.320512\,\text{K}\)

Present the final result

Therefore, the temperature difference between the front and back surfaces of the silicon chip in steady operation is about 0.32 K.