Question

Heat treatment is common in processing of semiconductor material. A 200 -mm- diameter silicon wafer with thickness of \(725 \mu \mathrm{m}\) is being heat treated in a vacuum chamber by infrared heat. The surrounding walls of the chamber have a uniform temperature of \(310 \mathrm{~K}\). The infrared heater provides an incident radiation flux of \(200 \mathrm{~kW} / \mathrm{m}^{2}\) on the upper surface of the wafer, and the emissivity and absorptivity of the wafer surface are 0.70. Using a pyrometer, the lower surface temperature of the wafer is measured to be \(1000 \mathrm{~K}\). Assuming there is no radiation exchange between the lower surface of the wafer and the surroundings, determine the upper surface temperature of the wafer. (Note: A pyrometer is a noncontacting device that intercepts and measures thermal radiation. This device can be used to determine the temperature of an object's surface.)

Short answer

Answer: The upper surface temperature of the wafer is approximately 1015 K.

Step-by-step solution

Calculate the power absorbed by the wafer's upper surface

To determine the power absorbed by the upper surface of the wafer, we can use the equation: \(P_{abs} = q \times A \times \alpha\) Where: \(P_{abs}\) - The power absorbed by the wafer's upper surface \(q\) - The incident radiation flux \(A\) - The area of the wafer's upper surface \(\alpha\) - The absorptivity of the wafer's surface First, we need to find the area of the wafer's upper surface using the diameter given: \(A = \pi \times (d/2)^{2}\) \(A = \pi \times (0.2/2)^{2}\) \(A = \pi \times (0.1)^{2}\) \(A = 0.0314 \text{ m}^2\) Now we can calculate the power absorbed: \(P_{abs} = 200 \text{ kW/m}^{2} \times 0.0314 \text{ m}^{2} \times 0.70\) \(P_{abs} = 4.388 \text{ kW}\)

Calculate the emitted radiation from the lower surface of the wafer

Due to the radiation exchange is absent, the lower surface of the wafer emits Stefan-Boltzmann radiation, which we can calculate with the equation: \(P_{emission} = \epsilon \times A \times \sigma \times T_{lower}^4\) Where: \(P_{emission}\) - The emitted radiation from the lower surface of the wafer \(\epsilon\) - The emissivity of the wafer's surface \(\sigma\) - Stefan-Boltzmann constant (5.67 x $10^{-8} \text{ W/m}^2\text{K}^4) \(T_{lower}\) - The temperature of the lower surface of the wafer (1000 K) \(P_{emission} = 0.70 \times 0.0314 \text{ m}^{2} \times 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \times (1000 \text{ K})^4\) \(P_{emission} = 3.978 \text{ kW}\)

Calculate the net power transfer between the upper and lower surfaces

Now that we have both the absorbed power from the infrared heater and the emitted power from the lower surface, we can calculate the net power transfer between the upper and lower surfaces. \(P_{net} = P_{abs} - P_{emission}\) \(P_{net} = 4.388 \text{ kW} - 3.978 \text{ kW}\) \(P_{net} = 0.41 \text{ kW}\)

Determine the upper surface temperature of the wafer

To find the upper surface temperature, we'll use the equation for radiation: \(P_{net} = \epsilon \times A \times \sigma \times (T_{upper}^4 - T_{lower}^4)\) Solve for \(T_{upper}\): \(T_{upper}^4 = (P_{net} / (\epsilon \times A \times \sigma)) + T_{lower}^4\) \(T_{upper}^4 = (0.41 \text{ kW} / (0.70 \times 0.0314 \text{ m}^{2} \times 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4)) + (1000 \text{ K})^4\) \(T_{upper}^4 = 5275.7 \times 10^3\) \(T_{upper} = \sqrt[4]{5275.7 \times 10^3}\) \(T_{upper} \approx 1015 \text{ K}\) The upper surface temperature of the wafer is approximately 1015 K.