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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 heater. 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 non-contacting device that intercepts and measures thermal radiation. This device can be used to determine the temperature of an object's surface.)

Short Answer

Expert verified
Based on the given information and the radiative heat transfer calculations, the upper surface temperature of the silicon wafer in the vacuum chamber is approximately 1099.6 K.

Step by step solution

01

Calculate radiative heat absorbed by the wafer

The radiative heat absorbed by the wafer can be calculated using the incident radiation flux and the absorptivity of the wafer surface. \(q_{absorbed} = q_{incident} \times A \times \alpha\) Where: \(q_{absorbed}\) = radiative heat absorbed by the wafer (W) \(q_{incident}\) = incident radiation flux (W/m²) \(A\) = wafer surface area (m²) \(\alpha\) = absorptivity of the wafer surface First, calculate the surface area of the wafer. \(A = \pi d^2 / 4\) \(A = \pi (0.2)^2 / 4\) \(A = 0.0314 \mathrm{m}^2\) Now, calculate the radiative heat absorbed by the wafer. \(q_{absorbed} = 200\times10^3 W/m^2 \times 0.0314 m^2 \times 0.70\) \(q_{absorbed} = 4396 \mathrm{~W}\)
02

Calculate radiative heat emitted by the wafer

We know that 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 this information, we can calculate the radiative heat emitted by the wafer using the Stefan-Boltzmann law. \(q_{emitted} = \epsilon \sigma A (T^4_{upper} - T^4_{lower})\) Where: \(q_{emitted}\) = radiative heat emitted by the wafer (W) \(\epsilon\) = emissivity of the wafer surface \(\sigma\) = Stefan-Boltzmann constant (\(5.67\times10^{-8} W/m^2K^4\)) \(T_{upper}\) = upper surface temperature (K) \(T_{lower}\) = lower surface temperature (K)
03

Equate the radiative heat absorbed and emitted by the wafer

Since there is no radiation exchange between the lower surface of the wafer and the surroundings, the radiative heat absorbed by the wafer must be equal to the radiative heat emitted by the wafer. \(q_{absorbed} = q_{emitted}\)
04

Solve for the upper surface temperature

Substitute the expressions for \(q_{absorbed}\) and \(q_{emitted}\) into the equation above and solve for the upper surface temperature \(T_{upper}\). \(4396 = 0.70 \times 5.67\times10^{-8} \times 0.0314 (T_{upper}^4 - 1000^4)\) Divide both sides by \(0.7\times 5.67\times10^{-8}\times 0.0314\): \(T_{upper}^4 - 1000^4 = 4396/(0.7\times 5.67\times10^{-8}\times 0.0314)\) Compute the right-hand side: \(T_{upper}^4 - 1000^4 = 2733480\) Add \(1000^4\) to both sides: \(T_{upper}^4 = 2733480 + 1000^4\) Take the fourth root of both sides: \(T_{upper} = \sqrt[4]{2733480 + 1000^4}\) \(T_{upper} = 1099.6 \mathrm{~K}\) The upper surface temperature of the wafer is \(1099.6 \mathrm{~K}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Treatment of Semiconductor
Heat treatment is a critical process in the manufacturing and conditioning of semiconductor materials. It generally involves the controlled heating and cooling of the semiconductor to alter its physical and perhaps structural properties, and is essential in processes such as annealing, doping, and oxidation. The 200-mm-diameter silicon wafer in the exercise undergoes such a treatment, with the precise control of temperature being crucial for achieving the desired material characteristics.

Uniform and controlled heating ensures minimal thermal stress and prevents defects in the crystal structure of the wafer. In a vacuum chamber, the presence of infrared heating helps to achieve these temperatures without physical contact, which can contaminate or otherwise harm the delicate semiconductor material.
Stefan-Boltzmann Law
The Stefan-Boltzmann law is a fundamental principle in thermodynamics and is critically applied in calculating the radiative heat transfer for the given semiconductor wafer. It states that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's absolute temperature. In the equation
ewline ewline \(\textbackslash q{\textbackslash text{\textunderscore emitted}} = \textbackslash epsilon \textbackslash sigma A (T{\textbackslash text{\textunderscore upper}}{\textbackslash text{\textasciicircum 4}} - T{\textbackslash text{\textunderscore lower}}{\textbackslash text{\textasciicircum 4}})\), the \textbackslash sigma represents the Stefan-Boltzmann constant, and the equation describes how much energy is emitted from the wafer. This law is critical in understanding the energy balance required to determine the upper surface temperature when the wafer is subject to infrared heating.
Pyrometer Temperature Measurement
Pyrometers are sophisticated instruments used to measure temperature without direct contact with the object being measured. They work by intercepting and quantifying thermal radiation produced by the object. This characteristic makes pyrometers particularly useful in situations where temperatures are too high for contact sensors or where contact with the sensor might introduce contamination or affect the reading. In the provided exercise, a pyrometer is used to measure the lower surface temperature of the silicon wafer during heat treatment. Accurate temperature measurement is vital, as the semiconductor's treatment process depends greatly on precise thermal conditions, and the pyrometer allows for the non-invasive monitoring of these conditions.
Infrared Heating
Infrared heating is a method of transferring energy via electromagnetic waves in the infrared spectrum. It's an efficient way to deliver heat, especially for purposes such as the heat treatment of semiconductors, because it can provide uniform heating across an object without direct contact. This heating method is particularly beneficial in controlled environments like vacuum chambers, where contamination is a concern, and semiconductors require precise temperature management. Infrared heaters emit radiation that is partly absorbed by the semiconductor's surface, is then converted into heat, and ultimately raises the temperature of the wafer's material to the required levels. The incident radiation flux mentioned in the problem is a measure of the power of this infrared radiation.

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Most popular questions from this chapter

Consider a person standing in a room maintained at \(20^{\circ} \mathrm{C}\) at all times. The inner surfaces of the walls, floors, and ceiling of the house are observed to be at an average temperature of \(12^{\circ} \mathrm{C}\) in winter and \(23^{\circ} \mathrm{C}\) in summer. Determine the rates of radiation heat transfer between this person and the surrounding surfaces in both summer and winter if the exposed surface area, emissivity, and the average outer surface temperature of the person are \(1.6 \mathrm{~m}^{2}, 0.95\), and \(32^{\circ} \mathrm{C}\), respectively.

Consider heat loss through the two walls of a house on a winter night. The walls are identical, except that one of them has a tightly fit glass window. Through which wall will the house lose more heat? Explain.

A cold bottled drink ( \(\left.m=2.5 \mathrm{~kg}, c_{p}=4200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(5^{\circ} \mathrm{C}\) is left on a table in a room. The average temperature of the drink is observed to rise to \(15^{\circ} \mathrm{C}\) in 30 minutes. The average rate of heat transfer to the drink is (a) \(23 \mathrm{~W}\) (b) \(29 \mathrm{~W}\) (c) \(58 \mathrm{~W}\) (d) \(88 \mathrm{~W}\) (e) \(122 \mathrm{~W}\)

An aluminum pan whose thermal conductivity is \(237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) has a flat bottom with diameter \(15 \mathrm{~cm}\) and thickness \(0.4 \mathrm{~cm}\). Heat is transferred steadily to boiling water in the pan through its bottom at a rate of \(1400 \mathrm{~W}\). If the inner surface of the bottom of the pan is at \(105^{\circ} \mathrm{C}\), determine the temperature of the outer surface of the bottom of the pan.

A soldering iron has a cylindrical tip of \(2.5 \mathrm{~mm}\) in diameter and \(20 \mathrm{~mm}\) in length. With age and usage, the tip has oxidized and has an emissivity of \(0.80\). Assuming that the average convection heat transfer coefficient over the soldering iron tip is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the surrounding air temperature is \(20^{\circ} \mathrm{C}\), determine the power required to maintain the tip at \(400^{\circ} \mathrm{C}\).

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