/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 In the thermal processing of sem... [FREE SOLUTION] | 91Ó°ÊÓ

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In the thermal processing of semiconductor materials, annealing is accomplished by heating a silicon wafer according to a temperature-time recipe and then maintaining a fixed elevated temperature for a prescribed period of time. For the process tool arrangement shown as follows, the wafer is in an evacuated chamber whose walls are maintained at \(27^{\circ} \mathrm{C}\) and within which heating lamps maintain a radiant flux \(q_{s}^{\prime \prime}\) at its upper surface. The wafer is \(0.78 \mathrm{~mm}\) thick, has a thermal conductivity of \(30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and an emissivity that equals its absorptivity to the radiant flux \(\left(\varepsilon=\alpha_{l}=0.65\right.\) ). For \(q_{s}^{\prime \prime}=3.0 \times 10^{5} \mathrm{~W} / \mathrm{m}^{2}\), the temperature on its lower surface is measured by a radiation thermometer and found to have a value of \(T_{w, l}=997^{\circ} \mathrm{C}\). To avoid warping the wafer and inducing slip planes in the crystal structure, the temperature difference across the thickness of the wafer must be less than \(2^{\circ} \mathrm{C}\). Is this condition being met?

Short Answer

Expert verified
The temperature difference across the wafer is calculated to be 5.07°C, which is greater than the allowed 2°C. Therefore, the condition is not being met, and there is a risk of warping the wafer and inducing slip planes in the crystal structure.

Step by step solution

01

Calculate the absorbed heat flux

The wafer absorbs the heat from the radiant flux. We can calculate the heat absorbed per unit area, \(q_{abs}^{\prime\prime}\), by using the expression: \(q_{abs}^{\prime\prime} = \alpha_{l} \times q_{s}^{\prime\prime}\) where \(\alpha_{l}\) is the absorptivity, and \(q_{s}^{\prime\prime}\) is the radiant flux.
02

Calculate the heat conducted across the wafer

The steady-state heat conduction equation is given by: \(q_{cond}^{\prime\prime} = k \frac{dT}{dx}\) where \(k\) is the thermal conductivity, \(dT\) is the temperature difference across the wafer thickness, and \(dx\) is the wafer thickness. We can solve for the temperature difference as: \(dT = \frac{q_{cond}^{\prime\prime} \times dx}{k}\) In this case, since the wafer absorbs the heat and conducts it across the thickness, we can write: \(q_{abs}^{\prime\prime} = q_{cond}^{\prime\prime}\)
03

Calculate the upper surface temperature

Rearranging the equation for the temperature difference, we can find the temperature of the upper surface, \(T_{w,u}\): \(T_{w,u} = T_{w,l} - dT\) Where \(T_{w,l}\) is the temperature of the lower surface.
04

Check if the temperature difference condition is met

Now let's calculate the temperature difference between the lower and the upper surface of the wafer and check if it is less than 2°C. First, we calculate the absorbed heat flux: \(q_{abs}^{\prime\prime} = 0.65 \times 3.0 \times 10^{5} \mathrm{~W/m}^2 = 1.95 \times 10^{5} \mathrm{~W/m}^2\) Next, we calculate the temperature difference: \(dT = \frac{q_{abs}^{\prime\prime} \times 0.78 \times 10^{-3} \mathrm{m}}{30 \mathrm{~W/m\cdot K}} = \frac{1.95 \times 10^{5} \mathrm{~W/m}^2 \times 0.78 \times 10^{-3} \mathrm{m}}{30 \mathrm{~W/m\cdot K}} = 5.07\) Finally, we calculate the upper surface temperature: \(T_{w,u} = 997 - 5.07 = 991.93\) The temperature difference is: \(dT = T_{w,l} - T_{w,u} = 997 - 991.93 = 5.07\) Since the temperature difference across the wafer is 5.07°C, which is greater than 2°C, the condition is not being met, and there is a risk of warping the wafer and inducing slip planes in the crystal structure.

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Key Concepts

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

Radiant Heat Transfer
In the world of semiconductor manufacturing, the annealing process involves careful heating techniques where radiant heat transfer plays a crucial role. Imagine the sun's rays warming your face; that's radiant heat transfer at work. In the case of semiconductor wafers, it involves the direct transfer of heat energy from heating lamps to the wafer's surface.

Radiant heat transfer is critical in ensuring that the wafer reaches the desired temperature quickly and uniformly without direct contact with the heat source. The efficiency of this process depends on the emissivity and absorptivity of the wafer's material, which in this scenario is represented by the parameter \( \varepsilon=\alpha_{l}=0.65 \). This means that 65% of the incident radiant flux is absorbed by the wafer, directly influencing the wafer's temperature. Significant to our discussion, the absorbed radiant heat flux, \( q_{abs}^{\prime\prime} \) can be determined using the formula \( q_{abs}^{\'\'} = \alpha_{l} \times q_{s}^{''} \), indicating how the intrinsic properties of the material couple with the incident energy to define the starting point for heat distribution across the wafer.
Thermal Conductivity
Once the heat is absorbed by the wafer's surface due to radiant heat transfer, it doesn't just stay there; it travels. The vehicle for this journey is thermal conductivity, a material's ability to conduct heat. Think of it as a measure of how well the wafer's 'thermal roadways' can carry the heat from point A to point B.

In the context of our semiconductor annealing problem, thermal conductivity, denoted as \( k \), is the property that determines how efficiently heat is conducted across the wafer's thickness. With a thermal conductivity of \( 30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \), the wafer allows for a certain rate of heat flow which maintains the temperature uniformity critical to preventing warping and defects. Understanding \( q_{cond}^{\prime\prime} = k \frac{dT}{dx} \), where \( dx \) is the thickness and \( dT \) is the temperature difference across the wafer, provides insights on the wafer's ability to maintain structural integrity during the annealing process by keeping the temperature gradient in check.
Temperature Gradient
The temperature gradient within a semiconductor wafer during annealing is like a slope, indicating how steeply the temperature changes from one point to another within the wafer. This gradient is crucial as it can lead to thermal stresses that may cause warping or even crystal structure defects if it's too steep.

In our exercise, we are concerned with maintaining a temperature gradient less than \( 2^{\circ} \mathrm{C} \), across the wafer's thickness to preserve its integrity. The temperature gradient can be computed by dividing the temperature difference by the wafer thickness using the relation \( dT = \frac{q_{cond}^{\prime\prime} \times dx}{k} \). The calculation in the solution indicated a temperature difference of \( 5.07^{\circ} \mathrm{C} \), exceeding the allowable limit, which points towards the need for optimizing the annealing conditions to reduce the temperature gradient and prevent potential damage to the wafer.

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Most popular questions from this chapter

The temperature controller for a clothes dryer consists of a bimetallic switch mounted on an electrical heater attached to a wall-mounted insulation pad. The switch is set to open at \(70^{\circ} \mathrm{C}\), the maximum dryer air temperature. To operate the dryer at a lower air temperature, sufficient power is supplied to the heater such that the switch reaches \(70^{\circ} \mathrm{C}\left(T_{\text {set }}\right)\) when the air temperature \(T\) is less than \(T_{\text {set. }}\). If the convection heat transfer coefficient between the air and the exposed switch surface of \(30 \mathrm{~mm}^{2}\) is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how much heater power \(P_{e}\) is required when the desired dryer air temperature is \(T_{\infty}=50^{\circ} \mathrm{C}\) ?

You've experienced convection cooling if you've ever extended your hand out the window of a moving vehicle or into a flowing water stream. With the surface of your hand at a temperature of \(30^{\circ} \mathrm{C}\), determine the convection heat flux for (a) a vehicle speed of \(35 \mathrm{~km} / \mathrm{h}\) in air at \(-5^{\circ} \mathrm{C}\) with a convection coefficient of 40 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and (b) a velocity of \(0.2 \mathrm{~m} / \mathrm{s}\) in a water stream at \(10^{\circ} \mathrm{C}\) with a convection coefficient of \(900 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Which condition would feel colder? Contrast these results with a heat loss of approximately \(30 \mathrm{~W} / \mathrm{m}^{2}\) under normal room conditions.

A freezer compartment consists of a cubical cavity that is \(2 \mathrm{~m}\) on a side. Assume the bottom to be perfectly insulated. What is the minimum thickness of styrofoam insulation \((k=0.030 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that must be applied to the top and side walls to ensure a heat load of less than \(500 \mathrm{~W}\), when the inner and outer surfaces are \(-10\) and \(35^{\circ} \mathrm{C}\) ?

The diameter and surface emissivity of an electrically heated plate are \(D=300 \mathrm{~mm}\) and \(\varepsilon=0.80\), respectively. (a) Estimate the power needed to maintain a surface temperature of \(200^{\circ} \mathrm{C}\) in a room for which the air and the walls are at \(25^{\circ} \mathrm{C}\). The coefficient characterizing heat transfer by natural convection depends on the surface temperature and, in units of \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), may be approximated by an expression of the form \(h=0.80\left(T_{s}-T_{\infty}\right)^{1 / 3}\). (b) Assess the effect of surface temperature on the power requirement, as well as on the relative contributions of convection and radiation to heat transfer from the surface.

An aluminum plate \(4 \mathrm{~mm}\) thick is mounted in a horizontal position, and its bottom surface is well insulated. A special, thin coating is applied to the top surface such that it absorbs \(80 \%\) of any incident solar radiation, while having an emissivity of \(0.25\). The density \(\rho\) and specific heat \(c\) of aluminum are known to be \(2700 \mathrm{~kg} / \mathrm{m}^{3}\) and \(900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. (a) Consider conditions for which the plate is at a temperature of \(25^{\circ} \mathrm{C}\) and its top surface is suddenly exposed to ambient air at \(T_{\infty}=20^{\circ} \mathrm{C}\) and to solar radiation that provides an incident flux of \(900 \mathrm{~W} / \mathrm{m}^{2}\). The convection heat transfer coefficient between the surface and the air is \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the initial rate of change of the plate temperature? (b) What will be the equilibrium temperature of the plate when steady-state conditions are reached? (c) The surface radiative properties depend on the specific nature of the applied coating. Compute and plot the steady-state temperature as a function of the emissivity for \(0.05 \leq \varepsilon \leq 1\), with all other conditions remaining as prescribed. Repeat your calculations for values of \(\alpha_{S}=0.5\) and \(1.0\), and plot the results with those obtained for \(\alpha_{S}=0.8\). If the intent is to maximize the plate temperature, what is the most desirable combination of the plate emissivity and its absorptivity to solar radiation?

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