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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

What is the thickness required of a masonry wall having thermal conductivity \(0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) if the heat rate is to be \(80 \%\) of the heat rate through a composite structural wall having a thermal conductivity of \(0.25 \mathrm{~W} / \mathrm{m}+\mathrm{K}\) and a thickness of \(100 \mathrm{~mm}\) ? Both walls are subjected to the same surface temperature difference.

A wall is made from an inhomogeneous (nonuniform) material for which the thermal conductivity varies through the thickness according to \(k=a x+b\), where \(a\) and \(b\) are constants. The heat flux is known to be constant. Determine expressions for the temperature gradient and the temperature distribution when the surface at \(x=0\) is at temperature \(T_{1}\).

Electronic power devices are mounted to a heat sink having an exposed surface area of \(0.045 \mathrm{~m}^{2}\) and an emissivity of \(0.80\). When the devices dissipate a total power of \(20 \mathrm{~W}\) and the air and surroundings are at \(27^{\circ} \mathrm{C}\), the average sink temperature is \(42^{\circ} \mathrm{C}\). What average temperature will the heat sink reach when the devices dissipate \(30 \mathrm{~W}\) for the same environmental condition?

Consider a surface-mount type transistor on a circuit board whose temperature is maintained at \(35^{\circ} \mathrm{C}\). Air at \(20^{\circ} \mathrm{C}\) flows over the upper surface of dimensions \(4 \mathrm{~mm} \times\) \(8 \mathrm{~mm}\) with a convection coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Three wire leads, each of cross section \(1 \mathrm{~mm} \times 0.25 \mathrm{~mm}\) and length \(4 \mathrm{~mm}\), conduct heat from the case to the circuit board. The gap between the case and the board is \(0.2 \mathrm{~mm}\). (a) Assuming the case is isothermal and neglecting radiation, estimate the case temperature when \(150 \mathrm{~mW}\) is dissipated by the transistor and (i) stagnant air or (ii) a conductive paste fills the gap. The thermal conductivities of the wire leads, air, and conductive paste are \(25,0.0263\), and \(0.12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. (b) Using the conductive paste to fill the gap, we wish to determine the extent to which increased heat dissipation may be accommodated, subject to the constraint that the case temperature not exceed \(40^{\circ} \mathrm{C}\). Options include increasing the air speed to achieve a larger convection coefficient \(h\) and/or changing the lead wire material to one of larger thermal conductivity. Independently considering leads fabricated from materials with thermal conductivities of 200 and \(400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), compute and plot the maximum allowable heat dissipation for variations in \(h\) over the range \(50 \leq h \leq 250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Under conditions for which the same room temperature is maintained by a heating or cooling system, it is not uncommon for a person to feel chilled in the winter but comfortable in the summer. Provide a plausible explanation for this situation (with supporting calculations) by considering a room whose air temperature is maintained at \(20^{\circ} \mathrm{C}\) throughout the year, while the walls of the room are nominally at \(27^{\circ} \mathrm{C}\) and \(14^{\circ} \mathrm{C}\) in the summer and winter, respectively. The exposed surface of a person in the room may be assumed to be at a temperature of \(32^{\circ} \mathrm{C}\) throughout the year and to have an emissivity of \(0.90\). The coefficient associated with heat transfer by natural convection between the person and the room air is approximately \(2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

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