Aluminum atoms are to be diffused into a silicon wafer using both predeposition and drive-in heat treatments; the background concentration of \(\mathrm{Al}\) in this silicon material is known to be \(3 \times 10^{19}\) atoms/m \(^{3}\). The drive-in diffusion treatment is to be carried out at \(1050^{\circ} \mathrm{C}\) for a period of \(4.0 \mathrm{~h}\), which gives a junction depth \(x_{j}\) of \(3.0 \mu \mathrm{m}\). Compute the predeposition diffusion time at \(950^{\circ} \mathrm{C}\) if the surface concentration is maintained at a constant level of \(2 \times 10^{25}\) atoms \(/ \mathrm{m}^{3}\). For the diffusion of \(\mathrm{Al}\) in \(\mathrm{Si}\), values of \(Q_{d}\) and \(D_{0}\) are \(3.41\). \(\mathrm{eV} /\) atom and \(1.38 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\), respectively.

Short Answer

Expert verified
For the predeposition process: \begin{equation} D_{1} = D_{0} \exp \left(-\frac{Q_{d}}{kT_{1}}\right) \end{equation} Using the given values, \(D_{0} = 4 \times 10^{-4}\) m²/s and \(Q_{d} = 3.69\) eV, we have: \begin{equation} D_{1} = 4 \times 10^{-4} \exp\left(-\frac{3.69}{8.62 \times 10^{-5} \times 1223}\right) \end{equation} \begin{equation} D_{1} \approx 3.33 \times 10^{-15}\, \text{m}^2\text{/s} \end{equation} For the drive-in process: \begin{equation} D_{2} = D_{0} \exp \left(-\frac{Q_{d}}{kT_{2}}\right) \end{equation} \begin{equation} D_{2} = 4 \times 10^{-4} \exp\left(-\frac{3.69}{8.62 \times 10^{-5} \times 1323}\right) \end{equation} \begin{equation} D_{2} \approx 1.11 \times 10^{-14}\, \text{m}^2\text{/s} \end{equation} Now that we have the diffusion coefficients for both processes, we can use the provided junction depth information to find the predeposition diffusion time.

Step by step solution

01

Set up the Arrhenius equation to calculate the diffusion coefficients

First, we need to calculate the diffusion coefficients for both predeposition and drive-in processes (\(D_{1}\) and \(D_{2}\)). For this, we will use the Arrhenius equation: \begin{equation} D = D_{0} \exp \left(-\frac{Q_{d}}{kT}\right) \end{equation} In this equation, \(k\) is the Boltzmann constant, which is \(8.62 \times 10^{-5}\) eV/K. Both \(Q_{d}\) and \(D_{0}\) are given in the problem. Recall that the temperatures for predeposition and drive-in processes are \(950^{\circ}C\) and \(1050^{\circ}C\), respectively. We first convert them to Kelvin (\(K = °C + 273\)): 1. For predeposition process: \(T_{1} = 950 + 273 = 1223K\) 2. For drive-in process: \(T_{2} = 1050 + 273 = 1323K\) Now, we can calculate the diffusion coefficients (\(D_{1}\) and \(D_{2}\)).

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Key Concepts

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

Arrhenius Equation
The Arrhenius equation is crucial for understanding how temperature affects the rate processes like diffusion in materials such as silicon. It tells us that the rate of a chemical process increases exponentially with temperature.

Mathematically expressed, the Arrhenius equation is: \[ D = D_0 \times e^{\frac{-Q_d}{kT}} \]In this formula,
  • \(D\) is the diffusion coefficient at the temperature \(T\),
  • \(D_0\) is the pre-exponential factor or the diffusion coefficient at infinite temperature,
  • \(Q_d\) is the activation energy for diffusion, and
  • \(k\) is the Boltzmann constant.
The \(e^{\frac{-Q_d}{kT}}\) term shows the exponential relationship between the diffusion rate and temperature. A higher temperature \(T\) means that the exponent is less negative, increasing the diffusion coefficient \(D\), and thus the rate at which atoms, such as Aluminum in our exercise, will spread out into the silicon lattice.

This relation is instrumental not just in theoretical calculations, but also for practitioners designing semiconductor devices where precise control of diffusion processes is necessary for functionality.
Predeposition Process
The predeposition process in silicon doping refers to the initial step where dopants (such as Aluminum) are introduced onto the silicon surface to achieve a desired surface concentration. This is typically done at a lower temperature than the subsequent drive-in process to allow dopants to settle on the surface without driving them deep into the silicon wafer.

A steady-state condition is assumed where the surface concentration remains constant during the process. The aluminum atoms will diffuse from the high concentration on the surface into the silicon to create a gradient, which is how materials naturally even out concentration differences.

Using the Arrhenius equation, you can calculate the required predeposition time to achieve a certain concentration at a specific depth by taking into account the diffusion coefficient \(D\) at the process temperature. In the exercise, a diffusion time calculation is required at \(950^{\text{\textdegree}}C\), with a known constant surface concentration, for the dopant atoms to settle just at the surface of the silicon wafer before the drive-in process ensues.
Drive-in Heat Treatment
After the predeposition phase, the 'drive-in' heat treatment follows. This is the step where the dopant atoms that have been deposited on the surface of the silicon are driven deeper into the material. This process uses a higher temperature than predeposition to increase the diffusion rate as per the Arrhenius equation.

The purpose of drive-in is to achieve a desired dopant profile throughout the silicon wafer, which is critical for creating junctions in semiconductor devices. The longer and hotter the drive-in process, the deeper the dopants diffuse into the silicon.

In our problem's context, the drive-in diffusion treatment at \(1050^{\text{\textdegree}}C\) for 4 hours results in a junction depth of \(3.0 \text{\textmu}m\). The relationship between time, temperature, and depth achieved is again predictable by the Arrhenius equation and understanding of the physics underlying diffusion in solid materials.
Diffusion Coefficient
The diffusion coefficient, denoted by \(D\), is a measure of how quickly atoms or molecules can move through a material. It's a critical parameter in the semiconductor manufacturing process because it influences how dopants spread within the silicon wafer.

In our exercise, the diffusion coefficient depends on the temperature and must be calculated for both predeposition and drive-in processes using the Arrhenius equation. The coefficient \(D\) changes with temperature because atomic vibrations and movements are more vigorous at higher temperatures, creating more opportunities for atoms to hop into adjacent positions in the crystalline silicon lattice.

The pre-exponential factor \(D_0\) and the activation energy for diffusion \(Q_d\) are material-specific constants that should be determined experimentally. They form part of the Arrhenius equation, which allows us to compute the diffusion coefficient for any temperature, providing that the temperature is above absolute zero and within a range where the material's properties do not change significantly.

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Most popular questions from this chapter

A sheet of BCC iron \(1 \mathrm{~mm}\) thick was exposed to a carburizing gas atmosphere on one side and a decarburizing atmosphere on the other side at \(725^{\circ} \mathrm{C}\). After reaching steady state, the iron was quickly cooled to room temperature. The carbon concentrations at the two surfaces of the sheet were determined to be \(0.012\) and \(0.0075 \mathrm{wt} \%\). Compute the diffusion coefficient if the diffusion flux is \(1.4 \times 10^{-8} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\). Hint: Use Equation \(4.9\) to convert the concentrations from weight percent to kilograms of carbon per cubic meter of iron.

Consider a diffusion couple composed of two semi-infinite solids of the same metal, and that each side of the diffusion couple has a different concentration of the same elemental impurity; furthermore, assume each impurity level is constant throughout its side of the diffusion couple. For this situation, the solution to Fick's second law (assuming that the diffusion coefficient for the impurity is independent of concentration) is as follows: $$ C_{x}=\left(\frac{C_{1}+C_{2}}{2}\right)-\left(\frac{C_{1}-C_{2}}{2}\right) \operatorname{erf}\left(\frac{x}{2 \sqrt{D t}}\right) $$ In this expression, when the \(x=0\) position is taken as the initial diffusion couple interface, then \(C_{1}\) is the impurity concentration for \(x<0\); likewise, \(C_{2}\) is the impurity content for \(x>0\). A diffusion couple composed of two silver-gold alloys is formed; these alloys have compositions of \(98 \mathrm{wt} \% \mathrm{Ag}-2 \mathrm{wt} \% \mathrm{Au}\) and \(95 \mathrm{wt} \% \mathrm{Ag}-5 \mathrm{wt} \%\) Au. Determine the time this diffusion couple must be heated at \(750^{\circ} \mathrm{C}\) (1023 K) in order for the composition to be \(2.5 \mathrm{wt} \% \mathrm{Au}\) at the \(50 \mu \mathrm{m}\) position into the 2 wt \(\%\) Au side of the diffusion couple. Preexponential and activation energy values for Au diffusion in \(\mathrm{Ag}\) are \(8.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) and \(202,100 \mathrm{~J} / \mathrm{mol}\), respectively.

(a) Briefly explain the concept of a driving force. (b) What is the driving force for steady-state diffusion?

The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of \(0.20 \mathrm{~m}^{2}\) at \(500^{\circ} \mathrm{C}\). Assume a diffusion coefficient of \(1.0 \times 10^{-8} \mathrm{~m}^{2} / \mathrm{s}\), that the concentrations at the high- and low-pressure sides of the plate are \(2.4\) and \(0.6 \mathrm{~kg}\) of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.

The diffusion coefficients for silver in copper are given at two temperatures: $$ \begin{array}{cc} \hline T\left({ }^{\circ} \mathrm{C}\right) & D\left(\mathrm{~m}^{2} / \mathrm{s}\right) \\ \hline 650 & 5.5 \times 10^{-16} \\ 900 & 1.3 \times 10^{-13} \\ \hline \end{array} $$ (a) Determine the values of \(D_{0}\) and \(Q_{d}\). (b) What is the magnitude of \(D\) at \(875^{\circ} \mathrm{C}\) ?

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