Antimony atoms are to be diffused into a silicon wafer using both predeposition and drive-in heat treatments; the background concentration of \(\mathrm{Sb}\). in this silicon material is known to be \(2 \times 10^{20}\) atoms/m^3. The predeposition treatment is to be conducted at \(900^{\circ} \mathrm{C}\) for \(1 \mathrm{~h}\); the surface concentration of \(\mathrm{Sb}\) is to be maintained at a constant level of \(8.0 \times 10^{25}\) atoms \(/ \mathrm{m}^{3}\). Drive-in diffusion will be carried out at \(1200^{\circ} \mathrm{C}\) for a period of \(1.75 \mathrm{~h}\). For the diffusion of \(\mathrm{Sb}\) in Si, values of \(Q_{d}\) and \(D_{0}\) are \(3.65 \mathrm{eV} /\) atom and \(2.14 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), respectively. (a) Calculate the value of \(Q_{0}\) - (b) Determine the value of \(x_{j}\) for the drive-in diffusion treatment. (c) Also, for the drive-in treatment, compute the position \(x\) at which the concentration of \(S b\) atoms is \(5 \times 10^{23}\) atoms/m \(^{3}\)

#Answers# #tag_title# Answer for (a) #tag_content# Calculating the values for the diffusivity: \(D_1 = 2.14 \times 10^{-5} \mathrm{m}^2/\mathrm{s} \times e^{-\frac{3.65 \mathrm{eV}}{8.6173 \times 10^{-5} \mathrm{eV/K} \times 1173.15 \mathrm{K}}} = 1.13 \times 10^{-14}\) m\(^2\)/s \(D_2 = 2.14 \times 10^{-5} \mathrm{m}^2/\mathrm{s} \times e^{-\frac{3.65 \mathrm{eV}}{8.6173 \times 10^{-5} \mathrm{eV/K} \times 1473.15 \mathrm{K}}} = 6.13 \times 10^{-14}\) m\(^2\)/s Now, solving for \(x\): \(\frac{x}{2\sqrt{D_1t_1}} = erf^{-1}(\frac{N_b}{N_0})\) \(x = 2\sqrt{D_1t_1} \times erf^{-1}(\frac{N_b}{N_0})\) With \(D_1 = 1.13 \times 10^{-14} \mathrm{m}^2/\mathrm{s}\), \(t_1 = 3600 \thinspace \mathrm{s}\), \(N_b = 2 \times 10^{20}\) atoms/m\(^3\), and \(N_0 = 5 \times 10^{23}\) atoms/m\(^3\), we can find \(x\): \(x = 2\sqrt{(1.13 \times 10^{-14} \mathrm{m}^2/\mathrm{s})(3600 \thinspace \mathrm{s})} \times erf^{-1}(\frac{2 \times 10^{20}}{5 \times 10^{23}}) \approx 2.78 \times 10^{-5}\) m Then, calculating the value of \(Q_0\): \(Q_0 = 5 \times 10^{23} \thinspace erf(\frac{2.78 \times 10^{-5}}{2\sqrt{(1.13 \times 10^{-14})(3600)}})- 2 \times 10^{20} \approx 4.52 \times 10^{23}\) atoms/m\(^3\) #tag_title# Answer for (b) #tag_content# Next, calculating the value of \(x_j\): \(\frac{x_j}{2\sqrt{D_2t_2}} = erf^{-1}(\frac{N_b}{Q_0})\) \(x_j = 2\sqrt{D_2t_2} \times erf^{-1}(\frac{N_b}{Q_0})\) Using \(D_2 = 6.13 \times 10^{-14} \mathrm{m}^2/\mathrm{s}\), \(t_2 = 6300 \thinspace \mathrm{s}\), \(N_b = 2 \times 10^{20}\) atoms/m\(^3\), and \(Q_0 = 4.52 \times 10^{23}\) atoms/m\(^3\), we find \(x_j\): \(x_j = 2\sqrt{(6.13 \times 10^{-14} \mathrm{m}^2/\mathrm{s})(6300 \thinspace \mathrm{s})} \times erf^{-1}(\frac{2 \times 10^{20}}{4.52 \times 10^{23}}) \approx 6.02 \times 10^{-4}\) m #tag_title# Answer for (c) #tag_content# Finally, calculating the value of \(x\) for a concentration of \(5 \times 10^{23}\) atoms/m\(^3\) during drive-in treatment: \(\frac{x}{2\sqrt{D_2t_2}} = erf^{-1}(\frac{5 \times 10^{23}}{Q_0})\) Using \(D_2 = 6.13 \times 10^{-14} \mathrm{m}^2/\mathrm{s}\), \(t_2 = 6300 \thinspace \mathrm{s}\), and \(Q_0 = 4.52 \times 10^{23}\) atoms/m\(^3\), we calculate \(x\): \(x = 2\sqrt{(6.13 \times 10^{-14} \mathrm{m}^2/\mathrm{s})(6300 \thinspace \mathrm{s})} \times erf^{-1}(\frac{5 \times 10^{23}}{4.52 \times 10^{23}}) \approx 1.37 \times 10^{-3}\) m