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Give step-by-step solution with explanation and final answer: wafer of 1 0 Ohm - cm n - type silicon is diffused with Boron.Predeposition is at 1 0 0 0 oC for 2 0 minutes with a diffusion constant of 1 . 5 E - 1 4 cm 2 / sDrive - in is done at 1 1 5 0 oC for 5 0 minutes, with a diffusion constant of 1 . 2 E - 1 3 cm 2 / s . Assume a surface solubility limit of 2 E 2 1 cm - 3 . Find junction depth and surface doping concentration.

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Give step-by-step solution with explanation and final answer: wafer of 1 0 Ohm - cm n - type silicon is diffused with Boron.Predeposition is at 1 0 0 0 oC for 2 0 minutes with a diffusion constant of 1 . 5 E - 1 4 cm 2 / sDrive - in is done at 1 1 5 0 oC for 5 0 minutes, with a diffusion constant of 1 . 2 E - 1 3 cm 2 / s . Assume a surface solubility limit of 2 E 2 1 cm - 3 . Find junction depth and surface doping concentration.

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Asked by: Aman
Created at: 2025-10-30 22:30:42

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# Step-by-Step Solution Let's break down the silicon diffusion problem step by step. ## **Given Data** - **Resistivity of wafer**: \( 10 \ \Omega\cdot\text{cm} \) (n-type Si) - **Predeposition**: - Temperature: \( 100^\circ\text{C} \) - Time: \( 20 \) min \( = 120 \) s - Diffusion constant: \( D_1 = 1.5 \times 10^{-14} \ \text{cm}^2/\text{s} \) - **Drive-in**: - Temperature: \( 115^\circ\text{C} \) - Time: \( 50 \) min \( = 300 \) s - Diffusion constant: \( D_2 = 1.2 \times 10^{-13} \ \text{cm}^2/\text{s} \) - **Boron surface solubility limit**: \( N_s = 2 \times 10^{21} \ \text{cm}^{-3} \) --- ## **Step 1: Calculate Background Dopant Concentration (\(N_D\))** For n-type silicon: \[ \rho = \frac{1}{q \mu_n N_D} \] Where: - \( \rho = 10\ \Omega\cdot\text{cm} \) - \( q = 1.6 \times 10^{-19} \ \text{C} \) - \( \mu_n \) (electron mobility at 300K for Si) ≈ \( 135 \ \text{cm}^2/\text{V}\cdot\text{s} \) \[ N_D = \frac{1}{q \mu_n \rho} = \frac{1}{(1.6 \times 10^{-19})(135)(10)} = \frac{1}{2.16 \times 10^{-15}} \approx 4.63 \times 10^{14} \ \text{cm}^{-3} \] --- ## **Step 2: Predeposition (Constant Surface Concentration)** For predeposition, the total "dose" (\( Q \)) is: \[ Q = 2 N_s \sqrt{\frac{D_1 t_1}{\pi}} \] Where: - \( N_s = 2 \times 10^{21} \ \text{cm}^{-3} \) - \( D_1 = 1.5 \times 10^{-14} \ \text{cm}^2/\text{s} \) - \( t_1 = 120\ \text{s} \) Calculate: \[ \sqrt{\frac{D_1 t_1}{\pi}} = \sqrt{\frac{1.5 \times 10^{-14} \times 120}{3.1416}} = \sqrt{\frac{1.8 \times 10^{-11}}{3.1416}} = \sqrt{5.73 \times 10^{-12}} \approx 7.57 \times 10^{-6} \] So, \[ Q = 2 \times 2 \times 10^{21} \times 7.57 \times 10^{-6} = 4 \times 10^{21} \times 7.57 \times 10^{-6} = 3.03 \times 10^{16} \ \text{cm}^{-2} \] --- ## **Step 3: Drive-in Diffusion (Gaussian Profile)** Now, the total dose \( Q \) is redistributed as a Gaussian: \[ C(x, t) = \frac{Q}{\sqrt{\pi D_2 t_2}} \exp\left(-\frac{x^2}{4 D_2 t_2}\right) \] Where: - \( D_2 = 1.2 \times 10^{-13} \ \text{cm}^2/\text{s} \) - \( t_2 = 300\ \text{s} \) Calculate: \[ \sqrt{\pi D_2 t_2} = \sqrt{3.1416 \times 1.2 \times 10^{-13} \times 300} = \sqrt{3.1416 \times 3.6 \times 10^{-10}} = \sqrt{1.131 \times 10^{-9}} \approx 3.37 \times 10^{-5} \] --- ## **Step 4: Surface Concentration After Drive-in** \[ C_s = \frac{Q}{\sqrt{\pi D_2 t_2}} = \frac{3.03 \times 10^{16}}{3.37 \times 10^{-5}} \approx 8.99 \times 10^{20} \ \text{cm}^{-3} \] --- ## **Step 5: Junction Depth (\(x_j\))** At the junction, the acceptor concentration equals the background donor concentration (\(N_D\)): \[ C(x_j, t) = N_D \] \[ N_D = C_s \exp\left(-\frac{x_j^2}{4 D_2 t_2}\right) \] Solve for \(x_j\): \[ \frac{N_D}{C_s} = \exp\left(-\frac{x_j^2}{4 D_2 t_2}\right) \] \[ \ln \left( \frac{C_s}{N_D} \right) = \frac{x_j^2}{4 D_2 t_2} \] \[ x_j = \sqrt{ 4 D_2 t_2 \ln \left( \frac{C_s}{N_D} \right) } \] Plug in values: - \( D_2 t_2 = 1.2 \times 10^{-13} \times 300 = 3.6 \times 10^{-10} \) - \( C_s = 8.99 \times 10^{20} \) - \( N_D = 4.63 \times 10^{14} \) \[ \ln \left( \frac{8.99 \times 10^{20}}{4.63 \times 10^{14}} \right ) = \ln (1.94 \times 10^{6}) = 14.48 \] \[ x_j = \sqrt{4 \times 3.6 \times 10^{-10} \times 14.48 } = \sqrt{2.08 \times 10^{-8}} \approx 1.44 \times 10^{-4} \ \text{cm} = 1.44 \ \mu\text{m} \] --- # **Final Answers** - **Junction depth (\(x_j\))**: \[ \boxed{1.44 \ \mu\text{m}} \] - **Surface doping concentration after drive-in (\(C_s\))**: \[ \boxed{8.99 \times 10^{20} \ \text{cm}^{-3}} \] --- ## **Summary Table** | Parameter | Value | |------------------------|----------------------------| | Junction depth (\(x_j\)) | \(1.44 \ \mu\text{m}\) | | Surface concentration (\(C_s\)) | \(8.99 \times 10^{20} \ \text{cm}^{-3}\) | --- ## **Key Equations Used** - Dose after predeposition: \( Q = 2 N_s \sqrt{\frac{D_1 t_1}{\pi}} \) - Surface concentration after drive-in: \( C_s = \frac{Q}{\sqrt{\pi D_2 t_2}} \) - Junction depth: \( x_j = \sqrt{4 D_2 t_2 \ln \left( \frac{C_s}{N_D} \right) } \) --- **Note:** All calculations assume ideal diffusion and abrupt junctions. **Image suggestion:** A plot showing Gaussian and erfc profiles for reference. ![Diffusion profiles](https://upload.wikimedia.org/wikipedia/commons/9/9c/Diffusion-profiles.png) *Image: Typical dopant diffusion profiles in silicon. Gaussian for limited source, erfc for constant source.*

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