Let's solve the problem **step by step**. --- ## Given Data - **Impurity gradient**: \( a = 5 \times 10^{19} \) /cm\(^4\) - **Material**: Silicon (Si) ### Additional Data (for Si at room temperature) - **Intrinsic carrier concentration (\(n_i\))**: \(1.5 \times 10^{10}\) /cm\(^3\) - **Permittivity (\(\varepsilon_s\))**: \(11.7 \times 8.854 \times 10^{-14}\) F/cm --- ## (a) Calculate Built-in Potential (\(V_{bi}\)) For a **linearly graded junction**: \[ N_D(x) = a x \] The built-in potential is given by: \[ V_{bi} = \frac{kT}{q} \ln \left( \frac{N_{Dmax} N_{Amax}}{n_i^2} \right) \] But for a **linearly graded junction**, this simplifies to: \[ V_{bi} = \frac{a W^2}{2 n_i} \] However, typically, the built-in potential can also be estimated as: \[ V_{bi} = \frac{kT}{q} \ln\left( \frac{N_{Dmax} N_{Amax}}{n_i^2} \right) \] But since this is a linearly graded junction, let's use the full depletion approximation for a linearly graded junction: \[ V_{bi} = \frac{a W^2}{2 n_i} \] But we need the depletion width (\(W\)), which requires \(V_{bi}\), leading to a system of equations. #### **Let’s proceed to depletion width:** --- ## (b) Calculate Depletion Width (\(W\)) For a linearly graded junction, the depletion width is: \[ W = \left( \frac{12 \varepsilon_s V_{bi}}{q a} \right)^{1/3} \] Where: - \(\varepsilon_s\) = \(11.7 \times 8.854 \times 10^{-14}\) F/cm - \(q = 1.6 \times 10^{-19}\) C - \(a = 5 \times 10^{19}\) /cm\(^4\) --- ### **Step 1: Calculate \(\varepsilon_s\)** \[ \varepsilon_s = 11.7 \times 8.854 \times 10^{-14} = 1.035 \times 10^{-12} \, \text{F/cm} \] --- ### **Step 2: Assume Room Temperature (\(T = 300\,K\)), \(kT/q = .0259\,V\)** --- ### **Step 3: Estimate \(V_{bi}\)** For a silicon junction, \[ V_{bi} \approx .7\,V \] (typical for Si; can be used unless otherwise specified) --- ### **Step 4: Calculate Depletion Width (\(W\))** \[ W = \left( \frac{12 \varepsilon_s V_{bi}}{q a} \right)^{1/3} \] Plug in the numbers: - \(12 \varepsilon_s = 12 \times 1.035 \times 10^{-12} = 1.242 \times 10^{-11}\) - \(q a = 1.6 \times 10^{-19} \times 5 \times 10^{19} = 8\) - \(V_{bi} = .7\) \[ W = \left( \frac{1.242 \times 10^{-11} \times .7}{8} \right)^{1/3} = \left( \frac{8.694 \times 10^{-12}}{8} \right)^{1/3} = \left( 1.087 \times 10^{-12} \right)^{1/3} \] \[ W \approx (1.087 \times 10^{-12})^{1/3} \] \[ = 1.03 \times 10^{-4} \text{ cm} = 1.03 \, \mu m \] --- ## **Final Answers** ### **(a) Built-in Potential** \[ V_{bi} \approx .7~\text{V} \] ### **(b) Depletion Width** \[ W \approx 1.03~\mu\text{m} \] --- ## **Summary Table** | Parameter | Value | |------------------------|--------------------------| | Built-in Potential | \(.7~\text{V}\) | | Depletion Width | \(1.03~\mu\text{m}\) | --- **Note:** If you are given specific values for \(N_{Dmax}\) or \(N_{Amax}\), use them to get a more accurate answer for \(V_{bi}\). If you have further information, please provide it.