Let's analyze and solve the problem step by step. --- ## **Problem Summary:** - **:** PN diode with a junction area \(A = 1\, \mu m^2\). - **Given Data:** The C–V characteristic is provided as a graph (capacitance \(C\) vs. voltage \(V\)). - **Objective:** Estimate the extent of the depletion region at \(V = \, V\). --- ## **Step 1: Understand the C–V characteristic** The graph shows \(C^{-3}\) (inverse cube of capacitance) versus \(V\). - This linear relation indicates the depletion approximation is valid. - The linear portion of the plot can be used to estimate depletion width parameters. --- ## **Step 2: Recall the relation between \(C\) and depletion width \(W\)** For a one-sided abrupt junction: \[ C = \frac{\varepsilon_s A}{W} \] Where: - \(\varepsilon_s\) is the permittivity of silicon (~\(11.7 \varepsilon_\)) - \(A\) is the junction area - \(W\) is the depletion width --- ## **Step 3: Use the C–V relation** For an abrupt junction, the relation between \(C\) and \(V\) is: \[ \frac{1}{C^3} \propto V \] The graph confirms this linearity. From the graph, at \(V = \, V\), the \(C^{-3}\) value is approximately **3 (fF\(^{-3}\))**. --- ## **Step 4: Find the depletion width at \(V=\, V\)** At \(V=\, V\), the depletion region extends throughout the entire depletion width \(W_\). The relation for depletion width at zero bias: \[ W_ = \sqrt{\frac{2 \varepsilon_s V_{bi}}{q N_A}} \] But since the question asks for the depletion region extent (or width) at zero bias, and the data is from the C–V plot, we can use the following relation: \[ C = \frac{\varepsilon_s A}{W} \] Rearranged as: \[ W = \frac{\varepsilon_s A}{C} \] --- ## **Step 5: Calculate \(C\) at \(V=\, V\)** - From the graph, \(C^{-3} = 3\, \text{fF}^{-3}\): \[ C = \left(\frac{1}{3}\right)^{1/3} \] - Convert \(C^{-3}\) to \(C\): \[ C = (3)^{-1/3} \] - Approximate: \[ 3^{1/3} \approx 1.442 \] So: \[ C \approx \frac{1}{1.442} \approx .693\, \text{fF} \] --- ## **Step 6: Convert units and compute \(W\)** - \(\varepsilon_s = 11.7 \times \varepsilon_ = 11.7 \times 8.854 \times 10^{-12}\, F/m \approx 1.036 \times 10^{-10}\, F/m\). - Area: \[ A = 1\, \mu m^2 = 1 \times 10^{-12}\, m^2 \] - Capacitance: \[ C \approx .693\, \text{fF} = .693 \times 10^{-15}\, F \] - Now, compute the depletion width: \[ W = \frac{\varepsilon_s A}{C} \] \[ W = \frac{1.036 \times 10^{-10}\, F/m \times 1 \times 10^{-12}\, m^2}{.693 \times 10^{-15}\, F} \] Calculate numerator: \[ 1.036 \times 10^{-10} \times 10^{-12} = 1.036 \times 10^{-22} \] Now: \[ W = \frac{1.036 \times 10^{-22}}{.693 \times 10^{-15}} \approx \frac{1.036}{.693} \times 10^{-7} \] \[ \approx 1.494 \times 10^{-7}\, m \] which is: \[ W \approx 149.4\, \text{nm} \] --- ## **## **Final Answer:** ### The estimated depletion width at \(V=\, V\) is approximately **149 nm**. --- Let me know if you'd like further clarification or details!