Let's break down the problem step by step. ### **Given Data** - **SiO₂ thickness \( t_{ox} \):** \( 100 \) nm \( = 100 \times 10^{-7} \) cm \( = 1 \times 10^{-5} \) cm - **Substrate carrier density \( N_A \):** \( 1.5 \times 10^{16} \) cm\(^{-3}\) (Assume p-type substrate unless otherwise specified) - **Oxide layer charge density \( Q_{ox} \):** \( 8 \times 10^{-8} \) C/cm\(^2\) - **Depletion layer charge density \( Q_{dep} \):** \( -6 \times 10^{-8} \) C/cm\(^2\) - **Permittivity of Si (\( \epsilon_{si} \)):** \( 11.7 \epsilon_0 \) - **Permittivity of SiO₂ (\( \epsilon_{ox} \)):** \( 3.9 \epsilon_0 \) - **\( \epsilon_0 \) (vacuum permittivity):** \( 8.854 \times 10^{-14} \) F/cm --- ## **a) Find the Depletion Width \( W \) (using \( Q_{dep} \))** The depletion charge per unit area is given by: \[ Q_{dep} = -q N_A W \] Where: - \( Q_{dep} \): depletion charge density (C/cm\(^2\)) - \( q \): elementary charge = \( 1.6 \times 10^{-19} \) C - \( N_A \): acceptor density (cm\(^{-3}\)) - \( W \): depletion width (cm) Solve for \( W \): \[ W = \frac{|Q_{dep}|}{q N_A} \] Plug in values: - \( |Q_{dep}| = 6 \times 10^{-8} \) C/cm\(^2\) - \( q = 1.6 \times 10^{-19} \) C - \( N_A = 1.5 \times 10^{16} \) cm\(^{-3}\) \[ W = \frac{6 \times 10^{-8}}{1.6 \times 10^{-19} \times 1.5 \times 10^{16}} \] \[ = \frac{6 \times 10^{-8}}{2.4 \times 10^{-3}} \] \[ = 2.5 \times 10^{-5} \text{ cm} \] \[ = 0.25 \ \mu \text{m} \] \[ = 250 \ \text{nm} \] --- ## **b) Find the Capacitance \( C \)** The equivalent capacitance per unit area for a MOS capacitor is: \[ \frac{1}{C} = \frac{1}{C_{ox}} + \frac{1}{C_{dep}} \] Where: - \( C_{ox} = \frac{\epsilon_{ox}}{t_{ox}} \) - \( C_{dep} = \frac{\epsilon_{si}}{W} \) **First, calculate \( C_{ox} \):** \[ \epsilon_{ox} = 3.9 \times 8.854 \times 10^{-14} = 3.453 \times 10^{-13} \ \text{F/cm} \] \[ t_{ox} = 1 \times 10^{-5} \ \text{cm} \] \[ C_{ox} = \frac{3.453 \times 10^{-13}}{1 \times 10^{-5}} = 3.453 \times 10^{-8} \ \text{F/cm}^2 \] --- **Now, calculate \( C_{dep} \):** \[ \epsilon_{si} = 11.7 \times 8.854 \times 10^{-14} = 1.035 \times 10^{-12} \ \text{F/cm} \] \[ W = 2.5 \times 10^{-5} \text{ cm} \] \[ C_{dep} = \frac{1.035 \times 10^{-12}}{2.5 \times 10^{-5}} = 4.14 \times 10^{-8} \ \text{F/cm}^2 \] --- **Now, find \( C \):** \[ \frac{1}{C} = \frac{1}{3.453 \times 10^{-8}} + \frac{1}{4.14 \times 10^{-8}} \] \[ = 2.895 \times 10^7 + 2.415 \times 10^7 = 5.31 \times 10^7 \] \[ C = \frac{1}{5.31 \times 10^7} \] \[ = 1.88 \times 10^{-8} \ \text{F/cm}^2 \] --- ## **c) Find Maximum Depletion Width and Bias Voltage at this Condition** **Maximum depletion width** (\( W_{max} \)) occurs at strong inversion (surface potential \( \psi_s = 2\psi_B \)), where \( \psi_B \) is the Fermi potential. First, calculate \( \psi_B \): \[ \psi_B = \frac{kT}{q} \ln \left( \frac{N_A}{n_i} \right) \] Assume: - \( n_i \) (intrinsic carrier concentration for Si at 300 K) \( = 1.5 \times 10^{10} \) cm\(^{-3}\) - \( kT/q = 0.0259 \) V (@ 300 K) \[ \psi_B = 0.0259 \ln \left( \frac{1.5 \times 10^{16}}{1.5 \times 10^{10}} \right) = 0.0259 \ln(10^6) = 0.0259 \times 13.82 = 0.358 \ \text{V} \] So at strong inversion, \( \psi_s = 2\psi_B = 0.716 \) V. --- **Maximum depletion width:** \[ W_{max} = \sqrt{ \frac{2 \epsilon_{si} 2\psi_B}{q N_A} } \] Plug in: \[ W_{max} = \sqrt{ \frac{2 \times 1.035 \times 10^{-12} \times 0.716}{1.6 \times 10^{-19} \times 1.5 \times 10^{16}} } \] Calculate denominator: \[ 1.6 \times 10^{-19} \times 1.5 \times 10^{16} = 2.4 \times 10^{-3} \] Calculate numerator: \[ 2 \times 1.035 \times 10^{-12} \times 0.716 = 1.482 \times 10^{-12} \] So, \[ W_{max} = \sqrt{ \frac{1.482 \times 10^{-12}}{2.4 \times 10^{-3}} } = \sqrt{ 6.175 \times 10^{-10} } = 2.484 \times 10^{-5} \ \text{cm} = 0.248 \ \mu\text{m} = 248 \ \text{nm} \] --- **Bias voltage at maximum depletion:** The threshold voltage: \[ V_{th} = V_{FB} + 2\psi_B + \frac{Q_{max}}{C_{ox}} \] Where: - \( Q_{max} = -\sqrt{2 q \epsilon_{si} N_A 2\psi_B} \) - \( V_{FB} \): Flatband voltage. If work function difference and fixed charges are neglected, \( V_{FB} = 0 \). Calculate \( Q_{max} \): \[ Q_{max} = -\sqrt{2 \times 1.6 \times 10^{-19} \times 1.035 \times 10^{-12} \times 1.5 \times 10^{16} \times 0.716} \] Calculate: - \( 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \) - \( 3.2 \times 10^{-19} \times 1.035 \times 10^{-12} = 3.312 \times 10^{-31} \) - \( 3.312 \times 10^{-31} \times 1.5 \times 10^{16} = 4.968 \times 10^{-15} \) - \( 4.968 \times 10^{-15} \times 0.716 = 3.56 \times 10^{-15} \) So, \[ Q_{max} = -\sqrt{3.56 \times 10^{-15}} = -5.97 \times 10^{-8} \ \text{C/cm}^2 \] Now, calculate the threshold voltage: \[ V_{th} = 0 + 0.716 + \frac{-5.97 \times 10^{-8}}{3.453 \times 10^{-8}} = 0.716 - 1.73 = -1.014 \ \text{V} \] --- ## **Summary Table** | Quantity | Value | |-----------------|-------------------------------| | **(a) Depletion width \( W \)** | \( 2.5 \times 10^{-5} \) cm \( = 250 \) nm | | **(b) Capacitance \( C \)** | \( 1.88 \times 10^{-8} \) F/cm\(^2\) | | **(c) Maximum depletion width (\( W_{max} \))** | \( 2.48 \times 10^{-5} \) cm \( = 248 \) nm | | **Bias voltage at maximum depletion** | \( -1.01 \) V | --- ### **Final Answers** **(a) Depletion width:** \[ \boxed{2.5 \times 10^{-5} \ \text{cm} \ (250 \ \text{nm})} \] --- **(b) Capacitance per unit area:** \[ \boxed{1.88 \times 10^{-8} \ \text{F/cm}^2} \] --- **(c) Maximum depletion width and bias voltage:** \[ \boxed{2.48 \times 10^{-5} \ \text{cm} \ (248 \ \text{nm})} \] \[ \boxed{-1.01 \ \text{V}} \] --- If you need further explanation or step breakdown for any part, let me know!