# CHEG 5023 – Microelectronics Materials ## Homework 2: Complete Step-by-Step Solutions --- ## Q1. Show that \( pn = N_c N_v e^{-E_gT} \). Prove that \( pn = n_i^2 \) ### Step 1: Electron Concentration \[ n = N_c e^{-(E_c - E_F)/kT} \] ### Step 2: Hole Concentration \[ p = N_v e^{-(E_F - E_v)/kT} \] ### Step 3: Multiply \( n \) and \( p \) \[ pn = N_c N_v e^{-(E_c - E_v)/kT} \] But, \( E_c - E_v = E_g \): \[ pn = N_c N_v e^{-E_g/kT} \] \[ \boxed{pn = N_c N_v e^{-E_g/kT}} \] ### Step 4: Definition of Intrinsic Carrier Concentration \[ n_i = \sqrt{N_c N_v} e^{-E_g/2kT} \] Squaring both sides: \[ n_i^2 = N_c N_v e^{-E_g/kT} \] \[ \boxed{pn = n_i^2} \] ✔ **Proved** --- ## Q2. Fraction of Conduction in Intrinsic Si Due to Electrons ### Given - Conductivity: \(\sigma = q(n\mu_n + p\mu_p)\) - Intrinsic silicon: \(n = p = n_i\) ### Step 1: Electron Fraction \[ \text{Electron fraction} = \frac{n\mu_n}{n\mu_n + p\mu_p} = \frac{\mu_n}{\mu_n + \mu_p} \] ### Step 2: Substitute Values \[ \mu_n = 190,\qquad \mu_p = 425 \] \[ \text{Electron fraction} = \frac{190}{190 + 425} = \frac{190}{2325} = .817 \] \[ \boxed{\text{Electron contribution} \approx 81.7\%} \] --- ## Q2(b). Boron Concentration to Increase Conductivity to \(.1~(\Omega\cdot\mathrm{cm})^{-1}\) ### Step 1: Conductivity in P-Type Si \[ \sigma = q p \mu_p \] ### Step 2: Solve for \( p \) \[ p = \frac{\sigma}{q\mu_p} \] \[ p = \frac{.1}{(1.6 \times 10^{-19}) \times 425} \] \[ p = 1.47 \times 10^{15}~\mathrm{cm}^{-3} \] \[ \boxed{N_A \approx 1.5 \times 10^{15}~\mathrm{cm}^{-3}} \] --- ## Q2(c). Dominant Carrier Since boron doping \(\rightarrow\) **p-type**, \[ \boxed{\text{Holes dominate}} \] --- ## Q3. Si Wafer with \(10^{14}~\mathrm{cm}^{-3}\) Dopants Given: \( n_i = 1.45 \times 10^{10}~\mathrm{cm}^{-3} \) --- ### (a) Boron Doped (p-type) #### Step 1: Carrier Concentrations \[ p \approx N_A = 10^{14} \] \[ n = \frac{n_i^2}{p} = \frac{(1.45 \times 10^{10})^2}{10^{14}} = 2.1 \times 10^{6} \] #### Step 2: Conductivity \[ \sigma = q(p\mu_p + n\mu_n) \] \[ \sigma \approx (1.6 \times 10^{-19})(10^{14})(425) = 6.8 \times 10^{-3} \] #### Step 3: Resistivity \[ \rho = \frac{1}{\sigma} = 147~\Omega\cdot\mathrm{cm} \] #### Step 4: Fermi Level \[ E_F - E_i = -kT \ln\left( \frac{p}{n_i} \right ) \] \[ = -.026 \ln \left( \frac{10^{14}}{1.45 \times 10^{10}} \right ) = -.23~\mathrm{eV} \] --- ### (b) Phosphorus Doped (n-type) \[ n \approx 10^{14} \] \[ p = \frac{n_i^2}{n} = 2.1 \times 10^{6} \] \[ \sigma = (1.6 \times 10^{-19})(10^{14})(190) = .0304 \] \[ \rho = 32.9~\Omega\cdot\mathrm{cm} \] \[ E_F - E_i = +.23~\mathrm{eV} \] --- ## Q4. Carrier Concentration, Conductivity & Fermi Level --- ### (a) Boron Concentration \(10^{15}~\mathrm{cm}^{-3}\) \[ p = 10^{15} \] \[ n = \frac{n_i^2}{p} = 2.1 \times 10^{5} \] \[ \sigma = (1.6 \times 10^{-19})(10^{15})(425) = .068 \] \[ E_F - E_i = -.026 \ln \left( \frac{10^{15}}{1.45 \times 10^{10}} \right ) = -.29~\mathrm{eV} \] --- ### (b) Phosphorus Concentration \(10^{15}~\mathrm{cm}^{-3}\) \[ n = 10^{15} \] \[ p = 2.1 \times 10^{5} \] \[ \sigma = (1.6 \times 10^{-19})(10^{15})(190) = .304 \] \[ E_F - E_i = +.29~\mathrm{eV} \] --- ## Final Summary Table | Case | Dominant Carrier | Conductivity \((\Omega^{-1}\mathrm{cm}^{-1})\) | Fermi Level Shift | | ------------ | ---------------- | ---------------------------------------------- | ----------------- | | Intrinsic | Electrons | – | At \(E_i\) | | B: \(10^{14}\) | Holes | .0068 | \(-.23~\mathrm{eV}\) | | P: \(10^{14}\) | Electrons | .030 | \(+.23~\mathrm{eV}\) | | B: \(10^{15}\) | Holes | .068 | \(-.29~\mathrm{eV}\) | | P: \(10^{15}\) | Electrons | .304 | \(+.29~\mathrm{eV}\) | --- If you want, I can also: - Convert this into **exam-ready answers** - Create a **one-page formula sheet** - Or explain **Fermi level shifts visually**