Question 2 of 6: Silicon Resistivity, Intrinsic Current Split and Phosphorus Doping

Question 2: Silicon Resistivity, Intrinsic Current Split and Phosphorus Doping (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Diamond-cubic Si, a = 5.4307×10^-8 cm (8 atoms/cell); μ_n = 1350 cm²/V·s, μ_p = 480 cm²/V·s; E_g = 1.11 eV; pure-Si resistivity ρ₁ = 2.3×10⁵ Ω·cm at T₁ = 27°C; P-doped wafer resistivity ρ_dop = 8.33×10^-3 Ω·cm at 27°C; k_B = 8.63×10^-5 eV/K.

Find. (a) Conductivity of pure Si at 200°C and the factor by which it changed from 27°C; (b) the hole current fraction in intrinsic Si; (c) the majority-carrier density, the P-to-Si atom ratio and the at% P of the doped wafer.

Approach. Intrinsic conductivity is dominated by n_i ∝ exp(−E_g/2k_BT), so the conductivity ratio between two temperatures follows directly from that exponential (mobility change neglected against the exponential swing). The hole current fraction reduces to a mobility ratio since both intrinsic carrier densities are equal. The doped wafer's majority-carrier density comes from σ = Nqμ_n (phosphorus donates electrons), compared against Si's own atomic density from its diamond-cubic cell.

1. (a) Conductivity at 200°C. With T₁ = 300.15 K and T₂ = 473.15 K, and σ_i ∝ exp(−E_g/2k_BT): $$\dfrac{\sigma_2}{\sigma_1}=\exp\left[\dfrac{E_g}{2k_B}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)\right]=\exp\left[\dfrac{1.11}{2(8.63\times10^{-5})}\left(\dfrac{1}{300.15}-\dfrac{1}{473.15}\right)\right]=\boxed{2525\times}$$ $$\sigma_1=\dfrac{1}{\rho_1}=\dfrac{1}{2.3\times10^5}=4.348\times10^{-6}\ (\Omega\cdot\text{cm})^{-1}$$ $$\sigma_2=\sigma_1\times2525=\boxed{1.098\times10^{-2}\ (\Omega\cdot\text{cm})^{-1}}$$
2. (b) Hole current fraction, intrinsic Si. Both carrier populations equal n_i and share the same field, so the current split is set purely by mobility: $$\dfrac{I_p}{I_{total}}=\dfrac{\mu_p}{\mu_n+\mu_p}=\dfrac{480}{1830}=\boxed{26.2\%}$$
3. (c) Majority-carrier density in the P-doped wafer. Phosphorus is a Group-V donor, so the majority carriers are electrons; with σ = N_Dqμ_n: $$N_D=\dfrac{1/\rho_{dop}}{q\mu_n}=\dfrac{1/(8.33\times10^{-3})}{(1.6\times10^{-19})(1350)}=\boxed{5.558\times10^{17}\ \text{cm}^{-3}}$$
4. Si's own atomic density and the P:Si ratio. Diamond-cubic Si carries 8 atoms/cell: $$V_{cell}=a^3=(5.4307\times10^{-8})^3=1.602\times10^{-22}\ \text{cm}^3,\qquad N_{Si}=\dfrac{8}{V_{cell}}=4.995\times10^{22}\ \text{cm}^{-3}$$ $$\dfrac{N_D}{N_{Si}}=\dfrac{5.558\times10^{17}}{4.995\times10^{22}}=1.113\times10^{-5}\ \Rightarrow\ at\%\ P=\boxed{1.113\times10^{-3}\ \text{at\%}}$$

A rise of under 200°C multiplies pure silicon's intrinsic conductivity by roughly 2500×, the same exponential sensitivity that made GaAs's conductivity double after only ~7°C in this discipline's other papers — both follow directly from E_g/2k_BT being large at room temperature.