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Intrinsic concentration and resistivity of Si

Short answer

Calculate Nc and Nv using Nc,v = 2(2πm*kT/h²)^(3/2) with given effective masses; then ni = √(Nc Nv) exp(–Eg/2kT) ≈ 1.05×10¹⁰ cm⁻³. Intrinsic conductivity σ = e ni (μe+μh) yields resistivity ρ = 1/σ ≈ 3.3×10⁵ Ω·cm.

Given that the density of states related effective masses of electrons and holes in Si are approximately 1.08mₑ and 0.60mₑ respectively and the electron and hole drift mobilities at room temperature are 1350 and 450 cm²V⁻¹s⁻¹ respectively. Calculate the intrinsic concentration and intrinsic resistivity of Si. The energy band gap for Si is 1.10eV. (T = 300K)

In an intrinsic semiconductor, n=p=nin = p = n_i with

ni=NcNv exp⁡(−Eg2kT),Nc=2(2πme∗kTh2)3/2,Nv=2(2πmh∗kTh2)3/2n_i = \sqrt{N_c N_v}\,\exp\left(-\frac{E_g}{2kT}\right), \quad N_c = 2\left(\frac{2\pi m_e^* kT}{h^2}\right)^{3/2}, \quad N_v = 2\left(\frac{2\pi m_h^* kT}{h^2}\right)^{3/2}

Data: me∗=1.08mem_e^* = 1.08m_e, mh∗=0.60mem_h^* = 0.60m_e, me=9.109×10−31m_e = 9.109\times10^{-31} kg, k=1.381×10−23k = 1.381\times10^{-23} J/K, h=6.626×10−34h = 6.626\times10^{-34} J s, T=300T = 300 K, Eg=1.10E_g = 1.10 eV, μe=1350\mu_e = 1350, μh=450\mu_h = 450 cm2^2V−1^{-1}s−1^{-1}. Then kT=0.02586kT = 0.02586 eV.

Effective densities of states

Nc=2[2π(1.08×9.109×10−31)(1.381×10−23)(300)(6.626×10−34)2]3/2=2.82×1025 m−3=2.82×1019 cm−3Nv=Nc(0.601.08)3/2=1.17×1019 cm−3\begin{aligned} N_c &= 2\left[\frac{2\pi (1.08 \times 9.109\times10^{-31})(1.381\times10^{-23})(300)}{(6.626\times10^{-34})^2}\right]^{3/2} \\ &= 2.82\times10^{25}\ \text{m}^{-3} = 2.82\times10^{19}\ \text{cm}^{-3} \\ N_v &= N_c\left(\frac{0.60}{1.08}\right)^{3/2} = 1.17\times10^{19}\ \text{cm}^{-3} \end{aligned}

Intrinsic concentration

Eg2kT=1.102×0.02586=21.27NcNv=2.82×1019×1.17×1019=1.813×1019 cm−3ni=1.813×1019×e−21.27=1.813×1019×5.80×10−10=1.05×1010 cm−3\begin{aligned} \frac{E_g}{2kT} &= \frac{1.10}{2 \times 0.02586} = 21.27 \\ \sqrt{N_c N_v} &= \sqrt{2.82\times10^{19} \times 1.17\times10^{19}} = 1.813\times10^{19}\ \text{cm}^{-3} \\ n_i &= 1.813\times10^{19} \times e^{-21.27} \\ &= 1.813\times10^{19} \times 5.80\times10^{-10} \\ &= 1.05\times10^{10}\ \text{cm}^{-3} \end{aligned}

Intrinsic resistivity

σi=eni(μe+μh)=1.602×10−19×1.052×1010×(1350+450)=3.03×10−6 Ω−1cm−1ρi=1σi=3.30×105 Ω cm\begin{aligned} \sigma_i &= e n_i (\mu_e + \mu_h) \\ &= 1.602\times10^{-19} \times 1.052\times10^{10} \times (1350 + 450) \\ &= 3.03\times10^{-6}\ \Omega^{-1}\text{cm}^{-1} \\ \rho_i &= \frac{1}{\sigma_i} = 3.30\times10^{5}\ \Omega\,\text{cm} \end{aligned}

Answer: ni≈1.05×1010n_i \approx 1.05\times10^{10} cm−3^{-3}; ρi≈3.3×105\rho_i \approx 3.3\times10^{5} Ω\Omega cm (3.3×1033.3\times10^{3} Ω\Omega m).

✓ Answer checked by Claude Opus 5.5 (high effort)

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