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Intrinsic Si concentration (hole mass 0.56me)

Short answer

Calculate effective densities of states: Nc = 2.82×10¹⁹ cm⁻³, Nv = 1.05×10¹⁹ cm⁻³. Then intrinsic concentration ni = sqrt(Nc Nv) exp(–Eg/(2kT)) = 1.0×10¹⁰ cm⁻³. Resistivity ρ = 1/[e ni (μe+μh)] = 3.47×10⁵ Ω cm.

The density of states related effective masses of electrons and holes in silicon are approximately 1.08mₑ and 0.56mₑ respectively. The electron and hole drift mobilities at room temperature are 1350 and 450cm²V⁻¹s⁻¹ respectively. Calculate intrinsic concentration and intrinsic resistivity of silicon. The energy band gap for silicon is 1.1ev.

For an intrinsic semiconductor,

ni=NcNvexp⁡(−Eg2kT),Nc,v=2(2πme,h∗kTh2)3/2n_i = \sqrt{N_cN_v}\exp\left(-\frac{E_g}{2kT}\right), \quad N_{c,v} = 2\left(\frac{2\pi m^*_{e,h}kT}{h^2}\right)^{3/2}

Data: me∗=1.08mem_e^* = 1.08m_e, mh∗=0.56mem_h^* = 0.56m_e, Eg=1.1E_g = 1.1 eV, μe=1350\mu_e = 1350, μh=450\mu_h = 450 cm2^2V−1^{-1}s−1^{-1}. Take T=300T = 300 K (room temperature), kT=0.02586kT = 0.02586 eV, me=9.109×10−31m_e = 9.109\times10^{-31} kg, h=6.626×10−34h = 6.626\times10^{-34} J s, k=1.381×10−23k = 1.381\times10^{-23} J/K.

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(0.561.08)3/2=1.05×1019 cm−3\begin{aligned} N_c &= 2\left[\frac{2\pi \times 1.08 \times 9.109\times10^{-31} \times 1.381\times10^{-23} \times 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{0.56}{1.08}\right)^{3/2} = 1.05\times10^{19}\ \text{cm}^{-3} \end{aligned}

Intrinsic concentration

NcNv=2.82×1019×1.05×1019=1.72×1019 cm−3Eg2kT=1.12×0.02586=21.27,e−21.27=5.80×10−10ni=1.72×1019×5.80×10−10=9.99×109 cm−3\begin{aligned} \sqrt{N_cN_v} &= \sqrt{2.82\times10^{19}\times1.05\times10^{19}} = 1.72\times10^{19}\ \text{cm}^{-3} \\ \frac{E_g}{2kT} &= \frac{1.1}{2\times0.02586} = 21.27, \quad e^{-21.27} = 5.80\times10^{-10} \\ n_i &= 1.72\times10^{19}\times5.80\times10^{-10} = 9.99\times10^{9}\ \text{cm}^{-3} \end{aligned}

Intrinsic resistivity

σi=eni(μe+μh)=1.602×10−19×9.99×109×1800=2.88×10−6 Ω−1cm−1ρi=1σi=3.47×105 Ω cm\begin{aligned} \sigma_i &= en_i(\mu_e+\mu_h) = 1.602\times10^{-19}\times9.99\times10^{9}\times1800 \\ &= 2.88\times10^{-6}\ \Omega^{-1}\text{cm}^{-1} \\ \rho_i &= \frac{1}{\sigma_i} = 3.47\times10^{5}\ \Omega\,\text{cm} \end{aligned}

Answer: ni≈1.0×1010n_i \approx 1.0\times10^{10} cm−3^{-3}; ρi≈3.47×105\rho_i \approx 3.47\times10^{5} Ω\Omega cm (3.47×1033.47\times10^{3} Ω\Omega m).

✓ Answer checked by Claude Opus 5.5 (high effort)

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