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Intrinsic conductivity and resistivity of GaAs

Short answer

σ_i = e n_i(μe+μh) = 1.6×10⁻¹⁹ × 1.8×10⁶ × 8900 = 2.56×10⁻⁹ S/cm; ρ_i = 1/σ_i ≈ 3.9×10⁸ Ω·cm.

Calculate the intrinsic conductivity and resistivity of GaAs at room temperature. The intrinsic concentration, electron mobility and hole mobility of GaAs are 1.8×10⁶ per cm⁻³, 8500 cm² v⁻¹s⁻¹ and 400 cm² v⁻¹s⁻¹ respectively at 300K.

For an intrinsic semiconductor n=p=nin = p = n_i, so

σi=eni(μe+μh),ρi=1σi\sigma_i = en_i(\mu_e+\mu_h),\qquad \rho_i = \frac{1}{\sigma_i}

Given (GaAs, 300 K): ni=1.8×106 cm−3n_i = 1.8\times10^{6}\ \text{cm}^{-3}, μe=8500\mu_e = 8500, μh=400 cm2V−1s−1\mu_h = 400\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}, e=1.6×10−19e = 1.6\times10^{-19} C.

Conductivity

σi=1.6×10−19×1.8×106×(8500+400)=1.6×10−19×1.8×106×8900=2.56×10−9 S cm−1 (=2.56×10−7 S m−1)\begin{aligned} \sigma_i &= 1.6\times10^{-19}\times1.8\times10^{6}\times(8500+400)\\ &= 1.6\times10^{-19}\times1.8\times10^{6}\times8900\\ &= 2.56\times10^{-9}\ \text{S cm}^{-1}\ (= 2.56\times10^{-7}\ \text{S m}^{-1}) \end{aligned}

Resistivity

ρi=12.563×10−9=3.90×108 Ω cm (=3.90×106 Ω m)\rho_i = \frac{1}{2.563\times10^{-9}} = 3.90\times10^{8}\ \Omega\,\text{cm}\ (= 3.90\times10^{6}\ \Omega\,\text{m})

Answer: σi≈2.56×10−9\sigma_i \approx 2.56\times10^{-9} S cm⁻¹ and ρi≈3.9×108 Ω\rho_i \approx 3.9\times10^{8}\ \Omega cm.

Although electrons in GaAs are about six times more mobile than in Si, the much larger band gap (1.42 eV) makes nin_i about 10410^4 times smaller, so intrinsic GaAs is nearly an insulator ("semi-insulating" GaAs is used as a substrate).

✓ Answer checked by Claude Opus 5.5 (high effort)

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