Question 4 of 6: Intrinsic GaAs — Conductivity, Thermal Sensitivity and Current Split

Question 4: Intrinsic GaAs — Conductivity, Thermal Sensitivity and Current Split (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. Intrinsic carrier density n_i = 1.5×10¹² m^-3 = 1.5×10⁶ cm^-3 (printed per m³, consistent with the SI mobility units and with the handbook n_i of GaAs, about 2×10⁶ cm^-3) at T₁ = 25°C (298.15 K); μ_n = 0.720 m²/V·s = 7200 cm²/V·s; μ_p = 0.020 m²/V·s = 200 cm²/V·s; E_g = 1.47 eV; k_B = 8.62×10^-5 eV/K.

Find. (a) The room-temperature intrinsic conductivity; (b) the temperature at which σ doubles; (c) the fraction of total current carried by holes.

Approach. Intrinsic conductivity uses both carrier types with the same n_i: σ_i = n_iq(μ_n+μ_p). Its temperature dependence is dominated by n_i ∝ exp(−E_g/2k_BT), so doubling σ (mobility change neglected over the small resulting ΔT) reduces to solving that exponential relation for a new T. The hole current fraction follows from the two carriers sharing the same field and the same density, so it reduces to a mobility ratio.

1. Room-temperature conductivity. Working in SI units first, $$\sigma_i=n_i\,q(\mu_n+\mu_p)=(1.5\times10^{12})(1.6\times10^{-19})(0.720+0.020)=1.776\times10^{-7}\ (\Omega\cdot\text{m})^{-1}$$ and since 1 m = 100 cm, $$\sigma_i=\dfrac{1.776\times10^{-7}}{100}=\boxed{1.776\times10^{-9}\ (\Omega\cdot\text{cm})^{-1}}$$ (Check in cm units: (1.5×10⁶)(1.6×10^-19)(7400) = 1.776×10^-9.)
2. Temperature for double conductivity. Since σ_i ∝ n_i ∝ exp(−E_g/2k_BT), doubling σ means $$\ln 2=\dfrac{E_g}{2k_B}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)\ \Rightarrow\ \dfrac{1}{T_2}=\dfrac{1}{T_1}-\dfrac{2k_B\ln2}{E_g}$$ $$\dfrac{1}{T_2}=\dfrac{1}{298.15}-\dfrac{2(8.62\times10^{-5})(0.6931)}{1.47}=3.2727\times10^{-3}\ \text{K}^{-1}$$ $$T_2=305.6\ \text{K}=\boxed{32.4^\circ\text{C}}\ \ (\Delta T\approx+7.4^\circ\text{C})$$
3. Fraction of current carried by holes. Both carrier populations equal n_i and share the same field, so the current split is set purely by mobility: $$\dfrac{I_p}{I_{total}}=\dfrac{\mu_p}{\mu_n+\mu_p}=\dfrac{200}{7400}=\boxed{2.70\%}$$

A rise of only about 7°C is enough to double GaAs's intrinsic conductivity — the ratio E_g/k_BT ≈ 57 at room temperature places GaAs deep in the exponential regime, so small temperature changes translate into large fractional swings in n_i.