Question 5 of 7: Extrinsic Semiconductor — Intrinsic Fermi Level, Carrier Type, Mass-Action Law, and Occupation Probability

Question 5: Extrinsic Semiconductor — Intrinsic Fermi Level, Carrier Type, Mass-Action Law, and Occupation Probability (20 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. Band gap $E_g=1.1\,\text{eV}$ (Figure P5); effective masses $m_e^*=1.2m$, $m_h^*=0.6m$; $T=300\,\text{K}$; extrinsic hole concentration $p=2.25\times10^3\,\text{cm}^{-3}$; extrinsic Fermi level $0.15\,\text{eV}$ below $E_c$; test level $0.1\,\text{eV}$ above $E_c$.

Find. (a) $\mu-E_v$ (intrinsic); (b) N or P type; (c) electron concentration $n$; (d) occupation probability $f(E)$.

Approach. (a) apply the paper's own eq. (21) for the intrinsic Fermi level; (b) compare the extrinsic $E_F$ position (given in part (d)) to the intrinsic level; (c) use the mass-action law $np=n_i^2$ with $n_i$ from (a)'s temperature; (d) apply the Fermi-Dirac distribution directly.

1. Part (a) — intrinsic Fermi level above $E_v$. The paper's eq. (21), measuring $\mu$ from $E_v$: $$\mu=\frac{E_g}2+\frac34k_BT\ln\!\left(\frac{m_h^*}{m_e^*}\right)$$ With $E_g=1.1\,\text{eV}$, $m_h^*/m_e^*=0.6/1.2=0.5$, $k_BT=0.02585\,\text{eV}$ at 300 K: $$\mu=0.550+\tfrac34(0.02585)\ln(0.5)=0.550-0.0134$$ $$\boxed{\mu\approx0.537\ \text{eV above }E_v}$$ slightly BELOW exact midgap ($0.550\,\text{eV}$) because the heavier electron effective mass ($1.2m>0.6m$) gives the conduction band more density of states, which pulls the intrinsic Fermi level down toward the valence band to keep $n=p$.
2. Part (b) — N-type or P-type. Part (d) states the extrinsic Fermi level sits $0.15\,\text{eV}$ below $E_c$, i.e. at $E_g-0.15=1.1-0.15=0.95\,\text{eV}$ above $E_v$ — far ABOVE the intrinsic level of $0.537\,\text{eV}$ found in part (a), and close to the conduction band edge. A Fermi level pulled up toward $E_c$ means the conduction band's electron occupation is greatly enhanced relative to the intrinsic case, i.e. donor impurities have contributed a large population of free electrons: $$\boxed{\text{N-type}}$$ (a Fermi level near $E_v$ instead would indicate acceptor-dominated P-type material).
3. Part (c) — electron concentration via the mass-action law. First find the intrinsic concentration $n_i$ at $T=300\,\text{K}$ from the paper's eq. (20)-style expression, $$n_i=2\left(\frac{k_BT}{2\pi\hbar^2}\right)^{3/2}(m_e^*m_h^*)^{3/4}\exp\!\left(-\frac{E_g}{2k_BT}\right)$$ Evaluating numerically gives $n_i\approx1.13\times10^{10}\,\text{cm}^{-3}$ — reassuringly close to real silicon's $n_i\approx1.5\times10^{10}\,\text{cm}^{-3}$ at 300 K, consistent with this problem's $E_g=1.1\,\text{eV}$ being essentially silicon's gap. The mass-action law $np=n_i^2$ holds regardless of doping, so with the given (minority-carrier, consistent with the N-type result of part (b)) hole concentration $p=2.25\times10^3\,\text{cm}^{-3}$: $$n=\frac{n_i^2}{p}=\frac{(1.13\times10^{10})^2}{2.25\times10^3}$$ $$\boxed{n\approx5.66\times10^{16}\ \text{cm}^{-3}}$$ a typical moderate donor concentration, and $n\gg n_i\gg p$ throughout — fully consistent with strongly N-type material.
4. Part (d) — occupation probability via Fermi-Dirac. The test level is $E=E_c+0.1\,\text{eV}$; the extrinsic Fermi level is $E_F=E_c-0.15\,\text{eV}$, so $E-E_F=0.1-(-0.15)=0.25\,\text{eV}$. The paper's eq. (15): $$f(E)=\frac1{\exp\!\left(\dfrac{E-E_F}{k_BT}\right)+1}$$ With $(E-E_F)/k_BT=0.25/0.02585=9.671$ (large, so the $+1$ is negligible — non-degenerate, Maxwell-Boltzmann-like tail): $$\boxed{f(E)\approx6.31\times10^{-5}}$$ a small but non-zero probability, exactly the tail behaviour expected an energy well above the Fermi level.