Question 5 of 7: Intrinsic Silicon — Band Gap, Effective Mass, Fermi Level, Carrier Concentration, Doping

Question 5: Intrinsic Silicon — Band Gap, Effective Mass, Fermi Level, Carrier Concentration, Doping (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. Intrinsic Si: $E_g=1.08\ \text{eV}$, electron effective mass $m_e^*=1.1m$, hole effective mass $m_h^*=0.56m$ ($m=$ free-electron rest mass). Paper's formulas: eq. (16), $\mu=\dfrac{E_g}{2}+\dfrac34k_BT\ln(m_h^*/m_e^*)$; eq. (15), $n_i=p_i=2\!\left(\dfrac{k_BT}{2\pi\hbar^2}\right)^{3/2}(m_e^*m_h^*)^{3/4}\exp\!\left(\dfrac{-E_g}{2k_BT}\right)$.

Find. (a)–(b) conceptual definitions; (c) $\mu$ at $T=300$ K; (d) $p_i$ at $T=320$ K; (e) effect of donor doping.

Part (a) — band gap. The band gap $E_g$ is the energy separation between the top of the highest fully-occupied electron energy band at $T=0$ (the valence band) and the bottom of the next, empty band (the conduction band). No allowed electron states exist inside this range in a perfect crystal; an electron must absorb at least $E_g$ of energy (thermally or optically) to cross from the valence band, where it leaves a hole, into the conduction band, where it becomes a mobile carrier. It is this gap that makes Si a semiconductor rather than a metal (no gap) or an insulator (very large gap).

Part (b) — effective mass. Near a band extremum, an electron (or hole) in the periodic crystal potential accelerates under an external force exactly as a free particle would, but with an effective mass $m^*$ that absorbs the effect of the periodic lattice potential, defined from the band curvature: $\dfrac1{m^*}=\dfrac1{\hbar^2}\dfrac{d^2E}{dk^2}$. It differs from the free-electron rest mass $m$ (here $m_e^*=1.1m,\ m_h^*=0.56m$ for Si) because the electron is also interacting with the periodic array of ion cores, and it lets all of the usual free-particle transport formulas ($v=\hbar k/m^*$, cyclotron/drift equations, density-of-states formulas) carry over unchanged once $m$ is replaced by $m^*$.

Approach (c, d). Substitute directly into the paper's own eqs. (16) and (15); convert $k_BT$ to eV for (c) and keep full SI units for (d) since the prefactor there is dimensionally a number density.

1. Part (c) — Fermi level at $T=300$ K. With $k_B=1.38062\times10^{-23}\ \text{J/K}=8.6173\times10^{-5}\ \text{eV/K}$, $k_BT=0.025852\ \text{eV}$ at 300 K, and $m_h^*/m_e^*=0.56/1.1=0.5091$: $$\mu=\frac{1.08}{2}+\frac34(0.025852)\ln(0.5091)=0.540+0.019389(-0.6754)$$ $$\boxed{\mu=0.527\ \text{eV (measured from the valence-band edge, via eq. 16)}}$$ Because $m_e^*>m_h^*$ (electrons have a "heavier," lower-mobility, higher-density-of-states band here than holes), $\mu$ sits slightly below exact midgap ($0.540\ \text{eV}$) — the intrinsic level shifts toward whichever band carries the smaller effective mass / lower density of states.
2. Part (d) — hole concentration at $T=320$ K. For an intrinsic semiconductor $p_i=n_i$, given directly by eq. (15). With $m_e^*=1.1(9.10956\times10^{-31})=1.0021\times10^{-30}\ \text{kg}$, $m_h^*=0.56(9.10956\times10^{-31})=5.1013\times10^{-31}\ \text{kg}$, $\hbar=1.05459\times10^{-34}\ \text{J}\cdot\text{s}$, $k_BT=1.38062\times10^{-23}(320)=4.418\times10^{-21}\ \text{J}$, and $E_g=1.08(1.60219\times10^{-19})=1.7304\times10^{-19}\ \text{J}$: $$p_i=2\left(\frac{4.418\times10^{-21}}{2\pi(1.05459\times10^{-34})^2}\right)^{3/2}\!\!\big[(1.0021\times10^{-30})(5.1013\times10^{-31})\big]^{3/4}\exp\!\left(\frac{-1.7304\times10^{-19}}{2(4.418\times10^{-21})}\right)$$ $$\boxed{p_i=n_i\approx6.01\times10^{16}\ \text{m}^{-3}=6.01\times10^{10}\ \text{cm}^{-3}}$$ This lands in the same $10^{10}\text{–}10^{11}\ \text{cm}^{-3}$ decade textbooks quote for intrinsic Si near room temperature, a reasonable check on the arithmetic (the exact figure depends sensitively on which $E_g,m^*$ are assumed, which is why it differs from the standard $1.5\times10^{10}\ \text{cm}^{-3}$ at 300 K quoted for real Si with its true density-of-states masses).

Part (e) — effect of donor doping. A donor impurity (e.g. phosphorus or arsenic substituting for Si) contributes one weakly-bound extra valence electron beyond what the four covalent Si–Si bonds need; this electron sits in a shallow level just below the conduction-band edge and ionizes (donates its electron to the conduction band) at ordinary temperatures with very little thermal energy required. A "high concentration of donor atoms" therefore adds a large population of conduction-band electrons far in excess of the small intrinsic carrier density found in part (d), directly raising the conductivity $\sigma=q(n\mu_n+p\mu_p)$ through the $n$ term (the material becomes n-type). Because the electron concentration $n$ now vastly exceeds $p$ (mass-action law $np=n_i^2$ still holds, so $p$ correspondingly drops), the Fermi level $\mu$ moves up from its intrinsic near-midgap position toward the conduction-band edge $E_c$, reflecting the much higher probability of finding an occupied state near $E_c$.