Question 5 of 7: Silicon Band Structure — Intrinsic Fermi Level, Carrier Concentrations, p–n Junction

Question 5: Silicon Band Structure — Intrinsic Fermi Level, Carrier Concentrations, p–n Junction (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. $E_g=1.08\ \text{eV}$ (Figures P5a–c); electron effective mass $m_e^*=1.1m$, hole effective mass $m_h^*=0.56m$; $T=300\ \text{K}$ ($k_BT=0.02586\ \text{eV}$). Figures P5b/P5c annotate the p-type and n-type Fermi levels relative to the intrinsic level $\mu_i$ (a common horizontal reference line drawn across all three panels): $\mu_i-\mu_p=0.467\ \text{eV}$ and $\mu_n-\mu_i=0.329\ \text{eV}$.

Find. (a) $\mu_i-E_v$; (b) intrinsic concentration $n_i$; (c) qualitative explanation; (d) hole concentration $p$ in the n-type sample; (e) the p–n junction band diagram.

[Figure not reproduced: Fig. 5 — Band-edge diagram as printed (Figures P5a–c): a single intrinsic Fermi level $\mu_i$ (dashed reference, computed in Part (a)) is shared across all three panels, with $\mu_p$ and $\mu_n$ offset from it by the given 0.467 eV and 0.329 eV. See the official exam paper.]

Approach. (a)/(b) apply the paper's own intrinsic-semiconductor formulas, eqs. (13)–(14); (c) qualitative doping argument; (d) combine the given $\mu_n-\mu_i$ offset with the mass-action law (eqs. 12/15); (e) sketch the standard equilibrium-junction band diagram with a common Fermi level.

1. Part (a) — intrinsic Fermi level above $E_v$. Eq. (14) gives the intrinsic level measured from $E_v$ directly: $$\mu_i=\frac{E_g}{2}+\frac{3}{4}k_BT\ln\!\left(\frac{m_h}{m_e}\right)=\frac{1.08}{2}+\frac34(0.02586)\ln\!\left(\frac{0.56}{1.1}\right)\ \text{eV}$$ $$\boxed{\mu_i-E_v\approx0.527\ \text{eV}}$$ Because $m_e^*>m_h^*$ here, $\ln(m_h/m_e)<0$ and $\mu_i$ sits below mid-gap (0.540 eV) — closer to $E_v$ than to $E_c$.
2. Part (b) — intrinsic carrier concentration. Eq. (13): $$n_i=2\left(\frac{k_BT}{2\pi\hbar^2}\right)^{3/2}(m_em_h)^{3/4}\exp\!\left(\frac{-E_g}{2k_BT}\right)$$ Substituting $m_e=1.1m$, $m_h=0.56m$, $E_g=1.08\ \text{eV}$, $T=300\ \text{K}$ (SI units throughout, then converting the final result to $\text{cm}^{-3}$): $$\boxed{n_i\approx1.47\times10^{10}\ \text{cm}^{-3}}$$ in close agreement with the accepted intrinsic carrier concentration of real silicon at room temperature ($\approx1.5\times10^{10}\ \text{cm}^{-3}$) — confirming $E_g=1.08\ \text{eV}$ and the given effective masses correspond to genuine Si data.
3. Part (c) — why $\mu_n$ sits close to $E_c$. Doping Si with a shallow donor (e.g. group-V As, P) adds states just below $E_c$ that ionize almost completely at room temperature, flooding the conduction band with electrons far in excess of the intrinsic value. To keep the Fermi–Dirac occupation of the conduction band consistent with this much larger electron population, the Fermi level $\mu_n$ must move up, away from mid-gap and toward $E_c$ (a small $E_c-\mu_n$ means a much larger $f(E_c)$ and hence $n$). The impact is that n-type Si becomes strongly electron-majority ($n\gg p\gg$ their intrinsic values would suggest), giving it dramatically higher conductivity than intrinsic Si, dominated by electron transport.
4. Part (d) — hole concentration in n-type Si. Eq. (15) gives the electron concentration referenced to the intrinsic level: $n=n_i\exp[(\mu_n-\mu_i)/k_BT]$. By the mass-action law (eq. 12, $np=n_i^2$ at fixed $T$), the corresponding hole concentration is $$p=\frac{n_i^2}{n}=n_i\exp\!\left(-\frac{\mu_n-\mu_i}{k_BT}\right)=n_i\exp\!\left(-\frac{0.329}{0.02586}\right)$$ $$\boxed{p\approx4.4\times10^{4}\ \text{cm}^{-3}}$$ many orders of magnitude below $n_i$, exactly as expected for holes as the minority carrier in an n-type sample.
5. Part (e) — p–n junction band diagram. Joining the p-type and n-type pieces, charge must redistribute (electrons diffuse from n to p, holes from p to n) until a single, flat, common Fermi level $E_F$ is established throughout the junction at equilibrium — this is the defining condition of thermal equilibrium. Because each side's bands sat at a different height relative to its own (pre-contact) Fermi level, forcing them onto a common $E_F$ bends $E_c(x)$ and $E_v(x)$ smoothly across the depletion region: they sit high on the p-side (small $\mu_p-E_v$ there means $E_c$ is far above $E_F$) and low on the n-side (small $E_c-\mu_n$ means $E_c$ sits just above $E_F$). The total band bending is the built-in potential $$eV_{bi}=(\mu_n-\mu_i)+(\mu_i-\mu_p)=0.329+0.467=\boxed{0.796\ \text{eV}}$$ sketched in Figure 6 below: $E_c$ and $E_v$ both drop by $eV_{bi}$ from the p-side bulk to the n-side bulk, while $E_F$ stays flat.