Semiconductor Physics, Impurity States & Two-Carrier Transport

1. Direct vs Indirect Bandgaps in Momentum Space

The fundamental optical and electronic properties of semiconductors are determined by the relative alignment of the conduction band minimum (CBM) and the valence band maximum (VBM) in the first Brillouin zone:

• Direct Bandgap Semiconductors (e.g. GaAs, InP, InAs, GaN): The conduction band minimum and valence band maximum occur at the exact same crystal wavevector, typically at the zone center ($\Gamma$-point, $\vec{k} = 0$). $$\vec{k}_{\text{CBM}} = \vec{k}_{\text{VBM}} = 0$$ An incoming photon carrying energy $E_{\text{ph}} = h\nu \ge E_g$ possesses negligible momentum ($q_{\text{ph}} = 2\pi/\lambda \sim 10^7\text{ m}^{-1} \ll \pi/a \sim 10^{10}\text{ m}^{-1}$). Because momentum is conserved directly, vertical optical transitions occur readily: $$E_c(\vec{k}) - E_v(\vec{k}) = h\nu$$ The optical absorption coefficient rises sharply as $\alpha(h\nu) \propto (h\nu - E_g)^{1/2}$, enabling efficient radiative recombination for solid-state lasers and LEDs.
• Indirect Bandgap Semiconductors (e.g. Si, Ge, AlAs, GaP): The valence band maximum occurs at the $\Gamma$-point, but the conduction band minima lie along high-symmetry axes away from $\vec{k} = 0$ (e.g. along the six equivalent $\Delta = [100]$ directions at $k \approx 0.85 \frac{2\pi}{a}$ in silicon, or at the four $L = [111]$ zone boundaries in germanium). $$\vec{k}_{\text{CBM}} \ne \vec{k}_{\text{VBM}}$$ An optical transition across the fundamental gap requires a change in electron crystal momentum $\Delta\vec{k} = \vec{k}_{\text{CBM}} - \vec{k}_{\text{VBM}}$. Because photons cannot provide this momentum, the transition must be mediated by the simultaneous emission or absorption of a lattice phonon with wavevector $\vec{q} \approx \Delta\vec{k}$ and energy $\hbar\Omega_{\vec{q}}$: $$h\nu = E_g \pm \hbar\Omega_{\vec{q}}$$ This second-order quantum perturbation process produces a much weaker absorption coefficient $\alpha(h\nu) \propto (h\nu - E_g \mp \hbar\Omega_{\vec{q}})^2$, making silicon inefficient for light emission but excellent for photovoltaics due to long carrier lifetimes.

1. Parabolic Band Edge Approximations

Near the band edges, the dispersion relations can be approximated quadratically:

$$E_c(\vec{k}) = E_c + \frac{\hbar^2 k^2}{2 m_e^*}, \quad E_v(\vec{k}) = E_v - \frac{\hbar^2 k^2}{2 m_h^*}$$

The 3D density of states per unit volume for electrons in the conduction band is:

$$g_c(E) = \frac{1}{2\pi^2} \left( \frac{2m_e^*}{\hbar^2} \right)^{3/2} \sqrt{E - E_c}, \quad E \ge E_c$$

Similarly, for holes in the valence band:

$$g_v(E) = \frac{1}{2\pi^2} \left( \frac{2m_h^*}{\hbar^2} \right)^{3/2} \sqrt{E_v - E}, \quad E \le E_v$$

2. Carrier Concentrations & The Mass Action Law

In non-degenerate semiconductors where the Fermi level lies inside the band gap at least $3k_B T$ away from both band edges ($E_c - E_F \gg k_B T$ and $E_F - E_v \gg k_B T$), the Fermi-Dirac distribution reduces to the classical Maxwell-Boltzmann tail:

$$f(E) = \frac{1}{e^{(E - E_F)/k_B T} + 1} \approx e^{-(E - E_F)/k_B T}$$

The total electron density in the conduction band is obtained by integrating $n_0 = \int_{E_c}^\infty g_c(E) f(E) dE$:

$$n_0 = N_c \exp\left( -\frac{E_c - E_F}{k_B T} \right)$$

where $N_c$ is the effective density of states of the conduction band:

$$N_c = 2 \left( \frac{2\pi m_e^* k_B T}{h^2} \right)^{3/2}$$

Similarly, the hole density in the valence band is $p_0 = \int_{-\infty}^{E_v} g_v(E) [1 - f(E)] dE$:

$$p_0 = N_v \exp\left( -\frac{E_F - E_v}{k_B T} \right), \quad N_v = 2 \left( \frac{2\pi m_h^* k_B T}{h^2} \right)^{3/2}$$

Multiplying $n_0$ and $p_0$ eliminates the Fermi energy $E_F$, yielding the fundamental Mass Action Law:

$$n_0 p_0 = N_c N_v \exp\left( -\frac{E_c - E_v}{k_B T} \right) = N_c N_v \exp\left( -\frac{E_g}{k_B T} \right) \equiv n_i^2$$

Crucially, the product $n_0 p_0 = n_i^2$ is an invariant constant at a given temperature, regardless of donor or acceptor doping!

1. Intrinsic Carrier Concentration

In an ultra-pure (intrinsic) semiconductor, thermal excitation across the gap creates equal numbers of electrons in the conduction band and holes in the valence band:

$$n_0 = p_0 = n_i$$

Using the mass action law $n_i^2 = N_c N_v e^{-E_g/k_B T}$:

$$n_i(T) = \sqrt{N_c N_v} \exp\left( -\frac{E_g}{2 k_B T} \right) = 2 \left( \frac{2\pi k_B T}{h^2} \right)^{3/2} (m_e^* m_h^*)^{3/4} \exp\left( -\frac{E_g}{2 k_B T} \right)$$

For silicon at $T = 300\text{ K}$, with $E_g = 1.12\text{ eV}$, $n_i \approx 1.0 \times 10^{10}\text{ cm}^{-3}$, compared to an atomic density of $5 \times 10^{22}\text{ cm}^{-3}$.

2. The Intrinsic Fermi Level Position

Equating $n_0 = p_0$:

$$N_c \exp\left( -\frac{E_c - E_{Fi}}{k_B T} \right) = N_v \exp\left( -\frac{E_{Fi} - E_v}{k_B T} \right)$$

Taking the natural logarithm and solving for $E_{Fi}$:

$$-\frac{E_c - E_{Fi}}{k_B T} = \ln\left(\frac{N_v}{N_c}\right) - \frac{E_{Fi} - E_v}{k_B T}$$ $$2 E_{Fi} = E_c + E_v + k_B T \ln\left( \frac{N_v}{N_c} \right)$$

Substituting $N_v / N_c = (m_h^* / m_e^*)^{3/2}$:

$$E_{Fi}(T) = \frac{E_c + E_v}{2} + \frac{3}{4} k_B T \ln\left( \frac{m_h^*}{m_e^*} \right)$$

At $T = 0\text{ K}$, the intrinsic Fermi level lies exactly at mid-gap: $E_{Fi} = (E_c + E_v)/2 = E_g/2$. As temperature rises, if $m_h^* > m_e^*$ (as in most semiconductors), $E_{Fi}$ shifts slightly upward toward the conduction band to maintain charge neutrality against the higher density of valence states.

1. The Hydrogenic Donor Impurity Model

Consider a group-V donor atom (e.g. Phosphorus or Arsenic) substituting for a host group-IV silicon atom in the crystal lattice. Four of its valence electrons participate in tetrahedral $sp^3$ covalent bonds with neighboring Si atoms. The fifth electron is attracted to the surplus positive nuclear charge $+e$ of the donor ion core.

Because the electron orbits at distances spanning many unit cells, the Coulomb potential is heavily screened by the static relative permittivity of the semiconductor crystal ($\epsilon_r \approx 11.7$ for Si, $13.1$ for GaAs), and the electron moves with effective mass $m_e^*$:

$$V(r) = -\frac{e^2}{4\pi \epsilon_r \epsilon_0 r}$$

This forms an effective hydrogen-like atom inside a dielectric medium. The effective Bohr radius of the donor ground state is:

$$a_d^* = a_0 \left( \frac{\epsilon_r}{m_e^* / m_0} \right)$$

where $a_0 = 0.529\text{ \AA}$ is the atomic Bohr radius. For silicon ($m_e^* \approx 0.26 m_0$, $\epsilon_r = 11.7$):

$$a_d^* \approx 0.529 \times \frac{11.7}{0.26} \approx 24\text{ \AA} = 2.4\text{ nm}$$

This vast orbit encloses thousands of host lattice atoms, justifying the continuum dielectric approximation.

2. Donor and Acceptor Ionization Energies

The ionization energy required to promote the bound donor electron into the conduction band is:

$$E_d = E_H \left( \frac{m_e^* / m_0}{\epsilon_r^2} \right)$$

where $E_H = 13.6\text{ eV}$ is the Rydberg ionization energy of atomic hydrogen. For silicon:

$$E_d \approx 13.6 \times \frac{0.26}{(11.7)^2} \approx 0.026\text{ eV} = 26\text{ meV}$$

Because $E_d \sim k_B T_{\text{room}} \approx 26\text{ meV}$, shallow donor levels lie just beneath the conduction band edge ($E_c - E_d$) and are nearly 100% ionized at room temperature!

Similarly, a group-III acceptor atom (e.g. Boron in Si) lacks one bonding electron, introducing a localized hole bound to a negative core with acceptor binding energy:

$$E_a = E_H \left( \frac{m_h^* / m_0}{\epsilon_r^2} \right) \sim 45\text{ meV}$$

1. Charge Neutrality & Fermi Level Evolution

In an $n$-type semiconductor with donor concentration $N_d$ and negligible acceptors ($N_a = 0$), overall charge neutrality demands:

$$n_0 = p_0 + N_d^+$$

where $N_d^+$ is the concentration of ionized donors given by Fermi statistics (including the factor of 2 for spin degeneracy of the donor ground state):

$$N_d^+ = \frac{N_d}{1 + 2 \exp\left(\frac{E_F - E_d}{k_B T}\right)}$$

2. The Three Distinct Temperature Regimes

• 1. Freeze-Out (Cryogenic) Regime ($T \to 0\text{ K}$, $k_B T \ll E_d$): Thermal energy is insufficient to ionize the donors. Most electrons remain bound to donor atoms ($N_d^+ \ll N_d$), and valence band hole generation is completely negligible ($p_0 \approx 0$). $$n_0 \approx N_d^+ \approx \sqrt{\frac{N_c N_d}{2}} \exp\left( -\frac{E_c - E_d}{2 k_B T} \right)$$ The Fermi level lies midway between the donor level and the conduction band: $$E_F(T) \approx \frac{E_c + E_d}{2} + \frac{k_B T}{2} \ln\left( \frac{N_d}{2 N_c} \right)$$ In this regime, carrier concentration rises exponentially with slope $-E_d / (2 k_B)$ on an Arrhenius plot $\ln n$ vs $1/T$.
• 2. Extrinsic / Exhaustion Regime (Room Temperature, $E_d \ll k_B T \ll E_g$): Virtually all donor atoms are completely ionized ($N_d^+ \approx N_d$), while thermal excitation across the band gap remains negligible ($n_i \ll N_d$). $$n_0 \approx N_d = \text{constant}$$ The carrier concentration is flat and independent of temperature. The Fermi level drops continuously as $T$ increases: $$E_F(T) = E_c - k_B T \ln\left( \frac{N_c(T)}{N_d} \right)$$ This is the standard operational regime for semiconductor devices.
• 3. Intrinsic Regime (High Temperatures, $k_B T \sim E_g$): Band-to-band thermal generation overwhelms the donor concentration ($n_i(T) \gg N_d$): $$n_0 \approx p_0 \approx n_i(T) \propto T^{3/2} \exp\left( -\frac{E_g}{2 k_B T} \right)$$ The Fermi level converges to the intrinsic level $E_{Fi} \approx E_g / 2$. Doping no longer controls the electrical behavior, causing semiconductor device failure.

1. Semiclassical Drift Conductivity & Mobility

In the presence of an electric field $\vec{\mathcal{E}}$, carrier momentum relaxes via collisions with acoustic phonons, optical phonons, and ionized impurities with relaxation time $\tau$. The drift velocity is:

$$\vec{v}_d = -\mu_e \vec{\mathcal{E}} = -\frac{e \tau_e}{m_e^*} \vec{\mathcal{E}}, \quad \vec{v}_{dh} = +\mu_h \vec{\mathcal{E}} = +\frac{e \tau_h}{m_h^*} \vec{\mathcal{E}}$$

where $\mu_e$ and $\mu_h$ are the electron and hole drift mobilities. The total conduction current density is the sum of electron and hole currents:

$$\vec{J} = \vec{J}_e + \vec{J}_h = (-e n \vec{v}_d) + (+e p \vec{v}_{dh}) = e (n \mu_e + p \mu_h) \vec{\mathcal{E}}$$

The electrical conductivity of the semiconductor is therefore:

$$\sigma = e (n \mu_e + p \mu_h)$$

2. Single-Carrier Hall Effect Formalism

Apply a longitudinal electric field $\mathcal{E}_x$ driving current density $J_x$, and a transverse magnetic field $\vec{B} = B_z \hat{z}$. The Lorentz force deflects mobile charges sideways along $y$:

$$\vec{F} = q (\vec{\mathcal{E}} + \vec{v} \times \vec{B})$$

In the steady state, charges accumulate on the lateral boundaries, generating a transverse Hall electric field $\mathcal{E}_y$ that exactly cancels the Lorentz deflection, enforcing zero net transverse current $J_y = 0$:

$$J_y = \sigma_0 \mathcal{E}_y - \mu B_z J_x = 0 \implies \mathcal{E}_y = \frac{1}{q n} J_x B_z$$

We define the Hall coefficient $R_H$:

$$R_H \equiv \frac{\mathcal{E}_y}{J_x B_z} = \frac{1}{q n}$$

• For an $n$-type semiconductor ($q = -e$): $$R_H = -\frac{1}{e n} < 0$$
• For a $p$-type semiconductor ($q = +e$): $$R_H = +\frac{1}{e p} > 0$$

The sign of $R_H$ unambiguously reveals the majority carrier type, and its magnitude provides a direct experimental measurement of the carrier concentration. The Hall mobility is defined as:

$$\mu_H \equiv |R_H| \sigma$$

1. Derivation of the Two-Carrier Hall Coefficient

In intrinsic or compensated semiconductors where both electrons ($n, \mu_e$) and holes ($p, \mu_h$) contribute simultaneously to transport, each carrier species experiences opposing Hall deflections.

From the Boltzmann transport equation in the low-field limit ($\mu B \ll 1$), the current densities in the $x$-$y$ plane are:

$$J_x = e (n \mu_e + p \mu_h) \mathcal{E}_x + e (n \mu_e^2 - p \mu_h^2) B_z \mathcal{E}_y$$ $$J_y = e (n \mu_e + p \mu_h) \mathcal{E}_y - e (n \mu_e^2 - p \mu_h^2) B_z \mathcal{E}_x$$

Imposing the open-circuit condition $J_y = 0$, we solve for the Hall field $\mathcal{E}_y$:

$$\mathcal{E}_y = \frac{p \mu_h^2 - n \mu_e^2}{n \mu_e + p \mu_h} B_z \mathcal{E}_x$$

Substituting $J_x \approx e (n \mu_e + p \mu_h) \mathcal{E}_x$, the two-carrier Hall coefficient is:

$$R_H = \frac{\mathcal{E}_y}{J_x B_z} = \frac{1}{e} \frac{p \mu_h^2 - n \mu_e^2}{(p \mu_h + n \mu_e)^2}$$

Remarkably:

• Even in a $p$-type material where $p > n$, the Hall coefficient can be negative ($R_H < 0$) if the electron mobility is sufficiently higher than the hole mobility ($n \mu_e^2 > p \mu_h^2$).
• The Hall coefficient vanishes ($R_H = 0$) exactly when: $$p \mu_h^2 = n \mu_e^2 \implies \frac{p}{n} = \left( \frac{\mu_e}{\mu_h} \right)^2$$ Since in silicon $\mu_e / \mu_h \approx 1400 / 450 \approx 3.1$, $R_H$ crosses zero when $p \approx 9.6 n$.

2. Cyclotron Resonance in Semiconductors

Cyclotron resonance provides the most precise direct technique for determining the effective mass tensor. A semiconductor sample is placed in a microwave cavity at cryogenic temperatures ($4.2\text{ K}$ to avoid thermal broadening $\omega_c \tau \gg 1$) with a static magnetic field $\vec{B}$.

When the microwave frequency $\omega$ matches the cyclotron frequency $\omega_c = e B / m^*$, resonant absorption of microwave power occurs:

$$P(\omega) \propto \frac{1}{1 + (\omega - \omega_c)^2 \tau^2}$$

In silicon, rotation of $\vec{B}$ relative to crystal axes splits the absorption peak into multiple lines, directly revealing the multi-valley spheroidal conduction band ellipsoids with longitudinal mass $m_l^* = 0.98 m_0$ and transverse mass $m_t^* = 0.19 m_0$.