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Chapter 8 • Theory & Derivations

Semiconductors, Junction Devices & Quantum Dots

Comprehensive semiconductor physics and low-dimensional quantum transport: direct vs indirect bandgaps, intrinsic carrier statistics, shallow donor and acceptor doping, drift and diffusion transport, Einstein relation, p-n junction electrostatics, Shockley diode equation, LEDs, photovoltaics, and quantum confinement in 2D quantum wells, 1D wires, and 0D quantum dots.

§8.1 Band Structures: Direct vs Indirect Semiconductors & Effective Masses

1. Direct vs. Indirect Band Gap Semiconductors

In crystalline semiconductors, the electrical and optical properties are governed by the band gap $E_g$ separating the highest occupied Valence Band Maximum (VBM) from the lowest unoccupied Conduction Band Minimum (CBM):

  • Direct Band Gap Semiconductors (e.g., $\text{GaAs}, \text{InP}, \text{GaN}$): Both the conduction band minimum and valence band maximum occur at the same crystal wavevector (typically the zone center $\Gamma$-point, $\vec{k} = 0$). Optical photon transitions ($\Delta \vec{k} \approx 0$) occur directly via vertical electric dipole absorption and rapid radiative electron-hole recombination ($h\nu = E_g$). Essential for efficient optoelectronics: LEDs, laser diodes.
  • Indirect Band Gap Semiconductors (e.g., $\text{Si}, \text{Ge}, \text{GaP}$): The conduction band minimum and valence band maximum occur at different wavevectors (e.g., in Silicon, the VBM is at $\Gamma$ while the CBM is near $0.85$ of the zone edge along $[100]$). Conservation of crystal momentum forbids first-order radiative transitions: photon absorption or emission requires simultaneous absorption or emission of a lattice phonon ($\hbar\vec{q}$):
    $$h\nu = E_g \pm \hbar\omega_{\text{phonon}}, \quad \vec{k}_{\text{initial}} + \vec{k}_{\text{photon}} \approx \vec{k}_{\text{final}} \pm \vec{q}$$

    Because second-order phonon-assisted transitions are far less probable, indirect semiconductors exhibit long carrier radiative lifetimes, making them ideal for photovoltaic solar cells and CMOS microelectronics, but inefficient as light emitters.

2. Kane Non-Parabolicity & Anisotropic Effective Masses

Near band extrema, expanding $E(\vec{k})$ to quadratic order yields anisotropic effective mass tensors:

$$E_c(\vec{k}) = E_c + \frac{\hbar^2}{2} \left( \frac{k_l^2}{m_l^*} + \frac{k_t^2 + k_t'^2}{m_t^*} \right)$$

where $m_l^*$ is the longitudinal mass and $m_t^*$ is the transverse mass (e.g., in Silicon, $m_l^* = 0.98 m_0$, $m_t^* = 0.19 m_0$). The density-of-states effective mass is $m_d^* = (m_l^* m_t^{*2})^{1/3}$, while the conductivity effective mass is $m_c^* = 3\left(\frac{1}{m_l^*} + \frac{2}{m_t^*}\right)^{-1}$.

§8.2 Intrinsic Carrier Statistics & The Law of Mass Action

1. Intrinsic Conduction Electron Density $n$ and Hole Density $p$

In a non-degenerate semiconductor ($E_c - E_F \gg k_BT$ and $E_F - E_v \gg k_BT$), the Fermi-Dirac distribution is accurately approximated by the classical Maxwell-Boltzmann tail:

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

Integrating over the parabolic conduction band density of states yields the thermal equilibrium electron concentration:

$$n = \int_{E_c}^\infty g_c(E) f(E) dE = N_c e^{-(E_c - E_F)/k_B T}$$

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

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

Similarly, the hole concentration in the valence band is:

$$p = \int_{-\infty}^{E_v} g_v(E) [1 - f(E)] dE = N_v e^{-(E_F - E_v)/k_B T}$$

where $N_v = 2 \left( \frac{2\pi m_h^* k_B T}{h^2} \right)^{3/2}$ is the effective density of states in the valence band.

2. The Law of Mass Action & Intrinsic Carrier Concentration $n_i$

Multiplying $n$ and $p$ eliminates the Fermi energy $E_F$ entirely, establishing the fundamental Law of Mass Action:

$$n \cdot p = N_c N_v e^{-(E_c - E_v)/k_B T} = N_c N_v e^{-E_g / k_B T} = n_i^2$$

where $n_i$ is the intrinsic carrier concentration:

$$n_i(T) = \sqrt{N_c N_v} e^{- E_g / 2 k_B T} \propto T^{3/2} e^{- E_g / 2 k_B T}$$

This product $n \cdot p = n_i^2$ holds strictly for any semiconductor in thermal equilibrium, regardless of whether it is intrinsic or heavily doped.

3. The Intrinsic Fermi Level $E_i$

In an undoped intrinsic semiconductor, charge neutrality requires $n = p = n_i$. Equating $n$ and $p$ reveals the position of the intrinsic Fermi level:

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

Because $\ln(m_h^*/m_e^*) \sim 0$, the intrinsic Fermi level lies virtually at the exact mid-gap ($E_g/2$).

§8.3 Extrinsic Doping: Donors, Acceptors & Carrier Transport

1. Shallow Hydrogenic Donor and Acceptor Impurities

Extrinsic semiconductors are created by intentionally introducing trace dopant atoms into the host lattice:

  • $n$-Type Doping (Group V in Group IV, e.g., P, As in Si): Donates an extra valence electron. The extra electron orbits the positively charged donor ion core ($+e$) in a hydrogen-like orbit screened by the large dielectric permittivity of the crystal ($\varepsilon_r \sim 12$):
    $$E_d = \frac{m_e^*}{m_0} \frac{1}{\varepsilon_r^2} E_H = \frac{m_e^*}{m_0} \frac{13.6\text{ eV}}{\varepsilon_r^2} \sim 10 - 50\text{ meV}$$

    The donor Bohr radius is expanded to $r_d = \varepsilon_r \frac{m_0}{m_e^*} a_0 \sim 30 - 80\text{ \AA}$. At room temperature ($k_BT \approx 26\text{ meV}$), virtually all donors are thermally ionized, so $n \approx N_D$.

  • $p$-Type Doping (Group III in Group IV, e.g., B, Ga in Si): Lacks one valence electron, creating a bound acceptor state near the valence band edge that accepts an electron, generating mobile positive holes ($p \approx N_A$).

2. Drift and Diffusion Transport

In the presence of an electric field $\vec{E}$ and carrier concentration gradients $\nabla n$ and $\nabla p$, total electron and hole current densities comprise both drift and diffusion components:

$$\vec{J}_n = n e \mu_n \vec{E} + e D_n \nabla n$$
$$\vec{J}_p = p e \mu_p \vec{E} - e D_p \nabla p$$

where $\mu_n, \mu_p$ are carrier mobilities and $D_n, D_p$ are diffusion coefficients. In thermal equilibrium ($\vec{J} = 0$), the drift and diffusion currents balance identically, deriving the universal Einstein Relation:

$$\frac{D_n}{\mu_n} = \frac{D_p}{\mu_p} = \frac{k_B T}{e}$$

The total electrical conductivity is given by $\sigma = e (n \mu_n + p \mu_p)$.

§8.4 The p-n Junction: Built-in Potential & Shockley Diode Equation

1. Electrostatics of the Equilibrium p-n Junction

When $p$-type and $n$-type semiconductor regions are brought into intimate metallurgical contact, the steep concentration gradient drives diffusion of majority electrons from $n$ to $p$ and holes from $p$ to $n$. This recombination uncovers fixed, ionized donor cores ($N_D^+$) on the $n$-side and ionized acceptor cores ($N_A^-$) on the $p$-side, creating a space-charge depletion region of width $W = x_p + x_n$.

The resulting internal electric field opposes further diffusion until equilibrium is established with a flat Fermi level ($dE_F/dx = 0$). The total electrostatic potential barrier across the junction is the built-in potential $V_{bi}$:

$$V_{bi} = \frac{k_B T}{e} \ln\left( \frac{N_A N_D}{n_i^2} \right)$$

Solving Poisson's equation $\frac{d^2 V}{dx^2} = - \frac{\rho(x)}{\varepsilon_s}$ under the depletion approximation yields the total depletion layer width under applied voltage $V_a$:

$$W = \sqrt{\frac{2 \varepsilon_s (V_{bi} - V_a)}{e} \left( \frac{1}{N_A} + \frac{1}{N_D} \right)}$$

2. The Shockley Ideal Diode Equation

Applying a forward bias voltage ($V_a > 0$) lowers the potential barrier to $V_{bi} - V_a$, exponentially increasing minority carrier injection across the depletion region edges. Integrating minority carrier diffusion currents yields the celebrated Shockley Diode Equation:

$$I(V_a) = I_0 \left( e^{e V_a / k_B T} - 1 \right)$$

where $I_0 = e A \left( \frac{D_n n_{p0}}{L_n} + \frac{D_p p_{n0}}{L_p} \right)$ is the reverse saturation current.

3. Optoelectronic Devices

  • Light-Emitting Diodes (LEDs): Under forward bias, electrons and holes are injected across the junction and recombine radiatively, emitting photons with peak energy $h\nu \approx E_g$ ($\lambda \approx hc/E_g$).
  • Photovoltaic Solar Cells: Photons with $h\nu > E_g$ generate electron-hole pairs within and near the depletion region. The built-in electric field sweeps electrons to the $n$-side and holes to the $p$-side, producing a photocurrent $I_{sc}$ and open-circuit photovoltage $V_{oc}$.

§8.5 Low-Dimensional Nanostructures: Quantum Wells, Wires & Quantum Dots

1. Quantum Confinement & Dimensionality Hierarchy

When the spatial dimensions of a semiconductor crystal are reduced below the exciton Bohr radius $a_B = \varepsilon_r \frac{m_0}{\mu} a_0 \sim 2 - 10\text{ nm}$, continuous electronic energy bands quantize into discrete sub-bands due to quantum confinement:

  1. 3D (Bulk): Confinement in $0$ dimensions; $g(E) \propto \sqrt{E}$.
  2. 2D (Quantum Well): Confinement along $1$ dimension ($L_z \sim \text{nm}$), free in $x, y$. Energy spectrum: $E(k_x, k_y, n_z) = E_{n_z} + \frac{\hbar^2(k_x^2 + k_y^2)}{2m^*}$. Density of states consists of discrete staircase steps: $g_{\text{2D}}(E) = \frac{m^*}{\pi\hbar^2} \sum_{n} \Theta(E - E_n)$.
  3. 1D (Quantum Wire): Confinement along $2$ dimensions ($L_x, L_y \sim \text{nm}$), free along $z$. Energy spectrum: $E(k_z, n_x, n_y) = E_{n_x, n_y} + \frac{\hbar^2 k_z^2}{2m^*}$. Density of states exhibits sharp $E^{-1/2}$ Van Hove singularities: $g_{\text{1D}}(E) \propto \sum_n (E - E_n)^{-1/2}$.
  4. 0D (Quantum Dot): Confinement along all $3$ spatial dimensions ($L_x, L_y, L_z \sim \text{nm}$). Completely discrete "artificial atom" spectrum: $E_{n_x, n_y, n_z} = \frac{\pi^2\hbar^2}{2m^*} \left(\frac{n_x^2}{L_x^2} + \frac{n_y^2}{L_y^2} + \frac{n_z^2}{L_z^2}\right)$. Density of states is a collection of Dirac delta-functions: $g_{\text{0D}}(E) = 2 \sum_i \delta(E - E_i)$.

2. Size-Dependent Bandgap Tuning in Quantum Dots (Brus Formula)

For a spherical semiconductor nanocrystal (quantum dot) of radius $R$, Louis Brus (1984) derived the quantum confinement energy shift using the effective mass approximation:

$$E_g(R) = E_{g,\text{bulk}} + \frac{\hbar^2 \pi^2}{2 \mu R^2} - \frac{1.786 e^2}{4\pi\varepsilon_0 \varepsilon_r R}$$

where $\frac{1}{\mu} = \frac{1}{m_e^*} + \frac{1}{m_h^*}$ is the exciton reduced mass.

  • The first correction term ($+ \hbar^2\pi^2 / 2\mu R^2 \propto 1/R^2$) represents kinetic quantum confinement energy, pushing the effective band gap upward.
  • The second correction term ($- 1.786 e^2 / 4\pi\varepsilon_0 \varepsilon_r R \propto -1/R$) represents screened Coulomb attraction between electron and hole.

By simply tuning the nanocrystal radius $R$ from $1.5\text{ nm}$ to $6.0\text{ nm}$, the emission color of cadmium selenide ($\text{CdSe}$) quantum dots can be tuned continuously across the entire visible spectrum from vibrant blue to deep red.

Solved Problem Example 8.1: Intrinsic Carrier Concentration and Resistivity of Silicon at 300 K

At $T = 300 \text{ K}$, Silicon has a band gap $E_g = 1.12 \text{ eV}$, effective conduction band density of states $N_c = 2.80 \times 10^{19} \text{ cm}^{-3}$, effective valence band density of states $N_v = 1.04 \times 10^{19} \text{ cm}^{-3}$, electron mobility $\mu_n = 1450 \text{ cm}^2/\text{V}\cdot\text{s}$, and hole mobility $\mu_p = 450 \text{ cm}^2/\text{V}\cdot\text{s}$. (a) Calculate the intrinsic carrier concentration $n_i$. (b) Determine the electrical conductivity $\sigma$ and electrical resistivity $\rho$ of pure intrinsic Silicon.

Step 1: Calculate the Product N_c N_v in SI Units
$$N_c N_v = (2.80 \times 10^{25}\text{ m}^{-3})(1.04 \times 10^{25}\text{ m}^{-3}) \approx 2.912 \times 10^{50}\text{ m}^{-6} \implies \sqrt{N_c N_v} \approx 1.7065 \times 10^{25}\text{ m}^{-3}$$

Convert densities from cm^-3 to m^-3 and compute their geometric mean.

Step 2: Evaluate the Boltzmann Exponential Factor
$$\frac{E_g}{2 k_B T} = \frac{1.12\text{ eV}}{2 \times (0.02585\text{ eV})} \approx \frac{1.12}{0.0517} \approx 21.663 \implies e^{-E_g / 2k_BT} = e^{-21.663} \approx 3.907 \times 10^{-10}$$

Evaluate the thermal excitation factor at 300 K.

Step 3: Calculate the Intrinsic Carrier Concentration n_i
$$n_i = \sqrt{N_c N_v} e^{-E_g/2k_BT} = (1.7065 \times 10^{25}\text{ m}^{-3}) \times (3.907 \times 10^{-10}) \approx 6.67 \times 10^{15}\text{ m}^{-3} = 6.67 \times 10^9\text{ cm}^{-3}$$

Compute n_i at 300 K.

Step 4: Compute Intrinsic Electrical Conductivity and Resistivity
$$\sigma_i = e n_i (\mu_n + \mu_p) = (1.6022 \times 10^{-19}\text{ C}) \times (6.67 \times 10^{15}\text{ m}^{-3}) \times (0.1450 + 0.0450\text{ m}^2/\text{V}\cdot\text{s}) = (1.069 \times 10^{-3}) \times (0.190) \approx 2.03 \times 10^{-4}\text{ }\Omega^{-1}\cdot\text{m}^{-1}$$

Convert mobilities to m^2/V*s and compute sigma_i.

Step 5: Determine Intrinsic Resistivity rho_i
$$\rho_i = \frac{1}{\sigma_i} = \frac{1}{2.03 \times 10^{-4}\text{ }\Omega^{-1}\cdot\text{m}^{-1}} \approx 4926\text{ }\Omega\cdot\text{m} \approx 4.93 \times 10^5\text{ }\Omega\cdot\text{cm}$$

Take the reciprocal of conductivity.

Final Answer & Physical Insight

n_i = 6.67 \times 10^9 \text{ cm}^{-3} \quad (6.67 \times 10^{15} \text{ m}^{-3}), \quad \sigma_i = 2.03 \times 10^{-4} \text{ }Ω^{-1}\cdot\text{m}^{-1}, \quad \rho_i = 4926 \text{ }Ω\cdot\text{m}

Solved Problem Example 8.2: Hydrogenic Donor Binding Energy and Bohr Radius in Silicon

Phosphorus ($Z = 15$) is added as a donor impurity to Silicon, which has a relative dielectric permittivity $\varepsilon_r = 11.7$ and an effective electron mass $m_e^* = 0.26 m_0$. Using the hydrogenic donor model: (a) Calculate the donor ionization energy $E_d$ in $\text{meV}$. (b) Calculate the effective donor Bohr radius $r_d$ in Angstroms, and determine how many Silicon unit cells ($a = 5.43 \text{ Å}$) are enclosed within the donor electron orbit.

Step 1: Calculate the Donor Ionization Energy E_d
$$E_d = \left(\frac{m_e^*}{m_0}\right) \frac{1}{\varepsilon_r^2} E_H = 0.26 \times \frac{13.606\text{ eV}}{(11.7)^2} = \frac{3.5376\text{ eV}}{136.89} \approx 0.02584\text{ eV} = 25.84\text{ meV}$$

Scale the hydrogen Rydberg energy (13.6 eV) by the effective mass and dielectric screening factor epsilon_r^2.

Step 2: Calculate the Effective Donor Bohr Radius r_d
$$r_d = \varepsilon_r \left(\frac{m_0}{m_e^*}\right) a_0 = 11.7 \times \left(\frac{1}{0.26}\right) \times (0.5292\text{ \AA}) \approx 11.7 \times 3.846 \times 0.5292\text{ \AA} \approx 23.81\text{ \AA}$$

Scale the atomic Bohr radius a_0 by epsilon_r and m_0 / m_e*.

Step 3: Calculate the Number of Enclosed Unit Cells
$$V_{\text{orbit}} = \frac{4}{3}\pi r_d^3 = \frac{4}{3}\pi (23.81\text{ \AA})^3 \approx 5.653 \times 10^4\text{ \AA}^3, \quad V_{\text{cell}} = a^3 = (5.431\text{ \AA})^3 \approx 160.2\text{ \AA}^3 \implies N_{\text{cells}} = \frac{5.653 \times 10^4}{160.2} \approx 353\text{ unit cells}$$

Divide the volume of the donor electron cloud by the unit cell volume of silicon.

Final Answer & Physical Insight

E_d = 25.8 \text{ meV}, \quad r_d = 23.8 \text{ Å}, \quad N_{\text{cells}} \approx 353 \text{ unit cells}

Solved Problem Example 8.3: Quantum Dot Size-Dependent Emission Tuning for Cadmium Selenide

Cadmium Selenide ($\text{CdSe}$) has a bulk band gap $E_{g,\text{bulk}} = 1.74 \text{ eV}$, dielectric constant $\varepsilon_r = 10.6$, electron effective mass $m_e^ = 0.13 m_0$, and hole effective mass $m_h^ = 0.45 m_0$. Using the Brus quantum confinement formula: (a) Calculate the reduced exciton mass $\mu$. (b) Determine the effective emission bandgap $E_g(R)$ and emission wavelength $\lambda$ for a spherical $\text{CdSe}$ quantum dot of radius $R = 2.0 \text{ nm}$.

Step 1: Calculate the Reduced Exciton Mass mu
$$\frac{1}{\mu} = \frac{1}{m_e^*} + \frac{1}{m_h^*} = \frac{1}{0.13 m_0} + \frac{1}{0.45 m_0} = \frac{7.692 + 2.222}{m_0} = \frac{9.914}{m_0} \implies \mu \approx 0.1009 m_0 \approx 9.19 \times 10^{-32}\text{ kg}$$

Compute the reduced mass of the electron-hole pair.

Step 2: Calculate the Kinetic Confinement Energy Shift Delta E_kin
$$\Delta E_{\text{kin}} = \frac{\hbar^2 \pi^2}{2 \mu R^2} = \frac{(1.0546 \times 10^{-34}\text{ J}\cdot\text{s})^2 \pi^2}{2 \times (9.19 \times 10^{-32}\text{ kg}) \times (2.0 \times 10^{-9}\text{ m})^2} = \frac{1.0978 \times 10^{-67}}{7.352 \times 10^{-49}}\text{ J} \approx 1.493 \times 10^{-19}\text{ J} \approx 0.932\text{ eV}$$

Evaluate the particle-in-a-sphere kinetic confinement energy in electron-volts.

Step 3: Calculate the Screened Coulomb Attraction Energy Shift Delta E_Coul
$$\Delta E_{\text{Coul}} = \frac{1.786 e^2}{4\pi\varepsilon_0 \varepsilon_r R} = \frac{1.786 \times (1.440\text{ eV}\cdot\text{\AA})}{10.6 \times (20.0\text{ \AA})} = \frac{2.5718}{212.0}\text{ eV} \approx 0.012\text{ eV}$$

Compute the electrostatic Coulomb correction term.

Step 4: Compute the Effective Quantum Dot Bandgap E_g(R)
$$E_g(R) = E_{g,\text{bulk}} + \Delta E_{\text{kin}} - \Delta E_{\text{Coul}} = 1.74\text{ eV} + 0.932\text{ eV} - 0.012\text{ eV} \approx 2.66\text{ eV}$$

Sum the bulk gap and quantum corrections.

Step 5: Determine the Emission Wavelength lambda
$$\lambda = \frac{h c}{E_g(R)} = \frac{1239.84\text{ eV}\cdot\text{nm}}{2.66\text{ eV}} \approx 466\text{ nm} \quad (\text{Vibrant Blue Emission})$$

Convert the band gap energy to emission wavelength. Note that while bulk CdSe emits in the deep red (713 nm), 2.0 nm quantum dots emit blue light at 466 nm.

Final Answer & Physical Insight

\mu = 0.101 m_0, \quad E_g(R=2\text{ nm}) = 2.66 \text{ eV}, \quad \lambda = 466 \text{ nm} \quad (\text{Blue Light})

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