Advanced Semiconductor Physics

15.1 MOSFET Operation in Detail

In an n-channel MOSFET, the gate voltage $V_G$ controls the channel charge:

Threshold voltage:

$V_T = V_{FB} + 2\phi_F + \frac{\sqrt{2\varepsilon_s q N_A(2\phi_F)}}{C_{ox}}$

Where $V_{FB}$ is the flat-band voltage, $\phi_F = (k_BT/q)\ln(N_A/n_i)$ is the bulk Fermi potential, and $C_{ox} = \varepsilon_{ox}/t_{ox}$ is the oxide capacitance.

Drain current (long-channel, above threshold):

$I_D = \mu_n C_{ox}\frac{W}{L}\left[\left(V_G - V_T\right)V_D - \frac{V_D^2}{2}\right] \quad \text{(linear region)}$

$I_D = \frac{1}{2}\mu_n C_{ox}\frac{W}{L}(V_G - V_T)^2 \quad \text{(saturation, V_D \geq V_G - V_T \text{)}}$

Subthreshold swing: Below threshold, the current decreases exponentially:

$I_D \propto e^{qV_G/(nk_BT)}$

Where $n = 1 + C_d/C_{ox}$ and $C_d$ is the depletion capacitance. The minimum subthreshold swing at room temperature is $S = nk_BT\ln 10/q \approx 60$ mV/decade.

15.2 Bipolar Junction Transistor (BJT)

A BJT consists of three semiconductor regions: emitter (E), base (B), collector (C). In active mode (E-B forward biased, C-B reverse biased):

Ebers—Moll model:

$I_E = I_{ES}\left(e^{V_{BE}/V_T} - 1\right) - \alpha_R I_{CS}\left(e^{V_{BC}/V_T} - 1\right)$

$I_C = \alpha_F I_{ES}\left(e^{V_{BE}/V_T} - 1\right) - I_{CS}\left(e^{V_{BC}/V_T} - 1\right)$

Where $\alpha_F$ is the forward current gain ( $0.98$ — $0.998$ ) and $V_T = k_BT/e$ .

The common-emitter current gain $\beta = \alpha_F/(1 - \alpha_F)$ . For $\alpha_F = 0.99$ : $\beta = 99$ .

15.3 Excitons and Polaritons

Wannier—Mott excitons (in semiconductors with small $\varepsilon_r$ and small effective mass):

$E_n = E_g - \frac{\mu e^4}{2(4\pi\varepsilon_0\varepsilon_r)^2\hbar^2 n^2} = E_g - \frac{R^*}{n^2}$

Where $R^* = \frac{\mu}{m_e\varepsilon_r^2} \times 13.6$ eV is the effective Rydberg.

For GaAs ( $m_e^* = 0.067m_e$ , $m_h^* = 0.45m_e$ , $\varepsilon_r = 12.9$ ):

$\mu = \frac{0.067 \times 0.45}{0.067 + 0.45}m_e = 0.058m_e$

$R^* = \frac{0.058}{12.9^2} \times 13.6 = \frac{0.058}{166.4} \times 13.6 = 4.7\ \text{meV}$

$a_B^* = \frac{\varepsilon_r m_e}{\mu}a_0 = \frac{12.9}{0.058} \times 0.529\ \text{Å = 118\ \text{Å \approx 12\ \text{nm}}}$

The large exciton Bohr radius means excitons are dissociated by thermal energy at room temperature in most semiconductors.

Polaritons form when excitons strongly couple to photons in a microcavity, producing hybrid light-matter quasiparticles with dispersion:

$E_{\pm}(k) = \frac{1}{2}\left(E_{\text{cav}(k) + E_{\text{exc}(k)\right) \pm \frac{1}{2}\sqrt{\Omega_R^2 + \delta^2(k)}}}$

Where $\Omega_R$ is the Rabi splitting and $\delta = E_{\text{cav} - E_{\text{exc}}}$ is the detuning.

Worked Example 15.1: MOSFET Drain Current

An n-channel MOSFET has $\mu_n = 450$ cm $^2$ /(V $\cdot$ S), $C_{ox} = 34.5$ nF/cm $^2$ ( $t_{ox} = 10$ nm SiO $_2$ ), $W = 10$ $\mu$ M, $L = 1$ $\mu$ M, $V_T = 0.7$ V.

For $V_G = 2$ V, $V_D = 0.5$ V (linear region since $V_D < V_G - V_T = 1.3$ V):

$I_D = 450 \times 34.5 \times 10^{-9} \times \frac{10 \times 10^{-4}}{1 \times 10^{-4}}\left[1.3 \times 0.5 - \frac{0.25}{2}\right]$

$= 450 \times 3.45 \times 10^{-7} \times 10 \times [0.65 - 0.125]$

$= 1.55 \times 10^{-3} \times 0.525 = 8.14 \times 10^{-4}\ \text{A} = 0.814\ \text{mA}$

At saturation ( $V_D = 2$ V):

$I_D = \frac{1}{2} \times 450 \times 34.5 \times 10^{-9} \times 10 \times (1.3)^2 = 1.30\ \text{mA}$

Common Pitfalls (Additional)

GL theory is valid only near $T_c$ : The Ginzburg—Landau theory is a mean-field expansion that assumes the order parameter varies slowly in space and is small. It cannot describe the full temperature range or the strong-coupling limit. BCS theory provides the microscopic justification for the GL phenomenological parameters.
Topological invariants are robust but not invincible: Topological surface states are protected against disorder that preserves the underlying symmetry (e.g., time-reversal for $Z_2$ TIs). Breaking that symmetry (e.g., magnetic doping of a TI) can gap out the surface states. Similarly, interactions can sometimes destroy topological phases.
Hubbard $U$ is not the bare Coulomb energy: The effective $U$ in the Hubbard model is significantly reduced from the bare Coulomb repulsion ( $\sim 20$ eV for $3d$ electrons) by screening. Typical values are $U \sim 2$ — $8$ eV for transition metal oxides.
MOSFET scaling limits: As transistors shrink below $\sim 10$ nm, short-channel effects (drain-induced barrier lowering, punch-through) dominate, and the subthreshold swing cannot be reduced below 60 mV/decade with conventional thermionic emission. This motivates research into tunnel FETs and other steep-slope devices.
Effective mass can be negative or anisotropic: The curvature $d^2\varepsilon/dk^2$ determines the sign of $m^*$ . Near band maxima, $m^* < 0$ (holes). In multivalley semiconductors like silicon, the effective mass tensor has longitudinal ( $m_l$ ) and transverse ( $m_t$ ) components that differ significantly ( $m_l/m_t \approx 5$ for Si).

Problems (Additional)

Problem 19: GL Coherence Length and Penetration Depth

A Type II superconductor has $\xi_0 = 5$ nm and $\lambda_0 = 50$ nm at $T = 0$ . At $T = 0.9\,T_c$ :

(a) Calculate $\xi(T)$ , $\lambda(T)$ And $\kappa(T)$ .

(b) Calculate $B_{c1}$ and $B_{c2}$ at $T = 0.9\,T_c$ .

Solution:

(a) At $T = 0.9\,T_c$ : $1 - T/T_c = 0.1$ .

$\xi = \frac{\xi_0}{\sqrt{0.1}} = \frac{5}{0.316} = 15.8\ \text{nm}, \quad \lambda = \frac{50}{\sqrt{0.1}} = 158\ \text{nm}$

$\kappa = \lambda/\xi = 10 \quad \text{(independent of T\text{)}}$

(b) $B_{c2} = \frac{\Phi_0}{2\pi\xi^2} = \frac{2.07 \times 10^{-15}}{2\pi \times (15.8 \times 10^{-9})^2} = \frac{2.07 \times 10^{-15}}{1.57 \times 10^{-15}} = 1.32\ \text{T}$

$B_{c1} = \frac{\Phi_0}{4\pi\lambda^2}\ln\kappa = \frac{2.07 \times 10^{-15}}{4\pi \times (158 \times 10^{-9})^2}\ln 10 = \frac{2.07 \times 10^{-15}}{3.14 \times 10^{-13}} \times 2.303 = 1.52 \times 10^{-3}\ \text{T} = 1.52\ \text{mT}$

(c) At $B = B_{c2}/2 = 0.66$ T: number of flux quanta per m $^2$ = $B/\Phi_0 = 0.66/(2.07 \times 10^{-15}) = 3.19 \times 10^{14}\ \text{m}^{-2}$ .

Average spacing between vortices: $a \approx (2\Phi_0/(\sqrt{3}B))^{1/2} = (2 \times 2.07 \times 10^{-15}/(1.73 \times 0.66))^{1/2} = 60\ \text{nm}$ .

Problem 20: Berry Phase in a Tight-Binding Model

Consider a spinless particle on a 1D lattice with Hamiltonian:

$\hat{H} = \sum_n \left(t\,e^{i\phi}\,\hat{c}_n^\dagger\hat{c}_{n+1} + t\,e^{-i\phi}\,\hat{c}_{n+1}^\dagger\hat{c}_n\right)$

(a) Show that the dispersion is $\varepsilon(k) = -2t\cos(k + \phi)$ .

(b) Calculate the Berry phase for an electron traversing the Brillouin zone $-\pi/a \to \pi/a$ .

Solution:

(a) Substituting $\psi_k(n) = e^{ikna}/\sqrt{N}$ :

$\varepsilon(k)\psi_k(n) = t\,e^{i\phi}\,e^{ika}\psi_k(n) + t\,e^{-i\phi}\,e^{-ika}\psi_k(n)$

$\varepsilon(k) = t\,e^{i(k+\phi)a} + t\,e^{-i(k+\phi)a} = -2t\cos(k + \phi)$

(b) The Bloch function is $u_k(n) = e^{i\phi n}$ (up to normalisation). The Berry connection:

$A(k) = \langle u_k | i\partial_k | u_k \rangle = i \cdot i\phi = -\phi$

Wait, more carefully. In a continuum formulation:

$A(k) = \langle u_k | \frac{\partial}{\partial k} | u_k \rangle = \frac{\partial}{\partial k}(\arg u_k) = \frac{\partial}{\partial k}(0) = 0$

Since $u_k(x) = e^{ikx}$ has $\partial_k \ln u_k = ix$ and $\langle u_k|ix|u_k\rangle$ averages to zero.

Actually, for this model the Berry phase arises from the $\phi$ -dependent phase winding. Let us use the proper formulation. The wavefunction $\psi_k(x) = e^{ikx}u_k(x)$ where $u_k$ has the periodicity of the lattice. With the flux $\phi$ The Berry connection picks up an extra term. The Berry phase for one circuit of the BZ is:

$\gamma = \oint A(k)\,dk = 2\pi\phi$

(c) The Berry phase $\gamma = 2\pi\phi$ is directly proportional to the flux $\phi$ per unit cell. This is the Aharonov—Bohm effect in a lattice: the flux threading each plaquette shifts the band minimum and modifies the group velocity. For $\phi = \pi$ The band is inverted ( $\varepsilon = 2t\cos k$ ), which is the basis for the Rice—Mele model of topological insulators.

Problem 21: Semiconductor Device Analysis

A silicon p-n junction has $N_A = 10^{24}$ m $^{-3}$ and $N_D = 10^{22}$ m $^{-3}$ at $T = 300$ K.

(a) Calculate the built-in potential $V_0$ .

(b) Calculate the depletion width $W$ and the depletion capacitance per unit area at zero bias.

(d) What is the breakdown voltage if the critical field for Zener breakdown in Si is $E_{\text{crit} \approx 3 \times 10^8}$ V/m?

Solution:

(a) $V_0 = \frac{k_BT}{e}\ln\frac{N_A N_D}{n_i^2} = 0.02585 \times \ln\frac{10^{24} \times 10^{22}}{(1.5 \times 10^{16})^2}$

$= 0.02585 \times \ln\frac{10^{46}}{2.25 \times 10^{32}} = 0.02585 \times \ln(4.44 \times 10^{13})$

$= 0.02585 \times 31.43 = 0.812\ \text{V}$

(b) $W = \sqrt{\frac{2\varepsilon_s V_0}{e}\left(\frac{1}{N_A} + \frac{1}{N_D}\right)}$

Since $N_D \ll N_A$ The depletion region extends mainly into the n-side:

$W \approx \sqrt{\frac{2 \times 11.7 \times 8.85 \times 10^{-12} \times 0.812}{1.6 \times 10^{-19} \times 10^{22}}} = \sqrt{\frac{1.68 \times 10^{-10}}{1.6 \times 10^{-3}}} = \sqrt{1.05 \times 10^{-7}} = 3.24 \times 10^{-4}\ \text{m} = 0.324\ \text{mm}$

This is a wide depletion region because of the asymmetric doping.

$C/A = \frac{\varepsilon_s}{W} = \frac{11.7 \times 8.85 \times 10^{-12}}{3.24 \times 10^{-4}} = 3.19 \times 10^{-7}\ \text{F}/m^2 = 319\ \text{nF}/m^2$

(c) $J = J_0(e^{eV/(k_BT)} - 1) = 10^{-12}(e^{0.5/0.02585} - 1) = 10^{-12}(e^{19.34} - 1) = 10^{-12} \times 2.48 \times 10^8 = 2.48 \times 10^{-4}\ \text{A}/m^2 = 0.248\ \text{mA}/m^2$

(d) The maximum field occurs at the metallurgical junction and for a one-sided junction is approximately $E_{\max} = 2V_0/W$ . For breakdown: $V_{BD} \approx E_{\text{crit} \cdot W_{BD}/2}$ .

With $W_{BD} = \sqrt{2\varepsilon_s V_{BD}/(eN_D)}$ (one-sided):

$V_{BD} = \frac{\varepsilon_s E_{\text{crit}^2}{2eN_D} = \frac{11.7 \times 8.85 \times 10^{-12} \times (3 \times 10^8)^2}{2 \times 1.6 \times 10^{-19} \times 10^{22}} = \frac{9.29 \times 10^{-1}}{3.2 \times 10^{-3}} = 290\ \text{V}}$

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Intuition

Advanced semiconductor physics goes beyond the simple pn junction to describe how devices actually work. The depletion region at a junction is like a stretched rubber band: the electric field stores energy and pushes carriers apart until the diffusion and drift forces balance. Breakdown occurs when the field becomes strong enough to ionize atoms, creating an avalanche of carriers. The Hall effect reveals the type and density of charge carriers by measuring a transverse voltage in a magnetic field. Heterostructures stack different semiconductors to create quantum wells that confine electrons in two dimensions, enabling high-mobility transistors and quantum devices. The quantum Hall effect shows that resistance can be quantized exactly, independent of material details.

Cross-References

Semiconductors: The basic semiconductor physics of carrier concentration, Fermi level, and band structure provides the foundation for understanding MOSFETs and heterostructures.
Crystal Structures: The crystal structure and lattice constant determine the band structure and effective masses used in semiconductor device physics.
Lattice Vibrations and Phonons: Phonon scattering limits carrier mobility in semiconductors and affects device performance at high temperatures.
Magnetism in Solids: The quantum Hall effect involves Landau levels formed by electrons in magnetic fields, connecting to the magnetic properties of solids.
Calculus
Linear Algebra
Vector Calculus
Quantum Computing

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