Chapter 18: Quantum-Well Devices ================================ Source: Hunsperger, sixth edition (2009), Problems 18.1--18.9, printed pages 398--399. Infinite-well convention ------------------------ For well width :math:`L_z`, a subband edge is :math:`E_n=\hbar^2\pi^2n^2/(2m^*L_z^2)`. Including spin degeneracy, each occupied 2D subband adds :math:`m^*/(\pi\hbar^2)` to density of states per area per energy. The number of states integrated up to energy has linear segments; the density of states itself has steps. Problem 18.1 — MQW versus superlattice -------------------------------------- **Working.** Multiple wells with thick barriers have nearly independent confined states: tunnelling between neighboring wells is weak. With thin barriers, adjacent wavefunctions overlap and the repeated potential forms coupled minibands along the growth direction. **Answer.** MQW normally denotes substantially isolated wells; superlattice emphasizes coherent coupling and miniband formation. **Check.** Merely counting the number of layers cannot distinguish them. Barrier thickness and electronic coupling are decisive. Problem 18.2 — GaAs subband steps --------------------------------- **Working.** State the material assumption :math:`m_e^*=0.067m_0`. With :math:`L_z=7` nm, .. math:: (E_1,E_2,E_3)=(0.114539,0.458154,1.03085)\ {\rm eV}. The step magnitude is .. math:: \Delta D_{2D}=\frac{0.067m_0}{\pi\hbar^2} =2.79880\times10^{13}\ {\rm eV^{-1}cm^{-2}}. **Answer.** Equal DOS jumps occur at those three energies. For the :math:`100\,\mu{\rm m}\times100\,\mu{\rm m}` area, each whole-well DOS step is :math:`2.79880\times10^9\,{\rm eV}^{-1}`. .. figure:: /_static/knowledge_base/worked_exercises/integrated_optics_6e/quantum_well_dos.svg :alt: Infinite-well subband energies scale as squared order while the two-dimensional density of states gains equal-height steps. :width: 100% The idealized first three GaAs conduction subbands. **Check.** Energies scale as :math:`n^2`, whereas each DOS step has the same height. The infinite-barrier assumption becomes poor for high subbands above a real barrier. Problem 18.3 — Fundamental interband quantum-well emission ---------------------------------------------------------- **Working.** Electron and hole confinement both increase transition energy. With stated approximations :math:`E_g=1.43` eV, :math:`m_e^*=0.067m_0`, and heavy-hole mass :math:`m_h^*=0.5m_0`, .. math:: E_\gamma=E_g+E_{1e}+E_{1h} =1.43+0.114539+0.0153482=1.55989\,{\rm eV}. **Answer.** :math:`\lambda_0=1.239841984/E_\gamma \simeq0.79483\,\mu{\rm m}`. **Check.** The hole term adds despite being plotted downward in an electron band diagram. Finite barriers, excitons, and the selected valence band shift the actual wavelength. Problem 18.4 — SEED optical feedback and logic ---------------------------------------------- **Working.** Absorbed light changes the current in an MQW p-i-n diode. In an appropriate electrical load circuit, that current changes diode voltage. The quantum-confined Stark effect makes absorption depend on that voltage, closing a feedback loop. Suitable load lines can yield two stable optical/electrical states. **Answer.** This self-consistent absorption-voltage feedback enables bistability and therefore potential binary switching or logic. **Check.** A bare passive MQW layer is not automatically bistable. The bias circuit, optical operating point, and feedback sign matter. Problem 18.5 — Six-nanometre well and twenty-well gain ------------------------------------------------------ **Working.** With :math:`E_g=1.52` eV and the same GaAs electron/heavy-hole masses as 18.3, the 6 nm infinite well gives :math:`E_{1e}+E_{1h}\simeq0.17679` eV. Thus :math:`\lambda_0\simeq1.239841984/1.69679=0.73070\,\mu{\rm m}`. For independent equally pumped wells, modal gain sums: :math:`g_{\rm modal}=\sum_j\Gamma_jg_{{\rm mat},j}`. **Answer.** (a) Approximately 730.7 nm within the infinite-well model. (b) If 80 inverse centimetres is each well's **modal contribution** and all overlaps remain equal, twenty contribute 1600 inverse centimetres. **Check.** If 80 is material gain, it does not automatically multiply by twenty. Optical confinement, carrier distribution, and total layer thickness must be specified. A high infinite-well level may not be bound in a finite AlGaAs barrier. Problem 18.6 — QCSE extinction for two propagation geometries ------------------------------------------------------------- **Working.** The absorption change is :math:`\Delta\alpha=2000-10=1990\,{\rm cm}^{-1}`. For perpendicular passage through 10 micrometres, .. math:: \frac{T_{\rm on}}{T_{\rm off}}=e^{\Delta\alpha(0.001)} =7.31553. For parallel passage over 50 micrometres with confinement 0.8, the exponent is :math:`1990(0.8)(0.005)=7.96`. **Answer.** Ratios are 7.316 (8.642 dB) and 2864.07 (34.57 dB), respectively. **Check.** Compare ratios of transmitted states, not absorbed fractions. The confinement factor multiplies the absorption-length product. Problem 18.7 — InGaAs subband density of states ----------------------------------------------- **Working.** Substitute the supplied electron mass :math:`0.045m_0` and 7 nm width: .. math:: (E_1,E_2,E_3)=(0.170535,0.682141,1.53482)\ {\rm eV}, \qquad \Delta D_{2D}=1.87979\times10^{13}\ {\rm eV^{-1}cm^{-2}}. **Answer.** Equal steps occur at those energies. Multiplying by the stated area gives :math:`1.87979\times10^9\,{\rm eV}^{-1}` per whole-well step. **Check.** The hole mass does not enter the requested conduction-band DOS. The lighter electron mass raises confinement energy and reduces DOS step height compared with GaAs. Problem 18.8 — Temperature-dependent threshold ---------------------------------------------- **Working.** Use the empirical characteristic-temperature law :math:`I_{\rm th}(T_2)/I_{\rm th}(T_1)=e^{(T_2-T_1)/T_0}`. Temperature differences have the same numerical value in kelvin and degrees Celsius: .. math:: I_{\rm th}(45^\circ{\rm C})=10\,{\rm mA}\,e^{24/425} =10.58095\,{\rm mA}. **Answer.** Approximately 10.58 mA. **Check.** :math:`T_0` is a fitted temperature scale, not a temperature to which 273.15 must be added in this ratio. Problem 18.9 — Why quantum wells can narrow laser linewidth ----------------------------------------------------------- **Working.** A step-like electronic DOS and strong differential gain can reduce the carrier-number change needed to alter stimulated emission. The associated amplitude-phase coupling and spontaneous-emission contribution to the lasing mode can be reduced in a well-designed QW laser. **Answer.** Quantum confinement can improve differential gain and noise/phase dynamics, producing a narrower line at comparable operating conditions. **Check.** Linewidth also depends on cavity loss, output power, linewidth-enhancement factor, temperature, and technical noise. Adding wells alone does not guarantee a narrower spectrum.