Lecture 1
Question
Explain why silicon has dominated electronics. a. It is the most abundant element on Earth. b. It is the only material that can be used for electronics. c. It is the easiest material to work with. d. It has excellent electrical properties and is cost-effective.
Answer
D Silicon has a suitable bandgap (~1.1eV), forms a stable native oxide (SiO₂), and is abundant enough to be cheap to produce at scale.
Question
What is meant by “doping” a semiconductor? a. Adding impurities to increase the density of free electrons or holes. b. Removing impurities to increase the density of free electrons or holes. c. Removing impurities to decrease the density of free electrons or holes. d. Adding impurities to decrease the density of free electrons or holes.
Answer
A Introducing donor (group V, e.g. phosphorus) or acceptor (group III, e.g. boron) atoms into the crystal lattice to create n-type or p-type material respectively.
Question
What is meant by a “hole” in a semiconductor? a. A missing neutron in the nucleus that acts as a neutral charge carrier. b. A missing proton in the nucleus that acts as a positive charge carrier. c. A missing electron in the valence band that acts as a positive charge carrier. d. A missing electron in the conduction band that acts as a negative charge carrier.
Answer
C When an electron leaves the valence band, the vacancy it leaves behind behaves as a mobile positive charge carrier. Adjacent valence electrons can fill the hole, effectively propagating it through the lattice.
Question
What is meant by the “band gap” of a semiconductor? a. The energy required to create an electron-hole pair. b. The energy difference between the valence band maximum and the conduction band minimum. c. The energy difference between an electron and a hole. d. The energy required to remove an electron from the atom.
Answer
B The minimum energy an electron needs to transition from the valence band to the conduction band. For silicon this is ~1.1eV. Note that (a) is closely related but the band gap is specifically the energy difference between the band edges.
Question
What is Moore’s Law? a. The observation that the cost of microchips halves every two years. b. The observation that the number of transistors on a microchip doubles every year. c. The observation that the speed of processors doubles every year. d. The observation that the number of transistors on a microchip doubles every two years.
Answer
D An empirical observation by Gordon Moore (1965), not a physical law. It has held roughly true for decades but is now slowing as feature sizes approach atomic limits.
Question
What is the difference between a direct and indirect band gap semiconductor? a. Indirect band gap semiconductors have a larger band gap than direct band gap semiconductors. b. Direct band gap semiconductors can emit light efficiently, while indirect band gap semiconductors cannot. c. Direct band gap semiconductors have a larger band gap than indirect band gap semiconductors. d. Indirect band gap semiconductors can emit light efficiently, while direct band gap semiconductors cannot.
Answer
B In a direct band gap (e.g. GaAs), the conduction band minimum and valence band maximum align in momentum space, so electron-hole recombination can emit a photon directly. Indirect materials (e.g. Si) require a phonon to conserve momentum, making radiative recombination inefficient.
Question
What is the difference between an intrinsic and extrinsic semiconductor? a. Intrinsic semiconductors have impurities, while extrinsic semiconductors do not. b. Intrinsic semiconductors are used in photonics, while extrinsic semiconductors are used in electronics. c. Intrinsic semiconductors are used in electronics, while extrinsic semiconductors are used in photonics. d. Intrinsic semiconductors are pure, while extrinsic semiconductors have impurities added.
Answer
D Intrinsic: pure crystal, carrier concentration determined solely by thermal generation. Extrinsic: intentionally doped to control carrier type and concentration.
Question
Which of the following semiconductors are transparent (or partially transparent) to visible light (400-700nm)? a. GaP and GaAs b. Si and GaAs c. GaN and GaP d. Si and GaN
Answer
C GaN (~3.4eV) and GaP (~2.26eV) both have band gaps above or near the visible spectrum energy range (~1.8-3.1eV), so they don’t absorb most visible wavelengths. Si (~1.1eV) and GaAs (~1.4eV) absorb visible light as their band gaps are below this range.
Question
Which semiconductors are mainly used to make LEDs and lasers? a. Silicon Carbide (SiC) and Germanium (Ge) b. Silicon and Gallium Phosphide (GaP) c. Gallium Arsenide (GaAs) and Gallium Nitride (GaN) d. Silicon and Germanium
Answer
C Both are direct band gap semiconductors, enabling efficient radiative recombination for light emission.
Question
Why has silicon not similarly dominated photonics? a. Silicon is too fragile for photonics. b. Silicon has poor optical properties for light emission. c. Silicon is too expensive for photonics. d. Silicon is not abundant enough for photonics.
Answer
B Silicon’s indirect band gap makes it inherently inefficient at emitting light, so III-V compounds like GaAs and GaN are used instead for photonic devices.
Mentimeter
Question
What is the primary reason silicon is used in electronics? a. It is the most abundant element on Earth. b. It has excellent electrical properties and is cost-effective. c. It is the only material that can be used for electronics. d. It is the easiest material to work with.
Answer
B Silicon has a suitable bandgap, forms a stable native oxide, and is abundant enough for cheap mass production. It’s not the most abundant element on Earth (that’s oxygen), though silicon is the most abundant semiconductor.
Question
Which of the following materials are commonly used for making LEDs and lasers? a. Silicon and Germanium b. Gallium Arsenide (GaAs) and Gallium Nitride (GaN) c. Silicon Carbide (SiC) and Indium Phosphide (InP) d. Silicon and Gallium Phosphide (GaP)
Answer
B Both are direct band gap III-V semiconductors, enabling efficient radiative recombination. GaAs covers infrared/red wavelengths, GaN covers blue/UV.
Question
What is the significance of the “band gap” in a semiconductor? a. The energy difference between the valence band maximum and the conduction band minimum. b. The energy required to remove an electron from the atom. c. The energy difference between the highest occupied and lowest unoccupied molecular orbitals. d. The energy required to create an electron-hole pair.
Answer
A The band gap defines the minimum photon energy for absorption/emission and determines the semiconductor’s electrical and optical behaviour. Note (d) is closely related but the band gap is specifically the energy difference between band edges. (c) is the HOMO-LUMO gap — a molecular chemistry concept, not solid-state.
Lecture 2
Question
Cleanroom classifications are based on what criteria? a. Type of equipment used b. Particle count per unit volume c. Size of the room d. Number of workers
Answer
B e.g. Class 10 = max 10 particles (≥0.5µm) per cubic foot. ISO 14644 uses a similar metric per cubic metre.
Question
Describe the key differences between a Schottky and an Ohmic contact. a. Schottky contacts are used in high-power applications; Ohmic contacts are used in low-power applications b. Schottky contacts have a rectifying behaviour; Ohmic contacts have a linear behaviour c. Schottky contacts are made of metal; Ohmic contacts are made of semiconductor d. Schottky contacts are more expensive; Ohmic contacts are cheaper
Answer
B Schottky: metal-semiconductor junction with a potential barrier, resulting in diode-like I-V characteristics. Ohmic: linear I-V (V=IR), achieved by heavy doping at the contact interface to allow carrier tunnelling.
Question
Describe the typical process steps involved in the production of a GaAs-based LED. a. Cleaning, doping, etching, and packaging b. Oxidation, diffusion, ion implantation, and annealing c. Deposition, planarisation, patterning, and testing d. Epitaxial growth, photolithography, etching, and metallisation
Answer
D Epitaxial growth (e.g. MOCVD) deposits the active layers, photolithography defines device geometry, etching removes unwanted material, and metallisation forms the electrical contacts.
Question
What are the differences between a positive and negative resist? a. Positive resist is more expensive; negative resist is cheaper b. Positive resist is used for etching; negative resist is used for deposition c. Positive resist becomes soluble when exposed to light; negative resist becomes insoluble d. Positive resist is used in UV lithography; negative resist is used in electron beam lithography
Answer
C Positive: exposed regions are removed by developer (pattern matches mask). Negative: exposed regions crosslink and remain (pattern is inverse of mask).
Question
What is the difference between a Class 10 and a Class 1000 cleanroom? a. Class 10 has fewer particles per cubic foot b. Class 1000 is more expensive to maintain c. Class 10 is larger d. Class 1000 has more workers
Answer
A The class number directly indicates the max allowable particles (≥0.5µm) per cubic foot. Class 10 is significantly cleaner and used for more sensitive fabrication steps.
Question
Why are clean rooms important for the production of semiconductor devices? a. To enhance lighting b. To prevent contamination c. To increase humidity d. To reduce noise levels
Answer
B Even a single particle on a wafer can cause defects in nanoscale features, reducing yield.
Question
Why are test structures added to mask sets? a. To increase the complexity of the design b. To monitor process parameters and ensure quality control c. To improve the aesthetic appearance of the wafer d. To reduce the cost of fabrication
Answer
B Placed in scribe lanes between dies, they allow measurement of parameters like sheet resistance, contact resistance, line width, and alignment accuracy without sacrificing device area.
Question
Why does electron beam lithography provide a higher spatial resolution? a. It uses a more precise alignment system b. It uses a different type of resist c. It operates at lower temperatures d. Electrons have a shorter wavelength than UV light
Answer
D De Broglie wavelength of electrons at typical accelerating voltages is sub-nanometre, far smaller than UV (~193-365nm), enabling finer feature definition. Tradeoff is it’s a serial write process so much slower.
Question
Why is the air pressure in a cleanroom usually higher than the outside air pressure? a. To maintain temperature b. To reduce energy consumption c. To prevent contaminants from entering d. To improve ventilation
Answer
C Positive pressure ensures air flows outward when doors open, preventing unfiltered outside air and particles from entering.
Question
Why is UV light typically used in optical lithography? a. It is cheaper than other light sources b. It provides higher resolution due to shorter wavelength c. It is safer to use in cleanrooms d. It is more energy-efficient
Answer
B Resolution is proportional to wavelength (Rayleigh criterion). Shorter UV wavelengths (e.g. 193nm DUV) enable smaller feature sizes than visible light.
Mentimeter
Question
What measures are used to keep a clean room clean? a. Regular vacuuming b. Use of air conditioning c. Strict protocols and specialised equipment d. Frequent painting
Answer
C Includes HEPA/ULPA filtered laminar airflow, gowning procedures (bunny suits, gloves, booties), airlocks/air showers, positive pressure, and restricted materials.
Question
What is meant by cleanroom classifications? a. Categories based on size b. Levels of cleanliness based on particle count c. Types of equipment used d. Number of workers allowed
Answer
B Classifications (e.g. ISO 14644 or Federal Standard 209E) define the maximum allowable number of particles of a given size per unit volume of air.
Question
What is the primary purpose of maintaining higher air pressure in a cleanroom? a. To cool the room b. To keep contaminants out c. To save energy d. To increase humidity
Answer
B Positive pressure ensures air flows outward through any gaps or door openings, preventing unfiltered external air from entering.
Question
Describe the types of hazards in cleanrooms. a. Chemical, biological, and physical hazards b. Electrical, thermal, and mechanical hazards c. Radiation, noise, and ergonomic hazards d. All of these
Answer
D Cleanrooms present a wide range of hazards: toxic/corrosive chemicals (HF, solvents), electrical equipment, UV/laser radiation, repetitive strain, and noise from vacuum pumps and gas systems.
Question
Why are test structures added to mask sets? a. To monitor process parameters and ensure quality control b. To increase the complexity of the design c. To reduce the cost of fabrication d. To improve the aesthetic appearance of the wafer
Answer
A Placed in scribe lanes between dies, they allow measurement of parameters like sheet resistance, contact resistance, line width, and alignment accuracy.
Question
Name and describe the key apparatus used in the fabrication process of a GaAs-based LED. a. Furnace, mask aligner, etcher, and wire bonder b. CVD reactor, stepper, plasma etcher, and probe station c. MOCVD reactor, mask aligner, etcher, and wire bonder d. Sputtering system, lithography tool, etcher, and tester
Answer
C MOCVD reactor for epitaxial growth of III-V layers, mask aligner for photolithography, etcher for pattern transfer, and wire bonder for making electrical connections during packaging.
Lecture 3-5
Question
What are rays called that make small angles with the lens’s or mirror’s axis (where sin θ ≈ θ)? a. Converging rays b. Marginal rays c. Paraxial rays d. Diverging rays
Answer
A, C Paraxial rays travel close to the optical axis at small angles, allowing the small-angle approximation. The question also accepted “converging rays” though paraxial is the standard term.
Question
Assuming a lens is in air (n_air ≈ 1), which of the following is the correct Lensmaker’s Equation for a thin lens? a. 1/f = (n-1)(1/R1 - 1/R2) b. 1/f = (n+1)(1/R1 - 1/R2) c. 1/f = (n+1)(1/R1 + 1/R2) d. 1/f = (n-1)(1/R1 + 1/R2)
Answer
D Note: the sign convention matters here. With the convention used in this course, 1/f = (n-1)(1/R1 + 1/R2). The more common form uses (1/R1 - 1/R2) with a different sign convention for R2.
Question
An object is placed at infinity in front of a converging lens with a focal length of 20 cm. What is the image distance (di) from the lens? a. -20 cm b. 40 cm c. 0 d. 20 cm
Answer
D From thin lens equation: 1/di = 1/f - 1/do. As do → ∞, 1/do → 0, so di = f = 20 cm. This is the definition of the focal point.
Question
According to the information provided, what is the relationship between the Numerical Aperture (NA) of a lens and its resolving power? a. A larger NA results in better resolving power. b. Resolving power is independent of wavelength and NA. c. NA has no impact on resolving power. d. A smaller NA results in better resolving power.
Answer
A Rayleigh criterion: minimum resolvable distance ∝ λ/NA. Larger NA → smaller minimum resolvable feature → better resolution.
Question
If the medium between an objective lens (NA=0.65) and the specimen is air, what is the maximum half-angle of the cone of light that can enter the lens? a. Approximately 40.5 degrees b. Approximately 75.8 degrees c. Approximately 60.1 degrees d. Approximately 29.9 degrees
Answer
A NA = n·sin(θ). In air n=1, so θ = arcsin(0.65) ≈ 40.5°.
Question
What is the fundamental difference between monochromatic (M) and chromatic (C) aberrations in optical systems? a. M: affects paraxial rays; C: affects all rays. b. M: occurs primarily in lenses; C: occurs in lenses, not mirrors. c. M: due to paraxial approximation; C: due to wavelength-dependent n. d. M: results in blurry images; C: results in colored fringes.
Answer
C Monochromatic aberrations (spherical, coma, astigmatism, etc.) arise because real rays deviate from the paraxial approximation. Chromatic aberration occurs because refractive index varies with wavelength (dispersion), so different colours focus at different points.
Question
Increasing NA to achieve higher resolution often comes with a trade-off. What is the primary disadvantage associated with using a very large NA? a. Increased lens thickness b. Increased spherical aberration c. Increased diffraction limit d. Increased chromatic aberration e. None of these
Answer
B Larger NA means collecting rays at steeper angles from the axis. These marginal rays deviate more from the paraxial approximation, increasing spherical aberration.
Question
According to the Rayleigh criterion, when are two images considered to be just resolvable? a. First minimum of one pattern coincides with the other’s centre. b. None of these. c. Centres separated by half the central max width. d. Edges of patterns are just touching. e. Central maxima of patterns overlap completely.
Answer
A The Rayleigh criterion defines the minimum resolvable separation as when the central maximum of one Airy pattern falls on the first minimum of the other.
Question
A compound microscope is made of an eyepiece with magnification Me=5 and an objective with magnification Mo=25. What is the overall magnification M? a. 30 b. 5 c. 25 d. 125
Answer
D M = Mo × Me = 25 × 5 = 125. Compound microscope total magnification is the product, not the sum.
Question
A compound microscope is made of an eyepiece with Me=5 and an objective with Mo=25. Eyepiece and objective are l=10 cm apart. (N=25 cm, eye focused at ∞.) What is the focal length of the eyepiece? a. 5 cm b. 125 cm c. 0.2 cm d. 2 cm e. 30 cm
Answer
A For eye focused at infinity: Me = N/fe, so fe = N/Me = 25/5 = 5 cm.
Question
A compound microscope is made of an eyepiece with Me=5 and an objective with Mo=25. Eyepiece and objective are l=10 cm apart. (N=25 cm, eye focused at ∞.) What is the focal length of the objective? a. 2 cm b. 0.2 cm c. 0.19 cm d. 1.9 cm
Answer
B, C Mo = L/fo where L = l - fe = 10 - 5 = 5 cm. So fo = L/Mo = 5/25 = 0.2 cm. The 0.19 cm answer likely accounts for a slightly different derivation.
Question
A compound microscope is made of an eyepiece with Me=5 and an objective with Mo=25. Eyepiece and objective are l=10 cm apart. (N=25 cm, eye focused at ∞.) What is the focal length of the eyepiece? a. 5.00 cm b. 2.00 cm c. 0.50 cm d. 0.20 cm
Answer
D This appears to be asking for the objective focal length (0.20 cm) despite the question text saying eyepiece. fo = L/Mo = 5/25 = 0.2 cm.
Question
A compound microscope (Me=5, Mo=25, l=10 cm, N=25 cm, eye relaxed). What is the NA if the space between object and objective is filled with oil n=1.5 and the diameter of the objective lens is D=2 cm? a. 2.00 b. 0.80 c. 1.47 d. 0.98
Answer
C NA = n·sin(θ). The half-angle θ = arctan(D/2 / fo) = arctan(1/0.2) = arctan(5) ≈ 78.7°. NA = 1.5 × sin(78.7°) ≈ 1.47.
Question
What is the fundamental principle behind bright field microscopy that allows for the visualisation of specimens? a. Interference of light waves that have passed through different parts b. Fluorescence of dyes bound to specific cellular components. c. Variations in light intensity due to absorption or reflection d. None of these. e. Dispersion of the refractive indices of different parts of the sample
Answer
C Bright field: specimen is illuminated from below/behind, contrast arises from regions that absorb or reflect light differently. Simple but limited contrast for transparent specimens.
Question
What is the key principle behind dark field microscopy that creates contrast in the image? a. Direct illumination of the specimen with high-intensity light. b. Utilising differences in refractive indices within the specimen. c. Employing fluorescent dyes to highlight specific structures. d. None of these. e. Blocking the central direct light and collecting only scattered light.
Answer
E A patch stop blocks the central beam so only light scattered by the specimen enters the objective. Features appear bright against a dark background, giving good contrast for unstained transparent specimens.
Lecture 6
Question
What is the main difference between spontaneous emission and stimulated emission? a. Stimulated emission is slower than spontaneous emission b. Spontaneous emission occurs randomly, while stimulated emission requires external stimulus c. Spontaneous emission is faster than stimulated emission d. Stimulated emission occurs naturally, while spontaneous emission requires external energy
Answer
B Spontaneous: electron drops to lower energy state randomly, emitting a photon in a random direction/phase. Stimulated: an incoming photon triggers emission of a second photon with identical wavelength, phase, and direction — the basis of laser operation.
Question
What is the main purpose of the cavity in semiconductor lasers? a. To reduce the cost of the laser b. To provide feedback for light amplification c. To change the colour of the laser d. To increase the size of the laser
Answer
B The optical cavity (e.g. cleaved facets acting as mirrors) reflects photons back through the gain medium, enabling multiple passes and stimulated emission to build up coherent light.
Question
When designing a semiconductor laser, which of the following is a key factor to develop a high efficiency device? a. Size of the laser b. Colour of the laser c. Material composition and cavity design d. Weight of the laser
Answer
C Material composition determines the band gap (emission wavelength) and lattice matching. Cavity design affects threshold current, mode selection, and optical confinement.
Question
Which of the following is a benefit of the MBE growth technique compared to MOCVD? a. More precise control over layer thickness and composition b. High growth rates c. Lower cost compared to MOCVD d. Simplicity of the process
Answer
A MBE operates under ultra-high vacuum with in-situ monitoring (e.g. RHEED), enabling monolayer-level control. Tradeoff is slower growth rate and higher cost than MOCVD.
Question
Which of the following is a characteristic of Vertical Cavity Surface Emitting Lasers (VCSELs)? a. Emission of light from the edge of the device b. High threshold current c. Emission of light perpendicular to the surface of the device d. Complex fabrication process
Answer
C VCSELs emit vertically through the wafer surface using DBR mirrors above and below the active region. Advantages include circular beam profile, low threshold current, and easy 2D array fabrication.
Question
Which of these is an approach used to manufacture single mode lasers? a. Increasing the size of the laser cavity b. Implementing distributed feedback structures c. Reducing the power output d. Using multiple quantum wells
Answer
B DFB lasers use a periodic grating structure within the cavity that selectively amplifies only one longitudinal mode via Bragg reflection.
Question
Why is alloying useful in semiconductors? a. It makes the semiconductor more brittle b. It increases the weight of the semiconductor c. It allows for tuning of the band gap and other properties d. It makes the semiconductor more flexible
Answer
C e.g. Al_xGa_(1-x)As allows continuous tuning of the band gap between GaAs and AlAs by varying x, enabling band gap engineering for specific wavelengths and heterostructure design.
Question
What are the benefits of using quantum wells as the active region in semiconductor lasers? a. They increase the size of the laser b. They allow for higher efficiency and lower threshold currents c. They reduce the lifespan of the laser d. They make the laser more expensive
Answer
B Quantum confinement modifies the density of states, concentrating carriers at specific energies. This increases gain per carrier and reduces the threshold current needed for lasing.
Question
What is a benefit of the MOCVD growth technique compared to MBE? a. More control over layer thickness b. Lower cost and easier to produce wafers in high volumes c. Ability to produce very pure materials d. Low growth rates
Answer
B MOCVD has higher throughput, can process multiple wafers simultaneously, and is the industry standard for mass production of III-V devices like LEDs.
Question
What is the main advantage of using Distributed Feedback (DFB) over Fabry-Perot (F-P) lasers? a. Lower cost of production b. Single longitudinal mode operation c. Higher output power d. Simpler design
Answer
B F-P lasers support multiple longitudinal modes, leading to broader spectral output. DFB’s integrated grating ensures only one mode satisfies the Bragg condition, critical for telecom applications where spectral purity matters.
Mentimeter
Question
What is meant by alloying in the context of a semiconductor? a. Mixing different metals to create a stronger material b. Combining different semiconductor materials to modify properties c. Adding impurities to a semiconductor d. Coating a semiconductor with a protective layer
Answer
B e.g. Al_xGa_(1-x)As — varying the composition ratio x tunes the band gap continuously between GaAs and AlAs. Distinct from doping (c), which adds trace impurities to control carrier concentration rather than fundamentally change the material.
Question
What is a benefit of the MOCVD growth technique compared to MBE? a. More control over layer thickness b. Lower cost and easier to produce wafers in high volumes c. Low growth rates d. Ability to produce very pure materials
Answer
B MOCVD has higher throughput, can handle multiple wafers simultaneously, and is the standard for industrial-scale III-V epitaxy. MBE wins on precision but not on cost or scalability.
Question
Why are quantum wells typically used in semiconductor lasers? a. Increasing the physical size of the device b. Enhancing the mechanical strength of the device c. Improving the efficiency of light emission d. Reducing the cost of manufacturing
Answer
C Quantum confinement concentrates the density of states at specific energies, increasing optical gain per carrier. This lowers threshold current and improves wall-plug efficiency.
Question
What is a key difference between F-P edge-emitting lasers and VCSELs? a. F-P lasers emit light vertically, while VCSELs emit light horizontally b. VCSELs emit light vertically, while F-P lasers emit light horizontally c. F-P lasers have a more complicated structure compared to VCSELs d. VCSELs are less efficient than F-P lasers
Answer
B F-P (edge-emitting) lasers emit from the cleaved facet parallel to the wafer surface. VCSELs emit perpendicular to the surface through DBR mirror stacks, enabling on-wafer testing and 2D array fabrication.
Lecture 7
Question
Describe one application of a Hall sensor. a. Measuring humidity in environmental monitoring b. Measuring magnetic field strength in industrial applications c. Measuring temperature in electronic devices d. Measuring pressure in mechanical systems
Answer
B Also used in brushless DC motor commutation, current sensing (clamp meters), and proximity/position detection.
Question
What is meant by the term “sheet resistance” and why is this a useful quantity? a. Sheet resistance is the resistance of a bulk material measured in ohms per cubic metre; it is useful for characterising bulk materials b. Sheet resistance is the resistance of a surface measured in ohms per square metre; it is useful for characterising surface coatings c. Sheet resistance is the resistance of a wire measured in ohms per metre; it is useful for characterising electrical conductors d. Sheet resistance is the resistance of a thin film measured in ohms per square; it is useful for characterising thin films and semiconductor layers
Answer
D Rs = ρ/t (resistivity/thickness). Units are Ω/□. Useful because the resistance of any square of the film is the same regardless of size, simplifying layout calculations.
Question
What is the impact of series resistance on an LED or solar cell? a. It improves the performance and reduces the power loss b. It reduces the efficiency and increases the power loss c. It has no impact on the efficiency and power loss d. It increases the efficiency and reduces the power loss
Answer
B Series resistance causes I²R losses. In solar cells it reduces fill factor and output power. In LEDs it causes resistive heating and reduces wall-plug efficiency.
Question
What properties of X-rays make them useful to characterise crystals? a. High energy and long wavelength b. Low energy and long wavelength c. High energy and short wavelength d. Low energy and short wavelength
Answer
C X-ray wavelengths (~0.1nm) are comparable to interatomic spacings, enabling diffraction from crystal planes (Bragg’s law: nλ = 2d·sinθ).
Question
What two properties of a semiconductor sample can Capacitance-Voltage (CV) profiling be used to measure? a. Doping concentration and depletion width b. Thermal conductivity and electrical resistivity c. Carrier mobility and lifetime d. Surface roughness and crystal orientation
Answer
A The depletion width is extracted from the capacitance (C = εA/W), and the doping profile from dN/d(1/C²) vs V.
Question
Describe the Hall Effect measurement. What parameters can be extracted from the Hall measurement? a. Hall Effect measures the voltage developed across a conductor when a magnetic field is applied; parameters include carrier concentration and mobility b. Hall Effect measures the current developed across a conductor when a magnetic field is applied; parameters include resistivity and conductivity c. Hall Effect measures the capacitance developed across a conductor when a magnetic field is applied; parameters include dielectric constant and permittivity d. Hall Effect measures the resistance developed across a conductor when a magnetic field is applied; parameters include temperature and pressure
Answer
A Current flows through the sample, a perpendicular B-field deflects carriers via the Lorentz force, generating a transverse Hall voltage. RH = 1/(nq), and combining with resistivity gives mobility μ = RH/ρ. The sign of VH also identifies carrier type.
Question
Describe why some of the diffraction orders are absent in body-centred and face-centred crystal structures. a. Due to the presence of impurities b. Due to constructive interference of scattered waves c. Due to the high temperature of the crystal d. Due to destructive interference of scattered waves
Answer
D The additional basis atoms in BCC/FCC unit cells scatter waves that destructively interfere for certain (hkl) reflections. e.g. BCC: absent when h+k+l is odd. FCC: absent when h,k,l are mixed odd/even.
Question
For a van der Pauw resistivity measurement, describe three requirements for the sample to ensure that the technique is valid. a. Sample must be thick, inhomogeneous, and have two ohmic contacts; typical geometry is a rectangular shape b. Sample must be flat, homogeneous, and have four ohmic contacts; typical geometry is a square or circular shape c. Sample must be thin, heterogeneous, and have three ohmic contacts; typical geometry is a triangular shape d. Sample must be curved, isotropic, and have five ohmic contacts; typical geometry is a hexagonal shape
Answer
B Also requires: uniform thickness, no isolated holes in the sample, and contacts placed at the periphery. The method works for arbitrary shape provided these conditions are met.
Question
In a Hall experiment, the Hall coefficient is found to change sign with increasing temperature. What does that indicate about the doping in the sample, and why? a. It indicates a change from p-type to n-type behaviour due to intrinsic carrier generation b. It indicates a change from n-type to p-type behaviour due to impurity scattering c. It indicates a change from p-type to n-type behaviour due to extrinsic carrier generation d. It indicates a change from p-type to n-type behaviour due to phonon scattering
Answer
A At high temperatures, intrinsic carrier generation dominates over extrinsic doping. Since electron mobility is typically higher than hole mobility in most semiconductors, the Hall coefficient can flip sign as the intrinsic electrons dominate the Hall voltage even in a p-type sample.
Question
Name two parameters that can be obtained from a PL spectrum. What information do these parameters give about the wafer? a. Intensity and absorption coefficient; they provide information about surface roughness and crystal orientation b. Peak wavelength and intensity; they provide information about bandgap energy and defect density c. Peak wavelength and absorption coefficient; they provide information about doping concentration and carrier mobility d. Peak wavelength and reflectivity; they provide information about thermal conductivity and electrical resistivity
Answer
B Peak wavelength → band gap energy (and therefore composition). PL intensity → material quality; lower intensity indicates more non-radiative recombination centres (defects). Linewidth also gives information about alloy disorder and inhomogeneity.
Mentimeter
Question
Which semiconductor properties can we measure with Photoluminescence? a. Band gap and material quality b. Carrier density and thermal conductivity c. Carrier mobility and doping density
Answer
A PL peak wavelength gives the band gap energy; PL intensity indicates material quality (higher intensity = fewer non-radiative defects). Linewidth provides info on alloy disorder.
Question
Which of the following is an application of a Hall Sensor? a. Measuring temperature b. Measuring magnetic field strength c. Measuring humidity
Answer
B The Hall voltage is directly proportional to the applied magnetic field, making it a straightforward magnetometer. Also used in current sensing, position detection, and brushless motor commutation.
Question
For the plane shown, which of the following are the correct Miller Indices? a. (001) b. (111) c. (100) d. (121)
Answer
A The shaded plane is parallel to a₁ and a₂ (intercepts at ∞) and intercepts a₃ at 1. Reciprocals: (1/∞, 1/∞, 1/1) = (001). This is the top face of the cubic unit cell.
Question
In a Hall experiment, the Hall coefficient changes sign with increasing temperature. What does that indicate about the doping in the sample? a. Change from p- to n-type behaviour due to intrinsic carrier generation b. Change from p- to n-type behaviour due to extrinsic carrier generation c. Change from p- to n-type behaviour due to impurity scattering d. Change from n- to p-type behaviour due to phonon scattering
Answer
A At high temperatures, thermally generated intrinsic carriers dominate over the extrinsic doping. Since electron mobility is typically higher than hole mobility, the Hall coefficient flips sign as intrinsic electrons dominate the Hall voltage.
Lecture 8
Question
What is one main disadvantage of ion implantation? a. High temperatures required b. Limited dopant types c. Poor control of doping d. High equipment cost and complexity
Answer
D Ion implanters require high vacuum, mass separation, and precise beam control. Also causes lattice damage requiring annealing.
Question
What is one primary advantage of ion implantation over diffusion doping? a. Higher temperatures required b. Simpler equipment c. Lower cost d. Better control of doping
Answer
D Ion implantation allows precise control of dose (via beam current × time) and depth profile (via accelerating voltage), independent of temperature.
Question
What is one purpose of rapid thermal annealing in the doping process? a. To repair crystal damage b. To introduce dopants c. To remove impurities d. To increase the diffusion coefficient
Answer
A RTA uses short high-temperature pulses to repair lattice damage from implantation and electrically activate dopants, while minimising unwanted diffusion.
Question
What is one significant reason for using an amorphous layer on top of the semiconductor during ion implantation? a. It simplifies the process b. It increases the ion dose c. It reduces the cost d. It prevents channelling
Answer
D In a crystalline target, ions can travel along open channels between lattice planes, penetrating much deeper than intended. An amorphous cap layer randomises the ion trajectories.
Question
What is the effect of local oxidation (LOCOS) on the diffusion coefficient of dopants in the underlying silicon? a. It fluctuates b. It decreases c. It increases d. It remains constant
Answer
C Oxidation injects silicon interstitials into the substrate (oxidation-enhanced diffusion), which increases the diffusion coefficient of dopants that diffuse via an interstitialcy mechanism.
Question
What is the effect of very high dopant concentrations on the diffusion coefficient? a. It remains constant b. It fluctuates c. It decreases d. It increases
Answer
D At high concentrations, the Fermi level shifts significantly, increasing the concentration of charged vacancies/interstitials and thus enhancing diffusion (concentration-dependent diffusion).
Question
What is the purpose of the analysis magnet in an ion implanter? a. To purify the ions b. To measure the ion dose c. To accelerate the ions d. To heat the ions
Answer
A The analysis magnet acts as a mass filter — ions are deflected by radius r = mv/(qB), so only the desired dopant species with the correct mass-to-charge ratio passes through the resolving slit.
Question
What is the term for the process of adding dopants during the growth of silicon wafers? a. Bulk doping b. Ion implantation c. Chemical vapour deposition d. Epitaxy
Answer
A Dopants are added to the melt during Czochralski or float-zone growth to set the background doping of the entire wafer.
Question
What is the term used to describe the movement of dopant atoms through a solid by occupying gaps created by missing atoms in the crystal lattice? a. Interstitial diffusion b. Surface diffusion c. Bulk diffusion d. Vacancy diffusion
Answer
D Vacancy diffusion: dopant atom moves into an adjacent vacant lattice site. Contrast with interstitial diffusion where atoms move between lattice sites through interstitial positions.
Question
What is the typical temperature range for diffusion doping in a furnace? a. 1200-1500°C b. 800-1200°C c. 400-600°C d. 600-800°C
Answer
B Diffusion requires sufficient thermal energy to mobilise dopant atoms. Below ~800°C diffusion is negligibly slow; above ~1200°C unwanted redistribution and defect generation become problematic.
Lecture 9
Question
What is one main problem with deep wet etching? a. High cost b. Slow etch rate c. Nearby features merging together d. Requires high temperature
Answer
C Wet etching is typically isotropic, so lateral etching undercuts the mask. For deep etches, this undercut becomes significant and adjacent features can merge.
Question
What is the advantage of temperature control in wet etching? a. Reduced contamination b. It controls etch rate c. Faster etch rate d. Increased chemical reaction
Answer
B Etch rate is exponentially dependent on temperature (Arrhenius behaviour). Precise temperature control ensures repeatable, uniform etching.
Question
What is the main component that sustains a plasma in dry etching? a. Radicals b. Ions c. Electrons d. Photons
Answer
C Electrons sustain the plasma through impact ionisation — they collide with neutral gas molecules to produce more ions and electrons, maintaining the discharge.
Question
What is the main disadvantage of using SF6 in plasma etching? a. Slow etch rate b. High cost c. Electron absorption d. Toxicity
Answer
C SF6 is electronegative and captures free electrons, which can destabilise or extinguish the plasma by reducing the electron density needed to sustain it.
Question
What is the main reason industry prefers dry etching over wet etching? a. Less toxic waste products b. Higher yield and reliability c. Lower cost d. Faster etch rate
Answer
B Dry etching offers anisotropic profiles with better dimensional control, enabling smaller feature sizes and more consistent results across wafers.
Question
What is the term that describes the probability of an interaction happening between species in a plasma? a. Electron affinity b. Ionisation potential c. Cross-section d. Dissociation rate
Answer
C Cross-section (σ) has units of area and represents the effective target size for a given collision process. Reaction rate ∝ density × velocity × cross-section.
Question
What process involves the transfer of momentum from an ion hitting a surface and knocking an atom out? a. Recombination b. Sputtering c. Dissociation d. Ionisation
Answer
B Physical sputtering is a purely momentum-transfer process. It’s non-selective (etches everything) and anisotropic since ions arrive directionally from the plasma sheath.
Question
What type of etching is characterised by the etch occurring at the same speed in all directions? a. Anisotropic etching b. Directional etching c. Reactive ion etching d. Isotropic etching
Answer
D Isotropic etching produces equal lateral and vertical etch rates, resulting in rounded profiles and undercutting beneath the mask. Typical of purely chemical wet etching.
Question
Which entity is the primary driver of reactions that occur in a plasma? a. Ions b. Electrons c. Radicals d. Photons
Answer
B Electrons have much lower mass so they gain significantly more energy from the RF field. Their high-energy collisions drive ionisation, dissociation, and excitation processes that generate all other reactive species.
Question
Which resist profile is ideal for lift-off? a. Flat profile b. Undercut profile c. Rounded profile d. Overexposed profile
Answer
B An undercut (re-entrant) profile creates a shadow during metal deposition, ensuring a clean break between metal on the substrate and metal on the resist, allowing clean removal during lift-off.
Lecture 10
Question
What describes the pressure dependence of the mean-free path in dry etching? a. It varies directly with the pressure b. It varies as the square of the pressure c. It varies inversely with the pressure d. It remains constant regardless of pressure
Answer
C λ = kT/(√2 · π · d² · P). Higher pressure → more molecules → shorter distance between collisions.
Question
What happens if the blocking capacitor is shorted out in an RIE machine? a. The plasma becomes more stable b. The sample cools down c. The machine starts arcing and sparking d. The etch rate increases
Answer
C Without the blocking capacitor, DC current flows directly through the plasma. This causes uncontrolled discharge, arcing, and potential damage to the system.
Question
What is the main disadvantage of using symmetrical electrodes in an RIE machine? a. The etch rate decreases b. The sample heats up c. The machine wears out quickly due to sputtering of the machine walls d. The machine becomes more expensive
Answer
C With symmetrical electrodes, both electrodes develop similar self-bias voltages, so ions sputter the chamber walls and the powered electrode equally. Asymmetric electrodes concentrate the bias on the smaller sample electrode.
Question
What is the main purpose of optical spectroscopy in dry etching? a. To measure the etch depth b. To increase the etch rate c. To cool the sample d. To monitor the species in the plasma
Answer
D Optical emission spectroscopy (OES) monitors specific spectral lines from plasma species. Changes in emission intensity indicate when a layer has been etched through (endpoint detection).
Question
What is the main reason for minimising the use of inert gases in dry etching? a. To stabilise the plasma b. To increase the etch rate c. To reduce the cost d. To minimise sputtering and damage
Answer
D Inert gas ions contribute only physical sputtering (no chemical selectivity), causing non-selective material removal and ion-induced lattice damage.
Question
What is the main reason for using a blocking capacitor in an RIE machine? a. To increase the voltage b. To reduce the pressure c. To cool the electrode d. To allow the sample electrode to build up a self-bias
Answer
D The blocking capacitor prevents DC current flow, allowing negative charge to accumulate on the sample electrode. This DC self-bias accelerates positive ions toward the sample for directional etching.
Question
What is the role of inert gases like helium and argon in dry etching? a. They make the gas mixture safer b. They react with the sample c. They stabilise the plasma d. They form radicals
Answer
C Inert gases are easily ionised and provide a stable electron population to sustain the plasma, particularly when the reactive gases are electronegative or difficult to maintain in a discharge.
Question
Which gas is difficult to sustain in a plasma but is commonly used to etch semiconductors due to its convenience? a. O2 b. N2 c. H2 d. SF6
Answer
D SF6 is electronegative (absorbs electrons), making it hard to sustain a plasma. However, it’s widely used because it produces fluorine radicals that etch silicon effectively and its etch products (SiF4) are volatile.
Question
Which species in a plasma are most effective at transferring momentum to the sample (i.e. sputtering it)? a. Electrons b. Neutrons c. Ions d. Photons
Answer
C Ions have much greater mass than electrons, so they transfer significant momentum on impact. They’re also accelerated directionally by the sheath electric field toward the sample surface.
Question
Why is iodine not commonly used in dry etching despite its effectiveness? a. It is too expensive b. It is too rare c. It contaminates and damages the etching machine d. It is not reactive enough
Answer
C Iodine is highly corrosive and its etch byproducts are not sufficiently volatile at typical process temperatures, leading to residue buildup and contamination of the chamber.