90+ free device-physics questions on energy bands, doping, drift and diffusion, PN and Schottky junctions, BJTs, the MOS capacitor, MOSFETs, FinFETs, power devices and reliability. Includes 18 worked answers, 8 solved numericals, a junction band diagram drawn step by step and a carrier and Fermi-level calculator.
580 questions with interview-ready answers and tips, 20 Python and SPICE device labs, 25 solved numericals and 30 fab and lab debugging scenarios.
Semiconductor device questions check whether you understand what's physically happening inside a transistor or diode. They make up most of the interview for device, process, TCAD, characterisation and reliability roles, and they come up in circuit and memory interviews whenever leakage, breakdown or ageing is on the table.
The best answers start with a picture (a band diagram, a carrier flow, a depletion region) and only then bring in the equation. Begin with the formula sheet and the calculator, skim the 50 questions by topic, then attempt each worked answer and numerical before opening it.
Bandgaps, the Fermi level, intrinsic and doped silicon, the mass-action law, drift and diffusion. Expect to draw a band diagram and place the Fermi level.
The PN junction under bias, the diode equation, breakdown, Schottky contacts, BJT action, the MOS capacitor, threshold voltage and short-channel effects.
FinFET and GAA, power and wide-bandgap devices, reliability mechanisms, process steps, C–V and Hall measurements, and "this lot came out wrong; why?"
Almost every junction, contact and MOS question becomes simple once Ec, Ev and EF are on the board. Interviewers often ask for it anyway.
No current means no Fermi-level gradient. This one rule fixes the shape of every equilibrium band diagram you'll draw.
Majority or minority? Drift or diffusion? Depletion region or neutral region? Diode current, for example, is minority-carrier diffusion in the neutral regions.
kT/q = 25.9 mV, ni ≈ 1010 cm−3, 60 mV per decade, Eg = 1.12 eV. A real number shows you can use the physics, not just describe it.
Positive or negative coefficient, activation energy near Eg or Eg/2: temperature behaviour separates mechanisms faster than anything else.
Bandgap to leakage and breakdown, mobility to drive current, traps to reliability. Show why the physics matters to a circuit or a fab.
Silicon at 300 K unless stated: kT/q = 25.9 mV, ni ≈ 1×1010 cm−3, εSi = 11.7ε0, εox = 3.9ε0.
Enter the doping and temperature. Uses ni(300 K) = 1×1010 cm−3, a temperature-dependent bandgap and a standard mobility model. Estimates, not a TCAD result.
"Why is GaN used for chargers?" or "why not make CPUs from GaAs?" both come down to these properties. Values are typical room-temperature figures.
| MATERIAL | BANDGAP (eV) | TYPE | ELECTRON MOBILITY (cm²/V·s) | BREAKDOWN FIELD (MV/cm) | WHERE IT WINS |
|---|---|---|---|---|---|
| Silicon | 1.12 | Indirect | ~1,400 | ~0.3 | Logic, memory, most analog; cheap, great native oxide |
| Germanium | 0.66 | Indirect | ~3,900 | ~0.1 | High hole mobility; SiGe strain and HBTs, IR detectors |
| GaAs | 1.42 | Direct | ~8,500 | ~0.4 | RF front ends, lasers, high-efficiency solar |
| 4H-SiC | 3.26 | Indirect | ~900 | ~2.5–3 | High-voltage power: EV inverters, chargers, grid |
| GaN | 3.4 | Direct | ~1,000 bulk, ~2,000 in a HEMT channel | ~3.3 | Fast power switching, RF power, blue and white LEDs |
A sample of what device interviewers ask most. The 580-question ebook goes much deeper on each topic and answers every question.
Why semiconductors behave as they do.
Controlling how many carriers there are.
How carriers move and disappear.
The structure under almost every device.
Metal junctions and light.
Still asked, especially in analog.
The heart of every chip.
What changed as transistors shrank.
Devices under stress.
Making and measuring devices.
580 questions in 27 chapters, all answered: 20 Python and SPICE device labs, 25 solved numericals, 30 debugging scenarios, plus optoelectronics, power devices, memories, process and TCAD.
Sketch your answer or say it out loud first, then open it. Each answer shows what a strong reply covers, what the interviewer is listening for, and the slip that costs candidates most.
The Fermi level EF is the energy at which a state has a 50% chance of being occupied by an electron. Its position relative to the band edges sets the carrier densities: the closer EF is to the conduction band, the more electrons; the closer to the valence band, the more holes.
In equilibrium EF is the same everywhere in a connected system. That rule is what fixes the shape of band diagrams for junctions and contacts.
In equilibrium, n·p = ni2, independent of doping. Thermal generation creates electron–hole pairs at a rate set by temperature and bandgap, and recombination removes them at a rate proportional to n·p. Equilibrium is where the two balance, which fixes the product.
Adding donors raises n. With more electrons available, holes recombine faster, so p falls until n·p is back to ni2. For silicon doped with 1016 donors, n ≈ 1016 and p ≈ 104 cm−3: a ratio of a trillion.
The two are linked by the Einstein relation D/μ = kT/q, because both come from the same random thermal motion and scattering. In an equilibrium PN junction, large drift and diffusion currents flow in opposite directions and cancel exactly, which is why no net current flows.
In a direct bandgap material (GaAs, GaN, InP), the conduction-band minimum and valence-band maximum occur at the same crystal momentum. An electron can drop straight down and emit a photon. In an indirect material (Si, Ge), they occur at different momenta, so the transition also needs a phonon to supply the momentum difference.
That three-particle process is far less likely, so in silicon, radiative recombination loses badly to non-radiative paths through defects. Silicon LEDs are very inefficient, which is why LEDs and lasers use III–V materials. Silicon still works well as a photodetector, because absorption only needs the photon energy to exceed the gap, helped by phonons.
| Zener (tunnelling) | Avalanche (impact ionisation) | |
|---|---|---|
| Mechanism | Electrons tunnel straight from valence to conduction band across a very thin depletion region | Carriers gain enough energy in a wide depletion region to knock out new electron–hole pairs, which multiply |
| Doping | Both sides heavily doped | At least one side lightly doped |
| Voltage | Low, below about 5 V in silicon | Higher, above about 7 V |
| Temperature coefficient | Negative: breakdown voltage falls as the gap shrinks with heat | Positive: more phonon scattering means carriers need a higher field |
Between about 5 and 7 V both mechanisms contribute and the temperature coefficients roughly cancel, which is why reference diodes near 5.6 V are so stable.
Forward bias lowers the junction barrier by qV. The number of majority carriers with enough energy to cross a barrier follows the Boltzmann tail of their energy distribution, so it grows as eqV/kT. Those carriers are injected as minority carriers and diffuse away, and that diffusion current is the diode current.
I = I_S·(e^(qV/nkT) − 1) At 300 K, kT/q = 25.9 mV ln(10) × 25.9 mV ≈ 59.6 mV → 10× current per ~60 mV (n = 1)
With recombination in the depletion region the ideality factor n approaches 2, giving about 120 mV per decade. At high current, series resistance and high-level injection bend the curve further.
For a p-type substrate:
At low frequency, the inversion electrons can be generated and recombined fast enough to follow the signal, so C returns to Cox. At high frequency they can't (thermal generation is slow), so only the depletion edge responds and C stays at its minimum. With a fast voltage sweep, even the DC inversion layer can't form and the device goes into deep depletion, with C falling below the minimum.
V_T = V_FB + 2φ_F + Q_dep/C_ox V_FB = φ_ms − Q_ox/C_ox
In short-channel devices, DIBL and VT roll-off also lower VT, so the effective value depends on length and drain voltage.
In an NPN transistor, forward-biasing the base–emitter junction injects electrons into the base. They diffuse across and are swept into the collector by the reverse-biased base–collector junction. Current gain depends on two things:
The limits: too thin or too lightly doped a base gives punch-through and a strong Early effect; too heavy an emitter causes bandgap narrowing that lowers efficiency. That's why HBTs use a wider-gap emitter instead.
A metal on a semiconductor usually forms a Schottky barrier, because Fermi-level pinning by surface states fixes the barrier height largely independent of the metal. Carriers then cross mainly by thermionic emission over the barrier, which rectifies.
To make an ohmic contact, the semiconductor under the metal is heavily doped (around 1020 cm−3). The depletion region becomes only a few nanometres thick, so carriers tunnel straight through it in both directions. Contact resistance then depends exponentially on the barrier height divided by √doping. A silicide such as NiSi is used to lower the resistance further.
A Schottky diode is faster than a PN diode because it conducts with majority carriers only, so there's no stored minority charge to remove at turn-off.
In a long transistor the gate alone controls the barrier between source and channel. In a short one, the source and drain depletion regions take up a large part of the channel, so less charge is left for the gate to control.
Controls: thinner gate dielectric (high-k), shallow source/drain extensions, halo or pocket implants, and, most effectively, multi-gate structures (FinFET, GAA) or fully depleted SOI, where the gate surrounds a thin channel.
Below roughly 25 nm, a planar gate on top of the channel couldn't stop the drain from influencing the deep parts of the channel. Leakage, DIBL and variability rose, and VDD couldn't keep scaling.
The idea throughout is to keep the channel thinner than the gate's "natural length", so the gate stays in charge of the barrier.
Measure reverse current at several temperatures and plot ln(I) against 1/T. The slope gives an activation energy Ea.
Then compare diodes of different area and perimeter. Leakage that scales with perimeter points to the edges: isolation, surface states or etch damage.
This is NBTI, negative bias temperature instability. A negative gate bias at high temperature creates interface traps and charges pre-existing traps in the dielectric, so |VT| rises and drive current falls. When the stress is removed, some of the trapped charge is released, so part of the shift recovers within micro- to milliseconds.
Each points to a different process step: clean chemistry for ions, the post-oxidation or forming-gas anneal for interface traps, the oxidation and deposition conditions for fixed charge.
Carriers generated inside the depletion region are swept out quickly by the field. Carriers generated in the neutral regions have to diffuse to the junction first, which can take microseconds. That slow diffusion tail dominates the response, especially for longer wavelengths that are absorbed deep in the silicon, below the depletion region.
It comes down to the temperature coefficient of the on-state voltage.
Not necessarily: this is expected. Oxide breakdown follows weakest-link statistics. A large capacitor is effectively many small ones in parallel, and it fails when its weakest spot fails, so time to breakdown falls with area.
Breakdown times follow a Weibull distribution. On a Weibull plot, data from different areas line up once scaled by area, using the Weibull slope β:
t₂ / t₁ = (A₁ / A₂)^(1/β)
So check whether the large and small structures fall on the same scaled line. If they do, the oxide is behaving normally, and the result is used to project lifetime to the full chip area. If the large structures fall well below the scaled line, or the plot is bimodal, there's an extrinsic defect population (particles, contamination, thinning at edges) that needs fixing.
Silicon at 300 K, ni = 1×1010 cm−3, kT/q = 0.0259 V. Work each on paper, then tap to check.
N_D ≫ n_i → n ≈ N_D = 5 × 10^16 cm^-3 p = n_i² / n = 10^20 / (5 × 10^16) = 2 × 10^3 cm^-3
Note: fourteen orders of magnitude between majority and minority carriers.
E_F − E_i = kT·ln(n/n_i)
= 0.0259 × ln(5 × 10^6)
= 0.0259 × 15.42 ≈ 0.40 eVNote: with Ec about 0.56 eV above Ei, EF sits roughly 0.16 eV below the conduction band.
ρ = 1 / (q·n·µ_n) = 1 / (1.6 × 10^-19 × 10^16 × 1350) = 1 / 2.16 ≈ 0.46 Ω·cm
Note: holes contribute nothing measurable here; p is only 104 cm−3.
V_bi = (kT/q)·ln(N_A·N_D / n_i²)
= 0.0259 × ln(10^33 / 10^20)
= 0.0259 × ln(10^13) = 0.0259 × 29.93 ≈ 0.78 VNote: Vbi grows only logarithmically, which is why it stays between about 0.6 and 1.0 V for silicon.
W = √(2·ε_s·(V_bi + V_R) / (q·N_D)) = √(2 × 1.04e-12 × 5.8 / (1.6e-19 × 1e16)) = √(7.5 × 10^-9 cm²) ≈ 8.7 × 10^-5 cm ≈ 0.87 µm
Note: almost all of it is on the lightly doped n side.
ΔV = n·(kT/q)·ln(10) n = 1 : ΔV = 0.0259 × 2.303 ≈ 60 mV → 0.66 V n = 1.5 : ΔV ≈ 89 mV → 0.69 V
Note: measuring the slope per decade is the quickest way to extract the ideality factor.
D_n = (kT/q)·µ_n = 0.0259 × 1200 ≈ 31 cm²/s L_n = √(D_n·τ) = √(31 × 10^-6) ≈ 5.6 × 10^-3 cm ≈ 56 µm
Note: a BJT base must be much thinner than this, and a solar cell should collect carriers from within about this distance of the junction.
C_ox = ε_ox / t_ox = (3.9 × 8.85e-14) / (2e-7 cm)
≈ 1.73 × 10^-6 F/cm² ≈ 17 fF/µm²
EOT = 4 nm × 3.9 / 20 ≈ 0.78 nmNote: high-k gives the capacitance of 0.78 nm of SiO2 with a physically thicker film, which cuts gate tunnelling leakage dramatically.
"Draw the band diagram of a PN junction" is probably the most common device-physics whiteboard question. Here is the method, so you can draw it for any junction or contact.
The net electron current is proportional to n·μn·dEF/dx, and similarly for holes. With no current flowing, the gradient of EF must be zero everywhere. A sloped Fermi level always means current is flowing.
The difference between the two Fermi levels before contact: qVbi = kT·ln(NAND/ni2). For NA = 1017 and ND = 1016 cm−3 that is about 0.78 eV, as in numerical N4 above.
The Fermi level splits into two quasi-Fermi levels, one for electrons and one for holes, separated by qV across the depletion region. Forward bias reduces the bending to q(Vbi − V) and narrows the depletion region; reverse bias increases it to q(Vbi + VR) and widens it.
One side is a metal, so its bands don't bend and all the bending happens in the semiconductor. The barrier seen from the metal side, ΦB, is fixed by the interface rather than by doping. That's why Schottky diodes have a lower turn-on voltage and no minority-carrier storage. The ebook has full chapters on junction electrostatics and metal–semiconductor contacts.
Written tests and screening calls love these. Answer each in a few seconds, then tap to check.
1.12 eV.
About 25.9 mV.
About 1010 cm−3 (modern value ≈ 9.7×109).
n·p = ni2 in equilibrium.
D/μ = kT/q.
Electrons.
Indirect.
Ten times.
Positive.
Near the conduction band.
Diffusion capacitance.
Heavy doping, so carriers tunnel through a thin barrier.
Less recombination, so higher gain and speed.
The minimum, Cmin.
Raises it.
VT, as VDS rises.
About 60 mV/decade.
A 2DEG from polarisation charge at the AlGaN/GaN interface.
PMOS.
The majority carrier type.
If nothing is biased, EF is flat. Drawing it bent across a junction is the single most common band-diagram error.
Diode, BJT and solar-cell currents are set by minority carriers diffusing. Answers that talk only about majority carriers miss the mechanism.
Interviewers want to hear what's happening first: a barrier lowering, a depletion region widening, carriers being swept or diffusing.
ni, leakage, mobility, breakdown and reliability all depend strongly on it. Many scenario questions are solved by asking what happens when it's hot.
Different doping, different voltage, opposite temperature coefficients. Get all three right.
cm versus m, cm−3 versus m−3. Quote typical values (1010, 1016, 1,400) so the interviewer can see your numbers are sane.
Device, process integration, TCAD, characterisation, reliability and yield roles test device physics in depth. Circuit, memory and analog design interviews ask the basics, and most core-company written tests for ECE graduates include them.
Mostly order-of-magnitude calculations: carrier densities, Fermi-level position, built-in potential, depletion width, diode current and threshold voltage. Interviewers care more about the physical picture and the right assumptions than about long derivations.
Energy bands and the Fermi level, doping and the mass-action law, drift and diffusion, the PN junction band diagram and diode equation, breakdown, the MOS capacitor and MOSFET threshold voltage, and why short-channel effects led to FinFETs.
For process and device roles, yes: oxidation, lithography, etch, implantation, annealing, deposition and CMP, and how each affects device parameters. For design roles, knowing the main steps and why they matter is usually enough.
Increasingly, yes, especially at power-electronics, automotive and compound-semiconductor companies. Know why SiC and GaN suit high voltage and high frequency, how a HEMT forms its channel, and how IGBTs and superjunction MOSFETs work.
This round is about how devices work inside: bands, carriers, junctions and transistor physics. The CMOS round uses those devices to build circuits, and asks about sizing, delay, power and memory cells.
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