68 free MCQs on Semiconductor Electronics with worked answers and explanations. Band theory, p-n junction, diodes, transistors, logic gates, and integrated circuits.
Take the timed Semiconductor Electronics chapterwise test →Below are 68 practice questions on Semiconductor Electronics, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Semiconductor Electronics notes.
Conductors have overlapping bands, semiconductors have a small energy gap that can be bridged by heat or doping, and insulators have a large gap that blocks conduction.
Q1.
A semiconductor has electrical conductivity:
Answer: C. Between metals and insulators
Why: Semiconductors have conductivity between metals (good conductors) and insulators. Examples: silicon, germanium.
Q2.
N-type semiconductor is doped with:
Answer: B. Pentavalent impurity (e.g., phosphorus)
Why: N-type: doped with pentavalent atoms (5 valence electrons like P, As, Sb). Extra electron becomes majority carrier.
Q3.
P-type semiconductor has _____ as majority carriers:
Answer: B. Holes
Why: P-type is doped with trivalent atoms creating holes (missing electrons). Holes are the majority carriers.
Q4.
In forward biased p-n junction:
Answer: C. Depletion region narrows and current flows
Why: Forward bias: positive voltage on p-side, negative on n-side. Depletion region narrows and current flows through the junction.
Q5.
In a p-n junction diode, current flows easily in:
Answer: B. Forward bias only
Why: P-n junction diode allows easy current flow in forward bias; reverse bias blocks current (only tiny leakage).
Q6.
Zener diode is used as:
Answer: C. Voltage regulator
Why: Zener diode is designed to operate in reverse breakdown at a specific voltage. Used as voltage regulator.
Q7.
LED emits light when:
Answer: C. In forward bias (electron-hole recombination)
Why: LED: forward bias causes electrons and holes to recombine at junction, emitting photons (light). Color depends on energy gap.
Q8.
NPN transistor has three terminals:
Answer: A. Emitter, Base, Collector
Why: Transistor terminals: Emitter (E), Base (B), Collector (C). BJT (Bipolar Junction Transistor).
Q9.
In common emitter configuration, output is taken from:
Answer: C. Collector
Why: In common emitter (CE) configuration, input at base, common emitter, output at collector. Most used configuration.
Q10.
NOT gate produces output that is:
Answer: B. Inverted (complement) of input
Why: NOT gate is an inverter: output = complement of input. If input is 1, output is 0 and vice versa.
Q11.
NAND gate is combination of:
Answer: B. NOT + AND
Why: NAND = NOT + AND. Output is 0 only when all inputs are 1; otherwise output is 1.
Q12.
Which gate is called a universal gate?
Answer: D. NAND
Why: NAND (and NOR) are universal gates: any logic function can be built using only NAND gates.
Q13.
Conductivity of semiconductor increases with temperature because:
Answer: B. More electron-hole pairs are created
Why: Higher temperature provides energy to break covalent bonds, creating more electron-hole pairs, increasing conductivity.
Q14.
Solar cell converts:
Answer: B. Light to electrical energy
Why: Solar cell (photovoltaic cell) converts light (photon) energy to electrical energy using photoelectric effect in p-n junction.
Q15.
For silicon, the energy gap is approximately:
Answer: B. 1.1 eV
Why: Silicon: Eg = 1.1 eV. Germanium: 0.72 eV. GaAs: 1.43 eV. Wider gap = less conduction at room temperature.
Q16.
The depletion region in a p-n junction is depleted of:
Answer: C. Both mobile charge carriers (electrons and holes)
Why: The depletion region forms where electrons and holes have recombined, leaving it depleted of free carriers.
Q17.
OR gate output is 1 when:
Answer: B. At least one input is 1
Why: OR gate: output = 1 if at least one input is 1. Output = 0 only if ALL inputs are 0.
Q18.
AND gate output is 1 when:
Answer: C. All inputs are 1
Why: AND gate: output = 1 only when ALL inputs are 1. If any input is 0, output is 0.
Q19.
A full-wave rectifier has _____ diodes:
Answer: D. 4 (bridge rectifier)
Why: Full-wave bridge rectifier uses 4 diodes arranged in a bridge circuit to rectify both halves of AC cycle.
Q20.
Intrinsic semiconductor at absolute zero behaves as:
Answer: B. Perfect insulator
Why: At 0 K, all bonds are intact, no free carriers. Intrinsic semiconductor acts as a perfect insulator at absolute zero.
Q21.
The band gap of silicon at room temperature is approximately:
Answer: B. 1.12 eV
Why: Silicon band gap Eg ≈ 1.12 eV at 300K. Germanium: 0.67 eV. GaAs: 1.42 eV.
Q22.
Minority carriers in n-type semiconductor are:
Answer: B. Holes
Why: N-type: majority carriers are electrons (from donor atoms). Minority carriers are holes.
Q23.
In forward-biased p-n junction, the depletion region:
Answer: B. Narrows (barrier reduces)
Why: Forward bias: external field opposes built-in field. Depletion region narrows. At sufficient voltage (~0.6V for Si), significant current flows.
Q24.
The Hall effect is used to determine:
Answer: B. Type of charge carrier (electron or hole) and carrier concentration
Why: Hall effect: transverse voltage in magnetic field. Sign of Hall voltage gives carrier type; magnitude gives carrier density.
Q25.
The threshold voltage of a MOSFET is the gate voltage at which:
Answer: B. Inversion layer forms (channel becomes conducting)
Why: Threshold voltage Vth: minimum gate-source voltage to create conducting channel (inversion layer) between source and drain.
Q26.
The breakdown mechanism in heavily doped p-n junctions at low reverse voltage:
Answer: B. Zener breakdown (quantum tunneling across thin depletion region)
Why: Zener breakdown: occurs in heavily doped junctions with thin depletion layer. Electrons tunnel from valence to conduction band. At < 5-6V typically.
Q27.
A transistor in saturation mode is used as:
Answer: B. Switch (ON state)
Why: Saturation: both junctions forward biased. Maximum current flows. Used as closed switch (ON state). Cut-off (both reverse): open switch (OFF state).
Q28.
The base of a BJT transistor is:
Answer: B. Very thin with low doping (lightly doped)
Why: BJT base: very thin (~1 micrometer), lightly doped. Most carriers injected from emitter pass through base to collector without recombining.
Q29.
Transconductance gm of a MOSFET is:
Answer: A. dI_D/dV_G (change in drain current per gate voltage)
Why: Transconductance: gm = dI_D/dV_GS. Measures how effectively gate voltage controls drain current. Key parameter in MOSFET amplifier design.
Q30.
In an intrinsic semiconductor, np product =
Answer: B. ni² where ni is intrinsic carrier concentration
Why: Mass action law: n × p = ni². This holds for both intrinsic and extrinsic semiconductors in thermal equilibrium.
Q31.
LED efficiency (quantum efficiency) is limited by:
Answer: B. Non-radiative recombination, optical extraction losses, and interface defects
Why: LED efficiency: quantum efficiency limited by non-radiative recombination (Auger, defect), light extraction (total internal reflection), series resistance.
Q32.
Ohmic contact between metal and semiconductor requires:
Answer: C. Heavily doped semiconductor (tunneling through thin barrier)
Why: Ohmic contact: formed on heavily doped semiconductor. Thin barrier allows tunneling. Resistance independent of current direction.
Q33.
The CMOS inverter uses:
Answer: B. Both NMOS and PMOS transistors complementarily
Why: CMOS (Complementary MOS): NMOS and PMOS in series. When input HIGH: NMOS on, PMOS off, output LOW. Extremely low static power consumption.
Q34.
Fermi level in n-type semiconductor shifts:
Answer: B. Toward conduction band
Why: N-type doping: extra electrons push Fermi level toward conduction band. P-type: Fermi level moves toward valence band.
Q35.
The built-in potential (contact potential) across a p-n junction:
Answer: B. Cannot be utilized as a battery (thermal equilibrium)
Why: Built-in potential (0.6-0.7V for Si): thermodynamic equilibrium quantity. Cannot be used as voltage source; no net current in equilibrium.
Q36.
Photovoltaic effect in solar cells converts:
Answer: B. Light to electricity (photons creating electron-hole pairs separated by p-n junction field)
Why: Photovoltaic: photons absorbed create electron-hole pairs. Junction built-in field separates carriers. Electrons go to n-side, holes to p-side creating current.
Q37.
Bipolar transistor current gain beta = hFE represents:
Answer: B. I<sub>C</sub>/I<sub>B</sub> (collector current divided by base current)
Why: Beta = hFE = I<sub>C</sub>/I<sub>B</sub>. Common emitter current gain. Typical values: 50-300 for BJTs. Small base current controls large collector current.
Q38.
Drift current in semiconductor is caused by:
Answer: B. Applied electric field
Why: Drift current: carriers move due to electric field. J = sigma × E. Diffusion current: carriers move due to concentration gradient.
Q39.
The I-V equation of a p-n junction diode (Shockley equation):
Answer: A. I = I₀(e<sup>V/VT</sup> - 1)
Why: Shockley diode equation: I = I₀(e<sup>V/V<sub>T</sub></sup> - 1) where I₀ = saturation current, V<sub>T</sub> = kT/e ≈ 26 mV at room temperature.
Q40.
In which region does MOSFET act as a voltage-controlled resistor?
Answer: C. Triode (linear/ohmic) region
Why: Triode (linear) region: V<sub>DS</sub> < V<sub>GS</sub> - V<sub>th</sub>. Channel resistance controlled by V<sub>GS</sub>. MOSFET acts as voltage-controlled resistor.
Q41.
The Einstein relation between mobility and diffusion coefficient is:
Answer: A. D = mu × kT/e
Why: Einstein relation: D/mu = kT/q = V<sub>T</sub> (thermal voltage). Fundamental relationship valid in equilibrium.
Q42.
In an npn BJT in active region: I<sub>E</sub> = I<sub>C</sub> + I<sub>B</sub>. The emitter injection efficiency gamma is:
Answer: A. I<sub>Cn</sub>/I<sub>E</sub>
Why: Emitter injection efficiency gamma = fraction of emitter current that is due to minority carriers injected into base. gamma = I<sub>En</sub>/I<sub>E</sub> for npn.
Q43.
Shockley-Read-Hall (SRH) recombination occurs via:
Answer: B. Trap levels in mid-gap
Why: SRH recombination: trap-assisted. Carrier captured by mid-gap trap, then recombines. Dominant in indirect-gap semiconductors like Si.
Q44.
The channel pinch-off in JFET/MOSFET occurs when:
Answer: B. V<sub>DS</sub> = V<sub>GS</sub> - V<sub>p</sub> (or V<sub>GS</sub> - Vth)
Why: Pinch-off: depletion region extends across channel. V<sub>DS</sub>,sat = V<sub>GS</sub> - Vth. Above V<sub>DS</sub>,sat: I<sub>D</sub> saturates (nearly constant).
Q45.
Quantum confinement in semiconductor nanostructures (quantum dots) causes:
Answer: B. Discrete energy levels and size-tunable bandgap
Why: Quantum confinement: when size < de Broglie wavelength, energy levels become discrete. Bandgap increases as size decreases. Used in tunable LEDs.
Q46.
The NMOS long-channel drain current in saturation (I<sub>D</sub>,sat) ∝
Answer: B. (V<sub>GS</sub> - Vth)²
Why: Long-channel MOSFET saturation: I<sub>D</sub> = (1/2) mu C<sub>ox</sub> (W/L) (V<sub>GS</sub> - Vth)². Quadratic dependence on overdrive voltage.
Q47.
Generation-recombination current in reverse-biased diode scales as:
Answer: B. exp(-E<sub>g</sub>/2kT) (via mid-gap traps)
Why: G-R current ~ ni (not ni²), scales as exp(-Eg/2kT). Dominant reverse current in wide-bandgap semiconductors at low temperature.
Q48.
In a CMOS circuit, the power consumption P<sub>dynamic</sub> =
Answer: B. C × V² × f (capacitance × supply voltage squared × frequency)
Why: CMOS dynamic power: P = alpha × C<sub>L</sub> × V<sub>DD</sub>² × f. Charging/discharging load capacitance each switching event. Key driver of CPU power limits.
Q49.
The heterojunction (type I): electrons tend to accumulate in:
Answer: B. Narrower bandgap material (lower conduction band minimum)
Why: Type-I heterojunction (straddled gap): both conduction and valence band offsets confine carriers to narrower bandgap material. Used in quantum well lasers.
Q50.
Negative differential resistance (NDR) is observed in:
Answer: B. Tunnel diodes (Esaki diodes) at forward bias
Why: Tunnel diode: at certain forward bias, tunneling current decreases as bias increases (NDR region). Used in oscillators and fast switches.
Q51.
The Zener diode breakdown voltage temperature coefficient is:
Answer: B. Negative for Zener (<5V), positive for avalanche (>7V)
Why: Zener (<5-6V): tunneling, negative TC. Avalanche (>7V): ionization, positive TC. At ~5.6V: compensate for near-zero TC in reference diodes.
Q52.
The depletion approximation in p-n junction analysis assumes:
Answer: B. Complete ionization within depletion region and zero carriers (sharp boundary)
Why: Depletion approximation: ionized donors/acceptors present in depletion region; mobile carrier density negligible. Creates abrupt space charge regions.
Q53.
Phonon dispersion in semiconductor: acoustic vs optical modes differ because:
Answer: A. Acoustic modes involve same-sub-lattice atoms moving in phase; optical involve opposite sublattice atoms out of phase
Why: Acoustic: atoms in same unit cell move in phase (sound waves). Optical: atoms in different sub-lattices move out of phase (interacts with IR light). Key for understanding thermal conductivity and electron scattering.
Q54.
The charge control model of BJT describes:
Answer: B. Transient response via stored charge in base (Q<sub>B</sub> = I<sub>C</sub> × tau_F)
Why: Charge control model: I<sub>C</sub> = Q<sub>B</sub>/tau_F where Q<sub>B</sub> is minority charge stored in base and tau_F is forward transit time. Foundation of BJT switching analysis.
Q55.
Channel length modulation in MOSFET causes:
Answer: B. Non-zero output conductance (dI_D/dV_DS) in saturation
Why: Channel length modulation (lambda parameter): effective channel shortens as V<sub>DS</sub> increases. I<sub>D</sub>,sat = (1/2)mu C<sub>ox</sub>(W/L)(V<sub>GS</sub>-Vth)²(1+lambda V<sub>DS</sub>). Finite output resistance.
Q56.
In a 2DEG (2D electron gas) at heterojunction, mobility is very high because:
Answer: B. Electrons spatially separated from ionized dopants, reducing Coulomb scattering
Why: 2DEG in HEMT: ionized donors in wide-gap layer, electrons accumulate in narrow-gap layer. Spatial separation reduces impurity scattering. Ultra-high mobility.
Q57.
The Gunn effect in GaAs: electrons in high electric field undergo:
Answer: B. Intervalley transfer from gamma valley (high mobility) to L valley (low mobility), causing NDR
Why: Gunn effect: high field transfers electrons from high-mobility central valley to low-mobility satellite valleys. Average velocity decreases. NDR → oscillations.
Q58.
Lightly doped drain (LDD) structure in short-channel MOSFETs is designed to:
Answer: B. Reduce hot carrier effects and drain-induced barrier lowering at drain edge
Why: LDD: graded doping near drain reduces peak electric field. Mitigates hot carrier injection into gate oxide, improving long-term reliability.
Q59.
The diffusion length of minority carriers L = sqrt(D × tau). In short devices (L<sub>device</sub> << L):
Answer: B. Most carriers reach the junction (low recombination loss)
Why: If device length much shorter than diffusion length, minority carriers traverse the base/region with minimal recombination. High efficiency (solar cells, BJT base).
Q60.
Threshold voltage roll-off in short-channel MOSFETs occurs because:
Answer: B. Drain depletion charge extends under channel, reducing effective Vth (drain-induced barrier lowering, DIBL)
Why: Short channel effects: drain depletion region reduces barrier for channel formation. Vth decreases as channel length shrinks. DIBL causes V<sub>th</sub> to decrease with V<sub>DS</sub>.
Q61.
When an intrinsic semiconductor is doped with a pentavalent impurity, it becomes:
Answer: A. n-type
Why: Pentavalent donors add free electrons, producing an n-type semiconductor.
Q62.
A p-n junction diode under forward bias:
Answer: A. conducts with low resistance
Why: Forward bias lowers the barrier so the diode conducts with low resistance.
Q63.
A Zener diode is normally used as a:
Answer: A. voltage regulator (in reverse breakdown)
Why: Operated in reverse breakdown, the Zener diode maintains a nearly constant voltage, acting as a regulator.
Q64.
A full-wave rectifier fed with a 50 Hz AC supply produces a ripple in the output at a frequency of:
Answer: C. 100 Hz
Why: A full-wave rectifier produces two output pulses per input cycle, so the ripple frequency is 2·50 = 100 Hz.
Q65.
Which logic gate is called a universal gate?
Answer: A. NAND gate
Why: The NAND gate (like NOR) can be combined to make any other logic gate, so it is universal.
Q66.
In a common-emitter transistor amplifier, the phase difference between the input and output voltages is:
Answer: C. 180°
Why: A common-emitter amplifier inverts the signal, giving a 180° phase difference.
Q67.
The approximate energy band gap of silicon at room temperature is:
Answer: C. 1.1 eV
Why: Silicon has a band gap of about 1.1 eV, typical of a semiconductor.
Q68.
For a two-input AND gate, the output when the inputs are 1 and 0 is:
Answer: A. 0
Why: An AND gate outputs 1 only when all inputs are 1; with inputs 1 and 0 the output is 0.