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Semiconductor Devices — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Semiconductor Devices MCQs with step-by-step solutions covering Semiconductors & Energy Bands, Intrinsic & Extrinsic Semiconductors, p-n Junction & Diode, Rectifiers & Filters, Special Purpose Diodes, Transistors & Logic Gates. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Semiconductors & Energy Bands · easy · theory
The energy gap in a solid is the separation between the:
A. Top of the valence band and the bottom of the conduction band ✓ Correct
B. Two halves of the valence band
C. Nucleus and the innermost electron shell
D. Two halves of the conduction band
Solution: No electron states exist within this forbidden gap, so an electron must acquire at least $E_g$ to become free.
Q2 — Semiconductors & Energy Bands · easy · theory
In a conductor, the valence band and the conduction band:
A. Overlap each other ✓ Correct
B. Are separated by more than $3\text{ eV}$
C. Are both completely empty
D. Are separated by about $1\text{ eV}$
Solution: With overlapping bands, electrons can move to higher states under even the feeblest field.
Q3 — Semiconductors & Energy Bands · easy · theory
In an insulator, the forbidden energy gap is typically:
A. About $1\text{ eV}$
B. Zero
C. Greater than about $3\text{ eV}$ ✓ Correct
D. Negative
Solution: Diamond, for example, has a gap of about $6\text{ eV}$, far beyond the reach of thermal energy at room temperature.
Q4 — Semiconductors & Energy Bands · easy · theory
In a semiconductor, the forbidden energy gap is typically:
A. About $100\text{ eV}$
B. Of the order of $1\text{ eV}$ ✓ Correct
C. Exactly zero
D. Greater than $5\text{ eV}$
Solution: A gap of this size lets a useful number of electrons cross it by thermal excitation at ordinary temperatures.
Q5 — Semiconductors & Energy Bands · easy · theory
The electrical resistivity of a semiconductor lies:
A. Below that of a conductor
B. Above that of an insulator
C. Between that of a conductor and that of an insulator ✓ Correct
D. Exactly equal to that of a conductor
Solution: This intermediate behaviour, controllable by doping, is what makes semiconductors so useful.
Q6 — Semiconductors & Energy Bands · easy · numerical
A semiconductor sample has resistivity $0.5\,\Omega\cdot\text{m}$. Its conductivity is:
A. $0.25\text{ S/m}$
B. $0.5\text{ S/m}$
C. $2\text{ S/m}$ ✓ Correct
D. $5\text{ S/m}$
Solution: $\sigma = \dfrac{1}{\rho} = \dfrac{1}{0.5} = 2\text{ S/m}$.
Q7 — Intrinsic & Extrinsic Semiconductors · easy · theory
In an intrinsic semiconductor, the numbers of free electrons and holes are:
A. Electrons far more numerous
B. Holes far more numerous
C. Equal ✓ Correct
D. Both zero at all temperatures
Solution: Every electron promoted to the conduction band leaves behind exactly one hole, so $n_e = n_h = n_i$.
Q8 — Intrinsic & Extrinsic Semiconductors · easy · theory
An extrinsic semiconductor is one that has been:
A. Placed in a strong magnetic field
B. Purified to remove all impurities
C. Heated above its melting point
D. Deliberately doped with a suitable impurity ✓ Correct
Solution: Doping in a proportion as small as one part in a million transforms the conductivity.
Q9 — Intrinsic & Extrinsic Semiconductors · easy · theory
An n-type semiconductor is produced by doping with a:
A. Divalent impurity such as magnesium
B. Tetravalent impurity such as carbon
C. Trivalent impurity such as boron
D. Pentavalent impurity such as phosphorus or arsenic ✓ Correct
Solution: The fifth valence electron is not needed for bonding and becomes a free conduction electron.
Q10 — Intrinsic & Extrinsic Semiconductors · easy · theory
A p-type semiconductor is produced by doping with a:
A. Pentavalent impurity such as antimony
B. Trivalent impurity such as boron or indium ✓ Correct
C. Tetravalent impurity such as silicon
D. Monovalent impurity such as sodium
Solution: With only three valence electrons, the impurity leaves a vacancy in one bond — a hole.
Q11 — Intrinsic & Extrinsic Semiconductors · easy · theory
In an n-type semiconductor, the majority and minority carriers are respectively:
A. Electrons and protons
B. Protons and electrons
C. Electrons and holes ✓ Correct
D. Holes and electrons
Solution: Donor atoms supply plenty of electrons while holes arise only from thermal generation.
Q12 — Intrinsic & Extrinsic Semiconductors · easy · theory
In a p-type semiconductor, the majority and minority carriers are respectively:
A. Holes and electrons ✓ Correct
B. Electrons and holes
C. Holes and protons
D. Protons and holes
Solution: Acceptor atoms create an abundance of holes, leaving electrons as the scarce minority carriers.
Q13 — Intrinsic & Extrinsic Semiconductors · easy · numerical
Germanium doped with phosphorus becomes:
A. An insulator
B. A p-type semiconductor
C. An n-type semiconductor ✓ Correct
D. An intrinsic semiconductor
Solution: Phosphorus is pentavalent, so it donates electrons.
Q14 — Intrinsic & Extrinsic Semiconductors · easy · numerical
Silicon doped with boron becomes:
A. An n-type semiconductor
B. A conductor
C. A p-type semiconductor ✓ Correct
D. An intrinsic semiconductor
Solution: Boron is trivalent, so it accepts electrons and creates holes.
Q15 — Intrinsic & Extrinsic Semiconductors · easy · numerical
In an intrinsic semiconductor the relation between electron, hole and intrinsic concentrations is:
A. $n_e = n_h = n_i$ ✓ Correct
B. $n_en_h = 0$
C. $n_e = 2n_h$
D. $n_h = 2n_e$
Solution: Thermal generation always produces electrons and holes in pairs.
Q16 — Intrinsic & Extrinsic Semiconductors · easy · numerical
The conductivity of an extrinsic semiconductor compared with that of the intrinsic material is:
A. Zero
B. Very much smaller
C. Very much greater ✓ Correct
D. Exactly the same
Solution: Doping supplies carriers far more numerous than those produced by thermal generation alone.
Q17 — p-n Junction & Diode · easy · theory
When a p-n junction is forward biased:
A. Only minority carriers move
B. The potential barrier is reduced and a large current flows ✓ Correct
C. The potential barrier is increased and no current flows
D. The depletion region widens
Solution: The external supply opposes the built-in field, letting majority carriers cross freely.
Q18 — p-n Junction & Diode · easy · theory
When a p-n junction is reverse biased:
A. The barrier increases and only a small leakage current flows ✓ Correct
B. The junction behaves as a perfect conductor
C. The barrier decreases and a large current flows
D. The depletion region disappears
Solution: The small reverse current is carried by thermally generated minority carriers.
Q19 — p-n Junction & Diode · easy · theory
A junction diode is described as a unidirectional device because it:
A. Conducts only at high frequency
B. Conducts appreciably in one direction only ✓ Correct
C. Never conducts at all
D. Conducts equally in both directions
Solution: This asymmetry is what makes rectification possible.
Q20 — p-n Junction & Diode · easy · numerical
The forward and reverse resistances of a junction diode are respectively:
A. Both very high
B. Both very low
C. Very high and very low
D. Very low and very high ✓ Correct
Solution: This large ratio, often $10^5$ or more, is what makes the diode an effective one-way valve.
Q21 — Rectifiers & Filters · easy · theory
A rectifier is a circuit that converts:
A. Low voltage into high voltage
B. Direct current into alternating current
C. Current into voltage
D. Alternating current into direct current ✓ Correct
Solution: It exploits the one-way conduction of a junction diode.
Q22 — Rectifiers & Filters · easy · theory
A half-wave rectifier uses:
A. Two diodes
B. One diode ✓ Correct
C. No diode at all
D. Four diodes
Solution: It passes only one half of each input cycle and blocks the other.
Q23 — Rectifiers & Filters · easy · numerical
A half-wave rectifier is supplied from a $50\text{ Hz}$ mains. The ripple frequency in the output is:
A. Zero
B. $100\text{ Hz}$
C. $25\text{ Hz}$
D. $50\text{ Hz}$ ✓ Correct
Solution: One output pulse per input cycle gives a ripple at the supply frequency.
Q24 — Rectifiers & Filters · easy · numerical
A full-wave rectifier is supplied from a $50\text{ Hz}$ mains. The ripple frequency in the output is:
A. $50\text{ Hz}$
B. $25\text{ Hz}$
C. $150\text{ Hz}$
D. $100\text{ Hz}$ ✓ Correct
Solution: Two pulses per input cycle double the ripple frequency.
Q25 — Rectifiers & Filters · easy · numerical
A half-wave rectifier is supplied at $60\text{ Hz}$. The ripple frequency is:
A. $60\text{ Hz}$ ✓ Correct
B. $120\text{ Hz}$
C. Zero
D. $30\text{ Hz}$
Solution: The ripple frequency of a half-wave rectifier equals the supply frequency.
Q26 — Rectifiers & Filters · easy · numerical
A full-wave rectifier is supplied at $60\text{ Hz}$. The ripple frequency is:
A. $180\text{ Hz}$
B. $60\text{ Hz}$
C. $120\text{ Hz}$ ✓ Correct
D. $30\text{ Hz}$
Solution: A full-wave circuit doubles the supply frequency in its output ripple.
Q27 — Special Purpose Diodes · easy · theory
A Zener diode is designed to operate in the:
A. Forward bias region before the knee
B. Unbiased state
C. Forward breakdown region
D. Reverse breakdown region ✓ Correct
Solution: Its breakdown voltage is sharply defined and stable, which is what makes voltage regulation possible.
Q28 — Special Purpose Diodes · easy · theory
The chief application of a Zener diode is as a:
A. Light source
B. Current amplifier
C. Voltage regulator ✓ Correct
D. Rectifier
Solution: It holds the voltage across a load constant despite changes in supply voltage or load current.
Q29 — Special Purpose Diodes · easy · theory
A light-emitting diode emits light when it is:
A. In the breakdown region
B. Unbiased
C. Reverse biased
D. Forward biased ✓ Correct
Solution: Electrons and holes injected across the junction recombine, releasing photons.
Q30 — Special Purpose Diodes · easy · theory
A light-emitting diode converts:
A. Light energy into electrical energy
B. Heat into light
C. Light into heat
D. Electrical energy into light energy ✓ Correct
Solution: The reverse conversion is performed by a photodiode or a solar cell.