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Semiconductor Electronics — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Semiconductor Electronics MCQs with step-by-step solutions covering Semiconductor and p-n Junction Diode, Transistors, Digital Circuits. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Semiconductor and p-n Junction Diode · easy · theory
An intrinsic semiconductor is converted into n-type extrinsic semiconductor by doping it with
A. phosphorous ✓ Correct
B. aluminium
C. silver
D. germanium
Solution: Doping with a pentavalent impurity such as phosphorus produces an n-type semiconductor.
Q2 — Semiconductor and p-n Junction Diode · easy · theory
The increase in the width of the depletion region in a p-n junction diode is due to
A. reverse bias only ✓ Correct
B. both forward bias and reverse bias
C. increase in forward current
D. forward bias only
Solution: In reverse bias the majority carriers are pulled away from the junction, widening the depletion region. Forward bias narrows it.
Q3 — Semiconductor and p-n Junction Diode · easy · theory
For a p-type semiconductor, which of the following statements is true?
A. Holes are the majority carriers and trivalent atoms are the dopants ✓ Correct
B. Holes are the majority carriers and pentavalent atoms are the dopants
C. Electrons are the majority carriers and pentavalent atoms are the dopants
D. Electrons are the majority carriers and trivalent atoms are the dopants
Solution: p-type semiconductors are made by adding trivalent impurities (B, Al, Ga, In), which create holes — the majority charge carriers.
Q4 — Semiconductor and p-n Junction Diode · easy · numerical
A diode D is connected to an external resistance R = 100 Ω and an emf of 3.5 V. If the barrier potential developed across the diode is 0.5 V, the current in the circuit will be
A. 30 mA ✓ Correct
B. 40 mA
C. 20 mA
D. 35 mA
Solution: Voltage across R $= 3.5 - 0.5 = 3.0$ V
$I = \frac{3}{100} = 0.03$ A $= 30$ mA
Q5 — Semiconductor and p-n Junction Diode · easy · theory
In a n-type semiconductor, which of the following statement is true?
A. Electrons are majority carriers and trivalent atoms are dopants
B. Electrons are minority carriers and pentavalent atoms are dopants
C. Holes are minority carriers and pentavalent atoms are dopants ✓ Correct
D. Holes are majority carriers and trivalent atoms are dopants
Solution: n-type semiconductors are doped with pentavalent atoms (e.g. phosphorus); electrons are the majority carriers and holes the minority carriers.
Q6 — Semiconductor and p-n Junction Diode · easy · theory
If a small amount of antimony is added to germanium crystal
A. the antimony becomes an acceptor atom
B. there will be more free electrons than holes in the semiconductor ✓ Correct
C. its resistance is increased
D. it becomes a p-type semiconductor
Solution: Antimony is pentavalent, so the crystal becomes an n-type semiconductor with more free electrons than holes.
Q7 — Semiconductor and p-n Junction Diode · easy · theory
In forward biasing of the p-n junction
A. the positive terminal of the battery is connected to n-side and the depletion region becomes thin
B. the positive terminal of the battery is connected to n-side and the depletion region becomes thick
C. the positive terminal of the battery is connected to p-side and the depletion region becomes thin ✓ Correct
D. the positive terminal of the battery is connected to p-side and the depletion region becomes thick
Solution: In forward bias the positive terminal goes to the p-side, and the depletion region becomes thin.
Q8 — Semiconductor and p-n Junction Diode · easy · theory
Which one of the following statement is false?
A. Pure Si doped with trivalent impurities gives a p-type semiconductor
B. Majority carriers in a n-type semiconductor are holes ✓ Correct
C. Minority carriers in a p-type semiconductor are electrons
D. The resistance of intrinsic semiconductor decreases with increase of temperature
Solution: In an n-type semiconductor the majority carriers are electrons, not holes — statement (b) is false.
Q9 — Semiconductor and p-n Junction Diode · easy · theory
For a cubic crystal structure which one of the following relations indicating the cell characteristic is correct?
A. $a \neq b \neq c$ and $\alpha \neq \beta$ and $\gamma \neq 90°$
B. $a \neq b \neq c$ and $\alpha \neq \beta = \gamma = 90°$
C. $a = b = c$ and $\alpha \neq \beta \neq \gamma = 90°$
D. $a = b = c$ and $\alpha = \beta = \gamma = 90°$ ✓ Correct
Solution: In cubic crystals the axes are mutually perpendicular ($\alpha = \beta = \gamma = 90°$) and the repeat interval is the same along all three axes ($a = b = c$).
Q10 — Semiconductor and p-n Junction Diode · easy · theory
Choose only false statement from the following
A. Substances with energy gap of the order of 10 eV are insulators
B. The conductivity of a semiconductor increases with increase in temperature
C. In conductors the valence and conduction bands may overlap
D. The resistivity of a semiconductor increases with increase in temperature ✓ Correct
Solution: The resistivity of a semiconductor decreases (not increases) with temperature — statement (d) is false.
Q11 — Semiconductor and p-n Junction Diode · easy · theory
Zener diode is used for
A. producing oscillations in an oscillator
B. amplification
C. stabilisation ✓ Correct
D. rectification
Solution: The Zener diode's reverse-bias characteristic makes it suitable for voltage regulation (stabilisation).
Q12 — Semiconductor and p-n Junction Diode · easy · numerical
If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be
A. 70.7 Hz
B. 100 Hz ✓ Correct
C. 25 Hz
D. 59 Hz
Solution: For a full-wave rectifier, ripple frequency $= 2 \times$ input frequency $= 2 \times 50 = 100$ Hz
Q13 — Semiconductor and p-n Junction Diode · easy · theory
For conduction in a p-n junction, the biasing is
A. high potential on n-side and low potential on p-side
B. high potential on p-side and low potential on n-side ✓ Correct
C. same potential on both p and n-sides
D. undetermined
Solution: Conduction requires forward bias — the p-side at higher potential and the n-side at lower potential.
Q14 — Semiconductor and p-n Junction Diode · easy · numerical
The number of atoms per unit cell in bcc lattice is
A. 1
B. 2 ✓ Correct
C. 4
D. 9
Solution: $N = N_b + \frac{N_f}{2} + \frac{N_c}{8} = 1 + 0 + \frac{8}{8} = 2$
Q15 — Semiconductor and p-n Junction Diode · easy · theory
In which of the following cases is the junction diode forward biased? (Each option gives the potentials at the p-side and n-side ends.)
A. p at 0 V, n at +2 V
B. p at 0 V, n at −2 V ✓ Correct
C. p at −2 V, n at 0 V
D. p at 2 V, n at 5 V
Solution: Forward bias needs the p-side at higher potential than the n-side; only 0 V > −2 V satisfies this.
Q16 — Semiconductor and p-n Junction Diode · easy · theory
In p-type semiconductor, the majority charge carriers are
A. holes ✓ Correct
B. electrons
C. protons
D. neutrons
Solution: In p-type semiconductors (doped with acceptor atoms), holes are the majority carriers and electrons the minority carriers.
Q17 — Semiconductor and p-n Junction Diode · easy · theory
In forward bias the width of depletion layer in a p-n junction diode
A. increases
B. decreases ✓ Correct
C. remains constant
D. first increases then decreases
Solution: The forward-bias field opposes the barrier field, so the depletion layer becomes thinner.
Q18 — Semiconductor and p-n Junction Diode · easy · theory
In a junction diode, the holes are due to
A. protons
B. extra electrons
C. neutrons
D. missing electrons ✓ Correct
Solution: A hole is the vacancy left in the valence band when an electron jumps out — effectively a missing electron with positive charge.
Q19 — Semiconductor and p-n Junction Diode · easy · theory
Which of the following when added as an impurity into silicon produces n-type semiconductor?
A. P ✓ Correct
B. Al
C. B
D. Mg
Solution: Phosphorus is pentavalent — doping silicon with it produces an n-type semiconductor.
Q20 — Semiconductor and p-n Junction Diode · easy · theory
A semiconducting device is connected in a series circuit with a battery and a resistance. A current is allowed to pass through the circuit. If the polarity of the battery is reversed, the current drops to almost zero. The device may be
A. a p-n junction ✓ Correct
B. an intrinsic semiconductor
C. a p-type semiconductor
D. an n-type semiconductor
Solution: Only a p-n junction conducts well in one direction (forward, mA) and almost not at all in the other (reverse, µA).
Q21 — Semiconductor and p-n Junction Diode · easy · theory
To obtain a p-type germanium semiconductor, it must be doped with
A. phosphorus
B. indium ✓ Correct
C. antimony
D. arsenic
Solution: A trivalent impurity gives p-type behaviour; among the options only indium is trivalent.
Q22 — Semiconductor and p-n Junction Diode · easy · theory
When arsenic is added as an impurity to silicon, the resulting material is
A. n-type semiconductor ✓ Correct
B. p-type semiconductor
C. n-type conductor
D. insulator
Solution: Arsenic is pentavalent; doping silicon with it provides free electrons, giving an n-type semiconductor.
Q23 — Semiconductor and p-n Junction Diode · easy · theory
Which of the following, when added as an impurity into the silicon, produces n-type semiconductor?
A. Phosphorus ✓ Correct
B. Aluminium
C. Magnesium
D. Both (b) and (c)
Solution: Phosphorus is pentavalent, so doping silicon with it forms an n-type semiconductor.
Q24 — Semiconductor and p-n Junction Diode · easy · theory
Diamond is very hard, because
A. it is covalent solid
B. it has large cohesive energy ✓ Correct
C. high melting point
D. insoluble in all solvents
Solution: The carbon-carbon bonds in diamond have a very large cohesive energy (about 7.3 eV), which makes diamond extremely hard.
Q25 — Semiconductor and p-n Junction Diode · easy · theory
Which one of the following is the weakest kind of the bonding in solids?
A. Ionic
B. Metallic
C. van der Waals' ✓ Correct
D. Covalent
Solution: van der Waals' forces — short-range attractions arising from charge fluctuations in nearby molecules — are the weakest bonding in solids.
Q26 — Semiconductor and p-n Junction Diode · easy · theory
For amplification by a triode, the signal to be amplified is given to
A. the cathode
B. the grid ✓ Correct
C. the glass-envelope
D. the anode
Solution: In a triode amplifier the weak signal is applied in the grid circuit; the amplified output appears across the load in the plate circuit.
Q27 — Semiconductor and p-n Junction Diode · easy · theory
p-n junction is said to be forward biased, when
A. the positive pole of the battery is joined to the p-semiconductor and negative pole to the n-semiconductor ✓ Correct
B. the positive pole of the battery is joined to the n-semiconductor and negative pole to the p-semiconductor
C. the positive pole of the battery is connected to n-semiconductor and p-semiconductor
D. a mechanical force is applied in the forward direction
Solution: Forward bias: positive terminal to p-side and negative terminal to n-side, which lowers the potential barrier and permits current flow.
Q28 — Semiconductor and p-n Junction Diode · easy · theory
At absolute zero, Si acts as
A. non-metal
B. metal
C. insulator ✓ Correct
D. None of these
Solution: At absolute zero no electrons can cross the energy gap into the conduction band, so silicon behaves as an insulator.
Q29 — Transistors · easy · theory
For transistor action, which of the following statements is correct?
A. Base, emitter and collector regions should have same size
B. Both emitter junction as well as the collector junction are forward biased
C. The base region must be very thin and lightly doped ✓ Correct
D. Base, emitter and collector regions should have same doping concentrations
Solution: For transistor action the base must be very thin and lightly doped; the emitter is heavily doped and the collector moderately doped and slightly larger.
Q30 — Transistors · easy · numerical
A transistor is operated in common-emitter configuration at $V_c = 2$ volt such that a change in the base current from 100 µA to 200 µA produces a change in the collector current from 5 mA to 10 mA. The current gain is
A. 75
B. 100
C. 150
D. 50 ✓ Correct
Solution: $\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{5 \times 10^{-3}}{100 \times 10^{-6}} = 50$