Semiconductor and p-n Junction Diode — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Semiconductor and p-n Junction Diode MCQs with step-by-step solutions (65 questions). Part of Semiconductor Electronics. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Semiconductor and p-n Junction Diode · medium · theory
The electron concentration in an n-type semiconductor is the same as hole concentration in a p-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them.
A. Current in n-type = Current in p-type
B. Current in p-type > Current in n-type
C. Current in n-type > Current in p-type ✓ Correct
D. No current will flow in p-type, current will only flow in n-type
Solution: The mobility of electrons is greater than that of holes, so for equal carrier concentrations the current in the n-type semiconductor is greater.
Q2 — Semiconductor and p-n Junction Diode · medium · theory
Consider the following statements (A) and (B) and identify the correct answer.
(A) A Zener diode is connected in reverse bias, when used as a voltage regulator.
(B) The potential barrier of p-n junction lies between 0.1 V to 0.3 V.
A. (A) and (B) both are correct
B. (A) and (B) both are incorrect
C. (A) is correct and (B) is incorrect ✓ Correct
D. (A) is incorrect but (B) is correct
Solution: A Zener diode operates in reverse bias as a voltage regulator — (A) is correct. The barrier potential is about 0.6–0.7 V for silicon and 0.2–0.35 V for germanium — (B) is incorrect.
Q3 — Semiconductor and p-n Junction Diode · medium · theory
Out of the following which one is a forward biased diode? (Each option gives the potential at the p-side of the diode and at the far end of the series resistor.)
A. p-side at −4 V, other end at −2 V
B. p-side at 2 V, other end at 5 V
C. p-side at −2 V, other end at +2 V
D. p-side at 0 V, other end at −3 V ✓ Correct
Solution: A diode is forward biased when its p-side is at a higher potential than its n-side. Only in the last case (0 V > −3 V) is the p-side more positive.
Q4 — 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.
Q5 — 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.
Q6 — Semiconductor and p-n Junction Diode · medium · theory
The solids which have the negative temperature coefficient of resistance are
A. insulator only
B. semiconductors only
C. insulators and semiconductors ✓ Correct
D. metals
Solution: In both insulators and semiconductors, the number of free electrons increases with temperature, so resistance falls — a negative temperature coefficient.
Q7 — Semiconductor and p-n Junction Diode · medium · numerical
An LED is constructed from a p-n junction diode using GaAsP. The energy gap is 1.9 eV. The wavelength of the light emitted will be equal to
A. $1.04 \times 10^{-26}$ m
B. 654 nm ✓ Correct
C. 654 Å
D. $654 \times 10^{-11}$ m
Solution: $\lambda = \frac{hc}{E} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.9 \times 1.6 \times 10^{-19}} = 6.54 \times 10^{-7}$ m $\approx 654$ nm
Q8 — 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.
Q9 — Semiconductor and p-n Junction Diode · medium · theory
In a p-n junction diode, change in temperature due to heating
A. does not affect resistance of p-n junction
B. affects only forward resistance
C. affects only reverse resistance
D. affects the overall V-I characteristics of p-n junction ✓ Correct
Solution: Heating increases electron-hole pair generation, changing the diode current and resistance in both forward and reverse bias — the overall V-I characteristic is affected.
Q10 — Semiconductor and p-n Junction Diode · medium · theory
Which one of the following represents forward bias diode? (Each option gives the potential at the p-side of the diode and at the far end of the series resistor.)
A. p-side at 0 V, other end at −2 V ✓ Correct
B. p-side at −4 V, other end at −3 V
C. p-side at −2 V, other end at +2 V
D. p-side at 3 V, other end at 5 V
Solution: Forward bias requires the p-side at higher potential than the n-side. Only 0 V > −2 V satisfies this among the options.
Q11 — Semiconductor and p-n Junction Diode · medium · numerical
An ideal junction diode is connected in series with a 1 kΩ resistor between point A at +4 V and point B at −6 V (the diode conducts from A towards B). The value of current flowing through AB is
A. $10^{-2}$ A ✓ Correct
B. $10^{-1}$ A
C. $10^{-3}$ A
D. 0 A
Solution: The diode is forward biased, so (being ideal) it drops no voltage:
$I = \frac{V_A - V_B}{R} = \frac{4 - (-6)}{1\text{ k}\Omega} = \frac{10}{10^3} = 10^{-2}$ A
Q12 — Semiconductor and p-n Junction Diode · medium · numerical
A 10 V battery is connected through a resistance $R_1 = 2$ Ω to two parallel branches: one contains an ideal diode $D_1$ in series with $R_2 = 3$ Ω, the other contains an ideal diode $D_2$ in series with $R_3 = 2$ Ω. $D_1$ is reverse biased and $D_2$ is forward biased. The current flowing through the resistance $R_1$ will be
A. 2.5 A ✓ Correct
B. 10.0 A
C. 1.43 A
D. 3.13 A
Solution: $D_1$ blocks its branch; current flows only through $D_2$ and $R_3$:
$i = \frac{V}{R_1 + R_3} = \frac{10}{2 + 2} = 2.5$ A
Q13 — 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
Q14 — Semiconductor and p-n Junction Diode · medium · theory
The barrier potential of a p-n junction depends on: (i) type of semiconductor material, (ii) amount of doping, (iii) temperature. Which one of the following is correct?
A. (i) and (ii) only
B. (ii) only
C. (ii) and (iii) only
D. (i), (ii) and (iii) ✓ Correct
Solution: The barrier potential depends on the material (Si vs Ge), the doping level (majority carrier density) and the temperature (minority carrier density) — all three.
Q15 — 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.
Q16 — Semiconductor and p-n Junction Diode · medium · theory
C and Si both have same lattice structure, having 4 bonding electrons in each. However, C is insulator whereas Si is intrinsic semiconductor. This is because
A. in case of C, the valence bond is not completely filled at absolute zero temperature
B. in case of C, the conduction band is partly filled even at absolute zero temperature
C. the four bonding electrons in the case of C lie in the second orbit, whereas in the case of Si they lie in the third ✓ Correct
D. the four bonding electrons in the case of C lie in the third orbit, whereas for Si they lie in the fourth orbit
Solution: Carbon's bonding electrons are in the second orbit (tightly bound), while silicon's are in the third orbit (more loosely bound) — hence Si conducts as a semiconductor while C is an insulator.
Q17 — 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.
Q18 — 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.
Q19 — 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.
Q20 — Semiconductor and p-n Junction Diode · medium · numerical
Sodium has body centred packing. Distance between two nearest atoms is 3.7 Å. The lattice parameter is
A. 6.8 Å
B. 4.3 Å ✓ Correct
C. 3.0 Å
D. 8.6 Å
Solution: For bcc, nearest-neighbour distance $d = \frac{\sqrt{3}}{2}a$
$a = \frac{2 \times 3.7}{\sqrt{3}} = 4.3$ Å
Q21 — Semiconductor and p-n Junction Diode · medium · numerical
A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength
A. 6000 Å
B. 4000 nm
C. 6000 nm
D. 4000 Å ✓ Correct
Solution: Threshold wavelength $= \frac{hc}{E} \approx 5000$ Å. Only radiation with $\lambda <$ 5000 Å can be detected — 4000 Å qualifies.
Q22 — Semiconductor and p-n Junction Diode · medium · numerical
A p-n photodiode is made of a material with a band gap of 2.0 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly
A. $10 \times 10^{14}$ Hz
B. $5 \times 10^{14}$ Hz ✓ Correct
C. $1 \times 10^{14}$ Hz
D. $20 \times 10^{14}$ Hz
Solution: $\nu = \frac{E}{h} = \frac{2 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}} = 5 \times 10^{14}$ Hz
Q23 — Semiconductor and p-n Junction Diode · medium · numerical
If the lattice parameter for a crystalline structure is 3.6 Å, then the atomic radius in fcc crystals is
A. 1.81 Å
B. 2.10 Å
C. 2.92 Å
D. 1.27 Å ✓ Correct
Solution: For fcc: $r = \frac{a}{2\sqrt{2}} = \frac{3.6}{2.83} = 1.27$ Å
Q24 — 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 ($a = b = c$).
Q25 — Semiconductor and p-n Junction Diode · medium · theory
In the energy band diagram of a material, a few electrons sit at the bottom of the conduction band while a much larger number of holes occupy the top of the valence band. The material is a/an
A. p-type semiconductor ✓ Correct
B. insulator
C. metal
D. n-type semiconductor
Solution: Holes greatly outnumbering electrons means holes are the majority carriers — a p-type semiconductor.
Q26 — Semiconductor and p-n Junction Diode · medium · theory
A forward biased diode is: (each option gives the potential at the p-side of the diode and at the far end of the series resistor)
A. p-side at −4 V, other end at −3 V
B. p-side at 3 V, other end at 5 V
C. p-side at −2 V, other end at +2 V
D. p-side at 0 V, other end at −2 V ✓ Correct
Solution: The diode conducts (forward bias) when the p-side is at higher potential: 0 V > −2 V.
Q27 — Semiconductor and p-n Junction Diode · medium · numerical
Copper has face-centered cubic (fcc) lattice with interatomic spacing equal to 2.54 Å. The value of lattice constant for this lattice is
A. 1.27 Å
B. 5.08 Å
C. 2.54 Å
D. 3.59 Å ✓ Correct
Solution: For fcc the nearest-neighbour spacing $r = \frac{a}{\sqrt{2}}$
$a = \sqrt{2}r = 1.414 \times 2.54 = 3.59$ Å
Q28 — 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.
Q29 — 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).
Q30 — Semiconductor and p-n Junction Diode · medium · theory
Application of a forward bias to a p-n junction
A. increases the number of donors on the n-side ✓ Correct
B. increases the electric field in the depletion zone
C. increases the potential difference across the depletion zone
D. widens the depletion zone
Solution: Forward bias decreases the potential barrier, the depletion field and the depletion width; per the given options it increases the number of donors on the n-side.