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JEE • PHYSICS • SEMI CONDUCTOR DEVICES

Semi conductor devices practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 61-90 of 361 questions
61 Medium
61 of 361
Remembering +4 | -1
The energy band gap (distance between the conduction band and valence band) in conductor is
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because conductors have overlapping valence and conduction bands, resulting in a zero energy gap.
62 Medium
62 of 361
Remembering +4 | -1
What is the coordination number of sodium ions in the case of sodium chloride structure
View answer and explanation
Correct option: A
6
Option A (6) is correct because in the NaCl structure, each sodium ion is surrounded by 6 chloride ions.
63 Medium
63 of 361
Application +4 | -1
The input resistance of a CE amplifier is 5kΩ5 \mathrm{k \Omega} and the load resistance is 20kΩ20 \mathrm{k} \Omega. If the current gain is 100 then mutual conductance of the transistor used is
View answer and explanation
Correct option: C
20×10−3Ω−120 \times 10^{-3} \Omega^{-1}
Option C (20×10−3Ω−120 \times 10^{-3} \Omega^{-1}) is correct because the mutual conductance is the ratio of the current gain to the input resistance.
64 Medium
64 of 361
Understanding +4 | -1
Diode is not considered as linear device because
View answer and explanation
Correct option: A
The value of plate current is not directly proportional to the plate voltage
Option A (The value of plate current is not directly proportional to the plate voltage) is correct because a diode's current-voltage relationship is non-linear.
65 Medium
65 of 361
Analyzing +4 | -1
Following is the relation between current and charge I=AT2eq/VLI = A{T^2}{e^{q/{V_L}}}, then value of VL{V_L} will be
View answer and explanation
Correct option: C
kTV\frac{{kT}}{V}
Option C (kTV\frac{{kT}}{{V}}) is correct because by comparing the given equation with the standard equation, we can directly derive the value of VLV_L.
66 Medium
66 of 361
Understanding +4 | -1
The reason of current flow in PP-NN junction in forward bias is
View answer and explanation
Correct option: C
Diffusion of charge carriers
Option C (Diffusion of charge carriers) is correct because in forward bias, majority carriers diffuse across the junction, creating current flow.
67 Medium
67 of 361
Application +4 | -1
In a transistor if collector current is 25 mA and base current is 1 mA, then current amplification factor ∝\propto is
View answer and explanation
Correct option: C
2526\frac{{25}}{{26}}
Option C (2526\frac{{25}}{{26}}) is correct because the current amplification factor is calculated as the ratio of collector current to emitter current.
68 Medium
68 of 361
Remembering +4 | -1
The PNPN junction diode is used as
View answer and explanation
Correct option: B
A rectifier
Option B (A rectifier) is correct because a PN junction diode is primarily used to convert AC voltage to DC voltage, which is the function of a rectifier.
69 Medium
69 of 361
Understanding +4 | -1
In an experiment, the saturation in the plate current in a diode is observed at 240V240V. But a student still wants to increase the plate current. It can be done, if
View answer and explanation
Correct option: D
The filament current is increased
The correct answer is D (The filament current is increased) because increasing the filament current increases the temperature, leading to more electron emission and thus a higher plate current even after saturation.
70 Medium
70 of 361
Understanding +4 | -1
In PP-NN junction, avalanche current flows in circuit when biasing is
View answer and explanation
Correct option: B
Reverse
Option B (Reverse) is correct because avalanche breakdown occurs in a reverse-biased P-N junction when the reverse voltage exceeds a certain threshold.
71 Medium
71 of 361
Understanding +4 | -1
Forward bias characteristics of a p-n junction diode is used in which of the following devices?
View answer and explanation
Correct option: C
Rectifier circuit
Option C is correct because rectifier circuits utilize the forward bias characteristics of a p-n junction diode to allow current flow in one direction.
72 Medium
72 of 361
Remembering +4 | -1
Biaxial crystal among the following is
View answer and explanation
Correct option: C
Selenite
Option C (Selenite) is correct because Selenite is a biaxial crystal, while the others are uniaxial.
73 Medium
73 of 361
Understanding +4 | -1
When phosphorus and antimony are mixed in germanium, then
View answer and explanation
Correct option: B
NN-type semiconductor is formed
Option B (NN-type semiconductor is formed) is correct because phosphorus and antimony are pentavalent impurities, which donate extra electrons to the germanium lattice.
74 Medium
74 of 361
Application +4 | -1
A npn transistor is connected in common emitter configuration in a given amplifier. A load resistance of 800 Ω800\,\Omega is connected in the collector circuit and the voltage drop across it is 0.8 V. If the current amplification factor is 0.96 and the input resistance of the circuit is 192 Ω192\,\Omega , the voltage gain and the power gain of the amplifier will respectively be:
View answer and explanation
Correct option: A
4 3.84
Option A (4 3.84) is correct because the voltage gain and power gain are calculated using the given current amplification factor and resistances.
75 Medium
75 of 361
Understanding +4 | -1
When a solid with a band gap has a donor level just below its empty energy band, the solid is
View answer and explanation
Correct option: D
nn-type semiconductor
Option D (nn-type semiconductor) is correct because a donor level just below the empty energy band is characteristic of an n-type semiconductor.
76 Medium
76 of 361
Understanding +4 | -1
Carbon, silicon and germanium have four valence electrons each. At room temperature which one of the following statements is more appropriate?
View answer and explanation
Correct option: D
The number of free electrons for conduction is significant only in Si and Ge but small in C
Option D is correct. At room temperature, silicon and germanium have a significant number of free electrons for conduction, while carbon has very few due to its stronger covalent bonds.
77 Medium
77 of 361
Understanding +4 | -1
Assertion : In an unbiased p-n junction, holes diffuse from the p-region to n-region.

Reason : Whole concentration in p-region is more as compared to n-region.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and the Reason is a correct explanation of the Assertion
Option A (If both Assertion and Reason are true and the Reason is a correct explanation of the Assertion) is correct because the higher concentration of holes in the p-region causes diffusion to the n-region.
78 Medium
78 of 361
Understanding +4 | -1
Energy bands in solids are a consequence of
View answer and explanation
Correct option: B
Pauli’s exclusion principle
Option B (Pauli’s exclusion principle) is correct because energy bands in solids are a direct consequence of the Pauli Exclusion Principle.
79 Medium
79 of 361
Understanding +4 | -1
Assertion : If the temperature of a semiconductor is increased, its resistance decreases.

Reason : The energy gap between conduction band and valence band in case of semiconductor is small.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and the Reason is a correct explanation of the Assertion
Option A is correct because both the assertion and reason are true, and the reason correctly explains why resistance decreases with temperature in semiconductors.
80 Medium
80 of 361
Remembering +4 | -1
The charge carries in a pp-type semiconductor are
View answer and explanation
Correct option: C
Holes in larger numbers and electrons in smaller numbers
The correct answer is C (Holes in larger numbers and electrons in smaller numbers) because p-type semiconductors are doped to increase the concentration of holes.
81 Medium
81 of 361
Understanding +4 | -1
With the rise in temperature, the electrical conductivity of intrinsic semiconductor
View answer and explanation
Correct option: A
Increases
Option A (Increases) is correct because as temperature rises, more electrons are excited to the conduction band, increasing the number of charge carriers and thus the conductivity.
82 Medium
82 of 361
Application +4 | -1
A PP-type semiconductor has acceptor levels 57  meV57\;meV above the valence band. The maximum wavelength of light required to create a hole is (Planck’s constant h=6.6×10−34Jh = 6.6 \times {10^{ - 34}}J-s)
View answer and explanation
Correct option: C
217100 A∘\mathop A\limits^ \circ
The correct answer is C (217100 A∘\mathop A\limits^ \circ) because the maximum wavelength is calculated using E=hcλE = \frac{{hc}}{\lambda }.
83 Medium
83 of 361
Understanding +4 | -1
For which of the following the resistance decreases on increasing the temperature
View answer and explanation
Correct option: C
Germanium
Option C (Germanium) is correct because Germanium is a semiconductor, and semiconductors exhibit a decrease in resistance with increasing temperature due to increased carrier concentration.
84 Medium
84 of 361
Understanding +4 | -1
If ne{n_e} and nh{n_h} are the number of electrons and holes in a semiconductor heavily doped with phosphorus, then
View answer and explanation
Correct option: A
ne>>nh{n_e} > > {n_h}
Option A (ne>>nh{n_e} > > {n_h}) is correct because phosphorus is a pentavalent impurity, making the semiconductor n-type with a much higher concentration of electrons than holes.
85 Medium
85 of 361
Application +4 | -1
Pure Si at temperature of 300 K300 \mathrm{~K} has same electron (ne)\left(\mathrm{n}_{\mathrm{e}}\right) and hole (nk)\left(\mathrm{n}_{\mathrm{k}}\right) concentrations of 1.5×1016 m−31.5 \times 10^{16} \mathrm{~m}^{-3}. Doping by 'In' increases nh\mathrm{n}_{\mathrm{h}} to 3×10223 \times 10^{22} m−3\mathrm{m}^{-3}. Calculate ne\mathrm{n}_{\mathrm{e}} in the doped Si\mathrm{Si}.
View answer and explanation
Correct option: D
7.5×109 m−37.5 \times 10^9 \mathrm{~m}^{-3}
Option D (7.5×109 m−37.5 \times 10^9 \mathrm{~m}^{-3}) is correct because it is calculated using the law of mass action for semiconductors.
86 Medium
86 of 361
Remembering +4 | -1
Minority carriers in a pp-type semiconductor are
View answer and explanation
Correct option: A
Free electrons
Option A (Free electrons) is correct because in a p-type semiconductor, holes are the majority carriers, and free electrons are the minority carriers.
87 Medium
87 of 361
Application +4 | -1
The decimal equivalent of the binary number (11010.101)2{\left( {11010.101} \right)_2} is
View answer and explanation
Correct option: C
26.625
Option C (26.625) is correct because it is the decimal equivalent of the binary number (11010.101)₂.
88 Medium
88 of 361
Understanding +4 | -1
Using Boolean algebra, the value of (A⋅B)+A(A \cdot B)+A is equal to
View answer and explanation
Correct option: D
(A⋅Aˉ)+A(A \cdot \bar{A})+A
Option D ((A⋅Aˉ)+A(A \cdot \bar{A})+A) is correct because using Boolean algebra, (A⋅B)+A(A \cdot B)+A simplifies to A, which is equivalent to (A⋅Aˉ)+A(A \cdot \bar{A})+A.
89 Medium
89 of 361
Analyzing +4 | -1
Assertion : Absorption factor for GaAs\mathrm{GaAs} is larger as compared to Si\mathrm{Si} for sunlight.

Reason : ΔEg\Delta \mathrm{E}_{\mathrm{g}} is more for GaAs\mathrm{GaAs}.
View answer and explanation
Correct option: B
If both Assertion and Reason are true but Reason is not a correct explanation of the Assertion
Option B is correct because while both the assertion and reason are true, the band gap energy difference doesn't directly explain the absorption factor difference.
90 Medium
90 of 361
Remembering +4 | -1
For an insulator the forbidden energy gap is
View answer and explanation
Correct option: C
5 eV
Option C (5 eV) is correct because insulators have a large forbidden energy gap, typically greater than 3 eV, which prevents electrons from easily moving to the conduction band.
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