JEE Physics Semiconductor Electronics MCQs Set 16

Physics Objective Questions and Answers: Semiconductor Electronics

Access targeted multiple-choice questions for Semiconductor Electronics designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question 1: Y is obtained by passing the inputs A and B through an OR gate and then ANDing its output with A. The truth table of Y is the same as that of
(a) Y = B
(b) Y = A
(c) Y = A + B
(d) Y = AB
Show Answer & Explanation

Answer: (b) Y = A

Explanation:
1. \( Y=(A+B)\cdot A=A+AB=A \) by the absorption law.
2. So Y equals A: 0, 0, 1, 1 for (A, B) = 00, 01, 10, 11.

Teacher's Note:
1. A + AB = A.

Question 2: An ideal operational amplifier is used with a 1 k\( \Omega \) input resistor and a 10 k\( \Omega \) feedback resistor in the inverting configuration (non-inverting input grounded). An input of +1 V gives an output of
(a) +10 V
(b) -10 V
(c) 0 V
(d) infinity
Show Answer & Explanation

Answer: (b) -10 V

Explanation:
1. \( V_{out}=-\frac{R_f}{R_{in}}V_{in}=-10\times1 \) V.
2. The output is -10 V.

Teacher's Note:
1. Inverting gain = -Rf/Rin.

Question 3: An ideal op-amp is used with a 1 k\( \Omega \) input resistor and a 100 k\( \Omega \) feedback resistor in the inverting configuration. The magnitude of the voltage gain is
(a) 10
(b) 100
(c) 1000
(d) 9.9
Show Answer & Explanation

Answer: (b) 100

Explanation:
1. Gain \( =-\frac{R_f}{R_{in}}=-\frac{100}{1}=-100 \).
2. Its magnitude is 100.

Teacher's Note:
1. Ideal op-amp has infinite gain and a virtual ground.

Question 4: Holes are charge carriers in
(a) intrinsic and p-type semiconductors only
(b) ionic solids and metals
(c) n-type semiconductors only
(d) intrinsic semiconductors, p-type semiconductors and metals
Show Answer & Explanation

Answer: (a) intrinsic and p-type semiconductors only

Explanation:
1. Holes arise in the valence band of intrinsic semiconductors (equal to electrons) and are majority carriers in p-type material.
2. Ionic solids and metals do not have hole conduction in the semiconductor sense.

Teacher's Note:
1. Holes are absent in n-type as majority carriers.

Question 5: Which pair of statements concerning the depletion zone of an unbiased p-n junction is true? (I) its width depends on the densities of the dopants; (II) the electric field in it is produced by ionised dopant atoms.
(a) Only (I) is true
(b) Only (II) is true
(c) Both (I) and (II) are true
(d) Neither is true
Show Answer & Explanation

Answer: (c) Both (I) and (II) are true

Explanation:
1. More heavily doped material gives a narrower depletion region, so (I) is true.
2. The field is produced by the fixed ionised donors and acceptors, so (II) is true.

Teacher's Note:
1. Heavier doping, thinner depletion region.

Question 6: A transistor is used in the common-emitter mode as an amplifier. Which pair of statements is correct? (I) the base-emitter junction is forward biased; (II) the input signal is connected in series with the voltage applied to bias the base-emitter junction.
(a) Only (I)
(b) Only (II)
(c) Both (I) and (II)
(d) Neither
Show Answer & Explanation

Answer: (c) Both (I) and (II)

Explanation:
1. In an amplifier the emitter-base junction is forward biased and the collector-base junction is reverse biased.
2. The a.c. signal is superimposed on (in series with) the base-emitter bias.

Teacher's Note:
1. The active region needs EB forward, CB reverse.

Question 7:
Assertion (A): In semiconductors, thermal collisions are responsible for taking a valence electron to the conduction band.
Reason (R): The number of conduction electrons goes on increasing with time as thermal collisions continuously take place.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. Thermal energy lifts valence electrons across the gap: A is true.
2. Recombination balances generation, so the number reaches a steady value; it does not keep increasing: R is false.

Teacher's Note:
1. Equilibrium carrier density is set by generation = recombination.

Question 8:
Assertion (A): A p-type semiconductor is a positive type crystal.
Reason (R): A p-type semiconductor is an uncharged crystal.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. The crystal is electrically neutral (holes are balanced by the negative acceptor ions), so A is false.
2. R is true.

Teacher's Note:
1. Doping does not add net charge.

Question 9:
Assertion (A): The diffusion current in a p-n junction is from the p-side to the n-side.
Reason (R): The diffusion current in a p-n junction is greater than the drift current when the junction is forward biased.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Holes diffuse from p to n and electrons from n to p, so the conventional diffusion current is from p to n: A is true.
2. In forward bias diffusion exceeds drift: R is true, but it does not explain the direction.

Teacher's Note:
1. Direction comes from carrier density gradients.

Question 10:
Assertion (A): Silicon is preferred over germanium for making semiconductor devices.
Reason (R): The energy gap in germanium is more than the energy gap in silicon.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. Silicon has a larger gap (1.1 eV) so it has far less thermal leakage, which is why it is preferred: A is true.
2. Germanium's gap is 0.67 eV, less than silicon, so R is false.

Teacher's Note:
1. Larger gap means higher operating temperature.

Question 11:
Assertion (A): A p-n junction with reverse bias can be used as a photodiode to measure light intensity.
Reason (R): In reverse bias the current is small but it is more sensitive to changes in incident light intensity.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Light creates extra minority carriers, which change the small reverse current by a large fraction, so the diode is a sensitive intensity detector.

Teacher's Note:
1. Photodiodes are operated in reverse bias.

Question 12:
Assertion (A): The number of electrons in a p-type silicon semiconductor is less than the number of electrons in a pure silicon semiconductor at room temperature.
Reason (R): It is due to the law of mass action.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Mass action gives \( n_in_h=n_i^2 \); extra holes from doping force the electron density below the intrinsic value.

Teacher's Note:
1. Product of electron and hole densities is fixed.

Question 13:
Assertion (A): A transistor amplifier in the common emitter configuration has a low input impedance.
Reason (R): The base to emitter region is forward biased.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A forward biased junction has a small dynamic resistance, giving a low input impedance.

Teacher's Note:
1. Forward bias gives low resistance; reverse bias high.

Question 14: Copper (monovalent, molar mass 63.54 g/mol, density 8.96 g/cm\( ^3 \)) has one conduction electron per atom. The number density of conduction electrons is
(a) \( 8.49\times10^{28} \) m\( ^{-3} \)
(b) \( 8.49\times10^{22} \) m\( ^{-3} \)
(c) \( 1.4\times10^{29} \) m\( ^{-3} \)
(d) \( 6.0\times10^{23} \) m\( ^{-3} \)
Show Answer & Explanation

Answer: (a) \( 8.49\times10^{28} \) m\( ^{-3} \)

Explanation:
1. Atoms per cm\( ^3 \) \( =\frac{8.96}{63.54}\times6.022\times10^{23}=8.49\times10^{22} \).
2. Converting to per m\( ^3 \) multiplies by \( 10^6 \): \( 8.49\times10^{28} \).

Teacher's Note:
1. 1 cm^3 = 10^-6 m^3.

Question 15: In a photodiode the conductivity increases when the material is exposed to light, and it changes only if the wavelength is less than 620 nm. The band gap is
(a) 2.0 eV
(b) 1.0 eV
(c) 3.1 eV
(d) 0.62 eV
Show Answer & Explanation

Answer: (a) 2.0 eV

Explanation:
1. \( E_g=\frac{hc}{\lambda}=\frac{1240}{620}=2.0 \) eV.

Teacher's Note:
1. Threshold wavelength corresponds to the band gap.

Question 16: A potential barrier of 0.3 V exists across a p-n junction whose depletion region is 1 \( \mu \)m wide. An electron of speed \( 5\times10^5 \) m/s approaches the junction from the n-side. The electric field in the region and the speed with which the electron enters the p-side are
(a) \( 3\times10^5 \) V/m and \( 3.8\times10^5 \) m/s
(b) \( 3\times10^4 \) V/m and \( 3.8\times10^5 \) m/s
(c) \( 3\times10^5 \) V/m and \( 6.2\times10^5 \) m/s
(d) \( 3\times10^5 \) V/m and \( 1.2\times10^5 \) m/s
Show Answer & Explanation

Answer: (a) \( 3\times10^5 \) V/m and \( 3.8\times10^5 \) m/s

Explanation:
1. Field \( E=\frac{0.3}{10^{-6}}=3\times10^5 \) V/m.
2. Initial kinetic energy \( \frac12(9.11\times10^{-31})(5\times10^5)^2=0.71 \) eV. It climbs the barrier of 0.3 eV, leaving 0.41 eV, so \( v=\sqrt{\frac{2\times0.41\times1.6\times10^{-19}}{9.11\times10^{-31}}}=3.8\times10^5 \) m/s.

Teacher's Note:
1. The electron loses 0.3 eV going up the barrier.

Question 17: The number density of conduction electrons in pure silicon at room temperature is about \( 10^{16} \) m\( ^{-3} \). By doping with phosphorus, the number is raised by a factor of a million (every donor gives one electron). Silicon has \( 5\times10^{28} \) atoms per m\( ^3 \). The fraction of silicon atoms that must be replaced by phosphorus is about
(a) \( \frac{1}{5\times10^6} \)
(b) \( \frac{1}{5\times10^{12}} \)
(c) \( \frac{1}{5\times10^{3}} \)
(d) \( \frac{1}{5\times10^{9}} \)
Show Answer & Explanation

Answer: (a) \( \frac{1}{5\times10^6} \)

Explanation:
1. Required donor density \( =10^{6}\times10^{16}=10^{22} \) m\( ^{-3} \).
2. Fraction \( =\frac{10^{22}}{5\times10^{28}}=\frac{1}{5\times10^6} \).

Teacher's Note:
1. One electron per phosphorus atom.

Question 18: Ge and Si diodes conduct at 0.3 V and 0.7 V respectively. A 12 V battery is connected in series with a 5 k\( \Omega \) resistor and a Ge diode (forward biased) in parallel with a Si diode connected in the reverse direction. The voltage across the resistor and the current are
(a) 11.7 V and 2.34 mA
(b) 11.3 V and 2.26 mA
(c) 12 V and 2.4 mA
(d) 0.3 V and 0.06 mA
Show Answer & Explanation

Answer: (a) 11.7 V and 2.34 mA

Explanation:
1. Only the Ge diode conducts, with a 0.3 V drop, so the voltage across the resistor is \( 12-0.3=11.7 \) V.
2. Current \( =\frac{11.7}{5000}=2.34 \) mA.

Teacher's Note:
1. The reversed diode carries no current.

Question 19: The current-voltage relation of an ideal p-n junction diode is \( i=i_0(e^{eV/kT}-1) \) with \( i_0=10\ \mu \)A and T = 300 K. The voltage at which \( e^{eV/kT}=100 \) is about
(a) 0.12 V
(b) 0.25 V
(c) 0.6 V
(d) 1.2 V
Show Answer & Explanation

Answer: (a) 0.12 V

Explanation:
1. \( V=\frac{kT}{e}\ln100=0.0259\times4.605 \).
2. So V = 0.12 V.

Teacher's Note:
1. kT/e is about 0.0259 V at 300 K.

Question 20: For the same diode (\( i=i_0(e^{eV/kT}-1) \), \( i_0=10\ \mu \)A, 300 K), for V much greater than 0.12 V the dynamic resistance is \( r=\frac{kT}{ei_0}e^{-eV/kT} \). The voltage at which the dynamic resistance is 0.2 \( \Omega \) is about
(a) 0.12 V
(b) 0.25 V
(c) 0.40 V
(d) 0.60 V
Show Answer & Explanation

Answer: (b) 0.25 V

Explanation:
1. \( e^{-eV/kT}=\frac{0.2\times10^{-5}}{0.0259}=7.7\times10^{-5} \).
2. \( \frac{eV}{kT}=9.47 \), so \( V=0.0259 imes9.47=0.245 \) V.

Teacher's Note:
1. Dynamic resistance is dV/di.

Question 21: A transistor is in common-emitter configuration with collector supply 8 V and a voltage drop of 0.5 V across an 800 \( \Omega \) resistor in the collector. If the current gain factor \( \alpha \) is 0.96, the base current is
(a) 26 \( \mu \)A
(b) 50 \( \mu \)A
(c) 0.625 mA
(d) 12 \( \mu \)A
Show Answer & Explanation

Answer: (a) 26 \( \mu \)A

Explanation:
1. Collector current \( I_C=\frac{0.5}{800}=0.625 \) mA; \( \beta=\frac{\alpha}{1-\alpha}=\frac{0.96}{0.04}=24 \).
2. Base current \( I_B=\frac{I_C}{\beta}=\frac{0.625}{24}=26 \) \( \mu \)A.

Teacher's Note:
1. First find beta from alpha.

Question 22: An npn transistor in common-emitter mode has collector current 4 mA, a collector supply of 8 V, collector-emitter voltage 4 V, base-emitter voltage 0.6 V and current gain \( \beta_{dc}=100 \). The load resistance \( R_L \) and base resistance \( R_B \) are
(a) \( R_L=1 \) k\( \Omega \), \( R_B=185 \) k\( \Omega \)
(b) \( R_L=185 \) k\( \Omega \), \( R_B=1 \) k\( \Omega \)
(c) \( R_L=2 \) k\( \Omega \), \( R_B=185 \) k\( \Omega \)
(d) \( R_L=1 \) k\( \Omega \), \( R_B=100 \) k\( \Omega \)
Show Answer & Explanation

Answer: (a) \( R_L=1 \) k\( \Omega \), \( R_B=185 \) k\( \Omega \)

Explanation:
1. Voltage across the load is 8 - 4 = 4 V, so \( R_L=\frac{4}{4\ \text{mA}}=1 \) k\( \Omega \).
2. Base current \( =\frac{4\ \text{mA}}{100}=40 \) \( \mu \)A and \( R_B=\frac{8-0.6}{40\ \mu\text{A}}=185 \) k\( \Omega \).

Teacher's Note:
1. Find the base current from Ic / beta.

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Multiple Choice Questions (MCQs) for JEE Physics Semiconductor Electronics

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