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Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Physics Worksheet for Class 12
Class 12 Physics students should refer to the following printable worksheet in Pdf in Class 12. This test paper with questions and solutions for Class 12 Physics will be very useful for tests and exams and help you to score better marks
Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Worksheet Pdf
Question. In the following circuit, the output Y for all possible inputs A and B is expressed by the truth table.
Answer: C
Question. What is the voltage gain in a common emitter amplifier, where input resistance is 3 Ω and load resistance 24 Ω, β = 0.6 ?
(a) 8 . 4
(b) 4 . 8
(c) 2 . 4
(d) 480
Answer: B
Question. In a common emitter transistor amplifier β = 60, Ro = 5000 Ω and internal resistance of a transistor is 500 Ω. The voltage amplification of amplifier will be
(a) 500
(b) 460
(c) 600
(d) 560
Answer: C
Question. The band gap in germanium and silicon in ev respectively is
(a) 1.1, 0
(b) 0, 1.1
(c) 1.1, 0.7
(d) 0.7, 1.1
Answer: D
Question. A transistor has three impurity regions. All the three regions have different doping levels. In order of increasing doping level, the regions are
(a) emitter, base and collector
(b) collector, base and emitter
(c) base, emitter and collector
(d) base, collector and emitter
Answer: D
Question. In a common base amplifier the phase difference between the input signal voltage and the output voltage is
(a) 0
(b) π/4
(c) π/2
(d) π
Answer: A
Question. The following circut represents
(a) OR gate
(b) XOR gate
(c) AND gate
(d) NAND gate
Answer: B
Question. The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap (in eV) for the semiconductor is
(a) 0.9
(b) 0.7
(c) 0.5
(d) 1.1
Answer: C
Question. The diagram of a logic circuit is given below. The output F of the circuit is represented by
(a) W . (X + Y)
(b) W . (X . Y)
(c) W + (X . Y)
(d) W + (X + Y)
Answer: C
Question. In the half wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple would be
(a) 25 Hz
(b) 50 Hz
(c) 70.7 Hz
(d) 100 Hz
Answer: B
Question. A transistor is operated in common emitter configuration at VC = 2V such that a change in the base current from 100 mA to 300 mA produces a change in the collector current from 10mA to 20 mA. The current gain is
(a) 50
(b) 75
(c) 100
(d) 25
Answer: A
Question. In a p-n junction having depletion layer of thickness 10–6 m the potential across it is 0.1 V. The electric field is
(a) 107 V/m
(b) 10–6 V/m
(c) 105 V/m
(d) 10–5 V/m
Answer: C
Question. In the diagram, the input is across the terminals A and C and the output is across B and D. Then the output is
(a) zero
(b) same as the input
(c) full wave rectifier
(d) half wave rectifier
Answer: C
Question. A half-wave rectifier is being used to rectify an alternating voltage of frequency 50 Hz. The number of pulses of rectified current obtained in one second is
(a) 50
(b) 25
(c) 100
(d) 2000
Answer: B
Question. What is the conductivity of a semiconductor if electron density = 5 × 1012/cm3 and hole density = 8 × 1013/cm3 (μe = 2.3 m2 V–1 s–1, μh = 0.01 m2V–1 s–1)
(a) 5.634
(b) 1.968
(c) 3.421
(d) 8.964.
Answer: B
Question. The cause of the potential barrier in a p-n diode is
(a) depletion of positive charges near the junction
(b) concentration of positive charges near the junction
(c) depletion of negative charges near the junction
(d) concentration of positive and negative charges near the junction
Answer: D
Question. The ratio of electron and hole currents in a semiconductor is 7/4 and the ratio of drift velocities of electrons and holes is 5/4, then the ratio of concentrations of electrons and holes will be
(a) 5/7
(b) 7/5
(c) 25/49
(d) 49/25
Answer: B
Question. If A is the atomic mass number of an element, N is the Avogadro number and a is the lattice parameter, then the density of the element, if it has bcc crystal structure, is
(a) A/Na3
(b) 2A/Na3
(c) √3A/Na3
(d) 2√2A/Na3
Answer: B
Question. The current gain in transistor in common base mode is 0.99. To change the emitter current by 5 mA, the necessary change in collector will be
(a) 0.196 mA
(b) 2.45 mA
(c) 4.95 mA
(d) 5.1 mA
Answer: C
Question. In a reverse biased diode when the applied voltage changes by 1 V, the current is found to change by 0.5 μA. The reverse bias resistance of the diode is
(a) 2 × 105 Ω
(b) 2 × 106 Ω
(c) 200 Ω
(d) 2 Ω.
Answer: B
Question. The output of the given circuit in figure given below,
(a) would be zero at all times
(b) would be like a half wave rectifier with positive cycles in output
(c) would be like a half wave rectifier with negative cycles in output
(d) would be like that of a full wave rectifier
Answer: C
Question. In the circuit shown in figure given below, if the diode forward voltage drop is 0.3 V, the voltage difference between A and B is
(a) 1.3 V
(b) 2.3 V
(c) 0
(d) 0.5 V
Answer: B
Question. The transistor are usually made of
(a) metal oxides with high temperature coefficient of resistivity
(b) metals with high temperature coefficient of resistivity
(c) metals with low temperature coefficient of resistivity
(d) semiconducting materials having low temperature coefficient of resistivity
Answer: A
Question. The truth table given below is for
(a) NOR
(b) AND
(c) XOR
(d) NAND
Answer: D
Question. In a full wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple would be
(a) 25 Hz
(b) 50 Hz
(c) 70.7 Hz
(d) 100 Hz
Answer: D
Question. The manifestation of band structure in solids is due to
(a) Bohr’s correspondence principle
(b) Pauli’s exclusion principle
(c) Heisenberg’s uncertainty principle
(d) Boltzmann’s law
Answer: B
Question. The frequency response curve of RC coupled amplifier is shown in figure. The band with of the amplifier will be (35 ex 2)
(a) f3 – f2
(b) f4 – f1
(c) f4 – f2/2
(d) f3 – f1
Answer: B
Question. A diode having potential difference 0.5 V across its junction which does not depend on current, is connected in series with resistance of 20Ω across source. If 0.1 A current passes through resistance then what is the voltage of the source?
(a) 1.5 V
(b) 2.0 V
(c) 2.5 V
(d) 5 V
Answer: C
Question. The difference in the variation of resistance with temperature in a metal and a semiconductor arises essentially due to the difference in the
(a) crystal structure
(b) variation of the number of charge carriers with temperature
(c) type of bonding
(d) variation of scattering mechanism with temperature
Answer: B
Question. Carbon, Silicon and Germanium atoms have four valence electrons each. Their valence and conduction bands are separated by energy band gaps represented by (Eg)C, (Eg)Si and (Eg)Ge respectively. Which one of the following relationship is true in their case?
(a) (Eg)C > (Eg)Si
(b) (Eg)C < (Eg)Si
(c) (Eg)C = (Eg)Si
(d) (Eg)C < (Eg)Ge
Answer: A
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