CBSE Class 12 Physics Semiconductor Electronics Materials Devices and Simple Circuits MCQs Set 05

Question: At absolute zero, Si acts as
(a) non metal
(b) metal
(c) insulator
(d) none of these. 
Answer: c

Question: Zener diode is used for
(a) amplification
(b) rectification
(c) stabilisation
(d) producing oscillations in an oscillator.
Answer: c

Question: To obtain a p-type germanium semiconductor, it must be doped with
(a) indium
(b) phosphorus
(c) arsenic
(d) antimony. 
Answer: a

Question: A Zener diode, having breakdown voltage equal to 15V, is used in a voltage regulator circuit shown in figure. The current through the diode is
(a) 5 mA
(b) 10 mA
(c) 15 mA
(d) 20 mA
Answer: a

Question: 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) 10.4 × 10^–26 m
(b) 654 nm
(c) 654 Å
(d) 654 × 10^–11 m
Answer: b

 Question: In a p-n junction
(a) high potential at n side and low potential at p side
(b) high potential at p side and low potential at n side
(c) p and n both are at same potential 
(d) undetermined. 
Answer: a

Question: Carbon, silicon and germanium atoms have four valence electrons each. Their valence and conduction bands are separated by energy band gaps represented by (E_g)_C, (E_g)_Si and (E_g)_Ge respectively. Which one of the following relationships is true in their case?
(a) (E_g)_C > (E_g)_Si
(b) (E_g)_C < (E_g)_Si
(c) (E_g)C = (E_g)_Si
(d) (E_g)_C < (E_g)_Ge
Answer: a

Question: Application of a forward bias to a p-n junction
(a) widens the depletion zone
(b) increases the potential difference across the depletion zone
(c) increases the number of donors on the n side
(d) decreases the electric field in the depletion zone.
Answer: d

Question: 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) 1 × 10¹⁴ Hz
(b) 20 × 10¹⁴ Hz
(c) 10 × 10¹⁴ Hz
(d) 5 × 10¹⁴ Hz
Answer: d 

 Question: If a small amount of antimony is added to germanium crystal
(a) it becomes a p-type semiconductor
(b) the antimony becomes an acceptor atom
(c) there will be more free electrons than holes in the semiconductor
(d) its resistance is increased.
Answer: c

 Question: The device that can act as a complete electronic circuit is
(a) junction diode
(b) integrated circuit
(c) junction transistor
(d) zener diode.
Answer: b

Question: In forward bias, the width of potential barrier in a p-n junction diode
(a) remains constant
(b) decreases
(c) increases
(d) first (a) then (b)
Answer: b

Question: In the energy band diagram of a material shown here, the open circles and filled circles denote holes and electrons respectively. The material is
(a) an insulator
(b) a metal
(c) an n-type semiconductor
(d) a p-type semiconductor.
Answer: d

Question: When arsenic is added as an impurity to silicon, the resulting material is
(a) n-type conductor
(b) n-type semiconductor
(c) p-type semiconductor
(d) none of these. 
Answer: b

Question:Barrier potential of a p-n junction diode does not depend on
(a) diode design
(b) temperature
(c) forward bias
(d) doping density
Answer: a

Question: In a p type semiconductor, the majority carriers of current are
(a) protons
(b) electrons
(c) holes
(d) neutrons
Answer: c

Question: Which of the following, when added as an impurity into the silicon produces n type semiconductor?
(a) B
(b) Al
(c) P
(d) Mg 
Answer: c

Question: In forward biasing of the p-n junction
(a) the positive terminal of the battery is connected to p-side and the depletion region becomes thick.
(b) the positive terminal of the battery is connected to n-side and the depletion region becomes thin.
(c) the positive terminal of the battery is connected to n-side and the depletion region becomes thick.
(d) the positive terminal of the battery is connected to p-side and the depletion region becomes thin.
Answer: d

Question: The cause of the potential barrier in a p-n junction diode is
(a) depletion of negative charges near the junction
(b) concentration of positive charges near the junction
(c) depletion of positive charges near the junction
(d) concentration of positive and negative charges near the junction.
Answer: d

Question:A semiconducting device is connected in a series circuit with a battery and a resistance. A current is found 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-type semiconductor
(b) an intrinsic semiconductor
(c) a p-n junction
(d) an n-type semiconductor
Answer: c 

 Question:The depletion layer in the p-n junction region is caused by
(a) drift of holes
(b) diffusion of charge carriers
(c) migration of impurity ions
(d) drift of electrons. 
Answer: b

Question:If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be
(a) 25 Hz
(b) 50 Hz
(c) 70.7 Hz
(d) 100 Hz 
Answer: d

 Question: In a junction diode, the holes are due to
(a) extra electrons
(b) neutrons
(c) protons
(d) missing of electrons
Answer: d

Question: 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) 4000 nm
(b) 6000 nm
(c) 4000 Å
(d) 6000 Å
Answer: c

Question: In a p-n junction photo cell, the value of the photoelectromotive force produced by monochromatic light is proportional to
(a) the barrier voltage at the p-n junction
(b) the intensity of the light falling on the cell
(c) the frequency of the light falling on the cell
(d) the voltage applied at the p-n junction.
Answer: b

Question: The barrier potential of a p-n junction depends on
(1) type of semiconductor material
(2) amount of doping
(3) temperature
Which one of the following is correct?
(a) (1) and (2) only
(b) (2) only
(c) (2) and (3) only
(d) (1), (2) and (3) 
Answer: d

Question: Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (E_g)_C, (E_g)_Si and (E_g)_Ge.Which of the following statements is true?
(a) (E_g)_Si < (Eg)_Ge < (Eg)_C
(b) (E_g)_C < (Eg)_Ge < (Eg)_Si
(c) (E_g)_C > (Eg)_Si > (Eg)_Ge
(d) (E_g)_C = (Eg)_Si = (Eg)_Ge
Answer: b

Question: In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) free electrons in the n-region attract them
(b) they move across the junction by the potential difference
(c) hole concentration in p-region is more as compared to n-region
(d) all of these. 
Answer: c

Question: Which of the following statements is correct?
(a) Hole is an antiparticle of electron.
(b) Hole is a vacancy created when an electron leaves a covalent bond.
(c) Hole is the absence of free electrons.
(d) Hole is an artificially created particle.
Answer: b

Question: If a small amount of antimony is added to germanium crystal
(a) its resistance is increased
(b) it becomes a p-type semiconductor
(c) there will be more free electrons than holes in the semiconductor
(d) none of these. 
Answer: c

Question: A potential barrier of 0.3 V exists across a p-n junction. If the depletion region is 1 mm wide, what is the intensity of electric field in this region?
(a) 2 × 10⁵ V m^–1
(b) 3 × 10⁵ V m^–1
(c) 4 × 10⁵ V m^–1
(d) 5 × 10⁵ V m^–1
Answer: b

Question: Which of the following statements is incorrect for the depletion region of a diode?
(a) There are mobile charges exist.
(b) Equal number of holes and electrons exist, making the region neutral.
(c) Recombination of holes and electrons has taken place.
(d) None of these. 
Answer: a

Question: If the energy of a proton of sodium light (λ = 589 nm) equals the band gap of semiconductor, the minimum energy required to create hole electron pair
(a) 1.1 eV
(b) 2.1 eV
(c) 3.2 eV
(d) 1.5 eV 
Answer: b

Question: Which of the following equations correctly represents the temperature variation of energy gap between the conduction and valence bands for Si?
(a) E_g(T) = 0.70 – 2.23 × 10–⁴T eV
(b) E_g(T) = 0.70 + 2.23 × 10–⁴T eV
(c) E_g(T) = 1.10 – 3.60 × 10–⁴T eV
(d) E_g(T) = 1.10 + 3.60 × 10–⁴T eV 
Answer: c

Question: A potential barrier of 0.50 V exists in a p-n junction. If the depletion region is 5.0 × 10–7 m thick, what is the electric field in this region?
(a) 1 × 10³ V m^–1
(b) 1.0 × 10⁶ V m^–1
(c) 1 × 10² V m^–1
(d) 1 × 10⁴ V m^–1 
Answer: b

Question: The maximum wavelength of electromagnetic radiation, which can create a hole-electron pair in germanium. (Given that forbidden energy gap in germanium is 0.72 eV)
(a) 1.7 × 10^–6 m
(b) 1.5 × 10^–5 m
(c) 1.3 × 10^–4 m
(d) 1.9 × 10^–5 m
Answer: a

Question: In the circuit shown if drift current for the diode is 20 mA, the potential difference across the diode is
(a) 2 V
(b) 4.5 V
(c) 4 V
(d) 2.5 V 
Answer: c

Question: The dominant mechanism for motion of charge carriers in forward and reverse biased silicon p-n junction are
(a) drift in forward bias, diffusion in reverse bias
(b) diffusion in forward bias, drift in reverse bias
(c) diffusion in both forward and reverse bias
(d) drift in both forward and reverse bias.
Answer: b

Very Short Answer Type Questions 

Question: Can the potential barrier across a p-n junction be measured by simply connecting a voltmeter across the junction?
Answer: No, the voltmeter should have a very high resistance as compared to the resistance of p-n junction, which is nearly infinite.

Question: In the case of n-type Si-semiconductor, the donor energy level is slightly below the bottom of conduction band whereas in p-type semiconductor, the acceptor energy level is slightly above the top of valence band. Explain, giving examples, what role do these energy levels play in conduction and valence bands.
Answer: In n-type extrinsic semiconductors, the number of free electrons in conduction band is much more than the number of holes in valence band. The donor energy level lies just below the conduction band. In p-type extrinsic semiconductor, the number of holes in valence band is much more than the number of free electrons in conduction band. The acceptor energy level lies just above the valence band.

Question: How does an increase in doping concentration affect the width of depletion layer of a p-n junction diode?
Answer: When there is an increase in doping concentration, the applied potential difference causes an electric field which acts opposite to the potential barrier. This results in reducing the potential barrier and hence the width of depletion layer decreases.

Question: Why cannot we use Si and Ge in fabrication of visible LEDs?
Answer: LED’s must have band gap in the order of 1.8 eV to 3 eV but Si and Ge have band gap less than 1.8 eV.