14.

Semiconductor Electronics; Materials, Devices and Simple Circuits

A Quick Recapitulation of the Chapter
1. Metals They have very low resistivity or high conductivity,
−2 −8 2 8 −1
ρ ∼ 10 −10 Ωm , σ ∼ 10 −10 Sm
2. Semiconductors They have resistivity or conductivity between metals and
insulators. i.e., ρ ∼ 10−5 −106 Ωm , σ ∼ 10+5−10−6 Sm−1
Types of Semiconductors
Types of semiconductors are given below
(i) Element semiconductors are available in natural form, e.g., silicon and
germanium.
(ii) Compound semiconductors are made by compounding the metals, e.g., CdS,
GaAs, CdSe, InP, anthracene, polyaniline, etc.
3. Insulators They have high resistivity or low conductivity.
i.e., ρ ∼ 1011 −1019 Ωm , σ ∼ 10−11−10−19 Sm−1
4. Fermi energy It is the maximum possible energy possessed by free electrons of a
material at absolute zero temperature (i.e., 0 K ¿ .
5. On the basis of purity, semiconductors are of two types
(i) Intrinsic semiconductor It is a pure semiconductor without any significant dopant
species present.
n e=nh=n i
where, n e and n h are number densities of electrons and holes respectively and ni is
called intrinsic carrier concentration.
An intrinsic semiconductor is also called an undoped semiconductor or i-type
semiconductor.
(ii) Extrinsic semiconductor Pure semiconductor when doped with the impurity is
known as extrinsic semiconductor.
Extrinsic semiconductors are basically of two types
(a) n -type semiconductor (b) p -type semiconductor
Note Both the types of semiconductors are electrically neutral.
6. In n -type semiconductor, majority charge carriers are electrons and minority charge
carriers are holes, i.e., n e >n h. Here, we dope Si or Ge with a pentavalent element,
then four of its electrons bond with the four silicon neighbours, while fifth remains
very weakly bound to its parent atom. 7. In p-type semiconductor, majority charge
carriers are holes and minority charge carriers are electrons, i.e., n h> ne. In a p-type
semiconductor, doping is done with trivalent impurity atoms i.e., those atoms which
have three valence electrons in their valence shell.
7. At equilibrium condition, n e nh=n2i
8. Minimum energy required to create a hole-electron pair, hv ≥ E g, where E g is energy
band gap.
9. A p-n junction is an arrangement made by a close contact of n -type semiconductor
and p-type semiconductor.
10. A semiconductor diode is basically a p−n junction with metallic contacts provided at
the ends for the application of an external voltage.
A p−n junction diode is represented as the symbol. The direction of arrow indicates
the conventional direction of current (when P y n the diode is under forward bias).
V
12. The DC resistance of a junction diode, r DC =
l
 ΔV
13. The dynamic resistance of junction diode, r AC =
Δ /¿ ¿
14. Diode as rectifier The process of converting alternating voltage or current into direct
voltage or current is called rectification. Diode is used as a rectifier for converting
alternating current or voltage into direct current or voltage.
There are two ways of using a diode as a rectifier i.e.,
(i) Diode as a half-wave rectifier Diode conducts corresponding to positive half cycle
and does not conduct during negative half cycle. Hence, AC is converted by diode
into unidirectional pulsating DC . This action is known as half-wave rectification.
(ii) Diode as a full-wave rectifier In the full-wave rectifier, two p - n junction diodes
D 1 and D2 are used. Its working is based on the principle that junction diode offers
very low resistance in forward bias and very high resistance in reverse bias.
15. Optoelectronic devices Semiconductor diodes in which carriers are generated by
photons. i.e., photo-excitation, such devices are known as optoelectronic devices.
These are as follows
(i) Light emitting diode (LED) It is a heavily doped p−n junction diode which
converts electrical energy into light energy.
Its symbol is given by
(ii) Photodiode A photodiode is a special type of junction diode used for detecting
optical signals. It is a reverse biased p−n junction made from a photosensitive
material. Its symbol is given by (iii) Solar cell Solar cell is a p−n junction diode which
converts solar energy into electrical energy.
Its symbol is given by

16. Zener diode is a reverse biased heavily doped p−n junction diode. It is operated in
breakdown region. Its symbol is given by and is used as voltage regulator.
17. A transistor is a combination of two p−n junction joined in series. A junction
transistor is known as bipolar junction transistor (BJT). Transistors are of two types
n−p - n and p−n− p. The central block thin and lightly doped is called 'Base' while
the other electrodes are emitter and collectors.
18. The emitter-base junction is forward biased while collector base junction is reversed
biased.
19. The transistor can be in three configurations, common emitter (CE), common
collector (CC) and common base (CB).
20. The plot between I C and V CE for fixed I B is called output characteristics while the
plot between I C and I B with V BE fixed is known as input characteristics.

21. Transistor parameters in CE configuration are Input resistance, r i= | |

Δ V BE
Δ I B V = constant
CE

| |
Output resistance, r O =
Δ V CE
Δ IC I B = constant
 Δ lC
Current amplification factor, β= v =¿ constant
Δ I B CE
22. A transistor can be used as an amplifier. The voltage gain of CE configuration is
VO RC
AV = =β
Vi RB
where, RC and R B are respectively resistances in collector and base sides of the
circuit.

23. In common base configuration, AV current gain is α = | |

Δ IC
Δ I E V = constant
CB

24. Transistor can be used as an oscillator as well as a switch (is cut-off or saturation
state).
25. A logic gate is a digital electronic circuit which follows a logical relationship between
its input and output. A logic gate may have one or more inputs but has only one
output. Logic gates follow Boolean algebra, which consists of three basic operations,
namely ¿ ⁡(A ⋅B=Y ), ¿( A +B=Y ) and NOT ⁡( Á=Y ).
(i) OR gate Boolean expression for OR gate is given by Y = A +B

(ii) AND gate

Boolean expression for AND gate is given by Y = A ⋅ B

(iii ) NOT gate

Boolean expression for NOT gate is given by Y = Á Logic symbol

(iv) NAND gate

NAND gate is the combination of AND gate and NOT gate. Boolean expression for
NAND gate is given by

(v) NOR gate

Boolean expression for NOR gate is given by
Y = A +B
 The NOR gate is the combination of OR and NOT gate given by

(vi) XOR gate

Boolean expression for output/input of XOR gate is Y = A ⊕ B= Á ⋅B+ A ⋅ B́
Logic symbol

[Objective Questions Based on NCERT Text

Topic 1 : Classification of Metals, Conductors and Semiconductors
1. For the flow of electrons in a vacuum tube, vacuum is required, because
(a) electrons are not ejected from cathode
(b) vacuum helps in extracting electrons from remaining gas molecules or atoms
(c) in vacuum work function of cathode is reduced
(d) electrons may lose their energy on collision with air molecules in their path
2. Semiconductor devices (diodes, transistors) are smaller than vacuum tubes because
(a) they are made from silicon /germanium crystals
(b) they have very high density
(c) large crystals of semiconductors have large resistance
(d) flow of charge carriers are within the solid itself
3. If a solid transmits the visible light and has a low melting point, it possesses
(a) metallic bonding
(b) ionic bonding
(c) covalent bonding
(d) van der Walls bonding
4. Bonding in a semiconductor is
(a) metallic
(b) ionic
(c) van der Walls
(d) covalent
5. The SI unit of conductivity is
(a) ¿
(b) Ω m−1
(c) Sm−1
(d) S
6. Correct one is
(a) σ semiconductor > σ insulator >σ metal
(b) σ metal > σ semiconductor >σ insulator
(c) σ semiconductor > σ metal >σ insulator
(d) σ insulator >σ semiconductor >σ metal
(here, σ represents conductivity.)
7. In a crystal, atomic separation is around 2 to 3 A . At this separation due to
interatomic interaction, energies of
(a) outermost electrons are changed
(b) innermost electrons are changed
(c) Both (a) and (b)
(d) None of the above
8. Following diagram shows energy band positions in a semiconductor at 0 K .

9. The splitting of 1 s and 2 s atomic energy levels when many atoms come together
to form a solid is best represented by

10. Forbidden energy gap in a semiconductor is nearly equal to

(a) 1 eV
(b) 6 eV
(c) 0 eV
(d) 3 eV
11. Which of these is a true graph showing a relation between resistivity ρ and
temperature for semiconductor?
 )

12. There is no hole current in good conductors, because they

(a) have large forbidden energy gap
(b) have no energy gap due to overlapping valence and conduction bands
(c) are full of electron gas
(d) have no valence band
13. A solid having upper most energy band partially filled with electrons is called
(a) insulator
(b) semiconductor
(c) conductor
(d) None of these
14. If the energy of a photon of sodium light (λ=580 nm) equals the band gap of
semiconductor, the minimum energy required to create hole electron pair.
(a) 1.5 eV
(b) 3.2 eV
(c) 2.1 eV
(d) 4.1 eV
Topic 2 : Intrinsic and Extrinsic Semiconductors
15. At elevated temperature, few of covalent bonds of Si or Ge are broken and a
vacancy in the bond is created. Effective charge of vacancy or hole is
(a) positive
(b) negative
(c) neutral
(d) sometimes positive and sometimes negative
16. In pure form, ¿ or Si, a semiconductor is called
(a) intrinsic semiconductor, n e=nh=n i
(b) extrinsic semiconductor, n e=nh=n i
(c) doped semiconductor
(d) None of the above
(Here, n e=¿ number of free electrons, n h=¿ number of free holes, ni =¿ intrinsic carrier
concentration)
17. If I is total current through an intrinsic semiconductor and I e is electron current and
I h is hole current, then
I
(a) I e = h
I
Ie
(b) =I
Ih
(c) I e −I h =I
(d) I e + I h=I
18. Energy band gap diagram for an intrinsic semiconductor at temperature T > 0 K is
(a)

19. In equilibrium condition, the rate of generation of electron-hole pairs

(a) is more than rate of recombination of electron and hole pairs
(b) is less than rate of recombination of electron and hole pairs
(c) equals to rate of recombination of electron and hole pairs
(d) is always zero
20. In intrinsic semiconductor at room temperature, number of electrons and holes are
(a) equal
(b) zero
(c) unequal
(d) infinite
21. A pure semiconductor behaves as a good conductor at
(a) room temperature
(b) low temperature
(c) high temperature
(d) Both (b) and (c)
22. At absolute zero, Si acts as
(a) non-metal
(b) metal
(c) insulator
(d) None of these
23. Si and Cu are cooled to a temperature of 300 K , then resistivity
(a) for Si increases and for Cu decreases
(b) for Cu increases and for Si decreases
 (c) decreases for both Si and Cu
(d) increases for both Si and Cu
24. The energy gap for silicon is 1.14 eV and for zinc sulphide it is 3.6 eV .
From the above data, we conclude that
(a) silicon is transparent and zinc sulphide is opaque
(b) silicon is opaque and zinc sulphide is transparent
(c) both ZnS and Si are transparent
(d) both ZnS and Si are opaque
25. Doping is
(a) a process of adding an impurity to a pure semiconductor
(b) a process of obtaining semiconductor from its ore
(c) melting of a semiconductor
(d) purification of a semiconductor
26. Doping of intrinsic semiconductor is done
(a) to neutralise charge carriers
(b) to increase the concentration of majority charge carriers
(c) to make it neutral before disposal
(d) to carry out further purification
27. An n -type and p-type silicon can be obtained by doping pure silicon with
(a) arsenic and phosphorous, respectively
(b) indium and aluminium, respectively
(c) phosphorous and indium, respectively
(d) aluminium and boron, respectively
28. Which of the following statement is correct for an n -type semiconductor?
(a) The donor level lies below the bottom of the conduction band
(b) The donor level lies closely above the top of the valence band
(c) The donor level lies at the halfway mark of the forbidden energy gap
(d) None of the above
29. Number of electrons present in conduction band due to doping
(a) shows a heavy increase with increase of temperature
(b) shows a heavy decrease with increase of temperature
(c) independent of change in ambient temperature
(d) reduces to zero at temperature above room temperature
30. To make a p-type semiconductor, germanium is doped with
(a) gallium
(b) boron
(c) aluminium
(d) All of these
31. In a p-type semiconductor, the majority and minority charge carriers are
respectively,
(a) protons and electrons
(b) electrons and protons
(c) electrons and holes
(d) holes and electrons
32. Which statement is correct?
(a) n-type germanium is negatively charged and p-type germanium is positively
charged
(b) both n -type and p-type germanium are neutrals
 (c) n -type germanium is positively charged and p-type germanium is negatively
charged
(d) both n -type and p-type germanium are negatively charged
33. If n e is number density of electrons in conduction band and n h is number density of
holes in valence band, then for an extrinsic semiconductor at room temperature, ¿
number density of intrinsic pairs ¿
ne 2
(a) =ni
nh
nh 2
(b) =ni
ne
(c) n e nh=n2i
(d) n e +n h=n2i
34. Carbon is more resistive than germanium and silicon. Then, order of energy gap is
(a) C> ¿> Si
(b) C> Si>¿
(c) Si> ¿>C
(d) C=Si=¿
35. Let a pure Si crystal has 5 ×1028 atoms m−3. It is doped by parts per million
concentration of pentavalent arsenic. If number of intrinsic pairs is 1.5 ×1016 m−3 , then
number of holes in doped crystal is
(a) 4.5 × 109 m−3
(b) ∼ 1016 m−3
(c) 2.25 ×1032 m−3
(d) 5 ×1022 m−3
36. If n e and n h are the number of electrons and holes in a semiconductor heavily doped
with phosphorous, then
(a) n e ≫ nh
(b) n e ≪nh
(c) n e ≤ nh
(d) n e=nh
37. n e and v d be the number of electrons and drift velocity in a semiconductor. When the
temperature is increased, then
(a) n e increases and v d decreases
(b) n e decreases and v d increases
(c) both n e and v d increase
(d) both n e and v d decrease
38. For extrinsic semiconductor,
(a) the conduction band and valence band overlap
(b) the gap between conduction band and valence band is more than 16 eV
(c) the gap between conduction band and valence band is near about 1 eV
(d) the gap between conduction band and valence band will be 100 eV and more
39. Three semiconductors are arranged in the increasing order of their energy gap as
follows. The correct arrangement is
(a) tin, germanium, silicon
(b) tin, silicon, germanium
(c) silicon, germanium, tin
(d) silicon, tin, germanium
40. When the electrical conductivity of semiconductor is due to the breaking of its
covalent bonds, then the semiconductor is said to be
(a) donor
(b) acceptor
(c) intrinsic
(d) extrinsic
41. The forbidden energy gap in the energy bands of germanium at room temperature
is about
(a) 1.1 eV
(b) 0.1 eV
(c) 0.67 eV
(d) 6.7 eV
42. A Ge specimen is doped with Al. The concentration of acceptor atom is ∼ 1021 atoms
per m3. Given that the intrinsic concentration of electron hole pairs is ∼ 1019 per m3,
the concentration of electrons in the specimen is
(a) 1017 per m3
(b) 1015 per m3
(c) 104 per m3
(d) 102 per m3
43. Which of the following has negative temperature coefficient of resistance?
(a) Metal
(b) Insulator
(c) Semiconductor
(d) All of these
44. The relation between the number of free electrons in semiconductors (n) and its
temperature (T ) is
(a) n ∝T 2
(b) n ∝T
(c) n ∝ √ T
(d) n ∝T 3 /2
45. In extrinsic p and n -type, semiconductor materials, the ratio of the impurity atoms
to the pure semiconductor atoms is about
(a) 1
(b) 10−1
(c) 10−4
(d) 10−7
46. The energy band diagrams for three semiconductor samples of silicon are as shown.
We can then assert that

(a) sample X is undoped while samples Y and Z have been doped with a third group
and a fifth group impurity, respectively
(b) sample X is undoped while both samples Y and Z have been doped with a fifth
 group impurity
(c) sample X has been doped with equal amounts of third and fifth group impurities
while samples Y and Z are undoped
(d) sample X is undoped while samples Y and Z have been doped with a fifth and a
third group impurity, respectively
47. In an n-type semiconductor, which of the following statement is true?
[NEET 2013]
(a) Electrons are majority charge carriers and trivalent atoms are the dopants
(b) Electrons are minority charge carriers and pentavalent atoms are the dopants
(c) Holes are minority charge carriers and pentavalent atoms are the dopants
(d) Holes are majority charge carriers and trivalent atoms are the dopants
48. In an n-type silicon, which of the following statements is correct?
(a) Electrons are majority charge carriers and trivalent atoms are the dopants
(b) Electrons are minority charge carriers and pentavalent atoms are the dopants
(c) Holes are minority charge carriers and pentavalent atoms are the dopants
(d) Holes are majority charge carriers and trivalent atoms are the dopants
49. The number of silicon atoms per m3 is 5 ×1028 . This is doped simultaneously with
5 ×10 atoms per m of arsenic and 5 ×10 per m atoms of indium. Given that
22 3 20 3

16 −3
ni =1.5× 10 m .
Number of electrons and holes (in per metre cube of sample) are respectively,
(a) 4.95 × 1022 , 4.54 ×109
(b) 4.54 ×109 , 4.54 × 109
(c) 4.54 ×109 , 4.95 ×10 22
(d) 4.95 × 1022 , 4.95 × 1022
50. A silicon specimen is made into a p-type semiconductor by doping on an average,
one indium atom per 5 ×107 silicon atoms. If the number density of atoms in the
silicon specimen is 5 ×1028 atoms ¿ m3, then the number of acceptor atoms in silicon
per cubic centimetre will be
(a) 2.5 ×1030 atoms per cm3
(b) 1.0 ×1013 atoms per cm3
(c) 1.0 ×1015 atoms per cm3
(d) 2.5 ×1036 atoms per cm3
51. The number of density of electrons and holes in pure silicon at 27∘ C are equal and
its value is 2.0 ×1016 m−3 on doping with indium the hole density increases to
−3
4.5 × 10 m , the electron density in doped silicon is
22

(a) 10 ×10 9 m−3

(b) 8.89 ×10 9 m−3
(c) 11×10 9 m−3
(d) 16.78 ×10 9 m−3
Topic 3
p−n Junction and Semiconductor Diode
52. A p - n junction contains
(a) a p-type semiconductor is joined with an n-type semiconductor by glue
(b) a p-type semiconductor is bolted with an n -type semiconductor
(c) a p-type semiconductor is kept in touch with an n-type semiconductor
(d) a p-type semiconductor is formed with an n-type semiconductor on same
semiconductor crystal wafer
53. Due to diffusion, the space charge region on either side of p - n junction is
developed. This space charge region is called
(a) dilution region
(c) depletion region
(b) diffusion region
(d) ionic region
54. Thickness of depletion region is of order of
(a) ∼ 10−7 m
(b) ∼ 10−10 m
(c) ∼ 10−9 m
(d) ∼ 10−3 m
55. Which of these graphs shows potential difference between p-side and n -side of a p -
n junction in equilibrium?

56. Potential difference of p and n -side which prevents diffusion of electrons is called
(a) potential gradient
(b) potential difference
(c) barrier potential
(d) depletion potential
57. Can we take one slab of p-type semiconductor and physically join it to another n -
type semiconductor to get p−n junction?
(a) Yes
(b) No
(c) It depends on the hole and electron concentrations on p and n -side
(d) Only when a p-type semiconductor is soldered with an n -type semiconductor
58. 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
59. 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?
 [CBSE AIPMT 2014]
(a) (i) and (ii)
(c) (ii) and (iii)
(b) (ii)
(d) (i), (ii) and (iii)
60. The electrical resistance of depletion layer is large because
(a) of strong electric field
(b) it has a large number of charge carriers
(c) it contains electrons as charge carriers
(d) it has holes as charge carriers
61. 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 moves across the junction by the potential difference
(c) hole concentration in p-region is more as compared to hole concentration in n -
region
(d) All of the above
62. A Si based p−n junction has a depletion layer of thickness 1 μ m and barrier potential
difference of n -side and p-side is 0.6 V .
The electric field in the depletion region is
(a) 0.6 Vm−1
(b) 6 ×10−4 Vm−1
(c) 6 ×10 5 Vm−1
(d) 6 ×10 4 Vm−1
63. A diode is a
(a) piece of a covalent crystal
(b) piece of a semiconductor crystal with metallic contacts provided at two ends
(c) p - n junction with metallic contacts provided at two ends
(d) piece of a metal which is sprayed over by a semiconductor
64. Symbol of a p−n junction diode is an arrow, its direction indicates

(a) nothing its just a symbol

(b) direction of flow of electrons
(c) direction of conventional current when it is forward biased
(d) direction of electric field
65. In the case of forward biasing of p−n junction, which one of the following figures
correctly depicts the direction of flow of charge carriers?
66. If V is applied potential difference in forward bias and V 0 is barrier potential of a p−n
junction, then effective barrier height under forward bias is
(a) V −V 0
(b) V 0−V
(c) V 0 +V
(d) V 0
67. In forward bias, forward current obtained from the p−n junction diode is
(a) due to injection of electrons in p-side
(b) due to injection of holes in n-side
(c) both (a) and (b)
(d) due to flow of electrons from negative terminal of supply to its positive terminal
68. In a p−n junction diode,
(a) the current in the reverse biased condition is generally very small
(b) the current in the reverse biased condition is small but that in forward biased
condition is independent of the bias voltage
(c) the reverse biased current is strongly dependent on the applied bias voltage
(d) the forward biased current is very small in comparison to reverse biased current
69. In a reverse biased p−n junction diode,
(a) current under reversed bias is not very much dependent on applied voltage
(b) current under reversed bias is directly proportional to applied voltage
(c) current initially depends on applied voltage, then it becomes independent
(d) no current flows in reversed bias
70. If reverse biasing potential is increased beyond a certain critical (breakdown) value,
then
(a) diode gets destroyed due to overheating
(b) no current flows through the diode
(c) after breakdown a heavy current flows from p to n -side
(d) potential barrier becomes zero
71. Characteristic curve of a p−n junction is (a)
72. Threshold or knee voltage for a forward biased germanium and silicon diodes have
respective values
(a) 0.2 V , 0.7 V
(b) 0.7 V ,1.1 V
(c) 1.2 V , 0.7 V
(d) 0.7 V ,0.2 V
73. Diode primarily allows the flow of current only in one direction (forward bias). The
forward bias resistance is low as compared to the reverse bias resistance. In a
circuit, a diode acts like a
(a) valve
(b) switch
(c) amplifier
(d) multi-way passage
74. Dynamic resistance of a diode is given by
ΔV
(a) r d =
ΔI
−Δ V
(b) r d =
ΔI
Threshold voltage
(c) r d =
Current
Breakdown voltage
(d) r d =
Current
75. V −I characteristics of a silicon diode is shown.

The ratio of resistance of diode at I D =15 mA and V D=−10 V , is

(a) 10−3
(b) 10−4
 (c) 10−5
(d) 10−6
76. If no external voltage is applied across p−n junction, there would be
(a) no electric field across the junction
(b) an electric field pointing from n -type to p-type side across the junction
(c) an electric field pointing from p-type to n -type side across the junction
(d) a temporary electric field during formation of p−n junction that would
subsequently disappear
77. For the given circuit of p−n junction diode, which of the following statement is
correct?

(a) In forward biasing, the voltage across R is V

(b) In forward biasing, the voltage across R is 2 V
(c) In reverse biasing, the voltage across R is V
(d) In reverse biasing, the voltage across R is 2 V
78. Which is reverse biased diode? (a)

79. In the circuit given below, the value of the current is

(a) 0 A
(b) 10−2 A
(c) 102 A
(d) 10−3 A
80. A semiconductor X is made by doping a germanium crystal with arsenic (Z=33). A
second semiconductor Y is made by doping germanium with indium (Z=49). The
 two are joined end to end and connected to a battery as shown. Which of the
following statements is correct?
(a) X is p-type, Y is n -type and the junction is forward biased
(b) X is n -type, Y is p-type and the junction is forward biased
(c) X is p-type, Y is n-type and the junction is reverse biased
(d) X is n -type, Y is p-type and the junction is reverse biased
81. The diode shown in the circuit is a silicon diode. The potential difference between
the points A and B will be

(a) 6 V
(c) 0.7 V
(b) 0.6 V
(d) 0 V
82. The current through an ideal p−n junction shown in the following circuit diagram will
be
(a) zero
(c) 10 mA
(b) 1 mA
(d) 30 mA

83. A potential barrier of 0.3 V exists across a p−n junction. If the depletion region is
1 μ m wide, what is the intensity of electric field in this region?
(a) 2 ×105 Vm−1
(b) 3 ×105 Vm−1
(c) 4 ×105 Vm−1
(d) 5 ×105 Vm−1
84. When the voltage drop across a p−n junction diode is increased from 0.65 V to
0.70 V , the change in the diode current is 5 mA . The dynamic resistance of diode is
(a) 5 Ω
(b) 10 Ω
(c) 20 Ω
(d) 25 Ω
85. The circuit shown in the figure contains two diodes each with a forward resistance
of 30 Ω and with infinite backward resistance. If the battery is 3 V , the current
through the 50 Ω resistance (in
 ampere) is
(a) 0
(b) 0.01
(c) 0.02
(d) 0.03
Topic 4
Application of Junction Diode as a Rectifier and Special Purpose p−n Junction
Diodes
86. If an alternating voltage is applied across a diode in series with a load, then
(a) a continuous DC voltage appears across load
(b) an AC voltage appears across load
(c) a pulsating DC voltage appears across load
(d) no voltage appears across load
87. Which of this is a half-wave rectifier circuit?

88. For any practical half-wave rectifier circuit,

(a) the reverse breakdown voltage of diode must be greater than peak AC voltage
(b) the reverse breakdown voltage of diode must be greater than rms AC voltage
(c) the reverse breakdown voltage must be greater than mean AC voltage
(d) the reverse breakdown voltage must be smaller than the rms AC voltage
89. Input to an half-wave rectifier is given as follows

Its output will be

90. In the course of rectification of the AC cycle when the voltage at A (upper diode
input) becomes negative with respect to centre tap, the voltage at B (lower diode
input) would be positive. This implies voltage drop between A and centre tap is half.
If a centre tap transformer is used with 2 diodes for full-wave rectification, then
output voltage of rectifier is

(a) 2 × secondary voltage of transformer

(b) 2/3 × secondary voltage of transformer
(c) 1/2 × secondary voltage of transformer
(d) 3/2 × secondary voltage of transformer
91. If two diodes are connected across two ends of secondary windings of a centre tap
transformer as shown in figure. If inputs at A and B are as shown

Then, output across load resistance will be (a)

92. Output of a full-wave rectifier is
(a) pure DC voltage
(b) pure AC voltage
(c) pulsating DC voltage
(d) pulsating AC voltage
93. Filters are used alongwith a full-wave rectifier to
(a) remove AC part from the output
(b) remove DC part from the output
(c) mix AC and DC
(d) None of the above
94. In the given circuit,

Capacitor C is used
(a) for storing potential energy
(b) as a bypass to DC component to get AC in R L
(c) to remove sparking
(d) as a bypass to AC component to get DC in R L
95. In a full-wave rectifier, input AC current has a frequency (v). The output frequency
of current is
(a) v /2
(b) v
(c) 2 v
(d) None of these
96. In comparison to a half-wave rectifier, the full-wave rectifier done by centre tapping
gives lower
(a) efficiency
(b) average
(c) output voltage
(d) None of these
97. In the figure alongside, 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 input
(c) full-wave rectified
(d) half-wave rectified

98. In case a single capacitor is connected in parallel with a load resistance of R L, it gets
discharged through the load. The rate of fall of voltage across the capacitor is
proportional to
(a) R L C
 C
(b)
RL
1
(c)
RL C
R
(d) L
C
99. What is the ratio of output frequencies of full-wave rectifier and a half-wave
rectifier, when an input of frequency 50 Hz is fed at input?
(a) 1 :2
(b) 2 :1
(c) 4 :1
(d) 1 :4
100. A zener diode which is used in reversed biased is used as a
(a) voltage regulator
(c) current regulator
(b) voltage rectifier
(d) current rectifier
101. A zener diode differs from a p - n junction that
(a) zener diode is made from very lightly doped p - n junction
(b) zener diode is made from a heavily doped p−n junction
(c) zener diode is made from a metal piece
(d) zener diode is made from a heavily doped p-type semiconductor
102. High current observed at breakdown of a zener diode due to emission and
movement of electrons from p to n-side is known as
(a) thermionic emission
(c) internal field emission
(b) external field emission
(d) photoemission
103. Correct circuit using a zener diode as a voltage regulator is
104. For a zener regulated power supply, a zener diode with zener voltage V z=6.0 V is
used for regulation. The load current is to be 4.0 mA and the unregulated input
10.0 V . The value of series resistor R s must be, it I Z / I 2=5

(a) 167 Ω
(b) 120 Ω
(c) 250 Ω
(d) 20 Ω
105. Optoelectronic devices are
(a) CFL's
(b) light based semiconductor diodes
(c) bulbs
(d) discharge tubes
106. A photodetector is a
(a) photodiode used for detecting optical signals
(b) LED's which are used for detection of infrared signals
(c) an evacuated tube consisting of a photosensitive cathode
(d) None of the above
107. A photodiode converts
(a) variation in intensity of light into current amplitude variation
(b) variation of current amplitude into variation in intensity of emitted light
(c) variation of voltage into variation of current
(d) variation of intensity of light into variation of volume
108. A photodiode in reverse biased is irradiated with light of suitable frequency and
current in circuit is measured.
 Characteristics of diode for different illumination intensities I 1 , I 2 , I 3 and I 4 are drawn
as follows.

Greatest intensity is
(a) I 1
(b) I 2
(c) I 3
(d) I 4
109. When LED is forward biased, then electrons move from n to p and electron-hole
combination occurs near junction plane. If E g is energy gap between conduction
band and valence band, then released energy (E) due to electron-hole combination
will be
(a) E=Eg
(b) E> E g
(c) E ≤ Eg
(d) E ≥ Eg
110. An LED cannot be used in reverse biased as a voltage regulator because
(a) reverse breakdown voltage is very low for them
(b) reverse breakdown voltage is very high for them
(c) they do not breakdown for any voltage
(d) None of the above
111. Semiconductors used to fabricate LED to produce visible light must have energy
gap E g such that
(a) 1.1 eV < E g
(b) E g >3 eV
(c) 1.8 eV < E g <3 eV
(d) 1.1 eV < E g <2.8 eV
112. Substance used to make red LEDs is
(a) silicon
(b) germanium
(c) gallium arsenide phosphide
(d) indium phosphide
113. A solar cell is
(a) photodetector
(b) photovoltaic device
(c) light emitting diode
(d) photogenerator
114. I −V characteristics of a solar cell is best represented by (a)
115. To fabricate solar cell, material used have an energy gap of
(a) around 0.7 eV
(b) less than 1 eV
(c) around 1.5 eV
(d) less than 0.7 eV
116. A p−n junction photodiode is fabricated from a semiconductor with a band gap of
2.8 eV . It can detect a wavelength nearing to
(a) 5200 Å
(b) 4400 Å
(c) 6200 Å
(d) 7500 Å
117. For a photodiode, the conductivity increases when a wavelength less than 620 nm is
incident on it. The band gap of crystal used to fabricate the diode is
(a) 1.12 eV
(b) 1.8 eV
(c) 2.0 eV
(d) 1.62 eV
118. In an LED, when it glows, electron moves from A to B, when an appropriate bias is
applied. A and B are respectively,
(a) conduction band, valence band
(b) valence band, conduction band
(c) conduction band, connecting wires
(d) connecting wires, conduction band
119. Photodetectors and LED's are used in
(a) road construction works
(b) optical telecommunication links
(c) power generation from falling water near dam
(d) radio transmitters
120. Two different semiconductors A and B are used to make 'red' and 'violet' LED's,
respectively. Then, ratio of energy gaps of semiconductors must be
EA
(a) >1
EB
EA
(b) <1
EB
(c) E A =E B
(d) E A >3 eV and E B <1.5 eV
121. LED's are not used for room lighting (although they are used for automobile bulbs
and in industrial lighting) because
(a) our eyes are not comfortable with very intense light
(b) our eyes are not comfortable with monochromatic light
 (c) LED's are much costlier than bulbs tubelights and CFL's
(d) LED manufacture in mass production will be a very polluting process 122. The
I −V characteristics of an LED is
[JEE Main 2013]

123. A p−n photodiode is made of a material with a band gap of 2 eV . The minimum
frequency of the radiation that can be absorbed by the material is nearly (Take
hc=1240 eV-nm)
Topic 5
Junction Transistor
125. A transistor has
(a) two doped regions forming a large p−n junction
(b) three doped regions forming two p - n junctions
(c) two p−n junctions connected by a conducting wire
(d) None of the above
126. For an n−p−n transistor shown below,

Regions marked I, II and III are respectively,

(a) emitter,collector, base
(b) base, collector, emitter
(c) emitter, base, collector
(d) collector, emitter, base
127. When a transistor is biased as follows.

Then, it is said to be in
(a) solid state
(b) active state
 (c) inactive state
(d) passive state
(a) 1 ×1014 Hz
(b) 20 ×1014 Hz
(c) 10 ×1014 Hz
(d) 5 ×1014 Hz
124. Identify the semiconductor devices whose characteristics are as given below in the
order (i), (ii), (iii), (iv). [JEE Main 2016]

128. Let ' ∙ ' shows an electron and ' O ' shows a hole, then which of the following shows
correct direction of motion of charge carriers?

(a) A and B
(b) B and C
(c) A and C
(d) B and D
129. In active state of a transistor, the emitter base junction acts as a ...A... resistance
and base-collector junction acts like a ...... resistance. Here, A and B refer to
(a) low, low
(b) low, high
(c) high, low
(d) high, high
130. 130. Correct circuit to study input-output characteristics of an n−p−n transistor in
CE configuration is
 (a) (b)

(c) (d)
131. For a transistor, which is correct?
(a) V CE =V CB + V BE
(c) V CB =V CE + V BE
(b) V BE=V CB +V CE
(d) V CE =V CB −V BE
132. For a silicon base transistor, V CE must be sufficiently larger than
(a) 21 V
(b) 0.7 V
(c) 0.1 V
(d) 20 V
133. In an n−p−n transistor in CE configuration, when V CE is increased, then
(a) I B increases and I C increases proportionally
(b) I B increases and I C remains constant
(c) effect on I B is negligible but I C increases
(d) Both I B and I C remain nearly constant
134. In a common-emitter (CE) amplifier having a voltage gain G , the transistor used has
transconductance 0.03 mho and current gain 25 . If the above transistor is replaced
with another one with transconductance 0.02 mho and current gain 20 , the voltage
gain will be
[NEET 2013]
2
(a) G
3
(b) 1.5 G
1
(c) G
3
5
(d) G
4
135. Input resistance ( r i ) of a transistor in CE configuration is
Δ V BE
(a) V
Δ I B CE
Δ V CE
(b)
Δ IB V
BE

Δ V BB
(c) V
Δ I B BE
Δ V BC
(d)
ΔIB V
CE
136. An n−p−n transistor is connected in common-emitter configuration in a given
amplifier. A load resistance of 800 Ω 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 circuits is 192 Ω, the voltage gain and the power gain of the
amplifier will respectively be [NEET 2016]
(a) 3.69 , 3.84
(b) 4,4
(c) 4, 3.69
(d) 4 , 3.84
137. If β DC for a transistor is
Δ IC
(a)
Δ IB
Δ IB
(b)
Δ IC
IC
(c)
IB
IB
(d)
IC
138.

The output characteristics of a typical n−p−n transistor in CE configuration are

shown. When V CE =10 V and I C =4.0 mA , then ratio of β AC and β DC is
(a) 1
(b) 2
(c) 3
(d) 4
139.

In above transfer characteristics of an n−p−n transistor in CE configuration; cut-off

region, active region, saturation region respectively, are
(a) II, III and I
(b) III, I and II
(c) III, II and I
(d) I, II and III
140. For an n−p−n transistor in CE configuration, correct graph showing variation of
output
voltage with variation of input voltage is (a)

141. For an n−p−n transistor used as amplifier, the power gain A P is given by ¿ voltage
gain ¿
2
(a) A P=( β AC ) × AV
1
(b) A P= A
β AC V
(c) A P=β AC × A V
1
(d) A P= 2
AV
( β AC )
142. For a CE transistor amplifier, the audio signal voltage across collector resistance of
2.0 kΩ is 2.0 V . Suppose the current amplification factor of the transistor is 100 .
What should be the value of R B in series with V BB supply of 2.0 V , if DC base current
has to be 10 times the signal current? ( V BE=0.6 V )
(a) 14 kΩ
(b) 24 kΩ
(c) 34 kΩ
(d) 44 kΩ
143. In an n−p−n transistor, the collector current is 24 mA . The possible emitter current
(in mA ) is

(a) 36
(b) 20
(c) 16
(d) 6
144. For tuned collector oscillator, using an n−p−n transistor, from rise and fall (or built
up) of I C , I E current graphs. It can be concluded
 (a) (b)

(c)
(a) both I C , I E increase initially
(b) both I C , I E decrease but I E decreases
(c) initially I C increases but I E decreases
(d) initially I C decreases but I E increases
145. Refer figure of Q. 144, after maximum collector current, there is no further change
in collector current, the magnetic field around T 2 ceases to grow. As soon as the
field becomes static, there will be no further feedback from T 2 to T 1. Without
continued feedback, the ...A... current begins to fall. Consequently, collector current
decreases from Y to Z . However, a decrease of collector current causes the
magnetic field to decay around the coil T 2. Thus, T 1 is now seeing a ...B... field in T 2
(opposite form what it saw when field was growing at the initial start position).
Here, A and B refer to
(a) emitter, rising
(b) emitter, decaying
(c) collector, rising
(d) collector, decaying
146. In a common emitter transistor, the current gain is 80 . If change in base current is
250 μ A , then change in collector current will be
(a) (80 × 250) μ A
(c) (250+ 80) μ A
(b) (250−80)μ A
(d) (250 /80) μ A
147. In case of an n−p−n transistor, the collector current is always less than the emitter

current because
(a) collector side is reverse biased and emitter side is forward biased

(b) after electrons are lost in the base and only remaining ones reach the emitter
back
(c) collector side is forward biased and emitter side is reverse biased
(d) collector being reverse biased attracts less electrons
148. In a transistor circuit shown here, the base current is 35 μ A .
The value of resistance Rb is
(a) 123.5 kΩ
(b) 257 kΩ
(c) 380.5 kΩ
(d) cannot be found from given data
149. For the given circuit,

Frequency of oscillation is

(a) f =
1
2π √ 1
( L 1 + L2 ) C
1
(b) f =
2 π √ ( L1−L2 ) C
1
(c) f =
2 π √ L1 L2 ⋅C
1
f=
(d)
2π
√L1 + L2
2
C

150. For the given circuit, if current amplification factor β=90 and V BE=0.7 V

Then, base resistance R B is

(a) 180 kΩ
(b) 185 kΩ
(c) 82 kΩ
(d) 190 kΩ
151. A circuit containing transistor is such that I B=10 μ A and I C =5 mA ,
(a) transistor can be used as amplifier with β DC =10
(b) transistor can be used as amplifier with β DC =100
(c) transistor can be used as amplifier with β DC =250
(d) transistor cannot be used as amplifier
152. For a common-emitter n−p−n transistor following is a true relationship between I B
and I C. In active region.

153. The current gain for a transistor working as common base amplifier is 0.96 . If the
emitter current is 7.2 mA , then the base current is
(a) 0.29 mA
(b) 0.35 mA
(c) 0.39 mA
(d) 0.43 mA
154. The power gain for common base amplifier is 800 and the voltage amplification
factor is 840 . The collector current when base current is 1.2 mA , is
(a) 24 mA
(b) 12 mA
(c) 6 mA
(d) 3 mA
155. The input signal given to a CE amplifier having a voltage gain of 150 is
π
V i=2 cos ⁡15t + . The corresponding output signal will be [CBSE AIPMT 2015]
3
π
(a) 300 cos ⁡15 t +
3
2π
(b) 75 cos ⁡15 t +
3
5π
(c) 2 cos ⁡15 t +
3
4π
(d) 300 cos ⁡15 t +
3
156. A transistor has a current gain of 30 . If the collector resistance is 6 kΩ , input
resistance is 1 kΩ, its voltage gain is
(a) 90
(b) 180
(c) 45
(d) 360
157. The input resistance of a transistor is 1000 Ω on charging its base current by 10 μ A ,
the collector current increases by 2 mA . If a load resistance of 5 kΩ is used in the
circuit, the voltage gain of the amplifier is
(a) 100
(b) 500
(c) 1000
(d) 1500
158. In an n−p - n circuit transistor, the collector current is 10 mA . If 80 % electrons
emitted reach the collector, then
(a) the emitter current will be 7.5 mA
(b) the base current will be 2.5 mA
(c) the base current will be 3.5 mA
(d) the emitter current will be 15 mA
159. When the voltage drop across a p−n junction diode is increased from 0.65 V to
0.70 V , then change in the diode current is 5 mA . The dynamic resistance of the
diode is
(a) 20 Ω
(b) 50 Ω
(c) 10 Ω
(d) 80 Ω
Topic 6 : Digital Electronics and Logic Gates
160. Analog signals are
(a) continuous waveforms
(b) discrete value signals
(c) intermittent signals
(d) erratic waveforms
161. Digital signals are
(a) continuous waveforms
(b) discrete value signals
(c) intermittent signals
(d) erratic waveforms
162. A NOT gate is called an invertor, because
(a) it produces an output which changes with time
(b) it produces 1 as output when input is 0 and vice-versa
(c) it produces no output for any input
(d) it has only a single input
163. Truth table for an OR gate is
164. Truth table for the logic gate whose symbol shown is

165. Inputs to a NAND gate are A and B are made as below.

Output of the NAND gate is

166. Consider the junction diode as ideal. The value of current flowing through AB is
[NEET 2016]

(a) 10−2 A
(b) 10−1 A
(c) 10−3 A
(d) 0 A
167. Inputs A and B are given to show combination of gates

Then, output C is

168. The output (X ) of the logic circuit shown in figure will be

[NEET 2013]
(a) X = Á ⋅ B́
(b) X =A ⋅B
(c) X =A ⋅B
(d) X =A + B
169. See the circuit shown in the figure. Name the gate that the given circuit resembles.
 (a) NAND
(b) AND
(c) OR
(d) NOR
170. In the given figure, 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
(b) 40 mA
(c) 20 mA
(d) 35 mA
171. To get output 1 for the following circuit, the correct choice for the input is
[NEET 2016]

(a) A=1 , B=0 ,C=0

(b) A=1 , B=1 , C=0
(c) A=1 , B=0 ,C=1
(d) A=0 , B=1 ,C=0
Special Format Questions
Assertion and Reason
Directions (Q. Nos. 172-176) In the following questions, a statement of assertion is
followed by a corresponding statement of reason. Of the following statements,
choose the correct one.
(a) Both Assertion and Reason are correct and Reason is the correct explanation of
Assertion.
(b) Both Assertion and Reason are correct but Reason is not the correct explanation
of Assertion.
(c) Assertion is correct but Reason is incorrect.
(d) Assertion is incorrect but Reason is correct.
172. Assertion The conductivity of an intrinsic semiconductor depends on its
temperature.
 Reason The conductivity of an intrinsic semiconductor is slightly higher than that of
a lightly doped p-type semiconductor.
173. Assertion A zener diode is used to obtain voltage regulation.
Reason When zener diode is operated in reverse bias after a certain voltage
(breakdown voltage), the current suddenly increases, but potential difference
across diode remains constant.
174. Assertion In a transistor, the base is made thin. Reason A thin base makes the
transistor stable.
175. Assertion In an oscillator, the feedback is in the same phase which is called as
positive feedback. Reason If the feedback voltage is in opposite phase, the gain is
greater than one.
176.Assertion

This circuit acts as OR gate.

Reason Truth table for two input OR gate is

Statement Based Questions Type I

Directions (Q. Nos. 177-187) In the following questions, a statement I is
followed by a corresponding statement II. Of the following statements,
choose the correct one.
(a) Both Statement I and Statement II are correct and Statement II is the correct
explanation of Statement I.
(b) Both Statement I and Statement II are correct but Statement II is not the correct
explanation of Statement I.
(c) Statement I is correct but Statement II is incorrect.
(d) Statement I is incorrect but Statement II is correct.
177. Statement I An electron on p-side of a p - n junction moves to n-side just an instant
after drifting of charge carriers occurs across junction plane.
Statement II Drifting of charge carriers reduces the concentration gradient across
junction plane.
178. Statement I In equilibrium condition, p-side of a p−n junction is at positive potential.
Statement II A p-type semiconductor contains more holes than electrons.
179. Statement I The applied voltage (in forward bias of a p−n junction) mostly drops
across the depletion region and the voltage drop across the p-side and n -side of the
junction is negligible.
Statement II Resistance of depletion region is large compared to resistance of n or p
-side.
180. Statement I In forward bias, as voltage increases beyond threshold voltage, the
forward current increases significantly.
Statement II By Ohm's law states V ∝ I .
181. Statement I A diode can be used to rectify alternating voltages.
Statement II A p - n junction allows current to pass only when it is in reverse bias.
182. Statement I To filter out AC ripple from a given pulsating DC voltage, an inductor is
connected in series and a capacitor is connected in parallel with load resistance.
Statement II For AC inductor has high reactance and a capacitor has a low
reactance when frequency is high.
183. Statement I A solar cell is made in wafer's shape (of large area).
Statement II By increasing area, work-function of electron is decreased.
184. Statement I When a light based p - n junction is radiated with light of frequency v
such that hv > E g, it produces an emf with p side becoming more negative than n -
side.
Statement II Junction field separates electron-hole pairs generated due to photon
absorption and sweep them to different regions.
185. Statement I In a typical transistor, I E =I C + I B ⇒ I E ≈ I C .
Statement II Base region of a transistor is very thin and lightly doped with I B current
small, collector current I C is large.
186. Statement I When an n−p−n transistor in CE configuration is being used as a switch,
it is operated in cut-off region or in saturation region.
Statement II In cut-off region, here V i is low but V o is high. In saturation region, here
V i is high but V o is low.
187. Statement I A logic gate is a digital circuit.
Statement II They are called gates, because they do not allow current through
them.
Statement Based Questions Type II
188. Semiconductor devices have the advantage over vacuum tubes of
I. small size.
II. long life and reliable.
III. low power use.
IV. low cost.
Advantages of semiconductor devices are
(a) I, II, III and IV
(b) II, III and IV
(c) I, III and IV
(d) I and IV
189. Consider four statements.
I. Inside crystal, because of difference of position, each electron has different
energy; this makes energy bands in crystals.
II. Energy levels of valence electrons are included in valence band.
III. Energy level above the valence band is conduction band.
IV. In metals, conduction band and valence band overlap.
Correct statements are
(a) I and II
(b) I, II and IV
(c) II and III
(d) I, II, III and IV
190. Due to diffusion of electrons from n to p-side,
I. electron-hole combination across p - n junction occurs.
II. an ionised acceptor is left in the p-region.
III. an ionised donor is left in the n -region.
 IV. electrons of n -side comes to p-side and electron-hole combination takes place in
p-side.
Correct options are
(a) I and II
(b) II and III
(c) II and IV
(d) II, III and IV
191. Which of the given statements are correct regarding unbiased p−n junction?
I. Drift and diffusion currents occur p to n -side.
II. Initially, diffusion current is large and drift current is small.
III. Finally, diffusion and drift currents grow to be equal in magnitude.
IV. Under equilibrium there is no net current across p - n junction plane.
(a) I and IV
(b) I, II and III
(c) II, III and IV
(d) All of these
192. Which of these are correct?
I. In forward biasing, holes from p-side crosses junction and reach n -side.
II. In forward biasing, electrons from n -side crosses junction and reach p-side. III. In n
-side, holes are minority charge carriers.
IV. In p-side, electrons are minority charge carriers.
(a) I, II and III
(b) I, III and IV
(c) II, III and IV
(d) I, II, III and IV
193. Consider the following statements I and II and identify the correct choice of the
given answers.
I. The width of the depletion layer in a p−n junction diode increases in forward
biased.
II. In an intrinsic semiconductor, the fermi energy level is exactly in the middle of
the forbidden energy gap.
(a) I is true and II is false
(b) Both I and II are false
(c) I is false and II is true
(d) Both I and II are true
194. Following of these are circuits used for full-wave rectification,

(a) I, II and III

(b) II, III and IV
(c) I, III and IV
(d) I, II and IV
195. For the circuit given, an n−p−n transistor is being used as a switch in CE
configuration.

Which of the following are correct?

I. V BB=I B R B +V BE
II. V CE =V CC −I C RC
III. V i=I B R B +V BE
IV. V o =V CC−I C RC
(a) I, II and IV
(b) II, III and IV
(c) I, II and III
(d) I, II, III and IV
196. In an oscillator,
I. we get AC output without any external input signal.
II. output is self sustained.
III. feedback can be achieved by inductive coupling (through mutual inductance) or
 L−C or R−C networks. Incorrect statement is
(a) Only I
(b) Only II
(c) Only III
(d) None of these
197. Which of these gates can be formed using a NOR gate?
I. AND
II. OR
III. NOT
IV. NAND
(a) I and II
(b) II and III
(c) I, II and IV
(d) All I, II, III, and IV
Matching Type
198. Before invention of transistor, vacuum tubes were used and these were named
according to number of electrodes they have.
Now, match these with number of electrodes
Column Column
I II

A. Pentode 1. 2

B. Tetrode 2. 3

C. Triode 3. 4

D. Diode 4. 5

A BC D
(a) 1 2 3 4
(b) 2 3 4 1
(c) 3 4 1 2
(d) 4 3 21
199. Match the elements or compounds with their respective energy gaps values.
(Energy gap existing between conduction and valence bands)

A B C D (a) 1 2 3 4
(b) 2 1 4 3
(c) 4 31 2
(d) 4 3 21
200. Following shows a plot of potential difference of n -side and p-side of a p - n junction
(battery, in forward biased)

Then, match the following columns.

Column I Column
II

A. Without battery 1. 1

B. Low potential
2. 2
battery

C. High potential
3. 3
battery
ABC ABC
(a) 1 2 3 (b) 2 1 3
(c) 2 3 1 (d) 1 3 2
201. Match the following columns.
Column
Column I
II

Moderate size and heavily 1

A. Base
doped

2
B. Very thin and lightly doped Collector

Moderately doped and of large 3

C. Emitter
size

ABC
(a) 1 2 3
(b) 1 3 2
(c) 3 1 2
(d) 3 2 1

202. Match the following symbols with their names.

203. Match the inputs of Column I with their respective outputs from Column II for a NOR
gate
Column Column
I II

1
A. 0,0 0

2
B. 0,1 1

C. 1,0

D. 1,1

ABCD ABCD
(a) 1 1 2 2 (b) 1 1 1 2
(c) 2 2 2 1 (d) 2 1 1 1
204. Match the following Column I with Column II.
 ABCD ABCD
(a) 3 4 2 1 (b) 4 2 1 3
(c) 2 4 3 1 (d) 4 3 2 1

Passage Based Questions

Directions (Q. Nos. 205-209) These questions are based on the following
situation. Choose the correct options from those given below.
The input and output resistances in a common-base amplifier circuits are 400 Ω and
400 kΩ, respectively. The emitter current is 2 mA and current gain is 0.98 .
205. The collector current is
(a) 1.84 mA
(b) 1.96 mA
(c) 1.2 mA
(d) 2.04 mA
206. The base current is
(a) 0.012 mA
(b) 0.022 mA
(c) 0.032 mA
(d) 0.042 mA
207. Voltage gain of transistor is
(a) 960
(b) 970
(c) 980
(d) 990
208. Power gain of transistor is
(a) 950
(b) 960
(c) 970
(d) 980
209. If peak voltage of input AC source is 0.1 V . The peak voltage of the output will be
(a) 9.8 V
(b) 98 V
(c) 980 V
(d) 970 V
Directions (Q. Nos. 210-211) These questions are based on the following
situation. Choose the correct options from those given below.
Three ideal p - n junction diodes D1 , D 2 and D3 are connected as shown in the circuit.
The potentials V A and V B can be changed.

210. If V A is kept at −10 V and V B at −5 V , the effective resistance between A and B

becomes
(a) R
(b) R/2
(c) 3 R
(d) 3 R/2
211. If V A =−5 V and V B=−10 V , then the resistance between A and B will be
(a) R
(b) R/2
(c) 3 R
(d) 3 R/2
More than One Option Correct
212. If the lattice constant of this semiconductor is decreased, then which of the
following are incorrect?

(a) All Ec , Eg , Ev increase

(b) Ec and E v increase, but E g decrease
(c) Ec and E v decrease, but E g increase
(d) All Ec , Eg , Ev decrease
213. Choose correct option(s) 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
214. Which of the following statement concerning the depletion zone of an unbiased p−n
junction is (are) true?
(a) The width of the zone is independent on the densities of the dopants (impurities)
(b) The width of the zone is dependent on the densities of the dopants
(c) The electric field in the zone is produced by the ionized dopant atoms
(d) The electric field in the zone is produced by the electrons in the conduction band
and the holes in the valence band
215. The impurity atoms with which pure silicon should be doped to make a p-type
semiconductor, are those of
 (a) phosphorus
(c) antimony
(b) boron
(d) aluminium
216. The diode used in the circuit shown in the figure has a constant voltage drop of
0.5 V at all current and a maximum power rating of 100 milliwatt. The value of
maximum current in the circuit is I when voltage across resistance R is V R and the
value of resistance in R , thus which of the following are correct?

(a) I =200 mA ,V R=1 V

(c) I =100 mA , V R=2 V
(b) I =200 mA , R=5 Ω
(d) I =100 mA , R=10 Ω
217. In an n - p - n transistor circuit, the collector current is 10 mA . If 90 % of the electrons
emitted reach the collector
(a) the emitter current will be 9 mA
(b) the base current will be 1 mA
(c) the emitter current will be 11 mA
(d) the base current will be −1 mA
218. Consider the following circuit. It V BE=0.7 V and β=90 , then which of the following are
correct?

(a) I C =2.5 mA
(c) R B=82 kΩ
(b) I B=27.8 mA
(d) I C =1.2 mA
[NCERT & NCERT Exemplar Questions
NCERT
219. When a forward bias is applied to a p - n junction. It
(a) raises the potential barrier
(b) reduces the majority carrier current to zero
(c) lowers the potential barrier
(d) None of the above
220. For transistor action, which of the following statements are correct?
(a) Base, emitter and collector regions should have similar size and doping
concentrations
(b) The base region must be very thin and lightly doped
(c) The emitter junction is forward biased and collector junction is reverse biased
(d) Both the emitter junction as well as the collector junction are forward biased
221. In an n-type silicon, which of the following statements is true?
(a) Electrons are majority charge carriers and trivalent atoms are the dopants
(b) Electrons are minority charge carriers and pentavalent atoms are the dopants
(c) Holes are minority charge carriers and pentavalent atoms are the dopants
(d) Holes are majority charge carriers and trivalent atoms are the dopants
222. 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 , ( Eg )Si and ( E g )¿. Which of the following statement is true?
(a) ( E g )Si < ( E g )¿ < ( E g )C
(b) ( E g )C < ( E g )¿ > ( E g )Si
(c) ( E g )C > ( E g ) Si > ( E g )¿
(d) ( E g )C =( Eg ) Si =( E g )¿
223. 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 the above
224. Identify the logical operations carried out by the two circuits given respectively are

(a) A-AND, B-NOT

(b) A-AND, B-OR
(c) A-NAND, B-NOT
(d) A-NOT, B-OR
225. For a transistor amplifier, the voltage gain
(a) remains constant for all frequencies
(b) is high at high and low frequencies and constant in the middle frequency range
(c) is low at high and low frequencies and constant at mid frequencies
(d) None of the above
226. For a CE-transistor amplifier, the audio signal voltage across the collector resistance
of 2 kΩ is 2 V . Suppose the current amplification factor of the transistor is 100 . Find
the input signal voltage and base current, if the base resistance is 1 kΩ
(a) V in =1 V , I B =5 μ A
(c) V in =1.5 V , I B =5 μ A
(b) V in =0.01 V , I B =10 μ A
(d) V in =1.3 V , I B =10 μ A
227. Two amplifiers are connected one after the other in series (cascaded). The first
amplifier has a voltage gain of 10 and the second has a voltage gain of 20 . If the
input signal is 0.01 V , calculate the output AC signal.
(a) 2 V
(b) 3 V
(c) 4 V
(d) 5 V
228. In an intrinsic semiconductor, the energy gap E g is 1.2 eV . Its hole mobility is much
smaller than electron mobility and independent of temperature. What is the ratio
between conductivity at 600 K and that at 300 K ?
Assume that the temperature dependence of intrinsic carrier concentration ni is
given by
Eg
ni =no exp− where n o is a constant.
2 k BT
(a) 0.5 ×10 2
(b) 2.1 ×103
(c) 1.1 ×105
(d) 3.2 ×104
229. In a p−n junction diode, the current I can be expressed as
eV
I =I 0 exp ⁡ −1
kB T
where, I 0 is called the reverse saturation current. V is the voltage across the diode
and is positive for forward bias and negative for reverse bias and I is the current
through the diode, k B is the Boltzmann constant ( 8.6 × 10−5 eV / K ) and T is the absolute
temperature. If for a given diode I 0=5 ×10−12 A and T =300 K , then what will be the
forward current at a forward voltage of 0.6 V ?
(a) 0.063 A
(b) 0.832 A
(c) 0.0763 A
(d) None of these
230. A hole is
(a) an anti-particle of electron
(b) a vacancy created when an electron leaves a covalent bond
(c) absence of free electrons
(d) an artificially created (particle)
NCERT Exemplar
231. Conductivity of a semiconductor increases with increase in temperature, because
(a) number density of free charge carriers increases
(b) relaxation time increases
(c) both number density of free charge carriers and relaxation time increases
(d) number density of free charge carriers increases relaxation time decreases but
effect of decrease in relaxation time is much less than increase in number density
232. Assuming diodes to be ideal,
 (a) D1 is forward biased and D2 is reverse biased, so current flows from A to B
(b) D2 is in forward bias and D1 is in reverse bias and hence no current flows from B
to A and vice-versa
(c) D1 and D2 both are in forward bias, so current flows from A to B
(d) D1 and D2 both are in reverse bias, so no current flows from A to B
233. A 220 V AC supply is connected between points A and B.

What will be potential difference across capacitor C ?

(a) 220 V
(b) 110 V
(c) 0 V
(d) 220 √ 2 V
234. Output of the circuit shown below will be

(a) zero all the times

(b) like half wave rectifier with positive cycles in output

(c) like half wave rectifier with negative cycles in output

(d) like that of a full wave rectifier
235. In the circuit shown, the voltage difference between A and B it the diode forward
voltage drop is 0.3 V

(a) 1.3 V
(b) 2.3 V
(c) 0
(d) 0.5 V
236. In the depletion region of a diode,
I. there are no mobile charges.
II. equal number of holes and electrons exist making the region neutral.
III. recombination of electron and holes takes place.
 IV. immobile charged ions exists.
The correct options are
(a) I, II and III
(b) II, III and IV
(c) I, II and IV
(d) I, III and IV
237. To reduce the ripples in a rectifier circuit with capacitor filter, which of these must
be done?
I. R L should be increased.
II. Input frequency should be decreased.
III. Input frequency should be increased.
IV. Capacitors with high capacitance should be used.
(a) I, II and III
(b) I, III and IV
(c) II, III and IV
(d) I, II and IV
238. What happens during regulation action of a zener diode?
I. The current and voltage across the zener remains constant.
II. The current through series resistance ( R S ) changes.
III. The zener resistance is constant.
IV. The resistance offered by zener changes.
Correct ones are
(a) I and IV
(b) II and III
(c) II and IV
(d) I and II
239. The breakdown in a reverse biased p−n junction diode is more likely to occurs due
to
I. large velocity of minority charge carriers, if the doping is small.
II. large velocity of minority charge carriers, if the doping is large.
III. strong electric field in depletion region, if the doping concentration is small.
IV. strong electric field in depletion region, if the doping concentration is large.
Correct ones are
(a) I and IV
(b) II and III
(c) I and III
(d) II and IV
240. Consider an n−p−n transistor with its base-emitter junction forward biased and
collector base junction reverse biased, which of these are correct?
I. Electrons cross over from emitter to collector.
II. Holes move from base to collector.
III. Electrons move from emitter to base.
IV. Electrons from emitter move out of base without going to collector.
(a) I and III
(b) I and II
(c) I and IV
(d) II and III 241. Base biased CE transistor has the transfer characteristics as shown
 Which of the following statements are correct?
I. At V i=0.4 V , transistor is in active state.
II. At V i=1 V , transistor can be used as an amplifier.
III. At V i=0.5 V , it can be used as a switch turned off.
IV. At V i=2.5 V , it can be used as a switch turned on.
(a) I, II and III
(b) II, III and IV
(c) I, II and IV
(d) I, III and IV
242. For given circuit, truth table is

243. In an n−p−n transistor in CE configuration, I C =10 mA . If 95 % of electrons from

emitter reaches the collector, which of these are correct?
I. I E =8 mA
II. I E =10.53 mA
III. I B=0.53 mA
IV. I B=2 mA
(a) I and IV
(b) II and IV
(c) II and III
(d) I and III
244. When an electric field is applied across a semiconductor, then
I. electrons move from lower energy level to higher energy level in the conduction
band.
II. electrons move from higher energy level to lower energy level in the conduction
band.
III. holes in the valence band move from higher energy level to lower energy level.
IV. holes in the valence band move from lower energy level to higher energy level.
Out of these, correct statements are
(a) I and III
(b) I and II
(c) II and IV
(d) I and IV
 Answers
1. (d) 2. (d) 3. (d) 4. (d) 5. (c) 6. (b) 7. (a) 8. (a) 9. (c) 10. (d)
11. (d) 12. (b) 13. (c) 14. (c) 15. (a) 16. (a) 17. (d) 18. (d) 19. (c) 20. (a)
21. (c) 22. (c) 23. (b) 24 (b) 25. (a) 26. (b) 27. (c) 28. (a) 29. (c) 30. (d)
31. (d) 32. (b) 33. (c) 34. (b) 35. (a) 36. (a) 37. (a) 38. (c) 39. (a) 40. (c)
41. (c) 42. (a) 43. (c) 44. (d) 45. (d) 46. (d) 47. (c) 48. (c) 49. (a) 50. (c)
51. (b) 52. (d) 53. (c) 54. (a) 55. (c) 56. (c) 57. (b) 58. (b) 59. (d) 60. (a)
61. (c) 62. (c) 63. (c) 64. (c) 65. (d) 66. (b) 67. (c) 68. (a) 69. (a) 70. (a)
71. (a) 72. (a) 73. (a) 74. (a) 75. (d) 76. (b) 77. (a) 78. (b) 79. (b) 80. (d)
81. (a) 82. (a) 83. (b) 84. (b) 85. (c) 86. (c) 87. (b) 88. (a) 89. (d) 90. (c)
91. (d) 92. (c) 93. (a) 94. (d) 95. (c) 96. (c) 97. (c) 98. (c) 99. (b) 100. (a)
101. (b) 102. (c) 103. (d) 104. (a) 105. (b) 106. (a) 107. (a) 108. (d) 109. (c)
110. (a) 111. (c) 112. (c) 113. (b) 114. (d) 115. (c) 116. (b) 117. (c) 118. (a) 119. (b)
120. (b)
121. (b) 122. (a) 123. (d) 124. (a) 125. (b) 126. (c) 127. (b) 128. (c) 129. (b) 130.
(a)
131. (a) 132. (b) 133. (c) 134. (a) 135. (a) 136. (d) 137. (c) 138. (a) 139. (d) 140.
(a)
141. (c) 142. (a) 143. (a) 144. (a) 145. (b) 146. (a) 147. (a) 148. (b) 149. (a) 150.
(c)
151. (c) 152. (b) 153. (a) 154. (a) 155. (d) 156. (b) 157. (c) 158. (b) 159. (c) 160. (a)
161. (b) 162. (b) 163. (d) 164. (b) 165. (b) 166. (a) 167. (d) 168. (c) 169. (b) 170.
(a)
171. (c) 172. (c) 173. (a) 174. (c) 175. (c) 176. (a) 177. (c) 178. (c) 179. (a) 180. (b)
181. (c) 182. (a) 183. (c) 184. (d) 185. (a) 186. (a) 187. (c) 188. (a) 189. (d) 190. (d)
191. (c) 192. (d) 193. (c) 194. (b) 195. (d) 196. (d) 197. (d) 198. (d) 199. (d) 200.
(a)
201. (c) 202. (d) 203. (d) 204. (a) 205 (b) 206. (d) 207. (c) 208. (b) 209. (b) 210. (d)
211. (a) 212. (a,b,d)213. (a,b,c)214. (b,c)215. (b,d)216. (a,b)217. (b,c)218.
(a,b,c)219. (c) 220. (b,c)
221. (c) 222. (c) 223. (c) 224. (b) 225. (c)226. (b) 227. (a) 228. (c) 229. (a) 230. (b)
231. (d) 232. (b) 233. (d) 234. (b) 235. (b) 236. (c) 237. (b) 238. (c) 239. (a) 240.
(a)

Hints and Explanations

1. (d) In a vacuum tube, the electrons are supplied by a heated cathode and the
controlled flow of these electrons in vacuum is obtained by varying the voltage
between its different electrodes.
Vacuum is required in the inter-electrode space; otherwise the moving electrons
may lose their energy on collision with the air molecules in their path.
2. (d) The supply and flow of charge carriers in the semiconductor devices are within
the solid itself. No external heating or large evacuated space is required by the
semiconductor devices, so they have small sizes.
3. (d) A van der Waals solid transmits light and has a low melting point.
4. (d) Semiconductors have 4 valency and so they form covalent bonds.
5. (c) The SI unit of conductivity is Siemen per metre ( Sm−1 ).
6. (b)
(i) Metal They possess very low resistivity (or high conductivity).
−2 −8
ρ ∼ 10 −10 Ω−m
σ ∼ 102−10 8 Sm−1
(ii) Semiconductor They have resistivity or conductivity intermediate to metal and
insulator.
−5 6
ρ ∼ 10 −10 Ω−m
σ ∼ 10 −10 Sm−1
5 −6

(iii) Insulator They have high resistivity (or low conductivity).

11 19
ρ ∼ 10 −10 Ω−m
σ ∼ 10−11−10−19 Sm−1
As, 108 >105 >10−19 , σ metal > σ semiconductor >σ insulator
7. (a) Due to atomic interactions, the energies of outermost electrons are changed in
larger amounts.
8. (a)

9. (c) Energy level splits into more finer levels and for many atoms they form nearly
continuous bands.
10. (d ) For a semiconductor, 0< E g ≤3 eV
For metal E g ∼ 0 and for insulator E g >3 eV .
Forbidden energy gap in semiconductor is maximum energy gap allowed.
11. (d) For a semiconductor, conductivity increases with temperature or more electron-
hole pairs are created due to thermal agitation.
1 1
So, ρ ∝ ⇒ σ ∝ T ⇒ ∝ T
T ρ
12. (b) In good conductors, which are metals there is no gap between valence band and
conduction band. Hence, no holes exist.
13. (c) In a conductor, uppermost band is occupied by conduction electrons. Uppermost
band is conduction band.
 −34 8
hc 6.6 ×10 × 3 ×10
14. (c) Using E=Eg = = −9
J
λ 580 ×10
−34 8
6.6 ×10 × 3 ×10
¿ −9 −19
eV =2.1 eV
580× 10 ×1.6 ×10
15. (a) In semiconductor as the temperature increases, more thermal energy becomes
available to electrons and some of these electrons may break away (becoming free
electrons contributing to conduction).
The thermal energy effectively ionises only a few atoms in the crystalline lattice and
creates a vacancy in the bond . These holes behave as positive charge carriers.
16. (a) Pure semiconductors are called intrinsic semiconductors n e=nn=n i.
17. (d) Total current is the sum of hole current and electron current in semiconductor.
Electrons move opposite to hole, but current in same direction.
I =I e + I h
18. (d)

19. (c) In equilibrium condition, there is no net current through the semiconductor. This
shows that the rate of generation of electron-hole pairs is equal to rate of
recombination of electron-hole pairs. 20. (a) An intrinsic semiconductor will behave
like an insulator at T =0 K . It is the thermal energy at higher temperature (T >0 K ),
which excites some electrons from the valence band to the conduction band. These
thermally excited electrons at T > 0 K , partially occupy the conduction band. These
have come from the valence band leaving equal number of holes there.
20. (c) Number density of electron-hole pairs is increased with temperature, so at high
temperature semiconductors have higher conductivity.
21. (c) Semiconductors have negative temperature coefficient of resistance i.e., the
resistance of a semiconductor decreases with the increase in temperature and vice-
versa. Silicon is actually an insulator at absolute zero of temperature but it becomes
a good conductor at high temperatures. Because on increasing temperatures of
semiconductor some of the electrons jumps from valence band to conduction band.
22. (a) Resistivity of a metal is directly proportional to temperature because its
temperature coefficient is positive and resistivity of semiconductor is inversely
proportional to temperature due to its negative temperature coefficient. This
implies that with decrease in temperature, resistivity of metal decreases while that
of semiconductor increases. So, resistivity of Si increases but that of Cu decreases.
23. (b) For visible region, 450 ≤ λ ≤ 750 nm
Hence, photon energy ranges from few 1.7 to 2.8 eV .
hc
using hc=1240 eV −nm , E=
λ
 As for silicon most of the photons have a higher energy, so they excite electrons
(electrons absorb photons) Hence, light cannot pass through silicon. Silicon is
opaque.
But as photons are not absorbed in ZnS, they pass through it and so zinc sulphide is
transparent.
25. (a) Process of adding an impurity is called doping.
26. (b) Doping increases concentration of majority charge carriers.
27. (c) To form an n-type semiconductor doping is done by using a pentavalent impurity
like phosphorous and to form a p-type semiconductor doping is done by using a
trivalent impurity like indium.
28. (a)

Donor energy level is slightly less than energy level lowest to conduction band.
29. (c) The number of electrons made available for conduction by dopant atoms
depends strongly upon the doping level and is independent of any increase in
ambient temperature.
30. (d) Gallium, boron and aluminium all are trivalent impurities. These impurities make
germanium p-type semiconductor.
31. (d) In a p-type semiconductor, holes are majority charge carriers and electrons are
minority charge carriers.
32. (b) In both n -type and p-type semiconductors, number of electrons is exactly equal
to number of protons. Both are neutrals.
33. (c) At room temperature, the density of holes in the valance band is predominantly
due to impurity in the extrinsic semiconductor. The electron and hole concentration
in a semiconductor in thermal equilibrium is given by
2
n e nh=ni
34. (b) The four bonding electrons of C , Si or Ge lie, respectively, in the second, third
and fourth orbit. Hence, energy required to take out an electron from these atoms
(i.e., ionisation energy E g ) will be least for ¿, followed by Si and highest for C .
35. (a) Total number of atoms ¿ 5 ×1028 m−3 For every 106 atoms, 1 A−s is doped. 1 A-s
contributely 1 e−¿¿
28 6 22
Ne=5× 10 /10 =5 ×10 n n =n 2 ¿
¿ e h i

The number of holes

n h=( 2.25 ×1032 ) / ( 5 ×10 22) =4.5 ×10 9 m−3
36. (a) Phosphorous is pentavalent, it donates electron.
∴ nh ≪ n e
37. (a) Increase of temperature causes more electrons to leave valance band and reach
conduction band, so n e increases. But increase of temperature causes lattice
vibrations and so v d decreases as number of collisions increases.
38. (c) Doping does not change energy gap, it is still around 1 eV .
39. (a) E g , Si=1.1 eV
E g ,≥¿ 0.7 eV ⇒ E g , Sn=0.45 eV
40. (c) In intrinsic semiconductor, conductivity occurs due to excitation of electrons
when they absorb energy and break covalent bonds.
41. (c) For Ge, the energy gap, E g is 0.7 eV .
42. (a) n2i =ne n h, for extrinsic semiconductor
19 2
⇒ ( 10 ) =ne ¿
43. (c) Conductivity of a semiconductor increases with temperature.
So, they have negative value of temperature coefficient α .
44. (d) Relation is found empirically it is, n ∝T 3 /2. 45. (d) Dopant concentration is usually
1 to 10 ppm (parts per million)
1 −7
⇒ 7 ≈ 10
10
46. (d) shows an undoped semiconductor, Y shows an n-type semiconductor, Z shows
X
a p-type semiconductor.
47. (c) The n-type semiconductor can be produced by doping an impurity atom of
valence 5 , i.e., pentavalent atoms. i.e., phosphorous.
48. (c) In an n-type silicon, dopants are pentavalent atoms, electrons are majority
charge carriers and holes are minority charge carriers.
49. (a) N=5 × 1028 atoms ¿ m3 , A=¿ acceptor, indium, D=¿ donor, arsenic
22 3
N A =0.05 ×10 atoms /m ¿ N =5 ×10 22 atoms /m 3 ¿ n =1.5× 1016 m−2 ¿⇒ ¿ ¿ N −N =n −n and n n =n2 ¿ ∴ ¿ ¿ N
D i D A e h e h i
¿
On substituting values, we get, n e=4.95 ×1022 /m3
2
n2i ( 1.5 ×1016 ) 9 −3
∴ nh= = =4.54 ×10 m
n e 4.95 ×1022
22
10 −6
Observe that doping level 28 =10 ppm nearly
10
50. (c) Number of Si atoms ¿ 5 ×1028 atoms ¿ m3
Number of indium atoms ¿ Number of indium atoms for 1 silicon atom × Total
number of Si atoms
28
5× 10 21 3 15 3
¿ 7
=1× 10 atoms /m =1 ×10 atoms /cm
5 ×10
2
51. (b) Using n e ×n h=ni
16 −3
here ni=2× 10 m
nh=4.5 ×1022 m−3
2
n2i ( 2× 1016 )
∴ ne = =
nh 4.5 × 1022
n e=8.89× 109 m−3
52. (d) Consider a thin p-type silicon ( p-Si) semiconductor wafer. By adding precisely a
small quantity of pentavalent impurity, part of the p-Si wafer can be converted into
n -Si. So, p−n junction is formed.
53. (c) When a hole diffuses from p →n due to the concentration gradient. It leaves
behind an ionised acceptor (negative charge) which is immobile. As the holes
continue to diffuse, a layer of negative charge (or negative space-charge region) on
the p-side of the junction is developed. Similarly, a layer of positive space-charge is
 developed on n-side because of electron departure. This space-charge region on
either side of the junction together is known as depletion region.
54. (a) Thickness of depletion region is around 10−7 m.
55. (c) p-side is at negative potential and n -side is at positive potential. Also, central
layer is at zero potential. Potential at small distance from junction on both sides
becomes constant.
56. (c) Potential tends to prevent the movement of electron from the n-region into the p
-region, it is often called barrier potential.
57. (b) No, A p - n junction is formed when a p-type semiconductor is formed alongwith
an n -type semiconductor on a single intrinsic semiconductor.
58. (b) It is caused by diffusion of charge carriers.
59. (a) E opposes movement of charge.
60. (c) Diffusion and drift current of electrons and holes is due to concentration
difference.

61. (c) Using | |dV

dr
=E , we get

or
V =Ed
V 0.6 V 5 −1
E= = =6 × 10 Vm
d 1 ×10−6 m
63. (c) A semiconductor diode is basically a p−n junction with metallic contacts
provided at the ends for the application of an external voltage. It is a two terminal
device.
64. (c) The direction of arrow indicates the conventional direction of current (when the
diode is under forward bias).
65. (d) In forward biasing, electrons from n -side cross depletion region and reach p-side.
66. (b) Effective potential barrier will be ( V 0 −V ). i.e., potential barrier decreases.
67. (c) Due to concentration gradient, the injected electrons on p-side diffuse from the
junction edge of p-side to the other end of p-side. Likewise, the injected holes on n-
side diffuse from the junction edge of n-side to the other end of n -side. This motion
of charged carriers on either side gives rise to current. The total diode forward
current is sum of hole diffusion current and conventional current due to electron
diffusion.
68. (a) In RB potential barrier increases hence movement of majority carrier decreases.
But strong E pushes the movement of minority carrier towards their respective side
and contributes small current.
69. (a) The diode reverse current is not very much dependent on the applied voltage.
Even a small voltage is sufficient to sweep the minority charge carriers from one
side of the junction to the other side of the junction. The current is not limited by
the magnitude of the applied voltage but is limited due to the concentration of the
minority charge carrier on either side of the junction.
70. (a) The current under reverse bias is essentially voltage independent upto a critical
reverse bias voltage, known as breakdown voltage ( V br ) . When V =V brr, the diode
reverse current increases sharply from n to p side.
Even a slight increase in the bias voltage causes large change in the current. If the
reverse current is not limited by an external circuit, the p - n junction will get
destroyed. Once it exceeds the rated value, the diode gets destroyed due to
overheating.
71. (a) Typical V −I characteristics of a silicon diode are as shown.

72. (a) In forward bias, the current first increases very slowly, almost negligibly till the
voltage across the diode crosses a certain value.
After the characteristic voltage, the diode current increases significantly
(exponentially), even for a very small increase in the diode bias voltage. This
voltage is called the threshold voltage or knee voltage ¿ for germanium diode and
∼ 0.7 V for silicon diode).
73. (a) By allowing current only in forward bias it acts like a one way valve.
74. (a) A p−n junction diode primarily allows the flow of current only in one direction
(forward bias). The forward bias resistance is low as compared to the reverse bias
resistance. For diodes, we define a quantity called dynamic resistance as the ratio
of small change in voltage Δ V to a small change in current Δ I ,
ΔV
rd =
ΔI
75. (d) From the graph, at I =20 mA
V =0.8 V I =10 mA , V =0.7 V ¿ Also, at ¿ r = Δ V = 0.1 V =10 Ω¿ V ¿=−10 V , I =−1 μ A ¿ ∴ ¿ r = 10 V =1×
¿ FB
Δ I 10 mA RB
1μA
76. (b) Without an external bias an electric field exists which points from n to p-side and
opposes any diffusion of electrons.
77. (a) In forward biasing a negligible potential drop occurs in diode, so potential drop
across resistance R is V .
78. (b) In reverse bias, V p -side −V n -side =¿ Negative
79. (b) Diode is in forward bias, so resistance ¿ 0
V 4−1 3 −2
So, I = = = =10 A
R 300 300
80. (d) Arsenic is pentavalent, X is n-type and indium is trivalent and Y is p-type.
So, the junction is in reverse bias.
81. (a) Diode is in reverse bias, current ¿ 0 , potential difference across R=0 ; V AB=6 V
82. (a) Current is zero as batteries cause p - n junction in reverse bias.
V 0.3 5 −1
83. (b) Electric field, E= = =3 ×10 Vm
d 1 ×10−6
ΔV
84. (b) Dynamic resistance is, r d =
ΔI
here, Δ V =0.7 V −0.65 V =0.05 V
−3
Δ I =5 mA=5 ×10 A
0.05
∴ r d= =10 Ω
5 ×10−3
85. (c) In the circuit, the upper diode Di is reverse biased and lower diode D2 is forward
biased. Thus, there will be no current across upper diode function. The effective
circuit will be shown as
 Total resistance,
R ¿ 50+30+ 70=150 Ω
V 3 V
Current in circuit, I ¿ = =0.02 A
R 150 Ω
86. (c) If an alternating voltage is applied across a diode in series with a load, a
pulsating voltage will appear across the load only during the half cycles of the AC
input during which the diode is forward biased.
87. (b) In a half-wave rectifier, the secondary of a transformer supplies the desired AC
voltage across terminals A and B. When the voltage at A is positive, the diode is
forward biased and it conducts. When A is negative, the diode is reverse biased and
it does not conduct.
The reverse saturation current of a diode is negligible and can be considered equal to
zero for practical purposes.

88. (a) The reverse breakdown voltage of the diode must be sufficiently higher than the
peak AC voltage at the secondary of the transformer to protect the diode from
reverse breakdown.
89. (d )

Input AC voltage and output waveforms from the half-wave rectifier circuit
90. (c) Due to centre tapping potential reaching the diodes is only half of secondary
voltage. It is clear from its circuit's diagram.

91. (d )
Fig. (a) Input waveform given to diode D1, at A and to diode D2 at B. Fig. (b) output
waveform across the load R L connected in the full-wave rectifier circuit
92. (c) The rectified voltage is in the form of pulses of the shape of half sinusoids.
Though, it is unidirectional, it does not have a steady value.
93. (a) To get steady DC output from the pulsating voltage normally, a capacitor is
connected across the output terminals (parallel to the load R L ).
One can also use an inductor in series with R L for the same purpose. Since, these
additional circuits appear to filter out the AC ripple and give a pure DC voltage, so
they are called filters.
1
94. (d) As, X C = , for AC component when ω is high, then X C is less and, so a
Cω
capacitor let AC part bypass through it, so only DC part reaches R L, the load
resistance.
Fig. (a) A full-wave rectifier with capacitor filter

Fig. (b) Input and output voltage of rectifier in (a)

95. (c) In full wave rectification, output frequency is double of that of input frequency.
96. (c) In this question, full wave rectification is done by using a centre tap transformer.
1
So, output voltage is that of an half wave rectifier.
2
v
+∙ v /2 diff =
2
∙
centre 0
v
−∙−v /2 } diff =
2
97. (c) During Ist half of cycle,

During IInd half of cycle,

Clearly, current through R L is unidirectional, in both halves of input AC.

98. (c) Rate of discharge is inversely proportional to time constant of the circuit. More
value of time constant implies slow discharge.
99. (b) For full-wave rectifier, frequency ¿ 2 × input frequency for half-wave rectifier
frequency ¿ input frequency.
2
∴ Ratio ¿
1
100. (a) Zener diode is a special purpose semiconductor diode. It is designed to operate
under reverse bias in the breakdown region and used as a voltage regulator. The
symbol for Zener diode is shown in figure.

101. (b) Zener diode is fabricated by heavily doping both p-sides and n-sides of the
junction. Due to this, depletion region formed is very thin ( ¿ 10−6 m ) and the electric
field of the junction is extremely high ( ∼5 ×10 6 V /m ) even for a small reverse bias
voltage.
102. (c) When the reverse bias voltage V =V z , then the electric field strength is high
enough to pull valence electrons from the host atoms on the p-side which are
accelerated to n -side. These electrons account for high current observed at the
breakdown. The emission of electrons from the host atoms due to the high electric
field is known as internal field emission or field ionisation.
103. (d) Zener diode must be attached in reverse bias

104. (a) The value of R S should be such that the current through the Zener diode is much
larger than the load current. This is to have good load regulation. I z =20 mA . The
total current through R s is, therefore, 24 mA . The voltage drop across R s is
10.0−6.0=4.0 V . This gives
R s=4.0 V / ( 24 ×10−3 ) A=167 Ω
105. (b) Semiconductor diodes in which charge carriers are generated by photons
(photo-excitation) are called optoelectronic devices. Optoelectronic devices are
(i) photodiodes used for detecting optical signal (photodetectors). Used near automatic
doors.
(ii) light Emitting Diodes (LED) which convert electrical energy into light.
(iii) photovoltaic devices which convert optical radiation into electricity (solar cells).
106. (a) A photodetector detects any change in intensity of light by changing either
potential difference across it or by changing current through it.
107. (a) The variation of intensity results in change in number of incident photons (per
second) and hence a corresponding change in generation rate of electron and holes
occurs. This causes a change in current amplitude.
108. (d) When intensity is increased, reverse saturation current also increases.
I - V characteristics of a photodiode for different illumination intensity I 4 > I 3 > I 2 > I 1.
109. (c) When the diode is forward biased, electrons are sent from n → p .
Thus, at the junction boundary on either side of the junction, excess minority charge
carriers are there which recombins with majority charge carriers near the junction.
On recombination, the energy equal to or slightly less than the band gap are
emitted.
110. (a) The reverse breakdown voltages of LEDs are very low. typically around 5 V . So,
care should be taken that high reverse voltages do not appear across them.
111. (c) As, emitted energy = Band gap value when an electron moves from conduction
band to valance band. If λ is wavelength of emitted radiation, then
hc
Eg ¿
λ
hc
λλ ¿
Eg
Taking hc=1240 eV −nm and 450 nm≤ λ ≤ 750 nm for visible region, we get
450 ≤ λ ≤750
hc
⇒ 450 ≤ ≤ 750
Eg
hc 1240 eV −nm
⇒ Eg≤ = ≈ 2.8 eV
450 450 nm
hc
and E g ≥
750
1240 eVnm
¿ ≈ 1.7 eV
750 nm
So,
1.8 ≤ E g ≤ 3 eV
112. (c) For red LEDs,
GaAs 0.6 P0.4 −¿ Gallium Arsenic Phosphide ( E g=1.9 eV ) is used. This corresponds to λ ≈ 700 nm
.
113. (b) Solar cells uses light energy (photons) to generate an emf. 114. (d) Solar cell
supplies current to load. So, I −V characteristics is drawn in fourth quadrant.
Fig. (a) Depletion region

Fig. (b) I-V characteristics of a solar cell

115. (c) Semiconductors with band gap closed to 1.5 eV are ideal materials for solar cell
fabrication. Solar cells are made with semiconductors like Si ( E g=1.1 eV ) ,
GaAs ⁡( E g=1.43 eV ) ⋅CdTe ⁡( E g=1.45 eV ) ,
CuInSe 2 ( E g=1.04 eV ) etc.
The important criteria for the selection of a material for solar cell fabrication are (i) band
gap ( 1.0 to 1.8 eV ¿, (ii) high optical absorption ( ∼ 10 4 cm−1 ).
116. (b) As, we know that wavelength
hc 1240 eV −nm
λ= = =440 nm ≈ 4400 Å
Eg 2.8 eV
hc 1240 eV −nm
117. (c) ∵ E g= = =2.0 eV
λ 620 nm
118. (a) Electron moves from conduction to valence band, in LED when it glows
119. (b) An optical telecommunication link,

hc
120. (b) In red LED, λ R =
ER
hc E R λ V
In violet, LED ⁡λV = ⇒ = < 1
E V E V λR
121. (b) Light from an LED is highly monochromatic.
122. (a) For same value of current, higher value of voltage is required for higher
frequency.
123. (d) Here, E g=2 eV
Wavelength of radiation corresponding to this energy,
hc 1240 eV −nm
λ= = =620 nm
Eg 2eV
8 −1
c 3 ×10 ms 14
Frequency, v= = −9
=5 ×10 Hz
λ 620× 10 m
125. (b) A transistor has three doped regions forming two p−n junctions between them.
Obviously, there are two types of transistors.
(i) n−p−n transistor
(ii) p - n−p transistor
126. (c) n - p - n transistor Here, two segments of n-type semiconductor (emitter and
collector) are separated by a segment of p-type semiconductor (base).
127. (b) The transistor works as an amplifier with emitter-base junction forward biased
and base collector junction reversed bias, a transistor is said to be in active state.
128. (c) For a p−n− p transistor, charge carrier motion is as

For n−p−n transistor,

129. (b) Low; High. When emitter-base junction is forward biased and base-collector
junction is reverse biased, then transistor, is said to be active state. In active state
of transistor the emitter-base junction acts as low resistance while the base
collector junction acts as high resistance.
130. (a) Observe that emitter (E) is connected to both batteries, so it is common. Emitter
E−n side is connected to negative of V BB battery, collector C−n side is connected to
positive of V CC battery, active state.

Circuit arrangement for studying the input and output characteristics of n−p−n transistor
in CE configuration
131. (a) For a transistor,
V CE =V CB + V BE
132. (b) V CE must be sufficiently larger than 0.7 V .
133. (c) I B does not depend on V CE . So, when V CE is increased, I B remains constant. I C
increases till saturation.
 RL ΔI Δ Ic
134. (a) As AV =β ∵ gm = c =
Ri Δ V Δ l B Ri
β
or G= R
Ri L
β
∵ g m=
Ri
⇒G=g m R L ⇒G ∝ g m
G 2 gm 0.02
∴ = 1
⇒G 2= ×G
G 1 gm
2
0.03
2
∴ Voltage gain, G2= G
3
135. (a) Input resistance ( r i ) This is defined as the ratio of change in base emitter voltage
( Δ V BE ) to the resulting change in base current ( Δ I B ) at constant collector-emitter
voltage ( V CE ). This is dynamic (ac resistance).
Δ V BE
r i=
Δ IB V
CE

136. (d) Given, resistance across load, R L=800 Ω

Voltage drop across load, V L=0.8 V
Input resistance of circuit, Ri=192 Ω
Collector current is given by
V L 0.8 8
I C= = = =1 mA
R L 800 8000
Output current I C
∵ Current amplification ¿ = =0.96
Input current IB
1 mA
⇒ I B= ∵ Voltage gain,
0.96
VL VL 0.8× 0.96
AV = = = −3 =4 ⇒ A V =4
V in I B Ri 10 ×192
and power gain,
¿
IC
137. (c) ∵ β DC =
IB
Δ IC
138. (a) ∵ β AC =
Δ I B CE
IC
⇒ β DC =
IB
Ratio =β AC : β DC =150: 150=1 :1
139. (d) It is said to be in cut-off state for small V i , active for intermediate V i, saturation
for large V i .
140. (a) With increasing V i ,V o decreases.
141. (c) Power gain, A P=β AC × A v
142. (a) The output AC voltage is 2.0 V . So, the AC collector current, i C =2.0 /2000=1.0 mA .
i 1.0 mA
The signal current through the base is, therefore given by i B = C = =0.010 mA .
β 100
The DC base current has to be 10 ×0.010=0.10 mA.
V BB ¿ V BE + I B R B ⇒ RB =( V BB −V BE ) /I B
Assuming V BE ¿ 0.6 V
RB ¿(2.0−0.6)/0.10=14 kΩ
143. (a) i E =i B +i C ⇒ i E >i B
So, according to the question, the possible emitter current is 36 .
144. (a) Initially, both I C and I E increase.
145. (b) Emitter current begins to fall, T 1 is seeing decaying field in T 2.
Δ ic
146. ( a) β=¿ Current gain ¿
Δ ib
⇒ Δ i c =β × Δ i b=(80 ×250) μ A
147. (a) i E =i B +i C ⇒ i C =i E −i B.
Emitter side is forward biased, collector side is reverse biased. 148. (b)V b=i b Rb
9
⇒ R b= −6
=257 kΩ
35× 10
149. (a) Inductors are in series,
∴ Leq=L1 + L2

and frequency of oscillator ¿

1
2π √ 1
=
1
Leq Ceq 2 π √( L1 + L2 ) C
150. (c) From Kirchhoff's loop rule in output loop,
9−4=I C R C
As, RC =2 kΩ,
5 I 2.5
⇒ IC ¿ 3
=2.5 mA ⇒ I B = C = =27.8 μ A
2× 10 β 90
V BE ¿ 0.7 V
From Kirchhoff, loop rule in input loop,
2.3 6
I B RB =3−0.7 ⇒ RB = ×10 =82 kΩ
27.8
153. (a) DC current gain in common base amplifier is given by
iC
α=
iE
where, i C is collector current and i E is emitter current.
Given, α =0.96 ,i E=7.2 mA
 ∴ iC =0.96 ×7.2 mA =6.91 mA
As , i E =i B +i C
As, i E =i B +i C
∴ Base current, i B =i E −iC =7.2−6.91=0.29 mA
power gain 800 20
154. (a) Current gain, α = = =
voltage gain 840 21
α 20/21
Now, β= = =20
1−α 1−(20 /21)
IC
As β= =I C =β I B=20 ×1.2=24 mA
IB
π
155. (d) Input signal of a CE amplifier, V i=2 cos ⁡15t +
3
Voltage gain, AV =150
As CE amplifier gives phase difference of π between input and output signals.
V0
AV ¿
Vi
⇒V 0 ¿ AV V i
So, π
V 0 ¿ 150 ×2 cos ⁡15 t + + π
3
4π
V ¿ 300 cos ⁡15t +
3
156. (b) Voltage gain ¿ Current gain × Resistance gain
RC 6
¿ Current gain × =30 × =180
RI 1
Δ IC
157. (c) Current gain, β=
Δ IB V CE

β × R out
and voltage gain, AV =
Rin
−5
here, Rin =1000 Ω, Δ I B=10 μ A=10 A
3
Rout =5 kΩ=5 ×10 Ω
−3
Δ I C =2 mA=2 ×10 A
−3
2× 10
β= −5
=200
10
Hence,
3
200× 5 ×10
AV = =1000
1000
80 I E
158. (b) Here, I C =80 % of I E =
100
or
I C 10
IE ¿ = =12.5 mA
0.8 0.8
IB ¿ I E −I C =12.5−10=2.5 mA
159. (c) Dynamic resistance,
ΔV 0.05 × 1000
⇒ rd =
rd = Ω=10 Ω
IΔ 5
160. (a) An analog signal is a continuous waveform as
164. (b) AND gate An AND gate has two or more inputs and one output. The output Y of
AND gate is 1 only when input A and input B are both 1 . The logic symbol and truth
table for this gate is given by

Input Outpu
t

A B Y

0 0 0

0 1 0

1 0 0

1 1 1

165. (b) For an NAND gate, truth table is

So, no output occurs when both inputs are at higher potentials (1). Till time t 4, output (1)
occurs because both inputs do not become (1) together. 166. (a) Let us assume that
current through the diode is I . From the given condition,
V A−V B 4−(−6) 10 −2
I= = = 3
=10 A
R 1 kΩ 1 ×10
167. (d) Given circuit is

Observing, this is NOR gate.

So, output waveform is option (d).
168. (c) X =AB= A ⋅B
(i.e., AND gate)
If the output X of NAND gate is connected to the input of NOT gate (made from NAND
gate by joining two inputs) from the given figure, then we get back an AND gate.
169. (b) It is an AND gate.
A B A B X =A + B X =Y
 0 0 1 1 1 0

0 1 1 0 1 0

1 0 0 1 1 0

1 1 0 0 0 1

171. (c) The resultant Boolean expression of the above logic circuit will be
Y =( A+ B)⋅C
So, we have seen that among the given options, only option (c) is the correct choice, i.e.,
Output, Y =1 only when inputs A=1 , B=0 and C=1.
172. (c) The conductivity of an intrinsic semiconductor is less than that of a lightly doped
p-type semiconductor.
173. (a) Zener diode is a special purpose diode. In reverse bias, after a certain voltage,
current suddenly increases in Zener diode. This property is used to obtain voltage
regulation.
174. (c) In a transistor, the base is made thin and lightly doped so that the majority
charge carriers coming from emitter may pass on to the collector and very few form
electron-hole combination in base.
175. (c) In an oscillator, the feedback is in the same phase, i.e., positive feedback. If the
feedback voltage is in opposite phase, i.e., negative feedback, the gain is less than
one and it can never work as oscillator. It will be an amplifier with reduced gain.
176. (a)

Truth table for given circuit is

A B X =A + B Y=X

0 0 1 0

0 1 0 1

1 0 0 1

1 1 0 1

Hence, the given circuit acts as OR gate.

177. (c) Due to diffusion of electrons, positive space-charge region on n-side of the
junction and negative space charge region on p-side of the junction, an electric field
directed from positive charge towards negative charge develops. Electric field is
from n -side to p-side. Due to this field, an electron on p-side of the junction moves
to n-side and a hole on n -side of the junction moves to p-side. The motion of charge
carriers due to the electric field is called drift. Thus, a drift current, which is opposite
in direction to the diffusion current starts. Concentration gradient is due to doping
of sides. It is not affected by drift of charge carriers.
178. (c) The loss of electrons from the n -region and the gain of electrons by the p-region
causes a difference of potential across the junction of the two regions. The

polarity of this potential is such as to oppose further flow of carriers so that a

condition of equilibrium exists.
179. (a) As resistance of depletion region is large, potential drop occurs mainly in
depletion region.
180. (b) In forward bias, forward current is obtained due to removal of barrier potential
by externally applied potential. Ohm's law states V ∝ I , not valid for diode.
181. (c) From the V −I characteristics of a junction diode, we see that it allows current to
pass only when it is forward biased. So, if an alternating voltage is applied across a
diode the current flows only in that part of the cycle when the diode is forward
biased.
182. (a) As, X L =Lω, inductive reactance is high at high frequency for AC and for
capacitor, X C =1/Cω.
183. (c) Solar cells are made wafer shape as a large area ensures more incident solar
power.

(a)

(b)
Typical p−n junction solar cell 184. (d) The generation of emf by a solar cell, when
light falls on, it is due to the following three basic processes generation, separation
and collection (i) generation of electron-hole pairs due to light (with hv > E g ) close to
the junction; (ii) separation of electrons and holes due to electric field of the
depletion region.
Electrons are swept to n-side and holes to p-side; (iii) the electrons reaching the n -side
are collected by the front contact and holes reaching p-side are collected by the
 back contact. Thus, p-side becomes positive and n-side becomes negative giving
rise to photovoltage.
186. (a) In the case of Si transistor, as long as input V i is less than 0.6 V , the transistor
will be in cut-off state and current I C will be zero.
Hence, V o =V CC
When V i becomes greater than 0.6 V , the transistor is in active state with some current
I C in the output path and the output V o decreases as the term I C RC increases. With
increase of V i , I C inreases almost linearly and, so V o decreases linearly till its value
becomes less than about 1.0 V
187. (c) A gate is a digital circuit that follows certain logical relationship between the
input and output voltages. Therefore, they are generally known as logic gates
because they control the flow of information. The five common logic gates used are
NOT, AND, OR, NAND and NOR.
188. (a) They are small in size, consume low power, operate at low voltages and have
long life and high reliability.
189. (d) In a solid, electron's energies are very different from that in an isolated atom.
Inside the crystal, each electron has a unique position and no two electrons see
exactly the same pattern of surrounding charges. Because of this, each electron will
have a different energy level.
These different energy levels with continuous energy variation form what are called
energy bands. The energy band which includes the energy levels of the valence
electrons is called the valence band. The energy band above the valence band is
called the conduction band. Some electrons from the valence band may gain
external energy to cross the gap between the conduction band and the valence
band. Then, these electrons will move into the conduction band.
At the same time they will create vacant energy levels in the valence band where other
valence electrons can move. Thus, the process creates the possibility of conduction
due to electrons in conduction band as well as due to vacancies in the valence
band. In metals, conduction band and valence band overlap.
190. (d) When an electron diffuses from n → p , it leaves behind an ionised donor (species
which has become ion by donating electron) on n -side.
This ionised donor (positive charge) is immobile as it is bonded to the surrounding
atoms. As the electrons continue to diffuse from n - p, a layer of positive charge (or
positive space-charge region) on n -side of the junction is developed. On p-side atom
receiving electrons are ionised acceptor.
191. (c) Initially, diffusion current is large and drift current is small. As the diffusion
process continues, the space-charge regions on either side of the junction extend,
thus increasing the electric field strength and hence drift current.
This process continues until the diffusion current equals the drift current. Thus, a p−n
junction is formed. In a p−n junction under equilibrium there is no net current.
192. (d) In forward biasing due to the applied voltage, electrons from n -side cross the
depletion region and reach p-side (where they are minority charge carriers).
Similarly, holes from p-side cross the junction and reach the n -side (where they are
minority charge carriers). This process under forward bias is known as minority
charge carrier injection. At the junction boundary, on each side, the minority charge
carrier concentration increases significantly compared to the locations far from the
junction.
193. (c) In forward bias, if depletion layer's width decreases, fermi energy level is in the
middle of forbidden gap in intrinsic semiconductor.
194. (b) The circuit using two diodes gives output rectified voltage corresponding to both
the positive as well as negative half of the AC cycle.
Hence, it is known as full-wave rectifier. There is another circuit of full-wave rectifier
which does not need a centre tap transformer but needs four diodes. It is called a
bridge rectifier.
195. (d) Applying Kirchhoff's voltage rule to the input and output sides of this circuit, we
get

We shall treat V BB as the DC input voltage V i and V CE as the DC output voltage V o . So, we
have
V ¿ I B R B+ V BE
and i
V o ¿ V CC −I C RC .
196. (d) In an oscillator, we get AC output without any external input signal. In other
words, the output in an oscillator is self-sustained. To attain this, an amplifier is
taken. A portion of the output power is returned back (feedback) to the input in
phase with the starting power (this process is termed as positive feedback) as
shown in figure. The feedback can be achieved by inductive coupling (through
mutual inductance) or LC or RC networks.

197. (d) A 'NOR' gate, like an 'NAND' gate is universal gate, is also universal gate. AND,
OR, NOT, NAND gates can be made using NOR gates.
198. (d) The given materials in decreasing order of conductivity are Al> ¿>Si >¿ Diamond
(C), so aluminium has least energy gap and carbon has largest. Diamond has
energy gap 6 eV . E g ¿ Germanium ¿=0.71 eV .
199. (d) Truth table for NOR gate is given by,
Input Outpu
t

A B Y

0 0 1
 0 1 0

1 0 0

1 1 0

205. (b) As, I C =α l E=0.98× 2=1.96 mA

206. (d) As, l B =l E −l C =2−1.96=0.04 mA
R ( 400 ×103
207. (c) As, AV =α O =0.98 × =980
Ri 400
208. (b) Power amplification, a P=α AV =0.98 ×980=960
209. (b) As, V o =V i × voltage gain ¿ 0.1 ×980=98 V
210. (d) When V A =−10 V and V B=−5 V , ideal diodes D1 and D3 are reverse biased and D2
is forward biased.

Since, ideal diode in reverse bias has infinite resistance (i.e., open circuited) and during
forward bias it has zero resistance (i.e., short-circuited) Therefore, the given circuit
may be shown as in figure.
R R 3R
∴ R AB= + R+ =
4 4 2

211. (a) When V A =−5 V and V B=−10 V

R R R
D2 is reverse biased and D1 and D3 get forward biased, then R AB= + + =R
4 2 4
212. (a ,b ,d ) If lattice constant of semiconductor is decreased, then Ec and E v decrease
but E g increases.
213. (a ,b ,c ) (a) In insulators, energy gap is of the order of 5 to 10 eV and it is practically
impossible to impart this much amount of energy to the electrons in valence band
so as to jump to conduction band. So, choice (a) is correct.
(b) In semiconductors, with the rise in temperature more electrons from valence band
jump to conduction band and this results in increase in conductivity. So, choice (b)
is correct.
(c) In conductors, the conduction band is either partially filled or the conduction band
overlaps on the valence band. So, choice (c) is correct.
(d) In semiconductors, resistivity decreases with increase in temperature. So, choice (d)
is wrong.
−3
P 100 ×10
216. (a ,b) Current in circuit, i= = ¿ voltage drop across diode ¿
Vd 0.5
−3
¿ 200 ×10 A
'
Voltage across resistance R , V =1.5−0.5=1.0 V
'
V 1
Thus, resistance R= = =5 Ω
i 200× 10−3
217. (b , c )I C =10 mA
I C ¿ 90 % of I E
90
10 ¿ I
100 E
∴ IE ¿11 mA
IE ¿ I B+ I C
IB ¿ 1 mA
218. (a ,b ,c ) Applying loop law at output port,
or
9−4 ¿ I C RC
IC ¿ 2.5 mA
I C 2.5 −5
IB ¿ = =2.78 ×10 A=27.8 μ A
β 90
Since, the transistor operates in active region therefore
V BE=0.7 V
3−0.7
Applying loop law at input port, I B=
RB
5
2.3 ×10
R B= =82 kΩ
2.78
219. (c) When a forward bias is applied across the p−n junction, the applied voltage
opposes the barrier voltage. Due to this, the potential barrier across the junction is
IC IC
lowered. 220. (b) For a transistor, β= ⇒ I B= base region is thin, so that majority
IB β
carrier of emitter will reach the collector.
V input V input
Rinput = = ⋅β
IB IC
1
i.e., Rinput ∝
IC
Therefore, Rinput is inversely proportional to the collector current. For high collector
current, the Rinput should be small for which the base region must be very thin and
lightly doped for a transistor action, the emitter junction is forward biased and
collector junction is reverse biased.
221. (c) In an n-type semiconductor, it is obtained by doping the Ge or Si with
pentavalent atoms. In n -type semiconductor, electrons are majority charge carriers
and holes are minority charge carriers.
222. (c) The energy band gap is largest for carbon, less for silicon and least for
germanium.
223. (c) In an unbiased p - n junction, the diffusion of charge carriers across the junction
takes place from higher concentration to lower concentration. Therefore, hole
concentration in p-region is more as compared to n -region.
224. (b)

So, this logic operation as AND gate.

(b) Split the gate

So, this logic operation resembles to OR gate.

225. (c) The voltage gain is low at high and low frequencies and constant at mid
frequencies.
226. (b) Given, collector resistance Routput =2 kΩ=2000 Ω
Current amplification factor of the transistor, β AC =100 Audio signal voltage, V output =2 V
Input (base) resistance, Rinput =1 kΩ=1000 Ω
V output R output
∵ Voltage gain, AV = =β AC
V input Rinput
V output
∴ Input signal voltage, V input =
β AC ( Routput / R input )
2
¿ =0.01 V
100(2000/1000)
V input 0.01 −6
Base (input) current, I B= = =10 ×10 A
R input 1000
¿ 10 μ A
227. (a) Given, voltage gain of first amplifier, AV =10
1

Voltage gain of second amplifier, AV =20 2

Input voltage, V i=0.01 V

Vo
Total voltage gain, AV = =A V × AV
Vi 1 2
 Vo
∴ ¿10 × 20
0.01
Vo ¿2 V
228. (c) Given, intrinsic carrier concentration ni =no e−E g /2 k B T
and energy gap E g=1.2 eV
−5
k B=8.62×10 eV / K
For T =600 K ,
−E g/ 2 k B × 600
n600 =n o e
For T =300 K ,
−Eg / 2k B × 300
n300 =n 0 e
Dividing Eq. (i) by Eq. (ii), we get
−Eg 1 1 Eg 1 1
n600 −
2k 600 300
−
2 k 300 600
¿e B
=e B

n300
¿ ¿
Let the conductivities are σ 600 and σ 300.
σ 600 n 600 5
= =1.1 ×10 ( ∵ σ=en μ e )
σ 300 n 300
229. (a) Given, I 0=5 ×10−12 A , T =300 K
−5
k B ¿ 8.6× 10 eV /K
¿ ¿
Given, voltage V =0.6 V
−19
eV 1.6 × 10 × 0.6
∴ = =23.26
k B T 8.6 × 10−5 ×1.6 ×10−19 ×300
The current I through a junction diode is given by
eV
−1
=5 ×10−12 ( e 23.26−1 )
2 k BT
I ¿ Io e
−12 10
¿ ¿ 5 ×10 ×1.259 ×10 =0.063 A
231. (d) When temperature increases, number density of free charge carriers increases
and mean relaxation time decreases due to increased lattice vibrations. Effect of
decrease in relaxation time is much less as compared to effect of increase in
number density.
232. (b) In the given circuit, p-side of p - n junction, D1 is connected to lower voltage and
n -side of D 1 to higher voltage. Thus, D1 is reverse biased. The p-side of p−n junction
D2 is at higher potential and n-side of D2 is at lower potential. Thus, D2 is forward
biased.
Hence, no current flows through the junction B to A .
233. (d) As p - n junction conducts during positive half cycle only, the diode connected
here will work in positive half cycle. Potential difference across C=¿ peak voltage of
the given AC voltage V 0=V rms √ 2=220 √ 2 V
234. (b) Each positive half passes through the diode and so output only contains positive
halves of voltage.
235. (b) Diode is forward biased, no current goes through side branch after capacitor is
charged.
−3 3
∴ V AB=I AB × R AB=0.2 ×10 × 10 ×10 =2 V Option (b) is most appropriate, extra 0.3 V occurs
due to diode.
236. (c) In depletion region, after equilibrium, no recombination can occur as electrons
and holes are not 'free'. In depletion region, there are immobile charged ions and
but no mobile charges; equal number of holes and electrons exists.
237. (b) Ripple factor (r ) of a full-wave rectifier using capacitor filter is given by
1
r=
4 √3 RL C
1
i.e., r∝
RL
1 1
⇒ r∝ ,r ∝
C v
Thus, to reduce r , RL should be increased, input frequency v should be increased and
capacitance C should be increased.
238. (c) Resistance of Zener diode changes at breakdown voltage and current through R s
increases after breakdown due to increased movement of minority carrier.
239. (a) In reverse biasing, the minority charge carriers will be accelerated due to
reverse biasing which on striking with atoms cause ionization resulting secondary
electrons and thus more number of charge carriers.
When doping concentration is large, there will be large number of ions in the depletion
region, which will give rise to a strong electric field.
241. (b) V i=1 V , active state
V i ¿ 0.5 V , cut-off region I C =0
Vi ¿ 2.5 V

The transistor circuit in active state can be used as an amplifier.

242. (c) It is a combination of AND gates.

A B Y1 Y2 Y3 Y
 0 0 1 0 0 0

0 1 1 0 1 1

1 0 0 0 0 0

1 1 0 1 0 1

243. (c) As, I E =I B + I C

100 % 10 % 95 %
95
So, I E × =I
100 C
100× 10
⇒ I E= =10.53 mA
95
Also, I B=10.53−10=0.53 mA
244. (a) When electric field is applied across a semiconductor, the electrons in the
conduction band get accelerated and acquire energy. They move from lower energy
level to higher energy level.
While the holes in valence band move from higher energy level to lower energy level,
where they will be having more energy.