C H A P T E R

14
Semiconductor
Electronics;
Materials, Devices
and Simple Circuits
A Quick Recapitulation of the Chapter
1. Metals They have very low resistivity or high
conductivity, ρ ~ 10−2-10−8 Ωm, σ ~102 - 108 Sm −1 5. On the basis of purity, semiconductors are of two types
2. Semiconductors They have resistivity or (i) Intrinsic semiconductor It is a pure semiconductor
conductivity between metals and insulators. without any significant dopant species present.
i.e., ρ ~ 10− 5 - 106 Ωm, σ ~ 10+ 5 -10− 6 Sm −1 ne = nh = ni
where, ne and nh are number densities of electrons and
Types of Semiconductors holes respectively and ni is called intrinsic carrier
Types of semiconductors are given below concentration.
(i) Element semiconductors are available in An intrinsic semiconductor is also called an undoped
natural form, e.g., silicon and germanium. semiconductor or i-type semiconductor.
(ii) Compound semiconductors are made by (ii) Extrinsic semiconductor Pure semiconductor when doped
compounding the metals, e.g., CdS, GaAs, with the impurity is known as extrinsic semiconductor.
CdSe, InP, anthracene, polyaniline, etc. Extrinsic semiconductors are basically of two types
3. Insulators They have high resistivity or low (a) n-type semiconductor (b) p-type semiconductor
conductivity. Note Both the types of semiconductors are electrically neutral.
− 11 − 19 −1 6. In n-type semiconductor, majority charge carriers are
i.e., ρ ~ 10 - 10 Ωm, σ ~ 10
11 19
- 10 Sm
electrons and minority charge carriers are holes, i.e.,
4. Fermi energy It is the maximum possible ne > nh . Here, we dope Si or Ge with a pentavalent element,
energy possessed by free electrons of a then four of its electrons bond with the four silicon
material at absolute zero temperature neighbours, while fifth remains very weakly bound to its
(i.e., 0 K). parent atom.
 7. In p-type semiconductor, majority charge carriers are (iii) Solar cell Solar cell is a p-n +
holes and minority charge carriers are electrons, i.e., junction diode which converts
p
nh > ne . In a p-type semiconductor, doping is done solar energy into electrical
with trivalent impurity atoms i.e., those atoms which energy. n
have three valence electrons in their valence shell. Its symbol is given by –
8. At equilibrium condition, nenh = ni2
9. Minimum energy required to create a hole-electron 16. Zener diode is a reverse biased heavily doped p-n
pair, hν ≥ E g , where E g is energy band gap. junction diode. It is operated in breakdown region.
10. A p-n junction is an arrangement made by a close Its symbol is given by and is used as
contact of n-type semiconductor and p-type voltage regulator.
semiconductor.
17. A transistor is a combination of two p-n junction
11. A semiconductor diode is basically a p-n junction joined in series. A junction transistor is known as
with metallic contacts provided at the ends for the bipolar junction transistor (BJT). Transistors are of
application of an external voltage. two types n-p-n and p-n-p. The central block thin and
A p-n junction diode is represented as the symbol. lightly doped is called ‘Base’ while the other
The direction of arrow indicates the p n electrodes are emitter and collectors.
conventional direction of current (when 18. The emitter-base junction is forward biased while
the diode is under forward bias). collector base junction is reversed biased.
V
12. The DC resistance of a junction diode, rDC = 19. The transistor can be in three configurations,
I
∆V common emitter (CE), common collector (CC) and
13. The dynamic resistance of junction diode, rAC =
∆I common base (CB).
14. Diode as rectifier The process of converting 20. The plot between IC and VCE for fixed IB is called
alternating voltage or current into direct voltage or output characteristics while the plot between IC and
current is called rectification. Diode is used as a IB with VBE fixed is known as input characteristics.
rectifier for converting alternating current or voltage
21. Transistor parameters in CE configuration are
into direct current or voltage.
∆ VBE
There are two ways of using a diode as a rectifier i.e., Input resistance, ri =
(i) Diode as a half-wave rectifier Diode conducts ∆ IB VCE = constant
corresponding to positive half cycle and does not
∆VCE
conduct during negative half cycle. Hence, AC is Output resistance, ro =
converted by diode into unidirectional pulsating ∆ IC IB = constant
DC. This action is known as half-wave rectification.
(ii) Diode as a full-wave rectifier In the full-wave  ∆I 
Current amplification factor, β =  C 
rectifier, two p-n junction diodes D1 and D2 are used.  ∆IB  V
CE = constant
Its working is based on the principle that junction
22. A transistor can be used as an amplifier. The voltage
diode offers very low resistance in forward bias and
gain of CE configuration is
very high resistance in reverse bias.
V  R
15. Optoelectronic devices Semiconductor diodes in AV =  o  = β C
which carriers are generated by photons.  Vi  RB
i.e., photo-excitation, such devices are known as where, RC and RB are respectively resistances in
optoelectronic devices. collector and base sides of the circuit.
These are as follows
23. In common base configuration, AV current gain is
(i) Light emitting diode (LED) It is a heavily doped ∆ IC
p-n junction diode which converts electrical energy α= .
into light energy. ∆ IE V = constant
CB
hν
+ – 24. Transistor can be used as an oscillator as well as a
Its symbol is given by switch (is cut-off or saturation state).
(ii) Photodiode A photodiode is a special 25. A logic gate is a digital electronic circuit which
type of junction diode used for detecting
follows a logical relationship between its input and
optical signals. It is a reverse biased p-n
output. A logic gate may have one or more inputs but
junction made from a photosensitive
has only one output.
material. Its symbol is given by
 Logic gates follow Boolean algebra, which consists of (iv) NAND gate
three basic operations, namely AND ( A ⋅ B = Y ), NAND gate is the combination of AND gate and NOT
OR ( A + B = Y ) and NOT ( A = Y ) . gate. Boolean expression for NAND gate is given by

(i ) OR gate Boolean expression for OR gate is given Y = A ⋅B

by Y = A + B Logic symbol
Logic symbol A
NAND Y
A B
Y
B
(v) NOR gate
(ii ) AND gate Boolean expression for NOR gate is given by
Boolean expression for AND gate is given by Y = A+B
Y = A ⋅B The NOR gate is the combination of OR and NOT
Logic symbol gate given by
A Logic symbol
Y
B A
NOR Y
B
(iii ) NOT gate
(vi) XOR gate
Boolean expression for NOT gate is given by Y = A
Boolean expression for output/input of XOR gate is
Logic symbol Y = A ⊕ B = A ⋅B + A ⋅B
Logic symbol
A Y
A
Y
B

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 4. Bonding in a semiconductor is
required, because (a) metallic (b) ionic
(a) electrons are not ejected from cathode (c) van der Walls (d) covalent
(b) vacuum helps in extracting electrons from remaining
gas molecules or atoms 5. The SI unit of conductivity is
(c) in vacuum work function of cathode is reduced (a) ( Ω m )−1 (b) Ωm−1 (c) Sm−1 (d) S
(d) electrons may lose their energy on collision with air
molecules in their path
6. Correct one is
(a) σ semiconductor > σ insulator > σ metal
2. Semiconductor devices (diodes, transistors) are (b) σ metal > σ semiconductor > σ insulator
smaller than vacuum tubes because (c) σ semiconductor > σ metal > σ insulator
(a) they are made from silicon /germanium crystals (d) σ insulator > σ semiconductor > σ metal
(b) they have very high density (here, σ represents conductivity.)
(c) large crystals of semiconductors have large resistance
7. In a crystal, atomic separation is around 2 to 3A. At this
(d) flow of charge carriers are within the solid itself
separation due to interatomic interaction, energies of
3. If a solid transmits the visible light and has a low (a) outermost electrons are changed
melting point, it possesses (b) innermost electrons are changed
(a) metallic bonding (b) ionic bonding (c) Both (a) and (b)
(c) covalent bonding (d) van der Walls bonding (d) None of the above
 8. Following diagram shows energy band positions 11. Which of these is a true graph showing a relation between
in a semiconductor at 0 K. resistivity ρ and temperature for semiconductor?
Empty 4N Ω-m Ω-m
states Ec ρ ρ
Ec Half filled
(a) Eg (b) Eg
Ev Half filled (a) (b)
4N Ev
Filled states
–1 –1
K K
Eg Empty 1000/T 1000/T
Filled
Ec Ec
Ω-m Ω-m
(c) Eg (d) Filled ρ ρ

Ev Ev
Empty Eg Empty
(c) (d)
9. The splitting of 1s and 2s atomic energy levels
when many atoms come together to form a solid 1000/T
K
–1
1000/T
K
–1

is best represented by
12. There is no hole current in good conductors, because they
(a) have large forbidden energy gap
2s 2s (b) have no energy gap due to overlapping valence and
Energy

Energy

(a) (b) conduction bands

1s 1s (c) are full of electron gas
Atomic separation Atomic separation
(d) have no valence band
13. A solid having upper most energy band partially filled
with electrons is called
2s (a) insulator (b) semiconductor
Energy

Energy

(c) (d) 2s
(c) conductor (d) None of these
1s
1s
14. If the energy of a photon of sodium light ( λ = 580 nm )
Atomic separation Atomic separation equals the band gap of semiconductor, the minimum
energy required to create hole electron pair.
10. Forbidden energy gap in a semiconductor is (a) 1.5 eV (b) 3.2 eV
nearly equal to (c) 2.1 eV (d) 4.1 eV
(a) 1 eV (b) 6 eV (c) 0 eV (d) 3 eV

Topic 2
Intrinsic and Extrinsic Semiconductors
15. At elevated temperature, few of covalent bonds of Si (c) doped semiconductor
or Ge are broken and a vacancy in the bond is created. (d) None of the above
Effective charge of vacancy or hole is (Here, ne = number of free electrons, nh = number of free
(a) positive holes, ni = intrinsic carrier concentration)
(b) negative 17. If I is total current through an intrinsic semiconductor
(c) neutral and I e is electron current and I h is hole current, then
(d) sometimes positive and sometimes negative Ih Ie
(a) I e = (b) =I
16. In pure form, Ge or Si, a semiconductor is called I Ih
(a) intrinsic semiconductor, ne = nh = ni (c) I e − I h = I (d) I e + I h = I
(b) extrinsic semiconductor, ne = nh = ni
18. Energy band gap diagram for an intrinsic 26. Doping of intrinsic semiconductor is done
semiconductor at temperature T > 0 K is (a) to neutralise charge carriers
Holes (b) to increase the concentration of majority charge carriers
Ec Ec
(c) to make it neutral before disposal
(d) to carry out further purification
(a) Eg Electrons (b) Eg 27. An n-type and p-type silicon can be obtained by
Ev doping pure silicon with
Ev
(a) arsenic and phosphorous, respectively
Ec
(b) indium and aluminium, respectively
Ec (c) phosphorous and indium, respectively
(d) aluminium and boron, respectively
(c) Eg (d) Holes Eg
28. Which of the following statement is correct for an
Ev Ev n-type semiconductor?
(a) The donor level lies below the bottom of the conduction
Electrons
Electrons band
(b) The donor level lies closely above the top of the
19. In equilibrium condition, the rate of generation of valence band
electron-hole pairs (c) The donor level lies at the halfway mark of the
(a) is more than rate of recombination of electron and hole forbidden energy gap
pairs (d) None of the above
(b) is less than rate of recombination of electron and hole pairs
(c) equals to rate of recombination of electron and hole pairs
29. Number of electrons present in conduction band due
(d) is always zero
to doping
(a) shows a heavy increase with increase of temperature
20. In intrinsic semiconductor at room temperature, (b) shows a heavy decrease with increase of temperature
number of electrons and holes are (c) independent of change in ambient temperature
(a) equal (b) zero (c) unequal (d) infinite (d) reduces to zero at temperature above room temperature
21. A pure semiconductor behaves as a good conductor at 30. To make a p-type semiconductor, germanium is
(a) room temperature (b) low temperature doped with
(c) high temperature (d) Both (b) and (c) (a) gallium (b) boron
22. At absolute zero, Si acts as (c) aluminium (d) All of these
(a) non-metal (b) metal 31. In a p-type semiconductor, the majority and minority
(c) insulator (d) None of these charge carriers are respectively,
23. Si and Cu are cooled to a temperature of 300 K, then (a) protons and electrons
(b) electrons and protons
resistivity
(c) electrons and holes
(a) for Si increases and for Cu decreases (d) holes and electrons
(b) for Cu increases and for Si decreases
(c) decreases for both Si and Cu 32. Which statement is correct?
(d) increases for both Si and Cu (a) n -type germanium is negatively charged and p-type
germanium is positively charged
24. The energy gap for silicon is 1.14 eV and for zinc (b) both n -type and p-type germanium are neutrals
sulphide it is 3.6 eV. (c) n -type germanium is positively charged and p-type
From the above data, we conclude that germanium is negatively charged
(a) silicon is transparent and zinc sulphide is opaque (d) both n -type and p-type germanium are negatively charged
(b) silicon is opaque and zinc sulphide is transparent 33. If ne is number density of electrons in conduction band
(c) both ZnS and Si are transparent
and nh is number density of holes in valence band,
(d) both ZnS and Si are opaque
then for an extrinsic semiconductor at room
25. Doping is temperature, (ni = number density of intrinsic pairs)
(a) a process of adding an impurity to a pure semiconductor ne nh
(a) = ni2 (b) = ni2
(b) a process of obtaining semiconductor from its ore nh ne
(c) melting of a semiconductor
(c) ne nh = ni2 (d) ne + nh = ni2
(d) purification of a semiconductor
34. Carbon is more resistive than germanium and 43. Which of the following has negative temperature
silicon. Then, order of energy gap is coefficient of resistance?
(a) C > Ge > Si (b) C > Si > Ge (a) Metal (b) Insulator
(c) Si > GE > C (d) C = Si = Ge (c) Semiconductor (d) All of these
35. Let a pure Si crystal has 5 × 10 28 atoms m −3 . It is 44. The relation between the number of free electrons in
doped by parts per million concentration of pentavalent semiconductors ( n) and its temperature (T ) is
arsenic. If number of intrinsic pairs is1.5 × 1016 m −3 , (a) n ∝ T 2 (b) n ∝ T
then number of holes in doped crystal is (c) n ∝ T (d) n ∝ T 3 / 2
9 −3 16 −3
(a) 4.5 × 10 m (b) ~ 10 m
(c) 2.25 × 1032 m−3 (d) 5 × 1022 m−3 45. In extrinsic p and n-type, semiconductor materials, the
ratio of the impurity atoms to the pure semiconductor
36. If ne and nh are the number of electrons and holes atoms is about
in a semiconductor heavily doped with (a) 1 (b) 10−1
phosphorous, then (c) 10−4 (d) 10−7
(a) ne >> nh (b) ne < < nh (c) ne ≤ nh (d) ne = nh
46. The energy band diagrams for three semiconductor
37. ne and v d be the number of electrons and drift
samples of silicon are as shown. We can then assert that
velocity in a semiconductor. When the temperature
is increased, then
Electron
(a) ne increases and vd decreases energy
(b) ne decreases and vd increases
(c) both ne and vd increase
(d) both ne and vd decrease X Y Z
(a) sample X is undoped while samples Y and Z have been
38. For extrinsic semiconductor,
doped with a third group and a fifth group impurity,
(a) the conduction band and valence band overlap
respectively
(b) the gap between conduction band and valence band
(b) sample X is undoped while both samples Y and Z have
is more than 16 eV
been doped with a fifth group impurity
(c) the gap between conduction band and valence band
is near about 1 eV (c) sample X has been doped with equal amounts of third and
(d) the gap between conduction band and valence band fifth group impurities while samples Y and Z are undoped
will be 100 eV and more (d) sample X is undoped while samples Y and Z have been
doped with a fifth and a third group impurity, respectively
39. Three semiconductors are arranged in the
increasing order of their energy gap as follows. The 47. In an n-type semiconductor, which of the following
correct arrangement is statement is true? [NEET 2013]
(a) tin, germanium, silicon (b) tin, silicon, germanium (a) Electrons are majority charge carriers and trivalent atoms
(c) silicon, germanium, tin (d) silicon, tin, germanium are the dopants
(b) Electrons are minority charge carriers and pentavalent
40. When the electrical conductivity of semiconductor
atoms are the dopants
is due to the breaking of its covalent bonds, then (c) Holes are minority charge carriers and pentavalent atoms
the semiconductor is said to be are the dopants
(a) donor (b) acceptor (c) intrinsic (d) extrinsic
(d) Holes are majority charge carriers and trivalent atoms are
41. The forbidden energy gap in the energy bands of the dopants
germanium at room temperature is about
48. In an n-type silicon, which of the following statements is
(a) 1.1 eV (b) 0.1 eV (c) 0.67 eV (d) 6.7 eV
correct?
42. A Ge specimen is doped with Al. The (a) Electrons are majority charge carriers and trivalent atoms
concentration of acceptor atom is ~ 10 21 atoms per are the dopants
m 3 . Given that the intrinsic concentration of (b) Electrons are minority charge carriers and pentavalent
atoms are the dopants
electron hole pairs is ~ 1019 per m 3 , the
(c) Holes are minority charge carriers and pentavalent atoms
concentration of electrons in the specimen is are the dopants
(a) 1017 per m3 (b) 1015 per m3 (d) Holes are majority charge carriers and trivalent atoms are
(c) 104 per m3 (d) 102 per m3 the dopants
49. The number of silicon atoms per m 3 is 5 × 10 28 . This then the number of acceptor atoms in silicon per cubic
centimetre will be
is doped simultaneously with 5 × 10 22 atoms per m 3
(a) 2.5 × 1030 atoms per cm3
of arsenic and 5 × 10 20 per m 3 atoms of indium.
(b) 1.0 × 1013 atoms per cm3
Given that ni = 1.5 × 1016 m −3 .
(c) 1.0 × 1015 atoms per cm3
Number of electrons and holes (in per metre cube of
(d) 2.5 × 1036 atoms per cm3
sample) are respectively,
(a) 4.95 × 1022 , 4.54 × 109 (b) 4.54 × 109 , 4.54 × 109 51. The number of density of electrons and holes in pure
9
(c) 4.54 × 10 , 4.95 × 10 22 22
(d) 4.95 × 10 , 4.95 × 10 22 silicon at 27°C are equal and its value is
2.0 × 1016 m −3 on doping with indium the hole
50. A silicon specimen is made into a p-type
density increases to 4.5 × 10 22 m −3 , the electron
semiconductor by doping on an average, one indium density in doped silicon is
atom per 5 × 10 7 silicon atoms. If the number density (a) 10 × 109 m−3 (b) 8.89 × 109 m−3
of atoms in the silicon specimen is 5 × 10 28 atoms/m 3 , (c) 11 × 109 m−3 (d) 16.78 × 109 m−3

Topic 3
p-n Junction and Semiconductor Diode
52. A p- n junction contains 56. Potential difference of p and n-side which prevents
(a) a p-type semiconductor is joined with an n -type diffusion of electrons is called
semiconductor by glue (a) potential gradient
(b) a p-type semiconductor is bolted with an n -type (b) potential difference
semiconductor (c) barrier potential
(c) a p-type semiconductor is kept in touch with an n -type (d) depletion potential
semiconductor
(d) a p-type semiconductor is formed with an n -type 57. Can we take one slab of p-type semiconductor and
semiconductor on same semiconductor crystal wafer physically join it to another n-type semiconductor to
53. Due to diffusion, the space charge region on either get p -n junction?
side of p-n junction is developed. This space charge (a) Yes
region is called (b) No
(a) dilution region (b) diffusion region (c) It depends on the hole and electron concentrations on p
and n -side
(c) depletion region (d) ionic region
(d) Only when a p-type semiconductor is soldered with an
54. Thickness of depletion region is of order of n-type semiconductor
(a) ~10−7 m (b) ~10−10 m (c) ~10−9 m (d) ~10−3 m 58. The depletion layer in the p-n junction region is
55. Which of these graphs shows potential difference caused by
between p-side and n-side of a p-n junction in (a) drift of holes
equilibrium? (b) diffusion of charge carriers
(c) migration of impurity ions
(d) drift of electrons
(a) p-side n-side (b) p-side 59. The barrier potential of a p- n junction depends on
n-side
Junction Junction
(i) type of semiconductor material
plane plane (ii) amount of doping
(iii) temperature
Which one of the following is correct?
(c) p-side (d) p-side
n-side n-side [CBSE AIPMT 2014]
(a) (i) and (ii) (b) (ii)
Junction Junction
plane plane (c) (ii) and (iii) (d) (i), (ii) and (iii)
60. The electrical resistance of depletion layer is large 66. If V is applied potential difference in forward bias and
because V0 is barrier potential of a p-n junction, then effective
(a) of strong electric field barrier height under forward bias is
(b) it has a large number of charge carriers (a) V − V0 (b) V0 − V (c) V0 + V (d) V0
(c) it contains electrons as charge carriers
67. In forward bias, forward current obtained from the
(d) it has holes as charge carriers
p-n junction diode is
61. In an unbiased p-n junction, holes diffuse from the (a) due to injection of electrons in p-side
p-region to n-region because (b) due to injection of holes in n -side
(a) free electrons in the n-region attract them (c) both (a) and (b)
(b) they moves across the junction by the potential (d) due to flow of electrons from negative terminal of
difference supply to its positive terminal
(c) hole concentration in p-region is more as compared to 68. In a p-n junction diode,
hole concentration in n-region (a) the current in the reverse biased condition is generally
(d) All of the above very small
(b) the current in the reverse biased condition is small but
62. A Si based p-n junction has a depletion layer of that in forward biased condition is independent of the
thickness 1µm and barrier potential difference of bias voltage
n-side and p-side is 0.6 V. (c) the reverse biased current is strongly dependent on the
The electric field in the depletion region is applied bias voltage
(a) 0.6 Vm−1 (b) 6 × 10−4 Vm−1 (d) the forward biased current is very small in comparison
to reverse biased current
(c) 6 × 105 Vm−1 (d) 6 × 104 Vm−1
69. In a reverse biased p-n junction diode,
63. A diode is a (a) current under reversed bias is not very much dependent
(a) piece of a covalent crystal on applied voltage
(b) piece of a semiconductor crystal with metallic contacts (b) current under reversed bias is directly proportional to
provided at two ends applied voltage
(c) p-n junction with metallic contacts provided at two ends (c) current initially depends on applied voltage, then it
(d) piece of a metal which is sprayed over by a becomes independent
semiconductor (d) no current flows in reversed bias

64. Symbol of a p-n junction diode is an arrow, its 70. If reverse biasing potential is increased beyond a
direction indicates 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
(a) nothing its just a symbol
(d) potential barrier becomes zero
(b) direction of flow of electrons
(c) direction of conventional current when it is forward 71. Characteristic curve of a p-n junction is
biased I (mA) I (mA)
(d) direction of electric field

65. In the case of forward biasing of p-n junction, which (a) (b)
V (volts) V (volts)
one of the following figures correctly depicts the
direction of flow of charge carriers? I (µA) I (µA)

p n p n I (mA) I (mA)

(a) e- e- (b) e- e- (c) (d)

V (volts) V (volts)
Vp Vp
I (µA) I (µA)
p n p n
(c) (d) 72. Threshold or knee voltage for a forward biased
e- e- e- e- germanium and silicon diodes have respective values
(a) 0.2 V, 0.7 V (b) 0.7 V, 1.1 V
Vp Vp
(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 79. In the circuit given below, the value of the current is
direction (forward bias). The forward bias resistance
+4 V P N 300 Ω +1 V
is low as compared to the reverse bias resistance. In a
circuit, a diode acts like a (a) 0 A (b) 10−2 A (c) 102 A (d) 10−3 A
(a) valve (b) switch
(c) amplifier (d) multi-way passage 80. A semiconductor X is made by
X Y
doping a germanium crystal with
74. Dynamic resistance of a diode is given by
arsenic ( Z = 33). A second
∆V ∆V
(a) rd = (b) rd = − semiconductor Y is made by doping
∆I ∆I germanium with indium ( Z = 49). The
Threshold voltage Breakdown voltage
(c) rd = (d) rd = two are joined end to end and connected to a battery
Current Current as shown. Which of the following statements is
75. V -I characteristics of a silicon diode is shown. correct?
I (mA) (a) X is p-type, Y is n-type and the junction is
30 forward biased
(b) X is n -type, Y is p- type and the junction is forward
Silicon biased
20 (c) X is p-type, Y is n-type and the junction is reverse
15 biased
(d) X is n-type, Y is p-type and the junction is reverse
10
biased
2Ω A S B
–10 V
81. The diode shown in the circuit is a
0 0.5 0.70.8 V (Volts) silicon diode. The potential
1 µA
difference between the points A
and B will be 6V
The ratio of resistance of diode at I D =15 mA and (a) 6 V (b) 0.6 V
VD = −10 V, is (c) 0.7 V (d) 0 V
(a) 10−3 (b) 10−4 (c) 10−5 (d) 10−6
82. The current through an ideal p-n p n 100 Ω
76. If no external voltage is applied across p-n junction, junction shown in the following
there would be circuit diagram will be 1V 2V
(a) no electric field across the junction (a) zero (b) 1 mA
(b) an electric field pointing from n-type to p-type side (c) 10 mA (d) 30 mA
across the junction
(c) an electric field pointing from p-type to n-type side 83. A potential barrier of 0.3 V exists across a
across the junction p-n junction. If the depletion region is 1µm wide,
(d) a temporary electric field during formation of p-n what is the intensity of electric field in this region?
junction that would subsequently disappear (a) 2 × 105 Vm−1 (b) 3 × 105 Vm−1
77. For the given circuit of p-n R (c) 4 × 105 Vm−1 (d) 5 × 105 Vm−1
junction diode, which of the
following statement is correct? 84. When the voltage drop across a p-n junction diode is
(a) In forward biasing, the voltage increased from 0.65 V to 0.70 V, the change in the
across R isV V diode current is 5 mA. The dynamic resistance of
(b) In forward biasing, the voltage across R is 2V diode is
(c) In reverse biasing, the voltage across R isV (a) 5 Ω (b) 10 Ω (c) 20 Ω (d) 25 Ω
(d) In reverse biasing, the voltage across R is 2V 30 Ω
85. The circuit shown in the figure
78. Which is reverse biased diode? contains two diodes each with a
30 Ω
forward resistance of 30 Ω and
(a) (b) – 20 V with infinite backward resistance. 50 Ω
5V If the battery is 3 V, the current
–10 V
through the 50 Ω resistance (in 3V

ampere) is
(c) 15 V (d) 20 V
(a) 0 (b) 0.01 (c) 0.02 (d) 0.03
10 V –5 V
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 diode input) would be positive. This implies voltage
series with a load, then drop between A and centre tap is half. If a centre tap
(a) a continuous DC voltage appears across load transformer is used with 2 diodes for full-wave
(b) an AC voltage appears across load rectification, then output voltage of rectifier is
(c) a pulsating DC voltage appears across load
(d) no voltage appears across load A D1 Centre tap

87. Which of this is a half-wave rectifier circuit?

B D2
RL
Output

RL Centre tap transformer

(a) (b) RL
(a) 2 × secondary voltage of transformer
AC input
AC input (b) 2/ 3 × secondary voltage of transformer
(c) 1/ 2 × secondary voltage of transformer
(c) RL (d) RL (d) 3 / 2 × secondary voltage of transformer

AC input AC input 91. If two diodes are connected across two ends of
secondary windings of a centre tap transformer as
88. For any practical half-wave rectifier circuit, shown in figure. If inputs at A and B are as shown
(a) the reverse breakdown voltage of diode must be greater VA
than peak AC voltage
(b) the reverse breakdown voltage of diode must be greater t
than rms AC voltage
(c) the reverse breakdown voltage must be greater than VB
mean AC voltage
(d) the reverse breakdown voltage must be smaller than the t
rms AC voltage
89. Input to an half-wave rectifier is given as follows Then, output across load resistance will be
VL
Voltage at input

VL

(a) (c)
t t
t
VL VL

Its output will be (b) (d)

t t
V0 V0
(a) (b) 92. Output of a full-wave rectifier is
t t (a) pure DC voltage
(b) pure AC voltage
V0 V0 (c) pulsating DC voltage
(c) (d) (d) pulsating AC voltage
t t
93. Filters are used alongwith a full-wave rectifier to
(a) remove AC part from the output
90. In the course of rectification of the AC cycle when (b) remove DC part from the output
the voltage at A (upper diode input) becomes negative (c) mix AC and DC
with respect to centre tap, the voltage at B (lower (d) None of the above
94. In the given circuit, 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
AC Rectifier RL DC
C (a) thermionic emission (b) external field emission
(c) internal field emission (d) photoemission

Capacitor C is used 103. Correct circuit using a zener diode as a voltage

(a) for storing potential energy regulator is
(b) as a bypass to DC component to get AC in RL +
(c) to remove sparking Unregulated
Rs
(d) as a bypass to AC component to get DC in RL (a) voltage
Vi Regulated
95. In a full-wave rectifier, input AC current has a RL voltage
– VZ
frequency (ν). The output frequency of current is
(a) ν/ 2 (b) ν +
(c) 2ν (d) None of these Input Regulated
(b) voltage RL voltage
96. In comparison to a half-wave rectifier, the full-wave Vi VZ
rectifier done by centre tapping gives lower –
(a) efficiency (b) average
(c) output voltage (d) None of these +
Unregulated Regulated
97. In the figure alongside, the B (c) voltage
RL
voltage
Vi VZ
input is across the terminals A
and C and the output is across B –
and D. Then, the output is A C +
(a) zero Unregulated
Rs
(b) same as input (d) voltage
Vi Regulated
(c) full-wave rectified D RL voltage
(d) half-wave rectified – VZ

98. In case a single capacitor is connected in parallel with 104. For a zener regulated power supply, a zener diode
a load resistance of R L , it gets discharged through the with zener voltage V z = 6.0 V is used for regulation.
load. The rate of fall of voltage across the capacitor is The load current is to be 4.0 mA and the unregulated
proportional to input 10.0 V. The value of series resistor R s must be,
(a) RL C (b)
C it I Z / I 2 = 5
RL
1 R RS
(c) (d) L
R LC C

99. What is the ratio of output frequencies of full-wave RL VZ

rectifier and a half-wave rectifier, when an input of
frequency 50 Hz is fed at input?
(a) 167 Ω (b) 120 Ω
(a) 1 : 2 (b) 2 : 1 (c) 4 : 1 (d) 1 : 4
(c) 250 Ω (d) 20 Ω
100. A zener diode which is used in reversed biased is
105. Optoelectronic devices are
used as a
(a) CFL’s
(a) voltage regulator (b) voltage rectifier
(b) light based semiconductor diodes
(c) current regulator (d) current rectifier
(c) bulbs
101. A zener diode differs from a p-n junction that (d) discharge tubes
(a) zener diode is made from very lightly doped 106. A photodetector is a
p-n junction
(a) photodiode used for detecting optical signals
(b) zener diode is made from a heavily doped p-n junction
(b) LED’s which are used for detection of infrared signals
(c) zener diode is made from a metal piece
(c) an evacuated tube consisting of a photosensitive cathode
(d) zener diode is made from a heavily doped p-type
semiconductor (d) None of the above
107. A photodiode converts 114. I -V characteristics of a solar cell is best represented by
(a) variation in intensity of light into current amplitude I I
variation VOC VOC
V V
(b) variation of current amplitude into variation in intensity (a) (b)
of emitted light
(c) variation of voltage into variation of current ISC ISC
(d) variation of intensity of light into variation of volume
I I
108. A photodiode in reverse Light VOC
V
VOC
V
biased is irradiated with (c) (d)
light of suitable frequency
and current in circuit is p n ISC ISC
measured.
µA 115. To fabricate solar cell, material used have an energy
Characteristics of diode
gap of
for different illumination R
(a) around 0.7 eV (b) less than 1 eV
intensities I 1 , I 2 , I 3 and
(c) around 1.5 eV (d) less than 0.7 eV
I 4 are drawn as follows.
mA 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
Reverse bias Volts (a) 5200 Å (b) 4400 Å (c) 6200 Å (d) 7500 Å
I1 117. For a photodiode, the conductivity increases when a
I2 wavelength less than 620 nm is incident on it. The
I3 band gap of crystal used to fabricate the diode is
I4 µA (a) 1.12 eV (b) 1.8 eV (c) 2.0 eV (d) 1.62 eV

Greatest intensity is 118. In an LED, when it glows, electron moves from A to

(a) I1 (b) I 2 (c) I 3 (d) I 4 B, when an appropriate bias is applied. A and B are
respectively,
109. When LED is forward biased, then electrons move (a) conduction band, valence band
from n to p and electron-hole combination occurs (b) valence band, conduction band
near junction plane. If E g is energy gap between (c) conduction band, connecting wires
conduction band and valence band, then released (d) connecting wires, conduction band
energy ( E ) due to electron-hole combination will be
(a) E = E g (b) E > E g (c) E ≤ E g (d) E ≥ E g 119. Photodetectors and LED’s are used in
(a) road construction works
110. An LED cannot be used in reverse biased as a voltage (b) optical telecommunication links
regulator because (c) power generation from falling water near dam
(a) reverse breakdown voltage is very low for them
(d) radio transmitters
(b) reverse breakdown voltage is very high for them
(c) they do not breakdown for any voltage 120. Two different semiconductors A and B are used to
(d) None of the above make ‘red’ and ‘violet’ LED’s, respectively. Then,
ratio of energy gaps of semiconductors must be
111. Semiconductors used to fabricate LED to produce
EA EA
visible light must have energy gap E g such that (a) >1 (b) <1
EB EB
(a) 1.1 eV < Eg (b) Eg > 3 eV
(c) 1.8 eV < Eg < 3 eV (d) 1.1 eV < Eg < 2.8 eV (c) E A = EB (d) E A > 3 eV and EB < 1.5 eV

112. Substance used to make red LEDs is 121. LED’s are not used for room lighting (although they
(a) silicon are used for automobile bulbs and in industrial
(b) germanium lighting) because
(c) gallium arsenide phosphide (a) our eyes are not comfortable with very intense light
(d) indium phosphide (b) our eyes are not comfortable with monochromatic light
(c) LED’s are much costlier than bulbs tubelights and
113. A solar cell is CFL’s
(a) photodetector (b) photovoltaic device (d) LED manufacture in mass production will be a very
(c) light emitting diode (d) photogenerator polluting process
122. The I -V characteristics of an LED is [JEE Main 2013]
(a) 1 × 1014 Hz (b) 20 × 1014 Hz
(c) 10 × 1014 Hz (d) 5 × 1014 Hz
RYGB
B
G 124. Identify the semiconductor devices whose
(a) I (b) YR characteristics are as given below in the order (i), (ii),
(iii), (iv). [JEE Main 2016]
O V O V I I
V O
- Red
- Yellow
(i) V (ii) V
I - Green
(c) I (d) R - Blue
Y
G
B
O V I I
Dark Resistance

123. A p-n photodiode is made of a material with a band

(iii) V (iv) V
gap of 2 eV. The minimum frequency of the radiation Intensity
of light
that can be absorbed by the material is nearly Illuminated
(Take hc = 1240 eV-nm)

Topic 5
Junction Transistor
125. A transistor has 128. Let ‘•’ shows an electron and ‘ ’ shows a hole, then
(a) two doped regions forming a large p-n junction which of the following shows correct direction of
(b) three doped regions forming two p-n junctions motion of charge carriers?
(c) two p-n junctions connected by a conducting wire p n p
(d) None of the above
A.
126. For an n-p-n transistor shown below,
+ – + –
n p n
I II III
p n p
B.

Regions marked I, II and III are respectively, + – + –

(a) emitter,collector, base

(b) base, collector, emitter
(c) emitter, base, collector C. n p n
(d) collector, emitter, base – + – +
127. When a transistor is biased as follows.

n p n D. n p n
– + – +
– + – +
(a) A and B (b) B and C (c) A and C (d) B and D
Then, it is said to be in
129. In active state of a transistor, the emitter base junction
(a) solid state
acts as a ...A... resistance and base-collector junction
(b) active state
acts like a ...B... resistance. Here, A and B refer to
(c) inactive state
(a) low, low (b) low, high
(d) passive state
(c) high, low (d) high, high
130. Correct circuit to study input-output characteristics of 135. Input resistance ( ri ) of a transistor in CE
an n-p-n transistor in CE configuration is configuration is
 ∆V   ∆V 
(a)  BE  (b)  CE 
+ R2  ∆I B  VCE  ∆I B  V BE
µA
(a) R1
VBE –  ∆V   ∆V 
+ + (c)  BB  (d)  BC 
VBB – – VCC  ∆I B  V BE  ∆I B  VCE

– + 136. An n-p-n transistor is connected in common-emitter

mA
configuration in a given amplifier. A load resistance
µA + R2 of 800 Ω is connected in the collector circuit and the
(b) R1 VCE voltage drop across it is 0.8V. If the current
VBE –
+
VBB – VCC 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
mA
µA R2 (c) 4, 3.69 (d) 4, 3.84
(c) R1 VCE
VBE
⊥–
137. If β DC for a transistor is
+
VBB – 1+ VCC
∆IC ∆I B IC IB
(a) (b) (c) (d)
– + ∆I B ∆IC IB IC
mA 138. IC (mA)
µA R2
(d) R1 VCE 12
VBE
– ⊥ V 10
VBB + 1 CC
8 60 µA= IB = base current
50 µA
6 40 µA
4.5 30 µA
131. For a transistor, which is correct? 4
3.0 20 µA
(a) VCE = VCB + VBE (b) VBE = VCB + VCE 2 10 µA
(c) VCB = VCE + VBE (d) VCE = VCB − VBE VCE (volts)
2 4 6 8 10 12 14
132. For a silicon base transistor, VCE must be sufficiently
larger than The output characteristics of a typical n-p-n transistor
(a) 21 V (b) 0.7 V in CE configuration are shown. When VCE =10 V and
(c) 0.1 V (d) 20 V I C = 40
. mA, then ratio of β AC and β DC is
133. In an n-p-n transistor in CE configuration, when VCE (a) 1 (b) 2 (c) 3 (d) 4
is increased, then 139. Vo
(a) I B increases and IC increases proportionally
I II III
(b) I B increases and IC remains constant
(c) effect on I B is negligible but IC increases
(d) Both I B and IC remain nearly constant
134. In a common-emitter (CE) amplifier having a voltage
gain G, the transistor used has transconductance 0.6 V
Vi
. mho and current gain 25. If the above transistor is
003 for Si transistor
replaced with another one with transconductance In above transfer characteristics of an n-p-n transistor
0.02 mho and current gain 20, the voltage gain will be in CE configuration; cut-off region, active region,
[NEET 2013] saturation region respectively, are
2 (a) II, III and I
(a) G (b) 15
. G
3 (b) III, I and II
1 5 (c) III, II and I
(c) G (d) G
3 4 (d) I, II and III
140. For an n-p-n transistor in CE configuration, correct 1
T1
graph showing variation of output voltage with 2 Mutual inductance
variation of input voltage is (Coupling through
magnetic field)
Vo Vo 3
C
(a) (b) n-p-n T2
T2′ Output
L
Vi Vi 4

S1 (Switch)
Vo Vo
(b)
(c) (d)
Y Y′
IC IE
Vi Vi
X Z X′ Z′
t t
(i) (ii)
141. For an n-p-n transistor used as amplifier, the power (c)
gainAP is given by (AV = voltage gain)
(a) both IC , I E increase initially
1
(a) A P = (β AC )2 × AV (b) AP = AV (b) both IC , I E decrease but I E decreases
β AC (c) initially IC increases but I E decreases
1 (d) initially IC decreases but I E increases
(c) AP = β AC × AV (d) AP = AV
(β AC )2 145. Refer figure of Q. 144, after maximum collector
142. For a CE transistor amplifier, the audio signal voltage current, there is no further change in collector current,
across collector resistance of 2.0 kΩ is 2.0 V. the magnetic field around T2 ceases to grow. As soon
Suppose the current amplification factor of the as the field becomes static, there will be no further
transistor is 100. What should be the value of R B in feedback from T2 to T1 . Without continued feedback,
series with VBB supply of 2.0 V, if DC base current the ...A... current begins to fall. Consequently, collector
has to be 10 times the signal current? (VBE = 0.6 V) current decreases from Y to Z. However, a decrease of
(a) 14 kΩ (b) 24 kΩ collector current causes the magnetic field to decay
(c) 34 kΩ (d) 44 kΩ around the coil T2 . Thus, T1 is now seeing a ...B... field
143. In an n-p-n transistor, the collector current is 24 mA. in T2 (opposite form what it saw when field was
The possible emitter current (in mA) is growing at the initial start position).
Here, A and B refer to
E B C
n p n (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 µA, then change in
collector current will be
(a) 36 (b) 20
(a) ( 80 × 250 ) µA (b) ( 250 − 80 ) µA
(c) 16 (d) 6
(c) ( 250 + 80 ) µA (d) ( 250 / 80 ) µ A
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. 147. In case of an n-p-n transistor, the E B C
n p n
It can be concluded collector current is always less
than the emitter current because
Input (a) collector side is reverse biased
Transistor Output and emitter side is forward biased
amplifier
(b) after electrons are lost in the base and only remaining
ones reach the emitter back
Feedback (c) collector side is forward biased and emitter side is
network
reverse biased
(a) (d) collector being reverse biased attracts less electrons
148. In a transistor circuit shown here, the base current is 152. For a common-emitter n-p-n transistor following is a
35 µA. true relationship between I B and I C . In active region.
IC IC nt
sta
VCC =constant
c on
(a) (b) E
=
RL VC
Rb
IB IB

9V IC VCE =constant IC
VC
The value of resistance R b is E=
co
(c) (d) ns
(a) 123.5 kΩ ta nt
(b) 257 kΩ IB IB
(c) 380.5 kΩ
(d) cannot be found from given data 153. The current gain for a transistor working as common
base amplifier is 0.96. If the emitter current is
149. For the given circuit,
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
L1 L2
current when base current is 1.2 mA, is
(a) 24 mA (b) 12 mA (c) 6 mA (d) 3 mA
C
155. The input signal given to a CE amplifier having a
Frequency of oscillation is  π
voltage gain of 150 is Vi = 2 cos 15 t +  . The
1 1 1  3
(a) f = (b) f =
2π ( L1 + L 2 ) C 2π ( L1 − L 2 ) C corresponding output signal will be [CBSE AIPMT 2015]
 π  2π 
1 1 (a) 300 cos 15 t +  (b) 75 cos 15 t + 
(c) f = (d) f =  3  3
2π L1 L 2 ⋅ C  L1 + L 2 
2π  C  5π   4π 
 2  (c) 2 cos 15 t +  (d) 300 cos 15 t + 
 3  3
150. For the given circuit, if current amplification factor 156. A transistor has a current gain of 30. If the collector
β = 90 andVBE = 0.7 V resistance is 6 kΩ, input resistance is 1 kΩ, its
voltage gain is
2 kΩ
(a) 90 (b) 180 (c) 45 (d) 360
157. The input resistance of a transistor is 1000 Ω on
RB + + charging its base current by10µA, the collector current
4V – 9V increases by 2 mA. If a load resistance of 5 kΩ is used in
– the circuit, the voltage gain of the amplifier is
3V
(a) 100 (b) 500 (c) 1000 (d) 1500
IB IC
158. In an n- p-n circuit transistor, the collector current is
10 mA. If 80% electrons emitted reach the collector,
Then, base resistance R B is then
(a) 180 kΩ (b) 185 kΩ (a) the emitter current will be 7.5 mA
(c) 82 kΩ (d) 190 kΩ (b) the base current will be 2.5 mA
(c) the base current will be 3.5 mA
151. A circuit containing transistor is such that I B =10µA (d) the emitter current will be 15 mA
and I C = 5 mA, 159. When the voltage drop across a p-n junction diode is
(a) transistor can be used as amplifier with β DC = 10 increased from 0.65 V to 0.70 V, then change in the
(b) transistor can be used as amplifier with β DC = 100 diode current is 5 mA. The dynamic resistance of the
(c) transistor can be used as amplifier with β DC = 250 diode is
(d) transistor cannot be used as amplifier (a) 20 Ω (b) 50 Ω (c) 10 Ω (d) 80 Ω
Topic 6
Digital Electronics and Logic Gates
160. Analog signals are 165. Inputs to a NAND gate are A and B are made as below.
(a) continuous waveforms (b) discrete value signals t1 t2 t3 t4 t5 t6
1
(c) intermittent signals (d) erratic waveforms
A
161. Digital signals are 0
(a) continuous waveforms
1
(b) discrete value signals
B
(c) intermittent signals 0
(d) erratic waveforms
Output of the NAND gate is
162. A NOT gate is called an invertor, because (a)
(a) it produces an output which changes with time
(b) it produces 1 as output when input is 0 and vice-versa (b)
(c) it produces no output for any input
(d) it has only a single input (c)

163. Truth table for an OR gate is (d)

t1 t2 t3 t4 t5 t6
(a) A B Y (b) A B Y
0 0 0 0 0 0 166. Consider the junction diode as ideal. The value of
1 0 0 1 0 1 current flowing through AB is [NEET 2016]
0 1 0 0 1 1
A 1 kΩ B
1 1 1 1 1 0
+4 V –6V
(c) A B Y (d) A B Y (a) 10−2 A (b) 10−1 A
0 0 1 0 0 0 (c) 10−3 A (d) 0 A
1 0 0 1 0 1
0 1 0 0 1 1 167. Inputs A and B are given to show combination of gates
1 1 0 1 1 1
A
A t
164. Truth table for the logic gate whose symbol shown is 1 2 3 4 5 C

B t B
1 2 3 4 5
A
Y
B Then, output C is
(a) A B Y (b) A B Y (a)
0 0 0 0 0 0 t
1 2 3 4 5
1 0 1 1 0 0
0 1 1 0 1 0 (b)
t
1 1 1 1 1 1
(c)
(c) A B Y (d) A B Y t
0 0 1 0 0 1 (d)
1 0 0 1 0 0 t
1 2 3 4 5
0 1 0 0 1 0
1 1 0 1 1 0
168. The output ( X ) of the logic circuit shown in figure 170. In the given figure, a diode D is D 100 Ω

will be [NEET 2013] connected to an external R

A
X resistance R =100 Ω and an emf
B of 3.5 V. If the barrier potential
(a) X = A ⋅ B (b) X = A ⋅ B developed across the diode is
3.5 V
(c) X = A ⋅ B (d) X = A + B 0.5 V, the current in the circuit
will be [CBSE AIPMT 2015]
169. See the circuit shown in the figure. Name the gate (a) 30 mA (b) 40 mA (c) 20 mA (d) 35 mA
that the given circuit resembles.
+5 V 171. To get output 1 for the following circuit, the correct
choice for the input is [NEET 2016]
D1 A
A V0 B Y
C
B
D2 (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
(a) NAND (b) AND (c) OR (d) NOR

Special Format Questions

I. Assertion and Reason 176. Assertion
A X
■ Directions (Q. Nos. 172-176) In the following Y
B
questions, a statement of assertion is followed by a
corresponding statement of reason. Of the following This circuit acts as OR gate.
statements, choose the correct one. Reason Truth table for two input OR gate is
(a) Both Assertion and Reason are correct and Reason is A B Y
the correct explanation of Assertion.
0 0 0
(b) Both Assertion and Reason are correct but Reason is
not the correct explanation of Assertion. 0 1 1
(c) Assertion is correct but Reason is incorrect. 1 0 1
(d) Assertion is incorrect but Reason is correct. 1 1 1

172. Assertion The conductivity of an intrinsic

semiconductor depends on its temperature. II. Statement Based Questions Type I
Reason The conductivity of an intrinsic ■ Directions (Q. Nos. 177-187) In the following
semiconductor is slightly higher than that of a lightly questions, a statement I is followed by a corresponding
doped p-type semiconductor. statement II. Of the following statements, choose the
173. Assertion A zener diode is used to obtain voltage correct one.
regulation. (a) Both Statement I and Statement II are correct
Reason When zener diode is operated in reverse bias and Statement II is the correct explanation of
Statement I.
after a certain voltage (breakdown voltage), the
(b) Both Statement I and Statement II are correct but
current suddenly increases, but potential difference Statement II is not the correct explanation of
across diode remains constant. Statement I.
174. Assertion In a transistor, the base is made thin. (c) Statement I is correct but Statement II is incorrect.
Reason A thin base makes the transistor stable. (d) Statement I is incorrect but Statement II is correct.

175. Assertion In an oscillator, the feedback is in the 177. Statement I An electron on p-side of a p - n
same phase which is called as positive feedback. junction moves to n-side just an instant after drifting
Reason If the feedback voltage is in opposite phase, of charge carriers occurs across junction plane.
the gain is greater than one. 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 187. Statement I A logic gate is a digital circuit.
junction is at positive potential. Statement II They are called gates, because they do
Statement II A p-type semiconductor contains more not allow current through them.
holes than electrons.
Statement Based Questions Type II
179. Statement I The applied voltage (in forward bias of a
p-n junction) mostly drops across the depletion region 188. Semiconductor devices have the advantage over
and the voltage drop across the p-side and n-side of vacuum tubes of
the junction is negligible. I. small size. II. long life and reliable.
Statement II Resistance of depletion region is large III. low power use. IV. low cost.
compared to resistance of n or p-side. Advantages of semiconductor devices are
180. Statement I In forward bias, as voltage increases (a) I, II, III and IV (b) II, III and IV
beyond threshold voltage, the forward current (c) I, III and IV (d) I and IV
increases significantly. 189. Consider four statements.
Statement II By Ohm’s law states V ∝ I. I. Inside crystal, because of difference of position, each
electron has different energy; this makes energy bands
181. Statement I A diode can be used to rectify alternating
in crystals.
voltages.
II. Energy levels of valence electrons are included in
Statement II A p-n junction allows current to pass valence band.
only when it is in reverse bias.
III. Energy level above the valence band is conduction band.
182. Statement I To filter out AC ripple from a given IV. In metals, conduction band and valence band overlap.
pulsating DC voltage, an inductor is connected in Correct statements are
series and a capacitor is connected in parallel with
(a) I and II (b) I, II and IV
load resistance.
(c) II and III (d) I, II, III and IV
Statement II For AC inductor has high reactance and
a capacitor has a low reactance when frequency is high. 190. Due to diffusion of electrons from n to p-side,

183. Statement I A solar cell is made in wafer’s shape I. electron-hole combination across p - n junction occurs.
(of large area). II. an ionised acceptor is left in the p-region.
Statement II By increasing area, work-function of III. an ionised donor is left in the n-region.
electron is decreased. IV. electrons of n-side comes to p-side and electron-hole
combination takes place in p-side.
184. Statement I When a light based p-n junction is
Correct options are
radiated with light of frequency ν such that
(a) I and II (b) II and III
hν > E g , it produces an emf with p side becoming
(c) II and IV (d) II, III and IV
more negative than n-side.
Statement II Junction field separates electron-hole 191. Which of the given statements are correct regarding
pairs generated due to photon absorption and sweep unbiased p-n junction?
them to different regions. I. Drift and diffusion currents occur p to n-side.
II. Initially, diffusion current is large and drift current is
185. Statement I In a typical transistor, I E = I C + I B
small.
⇒ I E ≈ IC . III. Finally, diffusion and drift currents grow to be equal in
Statement II Base region of a transistor is very thin magnitude.
and lightly doped with I B current small, collector IV. Under equilibrium there is no net current across p-n
current I C is large. junction plane.
186. Statement I When an n-p-n transistor in CE (a) I and IV (b) I, II and III
configuration is being used as a switch, it is operated (c) II, III and IV (d) All of these
in cut-off region or in saturation region. 192. Which of these are correct?
Statement II In cut-off region, here Vi is low but Vo I. In forward biasing, holes from p-side crosses junction
is high. In saturation region, here Vi is high but Vo is and reach n-side.
low. II. In forward biasing, electrons from n-side crosses
junction and reach p-side.
 III. In n-side, holes are minority charge carriers. 196. In an oscillator,
IV. In p-side, electrons are minority charge carriers. I. we get AC output without any external input signal.
(a) I, II and III (b) I, III and IV II. output is self sustained.
(c) II, III and IV (d) I, II, III and IV
III. feedback can be achieved by inductive coupling
193. Consider the following statements I and II and (through mutual inductance) or L-C or R-C networks.
identify the correct choice of the given answers. Incorrect statement is
I. The width of the depletion layer in a p-n junction diode (a) Only I
increases in forward biased. (b) Only II
II. In an intrinsic semiconductor, the fermi energy level is (c) Only III
exactly in the middle of the forbidden energy gap. (d) None of these
(a) I is true and II is false (b) Both I and II are false
197. Which of these gates can be formed using a NOR gate?
(c) I is false and II is true (d) Both I and II are true
I. AND II. OR
194. Following of these are circuits used for full-wave
III. NOT IV. NAND
rectification,
(a) I and II (b) II and III
(c) I, II and IV (d) All I, II, III, and IV
I. AC RL
mains
III. Matching Type
198. Before invention of transistor, vacuum tubes were
RL used and these were named according to number of
II. AC
mains electrodes they have.
Now, match these with number of electrodes
Column I Column II
III.
RL A. Pentode 1. 2

B. Tetrode 2. 3

C. Triode 3. 4
RL
D. Diode 4. 5
IV.
A B C D
(a) 1 2 3 4
(a) I, II and III (b) II, III and IV (b) 2 3 4 1
(c) I, III and IV (d) I, II and IV (c) 3 4 1 2
195. For the circuit given, an n-p-n transistor is being used (d) 4 3 2 1
as a switch in CE configuration. 199. Match the elements or compounds with their
RB respective energy gaps values. (Energy gap existing
B C IC
RC
between conduction and valence bands)
IB +
E
IE Vo
Vi – Column I Column II
VBB
VCC
A. Diamond 1. 1.1 eV

B. Aluminium 2. 0.71 eV
Which of the following are correct? C. Germanium 3. 0.03 eV
I. VBB = I B R B + VBE
II. VCE = VCC − I C RC D. Silicon 4. 6 eV

III. Vi = I B R B + VBE A B C D
IV. Vo = VCC − I C RC (a) 1 2 3 4
(a) I, II and IV (b) II, III and IV (b) 2 1 4 3
(c) I, II and III (d) I, II, III and IV (c) 4 3 1 2
(d) 4 3 2 1
200. Following shows a plot of potential difference of 203. Match the inputs of Column I with their respective
n-side and p-side of a p-n junction (battery, in outputs from Column II for a NOR gate
forward biased)
Column I Column II
1 A. 0, 0 1. 0
2 B. 0, 1 2. 1
Vn – Vp

3 C. 1, 0
D. 1, 1

A B C D A B C D
Then, match the following columns. (a) 1 1 2 2 (b) 1 1 1 2
(c) 2 2 2 1 (d) 2 1 1 1
Column I Column II
204. Match the following Column I with Column II.
A. Without battery 1. 1 Column I Column II
B. Low potential battery 2. 2
A. n-p-n transistor 1.
C. High potential battery 3. 3

A B C A B C
(a) 1 2 3 (b) 2 1 3 B. p-n-p transistor 2.
N
(c) 2 3 1 (d) 1 3 2
201. Match the following columns.
E C
Column I Column II C. Light emitting diode 3.

A. Moderate size and heavily doped 1. Base B

B. Very thin and lightly doped 2. Collector E C

D. Zener diode 4.
C. Moderately doped and of large size 3. Emitter

B
A B C
(a) 1 2 3 A B C D A B C D
(b) 1 3 2
(a) 3 4 2 1 (b) 4 2 1 3
(c) 3 1 2
(c) 2 4 3 1 (d) 4 3 2 1
(d) 3 2 1
202. Match the following symbols with their names. IV. Passage Based Questions
Column I Column II ■ Directions (Q. Nos. 205-209) These questions are
A. 1. OR based on the following situation. Choose the correct
options from those given below.
2. AND
B. The input and output resistances in a common-base
amplifier circuits are 400 Ω and 400 kΩ, respectively.
3. NAND The emitter current is 2 mA and current gain is 0.98.
C.
205. The collector current is
4. NOR
D. (a) 1.84 mA (b) 1.96 mA
(c) 1.2 mA (d) 2.04 mA
5. NOT
E. 206. The base current is
(a) 0.012 mA (b) 0.022 mA
A B C D E (c) 0.032 mA (d) 0.042 mA
(a) 1 2 3 4 5
(b) 3 1 2 4 5 207. Voltage gain of transistor is
(c) 5 1 2 5 4 (a) 960 (b) 970
(d) 5 1 2 3 4 (c) 980 (d) 990
208. Power gain of transistor is 214. Which of the following statement concerning the
(a) 950 (b) 960 (c) 970 (d) 980 depletion zone of an unbiased p-n junction is (are) true?
(a) The width of the zone is independent on the densities of
209. If peak voltage of input AC source is 0.1 V. The peak the dopants (impurities)
voltage of the output will be (b) The width of the zone is dependent on the densities of
(a) 9.8 V (b) 98 V (c) 980 V (d) 970 V the dopants
■ Directions (Q. Nos. 210-211) These questions are (c) The electric field in the zone is produced by the ionized
based on the following situation. Choose the correct dopant atoms
options from those given below. (d) The electric field in the zone is produced by the electrons
in the conduction band and the holes in the valence band
Three ideal p - n junction diodes D1 , D 2 and D3 are
connected as shown in the circuit. The potentials V A 215. The impurity atoms with which pure silicon should be
and V B can be changed. doped to make a p-type semiconductor, are those of
(a) phosphorus (b) boron
D1 R
(c) antimony (d) aluminium
D2 R 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
D3 R
maximum power rating of 100 milliwatt. The value of
R/4 R/4 maximum current in the circuit is I when voltage
across resistance R is VR and the value of resistance in
VA VB
R, thus which of the following are correct?
R D
i
210. If V A is kept at −10 V and VB at −5 V, the effective
resistance between A and B becomes
(a) R (b) R / 2 (c) 3R (d) 3R / 2
211. If V A = −5 V and VB = −10V, then the resistance
1.5 V
between A and B will be
(a) R (b) R / 2 (c) 3R (d) 3R / 2 (a) I = 200 mA,VR = 1 V (b) I = 200 mA, R = 5 Ω
(c) I = 100 mA,VR = 2 V (d) I = 100 mA, R = 10 Ω
V. More than One Option Correct 217. In an n- p- n transistor circuit, the collector current is
212. If the lattice constant of this semiconductor is 10 mA. If 90% of the electrons emitted reach the
decreased, then which of the following are incorrect? collector
(a) the emitter current will be 9 mA
Conduction
bandwidth Ec (b) the base current will be 1 mA
(c) the emitter current will be 11 mA
Band gap (d) the base current will be −1mA
Eg

Valence 218. Consider the following circuit. It VBE = 0.7 V

Ev
bandwidth

(a) All Ec , Eg , Ev increase

(b) Ec and Ev increase, but Eg decrease 2 kΩ
(c) Ec and Ev decrease, but Eg increase
(d) All Ec , Eg , Ev decrease RB +
4V 9V
213. Choose correct option(s) from the following.
(a) Substances with energy gap of the order of 10 eV are
–
3V
insulators
(b) The conductivity of a semiconductor increases with
increase in temperature
(c) In conductors the valence and conduction bands may
overlap and β = 90, then which of the following are correct?
(d) The resistivity of a semiconductor increases with (a) IC = 2.5 mA (b) I B = 27.8 mA
increase in temperature (c) RB = 82 kΩ (d) IC = 12
. mA
 NCERT & NCERT Exemplar Questions
NCERT 225. For a transistor amplifier, the voltage gain
219. When a forward bias is applied to a p- n junction. It (a) remains constant for all frequencies
(a) raises the potential barrier (b) is high at high and low frequencies and constant in the
(b) reduces the majority carrier current to zero middle frequency range
(c) lowers the potential barrier (c) is low at high and low frequencies and constant at mid
(d) None of the above frequencies
(d) None of the above
220. For transistor action, which of the following statements
are correct? 226. For a CE-transistor amplifier, the audio signal voltage
(a) Base, emitter and collector regions should have similar across the collector resistance of 2 kΩ is 2 V. Suppose
size and doping concentrations the current amplification factor of the transistor is
(b) The base region must be very thin and lightly doped 100. Find the input signal voltage and base current, if
(c) The emitter junction is forward biased and collector the base resistance is 1kΩ
junction is reverse biased (a) Vin = 1V, I B = 5 µA (b) Vin = 0.01V, I B = 10 µA
(d) Both the emitter junction as well as the collector (c) Vin = 15
. V, I B = 5 µA (d) Vin = 13
. V, I B = 10 µA
junction are forward biased
221. In an n-type silicon, which of the following 227. Two amplifiers are connected one after the other in
statements is true? series (cascaded). The first amplifier has a voltage gain
(a) Electrons are majority charge carriers and trivalent of 10 and the second has a voltage gain of 20. If the
atoms are the dopants input signal is 0.01 V, calculate the output AC signal.
(b) Electrons are minority charge carriers and pentavalent (a) 2 V (b) 3 V (c) 4 V (d) 5 V
atoms are the dopants 228. In an intrinsic semiconductor, the energy gap E g is
(c) Holes are minority charge carriers and pentavalent
atoms are the dopants
1.2 eV. Its hole mobility is much smaller than electron
(d) Holes are majority charge carriers and trivalent atoms mobility and independent of temperature. What is the
are the dopants ratio between conductivity at 600 K and that at
300 K?
222. Carbon, silicon and germanium have four valence
electrons each. These are characterised by valence Assume that the temperature dependence of intrinsic
and conduction bands separated by energy band gap carrier concentration ni is given by
respectively equal to ( E g )C , ( E g )Si and ( E g )Ge .  Eg 
Which of the following statement is true? ni = no exp −  
 2k B T 
(a) ( Eg ) Si < ( Eg ) Ge < ( Eg ) C (b) ( Eg ) C < ( Eg ) Ge > ( Eg ) Si
where no is a constant.
(c) ( Eg ) C > ( Eg ) Si > ( Eg ) Ge (d) ( Eg ) C = ( Eg ) Si = ( Eg ) Ge
(a) 0.5 × 102 (b) 2.1 × 103
223. In an unbiased p- n junction, holes diffuse from the
(c) 1.1 × 105 (d) 3.2 × 104
p-region to n-region because
(a) free electrons in the n-region attract them 229. In a p-n junction diode, the current I can be expressed as
(b) they move across the junction by the potential difference eV
(c) hole concentration in p-region is more as compared to
I = I 0 exp −1
kBT
n-region
(d) All of the above where, I 0 is called the reverse saturation current. V is
the voltage across the diode and is positive for
224. Identify the logical operations carried out by the two
forward bias and negative for reverse bias and I is
circuits given respectively are
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
(A) (B) of 0.6 V?
(a) 0.063A (b) 0.832A
(a) A-AND, B-NOT (b) A-AND, B-OR (c) 0.0763A (d) None of these
(c) A-NAND, B-NOT (d) A-NOT, B-OR
230. A hole is 236. In the depletion region of a diode,
(a) an anti-particle of electron I. there are no mobile charges.
(b) a vacancy created when an electron leaves a covalent bond II. equal number of holes and electrons exist making the
(c) absence of free electrons region neutral.
(d) an artificially created (particle) III. recombination of electron and holes takes place.
IV. immobile charged ions exists.
NCERT Exemplar The correct options are
231. Conductivity of a semiconductor increases with (a) I, II and III (b) II, III and IV
increase in temperature, because (c) I, II and IV (d) I, III and IV
(a) number density of free charge carriers increases
237. To reduce the ripples in a rectifier circuit with
(b) relaxation time increases
capacitor filter, which of these must be done?
(c) both number density of free charge carriers and
relaxation time increases I. R L should be increased.
(d) number density of free charge carriers increases II. Input frequency should be decreased.
relaxation time decreases but effect of decrease in III. Input frequency should be increased.
relaxation time is much less than increase in number IV. Capacitors with high capacitance should be used.
density
(a) I, II and III (b) I, III and IV
232. Assuming diodes to be ideal, (c) II, III and IV (d) I, II and IV
A 238. What happens during regulation action of a zener diode?
–10 V R D1 D2 B
I. The current and voltage across the zener remains
(a) D1 is forward biased and D2 is reverse biased, so constant.
current flows from A to B II. The current through series resistance ( R s ) changes.
(b) D2 is in forward bias and D1 is in reverse bias and III. The zener resistance is constant.
hence no current flows from B to A and vice-versa IV. The resistance offered by zener changes.
(c) D1 and D2 both are in forward bias, so current flows Correct ones are
from A to B
(a) I and IV (b) II and III (c) II and IV (d) I and II
(d) D1 and D2 both are in reverse bias, so no current flows
from A to B 239. The breakdown in a reverse biased p-n junction diode
233. A 220 V AC supply is connected between points A is more likely to occurs due to
and B. I. large velocity of minority charge carriers, if the doping
A is small.
II. large velocity of minority charge carriers, if the doping
~220 V AC C
is large.
B III. strong electric field in depletion region, if the doping
concentration is small.
What will be potential difference across capacitor C?
IV. strong electric field in depletion region, if the doping
(a) 220 V (b) 110 V (c) 0 V (d) 220 2 V concentration is large.
234. Output of the circuit shown below Correct ones are
Vm sin ωt

will be (a) I and IV (b) II and III

(a) zero all the times (c) I and III (d) II and IV
(b) like half wave rectifier with 240. Consider an n-p-n transistor with its base-emitter
positive cycles in output
junction forward biased and collector base junction
(c) like half wave rectifier with negative cycles in output
reverse biased, which of these are correct?
(d) like that of a full wave rectifier
I. Electrons cross over from emitter to collector.
235. In the circuit shown, the voltage difference between A II. Holes move from base to collector.
and B it the diode forward voltage drop is 0.3V
III. Electrons move from emitter to base.
IV. Electrons from emitter move out of base without going
to collector.
0.2 mA 5k Ω 5k Ω
(a) I and III (b) I and II
A B
(c) I and IV (d) II and III
(a) 1.3 V (b) 2.3 V (c) 0 (d) 0.5 V
241. Base biased CE transistor has the transfer (c) A B Y (d) A B Y
characteristics as shown 0 0 0 0 0 0
0 1 1 0 1 1
Vo
1 0 0 1 0 1
1 1 1 1 1 0

243. In an n-p-n transistor in CE configuration,

Vi
I C =10 mA. If 95% of electrons from emitter reaches
0 0.6 V 2V the collector, which of these are correct?
Which of the following statements are correct? I. I E = 8 mA II. I E = 10.53 mA
I. At Vi = 0.4 V, transistor is in active state. III. I B = 0.53 mA IV. I B = 2 mA
II. At Vi = 1 V, transistor can be used as an amplifier. (a) I and IV (b) II and IV
III. At Vi = 0.5V, it can be used as a switch turned off. (c) II and III (d) I and III
IV. At Vi = 2.5V, it can be used as a switch turned on. 244. When an electric field is applied across a
(a) I, II and III (b) II, III and IV semiconductor, then
(c) I, II and IV (d) I, III and IV I. electrons move from lower energy level to higher
242. For given circuit, truth table is energy level in the conduction band.
II. electrons move from higher energy level to lower
A energy level in the conduction band.
III. holes in the valence band move from higher energy
Y
level to lower energy level.
IV. holes in the valence band move from lower energy
B level to higher energy level.
Out of these, correct statements are
(a) A B Y (b) A B Y (a) I and III
0 0 1 0 0 1 (b) I and II
0 1 0 0 1 0 (c) II and IV
1 0 1 1 0 0 (d) I and IV
1 1 0 1 1 1

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, 213. (a,b, 214. (b,c 215. (b,d 216. (a,b 217. (b,c 218. (a,b, 219. (c)
220. (b,c 221. (c) 222. (c) 223. (c) 224. (b) 225. (c)
d) c) ) ) ) ) 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 9. (c) Energy level splits into more finer levels and for many
cathode and the controlled flow of these electrons in vacuum atoms they form nearly continuous bands.
is obtained by varying the voltage between its different 10. (d) For a semiconductor, 0 < Eg ≤ 3 eV
electrodes.
Vacuum is required in the inter-electrode space; otherwise For metal Eg ~ 0 and for insulator Eg > 3 eV.
the moving electrons may lose their energy on collision with Forbidden energy gap in semiconductor is maximum energy
the air molecules in their path. gap allowed.
2. (d) The supply and flow of charge carriers in the 11. (d) For a semiconductor, conductivity increases with
semiconductor devices are within the solid itself. temperature or more electron-hole pairs are created due to
No external heating or large evacuated space is required by thermal agitation.
the semiconductor devices, so they have small sizes. 1 1
So, ρ∝ ⇒ σ ∝T ⇒ ∝T
3. (d) A van der Waals solid transmits light and has a low T ρ
melting point. 12. (b) In good conductors, which are metals there is no gap
4. (d) Semiconductors have 4 valency and so they form between valence band and conduction band. Hence, no holes
exist.
covalent bonds.
13. (c) In a conductor, uppermost band is occupied by
5. (c) The SI unit of conductivity is Siemen per metre (Sm−1 ).
conduction electrons. Uppermost band is conduction band.
6. (b) hc 6.6 × 10−34 × 3 × 108
14. (c) Using E = Eg = = J
(i) Metal They possess very low resistivity (or high λ 580 × 10−9
conductivity).
6.6 × 10−34 × 3 × 108
ρ ~ 10−2 − 10−8 Ω -m = eV = 2.1eV
2 8 −1
580 × 10−9 × 1.6 × 10−19
σ ~ 10 − 10 Sm
15. (a) In semiconductor as the temperature increases, more
(ii) Semiconductor They have resistivity or conductivity thermal energy becomes available to electrons and some of
intermediate to metal and insulator. these electrons may break away (becoming free electrons
ρ ~ 10−5 − 106 Ω-m contributing to conduction).
σ ~ 105 − 10−6 Sm−1 The thermal energy effectively ionises only a few atoms in
the crystalline lattice and creates a vacancy in the bond .
(iii) Insulator They have high resistivity (or low These holes behave as positive charge carriers.
conductivity).
16. (a) Pure semiconductors are called intrinsic semiconductors
ρ ~ 1011 − 1019 Ω-m
n e = n n = n i.
σ ~ 10−11 − 10−19 Sm−1 17. (d) Total current is the sum of hole current and electron
As, 108 > 105 > 10−19 ,σ metal > σ semiconductor > σ insulator current in semiconductor. Electrons move opposite to hole,
but current in same direction.
7. (a) Due to atomic interactions, the energies of outermost
I = Ie + Ih
electrons are changed in larger amounts.
18. (d)
8. (a) Empty 4 N states
Ec Ec Ec
Intrinsic
Eg Eg
Electron energy

At 0 K, CB of SC semiconductor

Eg remain empty
Ev Ev

Ev
Filled 4 N states T=0 K (a) (b) T>0 K

Infinitely large number of 19. (c) In equilibrium condition, there is no net current through
states each occupied by the semiconductor. This shows that the rate of generation of
two electrons at 0 K electron-hole pairs is equal to rate of recombination of
electron-hole pairs.
20. (a) An intrinsic semiconductor will behave like an insulator 29. (c) The number of electrons made available for conduction
at T = 0 K. It is the thermal energy at higher temperature by dopant atoms depends strongly upon the doping level
(T > 0 K), which excites some electrons from the valence and is independent of any increase in ambient temperature.
band to the conduction band. These thermally excited 30. (d) Gallium, boron and aluminium all are trivalent impurities.
electrons at T > 0 K, partially occupy the conduction band. These impurities make germanium p-type semiconductor.
These have come from the valence band leaving equal
number of holes there. 31. (d) In a p -type semiconductor, holes are majority charge
21. (c) Number density of electron-hole pairs is increased with
carriers and electrons are minority charge carriers.
temperature, so at high temperature semiconductors have 32. (b) In both n-type and p-type semiconductors, number of
higher conductivity. electrons is exactly equal to number of protons. Both are
neutrals.
22. (c) Semiconductors have negative temperature coefficient of
resistance i.e., the resistance of a semiconductor decreases 33. (c) At room temperature, the density of holes in the valance
with the increase in temperature and vice-versa . Silicon is band is predominantly due to impurity in the extrinsic
actually an insulator at absolute zero of temperature but it semiconductor. The electron and hole concentration in a
becomes a good conductor at high temperatures. Because on semiconductor in thermal equilibrium is given by
increasing temperatures of semiconductor some of the ne nh = ni2
electrons jumps from valence band to conduction band.
34. (b) The four bonding electrons of C, Si or Ge lie,
23. (a) Resistivity of a metal is directly proportional to respectively, in the second, third and fourth orbit. Hence,
temperature because its temperature coefficient is positive energy required to take out an electron from these atoms
and resistivity of semiconductor is inversely proportional to (i.e., ionisation energy Eg ) will be least for Ge, followed by
temperature due to its negative temperature coefficient. This Si and highest for C.
implies that with decrease in temperature, resistivity of
metal decreases while that of semiconductor increases. 35. (a) Total number of atoms = 5 × 1028 m–3
So, resistivity of Si increases but that of Cu decreases. For every 106 atoms, 1 A-s is doped.
24. (b) For visible region, 450 ≤ λ ≤ 750 nm 1 A-s contributely 1e–

Hence, photon energy ranges from few 1.7 to 2.8 eV. Ne = 5 × 1028 / 106 = 5 × 1022
 hc Since, ne nh = ni2
 using hc = 1240 eV- nm, E = 
 λ The number of holes
As for silicon most of the photons have a higher energy, so nh = (2.25 × 1032 ) / (5 × 1022 ) = 4.5 × 109 m−3
they excite electrons (electrons absorb photons) Hence, light
cannot pass through silicon. Silicon is opaque. 36. (a) Phosphorous is pentavalent, it donates electron.
But as photons are not absorbed in ZnS, they pass through it ∴ nh << ne
and so zinc sulphide is transparent. 37. (a) Increase of temperature causes more electrons to leave
valance band and reach conduction band, so ne increases.
25. (a) Process of adding an impurity is called doping.
But increase of temperature causes lattice vibrations and so
26. (b) Doping increases concentration of majority charge vd decreases as number of collisions increases.
carriers.
38. (c) Doping does not change energy gap, it is still around 1 eV.
27. (c) To form an n-type semiconductor doping is done by
using a pentavalent impurity like phosphorous and to form a 39. (a) Eg , Si = 1.1 eV
p-type semiconductor doping is done by using a trivalent Eg , Ge = 0.7 eV ⇒ Eg , Sn = 0.45 eV
impurity like indium.
40. (c) In intrinsic semiconductor, conductivity occurs due to
28. (a) excitation of electrons when they absorb energy and break
covalent bonds.
Ec
ED
41. (c) For Ge, the energy gap, Eg is 0.7 eV.
≈ 0.01 eV
Electron energy

42. (a) ni2 = ne nh , for extrinsic semiconductor

Eg
1038
⇒ (1019 )2 = ne (10 )21 ⇒ ne = = 1017 m−3
Ev 1021
43. (c) Conductivity of a semiconductor increases with
temperature.
So, they have negative value of temperature coefficient α.
Donor energy level is slightly less than energy level lowest
to conduction band. 44. (d) Relation is found empirically it is, n ∝ T 3 / 2 .
45. (d) Dopant concentration is usually 53. (c) When a hole diffuses from p → n due to the concentration
1 to 10 ppm (parts per million) gradient. It leaves behind an ionised acceptor (negative
charge) which is immobile. As the holes continue to diffuse,
1
⇒ ≈ 10−7 a layer of negative charge (or negative space-charge region)
107 on the p-side of the junction is developed. Similarly, a layer
46. (d) X shows an undoped semiconductor, of positive space-charge is developed on n-side because of
Y shows an n-type semiconductor, electron departure. This space-charge region on either side
of the junction together is known as depletion region.
Z shows a p-type semiconductor.
54. (a) Thickness of depletion region is around 10−7 m.
47. (c) The n-type semiconductor can be produced by doping an
impurity atom of valence 5, i.e., pentavalent atoms. 55. (c) p -side is at negative potential and n-side is at positive
i.e., phosphorous. potential. Also, central layer is at zero potential. Potential at
small distance from junction on both sides becomes constant.
48. (c) In an n-type silicon, dopants are pentavalent atoms,
electrons are majority charge carriers and holes are minority 56. (c) Potential tends to prevent the movement of electron from
charge carriers. the n-region into the p-region, it is often called barrier
potential.
49. (a) N = 5 × 1028 atoms / m3 , A = acceptor, indium,
57. (b) No, A p-n junction is formed when a p-type
D = donor, arsenic semiconductor is formed alongwith an n-type semiconductor
N A = 0.05 × 1022 atoms / m3 on a single intrinsic semiconductor.
⇒ N D = 5 × 1022 atoms / m3 58. (b) It is caused by diffusion of charge carriers.

ni = 1.5 × 1016 m−2 60. (a) E opposes movement of charge.

⇒ N D − N A = ne − nh and ne nh = ni2 61. (c) Diffusion and drift current of electrons and holes is due
to concentration difference.
n2
∴ ND − N A = ne − i 62. (c) Using
dV
= E, we get
ne dr
⇒ ne2 − ( N D − N A ) ne − ni2 = 0 V = Ed
(N D − N A ) + (N D − N A ) + 2
4 ni2 V 0.6 V
or E= = = 6 × 105 Vm−1
⇒ ne = d 1 × 10−6 m
2
On substituting values, we get, ne = 4 .95 × 1022 / m3 63. (c) A semiconductor diode is basically a p-n junction with
metallic contacts provided at the ends for the application of an
n 2 (1.5 × 1016 )2 external voltage. It is a two terminal device.
∴ nh = i = = 4.54 × 109 m−3
ne 4.95 × 1022 64. (c) The direction of arrow indicates the conventional
1022 direction of current (when the diode is under forward bias).
Observe that doping level = 10−6 ppm nearly
1028 65. (d) In forward biasing, electrons from n-side cross depletion
region and reach p-side.
50. (c) Number of Si atoms = 5 × 1028 atoms / m 3
66. (b) Effective potential barrier will be (V0 − V ). i.e., potential
Number of indium atoms = Number of indium atoms for 1
silicon atom × Total number of Si atoms barrier decreases.
5 × 1028 67. (c) Due to concentration gradient, the injected electrons on
= = 1 × 1021 atoms / m3 = 1 × 1015 atoms /cm3 p-side diffuse from the junction edge of p-side to the other
5 × 107
end of p-side. Likewise, the injected holes on n-side diffuse
51. (b) Using ne × nh = ni2 from the junction edge of n-side to the other end of n-side.
here ni = 2 × 1016 m−3 This motion of charged carriers on either side gives rise to
current. The total diode forward current is sum of hole
nh = 4.5 × 1022 m−3 diffusion current and conventional current due to electron
ni2 ( 2 × 1016 )2 diffusion.
∴ ne = = 68. (a) In RB potential barrier increases hence movement of
nh 4.5 × 1022
majority carrier decreases. But strong E pushes the
ne = 8.89 × 109 m−3 movement of minority carrier towards their respective side
and contributes small current.
52. (d) Consider a thin p-type silicon (p-Si) semiconductor
wafer. By adding precisely a small quantity of pentavalent 69. (a) The diode reverse current is not very much dependent on
impurity, part of the p-Si wafer can be converted into n-Si. the applied voltage. Even a small voltage is sufficient to
So, p-n junction is formed. 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 76. (b) Without an external bias an electric field exists which
voltage but is limited due to the concentration of the points from n to p-side and opposes any diffusion of electrons.
minority charge carrier on either side of the junction.
77. (a) In forward biasing a negligible potential drop occurs in
70. (a) The current under reverse bias is essentially voltage diode, so potential drop across resistance R is V.
independent upto a critical reverse bias voltage, known as
breakdown voltage (V br ). When V = V brr , the diode reverse 78. (b) In reverse bias, V p -side − Vn -side = Negative
current increases sharply from n to p side.
79. (b) Diode is in forward bias , so resistance = 0
Even a slight increase in the bias voltage causes large change
in the current. If the reverse current is not limited by an V 4−1 3
So, I= = = = 10−2 A
external circuit, the p-n junction will get destroyed. Once it R 300 300
exceeds the rated value, the diode gets destroyed due to
80. (d) Arsenic is pentavalent, X is n-type and indium is
overheating.
trivalent and Y is p - type.
71. (a) TypicalV-I characteristics of a silicon diode are as shown. So, the junction is in reverse bias.
I (mA)
81. (a) Diode is in reverse bias, current = 0, potential difference
100 across R = 0;V AB = 6 V
80 82. (a) Current is zero as batteries cause p-n junction in reverse
60 bias.
40 V 0.3
83. (b) Electric field, E = = = 3 × 105 Vm−1
20 d 1 × 10−6
100 80 60 40 20
0.2 0.4 0.6 0.8 1.0
V (V) ∆V
–10 84. (b) Dynamic resistance is, rd =
Vb ∆I
–20
here, ∆V = 0.7V − 0.65V = 0.05 V
–30
I (µA) ∆I = 5 mA = 5 × 10−3 A
72. (a) In forward bias, the current first increases very slowly, 0.05
∴ rd = = 10 Ω
almost negligibly till the voltage across the diode crosses a 5 × 10−3
certain value. 30 Ω
85. (c) In the circuit, the upper diode 70 Ω
After the characteristic voltage, the diode current increases Di is reverse biased and lower
significantly (exponentially), even for a very small increase diode D2 is forward biased. Thus,
in the diode bias voltage. This voltage is called the threshold there will be no current across
voltage or knee voltage (~ 0.2 Vfor germanium diode and upper diode function. The +– 50 Ω
~ 0.7 Vfor silicon diode). effective circuit will be shown as
3V
73. (a) By allowing current only in forward bias it acts like a Total resistance,
one way valve. R = 50 + 30 + 70 = 150 Ω
74. (a) A p - n junction diode primarily allows the flow of V 3V
current only in one direction (forward bias). The forward Current in circuit, I = = = 0.02 A
R 150 Ω
bias resistance is low as compared to the reverse bias
resistance. For diodes, we define a quantity called dynamic 86. (c) If an alternating voltage is applied across a diode in
resistance as the ratio of small change in voltage ∆V to a series with a load, a pulsating voltage will appear across the
small change in current ∆I, load only during the half cycles of the AC input during
∆V which the diode is forward biased.
rd =
∆I 87. (b) In a half-wave rectifier, the secondary of a transformer
75. (d) From the graph, at I = 20 mA supplies the desired AC voltage across terminals A and B.
When the voltage at A is positive, the diode is forward
V = 0.8 V biased and it conducts. When A is negative, the diode is
and I = 10 mA,V = 0.7 V reverse biased and it does not conduct.
∆V 0.1 V The reverse saturation current of a diode is negligible and
rFB = = = 10 Ω
∆I 10 mA can be considered equal to zero for practical purposes.
Also, at V = − 10 V, I = − 1 µA Transformer
10 V A
rRB = = 1 × 107 Ω
1 µA
Primary Secondary RL
r 10
∴ Ratio = FB = 7 = 10−6
rRB 10 B
88. (a) The reverse breakdown voltage of the diode must be 1
94. (d) As, X C = , for AC component when ω is high, then
sufficiently higher than the peak AC voltage at the Cω
secondary of the transformer to protect the diode from X C is less and, so a capacitor let AC part bypass through it,
reverse breakdown. so only DC part reaches RL , the load resistance.
89. (d) X DC component
Voltage at A

IL
+
AC Rectifier – C RL DC
Input AC
t Y
AC component
bypass
across RL
Voltage

Output voltage Fig. (a) A full-wave rectifier with capacitor filter

AC
t input
Input AC voltage and output waveforms from
the half-wave rectifier circuit Output with
capacitor
90. (c) Due to centre tapping potential reaching the diodes is only input filter
half of secondary voltage. It is clear from its circuit’s diagram. Fig. (b) Input and output voltage of rectifier in (a)
Centre tap 95. (c) In full wave rectification, output frequency is double of
transformer Diode 1(D1) that of input frequency.
A
96. (c) In this question, full wave rectification is done by using a
Centre X centre tap transformer.
tap
B 1
So, output voltage is that of an half wave rectifier.
Diode 2(D2) RL Output 2
+ • v/ 2 v

Y  diff = 
centre • 0 2
 diff = v
91. (d) v 
− • − v / 2}diff =
Waveform Waveform

2 
at A

t
97. (c) During Ist half of cycle,
(i) B
at B

(ii)
(a) C RL
A
–
–
(across RL)

Due Due Due Due

waveform
Output

to to to to
D1 D2 D1 D2
(b) 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
During IInd half of cycle,
92. (c) The rectified voltage is in the form of pulses of the shape Clearly, current through RL is unidirectional, in both halves
of half sinusoids. Though, it is unidirectional, it does not of input AC.
have a steady value. B
93. (a) To get steady DC output from the pulsating voltage
normally, a capacitor is connected across the output terminals
(parallel to the load RL ).
– C
One can also use an inductor in series with RL for the same A RL
purpose. Since, these additional circuits appear to filter out
the AC ripple and give a pure DC voltage, so they are called
filters. + D
 98. (c) Rate of discharge is inversely proportional to time 107. (a) The variation of intensity results in change in number of
constant of the circuit. More value of time constant implies incident photons (per second) and hence a corresponding
slow discharge. change in generation rate of electron and holes occurs. This
99. (b) For full-wave rectifier, frequency = 2 × input frequency causes a change in current amplitude.
for half-wave rectifier frequency = input frequency. 108. (d) When intensity is increased, reverse saturation current
2 also increases.
∴ Ratio = mA
1
100. (a) Zener diode is a special purpose semiconductor diode.
It is designed to operate under reverse bias in the breakdown Reverse bias
region and used as a voltage regulator. The symbol for Volts
I1
Zener diode is shown in figure. I2
I3
I4
I4>I3>I2>I1 µA
101. (b) Zener diode is fabricated by heavily doping both p-sides
and n-sides of the junction. Due to this, depletion I-V characteristics of a photodiode for different illumination
region formed is very thin (< 10−6 m) and the electric intensity I 4 > I 3 > I 2 > I1 .
field of the junction is extremely high (~ 5 ×106 V/ m) 109. (c) When the diode is forward biased, electrons are sent
even for a small reverse bias voltage. from n → p.
Thus, at the junction boundary on either side of the junction,
102. (c) When the reverse bias voltage V = V z , then the electric
excess minority charge carriers are there which recombins
field strength is high enough to pull valence electrons from with majority charge carriers near the junction. On
the host atoms on the p-side which are accelerated to n-side. recombination, the energy equal to or slightly less than the
These electrons account for high current observed at the band gap are emitted.
breakdown. The emission of electrons from the host atoms
110. (a) The reverse breakdown voltages of LEDs are very low.
due to the high electric field is known as internal field
typically around 5 V. So, care should be taken that high
emission or field ionisation.
reverse voltages do not appear across them.
103. (d) Zener diode must be attached in reverse bias
111. (c) As, emitted energy = Band gap value when an electron
+ moves from conduction band to valance band.
Rs If λ is wavelength of emitted radiation, then
Unregulated hc
voltage (Vi) IL Eg =
Load Regulated λ
RL voltage (Vz) hc
– ⇒ λ=
Eg
Zener diode as DC voltage regulator
Taking hc = 1240 eV- nm and 450 nm ≤ λ ≤ 750 nm for
104. (a) The value of Rs should be such that the current through visible region, we get
the Zener diode is much larger than the load current. This is 450 ≤ λ ≤ 750
to have good load regulation. I z = 20 mA. The total current hc
through Rs is, therefore, 24 mA. The voltage drop across Rs ⇒ 450 ≤ ≤ 750
Eg
is 10.0 − 6.0 = 4.0 V. This gives
Rs = 4.0 V / (24 × 10−3 ) A = 167 Ω ⇒ Eg ≤
hc 1240 eV - nm
= ≈ 2.8 eV
105. (b) Semiconductor diodes in which charge carriers are 450 450 nm
generated by photons (photo-excitation) are called hc
and Eg ≥
optoelectronic devices. Optoelectronic devices are 750
(i) photodiodes used for detecting optical signal 1240 eV nm
= ≈ 1.7 eV
(photodetectors). Used near automatic doors. 750 nm
(ii) light Emitting Diodes (LED) which convert electrical So, 1.8 ≤ Eg ≤ 3 eV
energy into light.
112. (c) For red LEDs,
(iii) photovoltaic devices which convert optical radiation into
electricity (solar cells). GaAs 0.6 P0.4 – Gallium Arsenic Phosphide (Eg = 1.9 eV) is
used. This corresponds to λ ≈ 700 nm.
106. (a) A photodetector detects any change in intensity of light 113. (b) Solar cells uses light energy (photons) to generate an emf.
by changing either potential difference across it or by
changing current through it.
114. (d) Solar cell supplies current to load. So, I-V characteristics 122. (a) For same value of current, higher value of voltage is
is drawn in fourth quadrant. required for higher frequency.
IL 123. (d) Here, Eg = 2 eV
Wavelength of radiation corresponding to this energy,
hc 1240 eV-nm
λ= = = 620 nm
p n Eg 2eV
c 3 × 108 ms −1
Frequency, ν = = = 5 × 1014 Hz
Fig. (a) Depletion region λ 620 × 10−9 m
I 125. (b) A transistor has three doped regions forming two
VOC (open circuit voltage) p-n junctions between them. Obviously, there are two types
of transistors.
V
(i) n-p-n transistor (ii) p-n-p transistor
ISC 126. (c) n-p-n transistor Here, two segments of n-type
Short circuit current
semiconductor (emitter and collector) are separated by a
Fig. (b) I-V characteristics of a solar cell segment of p-type semiconductor (base).
115. (c) Semiconductors with band gap closed to 1.5 eV are ideal Emitter Collector
materials for solar cell fabrication. Solar cells are made with
semiconductors like Si (Eg = 1.1 eV),
GaAs (Eg = 1.43 eV). CdTe (Eg = 1.45 eV),
CuInSe 2 (Eg = 1.04 eV) etc.
Base
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 Emitter Base Collector
optical absorption (~ 104 cm−1 ).
127. (b) The transistor works as an amplifier with emitter-base
116. (b) As, we know that wavelength junction forward biased and base collector junction reversed
hc 1240 eV- nm bias, a transistor is said to be in active state.
λ= = = 440 nm ≈ 4400 Å
Eg 2.8 eV 128. (c) For a p-n-p transistor, charge carrier motion is as
hc 1240 eV - nm
117. (c)Q Eg = = = 2.0 eV
λ 620 nm
118. (a) Electron moves from conduction to valence band, in p
n
p
LED when it glows

A IE IC
Holes IB Electrons
hc
Eg= VEB VCB
λ

B VEE VCC
For n-p-n transistor,
p-base
119. (b) An optical telecommunication link, n-emitter region n-collector
1 4 2 4 3 1 4 2 4 3 1 4 2 4 3
Input Output
signal signal
LED
Optical Optical
p
fibre detector n n
hc
120. (b) In red LED, λ R =
ER
IE IC
hc ER λ V
In violet, LED λ V = ⇒ = <1 Electrons Holes
EV EV λR VEB VCB
IB
121. (b) Light from an LED is highly monochromatic.
VEE VCC
129. (b) Low; High. When emitter-base junction is forward Q Voltage gain,
biased and base-collector junction is reverse biased, then V V 0.8 × 0.96
AV = L = L = −3 =4 ⇒ AV = 4
transistor, is said to be active state. In active state of Vin I B Ri 10 × 192
transistor the emitter-base junction acts as low resistance
while the base collector junction acts as high resistance. and power gain,
2
130. (a) Observe that emitter (E) is connected to both batteries, so IC2 RL I  R 800
AP = =  C  ⋅ L = ( 0.96 )2 ×
it is common. Emitter E-n side is connected to negative of 2
I B Ri  I B  Ri 192
VBB battery, collector C-n side is connected to positive of
VCC battery, active state. AP = 3.84
I
IC 137. (c)Qβ DC = C
– + IB
IB C mA
 ∆IC 
+ – B + R2 138. (a) Qβ AC =   V CE
µA E VCE  ∆I B 
R1 –
VBE IE
IC
VBB
VCC ⇒ β DC =
IB
Ratio= β AC : β DC = 150:150 = 1:1
Circuit arrangement for studying the input and output characteristics
of n-p-n transistor in CE configuration 139. (d) It is said to be in cut-off state for small Vi , active for
131. (a) For a transistor, intermediateVi , saturation for large Vi .
VCE = VCB + VBE 140. (a) With increasing Vi ,Vo decreases.
132. (b)VCE must be sufficiently larger than 0.7 V. Cut-off Active
region region
133. (c) I B does not depend onVCE . So, when VCE is increased, I B Vo
remains constant. IC increases till saturation.
R  ∆I ∆I c 
134. (a) As AV = β L Q g m = c =  Saturation
Ri  ∆V ∆lB Ri  region
Av
β  β
or G =   RL Q g m = 
 Ri   Ri 
Vi
⇒ G = g m RL ⇒ G ∝ g m Transfer characteristics
G2 g m1 0.02 141. (c) Power gain, AP = β AC × A v
∴ = ⇒ G2 = ×G
G1 g m2 0.03 142. (a) The output AC voltage is 2.0 V. So, the AC collector
2 current, iC = 2.0 / 2000 = 10 . mA. The signal current through
∴ Voltage gain, G2 = G i 1.0 mA
3 the base is, therefore given by iB = C = = 0.010 mA.
135. (a) Input resistance ( ri ) This is defined as the ratio of β 100
change in base emitter voltage ( ∆VBE ) to the resulting The DC base current has to be 10 × 0.010 = 010 . mA.
change in base current ( ∆I B ) at constant collector-emitter VBB = VBE + I B RB ⇒ RB = (VBB − VBE ) / I B
voltage (VCE ). This is dynamic (ac resistance). AssumingVBE = 0.6 V
 ∆V  RB = ( 2.0 − 0.6 ) / 010
. = 14 kΩ
ri =  BE 
 ∆I B  V 143. (a) iE = iB + iC ⇒ iE > iB
CE

136. (d) Given, resistance across load, RL = 800 Ω So, according to the question, the possible emitter current is 36.
Voltage drop across load, VL = 0.8 V 144. (a) Initially, both IC and I E increase.
Input resistance of circuit, Ri = 192 Ω
145. (b) Emitter current begins to fall, T1 is seeing decaying field
Collector current is given by
V 0.8 8 in T2 .
IC = L = = = 1 mA ∆ic
RL 800 8000 146. (a) β = Current gain =
∆ib
Output current IC
QCurrent amplification = = = 0.96 ⇒ ∆ic = β × ∆ib = ( 80 × 250 ) µA
Input current IB
147. (a) iE = iB + iC ⇒ iC = iE − iB .
1 mA
⇒ IB =
0.96 Emitter side is forward biased, collector side is reverse biased.
148. (b)Vb = ib Rb  ∆IC 
157. (c) Current gain, β =  
9  ∆I B  V
⇒ Rb = = 257k Ω CE
35 × 10−6 β × Rout
and voltage gain, AV =
149. (a) Inductors are in series, Rin
∴ L eq = L1 + L 2 here, Rin = 1000 Ω , ∆I B = 10 µA = 10−5 A
1 1 1
and frequency of oscillator = = Rout = 5 kΩ = 5 × 103 Ω
2π L eq C eq 2π ( L1 + L 2 )C
∆IC = 2 mA = 2 × 10−3 A
150. (c) From Kirchhoff’s loop rule in output loop, 2 × 10−3
9 − 4 = IC RC β=
= 200
10−5
As, RC = 2kΩ, 200 × 5 × 103
5 I 2.5 Hence, AV = = 1000
⇒ IC = = 2. 5 mA ⇒ I B = C = = 27.8 µA 1000
2 × 103
β 90 80I E
158. (b) Here, IC = 80% of I E =
VBE = 0.7 V 100
From Kirchhoff, loop rule in input loop, IC 10
or IE = = = 12.5 mA
2.3 0.8 0.8
I B RB = 3 − 0.7 ⇒ RB = × 106 = 82 kΩ
27. 8 I B = I E − IC = 12.5 − 10 = 2.5 mA
153. (a) DC current gain in common base amplifier is given by 159. (c) Dynamic resistance,
iC ∆V 0.05 × 1000
α= rd = ⇒ rd = Ω = 10 Ω
iE I∆ 5
where, iC is collector current and iE is emitter current. 160. (a) An analog signal is a continuous waveform as
Given, α = 0.96, iE = 7.2 mA +
Voltage amplitude

∴ iC = 0.96 × 7.2 mA = 6.91 mA

As, iE = iB + iC t
Time
∴ Base current, iB = iE − iC = 7.2 − 6.91 = 0.29 mA
power gain 800 20
154. (a) Current gain, α = = =
voltage gain 840 21 –
α 20 / 21 164. (b) AND gate An AND gate has two or more inputs and
Now, β = = = 20 one output. The output Y of AND gate is 1 only when input
1 − α 1 − ( 20 / 21)
A and input B are both 1. The logic symbol and truth table
IC
As β= = IC = β I B = 20 × 12
. = 24 mA for this gate is given by
IB
A Y Input Output
 π
155. (d) Input signal of a CE amplifier, Vi = 2cos 15t +  B A B Y
 3 0 0 0
Voltage gain, AV = 150 0 1 0
As CE amplifier gives phase difference of π between input 1 0 0
and output signals. 1 1 1
V 165. (b) For an NAND gate, truth table is
So, AV = 0
Vi A
Y
B
⇒ V0 = AV Vi
 π  Input Output
V0 = 150 × 2cos 15t + + π
 3  A B Y
 4π  0 0 1
V = 300 cos 15t +  0 1 1
 3
1 0 1
156. (b) Voltage gain = Current gain × Resistance gain 1 1 0
RC 6
= Current gain × = 30 × = 180 So, no output occurs when both inputs are at higher
RI 1 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. 176. (a) A X
Y
From the given condition, B
V − VB 4 − ( −6 ) 10
I= A = = = 10−2 A Truth table for given circuit is
R 1 kΩ 1 × 103
A B X=A+ B Y = X
167. (d) Given circuit is 0 0 1 0
0 1 0 1
A
Y1 Y3 1 0 0 1
C1 1 1 0 1
Y5 Y6 Y7
Y2 Y4 Hence, the given circuit acts as OR gate.
B
177. (c) Due to diffusion of electrons, positive space-charge
Y1 Y2 Y3 Y4 y5 y6 y7
region on n-side of the junction and negative space charge
A B C
region on p-side of the junction, an electric field directed
0 0 0 0 1 1 1 0 0 1 from positive charge towards negative charge develops.
0 1 0 1 1 0 0 1 1 0 Electric field is from n-side to p-side. Due to this field, an
1 0 1 0 0 1 0 1 1 0
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
1 1 1 1 0 0 0 1 1 0 of charge carriers due to the electric field is called drift.
Observing, this is NOR gate. Thus, a drift current, which is opposite in direction to the
So, output waveform is option (d). diffusion current starts. Concentration gradient is due to
doping of sides. It is not affected by drift of charge carriers.
168. (c) X = AB = A ⋅ B (i.e., AND gate)
178. (c) The loss of electrons from the E
If the output X of NAND gate is connected to the input of n-region and the gain of electrons + –
NOT gate (made from NAND gate by joining two inputs) by the p-region causes a + –
from the given figure, then we get back an AND gate. difference of potential across the + –
169. (b) It is an AND gate. junction of the two regions. The n p
polarity of this potential is such
A B A B X=A + B X =Y as to oppose further flow of carriers so that a condition of
0 0 1 1 1 0 equilibrium exists.
0 1 1 0 1 0 179. (a) As resistance of depletion region is large, potential drop
1 0 0 1 1 0 occurs mainly in depletion region.
1 1 0 0 0 1 180. (b) In forward bias, forward current is obtained due to
removal of barrier potential by externally applied potential.
171. (c) The resultant Boolean expression of the above logic Ohm’s law statesV ∝ I, not valid for diode.
circuit will be 181. (c) From theV-I characteristics of a junction diode, we see
Y = ( A + B )⋅C that it allows current to pass only when it is forward biased.
So, we have seen that among the given options, only option So, if an alternating voltage is applied across a diode the
(c) is the correct choice, i.e., current flows only in that part of the cycle when the diode is
Output, Y = 1only when inputs A = 1, B = 0 and C = 1. forward biased.
172. (c) The conductivity of an intrinsic semiconductor is less 182. (a) As, X L = Lω , inductive reactance is high at high
than that of a lightly doped p-type semiconductor. frequency for AC and for capacitor, X C = 1/ Cω.
173. (a) Zener diode is a special purpose diode. In reverse bias, 183. (c) Solar cells are made wafer shape as a large area ensures
after a certain voltage, current suddenly increases in Zener more incident solar power.
diode. This property is used to obtain voltage regulation.
hν
174. (c) In a transistor, the base is made thin and lightly doped so
that the majority charge carriers coming from emitter may Metallised
pass on to the collector and very few form electron-hole Top surface finger electrode
combination in base. n
p
175. (c) In an oscillator, the feedback is in the same phase, i.e., n n
positive feedback. If the feedback voltage is in opposite p p
phase, i.e., negative feedback, the gain is less than one and it Back contact
can never work as oscillator. It will be an amplifier with (a) (b)
reduced gain. Typical p-n junction solar cell
184. (d) The generation of emf by a solar cell, when light falls continue to diffuse from n- p, a layer of positive charge (or
on, it is due to the following three basic processes positive space-charge region) on n-side of the junction is
generation, separation and collection (i) generation of developed. On p-side atom receiving electrons are ionised
electron-hole pairs due to light (with hν > Eg ) close to the acceptor.
junction; (ii) separation of electrons and holes due to 191. (c) Initially, diffusion current is large and drift current is
electric field of the depletion region. small. As the diffusion process continues, the space-charge
Electrons are swept to n-side and holes to p-side; (iii) the regions on either side of the junction extend, thus increasing
electrons reaching the n-side are collected by the front the electric field strength and hence drift current.
contact and holes reaching p-side are collected by the This process continues until the diffusion current equals the
back contact. Thus, p-side becomes positive and n-side drift current. Thus, a p-n junction is formed. In a p-n junction
becomes negative giving rise to photovoltage. under equilibrium there is no net current.
186. (a) In the case of Si transistor, as long as input Vi is less 192. (d) In forward biasing due to the applied voltage, electrons
than 0.6 V, the transistor will be in cut-off state and from n-side cross the depletion region and reach p-side (where
current IC will be zero. they are minority charge carriers).
Hence,Vo = VCC Similarly, holes from p-side cross the junction and reach the
When Vi becomes greater than 0.6 V, the transistor is in n-side (where they are minority charge carriers). This process
active state with some current IC in the output path and under forward bias is known as minority charge carrier
the outputVo decreases as the term IC RC increases. With injection. At the junction boundary, on each side, the minority
increase ofVi , IC inreases almost linearly and, soVo charge carrier concentration increases significantly compared to
decreases linearly till its value becomes less than about the locations far from the junction.
1.0 V. 193. (c) In forward bias, if depletion layer’s width decreases, fermi
187. (c) A gate is a digital circuit that follows certain logical energy level is in the middle of forbidden gap in intrinsic
relationship between the input and output voltages. semiconductor.
Therefore, they are generally known as logic gates 194. (b) The circuit using two diodes gives output rectified voltage
because they control the flow of information. The five corresponding to both the positive as well as negative half of
common logic gates used are NOT, AND, OR, NAND and the AC cycle.
NOR. Hence, it is known as full-wave rectifier. There is another
188. (a) They are small in size, consume low power, operate at circuit of full-wave rectifier which does not need a centre tap
low voltages and have long life and high reliability. transformer but needs four diodes. It is called a bridge
189. (d) In a solid, electron’s energies are very different from
rectifier.
that in an isolated atom. Inside the crystal, each electron 195. (d) Applying Kirchhoff’s voltage rule to the input and output
has a unique position and no two electrons see exactly the sides of this circuit, we get
same pattern of surrounding charges. Because of this, each IC
electron will have a different energy level.
C
These different energy levels with continuous energy RB B
variation form what are called energy bands. The energy RC +
IB
E Vo
band which includes the energy levels of the valence –
Vi VBB
electrons is called the valence band. The energy band IE
VCC
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
VBB = I B RB +VBE
the valence band. Then, these electrons will move into the
conduction band. and VCE = VCC − IC RC
At the same time they will create vacant energy levels in We shall treat VBB as the DC input voltage Vi andVCE as the
the valence band where other valence electrons can move. DC output voltageVo . So, we have
Thus, the process creates the possibility of conduction due Vi = I B RB + VBE
to electrons in conduction band as well as due to vacancies
and Vo = VCC − IC RC .
in the valence band. In metals, conduction band and
valence band overlap. 196. (d) In an oscillator, we get AC output without any external
input signal. In other words, the output in an oscillator is
190. (d) When an electron diffuses from n → p, it leaves
self-sustained. To attain this, an amplifier is taken. A portion of
behind an ionised donor (species which has become ion by the output power is returned back (feedback) to the input in
donating electron) on n-side. phase with the starting power (this process is termed as positive
This ionised donor (positive charge) is immobile as it is feedback) as shown in figure. The feedback can be achieved by
bonded to the surrounding atoms. As the electrons
 inductive coupling (through mutual inductance) or LC or RC 211. (a) When V A = − 5 VandVB = − 10 V
networks.
D2 is reverse biased and D1 and D3 get forward biased, then
Input R R R
Transistor R AB = + + = R
amplifier Output 4 2 4
212. (a, b, d) If lattice constant of semiconductor is decreased,
then Ec and Ev decrease but Eg increases.
Feedback
network 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
197. (d) A ‘NOR’ gate, like an ‘NAND’ gate is universal gate, is amount of energy to the electrons in valence band so as to
also universal gate. AND, OR, NOT, NAND gates can be jump to conduction band.
made using NOR gates. So, choice (a) is correct.
199. (d) The given materials in decreasing order of conductivity (b) In semiconductors, with the rise in temperature more
are Al > Ge > Si > Diamond (C), so aluminium has least electrons from valence band jump to conduction band and this
energy gap and carbon has largest. Diamond has energy gap results in increase in conductivity. So, choice (b) is correct.
6 eV. Eg (Germanium) = 0.71eV.
(c) In conductors, the conduction band is either partially filled
203. (d) Truth table for NOR gate is given by, or the conduction band overlaps on the valence band. So,
choice (c) is correct.
Input Output
(d) In semiconductors, resistivity decreases with increase in
A B Y temperature. So, choice (d) is wrong.
0 0 1 P 100 × 10−3
0 1 0 216. (a,b) Current in circuit, i = =
Vd 0.5
1 0 0
1 1 0
(Vd = voltage drop across diode)
= 200 × 10−3 A
205. (b) As, IC = αlE = 0.98 × 2 = 196
. mA Voltage across resistance R, V ′ = 15
. − 0.5 = 10
. V
206. (d) As, lB = lE − lC = 2 − 196
. = 0.04 mA V′ 1
Thus, resistance R = = = 5Ω
Ro ( 400 × 103 i 200 × 10−3
207. (c) As, AV = α = 0.98 × = 980
Ri 400 217. (b, c) IC = 10 mA
208. (b) Power amplification, aP = αAV = 0.98 × 980 = 960 IC = 90% of I E
209. (b) As, Vo = Vi × voltage gain = 0.1 × 980 = 98V 90
10 = IE
100
210. (d) When V A = − 10 VandVB = −5 V, ideal diodes D1 and D3
I E = 11 mA
are reverse biased and D2 is forward biased.
∴ I E = I B + IC
R
I B = 1mA
R
218. (a,b,c) Applying loop law at output port,
R 9 − 4 = IC RC
R/4 R/4
or IC = 2.5 mA
I 2.5
VA= – 10 V VB = – 5 V IB = C = = 2.78 × 10−5 A = 27.8 µA
Since, ideal diode in reverse bias has infinite resistance (i.e., β 90
open circuited) and during forward bias it has zero resistance Since, the transistor operates in active region therefore
(i.e., short-circuited) Therefore, the given circuit may be VBE = 0.7 V
shown as in figure. 3 − 0.7
R R 3R Applying loop law at input port, I B =
∴ R AB = + R + = RB
4 4 2
2.3 × 105
R RB = = 82 kΩ
2.78
R
219. (c) When a forward bias is applied across the p-n junction,
R
the applied voltage opposes the barrier voltage. Due to this,
R/4 R/4
the potential barrier across the junction is lowered.
VA=–5 V VB=–10 V
 IC I
220. (b) For a transistor, β = ⇒ I B = C base region is Input (base) resistance, Rinput = 1 kΩ = 1000 Ω
IB β
thin, so that majority carrier of emitter will reach the Voutput Routput
Q Voltage gain, AV = = β AC
collector. Vinput Rinput
Vinput Vinput
Rinput = = ⋅β Voutput
IB IC ∴ Input signal voltage, Vinput =
β AC ( Routput / Rinput )
1
i.e., Rinput ∝ 2
IC = = 0.01 V
100 ( 2000/ 1000 )
Therefore, Rinput is inversely proportional to the collector
current. For high collector current, the Rinput should be small Vinput 0.01
Base (input) current, I B = = = 10 × 10−6 A
for which the base region must be very thin and lightly Rinput 1000
doped for a transistor action, the emitter junction is forward
biased and collector junction is reverse biased. = 10 µA
221. (c) In an n-type semiconductor, it is obtained by doping the 227. (a) Given, voltage gain of first amplifier, AV 1 = 10
Ge or Si with pentavalent atoms. In n-type semiconductor, Voltage gain of second amplifier, AV 2 = 20
electrons are majority charge carriers and holes are minority
charge carriers. Input voltage, Vi = 0.01 V
222. (c) The energy band gap is largest for carbon, less for V
Total voltage gain, AV = o = AV 1 × AV 2
silicon and least for germanium. Vi
223. (c) In an unbiased p-n junction, the diffusion of charge Vo
carriers across the junction takes place from higher ∴ = 10 × 20
concentration to lower concentration. Therefore, hole 0.01
concentration in p-region is more as compared to n-region. Vo = 2 V
− Eg / 2k B T
224. (b) AND gate 1 AND gate 2 228. (c) Given, intrinsic carrier concentration ni = no e
NOT gate 1
A
Y
and energy gap Eg = 1.2 eV
B C D E
NOT gate 2 kB = 8.62 × 10−5 eV/K
A B C D E Y For T = 600 K,
− E g / 2 k B × 600
0 0 0 1 1 0 n600 = no e …(i)
0 1 0 1 1 0 For T = 300 K,
1 0 0 1 1 0 − E g / 2 k B × 300
n300 = n0 e …(ii)
1 1 1 0 0 1
Dividing Eq. (i) by Eq. (ii), we get
So, this logic operation as AND gate.
(b) Split the gate  Eg  1 1  Eg  1 1 
 −  −   − 
n600 2 k B  600 300   2 k B  300 600 
AND gate 1 NOT gate 1 = e =e
E n300
A AND gate 3
C 1.2  1 
G  
2 × 8.62 × 10 −5  600 
AND gate 2 Y =e
NOT gate 3
B F = e11.6 = (2.718)11.6 (Q e = 2.718)
D
5
NOT gate 2 = 1.1 × 10
A B C D E F G Y Let the conductivities are σ 600 and σ 300 .
0 0 0 0 1 1 1 0 σ 600 n 600
1 0 1 0 0 1 0 1 = = 1.1 × 105 (Qσ = enµ e )
σ 300 n 300
0 1 0 1 1 0 0 1
1 1 1 1 0 0 0 1 229. (a) Given, I 0 = 5 × 10−12 A, T = 300 K

So, this logic operation resembles to OR gate. kB = 8.6 × 10−5 eV/K

225. (c) The voltage gain is low at high and low frequencies and = 8.6 × 10−5 × 1.6 × 10−19 J/K
constant at mid frequencies.
Given, voltageV = 0.6 V
226. (b) Given, collector resistance Routput = 2 kΩ = 2000 Ω
eV 1.6 × 10−19 × 0.6
Current amplification factor of the transistor, β AC = 100 ∴ = = 23.26
kB T 8.6 × 10−5 × 1.6 × 10−19 × 300
Audio signal voltage, Voutput = 2 V
 The current I through a junction diode is given by atoms cause ionization resulting secondary electrons and
 eV 
− 1
thus more number of charge carriers.

I = I oe  2k B T 
= 5 × 10−12 ( e23.26 − 1) When doping concentration is large, there will be large
number of ions in the depletion region, which will give rise
= 5 × 10−12 (1.259 × 1010 − 1) to a strong electric field.
= 5 × 10−12 × 1.259 × 1010 = 0.063 A 241. (b)Vi = 1 V, active state
231. (d) When temperature increases, number density of free Vi = 0.5V, cut-off region IC = 0
charge carriers increases and mean relaxation time decreases Vi = 2.5 V
due to increased lattice vibrations. Effect of decrease in
relaxation time is much less as compared to effect of Cut-off
increase in number density. region
232. (b) In the given circuit, p-side of p-n junction, D1 is Vo Ac
tiv
connected to lower voltage and n-side of D1 to higher e
re
gi
voltage. Thus, D1 is reverse biased. The p-side of p-n on
Saturation
junction D2 is at higher potential and n-side of D2 is at lower region
potential. Thus, D2 is forward biased. Vi
S
Hence, no current flows through the junction B to A. 0.6 V
233. (d) As p-n junction conducts during positive half cycle The transistor circuit in active state can be used as an
only, the diode connected here will work in positive half amplifier.
cycle. Potential difference across C = peak voltage of the 242. (c) It is a combination of AND gates.
given AC voltage V0 = V rms 2 = 220 2 V
A Y2
234. (b) Each positive half passes through the diode and so output AND
only contains positive halves of voltage.
235. (b) Diode is forward biased, no current goes through side OR Y
branch after capacitor is charged.
Y1
∴ V AB = I AB × R AB = 0.2 × 10−3 × 10 × 103 = 2 V AND
B Y3
Option (b) is most appropriate, extra 0.3 V occurs due to
diode. A B Y1 Y2 Y3 Y
236. (c) In depletion region, after equilibrium, no recombination 0 0 1 0 0 0
can occur as electrons and holes are not ‘free’. In depletion 0 1 1 0 1 1
region, there are immobile charged ions and but no mobile
1 0 0 0 0 0
charges; equal number of holes and electrons exists.
1 1 0 1 0 1
237. (b) Ripple factor ( r ) of a full-wave rectifier using capacitor
1 243. (c) As, I E = I B + IC
filter is given by r =
4 3RLC 100% 10% 95%
95
1 So, IE × = IC
i.e., r∝ 100
RL
100 × 10
1 1 ⇒ IE = = 10.53 mA
⇒ r∝
,r∝ 95
C ν Also, I B = 10.53 − 10 = 0.53 mA
Thus, to reduce r, RL should be increased, input frequency ν
244. (a) When electric field is applied across a semiconductor,
should be increased and capacitance C should be increased.
the electrons in the conduction band get accelerated and
238. (c) Resistance of Zener diode changes at breakdown voltage acquire energy. They move from lower energy level to
and current through Rs increases after breakdown due to higher energy level.
increased movement of minority carrier. While the holes in valence band move from higher energy
239. (a) In reverse biasing, the minority charge carriers will be level to lower energy level, where they will be having more
accelerated due to reverse biasing which on striking with energy.