JEE Physics Semiconductor Electronics MCQs Set 27

Multiple Choice Questions (MCQs) for JEE Physics: Semiconductor Electronics

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Practice Semiconductor Electronics MCQs for JEE Physics

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Multiple Choice Questions

Question 1: In a common emitter amplifier, the phase difference between the input signal voltage and the output signal voltage is
(a) \( \pi/4 \)
(b) \( \pi \)
(c) 0
(d) \( \pi/2 \)
Show Answer & Explanation

Answer: (b) \( \pi \)

Explanation:
1. In the common emitter configuration an increase in base current raises the collector current and lowers the collector voltage.
2. So the output voltage is inverted: the phase difference is \( \pi \).

Teacher's Note:
1. T
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118.
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120. h
121. i
122. f
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124. .

Question 2: In a full-wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple would be
(a) 100 Hz
(b) 50 Hz
(c) 25 Hz
(d) 70.7 Hz
Show Answer & Explanation

Answer: (a) 100 Hz

Explanation:
1. A full-wave rectifier produces two output pulses for every cycle of the input.
2. So the ripple frequency is \( 2\times50=100 \) Hz.

Teacher's Note:
1. A
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81. b
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88. .

Question 3: An n-p-n transistor amplifier in common emitter configuration has a current gain of 100. If the collector current changes by 1 mA, the change in emitter current is
(a) 1.01 mA
(b) 1.1 mA
(c) 0.01 mA
(d) 10 mA
Show Answer & Explanation

Answer: (a) 1.01 mA

Explanation:
1. \( \beta=\Delta I_c/\Delta I_b=100 \), so \( \Delta I_b=0.01 \) mA.
2. \( \Delta I_e=\Delta I_c+\Delta I_b=1.01 \) mA.

Teacher's Note:
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130. .

Question 4: In a semiconducting material the mobilities of electrons and holes are \( \mu_e \) and \( \mu_h \) respectively. Which of the following is true?
(a) \( \mu_e=\mu_h \)
(b) \( \mu_e<\mu_h \)
(c) \( \mu_e>\mu_h \)
(d) \( \mu_e<0,\ \mu_h>0 \)
Show Answer & Explanation

Answer: (c) \( \mu_e>\mu_h \)

Explanation:
1. Electrons in the conduction band move freely, while holes move by successive jumps of electrons between bound states.
2. Hence electrons are two to three times more mobile than holes, so \( \mu_e>\mu_h \).

Teacher's Note:
1. T
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75. .

Question 5: A transistor has \( h_{FE}=95 \). The value of \( h_{FB} \) is
(a) 0.99
(b) 1.9
(c) 0.20
(d) 0.56
Show Answer & Explanation

Answer: (a) 0.99

Explanation:
1. \( h_{FB}=\alpha \) and \( h_{FE}=\beta \).
2. \( \alpha=\dfrac{\beta}{1+\beta}=\dfrac{95}{96}=0.99 \).

Teacher's Note:
1. A
2. l
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5. a
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97. .

Question 6: In an n-type silicon, which of the following statements is true?
(a) Electrons are majority carriers and trivalent atoms are the dopants
(b) Electrons are minority carriers and pentavalent atoms are the dopants
(c) Holes are minority carriers and pentavalent atoms are the dopants
(d) Holes are majority carriers and trivalent atoms are the dopants
Show Answer & Explanation

Answer: (c) Holes are minority carriers and pentavalent atoms are the dopants

Explanation:
1. An n-type semiconductor is formed by doping with pentavalent atoms (phosphorus, arsenic).
2. Each dopant atom donates a free electron, so electrons are the majority carriers and holes are the minority carriers.

Teacher's Note:
1. R
2. e
3. m
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11.
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112. .

Question 7: The correct relation between \( n_e \) and \( n_h \) in an intrinsic semiconductor at ordinary temperature is
(a) \( n_e=n_h \)
(b) \( n_e>n_h \)
(c) \( n_e<n_h \)
(d) \( n_e=n_h=0 \)
Show Answer & Explanation

Answer: (a) \( n_e=n_h \)

Explanation:
1. In an intrinsic semiconductor each electron that goes into the conduction band leaves behind a hole.
2. Hence the numbers of free electrons and holes are equal.

Teacher's Note:
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Question 8: The conductivity of a semiconductor increases with increase in temperature because
(a) the number density of carriers increases (its effect outweighs the decrease in relaxation time)
(b) the relaxation time increases
(c) both the number density of carriers and the relaxation time increase
(d) the number density of carriers decreases
Show Answer & Explanation

Answer: (a) the number density of carriers increases (its effect outweighs the decrease in relaxation time)

Explanation:
1. At higher temperature more electrons gain enough energy to cross the band gap, so the carrier density rises rapidly.
2. The relaxation time decreases slightly because of more collisions, but this effect is far smaller than the increase in carriers.

Teacher's Note:
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138. .

Question 9: The forbidden energy band gaps in conductors, semiconductors and insulators are \( E_{G1},E_{G2},E_{G3} \) respectively. The relation among them is
(a) \( E_{G1}<E_{G2}<E_{G3} \)
(b) \( E_{G1}>E_{G2}>E_{G3} \)
(c) \( E_{G2}>E_{G1}<E_{G3} \)
(d) \( E_{G3}>E_{G1}>E_{G2} \)
Show Answer & Explanation

Answer: (a) \( E_{G1}<E_{G2}<E_{G3} \)

Explanation:
1. In conductors the bands overlap or the gap is negligible.
2. Semiconductors have a small gap (about 1 eV) and insulators a large gap (more than about 3 eV).
3. So \( E_{G1}<E_{G2}<E_{G3} \).

Teacher's Note:
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Question 10: Wires P and Q have the same resistance at room temperature. When heated, the resistance of P increases and that of Q decreases. We conclude that
(a) P is a semiconductor and Q is a conductor
(b) P is a conductor and Q is a semiconductor
(c) P is an n-type semiconductor and Q is a p-type semiconductor
(d) none of the above
Show Answer & Explanation

Answer: (b) P is a conductor and Q is a semiconductor

Explanation:
1. For metals the resistance rises with temperature (positive temperature coefficient).
2. For semiconductors the resistance falls with temperature (negative temperature coefficient).

Teacher's Note:
1. T
2. e
3. m
4. p
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96. .

Question 11: In a semiconductor, the ratio of electron and hole currents is 7/4 and the ratio of the drift velocities of electrons and holes is 5/4. The ratio of the concentrations of electrons and holes is
(a) 5/7
(b) 7/5
(c) 25/49
(d) 49/25
Show Answer & Explanation

Answer: (b) 7/5

Explanation:
1. \( I=nAev_d \), so \( I_e/I_h=(n_e/n_h)(v_e/v_h) \).
2. \( \dfrac74=\dfrac{n_e}{n_h}\times\dfrac54 \), so \( n_e/n_h=7/5 \).

Teacher's Note:
1. C
2. u
3. r
4. r
5. e
6. n
7. t
8.
9. i
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11.
12. p
13. r
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73. a
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109.
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111. h
112. a
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114. g
115. e
116. .

Question 12: A piece of copper and another of germanium are cooled from room temperature to 80 K. Then
(a) the resistance of each will increase
(b) the resistance of copper will decrease while that of germanium will increase
(c) the resistance of each will decrease
(d) the resistance of copper will increase while that of germanium will decrease
Show Answer & Explanation

Answer: (b) the resistance of copper will decrease while that of germanium will increase

Explanation:
1. Copper is a metal: the lattice vibrations reduce on cooling, so its resistance decreases.
2. Germanium is a semiconductor: fewer carriers are excited on cooling, so its resistance increases.

Teacher's Note:
1. M
2. e
3. t
4. a
5. l
6. s
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22. c
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28. t
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31. i
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36. i
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66. o
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68. s
69.
70. w
71. o
72. r
73. s
74. e
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76. .

Question 13: The electrical conductivity of a semiconductor
(a) decreases with a rise in its temperature
(b) does not change with a rise in its temperature
(c) first increases and then decreases with a rise in temperature
(d) increases with a rise in its temperature
Show Answer & Explanation

Answer: (d) increases with a rise in its temperature

Explanation:
1. Heating creates more electron-hole pairs, increasing the number of charge carriers.
2. The conductivity therefore increases with temperature.

Teacher's Note:
1. T
2. h
3. i
4. s
5.
6. i
7. s
8.
9. t
10. h
11. e
12.
13. o
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24.
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26. h
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29. b
30. e
31. h
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43. e
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45. a
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60. d
61. e
62. f
63. i
64. n
65. i
66. n
67. g
68.
69. p
70. r
71. o
72. p
73. e
74. r
75. t
76. y
77.
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79. f
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82. e
83. m
84. i
85. c
86. o
87. n
88. d
89. u
90. c
91. t
92. o
93. r
94. s
95. .

Question 14: An n-type and a p-type silicon semiconductor can be obtained by doping pure silicon with
(a) phosphorus and boron respectively
(b) sodium and magnesium
(c) boron and phosphorus respectively
(d) indium and sodium respectively
Show Answer & Explanation

Answer: (a) phosphorus and boron respectively

Explanation:
1. Phosphorus is pentavalent and gives n-type silicon.
2. Boron is trivalent and gives p-type silicon.

Teacher's Note:
1. P
2. h
3. o
4. s
5. p
6. h
7. o
8. r
9. u
10. s
11.
12. h
13. a
14. s
15.
16. f
17. i
18. v
19. e
20.
21. v
22. a
23. l
24. e
25. n
26. c
27. e
28.
29. e
30. l
31. e
32. c
33. t
34. r
35. o
36. n
37. s
38. ,
39.
40. b
41. o
42. r
43. o
44. n
45.
46. h
47. a
48. s
49.
50. t
51. h
52. r
53. e
54. e
55. .
56.
57. T
58. h
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60.
61. f
62. i
63. r
64. s
65. t
66.
67. c
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69. e
70. a
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90.
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104. e
105. a
106. t
107. e
108. s
109.
110. h
111. o
112. l
113. e
114. s
115. .

Question 15: A silicon specimen is made p-type by doping, on average, one indium atom per \( 5\times10^7 \) silicon atoms. If the number density of atoms in the specimen is \( 5\times10^{28} \) atoms m\(^{-3}\), the number of acceptor atoms per cubic centimetre is
(a) \( 2.5\times10^{30}\ \text{cm}^{-3} \)
(b) \( 2.5\times10^{35}\ \text{cm}^{-3} \)
(c) \( 1.0\times10^{15}\ \text{cm}^{-3} \)
(d) \( 1.0\times10^{13}\ \text{cm}^{-3} \)
Show Answer & Explanation

Answer: (c) \( 1.0\times10^{15}\ \text{cm}^{-3} \)

Explanation:
1. Acceptor density \( =5\times10^{28}/(5\times10^7)=10^{21} \) atoms per m\(^3\).
2. In cm\(^{-3}\): \( 10^{21}\times10^{-6}=10^{15} \).

Teacher's Note:
1. D
2. i
3. v
4. i
5. d
6. e
7.
8. t
9. h
10. e
11.
12. h
13. o
14. s
15. t
16.
17. d
18. e
19. n
20. s
21. i
22. t
23. y
24.
25. b
26. y
27.
28. t
29. h
30. e
31.
32. d
33. i
34. l
35. u
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37. i
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39. n
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56. v
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61. c
62. u
63. b
64. i
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71. e
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124. i
125. n
126. u
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130. .

Question 16: When the electrical conductivity of a semiconductor is due to the breaking of its covalent bonds, the semiconductor is said to be
(a) acceptor
(b) intrinsic
(c) donor
(d) extrinsic
Show Answer & Explanation

Answer: (b) intrinsic

Explanation:
1. Conductivity from thermally broken covalent bonds produces equal numbers of electrons and holes.
2. This is intrinsic conductivity of a pure semiconductor.

Teacher's Note:
1. E
2. x
3. t
4. r
5. i
6. n
7. s
8. i
9. c
10.
11. c
12. o
13. n
14. d
15. u
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17. t
18. i
19. v
20. i
21. t
22. y
23.
24. c
25. o
26. m
27. e
28. s
29.
30. f
31. r
32. o
33. m
34.
35. a
36. d
37. d
38. e
39. d
40.
41. i
42. m
43. p
44. u
45. r
46. i
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52.
53. w
54. h
55. i
56. l
57. e
58.
59. i
60. n
61. t
62. r
63. i
64. n
65. s
66. i
67. c
68.
69. c
70. o
71. n
72. d
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80. y
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99. r
100. e
101.
102. m
103. a
104. t
105. e
106. r
107. i
108. a
109. l
110. .

Question 17: For a transistor amplifier, the voltage gain
(a) is low at high and low frequencies and constant at mid frequencies
(b) remains constant for all frequencies
(c) is high at high and low frequencies and constant in the middle range
(d) is zero at all frequencies
Show Answer & Explanation

Answer: (a) is low at high and low frequencies and constant at mid frequencies

Explanation:
1. At low frequencies the coupling and bypass capacitors reduce the gain.
2. At high frequencies the junction capacitances reduce the gain.
3. So the gain is constant only in the mid-frequency range.

Teacher's Note:
1. T
2. h
3. e
4.
5. g
6. a
7. i
8. n
9. -
10. f
11. r
12. e
13. q
14. u
15. e
16. n
17. c
18. y
19.
20. c
21. u
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25.
26. o
27. f
28.
29. a
30. n
31.
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34. p
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38. i
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40. r
41.
42. i
43. s
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45. f
46. l
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49.
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51. n
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55. e
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58. i
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60. d
61. l
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66. d
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75. n
76.
77. b
78. o
79. t
80. h
81.
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87. .

Question 18: A sinusoidal voltage of peak value 200 V is connected to a diode and resistor R so that half-wave rectification occurs. If the forward resistance of the diode is negligible compared with R, the rms voltage across R is approximately
(a) 200 V
(b) 100 V
(c) \( 200/\sqrt2\ \text{V} \)
(d) 280 V
Show Answer & Explanation

Answer: (b) 100 V

Explanation:
1. For half-wave rectification \( V_{rms}=V_0/2 \).
2. \( V_{rms}=200/2=100 \) V.

Teacher's Note:
1. F
2. o
3. r
4.
5. a
6.
7. f
8. u
9. l
10. l
11.
12. s
13. i
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17. w
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20. e
21.
22. t
23. h
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25.
26. r
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51. o
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73. r
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75. c
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81. a
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86.
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95. a
96. k
97.
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99. v
100. e
101. r
102.
103. t
104. w
105. o
106. .

Question 19: In a half-wave rectifier circuit operating from 50 Hz mains, the fundamental frequency in the ripple would be
(a) 50 Hz
(b) 25 Hz
(c) 70.7 Hz
(d) 100 Hz
Show Answer & Explanation

Answer: (a) 50 Hz

Explanation:
1. A half-wave rectifier passes one pulse per input cycle.
2. Hence the ripple has the same frequency as the input, 50 Hz.

Teacher's Note:
1. C
2. o
3. m
4. p
5. a
6. r
7. e
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9. w
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12. h
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21. l
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89. u
90. e
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92. c
93. y
94. .

Question 20: Two amplifiers are connected one after the other (cascaded). The first has a voltage gain of 10 and the second a voltage gain of 20. If the input signal is 0.01 V, the output AC signal is
(a) 3.6 V
(b) 4.2 V
(c) 5.2 V
(d) 2.0 V
Show Answer & Explanation

Answer: (d) 2.0 V

Explanation:
1. Total gain of the cascade \( =10\times20=200 \).
2. Output \( =200\times0.01=2.0 \) V.

Teacher's Note:
1. I
2. n
3.
4. a
5.
6. c
7. a
8. s
9. c
10. a
11. d
12. e
13.
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19. a
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31. y
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40. d
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42.
43. I
44. n
45.
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47. e
48. c
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50. b
51. e
52. l
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54.
55. t
56. h
57. e
58. y
59.
60. w
61. o
62. u
63. l
64. d
65.
66. a
67. d
68. d
69. .

Multiple Choice Questions (MCQs) for JEE Physics Semiconductor Electronics

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