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
(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
2. h
3. e
4.
5. c
6. o
7. m
8. m
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67.
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124. .
(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
2.
3. h
4. a
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7. -
8. w
9. a
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61. l
62. -
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81. b
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85.
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88. .
(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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63.
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65. h
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130. .
(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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74. e
75. .
(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
3. p
4. h
5. a
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97. .
(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
4. e
5. m
6. b
7. e
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10. N
11.
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112. .
(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:
1. E
2. q
3. u
4. a
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115. .
(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. .
(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:
1. T
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136. .
(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. .
(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
10. s
11.
12. p
13. r
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22. a
23. l
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25. t
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28. c
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31. r
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39. c
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63. e
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66. c
67. i
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70. .
71.
72. C
73. a
74. n
75. c
76. e
77. l
78.
79. t
80. h
81. e
82.
83. c
84. o
85. m
86. m
87. o
88. n
89.
90. f
91. a
92. c
93. t
94. o
95. r
96. s
97.
98. o
99. f
100.
101. a
102. r
103. e
104. a
105.
106. a
107. n
108. d
109.
110. c
111. h
112. a
113. r
114. g
115. e
116. .
(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
7.
8. i
9. m
10. p
11. r
12. o
13. v
14. e
15.
16. t
17. h
18. e
19. i
20. r
21.
22. c
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24. n
25. d
26. u
27. c
28. t
29. i
30. v
31. i
32. t
33. y
34.
35. w
36. i
37. t
38. h
39.
40. c
41. o
42. o
43. l
44. i
45. n
46. g
47. ,
48.
49. w
50. h
51. i
52. l
53. e
54.
55. s
56. e
57. m
58. i
59. c
60. o
61. n
62. d
63. u
64. c
65. t
66. o
67. r
68. s
69.
70. w
71. o
72. r
73. s
74. e
75. n
76. .
(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
14. p
15. p
16. o
17. s
18. i
19. t
20. e
21.
22. o
23. f
24.
25. t
26. h
27. e
28.
29. b
30. e
31. h
32. a
33. v
34. i
35. o
36. u
37. r
38.
39. o
40. f
41.
42. m
43. e
44. t
45. a
46. l
47. s
48.
49. a
50. n
51. d
52.
53. i
54. s
55.
56. t
57. h
58. e
59.
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.
78. o
79. f
80.
81. s
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. .
(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
59. e
60.
61. f
62. i
63. r
64. s
65. t
66.
67. c
68. r
69. e
70. a
71. t
72. e
73. s
74.
75. f
76. r
77. e
78. e
79.
80. e
81. l
82. e
83. c
84. t
85. r
86. o
87. n
88. s
89. ,
90.
91. t
92. h
93. e
94.
95. s
96. e
97. c
98. o
99. n
100. d
101.
102. c
103. r
104. e
105. a
106. t
107. e
108. s
109.
110. h
111. o
112. l
113. e
114. s
115. .
(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
36. t
37. i
38. o
39. n
40.
41. r
42. a
43. t
44. i
45. o
46. ,
47.
48. t
49. h
50. e
51. n
52.
53. c
54. o
55. n
56. v
57. e
58. r
59. t
60.
61. c
62. u
63. b
64. i
65. c
66.
67. m
68. e
69. t
70. r
71. e
72. s
73.
74. t
75. o
76.
77. c
78. u
79. b
80. i
81. c
82.
83. c
84. e
85. n
86. t
87. i
88. m
89. e
90. t
91. r
92. e
93. s
94.
95. b
96. y
97.
98. m
99. u
100. l
101. t
102. i
103. p
104. l
105. y
106. i
107. n
108. g
109.
110. b
111. y
112.
113. 1
114. 0
115.
116. t
117. o
118.
119. t
120. h
121. e
122.
123. m
124. i
125. n
126. u
127. s
128.
129. 6
130. .
(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
16. c
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
47. t
48. i
49. e
50. s
51. ,
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
73. u
74. c
75. t
76. i
77. v
78. i
79. t
80. y
81.
82. c
83. o
84. m
85. e
86. s
87.
88. f
89. r
90. o
91. m
92.
93. t
94. h
95. e
96.
97. p
98. u
99. r
100. e
101.
102. m
103. a
104. t
105. e
106. r
107. i
108. a
109. l
110. .
(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
22. r
23. v
24. e
25.
26. o
27. f
28.
29. a
30. n
31.
32. a
33. m
34. p
35. l
36. i
37. f
38. i
39. e
40. r
41.
42. i
43. s
44.
45. f
46. l
47. a
48. t
49.
50. i
51. n
52.
53. t
54. h
55. e
56.
57. m
58. i
59. d
60. d
61. l
62. e
63.
64. a
65. n
66. d
67.
68. f
69. a
70. l
71. l
72. s
73.
74. o
75. n
76.
77. b
78. o
79. t
80. h
81.
82. s
83. i
84. d
85. e
86. s
87. .
(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
14. n
15. e
16.
17. w
18. a
19. v
20. e
21.
22. t
23. h
24. e
25.
26. r
27. m
28. s
29.
30. v
31. a
32. l
33. u
34. e
35.
36. i
37. s
38.
39. p
40. e
41. a
42. k
43.
44. o
45. v
46. e
47. r
48.
49. r
50. o
51. o
52. t
53.
54. t
55. w
56. o
57. ,
58.
59. b
60. u
61. t
62.
63. h
64. a
65. l
66. f
67. -
68. w
69. a
70. v
71. e
72.
73. r
74. e
75. c
76. t
77. i
78. f
79. i
80. c
81. a
82. t
83. i
84. o
85. n
86.
87. g
88. i
89. v
90. e
91. s
92.
93. p
94. e
95. a
96. k
97.
98. o
99. v
100. e
101. r
102.
103. t
104. w
105. o
106. .
(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
8.
9. w
10. i
11. t
12. h
13.
14. t
15. h
16. e
17.
18. f
19. u
20. l
21. l
22. -
23. w
24. a
25. v
26. e
27.
28. r
29. e
30. c
31. t
32. i
33. f
34. i
35. e
36. r
37. ,
38.
39. w
40. h
41. e
42. r
43. e
44.
45. t
46. h
47. e
48.
49. r
50. i
51. p
52. p
53. l
54. e
55.
56. f
57. r
58. e
59. q
60. u
61. e
62. n
63. c
64. y
65.
66. i
67. s
68.
69. t
70. w
71. i
72. c
73. e
74.
75. t
76. h
77. e
78.
79. i
80. n
81. p
82. u
83. t
84.
85. f
86. r
87. e
88. q
89. u
90. e
91. n
92. c
93. y
94. .
(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.
14. t
15. h
16. e
17.
18. g
19. a
20. i
21. n
22. s
23.
24. m
25. u
26. l
27. t
28. i
29. p
30. l
31. y
32. ,
33.
34. n
35. o
36. t
37.
38. a
39. d
40. d
41. .
42.
43. I
44. n
45.
46. d
47. e
48. c
49. i
50. b
51. e
52. l
53. s
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. .
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Multiple Choice Questions (MCQs) for JEE Physics Semiconductor Electronics
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