ELECTRONICS AND COMMUNICATION ENGINEERING ESE TOPICWISE OBJECTIVE SOLVED PAPER-I
From (1991 – 2018)
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IES MASTER PUBLICATION F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 Phone : 011-26522064, Mobile : 8130909220, 9711853908 E-mail : info@iesmasterpublications.com, info@iesmaster.org Web : iesmasterpublications.com, iesmaster.org
All rights reserved. Copyright Š 2018, by IES MASTER Publications. No part of this booklet may be reproduced, or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior permission of IES MASTER, New Delhi. Violates are liable to be legally prosecuted.
First Edition
: 2017
Second Edition : 2018
Typeset at : IES Master Publication, New Delhi-110016
PREFACE It is an immense pleasure to present topic wise previous years solved paper of Engineering Services Exam. This booklet has come out after long observation and detailed interaction with the students preparing for Engineering Services Exam and includes detailed explanation to all questions. The approach has been to provide explanation in such a way that just by going through the solutions, students will be able to understand the basic concepts and will apply these concepts in solving other questions that might be asked in future exams. Engineering Services Exam is a gateway to a immensly satisfying and high exposure job in engineering sector. The exposure to challenges and opportunities of leading the diverse field of engineering has been the main reason for students opting for this service as compared to others. To facilitate selection into these services, availability of arithmetic solution to previous year paper is the need of the day. Towards this end this book becomes indispensable.
Mr. Kanchan Kumar Thakur Director–IES Master
Note: Direction of all Assertion Reasoning (A–R) type of questions covered in this booklet is as follows: DIRECTIONS: The following four items consist of two statements, one labelled as ‘Assertion A’ and the other labelled as ‘Reason R’. You are to examine these two statements carefully and select the answer to these two statements carefully and select the answer to these items using the codes given below: (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true. Note: Direction of all Statement-I and Statement-II type of questions covered in this booklet is as follows: DIRECTION: Following items consists of two statements, one labelled as ‘Statement (I)’ and the other as ‘Statement (II)’. You are to examine these two statements carefully and select the answers to these items using the code given below: (a) Both Statement : (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I). (b) Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I). (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true.
CONTENTS 1. Basic Electronic Engineering ...................................................................... 01–141 2. Materials and Components ....................................................................... 142–249 3. Measurement and Instrumentations ........................................................... 250–372 4. Network Theory ........................................................................................ 373–628 5. Analog Electronics ................................................................................... 629–791 6. Digital Electronics .................................................................................... 792–941 7. Microprocessor ........................................................................................ 942–989 8. Basic Electrical Engineering ................................................................... 990–1001
1 SYLLABUS Basics of semiconductors; Diode/Transistor basics and characteristics; Diodes for different uses; Junction & Field Effect Transistors (BJTs, JFETs, MOSFETs); Transistor amplifiers of different types, oscillators and other circuits; Basics of Integrated Circuits (ICs); Bipolar, MOS and CMOS ICs; Basics of linear ICs, operational amplifiers and their applications-linear/non-linear, Optical sources/ detectors; Basics of Opto electronics and its applications.
Contents 1.
Basic of Semiconductor ------------------------------------------------------------- 1–47
2.
p-n Junction Diode and Opto Electronics ------------------------------------ 48–82
3.
Bipolar Junction Transistors (BJTs) ----------------------------------------- 83–101
4.
Field Effect Transistors (FETs) --------------------------------------------- 102–122
5.
Silicon Controlled Rectifiers ------------------------------------------------- 123–132
6.
Integrated Circuits (ICs) ------------------------------------------------------ 133–141
1 IES – 2018 1.
(c) Decreases exponentially with temperature (d) Not altered with temperature.
Silicon devices can be employed for a higher temperature limit (190 ºC to 200 ºC) as compared to germanium devices (85 ºC to 100 ºC). With respect to this, which of the following are incorrect?
4.
(a) For insulators the band-gap is narrow as compared to semiconductors (b) For insulators the band-gap is relatively wide whereas for semiconductors it is narrow
1. Higher resistivity of silicon 2. Higher gap energy of siliocn 3. Lower intrinsic concentration of silicon
(c) The band-gap is narrow in width for both the insulators and conductors
4. Use of silicon devices in high-power applications Select the correct answer using the code given below:
2.
(a) 1, 2 and 4
(b) 1, 2 and 3
(c) 1, 3 and 4
(d) 2, 3 and 4
18
3
(b) 3.76 × 10 /m
18
3
(d) 9.94 × 1018/m3
5.
18
In an extrinsic semiconductor the conductivity significantly depends upon: (a) Majority charge carriers generated due to doping (b) Minority charge carriers generated due to thermal agitation (c) Majority charge carriers generated due to thermal agitation (d) Minority charge carriers generated due to impurity doping.
3
6.
The electrical conductivity of pure semiconductor is:
A Ge sample at room temperature has intrinsic carrier concentration, ni = 1.5 × 1013 cm–3 and is uniformly doped with acceptor of 3×1016 cm–3 and donor of 2.5×1015 cm–3. Then, the minority charge carrier concentration is:
(a) Proportional to temperature
(a) 0.918 × 1010 cm–3
(b) Increases exponentially with temperature
(b) 0.818 × 1010 cm–3
(c) 7.87 × 10 /m
IES – 2016 3.
(d) The band-gap is equally wide for both conductors and semiconductors.
A sample of germanium is made p-type be addition of indium at the rate of one indium atom for every 2.5×108 germanium atoms. Given, ni = 2.5×1019 /m3 at 300 K and the number of germanium atoms per m3 = 4.4 × 1028. What is the value of np? (a) 3.55 × 10 /m
Which one of the following statements is correct?
Basic Electronic Engineering
3
(c) 0.918 × 1012 cm–3 12
–3
(d) 0.818 × 10 cm 7.
8.
Assume that the values of mobility of holes and that of electrons in an intrinsic semiconductor are equal and the values of conductivity and intrinsic electron density are 2.32 / m and 2.5 × 1019/m3 respectively. Then, the mobility of electron/hole is approximately: (a) 0.3 m2/Vs
(b) 0.5 m2/Vs
(c) 0.7 m2/Vs
(d) 0.9 m2/Vs
A silicon sample A is doped with 1018 atom/ cm3 of Boron and another silicon sample B of identical dimensions is doped with 1018 atom/ cm3 of Phosphorous. If the ratio of electron to hole mobility is 3, then the ratio of conductivity of the sample A to that of B is: (a) (c)
9.
3 2
(b)
1 3
(d)
12.
13.
11.
EC – EV kT N – ln V 2 2 NC
The radius of the first Bohr orbit of electrons in hydrogen atom is 0.529 Å. What is the radius of the second Bohr orbit in singly ionized helium atom ? (a) 1.058 Å
(b) 0.264 Å
(c) 10.58 Å
(d) 0.0264 Å
For which one of the following materials, is the Hall coefficient closest to zero ? (b) Insulator (c) Intrinsic semiconductor (d) Alloy
14.
15.
At temperature of 298 Kelvin, Silicon is not suitable for most electronic applications, due to small amount of conductivity. This can be altered by (a) Gettering
(b) Doping
(c) Squeezing
(d) Sintering
(c) 0.034 m2 V–1 s–1
The energy gap in the energy band structure of a material is 9 eV at room temperature. The material is
(d) 0.044 m2 V–1 s–1
(a) Semiconductor
(b) Conductor
Doped silicon has Hall-coefficient of 3.68 × 10–4 m3C–1 and then its carrier concentration value is:
(c) Metal
(d) Insulator
(b) 0.024 m2 V–1 s–1
10.
(d)
(a) Metal
The Hall-coefficient of a specimen of doped semiconductor is 3.06 × 10–4 m–3C–1 and the resistivity of the specimen is 6.93 × 10–3 m . The majority carrier mobility will be: (a) 0.014 m2 V–1 s–1
EC EV kT N ln V 2 2 NC
IES – 2015
2 3
1 2
(c)
16.
By doping Germanium with Gallium, the types of semiconductors formed are :
(a) 2.0 × 1022 m–3
(b) 2.0 × 10–22 m–3
1. N type
2. P type
(c) 0.2 × 1022 m–3
(d) 0.2 × 10–22 m–3
3. Intrinsic
4. Extrinsic
The position of the intrinsic Fermi level of an undoped semiconductor (EFi) is given by:
Which of the above are correct ? (a) 1 and 4
(b) 2 and 4
EC – EV kT N (a) ln V 2 2 NC
(c) 1 and 3
(d) 2 and 3
(b)
EC EV kT N – ln V 2 2 NC
17.
An n-type of silicon can be formed by adding impurity of : 1. Phosphorous
2. Arsenic
Basic Electronic Engineering
21
ANSWER KEY 1. (b)
26. (b)
51. (b)
76. (a)
101. (d)
126. (c)
2. (a)
27. (b)
52. (a)
77. (b)
102. (a)
127. (d)
3. (b)
28. (c)
53. (b)
78. (a)
103. (d)
128. (b)
4. (b)
29. (b)
54. (c)
79. (d)
104. (b)
129. (c)
5. (a)
30. (d)
55. (d)
80. (d)
105. (a)
130. (c)
6. (b)
31. (c)
56. (c)
81. (d)
106. (c)
131. (c)
7. (a)
32. (b)
57. (b)
82. (b)
107. (b)
132. (b)
8. (c)
33. (b)
58. (d)
83. (d)
108. (c,d)
133. (b)
9. (d)
34. (a)
59. (a)
84. (c)
109. (c)
134. (a)
10. (a)
35. (b)
60. (c)
85. (d)
110. (b)
135. (a)
11. (c)
36. (a)
61. (c)
86. (d)
111. (b)
136. (a)
12. (a)
37. (d)
62. (b)
87. (b)
112. (b)
137. (c)
13. (a)
38. (d)
63. (b)
88. (c)
113. (d)
138. (a)
14. (b)
39. (c)
64. (a)
89. (d)
114. (d)
139. (c)
15. (d)
40. (a)
65. (d)
90. (c)
115. (d)
140. (a)
16. (b)
41. (a)
66. (c)
91. (c)
116. (d)
141. (a)
17. (a)
42. (b)
67. (b)
92. (b)
117. (b)
142. (d)
18. (b)
43. (b)
68. (b)
93. (b)
118. (b)
143. (c)
19. (d)
44. (d)
69. (d)
94. (c)
119. (a)
144. (a)
20. (c)
45. (b)
70. (c)
95. (d)
120. (a)
145. (d)
21. (c)
46. (a)
71. (c)
96. (c)
121. (b)
146. (b)
22. (b)
47. (b)
72. (b)
97. (a)
122. (b)
147. (d)
23. (a)
48. (b)
73. (c)
98. (d)
123. (d)
148. (c)
24. (a)
49. (b)
74. (a)
99. (d)
124. (b, c)
149. (b)
25. (a)
50. (b)
75. (b)
100. (d)
125. (a)
150. (c)
22
Sol–1:
ESE Topicwise Objective Solved Paper-I 1991-2018
(b)
ni = intrinsic carrier concentration • If the temperature of semiconductor increases, the concentration of charge carriers (electrons and holes) is also increased. Hence the conductivity of a semiconductor is increased accordingly. • The relation between temperature and concentration of charge carriers in pure semiconductor is given as
When silicon devices can be employed for higher temperature limit (190°C to 200°C) when compared to germanium devices (85°C to 100°C) implies that silicon devices can be used in high-power applications as they support flow of high amounts of currents. That is statement4 is true therefore statements 1, 2, 3 are incorrect as per options. Sol–2:
(a) Given that
n 2i = A 0 T 3 e –EG0 / KT
1 indium atom is doped for every 2.5 × 108 Ge atom intrinsic carrier concentration (ni) = 2.5 × 1019/m
Sol–4.
• For insulators, the energy band-gap is greater than 6 eV, whereas for semiconductors, the energy band gap is approximately equal to 1 eV.
number of germanium atoms = 4.4 × 1028 1 2.5 × 108 ? 4.4 × 1028
• Thus, for insulators, the band-gap is relatively wide, whereas for semiconductors, it is narrow.
Concentration of p-type impurities =
4.4 ×1028 2.5 ×10
where T is the temperature in Kelvin scale. (b)
Sol–5.
8
(a) In an extrinsic semiconductor,
= 1.76×1020 atom
Extrinsic Semicondcutor
We know (NA) (np) = n 2i (1.76×1020) (np) = (2.5 × 1019)2 (np) =
2.52 ×1038
p-type
n-type
Majority carrier concentration
pp = N A
nn = N D
Minority carrier concentration
np
Condutivity
1.76 ×1020
np = 3.551 × 1018 /m3 Sol–3.
p = no of holes
ni ND
A
p
Thus, conductivity of an extrinsic semiconductor significantly depends upon majority charge carrier, generated due to impurity doping.
• In a pure semiconductor the no of holes and electrons are equal and is given by where n = no of electrons
2
pn
p q p p p n p n n q n n n pn p qn n n qp p p n n pn p p n p n qN D n qN p
(b)
n = p = ni
2
ni NA
Sol–6.
(b) Given, Na = 3 ×1016 / cm3
Basic Electronic Engineering
23 Sol–9.
Nd = 2.5 ×1015 / cm 3
and
(d) Given,
Since, Na > Nd, semiconductor is of ptype.
R H = 3.06 ×10 –4 m –3C –1
Majority carrier (hole) concentration,
and 6.93 10 3 m
pp = Na – Nd
Resistivity,
= 2.75 ×1016 / cm 3
Minority carrier (electron) concentration,
2
n 2i 1.5 ×1013 / cm3 = np = pp 2.75 ×1016
Sol–10.
(a) RH = 3.68 10 4 m3C 1
(a)
1 1 Hall Coefficient, RH = nq v
Given, n = p i = 2.32 / m ni = 2.5 ×1019 / cm 3
Conductivity, i = n i q n + pi q p =
= 2ni qn n i = p i and n = p
i n = 2n q i
2.32 m2 V – s = 2 × 2.5 ×1019 ×1.6 ×10 –19
= 0.29 m2 V – s
n = p = 0.29m 2 V – s 0.3 m 2 V – s Sol–8.
(c) Given, pA = 1018 / cm3 and n B = 1018 / cm3
n p = 3
R H 3.06 10 4 0.044 m 2 V s 3 6.93 10
Given,
= 0.818 ×1010 / cm3 Sol–7.
1 n = R q H
1 3.68 10
4
Conductivity of sample B, B = nBqn Thus,
pAqp
=
A B
1 = 3
nBqn
1.6 10 19
n = 1.7 10 22 / m3
2 10 22 / m3 Sol–11.
(c) Intrinsic electron concentration, ni EC EFi = N C exp kT Intrinsic hole concentration, pi E Fi EV = N V exp kT For an intrinsic Semiconductor, ni = pi E EFi EFi EV NC exp C N V exp kT kT
Conductivity of sample A, A = pA q p
A B
1 RH
EFi =
EC E V kT N V ln 2 2 NC
Sol–12. (a) Radius of Bohr’s orbit of Hydrogen and Hydrogen like species is calculated as r =
n2h 2 2
4 me
2
1 , Z