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TRANSCRIPT
ELECTRONICS ANDCOMMUNICATION ENGINEERING
ESE TOPICWISE OBJECTIVE
SOLVED PAPER-I
From (1991 – 2018)
Office : Phone : F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 011-26522064Mobile : E-mail:
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iesmasterpublications.com, iesmaster.org
Typeset at : IES Master Publication, New Delhi-110016
All rights reserved.Copyright © 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 ofIES MASTER, New Delhi. Violates are liable to be legally prosecuted.
First Edition : 2017
Second Edition : 2018
IES MASTER PUBLICATIONF-126, (Lower Basement), Katwaria Sarai, New Delhi-110016
Phone : 011-26522064, Mobile : 8130909220, 9711853908
E-mail : [email protected], [email protected]
Web : iesmasterpublications.com, iesmaster.org
PREFACE
It is an immense pleasure to present topic wise previous years solved paper of EngineeringServices Exam. This booklet has come out after long observation and detailed interaction with thestudents 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 thesolutions, students will be able to understand the basic concepts and will apply these concepts insolving other questions that might be asked in future exams.
Engineering Services Exam is a gateway to a immensly satisfying and high exposure job in engineeringsector. The exposure to challenges and opportunities of leading the diverse field of engineeringhas been the main reason for students opting for this service as compared to others. To facilitateselection into these services, availability of arithmetic solution to previous year paper is the needof the day. Towards this end this book becomes indispensable.
Mr. Kanchan Kumar ThakurDirector–IES Master
Note: Direction of all Assertion Reasoning (A–R) type of questions coveredin this booklet is as follows:
DIRECTIONS:
The following four items consist of two statements, one labelled as ‘AssertionA’ and the other labelled as ‘Reason R’. You are to examine these twostatements carefully and select the answer to these two statements carefullyand 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 ofA
(b) Both A and R are individually true but R is not the correct explanationof 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 coveredin 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 statementscarefully and select the answers to these items using the code given below:
(a) Both Statement : (I) and Statement (II) are individually true andStatement (II) is the correct explanation of Statement (I).
(b) Both Statement (I) and Statement (II) are individually true butStatement (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.
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
CONTENTS
S YL L A BU SBasics of semiconductors; Diode/Transistor basics and characteristics; Diodesfor different uses; Junction & Field Effect Transistors (BJTs, JFETs, MOSFETs);Transistor amplifiers of different types, oscillators and other circuits; Basics ofIntegrated 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.
1
Contents1. 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
IES – 2018
1. Silicon devices can be employed for a highertemperature limit (190 ºC to 200 ºC) ascompared to germanium devices (85 ºC to100 ºC). With respect to this, which of thefollowing are incorrect?1. Higher resistivity of silicon2. Higher gap energy of siliocn3. Lower intrinsic concentration of silicon4. Use of silicon devices in high-power
applicationsSelect the correct answer using the code givenbelow:(a) 1, 2 and 4 (b) 1, 2 and 3(c) 1, 3 and 4 (d) 2, 3 and 4
2. A sample of germanium is made p-type beaddition of indium at the rate of one indiumatom for every 2.5×108 germanium atoms.Given, ni = 2.5×1019 /m3 at 300 K and thenumber of germanium atoms per m3 = 4.4 ×1028. What is the value of np?(a) 3.55 × 1018/m3 (b) 3.76 × 1018/m3
(c) 7.87 × 1018/m3 (d) 9.94 × 1018/m3
IES – 2016
3. The electrical conductivity of puresemiconductor is:(a) Proportional to temperature(b) Increases exponentially with temperature
(c) Decreases exponentially with temperature(d) Not altered with temperature.
4. Which one of the following statements iscorrect?(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 isnarrow
(c) The band-gap is narrow in width for boththe insulators and conductors
(d) The band-gap is equally wide for bothconductors and semiconductors.
5. In an extrinsic semiconductor the conductiv-ity 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.6. A Ge sample at room temperature has
intrinsic carrier concentration, ni = 1.5 ×1013 cm–3 and is uniformly doped with acceptorof 3×1016 cm–3 and donor of 2.5×1015 cm–3.Then, the minority charge carrierconcentration is:(a) 0.918 × 1010 cm–3
(b) 0.818 × 1010 cm–3
1
Basic Electronic Engineering 3
(c) 0.918 × 1012 cm–3
(d) 0.818 × 1012 cm–3
7. Assume that the values of mobility of holesand that of electrons in an intrinsicsemiconductor are equal and the values ofconductivity and intrinsic electron densityare 2.32 / m and 2.5 × 1019/m3 respectively.Then, the mobility of electron/hole isapproximately:(a) 0.3 m2/Vs (b) 0.5 m2/Vs(c) 0.7 m2/Vs (d) 0.9 m2/Vs
8. A silicon sample A is doped with 1018 atom/cm3 of Boron and another silicon sample B ofidentical dimensions is doped with 1018 atom/cm3 of Phosphorous. If the ratio of electronto hole mobility is 3, then the ratioof conductivity of the sample A to that of Bis:
(a) 32
(b) 23
(c) 13 (d) 1
29. The Hall-coefficient of a specimen of doped
semiconductor is 3.06 × 10–4 m–3C–1 and theresistivity of the specimen is 6.93 × 10–3 m .The majority carrier mobility will be:(a) 0.014 m2 V–1 s–1
(b) 0.024 m2 V–1 s–1
(c) 0.034 m2 V–1 s–1
(d) 0.044 m2 V–1 s–1
10. Doped silicon has Hall-coefficient of 3.68 ×10–4 m3C–1 and then its carrier concentrationvalue is:(a) 2.0 × 1022 m–3 (b) 2.0 × 10–22 m–3
(c) 0.2 × 1022 m–3 (d) 0.2 × 10–22 m–3
11. The position of the intrinsic Fermi level ofan undoped semiconductor (EFi) is given by:
(a) C V V
C
E – E NkT ln2 2 N
(b) C V V
C
E E NkT– ln2 2 N
(c) C V V
C
E E NkT ln2 2 N
(d) C V V
C
E – E NkT– ln2 2 N
IES – 2015
12. The radius of the first Bohr orbit of electronsin hydrogen atom is 0.529 Å. What is theradius of the second Bohr orbit in singlyionized helium atom ?(a) 1.058 Å (b) 0.264 Å(c) 10.58 Å (d) 0.0264 Å
13. For which one of the following materials, isthe Hall coefficient closest to zero ?(a) Metal(b) Insulator(c) Intrinsic semiconductor(d) Alloy
14. At temperature of 298 Kelvin, Silicon is notsuitable for most electronic applications, dueto small amount of conductivity. This can bealtered by(a) Gettering (b) Doping(c) Squeezing (d) Sintering
15. The energy gap in the energy band structureof a material is 9 eV at room temperature.The material is(a) Semiconductor (b) Conductor(c) Metal (d) Insulator
16. By doping Germanium with Gallium, thetypes of semiconductors formed are :1. N type 2. P type3. Intrinsic 4. ExtrinsicWhich of the above are correct ?(a) 1 and 4 (b) 2 and 4(c) 1 and 3 (d) 2 and 3
17. An n-type of silicon can be formed by addingimpurity of :1. Phosphorous 2. Arsenic
Basic Electronic Engineering 21
ANSWER KEY
1. (b)
2. (a)
3. (b)
4. (b)
5. (a)
6. (b)
7. (a)
8. (c)
9. (d)
10. (a)
11. (c)
12. (a)
13. (a)
14. (b)
15. (d)
16. (b)
17. (a)
18. (b)
19. (d)
20. (c)
21. (c)
22. (b)
23. (a)
24. (a)
25. (a)
26. (b)
27. (b)
28. (c)
29. (b)
30. (d)
31. (c)
32. (b)
33. (b)
34. (a)
35. (b)
36. (a)
37. (d)
38. (d)
39. (c)
40. (a)
41. (a)
42. (b)
43. (b)
44. (d)
45. (b)
46. (a)
47. (b)
48. (b)
49. (b)
50. (b)
51. (b)
52. (a)
53. (b)
54. (c)
55. (d)
56. (c)
57. (b)
58. (d)
59. (a)
60. (c)
61. (c)
62. (b)
63. (b)
64. (a)
65. (d)
66. (c)
67. (b)
68. (b)
69. (d)
70. (c)
71. (c)
72. (b)
73. (c)
74. (a)
75. (b)
76. (a)
77. (b)
78. (a)
79. (d)
80. (d)
81. (d)
82. (b)
83. (d)
84. (c)
85. (d)
86. (d)
87. (b)
88. (c)
89. (d)
90. (c)
91. (c)
92. (b)
93. (b)
94. (c)
95. (d)
96. (c)
97. (a)
98. (d)
99. (d)
100. (d)
101. (d)
102. (a)
103. (d)
104. (b)
105. (a)
106. (c)
107. (b)
108. (c,d)
109. (c)
110. (b)
111. (b)
112. (b)
113. (d)
114. (d)
115. (d)
116. (d)
117. (b)
118. (b)
119. (a)
120. (a)
121. (b)
122. (b)
123. (d)
124. (b, c)
125. (a)
126. (c)
127. (d)
128. (b)
129. (c)
130. (c)
131. (c)
132. (b)
133. (b)
134. (a)
135. (a)
136. (a)
137. (c)
138. (a)
139. (c)
140. (a)
141. (a)
142. (d)
143. (c)
144. (a)
145. (d)
146. (b)
147. (d)
148. (c)
149. (b)
150. (c)
22 ESE Topicwise Objective Solved Paper-I 1991-2018
Sol–1: (b)When silicon devices can be employedfor higher temperature limit (190°C to200°C) when compared to germaniumdevices (85°C to 100°C) implies thatsilicon devices can be used in high-powerapplications as they support flow of highamounts of currents. That is statement-4 is true therefore statements 1, 2, 3 areincorrect as per options.
Sol–2: (a)Given that1 indium atom is doped for every 2.5 ×108 Ge atomintrinsic carrier concentration (ni) = 2.5× 1019/mnumber of germanium atoms = 4.4 ×1028
1 2.5 × 108
? 4.4 × 1028
Concentration of p-type impurities
=28
84.4 ×102.5 ×10
= 1.76×1020 atomWe know
(NA) (np) = 2in
(1.76×1020) (np) = (2.5 × 1019)2
(np) = 2 38
202.5 ×101.76 ×10
np = 3.551 × 1018 /m3
Sol–3. (b)• In a pure semiconductor the no of
holes and electrons are equal and isgiven by
n = p = ni
where n = no of electronsp = no of holes
ni = intrinsic carrierconcentration• If the temperature of semiconductor
increases, the concentration of chargecarriers (electrons and holes) is alsoincreased. Hence the conductivity ofa semiconductor is increasedaccordingly.
• The relation between temperatureand concentration of charge carriersin pure semiconductor is given as
2in = G0/KT–E3
0A T ewhere T is the temperature in Kelvinscale.
Sol–4. (b)• For insulators, the energy band-gap
is greater than 6 eV, whereas forsemiconductors, the energy band gapis approximately equal to 1 eV.
• Thus, for insulators, the band-gap isrelatively wide, whereas forsemiconductors, it is narrow.
Sol–5. (a)In an extrinsic semiconductor,
Extrinsic Semicondcutor
Majority carrierconcentration
p-type n-type
p = Np A n = Nn D
Minority carrierconcentration
2i
pA
nnN
2i
nD
npN
Condutivity
p p p p n
p p
p p
p A p
q p nqp
p nqN
n n n n p
n n
n n
n D n
q n pqn
n pqN
Thus, conductivity of an extrinsicsemiconductor significantly dependsupon majority charge carrier, generateddue to impurity doping.
Sol–6. (b)Given,
Na = 16 33 ×10 / cm
Basic Electronic Engineering 23
and Nd = 15 32.5 ×10 / cm
Since, Na > Nd, semiconductor is of p-type.Majority carrier (hole) concentration,
pp = a dN –N
= 16 32.75 ×10 / cm
Minority carrier (electron) concentration,
np = 2132
i 316
p
1.5 ×10n = / cmp 2.75 ×10
= 10 30.818 ×10 / cmSol–7. (a)
Given, n = p
i = 2.32/ m
ni = 19 32.5 ×10 / cmConductivity, i = i n i pn q + p q
= i n2n q
i i n pn = p and =
n = i
i2n q
= 219 –192.32 m V – s
2 × 2.5 ×10 ×1.6 ×10= 20.29 m V – s
2 2n p= = 0.29m V – s 0.3 m V – s
Sol–8. (c)Given,
18 3Ap = 10 / cm and 18 3
Bn = 10 / cm
n
p
= 3
Conductivity of sample A, A A p= p q
Conductivity of sample B, B B n= n q Thus,
A
B
=
A p
B n
p qn q
A
B
=
13
Sol–9. (d)Given,
–4 –3 –1HR = 3.06 ×10 m C
and 36.93 10 m
Resistivity,
HR1
42H
3R 3.06 10 0.044m V s
6.93 10Sol–10. (a)
Given,
RH = 4 3 13.68 10 m C
Hall Coefficient, RH = v
1 1nq
n =H
1R q
= 4 191
3.68 10 1.6 10 n = 22 31.7 10 / m
22 32 10 / mSol–11. (c)
Intrinsic electron concentration, ni
=
C FiC
E EN expkT
Intrinsic hole concentration, pi
=
Fi VV
E EN expkT
For an intrinsic Semiconductor,ni = pi
C Fi Fi VC V
E E E EN exp N expkT kT
EFi =
C V V
C
E E NkT n2 2 N
l
Sol–12. (a)Radius of Bohr’s orbit of Hydrogen andHydrogen like species is calculated as
r =
2 2
2 2n h 1
Z4 me,
