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1991-2019
ELECTRONICS AND
COMMUNICATION ENGINEERING
ESE TOPICWISE OBJECTIVESOLVED PAPER-I
Typeset at : IES Master Publication, New Delhi-110016
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First Edition : 2017
Second Edition : 2018
Third Edition : 2019
IES MASTER PUBLICATIONF-126, (Lower Basement), Katwaria Sarai, New Delhi-110016
Phone : 011-26522064, Mobile : 8130909220, 9711853908
E-mail : [email protected]
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Engineering Services Examination is the gateway to an immensely satisfying job in the engineering sectorof India that offers multi-faceted exposure. The exposure to challenges and opportunities of leading thediverse field of engineering has been the main reason behind engineering students opting for EngineeringServices as compared to other career options. To facilitate selection into these services, availability ofarithmetic solution to previous years’ paper is the need of the day.
It is an immense pleasure to present previous years’ topic-wise objective solved papers of EngineeringServices Examination (ESE). This book is an outcome of regular and detailed interaction with the studentspreparing for ESE every year. It includes solutions along with detailed explanation to all questions. Theprime objective of bringing out this book is to provide explanation to each question in such a mannerthat just by going through the solutions, students will be able to understand the basic concepts and havethe capability to apply these concepts in solving other questions that might be asked in future exams.Towards the end, this book becomes indispensable for every ESE aspiring candidate.
IES Master PublicationNew Delhi
PREFACE
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...................................................................... 01–125
2. Materials and Components ....................................................................... 126–223
3. Measurement and Instrumentations............................................................ 224-338
4. Network Theory ......................................................................................... 339-570
5. Analog Electronics .................................................................................... 571-719
6. Digital Electronics ..................................................................................... 720-855
7. Basic Electrical Engineering ...................................................................... 856-868
CONTENT
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 othercircuits; Basics of Integrated Circuits (ICs); Bipolar, MOS and CMOS ICs; Basics of linear ICs, operationalamplifiers and their applications-linear/non-linear, Optical sources/detectors; Basics of Opto electronics andits applications.
CONTENTS
1. Basics of Semiconductor ................................................................................ 1-41
2. p-n Junction Diode and Opto Electronics ..................................................... 42-72
3. Bipolar Junction Transistors (BJTs) ............................................................... 73-89
4. Field Effect Transistors (FETs) ................................................................... 90-108
5. Silicon Controlled Rectifiers ....................................................................... 109-117
6. Integrated Circuits (ICs) ............................................................................. 118-125
UNIT-1Basic Electronic
Engineering
1 BASICS OFSEMICONDUCTORS
IES – 20181. Silicon devices can be employed for a higher
temperature limit (190 ºC to 200 ºC) as comparedto germanium devices (85 ºC to 100 ºC). Withrespect to this, which of the following areincorrect?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 – 20163. The electrical conductivity of pure semiconductor
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 is correct?(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 both
the insulators and conductors(d) The band-gap is equally wide for both
conductors and semiconductors.
5. In an extrinsic semiconductor the conductivitysignificantly 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 intrinsiccarrier concentration, ni = 1.5 × 1013 cm–3 andis uniformly doped with acceptor of 3×1016 cm–
3 and donor of 2.5×1015 cm–3. Then, the minoritycharge carrier concentration is:(a) 0.918 × 1010 cm–3
(b) 0.818 × 1010 cm–3
(c) 0.918 × 1012 cm–3
(d) 0.818 × 1012 cm–3
7. Assume that the values of mobility of holes andthat of electrons in an intrinsic semiconductorare equal and the values of conductivity andintrinsic electron density are 2.32 / m and2.5 × 1019/m3 respectively. Then, the mobility ofelectron/hole is approximately:(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 B is:
(a) 32
(b) 23
IES MASTER Publication
BASIC ELECTRONIC ENGINEERING 3E&T Engineering
(c) 13 (d) 1
2
9. The Hall-coefficient of a specimen of dopedsemiconductor 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 of anundoped 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 electrons inhydrogen atom is 0.529 Å. What is the radiusof the second Bohr orbit in singly ionized heliumatom ?(a) 1.058 Å (b) 0.264 Å(c) 10.58 Å (d) 0.0264 Å
13. For which one of the following materials, is theHall 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, due tosmall amount of conductivity. This can bealtered by(a) Gettering (b) Doping(c) Squeezing (d) Sintering
15. The energy gap in the energy band structure ofa material is 9 eV at room temperature. Thematerial is(a) Semiconductor (b) Conductor(c) Metal (d) Insulator
16. By doping Germanium with Gallium, the typesof semiconductors formed are :1. N type2. P type3. Intrinsic4. 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. Arsenic3. Boron 4. AluminiumWhich of the above are correct ?(a) 1 and 2 (b) 2 and 3(c) 3 and 4 (d) 1 and 4
18. According to Einstein’s relationship for asemiconductor, the ratio of diffusion constant tothe mobility of the charge carriers is(a) Variable and is twice the volt equivalent of
the temperature(b) Constant and is equal to the volt equivalent
of the temperature(c) Equal to two and is twice the volt equivalent
of the temperature(d) Equal to one and is equal to the volt
equivalent of the temperature
19. The number of holes in an N-type silicon withintrinsic value 1.5 × 1010 /cm3 and dopingconcentration of 1017/cm3, by using mass-actionlaw is(a) 6.67 × 106/cc (b) 4.44 × 10–25/cc(c) 1.5 × 10–24/cc (d) 2.25 × 103/cc
20. Statement (I): Hall voltage is given by
H HI.HV Rt
where I is the current, H is themagnetic field strength, t is the thickness ofprobe and RH is the Hall constant.Statement (II): Hall effect does not sense thecarrier concentration.
IES MASTER Publication
BASIC ELECTRONIC ENGINEERING 19E&T Engineering
EXPLANATIONS
Sol–1: (b)When silicon devices can be employed forhigher temperature limit (190°C to 200°C)when compared to germanium devices (85°Cto 100°C) implies that silicon devices canbe used in high-power applications as theysupport flow of high amounts of currents.That is statement-4 is true thereforestatements 1, 2, 3 are incorrect as peroptions.
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 is given byn = 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 of asemiconductor is increased accordingly.
• The relation between temperature andconcentration of charge carriers in puresemiconductor is given as
2in = G0/KT–E3
0A T ewhere T is the temperature in Kelvin scale.
Sol–4. (b)• For insulators, the energy band-gap is
greater than 6 eV, whereas forsemiconductors, the energy band gap isapproximately 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 depends uponmajority charge carrier, generated due toimpurity doping.
Sol–6. (b)Given,
Na = 16 33 ×10 / cm
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 / cm
20 ESE Topicwise Objective Solved Paper-I 1991-2019 E&T Engineering
Sol–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,
where n = principal quantum number oforbit and Z = Atomic numberFor Hydrogen,
rH=
2 2
2 2h 1 0.529Å
14 me...(i)
For Singly ionized Helium (He+) :-n = 2 (for second Bohr orbit)
and Z = 2
rHe =
2 2
2 2h 2
24 me= 0.529 × 2Å
( from Eq. (i),
2
2 2h
4 me= 0.529)
= 1.058Å.Sol–13. (a)
Hall coefficient,
RH =1
nqFor metals, ‘n’ is very large. Hence RH 0.
Sol–14. (b)Doping is the process of adding impurity toan intrinsic (Pure) semiconductor toincrease its conductivity. Doping introduces