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GATEPRACTICE BOOKLET

Electronics & Communication Engineering

(Volume - II)Electronic Devices & Circuits, Electromagnetics, Communication Systems,

Engineering Mathematics & General Aptitude (Verbal & Numerical Ability)

1116 Expected Questions with Solutions

Copyright © ACE Engineering Publications 2018

All rights reserved.

Published at :

Authors : Subject experts of ACE Engineering Academy, Hyderabad

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ForewordGATE in Electronics & Communication Engineering

GATE PRACTICE BOOKLET (Bits & Bytes) (VOLUME - II )

Dear Students,

Solutions of all previous GATE Questions are already available. Every year about

20% of questions will have repetitive nature. However, rest of the questions are from

untapped areas (never asked areas) and few from Previous Engineering Services & Civil

Services Questions. Keeping this in view, possible questions are prepared in various

subjects (chapter wise) along with their hints/solutions. The student is advised to

practice the questions systematically so that their chances of getting high score in GATE Exam will

increase.

The student is advised to solve the problems without referring to the solutions. The student has to

analyze the given question carefully, identify the concept on which the question is framed, recall

the relevant equations, find out the desired answer, verify the answer with the final key such as (a),

(b), (c), (d), then go through the hints to clarify his answer. This will help to face numerical answer

questions better. The student is advised to have a standard text book ready for reference to strengthen

the related concepts, if necessary. The student is advised not to write the solution steps in the space

around the question. By doing so, he loses an opportunity of effective revision.

As observed in the GATE Exam, number of sets may be possible, being online exams. Hence, don’t

skip any subject. All are equally important.

It is believed that this book is a Valuable aid to the students appearing for competitive exams like IES,

ISRO and Other PSU’s. This book can also be used by fresh Teachers in Engineering in improving

their Concepts.

With best wishes to all those who wish to go through the following pages.

Y.V. Gopala Krishna Murthy,M Tech. MIE,

Chairman & Managing Director,ACE Engineering Academy,

ACE Engineering Publications.

Electronics & Communication Engineering GATE PRACTICE BOOKLET (Bits & Bytes)

(VOLUME - II )

MAIN INDEX

S.No. Name of the Subject Page No.

1 Electronic Devices & Circuits 1 - 76

2 Electromagnetics 77 - 166

3 Communication Systems 167 - 267

4 Engineering Mathematics 268 - 325

5 General Aptitude (Verbal & Numerical Ability) 326 - 363

ACE Engg. Publications Hyderabad|Delhi|Bhopal |Pune |Bhubaneswar |Bengaluru |Lucknow |Patna|Chennai |Vijayawada |Vizag | Tirupati |Kukatpally | Kolkata

01. An n-type silicon bar of 0.1 cm long and

100 m2 in cross sectional area has a majority carrier concentration of

5 1020/m3 and the carrier mobility is

0.13 m2/V-s at 300 K. If the charge of an electron is 1.610–19 coulomb, then the resistance of the bar is

(a) 108 (b) 104

(c) 961.5 k (d) 960 M

02. The electron and hole concentrations in an intrinsic semiconductor are ni and pi

respectively. When doped with a p-type material, these change to n and p, respectively. Then

(a) n + p = ni + pi (b) n + ni = p + pi (c) npi = ni p (d) np = ni pi 03. Two initially identical samples A and B

of pure germanium are doped with donors concentrations of 1 1020 m-3 and 3 1020 m–3 respectively. If the hole concentration in A is 9 1012 m-3, then the hole concentration in B at the same temperature will be

(a) 3 1012 m-3 (b) 7 1012 m-3

(c) 11 1012 m-3 (d) 27 1012 m-3

04. Silicon contains 5 1028 atoms per cubic metre. If it is doped with two parts per million of arsenic, then the electron density at room temperature will be approximately

(a) 4 1023 m-3 (b) 1023 m-3

(c) 2 106 m-3 (d) 104 m-3

05. The mobilities of electron and hole in silicon are respectively 1350 cm2/V-sec and 480 cm2/V-sec and intrinsic carrier concentration of silicon is 1.5 1010/cm3. Then find the majority and minority concentrations of a p-type semiconductor which is having resistivity of 10 -cm.

(a) 1.31015 cm-3 and 1.73105 cm-3

(b) 1.31016 cm-3 and 1.73104 cm-3

(c) 1.31014 cm-3 and 1.73106 cm-3

(d) 1.31015 cm-3 and 1.73108 cm-3

06. In a semiconductor, it is observed that

(3/4)th of the current is carried by electrons and (1/4)th by holes. If the drift speed of electrons is two times that of holes, the relation between electron and hole concentrations is

(a) n = 1.5 p (b) n = 2 p (c) n = 3 p (d) p = 1.5 n

07. The intrinsic carrier concentration of silicon

(Eg = 1.12eV) at 300K is 1.51010cm-3,

then the value of ni for Si at 400K (Boltzmann constant K = 8.6210-5eV/K).

(a) 1.51010cm-3 (b) 3.21011cm

(c) 5.21012cm-3 (d) 6.81013cm-3

08. Find the conductivity of the n-type silicon, which is doped with ND = 1017cm-3. Assuming 50 percent of the donors are ionized at 300K.

(n(si) = 1000, p(si) = 350, n(Ge) = 3500 and p(Ge) = 1500 are in cm2/V-sec)

(a) 8(-cm)-1 (b) 12(-cm)-1

(c) 16(-cm)-1 (d) 10(-cm)-1

Volume‐II(1116–Questions)

SemiconductorPhysics 1Chapter

: 4 : Electronic Devices


09. The current density of n-type germanium is 100 A/m2 and the resistivity of 0.5 –m. If the mobility of electron is 0.4 m2/V-sec. The drift velocity is _______m/sec

10. n-type devices are preferred over p-type

devices because of the following reason. (a) Mobility of hole > Mobility of electron (b) Mobility of hole = Mobility of electron (c) For same doping n type materials give

less current than p type materials (d) For same doping n type materials give

more current than p type materials

11. The majority carriers in an n-type

semiconductor have an average drift velocity() in a direction perpendicular to a uniform magnetic field B. The electric field E induced due to Hall effect acts in the direction

(a) B (b) B

(c) along (d) opposite to

12. Mobility of extrinsic semi conductor

depends on (a) temperature (b) electric field intensity (c) both (d) none 13. In an extrinsic semiconductor if (a) The resistivity decreases (b) The temperature increases (c) The photo conductivity is low 1. The doping concentration is low 2. The length of the semi - conductor is

reduced Resistivity is negative 3. The band gap is high

4. Resistance decreases

5. The doping concentration is increased

Codes:

A B C (a) 5 4 3 (b) 4 3 5 (c) 5 1 2 (d) 5 2 3

14. The current produced in a Ge sample of area 1 cm 2 and length 0.3 mm when a potential difference of 2V is applied across it [Assume n i = 2 10 19 m – 3,

n= 0.35 m 2 / V-sec and p = 0.17 m2/ V-sec] (a) 2 A (b) 1.11 A (c) 0.5 A (d) 100 mA 15. The diffusion current in a sample of Ge

having concentration gradient for electrons of 1.5 1022 electrons/m4 is ______ A/m2

[Assume the diffusion constant for electrons = 0.00120 m2/sec]

16. In a semiconductor sample the electron and hole mobilities are 0.15m2/V–S and 0.06m2/V–S respectively. The diffusion constant for electrons is 3.75 10-3 m2/sec. The diffusion constant for holes is

(a) 1.55 10-3 m2/s (b) 15.5 10-3 m2/s (c) 1.55 10-3 cm2/s (d) 155 10-3cm2/s

17. 2 volts supply is applied to the semiconductor bar of length 2 m. Then magnitude of electric field (in V/m) at x = 0.5 m is

(a) 4 106 (b) 106 (c) 2 106 (d) 3 106 18. In an N - type semiconductor, the position

of the Fermi level (a) is lower than the center of the energy

gap (b) is at the center of the energy gap (c) is higher than the center of the energy gap (d) can be anywhere depending upon the

doping concentration

: 5 : Semiconductor Physics


19. The current in a forward biased p+n junction diode is entirely due to diffusion of holes from p-side to n-side of the junction over the distance L = 10–3 cm. The incremental change in the hole concentration on n-side at the junction is p(0) = 1012 cm-3. The current density in the diode assuming that the diffusion coefficient of holes is 12 cm2/sec ______ mA/m2.

20. Assuming the Fermi level EF to be

independent of temperature, EF may be defined as the level with an occupancy probability of

(a) 0 % (b) 50 % (c) 75 % (d) 100 % 21. The effect of doping in intrinsic

semiconductors is to (a) move the Fermi level away from the

center of the forbidden band (b) move the Fermi level towards the center the forbidden band (c) change the crystal structure of the semiconductor (d) keep the Fermi level at the middle of the forbidden band

22. Which of the following statements are wrong. 1. Fermi level is closer to conduction band in n-type. 2. Fermi level is not used for intrinsic semi conductor. 3. Fermi level in P side of tunnel diode is inside conduction band 4. Fermi level comments on 50% occupancy.

(a) 1 & 4 (b) 2 & 3 (c) 1 & 2 (d) 3 & 4

23. Given Donor concentration as

2.63 1019cm–3 in silicon sample, then the temperature at which Fermi level coincides with edge of conduction band is [Assume mn = m]

(a) 310C (b) 10K (c) 310K (d) 300K

24. An n-type semiconductor has its Fermi level at 0.25 eV below the conduction band edge. If the density of semiconductor in conduction band at room temperature is 2.511025/m3, the majority carrier concentration is

(a) 1.6 1011 /m3 (b) 1.6 1021 /m3

(c) 2.5 1022 /m3 (d) 1.6 1014 /m3

25. Match List – I (Parameter for silicon) with List – II(values in MKS units) and select the correct answer using the codes given below the lists:

List – I A. Electron mobility B. Electron diffusion constant C. Diffusion length (hole) D. Product of electron and hole concentration

List – II 1. 5 1028

2. 0.13

3. 2.25 1032

4. 0.0035

5. 0.002

Codes: A B C D

(a) 2 4 5 1 (b) 4 2 1 3 (c) 2 4 5 3 (d) 4 2 3 5

26. Match List – I (Semiconductor parameters)

with List – II (Physical processes) and select the correct answer using the codes given below the lists:

List – I

A. Impurity Concentration B. Carrier mobility C. Carrier lifetime D. Intrinsic carrier concentration



BZ FY EY

Ix

dY

Y

X WLx

+ VH –

Z

List – II 1. Recombination 2. Band to Band transition 3. Scattering 4. Ion implantation

Codes: A B C D

(a) 3 4 2 1 (b) 4 3 2 1 (c) 3 4 1 2 (d) 4 3 1 2

27. The magnitude of induced hall voltage VH

in the given Figure is (Given n-type Germanium bar with

doping = 1.5 106 cm-3, BZ = 0.5 wb/m2,

dy = 4 10-3m, Ex = 700 v/m,

n = 0.38 m2 /vsec, Wz = 1 10-3 m)

(a) 133 mV (b) 133 V (c) 532V (d) 532 mV

28. Calculate majority and minority carrier

concentrations to convert an intrinsic Silicon into an n- type silicon of resistivity of 10 ohm-cm. [n = 1350 cm2/V-sec, ni = 1.5 1010 cm–3]

(a) 4.631015 cm-3 and 4.86105 cm-3

(d) 4.631016 cm-3 and 4.86104 cm-3

(d) 4.631014 cm-3 and 4.86105 cm-3

(d) 4.631015 cm-3 and 4.86108 cm-3

29. A sample of Si is doped with 1.0 1017 phosphorous atoms/cm3. If the applied current is 1 mA and magnetic field strength is 10–5 Wb/cm2, the magnitude of Hall voltage in a sample of 100 m thick is _____ V.

30. A Si sample is doped with Arsenic. Assume effective densities of states at edge of conduction band NC = 8.851018 cm-3 & KT = 0.026 (EC, EV are edges of conduction band, valence band respectively)

Find location of Fermi level (EF) if doping concentration is 4.41 1014 cm-3

(a) EF coincides with EC

(b) EF lies 0.257 eV above EC

(c) EF lies 0.257 eV below EC

(d) EF lies 0.257 eV below intrinsic Fermi level

31. As per Hall effect, if any specimen carrying a current I is placed in a transverse magnetic field B then an electric field `E’ is induced in the Specimen in a direction

(a) parallel to I

(b) perpendicular to B and parallel to I

(c) parallel to I and B

(d) perpendicular to both I and B

32. Which of the following is the unit of mobility to diffusion constant ratio?

(a) Per Volt (b) Volt (c) cm2/V-s (d) V/cm2

33. The relationship between mobility (), conductivity () and Hall coefficient (RH)

(a) = RH (b) = RH

(c) RH = (d) 2HR



34. In Hall effect, which of the following is true

(a) Magnetic Field & Electric Field are induced

(b) Magnetic Field is induced & electric field is applied

(c) Magnetic field is applied & electric field is induced

(d) Hall voltage has same polarities for both n & p type semiconductors

35. Determine the percentage of si atoms that

contribute a conduction electron hole pair at 300K, (given silicon concentration as 4.991022 atoms/cm3, ni = 1.48 1010cm–3). Assume that a covalent bonded semiconductor contributes approximately one electron hole pair per atom to the current conduction process.

(a) 0.297 10- 10 (b) 3.367 10- 10

(c) 0.438 10 – 2 (d) 2.27 10- 2 36. Find the density of valence band electrons

available to form conduction current. For a copper specimen given density = 8.92

gm/cm3, Atom weight = 63.54 g/mole and Avagadro’s number = 6.023 1023.

(a) 429.03 1022 cm- 3

(b) 858.06 1022 cm- 3

(c) 16.9 1022 cm- 3

(d) 8.45 1022 cm- 3

37. Assume that we have an infinitely long

n-type semiconductor bar at 300oK with an electric field of 1000 V/cm in the positive X-direction. At t = 0, electron-hole pairs are generated at x = 0 by a light pulse. At t = 1 sec, the charge concentration maximum is measured as a function of x, and it occurs at xp = 0.5 cm. The mobility (in cm2/V-sec) and diffusion constant (in cm2/sec) of holes respectively are

(a) 500, 13 (b) 1000, 50

(c) 250, 5 (d) 350, 20

38. The Hall coefficient of a specimen of doped silicon is found to be 3.6610-4 m3/C; the

resistivity of the specimen is 8.9310-3-m. Assuming single carrier conduction, then the mobility of the charge carriers_______ cm2 / V-s

39. At 50 K or 100K charge carriers are generated in an extrinsic semiconductor.

Which of the following reason is true ?

(a) Band to band transition

(b) Impurity Ionization

(c) EHP generation

(d) Lattice defects will not get introduced into band structure

40. Find the concentration of atoms in an

intrinsic germanium given

A0 = 6.021023 atoms/mole,

Atomic Weight, A = 72.6 gm/mole

& density, d = 5.32 gm/cm3

(a) 4.41 1022 atoms/cm3

(b) 82.15 1023 atoms/cm3

(c) 0.2261022 atoms/cm3

(d) 0.0121023 atoms/cm3

41. Two semiconductor materials have same properties except that material 1 has band gap at 0 K as 1.1eV and material 2 has a band gap at 0K as 1.21 eV. The ratio of square of intrinsic concentration of material 2 to material 1 is

(a) 0.1204 (b) 0.0145

(c) 68.71 (d) 8.289



42. Statement (I): Hall crystal can be used as a multiplier of two signals.

Statement (II): Hall voltage is proportional to the currents and voltages applied in perpendicular directions across the Hall crystal.

43. Statement (I): An n-type semiconductor behaves as an intrinsic semiconductor at very high temperatures

Statement (II): The breaking of the covalent bonds becomes a significant phenomenon at high temperatures.

44. The minimum value of for silicon is

(a) 3.87 10–6 ( - cm)-1

(b) 1.935 10–6 ( - cm)-1

(c) 2.58 10–6 ( - cm)-1

(d) 1.29 10–6 ( - cm)-1


01. Ans: (c) Sol: Conductivity of si due to majority carriers is n = ND qn = 51020 1.6 10–19 0.13 n = 10.4 (–m)–1

Resistivity = n

1

= 0.09615 -m

Resistance = R =A

l

= 12

2

10100

101.009615.0

R = 961.5 k 02. Ans: (d) Sol: intrinsic carrier concentrations are, ni = pi according to mass action law

n.p = 2in

n.p = ni (pi) n = number of electrons in p-doped p = number of holes in p-doped

03. Ans: (a) Sol: For sample A donor concentration ND = 11020m–3 hole concentration p = 91012 By mass action law

NDp = 2in 12202

i 109101n

322i 109n

For sample B donor concentration ND = 31020m–3

hole concentration = D

2i

N

n =

20

32

103

109

= 31012 m–3

04. Ans: (b) Sol: Number of silicon atoms = 51028 atoms/m3 Doped with 2 parts per million of arsenic

means 2 arsenic atoms added for each 106 silicon atoms

Number of arsenic atoms = 210

1056

28

= 1023 atoms/m3 Each arsenic atom will give one e– at room

temperature. Assuming all the atoms are ionized.

Number of electrons = 1023/m3

05. Ans: (a) Sol: Given resistivity () = 10 ohm –cm Conductivity = 0.1 (ohm - cm)–1 Given intrinsic silicon to p-type silicon

conversion For p-type majority carriers = holes = NA

Minority carriers = ND = A

2i

N

n

= nqn + pqp 0.1 = 1.6 10–19 (ND n +NA p)

0.1 =

pAnA

2i19 N

N

n106.1

0.1 = 1.6 10–19 (NA480)

0106.1

N

n 19n

A

2i

NA = 1.3 1015cm–3

ND = 15

20

A

2i

103.1

1025.2

N

n

= 1.73 105 cm–3

06. Ans: (a) Sol: I due to electrons In = nqn EA I due to holes IP = pq P EA

Given In = 4

3 (I)

Volume‐II(1116–Solutions)

SemiconductorPhysics 1Chapter



)I(4

1IP

p

n

p

n

P

n

p

n

EApq

EAnq3

I

I

dn

dp

n

p 33

p

n [ d = E]

2

13

p

n = 1.5

07. Ans: (c) Sol: Given Eg = 1.12 eV. kT/E3

02i

GeTAn

21

G

T

1

T

1

k2

E2/3

2

1

2i

1i eT

T

n

n

400

1

300

1

21062.8

12.12/3

2i

105

e400

300

n

105.1

4.52/3

e4

3

3

2i

10

109.2n

105.1

ni2 = 5.18 1012 / cm3

08. Ans: (a) Sol: According to charge neutrality principle. n + NA = p + ND n = number of electrons in conduction band p = number of holes in valence band. NA = number of acceptors ionized ND = number of donors ionized

n+0 = 17102

10

n = 0.5 1017 conductivity = nnq conductivity of si sample = si = nn q

= 0.5 1017 (1000)1.6 10-19 = 8 (-cm)-1 09. Ans: 20 Sol: n-type germanium current density Jd = nqn E =E = 100 A/m2

Conductivity 5.0

1

sistivityRe

1

= 2(-m)-1 100 A/m2 = 2(-m)-1 E E = 50 V/m Drift velocity vd = n E

vd = 0.4 50 vd = 20 m/sec.

10. Ans: (d) Sol: The current and mobility are related as EJ E)pene(J pn

Current, I= J.A Hence I The current through a semiconductor device

directly proportional to the mobility of charge carriers.

We know mobility of electrons is more than that of holes pn

Hence for the same doping levels, n-type materials give more current than p-type materials.

11. Ans: (b) Sol: Induced electric force Fn = –q(vn B) (for n-type)

Bvq

FE n

nn

BvE nn

E = B v

12. Ans: (c) Sol: For an extrinsic semiconductor mobility Tm and drift velocity d = E E = Electric field When, drift velocity saturated at = 107 cm/sec

Then E

E

1



So, in an extrinsic semiconductor mobility depends on both temperature and electric field

13. Ans: (a) Sol: In an extrinsic semiconductor, (A) If doping concentration is increased,

conductivity is increased so the resistivity decreases.

(B) As the temperature increases, conductivity increases and resistivity decreases. So, it exhibits negative temperature co-efficient of Resistivity.

(C) By using light energy, if electrons excited to conduction band from valence band, this gives photo conductivity. The photo conductivity is low means the available light energy is not enough to excite the electrons through band gap. i.e. semiconductor band gap is very high to excite the electrons with available light energy.

14. Ans: (b) Sol: AEqpqnI pn

Given pure Ge sample n = p = ni I = ni (n + p) qAE

3

41919

103.0

210106.117.035.0102

I = 1.11 A

15. Ans: 2.88 Sol: Given Concentration gradient

= 422 m/electrons105.1dx

dn

Diffusion current density = dx

dnDq n

2219 105.100120.0106.1 ID = 2.88 A/m2

16. Ans: (a) Sol: In a semiconductor at a temperature (TK) From the Einstein’s relation

eVKTDD

n

n

p

P

.

Given, n = 0.15 m2/V-s

p = 0.06 m2/V-s

Dn = 3.75 103 m2/sec

sec/m105.1DD 23n

n

pP

17. Ans: (b) Sol: Given that V = 2V, L = 210–4 cm

m/V10cm/V10102

2

L

VE 64

4

18. Ans: (c) Sol: In an N-type semiconductor position of

Fermi level.

D

CcF N

NlnKTEE

n

Centre of the energy gap = 2

EE VC

So Fermi level is above the centre of energy gap in N-type semiconductor.

19. Ans: 1.92 Sol:

EC Conduction band

nFE Fermi level

2

EE VC

EV Valence band

p(x)

x = 0

0np

p-side n-side

Incremental change in holes at junction p(0)

Minority hole



Diffusion current density of holes on n-side

= dx

)x(dpqDp

p(x) on n-side is p(x) = P

0

L/xn e)0('pp

0np = thermally generated holes

Lp = diffusion length of hole = 10–3 cm, given

Dp = 12 cm2/sec p(x) = P

0

L/xn e)0('pp

Lp

1e)0('p

dx

)x(dp pL/x

Lp

1)0('p

dx

)x(dp

)0x(

4153

12 mc1010

110

Diffusion current density

Jn = –qDp 0xdx

)x(dp

= –1.610–19(12) (–1015) JD = 1.92 mA/cm2

20. Ans: (b)

Sol: f(E) = KT/)EE( Fe1

1

If EF is independent of temperature

(i) E >> EF KT/)EE( Fe tends to

f(E) = 1

1 = 0

(ii) E << EF KT/)EE( Fe tends to 0

f(E) = 01

1

= 1

(iii) E = EF f(E) = 2

1

11

1

f(E) = Probability of electron occupying energy level ‘E’ from, the above for the energy levels above fermi energy level probability of electron occupancy = 0

For energy levels below fermi energy level, probability of electron occupancy is 100%.

But, at energy level equal to Fermi energy level probability of electron occupancy is 50%.

21. Ans: (a) Sol: For n-type doping position of Fermi level nFE = Ec – KT ln(NC/ND)

If we increase doping, ND increases KT ln(NC/ND) decreases, then nFE moves

more closer to conduction band. So, it is moved away from centre of forbidden band.

For p-type doping. pFE = Ev + KT ln(NC/NA)

If we increase doping, NA increase, KT ln(NC/NA) decreases therefore pFE

moves closer to EV. So it is moving away from the centre of for bidden gap.

So in any extrinsic semiconductor as doping increases Fermi level moving away from centre of forbidden gap.

22. Ans: (b) Sol: For an n-type semiconductor

EC – EF =

D

C

N

NnKT l

Fermi level is closer to conduction band in n-type material

Fermi level exist in all types of semiconductor.

Fermi level is the energy level where the probability of occupation of electron is 0.5

In a tunnel diode ND > NC ; NA> NV

hence EF – EV =

0A

V

N

NKT nl

EF – EV is negative in tunnel diode Hence Fermi level lies below valance band. 23. Ans: (c)

Sol: EFn = EC – KT ln C

D

N

N

EFn = EC KT ln C

D

N0

N



BZ FY EY

Ix

dY

Y

X WLx

+ VH –

Z

NC = ND

4.82 1015 3

23nm T

cmm

=2.63 1019 cm–3

Assume mn = m Then T = 310K

24. Ans: (b) Sol: Position of fermilevel in an n-type

semiconductor with respect to conduction band gives as

D

CCF N

NnKTEE

n

Given EC EFn= 0.25 eV = KT ln (NC/ND)

D

25

N

1051.2n026.0eV25.0

ND = 1.6 1021/m3

25. Ans: (c) Sol: For silicon

(a) electron mobility = 1300 cm2/V-sec = 0.13 m2/ V-sec

(b) electron diffusion constant = D = VT = (0.13) (0.0254) = 0.0035 m2/sec

(c) diffusion length of hole = 0.002 m (d) product of electron and hole concentration

np = 2102i 105.1n = 2.251020 /cm6

np = 6322i m1025.2n

26. Ans: (d) Sol: (a) Impurity concentration is added in the

process of ion-implantation by adding dopants.

(b) when the temperature increases lattice vibrations increases, so the free e– present in crystal collides with atoms and mobility of carriers changes. This is called scattering mechanism.

(c) when the carrier moves freely, after it’s life time it falls from conduction band and recombination takes place.

(d) when the temperature increases, band to band transition takes place and electron hole pairs generated which is known as intrinsic carrier concentration.

27. Ans: (d) Sol:

W.

B.IV zx

H

xx E.J

A

I

Ix = Ex .A A = WdY [from the given figure] = n Ix n Ex. W.dY

W.

B.d.W.E..V ZYXn

H

VH = n Ex dYBZ

VH = 0.38 × 700 × 4 × 10–3 × 0.5 = 532×10–3 V

28. Ans: (c) Sol: For intrinsic to n-type conversion Majority carrier concentration = ND

Minority carrier concentration = NA = D

2i

N

n

= ND n q 0.1 = ND 13501.6 10–19 ND = 4.63 1014 cm–3

14

20

D

2i

A 1063.4

1025.2

N

nN

35A cm1086.4N

29. Ans: 62.5

Sol: Hall voltage W

BIVH

= charge density = ND q = 11017 1.6 1019 = 1.6 102 C/cm3

42

35

H 10100106.1

10110V

= 0.625 104 VH = 62.5 V



30. Ans: (c) Sol: As the Si sample is doped with Arsenic

(V group element), it would form a n-type semiconductor

For a n-type material the relation between

Fc E&E is given by

D

cFc N

NnkTEE l here DN is dopant

concentration

eV1041.4

1085.8n026.0EE

14

18

FC

l = 0.026 eV)102(n 4l

EF lies 0.257 ev below Ec

31. Ans: (d) Sol: As per Hall effect it is stated that if any

specimen carrying a current I is placed in a transverse magnetic field B then an electric field ‘E’ is induced in specimen in a direction perpendicular to both B and I.

Direction of E = IB

32. Ans: (a) Sol: drift velocity vd = E cm/sec.

=E

vd

sec/

sec/ 2

V

cm

cmv

cm

Diffusion constant Tpp VD

p

pT D

V

sec/cm

sec.V/cm2

2

units of p

p

D

is

Volt

1

33. Ans: (a)

Sol: The Hall constant ne1

BI

WVR H

H

and ne

HR

1

HR

By using this equation we can calculate the mobility of charge carriers by Hall-effect

34. Ans: (c) Sol: Hall Effect: A metal (or) semiconductor

specimen carrying a current along its length when placed in a Transverse magnetic field; an “Electric field” will be induced in a direction perpendicular to both current and magnetic field. This is called “Hall Effect” so in Hall effect magnetic field is applied and electric field is induced

35. Ans: (a)

Sol: Given ni = 1.48 1010 cm-3 , the number of Si atoms contributing to a conducting electron hole pair.

Total number of Si atoms

N= 4.99 1022 atoms/cm3 Now percentage of Si atoms contributing

current conduction is given by

1001099.4

1048.1100

N

n22

10i

= 0.297 10-10

36. Ans: (d) Sol: Avagadro’s number (A0)

= 6.023 1023 atoms/mole Density (d) = 8.92 gm/cm3 Atomic weight (A) = 63.54 gm/mole

Atomic concentration = A

dA0

= 54.63

92.810023.6 23

= 8.45 1022 atoms/cm3 Density of valence band electrons available

for copper = (Atomic concentration of Copper

number of free electrons per atom)

= 8.45 1022 atoms/cm3



37. Ans: (a) Sol: Given electric field E = 1000 V/cm At t = 0 EHP generated at X = 0 At t = 1 microsecond, maximum occurs at xp = 0.5 cm

drift velocity s01

0cm5.0

TT

XX

12

12

= 0.5 106 cm/sec

Mobility E

vd

1000

105.0 6

= 500 cm2/V-sec Diffusion constant (D) by Einstein’s

relation

TVD

= 0.026 V at 300K D = VT = (500) (0.026) D = 13 cm2/sec

38. Ans: 414.4 Sol: RH = 3.66 10-4 m3/c. Resistivity () = 8.93 10-3 - m Assuming single carrier conduction

RH =densityeargch

1

Charge density = 4

H 1066.3

1

R

1

= 2.7 103

Conductivity = yresictivit

1

= 31093.8

1

= 111.9

= (charge density )(mobility) 111.9 = 2.7 103 = 414.4 cm2 /v- sec 39. Ans: (b)

Sol: The temperature 50K (or) 100K is capable of exciting the donar atoms (or) Acceptor

atoms in a extrinsic semiconductor, because they need very low energy of the order of 0.03 eV for excitation.

For Ge Eg(0) = 0.01 eV

For Si Eg(0) = 0.05 eV 40. Ans: (a)

Sol: Concentration of atoms =A

dAo

=6.72

32.51002.6 23

= 4.41 1022 atoms/ cm3

41. Ans: (b) Sol: kT/E3

o2i

GoeTAn

kT/E3o

kT/E3o

21i

22i

01G

2Go

eTA

eTA

n

n

0145.0e kT/EE 01G2Go

42. Ans: (c)

Sol: Hall voltage is given by,

W

BIRV H

H

For a given current, hall voltage VH is proportional to magnetic field energy. i.e.,VH B. If current I made proportional to one input and if magnetic field density, B is proportional to second input then hall voltage is proportional to the product of two signals.

43. Ans: (a)

Sol: At very high temperatures band - band transition (breaking of the covalent bonds

C.B

Eg(0) = 0.01 eV EC

Eg

Ev V.B

n-type Germanium Semiconductor



at high temperatures) dominates impurities ionization.

Hence n p Intrinsic semiconductor.

44. Ans: (a)

Sol: Now 2

1

n

pi

p

2i

p np

nn

and q)np( npppmin

q)nn( nn

pip

p

nimin

q)nn( pnipnimin

q.n2 pnimin

1910min 106.15001300105.12

19min ).(1088.3869 cm

16min ).(10869.3 cm

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