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Eighth Edition

GATEELECTRONICS & COMMUNICATION

Electronics DevicesVol 4 of 10

R. K. Kanodia Ashish Murolia

NODIA & COMPANY

GATE Electronics & Communication Vol 4, 8eElectroncis DevicesRK Kanodia & Ashish Murolia

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To Our Parents

Preface to the Series

For almost a decade, we have been receiving tremendous responses from GATE aspirants for our earlier books: GATE Multiple Choice Questions, GATE Guide, and the GATE Cloud series. Our first book, GATE Multiple Choice Questions (MCQ), was a compilation of objective questions and solutions for all subjects of GATE Electronics & Communication Engineering in one book. The idea behind the book was that Gate aspirants who had just completed or about to finish their last semester to achieve his or her B.E/B.Tech need only to practice answering questions to crack GATE. The solutions in the book were presented in such a manner that a student needs to know fundamental concepts to understand them. We assumed that students have learned enough of the fundamentals by his or her graduation. The book was a great success, but still there were a large ratio of aspirants who needed more preparatory materials beyond just problems and solutions. This large ratio mainly included average students.

Later, we perceived that many aspirants couldn’t develop a good problem solving approach in their B.E/B.Tech. Some of them lacked the fundamentals of a subject and had difficulty understanding simple solutions. Now, we have an idea to enhance our content and present two separate books for each subject: one for theory, which contains brief theory, problem solving methods, fundamental concepts, and points-to-remember. The second book is about problems, including a vast collection of problems with descriptive and step-by-step solutions that can be understood by an average student. This was the origin of GATE Guide (the theory book) and GATE Cloud (the problem bank) series: two books for each subject. GATE Guide and GATE Cloud were published in three subjects only.

Thereafter we received an immense number of emails from our readers looking for a complete study package for all subjects and a book that combines both GATE Guide and GATE Cloud. This encouraged us to present GATE Study Package (a set of 10 books: one for each subject) for GATE Electronic and Communication Engineering. Each book in this package is adequate for the purpose of qualifying GATE for an average student. Each book contains brief theory, fundamental concepts, problem solving methodology, summary of formulae, and a solved question bank. The question bank has three exercises for each chapter: 1) Theoretical MCQs, 2) Numerical MCQs, and 3) Numerical Type Questions (based on the new GATE pattern). Solutions are presented in a descriptive and step-by-step manner, which are easy to understand for all aspirants.

We believe that each book of GATE Study Package helps a student learn fundamental concepts and develop problem solving skills for a subject, which are key essentials to crack GATE. Although we have put a vigorous effort in preparing this book, some errors may have crept in. We shall appreciate and greatly acknowledge all constructive comments, criticisms, and suggestions from the users of this book. You may write to us at [email protected] and [email protected].

Acknowledgements

We would like to express our sincere thanks to all the co-authors, editors, and reviewers for their efforts in making this project successful. We would also like to thank Team NODIA for providing professional support for this project through all phases of its development. At last, we express our gratitude to God and our Family for providing moral support and motivation.

We wish you good luck ! R. K. KanodiaAshish Murolia

SYLLABUS

GATE Electronics & CommunicationsEnergy bands in silicon, intrinsic and extrinsic silicon. Carrier transport in silicon: diffusion current, drift current, mobility, and resistivity. Generation and recombination of carriers. p-n junction diode, Zener diode, tunnel diode, BJT, JFET, MOS capacitor, MOSFET, LED, p-I-n and avalanche photo diode, Basics of LASERs. Device technology: integrated circuits fabrication process, oxidation, diffusion, ion implantation, photolithography, n-tub, p-tub and twin-tub CMOS process.

IES Electronics & Telecommunication

Electrons and holes in semiconductors, Carrier Statistics, Mechanism of current flow in a semiconductor, Hall effect; Junction theory; Different types of diodes and their characteristics; Bipolar Junction transistor; Field effect transistors; Power switching devices like SCRs, GTOs, power MOSFETS; Basics of ICs - bipolar, MOS and CMOS types; basic of Opto Electronics.

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CONTENTS

CHAPTER 1 SEMICONDUCTORS IN EQULIBRIUM

1.1 INTRODUCTION 1

1.2 SEMICONDUCTOR MODELS 1

1.2.1 Bonding Model 1

1.2.2 Energy Band Model 2

1.3 CARRIERS 2

1.3.1 Carrier Properties 3

1.4 INTRINSIC SEMICONDUCTOR 3

1.5 DOPING 3

1.5.1 n -type Semiconductor 4

1.5.2 p-type Semiconductor 4

1.6 COMPENSATED SEMICONDUCTOR 5

1.7 FERMI FUNCTION 5

1.7.1 Energy Dependence of Fermi Function 5

1.8 EQUILIBRIUM CARRIER CONCENTRATIONS 7

1.8.1 Intrinsic Carrier Concentration 9

1.8.2 Extrinsic Carrier Concentration 9

1.9 ENERGY BAND DIAGRAM FOR INSULATOR, SEMICONDUCTOR, AND METAL 10

1.9.1 Insulator 10

1.9.2 Semiconductor 10

1.9.3 Metal 10

1.10 POSITION OF FERMI ENERGY LEVEL 10

1.10.1 Fermi Energy Level for n -type Semiconductor 11

1.10.2 Fermi Energy Level for p-type Semiconductor 12

1.10.3 Variation of Fermi Level with Temperature 12

1.11 CHARGE NEUTRALITY 13

1.11.1 Determination of Thermal Equilibrium Electron Concentration as a Function of Impurity Doping Concentration 13

1.11.2 Determination of Thermal Equilibrium Hole Concentration as a Function of Impurity Doping Concentration 13

1.12 DEGENERATE AND NON DEGENERATE SEMICONDUCTORS 14

1.12.1 Non-degenerate Semiconductor 14

1.12.2 Degenerate Semiconductor 14

1.13 IMPORTANT PROPERTIES AND STANDARD CONSTANTS 15

EXERCISE 1.1 17

EXERCISE 1.2 24

EXERCISE 1.3 27

SOLUTIONS 1.1 31

SOLUTIONS 1.2 47

SOLUTIONS 1.3 60

CHAPTER 2 SEMICONDUCTORS IN NON EQUILIBRIUM

2.1 INTRODUCTION 65

2.2 CARRIER DRIFT 65

2.2.1 Motion of Carriers in a Crystal 65

2.2.2 Drift Current 66

2.3 CARRIER MOBILITY 67

2.3.1 Mobility due to Lattice Scattering 68

2.3.2 Mobility due to Ionized Impurity Scattering 68

2.3.3 Mobility Variation Due to Electric field 69

2.4 CONDUCTIVITY 69

2.5 RESISTIVITY 69

2.6 CARRIER DIFFUSION 70

2.6.1 Diffusion Current Density for Electron 70

2.6.2 Diffusion Current Density for Hole 70

2.6.3 Diffusion Length 70

2.7 TOTAL CURRENT DENSITY 71

2.8 THE EINSTEIN RELATION 71

2.9 BAND BENDING 72

2.10 QUASI-FERMI LEVELS 73

2.11 OPTICAL PROCESSES IN SEMICONDUCTORS 74

2.11.1 Absorption 74

2.11.2 Emission 74

2.12 AMBIPOLAR TRANSPORT 74

2.13 HALL EFFECT 75

2.13.1 Hall Field 75

2.13.2 Hall Voltage 76

2.13.3 Hall Coefficient 76

2.13.4 Applications of Hall effect 76

EXERCISE 2.1 77

EXERCISE 2.2 86

EXERCISE 2.3 92

SOLUTIONS 2.1 98

SOLUTIONS 2.2 111

SOLUTIONS 2.3 125

CHAPTER 3 PN JUNCTION DIODE

3.1 INTRODUCTION 133

3.2 BASIC STRUCTURE OF THE pn -JUNCTION 133

3.2.1 Space Charge Region in pn junction 134

3.3 ZERO APPLIED BIAS 134

3.3.1 Energy Band Diagram for Zero Biased pn junction 134

3.3.2 Built-in Potential Barrier for Zero Biased pn junction 135

3.3.3 Electric Field in Space Charge Region 135

3.3.4 Space Charge Width 136

3.4 REVERSE APPLIED BIAS 136

3.4.1 Energy Band Diagram for Reverse Biased pn Junction 136

3.4.2 Potential Barrier for Reverse Biased pn Junction 137

3.4.3 Space Charge Width 137

3.4.4 Electric Field 137

3.4.5 Junction Capacitance 137

3.5 FORWARD APPLIED BIAS 138

3.5.1 Energy Band Diagram for Forward Biased pn Junction 138

3.5.2 Excess Carrier Concentration 138

3.5.3 Ideal pn Junction Current 138

3.5.4 Ideal Current-Voltage Relationship 139

3.6 SMALL-SIGNAL MODEL OF THE pn JUNCTION 139

3.6.1 Diffusion Resistance 140

3.6.2 Small-Signal Admittance 140

3.7 COMPARISON BETWEEN PN JUNCTION CHARACTERISTICS FOR ZERO BIAS, REVERSE BIAS, AND FORWARD BIAS 140

3.8 JUNCTION BREAKDOWN 141

3.8.1 Zener Breakdown 141

3.8.2 Avalanche Breakdown 141

3.9 TURN-ON TRANSIENT 141

3.10 SOME SPECIAL PN JUNCTION DIODE 142

3.10.1 Tunnel Diode 142

3.10.2 PIN Diode 144

3.10.3 Varactor Diode 144

3.10.4 Schottky Diode 145

3.11 THYRISTORS 146

3.11.1 Silicon Controlled Rectifier (SCR) 146

3.12 TRIAC 150

3.13 DIAC 151

EXERCISE 3.1 153

EXERCISE 3.2 163

EXERCISE 3.3 169

SOLUTIONS 3.1 182

SOLUTIONS 3.2 202

SOLUTIONS 3.3 220

CHAPTER 4 BJT


4.2 BASIC STRUCTURE OF BJT 233

4.2.1 Typical Doping Concentrations for BJT 234

4.2.2 Depletion Region 234

4.3 TRANSISTOR BIASING 235

4.3.1 Active Region 236

4.3.2 Saturation Region 236

4.3.3 Cut-off Region 236

4.3.4 Reverse Active Region or Inverse Region 237

4.4 OPERATION OF BJT IN ACTIVE MODE 237

4.4.1 Transistor Current Relation 238

4.5 MINORITY CARRIER DISTRIBUTION 240

4.5.1 Minority Carrier Distribution in Forward Active mode 241

4.5.2 Minority Carrier Distribution in Cut-Off Mode 241

4.5.3 Minority Carrier Distribution in Saturation Mode 242

4.5.4 Minority Carrier Distribution in Reverse Active Mode 242

4.6 CURRENT COMPONENTS IN BJT 243

4.6.1 DC Common-Base Current Gain 243

4.6.2 Small Signal Common Base Current Gain 243

4.6.3 Common Emitter Current Gain 245

4.7 EARLY VOLTAGE 245

4.8 BREAKDOWN VOLTAGE 246

4.8.1 Punch-Through Breakdown 246

4.8.2 Avalanche Breakdown 246

4.9 IMPORTANT PROPERTIES AND STANDARD CONSTANTS 247

EXERCISE 4.1 249

EXERCISE 4.2 261

EXERCISE 4.3 266

SOLUTIONS 4.1 274

SOLUTIONS 4.2 294

SOLUTIONS 4.3 308

CHAPTER 5 MOSFET


5.2 TWO TERMINAL MOS STRUCTURE 317

5.3 ENERGY BAND DIAGRAM FOR MOS CAPACITOR 318

5.3.1 Energy Band Diagram for MOS Capacitors with the p-type Substrate 318

5.3.2 Energy Band Diagram for MOS Capacitors with the n -type Substrate 319

5.4 DEPLETION LAYER THICKNESS 320

5.4.1 Space Charge Width for p-type MOSFET 320

5.4.2 Space Charge Width for n -type MOSFET 320

5.5 WORK FUNCTION DIFFERENCES 321

5.5.1 Work Function Difference for p-type MOS Capacitors 321

5.5.2 Work Function Difference for n -type MOS Capacitors 321

5.6 FLAT BAND VOLTAGE 322

5.7 THRESHOLD VOLTAGE 323

5.7.1 Threshold Voltage for MOS Structure with p-type Substrate 323

5.7.2 Threshold Voltage for MOS Structure with n -type Substrate 323

5.8 DIFFERENTIAL CHARGE DISTRIBUTION FOR MOS CAPACITOR 324

5.8.1 Differential Charge Distribution in Accumulation Region 324

5.8.2 Differential Charge Distribution in Depletion Region 324

5.8.3 Differential Charge Distribution in Inversion Region 325

5.9 CAPACITANCE-VOLTAGE CHARACTERISTICS OF MOS CAPACITOR 325

5.9.1 Frequency Effects on C -V Characteristics 326

5.10 MOSFET STRUCTURES 327

5.11 CURRENT-VOLTAGE RELATIONSHIP FOR MOSFET 329

5.11.1 n -channel Enhancement Mode MOSFET for V V<GS T 329

5.11.2 n -channel Enhancement Mode MOSFET for V V>GS T 329

5.11.3 Ideal Current-Voltage Relationship for MOSFET 330

5.11.4 Transconductance 332

5.12 IMPORTANT TERMS 332

5.13 IMPORTANT CONSTANTS AND STANDARD NOTATIONS 334

EXERCISE 5.1 335

EXERCISE 5.2 345

EXERCISE 5.3 351

SOLUTIONS 5.1 355

SOLUTIONS 5.2 372

SOLUTIONS 5.3 389

CHAPTER 6 JFET


6.2 BASIC CONCEPT OF JFET 393

6.2.1 n -channel JFET 393

6.2.2 p-channel JFET 394

6.3 BASIC JFET OPERATION 394

6.3.1 JFET Operation for Constant VDS and Varying VGS 394

6.3.2 JFET Operation for V 0GS = and Varying VDS 395

6.4 DEVICE CHARACTERISTIC 397

6.4.1 n -channel JFET Characteristic 397

6.4.2 p-channel JFET Characteristic 398

6.5 IDEAL DC CURRENT-VOLTAGE RELATIONSHIP FOR DEPLETION MODE JFET 398

6.6 TRANSCONDUCTANCE OF JFET 399

6.7 CHANNEL LENGTH MODULATION 399

6.7.1 Depletion Legnth 399

6.7.2 Small Signal Output Impedance 399

6.8 EQUIVALENT CIRCUIT AND FREQUENCY LIMITATIONS 399

6.8.1 Small-Signal Equivalent Circuit 399

6.8.2 Frequency Limitation Factors and Cutoff Frequency 400

EXERCISE 6.1 401

EXERCISE 6.2 405

EXERCISE 6.3 407

SOLUTIONS 6.1 412

SOLUTIONS 6.2 422

SOLUTIONS 6.3 428

CHAPTER 7 INTEGRATED CIRCUIT


7.2 BASIC MONOLITHIC INTEGRATED CIRCUIT 433

7.3 FABRICATION OF A MONOLITHIC CIRCUIT 434

7.4 EPITAXIAL GROWTH 436

7.5 OXIDATION 436

7.5.1 Dry oxidation 436

7.5.2 Wet oxidation 436

7.6 MASKING AND ETCHING 436

7.7 DIFFUSION OF IMPURITIES 437

7.7.1 Diffusion Law 438

7.7.2 Complementary Error Function 438

7.7.3 The Gaussian Distribution 438

7.8 ION IMPLANTATION 439

7.9 THIN FILM DEPOSITION 440

7.9.1 Evaporation 440

7.9.2 Sputtering 440

7.9.3 Chemical Vapour Deposition (CVD) 441

7.10 PN JUNCTION DIODE FABRICATION 441

7.11 TRANSISTOR CIRCUIT 442

7.11.1 Monolithic Integrated Circuit Transistor 442

7.11.2 Discrete Planar Epitaxial Transistor 442

EXERCISE 7.1 444

EXERCISE 7.2 452

EXERCISE 7.3 453

SOLUTIONS 7.1 458

SOLUTIONS 7.2 464

SOLUTIONS 7.3 465

***********

Chap 1

Semiconductors in Equilibrium

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GATE STUDY PACKAGE Electronics & Communication

Sample Chapter of Electronic Devices (Vol-4, GATE Study Package)

1.1 INTRODUCTION

Equilibrium or thermal equilibrium, implies that no external forces such as voltages, electric fields, magnetic fields, or temperature gradients are acting on the semiconductor. This chapter deals with the semiconductor in equilibrium. Following topics are included in this chapter: • Semiconductor Models: bonding model, energy band model.

• Electron and hole carriers; its properties: charge, effective mass.

• Intrinsic semiconductor

• Extrinsic semiconductor: n -type semiconductor, p-type semiconductor, compensated semiconductor

• Effect of donor and acceptor impurities

• Fermi function: energy dependence

• Equilibrium carrier concentration: electron concentration in conduction band, hole concentration in valence band.

• Energy band diagram for insulator, semiconductor, and conductor

• Fermi energy level: position of Fermi energy level, variation of Fermi energy level with temperature

• Charge neutrality

• Degenerate and non-degenerate semiconductors

1.2 SEMICONDUCTOR MODELS

In this section, we introduce and describe two very important models that are used extensively in the analysis of semiconductor devices.

1.2.1 Bonding Model

The isolated Si atom, or a Si atom not interacting with other atoms, was found to contain four valence electrons. The implication here is that, in going from isolated atoms to the collective crystalline state, Si atoms come to share one of their valence electrons with each of the nearest neighbours. This results in covalent bonding, or equal sharing of valence electrons with nearest neighbors. The bonding model is shown in Figure 1.1.

CHAPTER 1SEMICONDUCTORS IN EQULIBRIUM

Chap 1Semiconductors in Equilibrium

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Figure 1.1: Bonding Model of Silicon

1.2.2 Energy Band Model

Starting with N-isolated Si atoms and conceptually bringing the atoms closer and closer together, one finds the inter atomic forces lead to a progressively spread in the allowed energies.1. The spread in energies give rise to closely spaced sets of allowed states

known as energy bands.

2. The distribution of allowed states consists of two bands separated by an intervening energy gap.

3. The upper band of allowed states is called the conduction band; the lower band of allowed state is called the valence band; and the intervening energy gap is called the forbidden gap or band gap.

4. The valence band is completely filled and the conduction band is completely empty at temperature approaching KT 0= .

Figure 1.2: Simplified Version of the Energy Band Model

1.3 CARRIERS

We are now in a position to introduce and to visualize the current carrying entities within semiconductor. In a semiconductor, the two carriers are:1. Electrons: In bonding model of semiconductor when Si-Si band is broken

and the associated electron is free to wander about the lattice, the released electron is a carrier. Equivalently in term of the energy band

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model, excitation of valence band electrons into the conduction band creates carriers; that is electron in the conduction band are carrier.

2. Holes: The breaking of a Si-Si bonds also creates a missing bond or void in the bonding structure and in term of the energy band model where the removal of an electron from the valence band create an empty state. The missing band in the bonding scheme, the empty state in the valence band, is the second type of carrier found in semiconductors called the hole.

1.3.1 Carrier Properties

Having formally introduced the electron and hole in this section, we study about the nature of these carriers.

Charge

Both electrons and holes are charge entities. Electrons are negatively charged, holes are positively charged, and the magnitude of the carrier charge is

q . C1 6 10 19#= -

Effective Mass

When an external field is applied to a crystal, the free electron or hole in the crystal responds, as if its mass is different from the true mass. This mass is called the effective mass of the electron or the hole. Following are some important points about effective mass:

POINTS TO REMEMBER

1. By considering this effective mass, it will be possible to remove the quantum features of the problem.

2. The effective mass allows us to use Newton’s law of motion to determine the effect of external forces on the electrons and holes within the crystal.

1.4 INTRINSIC SEMICONDUCTOR

A semiconductor is said to be intrinsic if it contains no impurities and no crystalline defects. In an intrinsic semiconductor, the equilibrium concentration n0 of electrons in the conduction band is the same as the equilibrium concentration of holes p0 in the valence band. i.e.

n0 p ni0= =where n0 = number of electron/cm3

p0 = number of holes/cm3

1.5 DOPING

Doping, in semiconductor terminology is the addition of controlled amounts of specific impurity atoms with the express purpose of increasing either the electron or the hole concentration. Depending on the characteristic of dopants, semiconductors are classified as n -type and p-type semiconductors.


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1.5.1 n -type Semiconductor

To increase the electron concentration, one can add either phosphorus, arsenic or antimony atoms to the Si crystal, these are the donor (electron-increasing) dopants or n-type impurity. The n -type semiconductor is created by introducing the donor impurities in an intrinsic semiconductor.

EFFECT OF DONOR IMPURITIES

1. If a pentavalent impurity (phosphorus, arsenic or antimony) is added to an intrinsic semiconductor, the four covalent bonds are still present. However, it creates an additional fifth electron due to the impurity atom. This remaining electron is relatively free to move within the material.

2. When donor impurities are added to a semiconductor, allowable energy levels are introduced a very small distance below the conduction band . The energy band diagram of n -type semiconductor is shown in figure below.

3. If intrinsic semiconductor material is “doped” with n -type impurities, not only does the number of electron increase, but the number of hole decreases below that which would be available in the intrinsic semiconductor. The reason for the decrease in the number of holes is that the larger number of electron present increases the rate of recombination of electron with holes.

1.5.2 p-type Semiconductor

To increase the hole concentration, one can add either boron, gallium or indium atoms to the Si crystal, these are the acceptor (hole-increasing) dopants or p-type impurity. The p-type semiconductor is created by introducing the acceptor impurities in an intrinsic semiconductor.

EFFECT OF ACCEPTOR IMPURITIES

1. If a trivalent impurity (boron, gallium or indium) is added to an intrinsic semiconductor, only three of the covalent bonds can be filled; and the vacancy that exists in the fourth bond constitutes a hole.

2. When acceptor or p-type, impurities are added to the intrinsic semiconductor, they produce an allowable discrete energy level which is just above the valence band. The energy band diagram of p-type semiconductor is shown in figure below.

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MAJORITY AND MINORITY CARRIERS

1. In n -type semiconductor, the electrons are called the majority carriers, and the holes are called the minority carriers.

2. In p-type material, the holes are the majority carriers and the electrons are the minority carriers.

1.6 COMPENSATED SEMICONDUCTOR

A compensated semiconductor is one that contains both donor and acceptor impurity atoms in the same region. We classify the compensated semiconductors as1. An n -type compensated semiconductor occurs when N N>d a .

2. A p-type compensated semiconductor occurs when N N>a d .

3. If N Na d= , we have a completely compensated semiconductor that has the characteristics of an intrinsic material.

1.7 FERMI FUNCTION

The Fermi function f E^ h specifies how many of the existing states at the energy E will be filled with an electron.

orf E^ h specifies, under equilibrium conditions, the probability that an available state at an energy E will be occupied by an electron.

Mathematically, the Fermi function is simply a probability distribution function, defined as

f E^ h e1

1/E E kTF

=+ -^ h

where E = Any energy level

EF = Fermi energy or Fermi level

k = Boltzmann constant ( . /eV kk 8 617 10 5#= - )

T = Temperature in kelvin (K )

1.7.1 Energy Dependence of Fermi Function

We analyse the energy dependence of Fermi function for the following two cases:


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CASE I: KT 0"Let us begin by investigating the Fermi functions energy dependence for T 0" .

For all energies E E< F ,

kTE EF- " 3-

and f E E< F^ h e11 1"

+=3-

For all energies E E> F ,

kTE EF- " 3+

and f E E> F^ h / e1 1 0" + =3

This result is plotted in figure below.

Figure 1.3: Energy Dependence of Fermi Function for KT 0"

CASE II: KT 0>Examining the Fermi function, we make the following observations.1. If E EF= , then

f EF^ h /1 2=

2. If E kTE 3F$ + , then

/e kTE EF-^ h 1>> ;

f E^ h e /E E kTF, - -^ h

Consequently, above kTE 3F + , the Fermi function or filled-state probability decays exponentially to zero with increasing energy moreover, most-states at energies kT3 or more above EF will be empty.

3. If E kTE 3F# - , then

e /E E kTF-^ h 1<< ;

f E^ h e1 /E E kTF- - -^ h

Below kTE 3F - , therefore, f E1 - ^ h6 @, the probability that a given state will be empty, decays exponentially to zero.

For the above case, the Fermi function is plotted in figure below.

Figure 1.4: Energy Dependence of Fermi Function for KT 0>

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Equilibrium Distribution of Carrier

In the following diagram, the equilibrium distribution of carrier is illustrated.

Figure 1.5: Equilibrium Distribution of Carrier

1.8 EQUILIBRIUM CARRIER CONCENTRATIONS

The distribution (with respect to energy) of electrons in the conduction band is given by the density of allowed quantum states times the probability that a state is occupied by an electron, i.e.

n E^ h g E f Ec= ^ ^h h ...(1.1)

where f E^ h = Fermi-Dirac probability function

and g Ec ^ h = Density of quantum states in the conduction bandSimilarly, the distribution (with respect to energy) of holes in the valence


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band is the density of allowed quantum states in the valence band multiplied by the probability that a state is not occupied by an electron, i.e.

p E^ h g E f E1v= -^ ^h h6 @ ...(1.2)Integrating the above two equations [(1.1) and (1.2)], we obtain the

electron and hole concentrations respectively.

Electron Concentration in Conduction Band

The total electron concentration per unit volume in the conduction band is found by integrating the distribution function given in equation (1.1), i.e.

n0 g E f E dEc= ^ ^h h#

where f E^ h e1

1/E E kTF

=+ -^ h

e /E E kTF- - -^ h

So, the thermal-equilibrium density of electron in the conduction band is obtained as

n0 hm

E E e dE4 2 * /

/nc

E E kT

E3

3 2F

c

p = -

3 - -^ ^h h#

h

m kT e2 2 * //n E E kT

2

3 2Fcp = - -^^ h h

; E

or n0 N e / /c

E E kTFc= -^ h

where the parameter Nc is called the effective density of states function in the conduction band, given as

Nc hm kT2 2 * /

n2

3 2p = ; E

NOTE :Since g E dEc _ i represents the number of conduction band states/cm3 lying in the E to E dE+ energy range, and f E_ i specifies the probability that an available state as an energy E will be occupied by an electron. So, g E f E dEc _ _i i gives the number of conduction band electrons/cm3 lying in the E to E dE+ energy range. Thus, g E f E dEc _ _i i integrated over all conduction band energies must yield the total number of electrons in the conduction band. A similar statement can be made relative to the hole concentration.

Hole Concentration in Valence Band

The thermal equilibrium concentration of holes in the valence band is found by

p0 g E f E dE1v F= -^ ^h h6 @#

where f E1 E- ^ h e1

1/E E kTF

=+ -^ h

e kTE EF

- - -b l

So, the thermal equilibrium concentration of holes in the valence band is

p0 hm

E E e dE4 2 * /

pv

EkT

E E

3

3 2Fv p

= -3-

- -^ bh l#

-

hm kT

e22 * /

p kTE E

2

3 2F vp =-

eb

ol

or p0 N e /v

E E kTF v= - -^ h

where Nv is called the effective density of states function in the valence band, given as

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Nv hm kT

22 * /

p2

3 2p = e o

1.8.1 Intrinsic Carrier Concentration

For an intrinsic semiconductor, the concentration of electrons in the conduction band is equal to the concentration of holes in the valence band. These parameters are usually referred to as the intrinsic electron concentration and intrinsic hole concentration, i.e.

ni pi=Fermi energy level for the intrinsic semiconductor is called the intrinsic

Fermi energy, or E EF Fi= . So, we have

n0 n N e /i c

E E kTFic= = - -^ h ...(1.3)

and p0 p n N e /i i v

E E kTFi v= = = - -^ h ...(1.4)Hence, multiplying equations (1.3) and (1.4), we get

ni2 N N e e/ /

c vE E kT E E kTFi Fi vc= - - - -^ ^h h

or ni2 N N e N N e/ /

c vE E kT

c vE kTv gc= =- - -^ h ...(1.5)

or ni N N e / kTc v

E 2G= -

Temperature Dependence of Intrinsic Concentration

Substituting the values of Nc and Nv in equation (1.5), we have

ni2 .

mm m

T e2 33 10* * /

/n p E kT43

02

3 23 G#= -

e o ...(1.6)

Since, EG E TG0 b = -where EG0 is the magnitude of energy gap at 0 K. So, by substituting this relation into equation (1.6), we get

ni2 A T e /E kT

03 G0= -

where A0 .m

m me2 33 10

* * //n p k43

2

3 2

#= b -^ eh o

and b has the dimension of electron volt per degree kelvin.

1.8.2 Extrinsic Carrier Concentration

Let us derive a general form of equations for the thermal-equilibrium concentration of electrons and holes in extrinsic semiconductors. For an extrinsic semiconductor, we have

n0 N e /c

E E kTFc= - -^ h

If we add and subtract an intrinsic Fermi energy ( )EFi in the exponent of above equation, we can write

n0 N ec kTE E E EFi F Fic

=- - + -^ ^

ch h

m

or n0 N e e/ /c

E E kT E E kTFi F Fic= - - -^ ^h h ...(1.7)Since, the intrinsic carrier concentration is given by [from equation (1.3)]

ni N e /c

E E kTFic= - -^ h

Hence, by substituting it in equation (1.7), we get the thermal equilibrium electron concentration as

n0 n e /i

E E kTF Fi= -^ h

Similarly, we obtain the thermal equilibrium hole concentration as

p0 n e /i

E E kTF Fi= - -^ h


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1.9 ENERGY BAND DIAGRAM FOR INSULATOR, SEMICONDUCTOR, AND METAL

Figure 1.6 shows the energy band diagram of an insulator, a semiconductor, and a metal. The characteristic of these materials are described in the following texts.

Figure 1.6: Energy Band Diagram of (a) an Insulator (b) a Semiconductor (c) a Metal

1.9.1 Insulator

A very poor conductor of electricity is called an insulator. For insulator, the width of the forbidden energy region is high (6 eV).

1.9.2 Semiconductor

A substance whose conductivity lies between the metal and insulator is called semiconductor. The width of the forbidden energy region for semiconductor is relatively small eV1.^ h. The energy bandgap for silicon and germanium semiconductor are tabulated below.

Table 1.1: Energy Bandgap of Semiconductors

Semiconductors Energy bandgap at temperature (T ) Energy bandgap at room temperature ( )KT 300=

Silicon . . TE T 1 21 3 60 10G4

#= - -^ h EG . eV1 1=

Germanium . . TE T 0 785 2 23 10G4

#= - -^ h Eg . eV0 72=

1.9.3 Metal

An excellent conductor is a metal. The band structure of a metal contains overlapping valence and conduction bands, as shown in Figure 1.6 (c).

1.10 POSITION OF FERMI ENERGY LEVEL

We can now determine the position of the Fermi energy level as a function of the doping concentration. The position of the Fermi energy level within the bandgap can be determined by using the equations already developed for the thermal equilibrium electron and hole concentration.

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1.10.1 Fermi Energy Level for n -type Semiconductor

For an n -type semiconductor, we define the carrier concentration as

n0 N e /c

E E kTFc= - -^ h

or E Ec F- lnkT nNc

0= b l ...(1.8)

If the donor concentration, Nd n>> i

then, n0 Nd-Therefore, equation (1.8) becomes

E Ec F- lnkT NN

d

c= b l

Figure 1.7 (a) shows the position of Fermi energy level for n -type semiconductor. Now, we may develop a slightly different expression for the position of the Fermi level. Since,

n0 n e /i

E E kTF Fi= -^ h

So, E EF Fi- lnkT nn

i

0= a k

Figure 1.7: Illustration of Fermi Energy Level for (a) n -type Semiconductor, (b) p -type Semiconductor

Following are some important points about position of Fermi level in an n -type semiconductor.

CHARACTERISTICS OF FERMI LEVEL IN n -TYPE SEMICONDUCTOR

1. The distance between the bottom of the conduction band and the Fermi energy is a logarithmic function of the donor concentration.

2. As the donor concentration increases, the Fermi level moves closer to the conduction band.

3. If the Fermi level moves closer to the conduction band, then the electron concentration in the conduction band is increasing.

4. If we have a compensated semiconductor, then the Nd term in equation is simply replaced by N Nd a- , or the net effective donor concentration.

5. The difference between the Fermi level and the intrinsic Fermi level is a function of the donor concentration.

6. If the net effect donor concentration is zero, i.e. N N 0d a- = then n ni0 = and E EF Fi= .

7. A completely compensated semiconductor has the characteristics of an intrinsic material in term of carrier concentration and Fermi level position.

8. For an n -type semiconductor, n n> i0 , E F>F Fi , and therefore the Fermi level is above EFi .


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1.10.2 Fermi Energy Level for p-type Semiconductor

For a p-type semiconductor, the carrier concentration is

p0 N e /v

E E kTF v= - -^ h

or E EF v- lnkT pNv

0= b l ...(1.9)

If the acceptor concentration, Na n>> i

then, p0 Na,Therefore, equation (1.9) becomes

E EF v- lnkT NN

a

v= b l

Figure 1.7 (b) shows the position of Fermi energy level for p-type semiconductor. Now, we may develop a slightly different expression for the position of the Fermi level. Since,

p0 n e /i

E E kTF Fi= - -^ h

So, E EFi F- lnkT np

i

0= b l

Following are some important points about position of Fermi level in a p-type semiconductor.

CHARACTERISTICS OF FERMI LEVEL IN p -TYPE SEMICONDUCTOR

1. The distance between the Fermi level and the top of the valence band energy for a p-type semiconductor is a logarithmic function of the acceptor concentration.

2. As acceptor concentration increases, the Fermi level moves closer to the valence band.

3. If we have a compensated p-type semiconductor, then the Na term in equation is replaced by N Na d- or the net effective acceptor concentration.

4. We can also derive an expression for the relationship between the Fermi level and the intrinsic Fermi level in term of the hole concentration.

p0 n e /i

E E kTF Fi= - -^ h

E EFi F- lnkT np

i

0= b l

5. For a p-type semiconductor, p n> i0 , E E>Fi F , and therefore the Fermi level is below EFi .

1.10.3 Variation of Fermi Level with Temperature

The Fermi energy level EF for a semiconductor varies with the temperature in following manner:1. The intrinsic concentration ni is a strong function of temperature, so

that EF is a function of temperature also.

2. As the temperature increases, ni increases, and EF moves closer to the intrinsic Fermi level.

3. At high temperature, the semiconductor material begins to lose its extrinsic characteristics and begins to behave more like an intrinsic semiconductor.

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4. At the very low temperature, freeze-out occurs; the fermi level goes above Ed for the n -type material and below Ea for the p-type material.

5. At absolute zero degrees, all energy states below EF are full and all the energy states above EF are empty.

1.11 CHARGE NEUTRALITY

In thermal equilibrium, the semiconductor crystal is electrically neutral. The electrons are distributed among the various energy states, creating negative and positive charges but the net charge density is zero. This charge neutrality condition is used to determine the thermal equilibrium electron and hole concentration as a function of the impurity doping concentration.

1.11.1 Determination of Thermal Equilibrium Electron Concentration as a Function of Impurity Doping Concentration

We assume complete ionization. The charge neutrality condition is expressed by equating the density of negative charges to the density of positive charges, i.e.

n Na0 + p Nd0= +or n N p Na d0 0+ - - 0= ...(1.10)

From mass action law, we have

p0 nni

0

2

=

Substituting it in equation (1.10), we get

n N N nn

a di

00

2

+ - - 0=

or n N N n nd a i02

02- - -^ h 0=

Thus, by solving the above quadratic equation, we obtain the electron concentration as

n0 N N N N n

24d a d a i

2 2

=- + - +^ ^h h

CASE I: N N 0a d= =Substituting N N 0a d= = in above expression, we get

n0 ni!=Since, the electron concentration must be a positive quantity, so

n0 ni=

CASE II: N 0a =For N 0a = , the electron concentration becomes

n0 N N n

24d d i

2 2

= + +

1.11.2 Determination of Thermal Equilibrium Hole Concentration as a Function of Impurity Doping Concentration

Again, from mass action law, we have

n0 pni

0

2

=

Substituting it in the equation (1.10), we obtain the hole concentration as


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p N N p na d i02

02- - -^ h 0=

or p0 N N N N n

24a d a d i

2 2

=- + - +^ ^h h

CASE I: N N 0a d= =Substituting N N 0a d= = in above expression, we get

p0 ni!=Since, the hole concentration must be a positive quantity, so

p0 ni=

CASE II: N 0d =For N 0d = , the hole concentration becomes

p0 N N n

24a a i

2 2

= + +

1.12 DEGENERATE AND NON DEGENERATE SEMICONDUCTORS

We may define the degenerate and non-degenerate semiconductors in following ways:

1.12.1 Non-degenerate Semiconductor

When the concentration of dopant atoms added is small compared to the density of host or semiconductor atoms, the impurities introduce discrete, non-interacting donor energy states in the n -type semiconductor and discrete non-interacting acceptor states in the p-type semiconductor. These types of the semiconductors are referred to as non-degenerate semiconductor.

1.12.2 Degenerate Semiconductor

When the donor concentration is increased, the band of donor states widens and may overlap the bottom of the conduction band. This overlap occurs when the donor concentration becomes comparable with the effective density of states. The two types of degenerate semiconductors are defined as

Degenerate n -Type Semiconductor

When the concentration of electrons in the conduction band exceeds the density of states Nc , the Fermi energy lies within the conduction band. This type of semiconductor is called a degenerate n -type semiconductor. Figure 1.8 (a) shows the energy band diagram for a degenerate n -type semiconductor.

Figure 1.8: Energy Band Diagram for (a) Degenerate n -type Semiconductor, (b) Degenerate p -type Semiconductor

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Degenerate p-Type Semiconductor

In a similar way, as the acceptor doping concentration increases in a p-type semiconductor, the discrete acceptor energy states will split into a band of energies and may overlap the top of the valence band. The Fermi energy will lie in the valence band when the concentration of holes exceeds the density of states Nv . This type of semiconductor is called a degenerate p-type semiconductor. Figure 1.8 (b) shows the energy band diagram for a degenerate p-type semiconductor.

1.13 IMPORTANT PROPERTIES AND STANDARD CONSTANTS

Following are some important properties and standard values used in determination of semiconductor parameters in the exercises of the chapter.

Table 1.2: Some Standard Constants

Avogadro’s number .N 6 02 10A23

#= + atoms per gram molecular weight

Boltzmann’s Constant . / . /J K eV Kk 1 38 10 8 62 1023 5# #= =- -

Electronic Charge (Magnitude) . Ce 1 60 10 19#= -

Free Electron Rest Mass . kgm 9 11 10031

#= -

Permeability of Free Space /H m4 1007

#m p= -

Permittivity of Free Space . / . /F cm F me 8 85 10 8 85 10014 12

# #= =- -

Planck’s Constant . .J s eV sh 6 625 10 4 135 1034 15# #- -= =- -

Thermal Voltage KT 300=^ h . voltV ekT 0 0259t = = , . eVkT 0 0259=

Table 1.3 : Properties of Silicon, Gallium Arsenide, and Germanium KT 300=_ i

Property Si GaAs Ge

Atoms cm 3-^ h .5 0 1022

# .4 42 1022# .4 42 1022

#

Dielectric Constant 11.7 13.1 16.0

Bandgap Energy (eV) 1.12 1.42 0.66

Effective Density of States in Conduction Band, cmNc

3-^ h

.2 8 1019# .4 7 1017

# .1 04 1019#

Effective Density of States in Valence Band, cmNv

3-^ h

.1 04 1019# .7 0 1018

# .6 0 1018#

Intrinsic Carrier Concentration cm 3-^ h

.1 5 1010# .1 8 106

# .2 4 1013#

Electron Mobility, nm /cm V s2-^ h 1350 8500 3900

Hole Mobility, pm /cm V s2-^ h 480 400 1900

Effective Mass (Density of States) 1.08 0.067 0.55


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Table 1.4: Carrier Modeling Equation Summary

Density of States and Fermi Function

g Eh

m m E E2* *

cn n c

2 3p =

-^

^h

h, E Ec$

g Eh

m m E E2* *

vp p v

2 3p =

-^

^h

h, E Ev#

f Ee1

1/E E kTF

=+ -^

^h

h

Carrier Concentration Relationships

Nh

m kT22

* /

cn

2

3 2

p = ; E

Nh

m kT2

2

* /

vp

2

3 2

p = = G

n N e /c

E E kTF c= -^ h

p N e /v

E E kTv F= -^ h

n n e /i

E E kTF i= -^ h

p n e /i

E E kTi F= -^ h

ni , np -Product, and Charge Neutrality

n N N e /i c v

E kT2G= - np ni2= p n N N 0d a- + - =

n , p , and Fermi Level Computational Relationships

n N N N N n2 2/

d a d ai

22

1 2= - + - +b l; E lnE E E kT

mm

2 43

*

*

ic v

n

p= + + e o

For N N>>d a , N n>>d i ;n Nd-

p n Ni d2-

/ /ln lnE E kT n n kT p nF i i i- = =-^ ^h h

For N N>>a d , N n>>a i ; p Na-

/n n Ni a2-

/lnE E kT N nF i d i- = ^ h ,N N N n>> >>d a d i

/lnE E kT N ni F a i- = ^ h ,N N N n>> >>a d a i

***********

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EXERCISE 1.1

MCQ 1.1.1 In intrinsic semiconductor at 300 K, the magnitude of free electron concentration in silicon is about(A) per cm15 104 3

# (B) per cm5 1012 3#

(C) . per cm1 45 1010 3# (D) . per cm1 45 106 3

#

MCQ 1.1.2 Two initially identical samples A and B of pure germanium are doped with donors to concentrations of 1 1020

# and 3 1020# respectively. If the hole

concentration in A is 9 1012# , then the hole concentration in B at the same

temperature will be(A) m3 1012 3

#- (B) m7 1012 3

#-

(C) m11 1012 3#

- (D) m27 1012 3#

-

MCQ 1.1.3 Given the effective masses of holes and electrons in silicon respectively as

m *p . m0 56 o= , .m m1 08*

n o=What will be the position of the intrinsic Fermi energy level with respect to the center of the bandgap for the semiconductor at 300 KT = ?(A) 0.029 eV0 below the centre

(B) 0.0128 eV above the centre

(C) 0.0128 eV below the centre

(D) 0.029 eV0 above the centre

MCQ 1.1.4 What will be the position of Fermi energy level, EFi with respect to the center of the bandgap in silicon for 200 KT = ?(A) 0.0 eV085 below the centre

(B) 0.0128 eV above the centre

(C) 0.0128 eV below the centre

(D) 0.0 eV085 above the centre

Common Data For Q. 5 and 6The electron concentration in silicon at 300 KT = is 5 10 cmn0

4 3#= - .

MCQ 1.1.5 What will be the hole concentration (in cm 3- ) in silicon ?(A) 109 15# (B) 3 10 9#

(C) .4 5 1015# (D) 3 10 5#

SAN. SHARMA428/21

B.L.THERAJA37/23


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MCQ 1.1.6 The material is(A) p-type (B) n-type

(C) intrinsic (D) can’t be determined

MCQ 1.1.7 If the Fermi energy in silicon is 0.22 eV above the valence band energy, what will be the values of n0 and p0 for silicon at 300 KT = ? n0 (in cm 3- ) p0 (in cm 3- )

(A) 2.27 104# 2.13 1015

#

(B) 2.13 10 cm15 3#

- 2.27 104#

(C) . 10 cm1 04 4 3#

- 2.8 1015#

(D) 2.8 1015# 1.04 104

#

Common Data For Q. 8 and 9Consider 0.25 eVE Ec F- = in gallium arsenide (GaAs) at 400 KT = .

MCQ 1.1.8 What will be the electron and hole concentrations in the material at 400 KT = ?

n0 (in cm 3- ) p0 (in cm 3- )

(A) 2.27 104# 2.13 1015

#

(B) 2.13 10 cm15 3#

- 2.27 104#

(C) . 10 cm5 19 14 3#

- 2. 8 100 4#

(D) 2. 8 100 4# . 105 19 14

#

MCQ 1.1.9 If the value of n0, obtained in above question, remains constant then, what will be the hole concentration at 300 KT = ?(A) 10 cm7 3 3

#- -

(B) 9.67 10 cm3 3#

- -

(C) 96.7 10 cm3 3#

- -

(D) . 10 cm2 08 4 3#

-

MCQ 1.1.10 If a germanium semiconductor is doped with the donor and acceptor concentrations respectively as

Nd 5 10 cm15 3#= - , N 0a = .What will be the thermal equilibrium concentrations, n0 and p0 at 300 KT = in the material ? n0 (in cm 3- ) p0 (in cm 3- )

(A) 2. 8 100 4# 2.13 1015

#

(B) .1 10 cm1 5 11 3#

- . 105 0 15#

(C) 2.13 1015# 2. 8 100 4

#

(D) . 105 0 15# .1 10 cm1 5 11 3

#-

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MCQ 1.1.11 The doping concentrations in silicon semiconductor are Nd 10 cmNa15 3= = - .

What will be the concentrations of n0 and p0 in the material at 300 KT = ?

n0 (in cm 3- ) p0 (in cm 3- )

(A) 1015 1015

(B) . 10 cm2 25 10 3#

- 1015

(C) . 101 5 10# . 101 5 10

#

(D) 1015 . 10 cm2 25 10 3#

-

MCQ 1.1.12 Assume that gallium arsenide has dopant concentrations of 1 10 cmNd13 3

#= - and 2.5 10 cmNa

13 3#= - at 300 KT = . The material is

(A) p-type with . , 0.216cm cmp n1 5 10013 3

03

#= =- -

(B) p-type with 0.216 , 1.5 10cm cmp n03

013 3

#= =- -

(C) n-type with 0.216 , 1.5 10cm cmp n03

013 3

#= =- -

(D) n-type with . , 0.216cm cmp n1 5 10013 3

03

#= =- -

MCQ 1.1.13 A sample of silicon at 450 KT = is doped with boron at a concentration of 1.5 10 cm15 3

#- and with arsenic at a concentration of 8 10 cm14 3

#- . The

material is(A) p-type with 4.23 10 , 7 10cm cmp n0

11 30

14 3# #= =- -

(B) p-type with 10 , 4.23 10cm cmp n7014 3

011 3

# #= =- -

(C) n-type with 4.23 10 , 7 10cm cmp n011 3

014 3

# #= =- -

(D) n-type with 10 , 4.23 10cm cmp n7014 3

011 3

# #= =- -

MCQ 1.1.14 A particular semiconductor material is doped at 2 10 cmNd13 3

#= - , N 0a =, and the intrinsic carrier concentration is 2 10 cmni

13 3#= - . The thermal

equilibrium majority and minority carrier concentrations will be, respectively (Assume complete ionization)(A) 1. 10 , 0.216cm cmp n230

13 30

3#= =- -

(B) 0.216 , 3.24 10cm cmp n03

013 3

#= =- -

(C) . , 3.24 10cm cmp n1 23 10013 3

013 3

# #= =- -

(D) 3.24 10 , .2cm cmp n 1 3 10013 3

013 3

##= =- -

MCQ 1.1.15 Consider germanium with an acceptor concentration of 10 cmNa15 3= - and

a donor concentration of N 0d = at 200 KT = . The Fermi energy of the material will be(A) 0. 85 eV1 5 below the intrinsic Fermi level

(B) 0.0128 eV above the intrinsic Fermi level

(C) 0.0128 eV below the intrinsic Fermi level

(D) 0. 85 eV1 5 above the intrinsic Fermi level


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MCQ 1.1.16 Consider germanium at 300 KT = with doping concentrations of 10 cmNd

14 3= - and N 0a = . What will be the position of Fermi energy level with respect to the intrinsic Fermi level for these doping concentrations ?(A) 0. 85 eV1 5 below the intrinsic Fermi level

(B) 0.0 8 eV3 2 above the intrinsic Fermi level

(C) 0.0 8 eV3 2 below the intrinsic Fermi level


MCQ 1.1.17 If silicon is doped with phosphorus atoms at a concentration of 10 cm15 3- then, what will be the position of the Fermi level with respect to the intrinsic Fermi level in silicon at 300 KT = ?(A) 0. 85 eV1 5 below the intrinsic Fermi level

(B) 0.2877 eV above the intrinsic Fermi level

(C) 0.2877 eV below the intrinsic Fermi level


MCQ 1.1.18 Gallium arsenide at 300 KT = contains acceptor impurity atoms at a density of 10 cm15 3- . Additional impurity atoms are to be added so that the Fermi level is 0.45 eV below the intrinsic level. The concentration and type of additional impurity atoms will be respectively (A) 9.368 10 cmNa

14 3#= - , acceptor

(B) 6.32 10 cmNa13 3

#= - , acceptor

(C) 9.368 10 cmNd14 3

#= - , donor

(D) 6.32 10 cmNd13 3

#= - , donor

Common Data For Q. 19 and 20For a particular semiconductor, the effective mass of electron is .m m1 4*

n = (where m is electron mass at rest).

MCQ 1.1.19 What is the effective density of states in the conduction band at KT 300c= .(A) . m4 15 1025 3

#-

(B) . m2 08 1025 3#

-

(C) . m4 27 1026 3#

-

(D) . cm4 15 1020 3#

-

MCQ 1.1.20 If . eVE E 0 25C F- = at KT 300c= , then what is the concentration of electrons in the semiconductor ?(A) . m1 33 1021 3

#-

(B) . m2 67 1021 3#

-

(C) . cm2 67 1015 3#

-

(D) . m2 67 1022 3#

-

MILMAN90/4.3

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MCQ 1.1.21 The effective masses of electron and hole in germanium are .m m0 55*n = and

.m m0 37*p = (where m is the electron rest mass) what will be the position

of the intrinsic Fermi energy lavel with respect to the centre of the bandgap for the Germanium semiconductor at KT 300= ?(A) 0.0154 eV above the centre

(B) 0.0154 eV below the centre

(C) 0.0077 eV above the centre

(D) 0.0077 eV below the centre

Common Data For Q. 22 to 24For a particular material, . cmN 1 5 10C

18 3#= - , . cmN 1 3 10V

19 3#= - and

bandgap . eVE 1 43G = at KT 300c= .

MCQ 1.1.22 What is the position of the Fermi level with respect to the top of the valence band EV ?(A) 0.028 eV above the valence band edge EV

(B) 0.743 eV below the valence band edge EV

(C) 0.028 eV below the valence band edge EV

(D) 0.743 eV above the valence band edge EV

MCQ 1.1.23 What is the position of the Fermi level with respect to the conduction band edge EC ?(A) 0.687 eV above EC

(B) 0.687 eV below EC

(C) 0.743 eV below EC

(D) 0.743 eV above EC

MCQ 1.1.24 What are the effective masses m *n and m *

p of electron and hole respectively ? m *

n m *p

(A) . kg1 35 10 31#

- . kg6 46 10 31#

-

(B) . kg1 23 10 21#

- . kg6 46 10 31#

-

(C) . kg1 35 10 31#

- . kg5 88 10 31#

-

(D) . kg1 23 10 31#

- . kg5 88 10 31#

-

MCQ 1.1.25 The probability that an energy state is filled at E KTC + , is equal to the probability that a state is empty at E KTC + . Where is the Fermi level EF^ h located ?(A) E E KT2F C= +

(B) E E KT2F C= -(C) E E KTF C= +

(D) E E KT2F C= -

MILMAN93/4.4

MILMAN95/4.5

MILMAN95/4.5

MILMAN95/4.5(C)

PEIRRET71/2.6(C)


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MCQ 1.1.26 A silicon wafer is uniformly doped p-type with acceptor impurity .per cmN 10A

15 3= At KT 0c, what is the equilibrium electron concentration ?(A) /cm1015 3 (B) /cm105 3

(C) /cm1010 3 (D) 0,

MCQ 1.1.27 In a non-degenerate germanium sample maintained under equilibrium conditions near room temperature, it is known that intrinsic concentration

/ ,cmn 10i13 3= n p2= and N 0A = . What are the values of n (electron

concentration) and ND (Donor concentration) ?

n ND

(A) . per cm7 07 1012 3# .1 414 1013

# per cm3

(B) . per cm1 414 1013 3# . per cm0 707 1013 3

#

(C) . per cm2 828 1013 3# . per cm0 707 1013 3

#

(D) .1 414 1013# per cm3 . per cm1 414 1013 3

#

MCQ 1.1.28 Which of the following sketches best describes the DN versus ND dependence of electrons in silicon at room temperature ?

Common Data For Q. 29 and 30At room temperature KT 300=^ h, the probability that an energy state in the conduction band edge EC^ h of silicon is 10 4- .

MCQ 1.1.29 The type of semiconductor is(A) n -type (B) p-type

(C) intrinsic (D) can’t be determine

MCQ 1.1.30 Assume the effective density of states function . cmN 2 86 10,C V19 3

#= - . What is the value of doping concentration ?(A) . cmN N 1 26 10D A

10 3#- = -

(B) . cmN N 1 26 10A D10 3

#- = -

(C) . cmN N 2 8 10D A15 3

#- = -

(D) . cmN N 2 8 10A D15 3

#- = -

PIERRET73/2.16(A)

PIERRET73/2.16(C)

PIERRET183/7

ANDERSON109/2.34

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Common Data For Q. 31 and 32An unknown semiconductor has bandgap . eVE 1 1g = and N Nc v= . It is doped with cm1015 3- donors, where the donor level is 0.2 eV below EC . Given that EF is 0.25 eV below EC at KT 300= .

MCQ 1.1.31 What is the concentration of electrons ?(A) . cm8 733 1014 3

#-

(B) . cm1 27 1014 3#

-

(C) cm1015 3-

(D) cm105 3-

MCQ 1.1.32 What is the value of effective density of electron NC^ h ?(A) . cm1 97 1018 3

#- (B) . cm1 359 1019 3

#-

(C) . cm4 82 1018 3#

- (D) . cm2 85 1019 3#

-

MCQ 1.1.33 What is the value of intrinsic carrier concentration ?(A) . cm3 10 109 3

#-

(B) . cm6 62 1019 3#

-

(C) . cm7 59 104 3#

-

(D) . cm8 142 109 3#

-

Common Data For Q. 34 and 35A piece of intrinsic silicon at room temperature is kept at thermal equilibrium. The position of energy level, Ex is set exactly 0.6 eV above the intrinsic level and band gap of intrinsic silicon . eVE 1 1g =^ h .

MCQ 1.1.34 What is the type of semiconductors if the probability of capture of an energy state by an electron at Ex is 50%.(A) p-type, non degenerate

(B) n -type, non degenerate

(C) p-type, degenerate

(D) n -type, degenerate

MCQ 1.1.35 What is the doping concentration(A) . cm1 725 1016 3

#-

(B) . cm1 725 1020 3#

-

(C) . cm2 879 1019 3#

-

(D) . cm2 879 1016 3#

-

***********

STREETMAN V106/3.20

BHATTACHA-RYA 119/3.14


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EXERCISE 1.2

QUES 1.2.1 The intrinsic carrier concentration, ni at 200 KT = for silicon is_____cm104 3

#- .

QUES 1.2.2 The intrinsic carrier concentration, ni for germanium at 400 KT = is ____ cm104 3

#- .

QUES 1.2.3 The intrinsic carrier concentration, ni at 600 KT = for GaAs is____cm1012 3

#- .

QUES 1.2.4 The intrinsic carrier concentration in silicon is to be no greater than 1 10 cmni

12 3#= - . What will be the maximum temperature (in K) allowed

for the silicon ?

QUES 1.2.5 Two semiconductor materials have exactly the same properties except that material A has a bandgap energy of 1.0 eV and material B has a bandgap energy of 1.2 eV. The ratio of intrinsic concentration of material A to that of material B for 300 KT = will be _____

QUES 1.2.6 The hole concentration in silicon at 300 KT = is 10 cm15 3- . The concentration of electrons in the material will be_____ cm104 3

#- .

QUES 1.2.7 Given the acceptor and donor concentrations in a germanium semiconductor respectively as Na 10 cm13 3= - , N 0d = . The thermal equilibrium hole concentration in the material at 300 KT = will be_____ cm1013 3

#- .

QUES 1.2.8 A silicon semiconductor is doped with the donor and acceptor concentrations respectively as Nd 2 10 cm15 3

#= - and N 0a = . The thermal equilibrium hole concentration in the material at 300 KT = will be_____ cm105 3

#- .

QUES 1.2.9 A silicon semiconductor has the dopant concentrations Nd 0= , 10 cmNa14 3= - .

The thermal equilibrium electron concentration in the material at 00 KT 4= will be_____ cm1010 3

#- .

Chap 1


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QUES 1.2.10 In a sample of GaAs at 200 KT = , we have experimentally determined that n p50 0= and N 0a = . What will be the concentration of electrons (in cm 3-

) in the material ?

QUES 1.2.11 Silicon at 300 KT = is uniformly doped with phosphorus atoms at a concentration 2 10 cm15 3

#- and boron atoms at a concentration of

3 10 cm16 3#

- . The thermal equilibrium concentration (in cm 3- ) of minority carriers will be_____ cm103 3

#- .

QUES 1.2.12 In silicon at 300 KT = , we have experimentally found that 4.5 10 cmn04 3

#= - and 5 10 cmNd

15 3#= - . The acceptor impurity concentration in the material

will be _____ cm1016 3#

- .

QUES 1.2.13 A GaAs device is doped with a donor concentration of 3 10 cm15 3#

- . For the device to operate properly, the intrinsic carrier concentration must remain less than 5 percent of that electron concentration. What is the maximum temperature (in K) that the device may operate ?

QUES 1.2.14 Silicon at 300 KT = contains acceptor atoms at a concentration of 5 10 cmNa

15 3#= - . Donor atoms are added forming an n -type compensated

semiconductor such that the Fermi level is 0.215 eV below the conduction band edge. The concentration of donor atoms added is ____ cm1016 3

#- ?

Common Data For Q. 15 and 16Silicon at 300 KT = is doped with acceptor atoms at a concentration of

7 10 cmNa15 3

#= - .

QUES 1.2.15 The difference between Fermi energy and valence band energy, E EF v- equals to_____eV.

QUES 1.2.16 The concentration of additional acceptor atoms that must be added to move the Fermi level a distance kT closer to the valence-band edge will be _____ cm1016 3

#- .

QUES 1.2.17 If silicon is doped with boron atoms at a concentration of 10 cm15 3- then, the change in Fermi level, E EF Fi- will be_____eV.

QUES 1.2.18 Consider intrinsic germanium at room temperature (300 K). By what percent does the conductivity increase per degree rise in temperature ? MILINANS

87/4.1


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QUES 1.2.19 In an n -type silicon, the donor concentration is 1 atom per 2 108# silicon

atoms. Assume that the effective mass of the electron equals the true mass and the density of atoms in the silicon is /atoms cm5 1022 3

# . At what temperature in Kc^ h will the Fermi level coincide with the edge of the conduction band ?

QUES 1.2.20 What is the ratio of the probability that a state is filled at the conduction band edge EC^ h to the probability that a state is empty at the valence band edge EV^ h if the Fermi level is positioned at midgap ?

QUES 1.2.21 If Fermi energy level EF is positioned at EC (edge of the conduction band), then the probability of finding electrons in states at the E KTC + will be___

QUES 1.2.22 For a non-degenerate semiconductor, the peak in the electron distribution versus energy inside the conduction band occurs at /E KT 2C + . What is the ratio of the electron population in a non-degenerate semiconductor at E E KT5C= + to the electron population at the peak energy ?

QUES 1.2.23 For a silicon sample maintained at KT 300= , the Fermi level is located 0.259 eV above the intrinsic level and intrinsic concentration per cmn 10i

10 3= . The hole concentration is_____ per cm105 3

# .

QUES 1.2.24 The probability of occupancy of a state at the bottom of the conduction band in intrinsic silicon at room temperature is _____ 10 10

#- (assume at

room temperature . VKT 0 026= ).

QUES 1.2.25 Two semiconductors A and B have the same density of states effective masses. Semiconductor A has a bandgap energy of . eV1 0 and semiconductor B has a bandgap energy of eV2 . The ratio of intrinsic concentration of semiconductor A to semiconductor B for KT 300= will be_____ 108

# .

QUES 1.2.26 For a piece of GaAs (Gallium Arsenide) having a band gap . eVE 1 43g =. The minimum frequency of an incident photon that can interact with a valence electron and elevate the electron to the conduction band is_____

Hz1014# .

QUES 1.2.27 A piece of intrinsic silicon at room temperature is kept at thermal equilibrium. The position of some random level Ex is to be fixed at . eV0 9 above the valence band edge. The doping concentration such that the probability of capture of an energy state by an electron at Ex is 50% is ___ cm1016 3

#- ?

***********

MILMAN102/4.8

PEIRRET44/2.2

PEIRRET71/2.6

PIERRET71/2.8

PIERRET73/2.16(D)

ANDERSON76/2.2

ANDERSON109/2.36



Chap 1


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EXERCISE 1.3

MCQ 1.3.1 The mobility of charge-carriers has the unit(A) / secm volt 2

- (B) secm volt-

(C) / secm volt3- (D) / secm volt2

-

MCQ 1.3.2 The total energy of a revolving electron in an atom can(A) have any value above zero

(B) never be positive

(C) never be negative

(D) not be calculated

MCQ 1.3.3 Electronic distribution of an Si atom is(A) 2, 10, 2

(B) 2, 8, 4

(C) 2, 7, 5

(D) 2, 4, 8

MCQ 1.3.4 Major part of the current in an intrinsic semi-conductor is due to(A) conduction-band electrons

(B) valence-band electrons

(C) holes in the valence band

(D) thermally-generated electron

MCQ 1.3.5 Conduction electrons have more mobility than holes because they(A) are lighter

(B) experience collisions less frequently

(C) have negative charge

(D) need less energy to move them

MCQ 1.3.6 Current flow in a semiconductor depends on the phenomenon of (A) drift

(B) diffusion

(C) recombination

(D) all of the above

SAN. SHARMA26/4

B.L.THERAJA36/1

B.L.THERAJA36/4

B.L.THERAJA36/8

B.L.THERAJA36/9

B.L.THERAJA36/11


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MCQ 1.3.7 The process of adding impurities to a pure semiconductor is called(A) mixing (B) doping

(C) diffusing (D) refining

MCQ 1.3.8 Hall effect is observed in a specimen when it (metal or a semiconductor) is carrying current and is placed in a magnetic field. The resultant electric field inside the specimen will be in(A) a direction normal to both current and magnetic field

(B) the direction of current

(C) a direction antiparallel to the magnetic field

(D) an arbitary direction depending upon the conductivity of the specimen

MCQ 1.3.9 An n -type semiconductor as a whole is(A) positively charged

(B) negatively charged

(C) positively or negatively charged depending upon doping

(D) electrically neutral

MCQ 1.3.10 In p-type semiconductor, there are(A) no majority carriers

(B) electrons as majority carriers

(C) immobile negative ions

(D) immobile positive ions

MCQ 1.3.11 Fermi level represents the energy level with probability of its occupation of (A) 0 (B) 50%

(C) 75% (D) 100%

MCQ 1.3.12 The resistivity of a semiconductor depends on the(A) shape of the semiconductor

(B) atomic nature of the semiconductor

(C) width of the semiconductor

(D) length of the semiconductor

MCQ 1.3.13 An electron in conduction band has(A) no charge

(B) higher energy than electron in the valance band

(C) lower energy than the electron in the valance band

(D) All of these

B.L.THERAJA36/12

B.L.THERAJA37/24

SANJEE GUPTA2.24/6

SANJEE GUPTA2.24/10

SANJEE GUPTA3.29/1

S SALIVAHANA2.76/16

B.P.SINGH49/4

Chap 1


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MCQ 1.3.14 In an intrinsic silicon the band gap is(A) 1.12 eV (B) 0.7 eV

(C) 2 eV (D) 0.2 eV

MCQ 1.3.15 Mobility of holes in intrinsic Si is(A) . /m Vs0 048 2

(B) . /m Vs0 135 2

(C) /m Vs1350 2

(D) /m Vs480 2

MCQ 1.3.16 Fermi level in the intrinsic Si or Ge is(A) in the middle of the band gap

(B) near the valance band

(C) near the conduction band

(D) none of these

MCQ 1.3.17 The diffusion constant and the mobility of electron are related as(A) / /D KT qn nm =

(B) / /D q KTn nm =(C) /D KTqn nm =

(D) /D qKTn nm =

MCQ 1.3.18 Electron population in silicon is not(A) zero in the forbidden band

(B) zero in the conduction band at 0 K

(C) zero at the conduction band edge EC

(D) zero in the conduction band at room temperature

MCQ 1.3.19 Acceptor impurity atom in germanium results in(A) increased for bidden energy gap

(B) reduced for bidden energy gap

(C) new narrow energy band slightly above the valence level

(D) new discrete energy level slightly above the valence level

MCQ 1.3.20 If ND and NA are the donor and acceptor concentrations respectively and N N>D A, then net impurity concentration is(A) N ND A- (B) N ND A+(C) N NA D- (D) .N ND A

B.P.SINGH51/38

B.P.SINGH52/45

B.P.SINGH52/57

B.P.SINGH55/100

B.P.SINGH55/105

G.K. MITHAL84/2.10

G.K. MITHAL85/2.24


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MCQ 1.3.21 If donor concentration ND equals acceptor concentration NA, the resulting semiconductor is(A) n -type (B) p-type

(C) both p and n type (D) intrinsic

MCQ 1.3.22 In an intrinsic semiconductor, the concentration of free electrons in the conduction band is given by(A) N e /

cE E kTc F-^ h (B) N e /

cE E kTF c-^ h

(C) N e /c

E E kTF v-^ h (D) N e /c

E E kTv F-^ h

MCQ 1.3.23 In an intrinsic semiconductor, the concentration of holes in the valence band is given by(A) N e /

vE E kTv F-^ h (B) N e /

vE E kTF v-^ h

(C) N e /v

E E kTV F-^ h (D) N e /v

E E kTF v-^ h

MCQ 1.3.24 In an intrinsic semiconductor, the concentration of charge carriers equals(A) A T e /E kT

02 G0 (B) A T e /E kT

03 G0

(C) A Te /E kT0

G0 (D) A T e/ /E kT0

3 2 G0

MCQ 1.3.25 In an intrinsic semiconductor, forbidden energy gap EG equals(A) E T /

G01 2b - (B) E TG0 b +

(C) E TG0 b - (D) E T /G0

1 2b +where b is a positive number

***********

B.P.SINGH86/2.27

G.K. MITHAL131/3.5

G.K. MITHAL131/3.6

G.K. MITHAL131/3.7

G.K. MITHAL131/3.8

Chap 1


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SOLUTIONS 1.1

SOL 1.1.1 Correct option is (C).At 300 K, intrinsic carrier concentration is

ni . /cm1 5 1010 3#=

Hence, free electron concentration is

ne . /cm1 5 1010 3#=

SOL 1.1.2 Correct option is (A).Since, sample A and B are identical. So, by using mass action law, we have

n pA A n pB B=The donor impurity in sample A is

NA m1 1020 3#= -

Donor impurity in sample B is

NB m3 1020 3#= -

The hole concentration in sample A is

pA 9 1012#=

So, the hole concentration in sample B is

pB 3 10

1 10 9 1020

20 12

#

# # #= ^ ^h h

m3 1012 3#= -

SOL 1.1.3 Correct option is (C).The concentration of electrons and holes are defined as

n0 expN kTE E

cc F= --^ h

; E

p0 expN kTE E

vF v= --^ h

; E

At Fermi level position, the electron and hole concentrations are equal. So, we have

expN kTE E

cc F--^ h

; E expN kTE E

vF v= --^ h

; E

If we take natural log of both sides, then

EF lnE E kT NN

21

21

c vc

v= + +^ bh l ...(1)

Also, we know that

E E21

c v+^ h Emidgap=

Substituting it in equation (1), we get

E EmidgapF - lnkT NN

21

c

v= b l ...(2)


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Again, we have

Nc hm kT2 2 * /

n2

3 2p = c m

and Nv hm kT

22 * /

p2

3 2p = e o

Therefore, we obtain

NN

c

v mm

*

* /

n

p3 2

= e o

Thus, by substituting the above expression in equation (2), we get

E EmidgapF - lnkTmm

43

*

*

n

p= e o

..lnkT m

m43

1 080 56

o

o= c m

. .e43 0 0259 0 656# #= -^ h

. eV0 0128=-The negative sign indicate the Fermi Energy level is . eV0 0128 below the centre of the band gap.

SOL 1.1.4 Correct option is (A).The concentration of electrons and holes are defined as

n0 expN kTE E

cc F= --^ h

; E

p0 expN kTE E

vF v= --^ h

; E

At Fermi level position, the electron and hole concentration are equal, i.e.

expN kTE E

cc F--^ h

; E expN kTE E

vF v= --^ h

; E

If we take natural log of both sides, then

EF lnE E kT NN

21

21

midgap

c vc

v= + +^ bh l1 2 344 44

or E EmidgapF - lnkT NN

21

c

v= b l ...(1)

At T 300c= for silicon, we have

Nc .2 8 1019#=

and Nv .1 04 1019#=

Therefore, we obtain

NN

c

v ..2 81 04=

Since the ratio does not depend on temperature, so at T 200= , we get

NN

c

v ..2 81 04=

Substituting it in equation (1), we have

E EmidgapF - . ..lne2

1 0 0259 300200

2 81 04

#= b bl l

. eV0 0085=-Thus, the intrinsic Fermi level is . eV0 0085 below the centre of the bandgap.

Chap 1


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SOL 1.1.5 Correct option is (C).At KT 300= the electron concentration is

n0 cm5 104 3#= -

At KT 300= intrinsic carrier concentration for silicon is

ni .1 5 1010#=

By using mass action law, we obtain the hole concentration as

n p0 0 ni2=

or p0 nni

0

2

=

.5 10

1 5 104

10 2

#

#= ^ h

. cm4 5 1015 3#= -

SOL 1.1.6 Correct option is (A).We have the electron and hole concentrations for the silicon as

n0 cm5 104 3#= -

and p0 . cm4 5 1015 3#= -

So, we conclude that

p0 n> 0

i.e. the concentration of hole is greater the concentration of electron. It means hole are in majority in this material, hence it is p-type material.

SOL 1.1.7 Correct option is (A).Given that Fermi energy in silicon is 0.22 eV above the valence band energy, i.e.

E EF v- . eV0 22=So, we obtain the hole concentration as

p0 expN kTE E

vF v

= --^ h

; E

. ..exp e

e1 04 10 0 02590 2219

#= -: D

. cm2 13 1015 3#= -

Now, the energy bandgap for silicon is 1.12 eV, i.e.

Eg . eVE E 1 12c v= - =Therefore, we obtain

E E E Ec F F v- + -^ h . eV1 12=or E Ec F- . . . eV1 12 0 22 0 90= - =Hence, the hole concentration is

n0 expN kTE E

cc F

= --^ h

; E

. ..exp e

e2 8 10 0 02590 9019

#= -: D

. cm2 27 104 3#= -


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SOL 1.1.8 Correct option is (C).Given the difference between conduction energy level and Fermi energy level as

E Ec F- . eV0 25=At KT 300= , for GaAs,

Nc .4 7 1017#=

So, at KT 400= , we obtain

Nc . T4 7 10 300/

173 2

#= b l

Therefore, the electron concentration is

n0 expN kTE E

cc F= --^ h

; E

..

.expe

e4 7 10 300400

0 025 300400

0 25/17

3 2

# ##

= -bb

ll> H

. ..exp7 24 10 0 03453

0 02517#= -b l

. cm5 19 1014 3#= -

Again, we have the energy bandgap for GaAs

Eg .E E 1 42c v= - =So, E E E Ec F F v- + - .1 42=or E EF v- . . . eV1 42 0 25 1 17= - =Also, At KT 400c= , we obtain

Nv . T7 0 10 300/

183 2

#= b l

.7 0 10 300400 /

183 2

# #= b l

. cm1 08 1019 3#= -

Therefore, the hole concentration is obtained as

p0 expN kTE E

vF v= --^ h

; E

..

.expe

e1 08 100 0259 300

4001 1719

##

= -> H

. cm2 08 104 3#= -

SOL 1.1.9 Correct option is (B).Since, the electron concentration n0 obtained in previous question should remain constant, so we have

n0 . cm5 19 1014 3#= -

At KT 300= ,

kT . eV0 0259=and Nc .4 7 1017

#=Since, the electron concentration is defined as

n0 expN kTE E

cc F= --^ h

; E

So, E Ec F- lnkT nNc

0= b l

Chap 1


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...ln0 0259

5 19 104 7 10

14

17

##= c m

. eV0 176=Therefore, we get

E EF v- E E Eg c F= - -^ h

. .eV eV1 42 0 176= - . eV1 244=Hence, the hole concentration is given as

p0 ..exp eV

eV7 10 0 02591 24418

#= -: D

. cm9 67 10 13 3#= - -

SOL 1.1.10 Correct option is (D).Given the doping concentrations,

Nd cm5 1015 3#= - , N 0a =

At thermal equilibrium, we have

N pd 0+ N na 0= + ...(1) From mass action law, we have

p0 nni

0

2

=


N nn

di

0

2

+ n0= (given N 0a = )

or n n N nd i02

02- - 0=

Solving the quadratic equation, we obtain the electron concentration as

n0 N N n

24D D i

2 2!= +

.

25 10 5 10 4 2 4 1015 15 2 13 2# # #=

+ +^ ^h h

cm5 1015 3#= -

Hence, the hole concentration is given as

p0 .

nn

5 102 4 10i

0

2

15

13 2

#

#= = ^ h

. cm1 15 1011 3#= -

SOL 1.1.11 Correct option is (C).At thermal equilibrium, we have

N pd 0+ N na 0= +

or N nn

di

0

2

+ N na 0= + (mass action law, p0 /n ni2

0= )

or n N N n nd a i02

02- - -^ h 0=

Solving the quadratic equation, we have the electron concentration as

n0 N N N N n

24d a d a i

2 2

=- + - +^ ^h h

Since, the doping concentrations are same, i.e.

Nd Na=


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Therefore, we get

n0 ni=and p0 ni=For silicon, the intrinsic concentration at 300 KT = is

ni . cm1 5 1010 3#= -

Hence, n0 . cmp n 1 5 10i010 3

#= = = -

SOL 1.1.12 Correct option is (A).Given the dopant concentrations for GaAs (Gallium arsenide) as

Nd cm1 1013 3#= -

and Na .2 5 1013#=

Since, the acceptor concentration is greater than donor concentration, i.e.

Na N> d

So, the material is p-type.Now, the intrinsic concentration of GaAs at T K300= is

ni .1 8 106#=

Again, at thermal equilibrium, we have

N Na d-^ h n>> i

Therefore, the hole concentration is obtained as

p0 N N N N n

24a d a d i

2 2

=- + - +^ ^h h

(N N n>>a d i- )

N Na d= - .2 5 10 1 1013 13

# #= - . cm1 5 1013 3

#= -

Also, we obtain the electron concentration using mass action law as

n0 ..

pn

1 5 101 8 10i

0

2

13

6 2

#

#= = ^ h

. cm0 216 3= -

SOL 1.1.13 Correct option is (B).Boron is acceptor atom and arsenic is donor atom. So, we have the dopant concentrations in silicon as

Na . cm1 5 1015 3#= -

and Nd cm8 1014 3#= -

Therefore, we conclude that

Na N> d

i.e. the material is p-type.Now, for silicon at T 300= , we have

Nc .2 8 1019#=

and Nv .1 04 1019#=

So, the intrinsic concentration of silicon at T 450= is given as

ni2 N N ec v kT

Eg

= -

. . T e2 8 10 1 04 10 300 . /e TE

19 190 0259 300

g

# # # #= #-

b ^l h

Chap 1


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. . e2 8 10 1 04 10 300450

.. eV

e19 19

2

0 02591 12

300450# # # # #= #

-b l

or ni . cm1 72 1013 3#= -

Hence, we obtain the hole concentration as

p0 N N N N n

2a d a d i

2 2

=- + - +^ ^h h

N Na d= -^ h (N N n>>a d i- )

cm7 1014 3#= -

Using mass action law, the electron concentration ( )n0 is obtained as

n0 .

pn

7 101 72 10i

0

2

14

13 2

#

#= = ^ h

. cm4 23 1011 3#= -

SOL 1.1.14 Correct option is (C).Given the doping concentration for the semiconductor,

Nd cm2 1013 3#= - , N 0a =

So, the donor atom impurity is greater than acceptor impurity, i.e.

Nd N> a

i.e. this is the n -type material.Again, we have the intrinsic concentration of the semiconductor as

ni cm2 1013 3#= -

Therefore, the electron concentration (majority carrier) is obtained as

n0 N N n

24d d i

2 2

= + +

22 10 2 10 4 2 1013 13 2 13 2# # # #=

+ +^ ^h h

. cm3 24 1013 3#= -

Hence, the thermal equilibrium concentration of minority carriers (holes) is

p0 .nn

3 24 102 10i

0

2

13

13 2

#

#= = ^ h

. cm1 23 1013 3#= -

SOL 1.1.15 Correct option is (A).At T K200= , we have

kT . eV0 0259 300200= b l

. eV0 01727=So, the intrinsic concentration is obtained as

ni2 N N ec v kT

Eg

= -

. . e1 04 10 6 0 10 300200

..19 18

30 01727

0 66# # #= -

b l

or ni . cm2 16 1010 3#= -

Now, we have the concentration of dopants as

Na 1015= , N 0d =So, Na n>> i


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Therefore, the hole concentration in the germanium atom is given as

p0 N N N N n

24a d a d i

2

=- + - +^ ^h h

N N n2

4a a i2 2

= + +

Na= (Na n>> i)

cm1015 3= -

Again, the hole concentration is defined as

p0 expN kTE E

vF v

= --^ h

; E

expN kTE E E E

vFi F Fi v=- - -^ h

; E

exp expN kTE E

kTE E

vFi v Fi F= -- -^ h

; :E D

or p0 expn kTE E

iFi F= -

b l Put expn N kTE E

i vFi F

= --^ h

; E

or E EFi F- lnkT np

i

0= b l ..

ln0 017272 16 10

106

15

#= c m

Hence, E EFi F- . eV0 1855=i.e. Fermi energy is . eV0 1855 below the intrinsic Fermi level.

SOL 1.1.16 Correct option is (B).For germanium, the intrinsic concentration at T K300= is

ni . cm2 4 1013 3#= -

Given the doping concentrations,

Nd cm1014 3= -

Na 0=So, Nd ni-

Therefor, we obtain the electron concentration as

n0 N N n

24d d i

2 2

= + +

.

210 10 4 2 4 1014 14 2 13 2

#=+ +^ ^h h

. cm1 05 1014 3#= -

Again, we defined the electron concentration as

n0 N e /c

E E kTc F= - -^ h

N e /c

E E E E kTc Fi F Fi= - - + -^ ^h h6 @

N e e/ /c

E E kT

n

E E kTc Fi

i

F Fi= - - -^ ^h h

1 2 3444 444

or n0 n e /i

E E kTF Fi= -^ h (n N e /i c

E E kTc Fi= - -^ h )


i

0= a k

...ln0 02592 4 101 05 10

13

14

##= c m

. eV0 0382=i.e. position of Fermi energy level is . eV0 0382 above the intrinsic Fermi level.

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SOL 1.1.17 Correct option is (B).Since. the phosphorus atom is a donor atom, so the donor concentration is

Nd cm1015 3= -

For silicon, the intrinsic concentration at KT 300= is

ni . cm1 5 1010 3#= -

So, Nd n>> i

Therefore, the electron concentration is

n0 Nd,

cm1015 3= -

Since, we define the electron concentration as

n0 n e /i

E E kTF Fi= -^ h


i

0= a k

..

ln0 02591 5 10

1010

15

#= c m

. eV0 2877=

i.e. position of Fermi level is . eV0 2877 above the intrinsic Fermi level.

SOL 1.1.18 Correct option is (D).Given the concentration of acceptor atom,

Na 1015=and E EFi F- . eV0 45=For GaAs, the intrinsic concentration at KT 300= is

ni .1 8 106#=

So, we obtain the hole concentration as

E EFi F- lnkT np

i

0= b l

or . eV0 45 . ln eVnp0 0259

i

0= b l

or p0 n e ..

i 0 02590 45

= b l

. cm6 32 1013 3#= -

Hence, p0 N< a

Therefore, the donors must be added with concentration

Nd N pa 0= - .10 6 32 1015 13

#= -

. cm9 368 1014 3#= -

SOL 1.1.19 Correct option is (A).The effective density of states in the conduction band is given by

NC hm KT

22 * /

n2

3 2p = ; E ...(1)

where m *n is effective mass of electron, given as

m *n . . .m kg1 4 1 4 9 1 10 31

# #= = -

and h is the Plank’s constant, given as


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h . secJ6 62 10 34#= --

At KT 300c= , we have

KT . eV0 0259=So, by substituting the values in equation (1), we get

NC .

. . . eV2

6 62 102 1 4 9 1 10 0 0259

/

34 2

31 3 2

#

# # #p = -

-

^

^

h

h> H

.

. . . .26 62 10

2 1 4 9 1 10 0 0259 1 6 10 /

34 2

31 19 3 2

#

# # # # # #p = -

- -

^

^

h

h> H

. m4 15 1025 3#= -

SOL 1.1.20 Correct option is (B).The concentration of electrons in the semiconductor is given by

no expN KTE E

CC F= --^ h

; E

In previous question, we have determined

NC . m4 15 1025 3#= -

Also, we have

E EC F- . eV0 25=and KT . eV0 0259=Hence, the concentration of electrons in the semiconductor is

no . ..exp eV

eV4 15 10 0 02590 2525

#= -: D

. m2 67 1021 3#= -

SOL 1.1.21 Correct option is (D).The effective density of states in the conduction band is given by

NC hm KT

22 * /

n2

3 2p = ; E ...(1)

The effective density of state in the valence band is given by

NV hm KT

22 * /

p2

3 2p = = G ...(2)

So, from equations (1) and (2), we get

NN

V

C mm

*

* /

p

n3 2

= e o

In case of intrinsic material, we have

ni pi=

or expN KTE E

CC F--^ h

; E expN KTE E

VF V= --^ h

; E

Taking the logarithm of both sides, we obtain

ln NN

V

Cc m KTE E E2C V F= + -

or EF lnE E KTNN

2 2C V

V

C= + - c m

Since, for intrinsic material, we know

Emidgap E E

2C V= +

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Hence, the position of Fermi level with respect to the middle of the bandgap is

E EmidgapF - lnKTNN

2 V

C=- c m

ln lnKTmm KT

mm

2 43

*

* /

*

*

p

n

p

n3 2

=- =-e eo o

...

.lneVmm

eV43 0 0259

0 370 55

7 7 10 3# #=- =- -b l

i.e. the Fermi level is located at . eV7 7 10 3#

- below the middle of the forbidden gap the intrinsic germanium at K300c .

SOL 1.1.22 Correct option is (D).The Fermi energy level is defined as

EF lnE E KTNN

2 2C V

V

C= + - c m

lnE E E KTNN

22

2C V V

V

C= - + - c m

lnE E E KTNN

2 2C V

VV

C= - + - c m

Substituting bandgapE E EG C V= -^ h in the above equation, we obtain

E EF V- lnE KTNN

2 2G

V

C= - c m

...lneV KT

21 43

2 1 3 101 5 10

19

18

##= - d n

. . . eV0 715 0 028 0 743= + =Hence, the Fermi level is located at 0.743 eV above the valence band edge EV .

SOL 1.1.23 Correct option is (B).Again, the Fermi energy level is given as

EF lnE E KTNN

2 2C V

V

C= + - c m

lnE E E KTNN

22

2C V C

V

C= + - - c m

lnE E KTNN

2 2CG

V

C= - - c m

So, we obtain

E EF C- lnE KTNN

2 2G

V

C=- - c m

. .eV eV21 43 0 028=- +

. eV0 687=-which implies that the Fermi level is located at 0.687 eV below EC .

SOL 1.1.24 Correct option is (D).The effective density of states in the conduction band is given by


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NC h

m KT2 2 * /n2

3 2p = < F

. .T mm4 82 10 /

* /n21 3 2

3 2

#= c m

Hence, we obtain the effective mass of electron as

m *n .

NT m4 82 101

/C

21

2 3

#= c m) 3

..

m4 82 101 5 10

31 00

/

21

24 2 3

##= c m) 3

. . .m0 135 0 135 9 1 10 31# #= = -

. kg1 23 10 31#= -

Similarly, we define the effective density of states in the valence band as

NV . .T mm

4 82 10 /* /p21 3 2

3 2

#= c m

Hence, the effective mass of hole is

m *p .

NT m4 82 101

/V

21

2 3

#= c m( 2

.. m4 82 10

1 3 103001/

21

25 2 3

##= d n) 3

. . .m0 646 0 646 9 1 10 31# #= = -

. kg5 88 10 31#= -

SOL 1.1.25 Correct option is (C).The probability that an energy state is filled at E KTC + is given by

f E KTC +^ h /exp E KT E KT1

1C F

=+ + -^ h8 B

...(1)

The probability that a state is empty at E KTC + is given by

f E KT1 C- +^ h /exp E KT E KT

11

1C F

= -+ + -^ h8 B

...(2)

Given that the two probabilities are equal, i.e.

f E KTC +^ h f E KT1 C= - +^ h

or e1

1/E KT E KTC F+ + -^ h

e

11

1/E KT E KTC F

= -+ + -^ h

[From eqs (1) and (2)]

or e1


e

e1 /

/

E KT E KT

E KT E KT

C F

C F

=+ + -

+ -

^

^

h

h

or e1


e1

1/E E KT KTF C

=+ - +^ h6 @

or KTE KT EC F+ - KT

E E KTF C= - +^ h

or E2 F E KT2 C= +^ h

Hence, EF E KTC= +

SOL 1.1.26 Correct option is (D).At temperature KT 0" c , the thermal energy available in the system is insufficient to release the weakly bound electron or hole in donor or acceptor

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atom. Hence, the electron concentration is

n 0= at KT 0c=

SOL 1.1.27 Correct option is (B).From mass action law, we have

np ni2= ...(1)

Given the intrinsic concentration,

ni /cm1013 3=and the relation,

n p2=or p /n 2=Substituting it in equation (1), we get

n n2^ ah k ni

2=

or n n2 2 10i13

#= = . per cm1 414 1013 3

#=Now, we have the charge neutrality relationship as

p ND+ n NA= +

or n N2 D+ n 0= + (Given /p n 2= , N 0A = )

or ND n2=

n22 i= (put n n2 i= )

n2i= . per cm

210 0 707 10

1313 3

#= =

SOL 1.1.28 Correct option is (C).

The graph exhibits the observed doping dependence of the electron and hole mobilities in semiconductor. At low doping concentrations below approximately /cm1015 3 in Si, the carrier mobilities are essentially independent of doping concentration. For dopings in excess of /cm1015 3-

, the mobilities monotonically decrease with increasing NA or ND , and we know that


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DN qKT

nm = c m

Hence, the DN versus ND curve is same as nm versus ND .

SOL 1.1.29 Correct option is (A).Given, the probability that an energy state in the conduction band edge EC^ h of silicon,

f EC^ h 10 4= - ...(1)Since, the probability that an energy state in the conduction band edge EC^ h of silicon can also be expressed as

f EC^ h /exp E E KT1

1C F

=+ -^ h8 B

...(2)

So, from equations (1) and (2), we get

E EC F- lnKT101 14= --c m

. , .lneV eV0 026 9 999 0 24= =^ h

Now, the bandgap of silicon at room temperature is

Eg . eV1 12=

So, E2

g . eV0 56=

Therefore, we may sketch the energy band diagram for the semiconductor as

From the energy band diagram, we observe that the Fermi level lies above the midgap energy level, hence it is an n -type semiconductor.

SOL 1.1.30 Correct option is (C).It is n -type semiconductor and Fermi level is well above the intrinsic level. So, we have

N ND A- n>> i

Therefore, we may express

N ND A- /expn N E E KTo C C F= = - -^ h8 B

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. ..exp eV

eV2 86 10 0 0260 2419

#= -: D

. cm2 8 1015 3#= -

SOL 1.1.31 Correct option is (A).For the given problem, we sketch the energy band diagram as

The energy band diagram shows the incomplete ionization. So, we have

f ED^ h

.. .

exp eVeV1 0 0259

0 051 0 1267=

+=

b l

Therefore, the concentration of electron is

n [ ( )]f E N1 d d= - . . cm1 0 1267 10 8 733 1015 14 3

#= - = -^ h

SOL 1.1.32 Correct option is (B).The electron concentration is defined as

n N e /C

E E KTC F= - -^ h

Hence, we get

NC ne /E E KTC F= -^ h

. e8 733 10 . / .14 0 25 0 0259#=

. cm1 359 1019 3#= -

SOL 1.1.33 Correct option is (D).In previous solution, we have obtained

N NC V= . cm1 359 1019 3#= -

Hence, the hole concentration is

p N e /V

E E KTF V= - -^ h

. e1 359 10 . . / .eV eV19 1 1 0 25 0 0259#= - -^ h

. cm7 591 104 3#= -

Using mass action law, we obtain

ni2 np=

or ni2 ( . ) ( . ) .8 733 10 7 591 10 6 62 1014 4 19

# # # #= =or ni . cm8 142 109 3

#= -

SOL 1.1.34 Correct option is (D).Probability of capture of an energy state by an electron at Ex is 50% means that Ex is actually the Fermi level. Hence, the Fermi level is 0.6 eV above


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intrinsic level. Therefore, we may easily analyze that the Fermi level has moved into the conduction band, as shown below.

Thus, the semiconductor is n -type degenerate semiconductors.

SOL 1.1.35 Correct option is (B).The doping concentration for n -type degenerate semiconductor is equal to the electron concentration, i.e.

Doping concentration Electron concentration=Since, we have the energy band diagram for the semiconductor as

Hence, we obtain

no /expn E E KTi F i= -^ h8 B

. ..exp eV

eV1 5 10 0 02590 610

#= : D

. cm1 725 1020 3#= -

***********

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SOLUTIONS 1.2

SOL 1.2.1 Correct answer is 7.68.Intrinsic carrier concentration is given by

ni2 expN N kT

Ec v

g= -b l

For silicon, at KT 300c= Nc = effective density of state in conduction band cm 3-

^ h

. cm2 8 1019 3#= -

Nv = effective density of states in valence band cm 3-^ h

. cm1 04 1019 3#= -

Eg = bandgap energy . eV1 12=

qkT . mVe

kT 25 9= =

Intrinsic carrier concentration is a very strong function of temperature. So, we obtain the parameters at T C200c= as

ekT . m25 9 300

200#=

Nc .2 8 10 300200 /

193 2

#= b l , .N 1 04 10 300200 /

v19

3 2

#= b l

Hence, the intrinsic carrier concentration, at 200 KT = for silicon is obtained as

ni2 . .

..exp

m2 8 10 1 04 10 300

20025 9 300

2001 1219 19

3

# # ##

= -bf

lp

.. /

.expm

0 2876 1025 9 2 3

1 1238#

#= -c m

or ni . .0 8628 10 6 753 1038 29# # #= -

. cm7 68 104 3#= -


ni2 expN N kT

Ec v

g= -b l

For Ge (germanium), at KT 300c= (room temperature)

Nc = effective density of state in conduction band (cm 3- )

. cm1 04 1019 3#= -

Nv = effective density of state in valence band

. cm6 0 1018 3#= -

Eg = bandgap energy

. eV0 66= kT . eV0 0259=


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At T K400= , we have

kT . e0 0259 300400

#=

So, the intrinsic concentration at T K400= for Ge is obtained as

ni2 . .

.

.expe

e1 04 10 6 0 10 300400

0 0259 300400

0 6619 183

# # ##

= -bf

lp

. .6 24 10 300400 5 009 1037

39

# # # #= -b l

.7 409 1029#=

Hence, ni . cm8 608 1014 3#= -


ni2 N N ec v kT

Eg

= -

For GaAs (Gallium arsenide), at KT 300c= Nc = effective density of state in conduction band

. cm4 7 1017 3#= -

Nv = effective density of state in valence band cm 3-^ h

. cm7 0 1018 3#= -

^ h

Eg .bandgap energy eV1 42= = kT . eV0 0259=At KT 600= , we have

kT . e0 0259 300600

#=

So, the intrinsic concentration at T K600= for GaAs is obtained as

ni2 . . e4 7 10 7 0 10 300

600.

.

e

e17 183

0 0259300600

1 42# # # #= #

-b l

. .2 632 10 1 24 1037 12# # #= -

.3 2636 1025#=

Hence, ni . cm5 72 1012 3#= -

SOL 1.2.4 Correct answer is 381.Intrinsic carrier concentration is given by

ni2 expN N kT

Ec v

g= -b l ...(1)

For silicon, we obtain

Nc . T2 8 10 300/

193 2

# #= b l

Nv . T1 04 10 300/

193 2

# #= b l

kT . e T0 0259 300#=

Given the intrinsic concentration for silicon,

ni cm1 1012 3#= -

Substituting the values in equation (1), we get

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1012 2^ h . .

.

.expT

e Te2 8 10 1 04 10 300 0 0259 300

1 1219 193

# # # ##

= -bf

lp

or ..exp T0 0259

1 12 300#

#b l . T2 912 300 10

314

# #= b l

Now, we solve the above equation using hit and trial method. Checking for the given options, we get

T K381=


ni2 N N ec v kT

Eg

= -

At constant temperature,

N Nc v = constantTherefore, we have

n Bn A

i

i2

2

^

^

h

h

..

..

exp

exp

N N ee

N N ee

0 02591 2

0 02591 0

c v

c v

=-

-b

b

l

l

. ..exp 0 0259

10 0259

1 2= - -b l; E

..exp 0 0259

0 2= b l

.2257 48=

or n Bn A

i

i

^

^

h

h .47 5=

SOL 1.2.6 Correct answer is 4.9.Given the hole concentration at T K300= as

n0 1015=At T K300= , for silicon, we have

kT . eV0 0259=and Nv .1 04 1019

#=Since, the hole concentration is defined as

p0 N e /v

E E kTF v= - -^ h

So, E EF v- lnkT pNv

0= b l

. .lne0 025910

1 04 1015

19#= c m

. eV0 24=Therefore, we obtain

E Ec F- E E Eg F v= - -^ h

. .1 12 0 24= - . eV0 88=Hence, the concentration of electron is

n0 expN kTE E

cc F= --^ h

; E


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. ..exp e

e2 8 10 0 02590 8819

#= -b l

. cm4 9 104 3#= -

SOL 1.2.7 Correct answer is 2.95.Given the acceptor and donor concentrations,

Na 1013= , N 0d =At thermal equilibrium the magnitude of the positive charge density must equal that of the negative concentration, i.e.

N pd 0+ N na 0= + ...(1)

where Nd = Donor concentration

Na = acceptor concentration

n0 = electron concentration

p0 = hole concentrationSubstituting N 0d = , equation (1) becomes

p0 n Na0= + ...(2)From mass action law, we have

n0 pni

0

2

=

Again, substituting it in equation (2), we obtain

p0 pn Ni

A0

2

= +

or p N p na i02

02- - 0=

Solving the quadratic equation, we get

p0 N N n

24a a i

2 2!= +

.

210 10 4 2 4 1013 13 2 13 2

#=+ +^ ^h h

. cm2 95 1013 3#= -

SOL 1.2.8 Correct answer is 1.125.

Given the doping concentrations,

Nd 2 1015#= and N 0a =

At KT 300= , intrinsic concentration for silicon is

ni .1 5 1010#=

Since, we have

Nd n>> i

Therefore, the electron concentration will be same as donor concentration, i.e.

n0 cmN 2 10d15 3

#= = -

Hence, using mass action law, we obtain

p0 nni

0

2

=

.

. cm2 10

1 5 101 125 1015

10 25 3

#

##= = -^ h

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SOL 1.2.9 Correct answer is 5.66.For silicon, we have the energy bandgap

Eg . eV1 12=At KT 400= , we obtain

kT . .e T e0 0259 300 0 0259 300400

# #= =b l

. eV0 0345=So, the intrinsic concentration at KT 400= is given as

ni2 expN N kT

Ec v

g= -b l

. . ..exp e

e2 8 10 1 04 10 300400

0 03451 1219 19

3

# #= -^ ^ bh h l : D

or ni . cm2 38 1012 3#= -

Now, for thermal equilibrium, we have

N na 0+ N pd 0= +or N na 0+ p0= (given N 0d = )

or N pn

ai

0

2

+ p0= ,mass action law n pni

00

2

=c m

or p N p na i02

02- - 0=

Hence, we obtain the hole concentration as

p0 N N n

24a a i

2 2

= + +

.

210 10 4 2 38 1014 14 2 12 2

#=+ +^ ^h h

. cm1 0 1014 3#= -

Again, using mass action law, we obtain the electron concentration as

n0 pni

0

2

=

.

102 38 10

14

12 2#= ^ h

. cm5 66 1010 3#= -

SOL 1.2.10 Correct answer is 3.09.At T K300= , for GaAs, we have

kT . eV0 0259= ;

Nc .4 7 1017#= ;

Nv 7 1018#=

So, we obtain these parameters for GaAs at T K200= as

kT . eV0 0259 300200= b l . eV0 01727=

Nc .4 7 10 300200 /

173 2

# #= b l

Nv 7 10 300200 /

183 2

# #= b l

Therefore, the intrinsic concentration at T K200= is obtained as

ni2 N N ec v kT

Eg

= -

. e4 7 10 7 10 300200

..17 18

30 01727

1 42# # #= -

b l


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or ni . cm1 38 3= -

Now, from mass action law, we have

n p0 0 ni2=

or p5 02 ni

2= (given, n0 p5 0= )

or p0 n5i=

or p0 . cm0 617 3= -

Hence, the electron concentration is given as

n0 p5 0= . cm3 09 3= -

SOL 1.2.11 Correct answer is 8.04.For silicon, the intrinsic concentration at T 300= K is

ni .1 5 1010#=

Since, boron is acceptor atom and phosphorous is donor atom, so we have the doping concentrations as

Na cm3 1016 3#= -

and Nd cm2 1015 3#= -

So, Na N> d

Therefore, the material is p-type. Hence, the concentration of majority carrier (hole concentration) is given as

p0 N Na d= - (N Na d- n>> i)

3 10 2 1016 15# #= -

. cm2 8 1016 3#= -

So, by using mass action law, we obtain the concentration of minority carriers (electron concentration) as

n0 pni

0

2

=

..2 8 101 5 10

16

10 2

#

#= ^ h

. cm8 04 103 3#= -

SOL 1.2.12 Correct answer is 1.For silicon, the intrinsic concentration at T K300= is

ni .1 5 1015#=

Given the electron concentration,

n0 . cm4 5 104 3#= -

It is very low compared to intrinsic concentration. So, majority (hole) concentration will be high with respect to intrinsic concentration, i.e.

p0 n>> i

Hence, by using mass action law, we obtain

p0 pni

0

2

=

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..4 5 101 5 10

4

10 2

#

#= ^ h

cm5 1015 3#= -

Therefore, we obtain the concentration of acceptor atoms in the semiconductor as

p0 N Na d= -or 5 1015

# N 5 10a15

#= -or Na cm1016 3= -

SOL 1.2.13 Correct answer is 762.Given the donor concentration,

Nd cm3 1015 3#= -

For maximum temperature, we have

ni . n0 05 0= ...(1)So, the electron concentration is given as

n0 N N n

24d d i

2 2

= + +

.N N n

24 0 05d d

20

2

=+ + ^ h

or n N2 d02-^ h .N n0 0025d

202= +

or n N n N4 4d d02 2

0+ - .N n4 0 0025d2

02

#= +or . n n N3 99 4 d0

20- 0=

or .n n N3 99 4 d0 0 -^ h 0=

So, n0 . N3 994

d=

. cm3 0075 1015 3#= -


ni . . cmn0 05 1 504 10014 3

#= = -

Since, the intrinsic concentration is defined as

ni2 expN N kT

Ec v

g= -b l

or .1 504 1014 2#^ h .

.

.expTT4 7 10 7 10 300 0 0259 300

1 4217 183

# # # ##

= -bf

lp

Checking the above equation for given options, we get

T K762-

SOL 1.2.14 Correct answer is 1.2.

At T K300= , for silicon, we have

Nc .2 8 1019#=

and kT . eV0 0259=Given that

E Ec F- . eV0 215=So, the electron concentration is given as

n0 expN kTE E

cc F= --^ h

; E


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. ..exp eV

eV2 8 10 0 02590 21519

#= -b l; E

. cm6 95 1015 3#= -

Hence, we obtain the donor concentration as

n0 N Nd a= -or Nd n Na0= + .6 95 10 5 1015 15

# #= + . cm1 2 1016 3

#= -

SOL 1.2.15 Correct answer is 0.189.For silicon, the intrinsic concentration at T K300= is

ni .1 5 1015#=

and Nv .1 04 1019#=

Given the acceptor concentration,

Na cm7 1015 3#= -

So, Na n>> i

Therefore, the hole concentration is

p0 Na,

p0 7 1015#=

Since, the hole concentration is also given as

p0 expN kTE E

vF v= --^ h

; E

So, we obtain

E EF v- lnkT pNv

0= b l

. .lneV0 02597 10

1 04 1015

19

###= c m

. .ln0 0259 1 49 103#= ^ h

. eV0 189=

SOL 1.2.16 Correct answer is 1.2.In the previous problem, we obtained the difference between Fermi energy level and valence band energy as

E EF v- . eV0 189=Now, Fermi level is moved a distance kT closer to the valence-band edge, so the difference between Fermi energy and valence band energy becomes

E EF v- . eV kT0 189= - . .e e0 189 0 0259= - . eV0 1633=Therefore, the hole concentration is obtained as

p0 expN N kTE E

a cF v= = --^ h

; E

or Na . ..exp e

e1 04 10 0 02590 163319

#= -b l

. cm1 90 1016 3#= -

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Hence, concentration of acceptor impurities that must be added to get the desired value, is

NaT .1 90 10 7 1016 15# #= -

. cm1 2 1016 3#= -

SOL 1.2.17 Correct answer is .0 2877- .For silicon, the intrinsic concentration at KT 300= is

ni . cm1 5 1010 3#= -

Since, boron is an acceptor atom, so we have

Na cm1015 3= -

or Na n>> i

Therefore, the hole concentration is

p0 Na=Hence, we obtain

E EFi F- lnkT np

i

0= b l

lnkT nN

i

a= b l

..

ln0 02591 5 10

1010

15

#= c m

. eV0 2877=or E EF Fi- . eV0 2877=-

SOL 1.2.18 Correct answer is 5.56.The intrinsic concentration is defined as

ni2 A T eo KT

E3 Go

= -

or ni A T e/o KT

E3 2

2Go

= -

So, the conductivity of an intrinsic semiconductor is

is e nn p im m= +^ h

e A T e/n p o KT

E3 2

2Go

m m= + -^ h

Taking logarithm of both sides, we have

log is log loge A T KTE

23

2n p oGom m= + + -^ h$ .

Again, differentiating above expression with respect to T , we get

di

i

ss T dT

KTE dT0 2

32

G2

o= + +

or dT

di

i

ss

b l T KT

E23

2G

2o= + ...(1)

At KT 300c= , we have

KT . eV0 0259=and forbidden energy or bandgap energy of germanium at 0 K is

EGo . eV0 785=So, by substituting these values in equation (1), we get


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dT

di

i

ss

b l

.. . /

K eV KeV K

2 3003

2 0 0259 3000 785 0 0556

# # #c cc= + =

or %dT

d

100i

i

#ss

b l . %5 56= per degree (Kelvin)

i.e, conductivity of intrinsic germanium increase by 5.56% per degree (Kelvin) rise in temperature.

SOL 1.2.19 Correct answer is 0.14.The concentration of donor atom per cm3 is given by

ND /atoms cm2 10

1 5 10822 3

## #= ^ h

. /atoms cm2 5 1014 3#=

Since, the Fermi level coincides with the edge of the conduction band, so we have

E EF C- lnKT NN

0D

C= = b l

or NC ND=or NC . .cm m2 5 10 2 5 1014 3 20 3

# #= =- -

Given that the effective mass is equal to true mass, i.e.

m *n m=

So, we get

NC . T4 82 10 /21 3 2#=

or T /3 2 .

N4 82 10

C21

#=

Hence, T .. . K

4 82 102 5 10 0 14

/

21

20 2 3

##= =d n

This is the temperature at which the Fermi level coincides with the edge of the conduction band.

SOL 1.2.20 Correct answer is 1.The Fermi function, f E^ h specifies the probability of electrons occupying states at a given energy E^ h. The probability that a state is empty (not filled) at a given energy E is equal to f E1 - ^ h. The Fermi function, f E^ h is expressed as

f E^ h exp KT

E E1

1F

=+

-^ h; E

or f EC^ h exp KT

E E1

1C F

=+

-^ h; E

or f E1 V- ^ h exp KT

E E1

1

1V F

= -+

-^ h; E

/

/

exp

exp

E E KT

E E KT

1 V F

V F=

+ -

-

^

^

h

h

8

8

B

B

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/exp E E KT1

1F V

=+ -^ h8 B

...(1)

Given that the Fermi level is positioned at midgap, i.e.

EF E E2

C V= +

So, by substituting it in equation (1), we get

f E1 V- ^ h exp kT

E E E1

21

F F C=

+- -^ h

; E

or f E1 V- ^ h /exp E E KT1

1C F

=+ -^ h6 @

or f EC^ h f E1 V= - ^ h

or f E

f E1 V

C

- ^

^

h

h 1=

Hence, we have

( )( )

probability that a state is empty at the valence band edgeprobability that a state is filled at the conduction band edge

EEV

C 1=

SOL 1.2.21 Correct answer is 0.269.The desired probability is given by the Fermi function,

F E^ h exp KT

E E1

1F

=+

-^ h; E

At, E E KTC= + , we have

f E KTC +^ h /exp E KT E KT1

1C F

=+ + -^ h8 B

...(1)

Given that EF is positioned at EC , i.e.

EF EC=Applying it to equation (1), we get

f E KTC +^ h / ( )

.exp expE KT E KT1

11 1

1 0 269C C

=+ + -

= + =^ h8 B

SOL 1.2.22 Correct answer is 0.0351.The electron population at any energy level is given by

( )n E g E f EC= ^ ^h h

Since, the semiconductor is non degenerate, so we have

f E^ h e

e1

1/

/E E KT

E E KTF

Fb=+ -

- -

^

^

h

h for all E E> C

Therefore, the electron population at E E KT5C= + is

( )n E KT5C + g E KT f E KT5 5C C C= + +^ ^h h

Again, the peak in the electron distribution occurs at /E KT 2C + . So, we have the peak electron population as

( / )n E KT 2C + g E KTf E KT

2 2C C C= + +b bl l

Hence, the required ratio is


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( / )( )

n E KTn E KT

25

C

C

++

g E KT f E KTg E KT f E KT

2 2

5 5

C C C

C C C=+ +

+ +

b b

^ ^

l l

h h

.KTKT

e

e e

2

5 10 0 0351/

/.

E KT E KT

E KT E KT

2

54 5

C F

C F

= = =- + -

- + --

b

^

l

h

SOL 1.2.23 Correct answer is 4.54.Given that the Fermi level is located 0.259 eV above the intrinsic level, i.e.

E EF i- . eV0 259=At KT 300= , we have

KT . eV0 0259=Hence, the hole concentration is

p expn KTE E

iF i= --^ h

; E

..exp eV

eV100 02590 25910= -^ h : D

. per cm4 54 105 3#=

SOL 1.2.24 Correct answer is 4.4.The band gap of silicon at room temperature is

Eg . eV1 12=We have to determine the probability of occupancy of a state at the bottom of the conduction band, i.e.

E EC=Since, the Fermi level is approximately at midgap for intrinsic materials, so we may write

E EC f- . .eV eVE2 2

1 12 0 56gb = =

Now, the probability of electron to occupy any energy band E^ h is

f E^ h e1

1/E E KTF

=+ -^ h

Therefore, the probability of occupancy of a state at bottom of the conduction band E EC=^ h is given by

f EC^ h /exp E E KT1

1C F

=+ -^ h8 B

. / .exp eV eV1 0 56 0 026

1= + 6 @

.1 2 26 10

19

#=

+ .4 4 10 10

#= -

SOL 1.2.25 Correct answer is 2.2.Intrinsic concentration in semiconductor is given by

ni /expN N E KT2C V g= -6 @

Given the bandgap energy,

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EgA eV1= EgB eV2=So, the ratio of intrinsic concentration of semiconductor A to semiconductor B is obtained as

nn

iB

iA /

/

exp

exp

N N E KT

N N E KT

2

2

C V gB

C V gA=

-

-

6

6

@

@

/exp E E KT2gA gB= - -^ h8 B

.exp eVeV

2 0 0261 2

#= - -^ h

; E

.2 2 108#=

SOL 1.2.26 Correct answer is 3.454.By using Plank’s equation, we have

E hn =Hence, the frequency of incident photon is

n .

. .hE

6 62 101 43 1 6 10

34

19

## #= = -

-

. Hz3 454 1014#=

SOL 1.2.27 Correct answer is 1.109.Probability of capture of an energy state by an electron at Ex is 50% means that Ex is actually the Fermi level. The bandgap of Si is 1.1 eV. So, we sketch the energy band diagram as

From the energy band diagram, we observe that the Fermi level is 0.2 eV below the bottom of conduction band edge, i.e.

E EF i- . eVE

0 9 2g= -

. .eV eV0 9 21 1= -

. . .eV eV eV0 9 0 55 0 35= - =Also, the intrinsic concentration of silicon at room temperature is

ni .1 5 1010#=

Hence, the electron concentration is obtained as

no /expn E E KTi F i= -^ h8 B

. ..exp eV

eV1 5 10 0 02590 3510

#= : D

. cm1 109 1016 3#= -

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SOLUTIONS 1.3

SOL 1.3.1 Correct option is (D).As we know that,

vd Em =

So, m Evd=

// sec

V mm=

/m V s2-=

SOL 1.3.2 Correct option is (B).Total energy of a revolving electron in an atom is

E . eVn

13 62=- ^ h

Hence, it can never be positive.

SOL 1.3.3 Correct option is (B).The atomic number of Si is 14.Hence, electronic distribution of an Si atom is 2, 8, 4.

SOL 1.3.4 Correct option is (A).

SOL 1.3.5 Correct option is (D).

SOL 1.3.6 Correct option is (D).Under normal operating conditions, the three primary types of carrier action occurring inside semiconductors are drift, diffusion and recombination-generation.Hence, current flow in a semiconductor depends on all the three effect drift, diffusion and recombination-generation.

SOL 1.3.7 Correct option is (B).The electrical conductivity of intrinsic semiconductor can be increased by adding some impurity in the process of crystallization. The added impurity is very small. Such semiconductor is called impurity or extrinsic semiconductor. The process of adding impurity to a semiconductor is known as doping.

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SOL 1.3.8 Correct option is (A).When a specimen (metal or semiconductor) carrying a current I is placed in a transverse magnetic field B , then an electric field E is induced in the direction perpendicular to both I (current) and B (magnetic field).

SOL 1.3.9 Correct option is (D).Although n -type semiconductor has excess of electrons but it is electrically neutral. This is due to the fact that electrons are created by the addition of neutral pentavalent impurity atoms to the semiconductor there is no addition of either negative chargers or positive charges.

SOL 1.3.10 Correct option is (C).A p-type semiconductor is formed by the addition of trivalent atoms as aluminium gallium, indium, etc. This result in an additional holes in the crystal lattice. The holes move away from the parent atoms and the atoms acquire negative charges. A negatively charged atom is known as acceptor atom and its position is fixed in the crystal lattice. The acceptor ion (immobile ion) can not take part in conduction. Therefore, p-type semiconductor may be regarded as it consists of acceptor ions, i.e. immobile and holes.

SOL 1.3.11 Correct option is (B).By using Fermi-Dirac distribution function,

f E^ h e1

1/E E kTF

=+ -^ h

At Fermi level,

E EF=

f E^ h /e1

1 1 20=+

=

Hence, Fermi level represents the energy level with probability of its occupation of 50%.


SOL 1.3.13 Correct option is (B).


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From the figure, we can deduce that an electron in conduction band has higher energy than electron in the valence band.

SOL 1.3.14 Correct option is (A).Band gap of intrinsic silicon is

Eg . . eVT1 21 3 6 10 4#= - -

^ h

At KT 0= , the bandgap of Si is . eVE 1 21g = . At room temperature KT 300= , the bandgap of Si is . eVE 1 1g = .

SOL 1.3.15 Correct option is (A).In intrinsic Si, the hole mobility is

nm /cm V s500 2-=

. / secm V0 05 2-=

SOL 1.3.16 Correct option is (A).Fermi level in the intrinsic Si or Ge is in the middle of the band gap.

SOL 1.3.17 Correct option is (A).By using Einstein relation,

Dm q

kT=

For electrons, we may write

Dn

n

m qkT=


SOL 1.3.19 Correct option is (D).Acceptor impurity atom in germanium results in new discrete energy level slightly above the valence band, as shown in figure below.

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SOL 1.3.20 Correct option is (A).A compensated semiconductor is one that contains both donor and acceptor impurity atoms in the same region, if N Na d= , we have a completely compensated semiconductor that has the characteristics of an intrinsic material. For N N>D A, then net impurity concentration is

N ND A-

SOL 1.3.21 Correct option is (D).If donor concentration ND equals acceptor concentration NA, the resulting semiconductor is an intrinsic semiconductor.





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