chapter 1 - energy bands and

7/31/2019 Chapter 1 - Energy Bands and

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Dr. Siti Fazlili Abdullah

(office: BA-3-072)[email protected]

You can upload your notes here:http://moodle.uniten.edu.my/moodle/

course/category.php?id=133

mailto:[email protected]

http://moodle.uniten.edu.my/moodle/course/category.php?id=133




mailto:[email protected]



Assignment



Topic 1: ENERGY BANDS AND CARRIER

CONCENTRATION

SUBTOPICS

1. Semiconductor Materials 2. Crystal structures

3. Quantum Mechanics 4. Energy Bands

5. Density of States

6. Intrinsic Carrier Concentration

7. Donors and Acceptors



1. Semiconductor Materials

Def: “Semiconductors are a group of materialshaving conductivities/resistivity between those

of metals and insulators”.

Solid State Materials

As Electrical/Electronic Engineers, we are interested in

electrical/electronic of solid state materials. Electrical/Electronic properties of solids are determined by:

Structural (crystal structure)

Physical (atomic properties, optical properties, etc.)

Chemical (composition of compounds)

Nature of materials as well as by

Environmental conditions (i.e. temperature, radiation, etc.)



Semiconductor Materials

• High σ, conductivity/lowresistivity to current flow, I .

Metals

• Very poor σ. Insulators

• insulator <σ< metal. Semiconductors= (1/Ω-cm)

]



Group

Period

3

2

4

56

A portion of the periodic table showingelements used in semiconductors materials



Compound Semiconductor

III-V GaAs

AlAs

InP

II-VICdS

HgTe

CdTe



States of Matter

• Matter in the universe exists in 3 basicstates: solid, liquid and gas. A fourth state is

plasma.

Properties of Materials

• Material properties are the physical and

chemical characteristics that describe itsunique identity



3 types of solids, classified according to atomic

arrangement

(a) Amorphous (b) polycrystalline (c) Single crystal



2. Crystal Structures: Bravais Lattice

• Collection of points that fill up space.

Lattice • Consists of one or more atoms.Basis• A crystal is produced by attaching a

basis to every lattice point.Crystal

LATTICE BASISCrystal

structure



• Define 3 vectors: a1,a2 & a3 such that any lattice point R ’ can be obtained from any other lattice point R by a translation. m1-3 are integers.

Importantproperty

Crystal Structures: Bravais Lattice

R’ R m1a1 m2a2 m 3a 3



• If the volume formed by a1,a2 & a3 is thesmallest possible.

Primitive

translation vectors

• The basic building block of the crystal.

• A unit cell is a small volume of the crystalthat can be used to reproduce the entirecrystal.

Unit cell

Primitive Cells

R’ R m1a1 m2a2 m 3a 3



Cubic Lattices

SimpleCubic

(sc)

Body -centered

Cubic(bcc)

Face-centered

Cubic(fcc)



Simple Cubic (SC)

Generated by primitive vectors:a1 x, a2 ŷ, a3z

x, ŷ , z = unit vectors a = lattice constant of SC.

a1 a2

a3

ˆ ˆ

ˆ ˆ

ˆ ˆ



Body- centered Cubic (bcc)

Generated from simple

cubic by placing an atomat the center of cube.

a1

a2

a3

d b (f )



Face- centred Cubic (fcc)

Generated from simple

cubic by placing an atomat the center of eachsquare face.



Additional: DIAMOND (Si & Ge) & ZINC BLENDE STRUCTURES (GaAs)



D I A M O

N D• 2 atoms on the

basis identical.• Si, Ge, C etc.

• Elemental SC

Z I N C B L E N D E

• Atoms on basis– different.

• GaAs, AlAs,CdS etc.

• Compound SC.

E l 1



(a) fcc: 8 corner atoms × 1/8 = 1 atom6 face atoms × ½ = 3 atomsTotal of 4 atoms per unit cell

(b) bcc: 8 corner atoms × 1/8 = 1 atom1 enclosed atom = 1 atom

Total of 2 atoms per unit cell

(c) Diamond (Si): 8 corner atoms × 1/8 = 1 atom6 face atoms × ½ = 3 atoms4 enclosed atoms = 4 atoms

Total of 8 atoms per unit cell *

1. Determine the number of atoms per unit cell in(a)face-centered cubic(b)body-centered cubic(c)diamond lattice

Example 1

A body-centeredcubic (bcc)lattice has twopattern per unitcell: the patternsat each vertexare shared

between 8neighbouringcells

Example 2

http://en.wikipedia.org/wiki/File:CCC_crystal_cell_(opaque).svg






1. The lattice constant of GaAs is, a0 = 5.65Ǻ. Determine the number of Ga(fcc) atoms and As (fcc) atoms per cm3.

Example 2

4 Ga atoms per unit cell, density = 4 ⇒ 2.22 x 1022 atoms/cm3

(5.65 x 10-8)3

4 As atoms per unit cell, density = 4 ⇒ 2.22 x 1022 atoms/cm3

(5.65 x 10-8)3

2. Determine the volume density of germanium atoms in a germanium (diamond)semiconductor. The lattice constant of germanium is 5.65Ǻ

8 Ge atoms per unit cell, density = ⇒ 8 ⇒ 4.44 x 1022 atoms/cm3

(5.65 x 10-8)3

3.

E l



Nsi = (5x1022

atoms/cm3

)*(10x10-12

cm3

)= 5x1011 atoms

NGa = N As = (2.22x1022 atoms/cm3)*(104x10-12 cm3)

= 2.22x1014 atoms

* info. fr. example 2 (no.3 & 1)

Example

In semiconductor technology, an Si device on a VLSI chip isone the smallest devices while a GaAs laser is one of thelarger devices. Consider a Si device with dimensions(5x2x1)µm 3 and a GaAs semiconductor laser withdimensions (200x10x5)µm 3. Calculate the number of atomsin each device.



In the (001) surfaces, the top atoms are either Ga or As, leading tothe terminology Ga-terminated (or Ga-stabilized) and As-

terminated (or As-stabilized), respectively. A square of area a2

hasfour atoms on the edges of the square and one atom at the center of the square. The atoms on the square edges are shared by a total of four squares. The total number of atoms per square is

N (a2 ) = 4/4 + 1 = 2

The surface density = σ = 2/ a2 = 2/ (5.65 x 10-8)2 = 6.26x1014cm-2

Example

a)Calculate the surface density of Ga atoms on a Ga-

terminated (001) GaAs surface.

Note: APF = atomic packing fraction (APF) – Final Sem 3 11/12



Note: APF = atomic packing fraction (APF) – Final Sem 3 11/12

In crystallography, atomic packing factor (APF) or packing fraction is thefraction of volume in a crystal structure that is occupied by atoms. It is dimensionless andalways less than unity. For practical purposes, the APF of a crystal structure is determinedby assuming that atoms are rigid spheres. The radius of the spheres is taken to be the

maximal value such that the atoms do not overlap.

1.Fcc/hcp (hexagonal close-packed)Example: fcc lattice of identical atoms with a lattice constant, “a” of 8 x 10 -8 cm,

calculate the maximum atomic packing fraction (APF) and the radius of the atomstreated as hard spheres with the nearest neighbours touching.

Method 1r = √[(a/2)2+(a/2)2]/2= √(a2/4+a2/4)/2 = √2(a/4) = √2(8x10-8cm)/4 = 2.83x10-8 cm

no. of atoms for fcc = 6 x ½ + 8 x 1/83+1 = 4 atoms

Hence APF = 4 x 4/3 πr 3/a3 = 4 x 4/3 π (2.83x10-8 cm)3/(8x10-8cm)3

= 0.74

2. Bcc (APF = 0.68)

3. Sc (APF = 0.52)

3 3 3

33

2

4 2( )

2 4

4 2

4 44 4 23 3

0.74(3)(4)4

2

r a

r a fcc

a r r or a

r r

PF a r

Method 2

a

a

a√2

Miller Indices

http://en.wikipedia.org/wiki/Crystallography

http://en.wikipedia.org/wiki/Crystal_structure

http://en.wikipedia.org/wiki/Atom

http://en.wikipedia.org/wiki/Atom

http://en.wikipedia.org/wiki/Crystal_structure

http://en.wikipedia.org/wiki/Crystallography



Miller Indices In a cubic system, Miller indices of a plane are the same

as the direction perpendicular to the plane.



Example

Calculate the surface density of atoms on a particular plane in a crystal. Consider the

fcc structure and the (110) plane shown. Assume the atoms can be represented as

hard spheres with the closest atoms touching each other and that the lattice

constant is a0 = 4.5Ǻ = 4.5 x 10-8 cm.

σ = 2 atoms/[(a0) (a0 √2)] = 2/ [(4.5 x 10-8 )2 (√2)] = 6.98 x 1014 atoms/cm2

Calculate the surface density of atoms on a particular plane in a crystal. Consider the bcc



y p p y

structure plane of (100) and (110). Assume the atoms can be represented as hard spheres

with the closest atoms touching each other and that the lattice constant is a0 = 4.75Ǻ =

4.75 x 10-8 cm.

Ans: (a) 4.43 x 1014 cm-2, (b) 6.27 x 1014 cm-2

ExampleConsider the (100), (110) and (111) planes in Si.a) Which plane has the highest surface density of atoms?b) What is that density?c) Which plane has the smallest surface density of atoms?

d) What is that density?(100) plane: Density =No atoms/ a0

2 = 2/(5.43 x10-8)2 = 6.78 x1014 cm -2

(110) plane: Density = No atoms/ [a02 √2)]= 4/[(5.43 x10-8)2 √2 ]= 9.59 x1014 cm -2

(111) plane: Density = No atoms/ [a02 (√3/2)]=

[(1/6) ∙ 3 + (1/2) ∙ 3] / [(5.43 x10-8)2 ∙ (√3/2) ]= 7.83 x1014 cm -2

(a), (b) highest surface density: (110) plane and 9.59 x1014 cm -2

(c), (d) lowest surface density: (100) plane and 6.78 x1014 cm -2



Miller Indices



1. Miller Indices is a set of parallel planes in a crystal by three numbers (h k l).These lattice planes can be chosen in different ways in a crystal.

2. Planes:-The crystal may be regarded as made up of an aggregate of a set of parallel equidistantplanes, passing through the lattice points which are known as Lattice planes.

3. Steps to determine miller Indices of given set of parallel planes.“Miller indices may be defined as the reciprocals of the intercepts made by the plane onthe crystallographic axes when reduced to smallest numbers”.

Consider a plane ABC which is one of the planes belonging to the set of parallel planes

with miller indices (h k l).

Let x, y and z be the intercepts made by the plane along the three crystallographic axesX, Y and Z respectively.

a)Determine the coordinates of the intercepts made by the plane along the threecrystallographic axes.

b) Express the intercepts as multiples of the unit cell dimensions, or lattice parametersalong the axes

c) Determine the reciprocals of these numbersd) Reduce them into the smallest set of integral numbers and enclose them in simple

brackets. (No commas to be placed between indices)

z



Direction OA = [2 2]Direction OB = [3 1] Direction OA = [0 1 1]

Direction OB = [1 1 0]

Different Lattice Planes

O

A

B

A

B

O

xy

z



Miller Indices



Miller Indices

1.• Find the intercepts of the plane on the 3 Cartesian

coordinates in terms of the lattice constant.

2.• Take the reciprocals of these numbers and reduce

them to the smallest 3 integers having the same ratio.

3.• Enclose the result in parentheses (hkl) as the Miller

indices for a single plane.

The indices can be obtained via 3 steps:

Exercise 2: If a plane has intercepts at 2a 3a and 4a along the



Exercise 2: If a plane has intercepts at 2a, 3a and 4a along the 3 Cartesian coordinates, where a is the lattice constant, findthe Miller indices of the planes.

1.• Find the intercepts of the plan on the 3 Cartesian coordinates in

terms of the lattice constant: 2a, 3a and 4a

2.

• Take the reciprocals of these numbers and reduce them to the

smallest 3 integers having the same ratio.

3. • Enclose the result in parentheses (hkl) as the Miller indices for asingle plane: The plane is referred to as (643) plane.

1 1 1

, , 12 6,4,32 3 4

Test question:



Using the established steps to obtain the Miller index for a plane in a

semiconductor cubic crystal, draw the corresponding plane designed by

(123). (5 marks)

Step 1: Take Miller indexes (123) Step 2: Invert (123) ->

Step 3: Multiply by 6 -> (632)

Step 4: Multiply by ‘a’, lattice constant (6a, 3a, 2a)

1 1 1

1 2 3

6a

Example 1Describe this lattice direction (Fig 1) and the corresponding lattice plane that is perpendicular ( ) to



Describe this lattice direction (Fig. 1) and the corresponding lattice plane that is perpendicular ( ) tothis direction.

Ans:

The corresponding vector: p = 2, q = 4 and s = 1 or [241] direction or (241) plane.

The intercepts of the plane are then found by taking the reciprocal of (241) andmultiplying by the least common denominator.(1/2, 1/4, 1/1) X 4 → (2, 1, 4) or p = 2, q = 1 and s = 4 or [214] direction or (214) plane.This plane is to the corresponding plane (Fig 2).

Fig. 1 Fig. 2

Example 2



Describe this lattice direction (Fig. 1) and the corresponding lattice plane that isperpendicular () to this direction.

Ans: p = 1, q = 1 and s = 3 or [113] direction or (113) plane.

Corresponding: p = 3, q = 3 and s = 1 or [331] direction or (331) plane.

Fig. 1

Example 3Consider a 3D cubic lattice with a lattice constant = a



Consider a 3D cubic lattice with a lattice constant = a0

a)Find the value of the corresponding p , q and s in the following planes.(i) (100) Ans: p = 1, q = ∞, s = ∞ (ii) (130) Ans: p = 3, q = 1, s = ∞ (iii) (203) Ans: p = 3, q = ∞, s = 2

b)Find the value of p , q and s in the following directions.(i) [110] Ans: p = 1, q = 1, s = 0(ii) [311] Ans: p = 3, q = 1, s = 1(iii) [123] Ans: p = 1, q = 2, s = 3

c)Determine the Miller indices for the plane in Fig.1.

Fig. 11 , 1 , 11 3 1( ) (313)(a)

(b)

1 , 1 , 1

4 2 4( )(121)

Final : Sem 2 08/09



1 1 1

2 3 2

•Draw the crystal plane for the Miller indices (232). [3 marks]

Step 1: Invert (232) -> and multiply by 6 -> (323)

Step 2: Multiply by a, lattice constant (3a, 2a, 3a)

3a

2a

3a

2 useful relationships in Miller indices( bi t l)



(cubic crystal)

a) d = distance between 2 adjacent planes labeled (hkl ):

d = a / (h 2 + k 2 + ℓ 2)½b) The angle between directions [h1k1ℓ1] and [h2k2ℓ2] is given by:

Cos (θ ) = (h 1h 2 + k 1k 2 + ℓ 1ℓ 2 )

[(h 12 + k 12 + ℓ 12 ) (h 22 + k 22 + ℓ 22 ) ]½

Fig.1Final :Sem 1 09/10



Fig. 1 shows direction of the plane (111). What are theother 4 possibilities of the direction of this plane.

[4 marks]

g

Y

Z

X

Solutions

Fig. 1

1 1 1, ,

3 3 3

1 1 1, ,

3 3 3

1 1 1, ,

3 3 3

1 1 1, ,

3 3 3

2 2 2 2 2 2

1 1:

31 1 1

1 1 1 1. 111 , ,

3 3 3 3

anote d

h k l

Test 1 :Sem 1 09/10



1 1, , 02 2

1 1, , 0

2 2

1 1, , 0

2 2

1 1, , 0

2 2

Fig.1

•What are the other 4 possibilities of the direction of the plane (110) as shownin Fig. 1. [4 marks]

ExampleThe lattice constant of a sc cell is a0 = 5 25 x 10-8 cm Calculate the distance



The lattice constant of a sc cell is a0 5.25 x 10 cm. Calculate the distancebetween the nearest parallel (//)

(a) (100) plane. Ans: d = D = a0 = 5.25 x 10-8 cm

(b) (110) plane. Ans: d = (a0 √2)/2 *= 3.71 x 10-8 cm

(a) (111) plane. Ans: d = (a0 √3)/3 = 3.03 x 10-8 cm

d

d

a0

a0

a0√2 * d

3. Quantum Mechanics



ELECTRONS IN A (SOLID) CRYSTAL

3. Quantum Mechanics

De Broglie’s Hypothesis



De Broglie s Hypothesis A particle was given a wave character by assigning a

wavelength to it.

Where p is the particle momentum.

When the wavelength of a particle is small compared to thedistance potential energy changes, classical concepts are quite valid.

When the potential energy changes over distances arecomparable or smaller than the λ , use Quantum Mechanics.

h

p

3. Quantum Mechanics of Free



ParticlePronounce “sai” (Psi)

Shrödinger Equation



g q



ELECTRONS IN AN ATOM



4

o

2 2 2 2o

o

o

-m q -13.6E= = eV

8ε h n n

where

m =free-electronmass

q=electron charge

ε =free-electron mass

h = Planck constant

n = pricipal quantum number

(The lowest energy state=ground state (n=1))

4. ENERGY BANDS



FILLING OF ENERGY BANDS: FERMI-DIRAC DISTRIBUTIONThe distribution fn f(E) means the probability that an allowed level at energy E is



The distribution fn, f(E) means the probability that an allowed level at energy, E isoccupied, where the most one electron can occupy an allowed state (this principle iscalled the Pauli Exclusion Principle).

E F =Fermi Energy= ½ ∙f(E)

METALS, INSULATORS, AND SEMICONDUCTORS



METALS, INSULATORS, AND SEMICONDUCTORS

Metal Insulator/SemiconductorHighest occupied band ispartially filled

Highest occupied band iscompletely filled at T = 0

Conductor Semiconductor Insulator



Conductor Semiconductor Insulator

Ev

Ec

EC

Ev

O v e r l a p

Electrons in a Semiconductor



Ec

Ev

Bond Model



(a) 2D representation of thecovalent bond in a single-crystal Si lattice at T = 0 K,surrounded by 8 valence

electrons.

(b) 2D representation of thebreaking of a covalent bond asthe temp. increases above 0 K, afew valence band e- may gain

enough thermal or opticalenergy to break the covalentbond and jump into theconduction band.

(a) (b)



(c)

(c) Simple line representation of the energy-bandmodel showing the same bond-breaking effect.

5. Density of States (N (E))



1. To determine the current-voltage characteristics of semiconductor devices.

2. Current = Result of the flow of charge.

3. Charge Carriers = charges in a semiconductor thatcan move when forces are applied.a)Electron = negative charge particle

(interested in e- at the bottom of a conduction band)dE = F dx = F v dt

(E = increasing energy, F = applied force, x = distance, v = velocity, t = time)

b)Hole = “positive charge particle”

Electrons and holes



Characteristics of Semiconductor Materials



Effective mass



“m 0 = effective mass(particle mass + masseffect of the internalforces)”



The k -Space DiagramShows the E (energy) vs k (crystal momentum parameter)



Shows the E (energy) vs k (crystal momentum parameter)

2

2

2 2 2 2

2 2 2

V( )2

V ( ) ( )2

o

r E m

r E r m x y z

The energy of an electron in a uniform potential Vo:

2 2

Vo

o

k E

m

346.626 101.055

2 2

h Jswhere Js

Dispersion relation energy –momentum/ E-k relation for electrons

Shrödinger Equation

2 2k



The parabolic E vs k curve for a free electron

Vo

o

k E

m

E =p 2 /2m

Direct and Indirect Semiconductors(a)direct transition with

accompanying photon



Direct bandgap semiconductor = The band structure of some semiconductors that has a minimumin the conduction band and a maximum in the valence band for the same k value (k = 0) (e.gGaAs).Indirect bandgap semiconductor = The band structure of some semiconductors that have itsvalence band maximum at a different value of k than its conduction band minimum (e.g. Si)Thus an electron making a smallest-energy transition from the conduction band to the valence bandin GaAs can do so without a change in k value; on the other hand, a transition from the minimumpoint in the Si conduction band to the maximum point of the valence band requires some change ink. An indirect transition, involving a change in k, requires a change of momentum for the electron.

accompanying photonemission; (i.e: GaAs)

(b)indirect transition via a

defect level. (i.e: Si)

What is the meaning of direct and in-direct semiconductor?

Final :Sem 1 09/10



g[4 marks]

∆k = 0

Direct Semiconductor In-Direct Semiconductor

∆p ≠ 0

EC min

∆k ≠ 0

∆p = 0

EV max

k = crystal momentumparameter

p = momentum of theelectron

EC min = Min. value inconduction band

EV max = Max. valuein valence band

BANDSTRUCTURE NEAR BANDEDGES Behavior of most electrons near the band edges



Behavior of most electrons near the band edgesdetermines most device properties. Near theband edges, the electrons can be described by

simple effective mass pictures, i.e. the electronsbehave as if they are in free space except theirmasses are m*.

(Right) Schematic of the valence band, directband gap, and indirect band gap conductionbands. The conduction band of the direct gapsemiconductor is shown in the solid line, whilethe conduction band of the indirectsemiconductor is shown in the dashed line.The curves I & II in the valence band are calledheavy hole and light hole respectively.

Direct band gap: GaAs, InP, InGaAs etc.

Indirect band gap: Si, Ge, AlAs etc

BANDSTRUCTURE: SILICON



BANDSTRUCTURE: GaAs



The bandgap at 0 K is 1.51 eV and at 300 K it is 1.43 eV. The bottomof the conduction band is at k = (0,0,0), i.e., the G-point. The upper

conduction band valleys are at the L-point.



(a) Band structure of Ge.

(b) Band structure of AlAs.(c) Band structure of InAs.(d) Band structure of InP.

InP is a very important material for

high speed devices as well as asubstrate and barrier layer material for semiconductor lasers.

2 2

*( )

2c

k E k E

m

For conduction band with k=0,2 2

*( )

2v

hh

k E k E

m

For valence band,

Heavy hole band:



2m

m0 = m* is the effective mass.The electron responds to the

external force as if It has this mass.

2 hhm

2 2

*( ) 2vlh

k

E k E m

Light hole band:

3/ 2 1/ 2

2 3

2 * ( )( ) ;C

C

m E E N E E E

* 2 / 3 * *2 1/ 36 ( )

dos l t m m m

3/ 2 1/ 2

2 3

2 * ( )( ) ;v

v

m E E N E E E

* 3 / 2 *3 / 2 *3 / 2

dos hh lhm m m

*3/ 2 1/ 2 *3/ 2 1/ 2

2 3 2 3

*3/ 2 1/ 2

2 3

2 ( ) 2 ( )

( )

2 ( )( )

hh v lh v

dos v

m E E m E E

N E

m E E N E

For Si:

ELECTRONIC PROPERTIES OF SOME SEMICONDUCTORS



Note: D and I stand for direct and indirect gap, respectively. The data are at 300 K.

HOLES IN SEMICONDUCTORS: WHAT ARE HOLES?

In a filled band (valence band) no current can flow. The electrons can “move” if there is an empty state available. The empty states in the valence band are called



p y p yholes.

Diagram illustrating the wavevector of the missing electron k e . The

wavevector of the system with the missing electron is –k e , which is

associated with the hole.

0

i e

i i e

k k

k k k

When all the valence band states are occupied, the

sum over all wavevector states is zero, i.e.

When the electron at wavevector ke is missing,The total wavevector is

i e

i e

k k

k k

HOLES IN SEMICONDUCTORS: HOW DO HOLES MOVE?



The movement of an empty electron state, i,e,. a hole under an electric field.The electrons move in the direction opposite to the electric field so that thehole moves in the direction of the electric field thus behaving as if it werepositively charged, as shown in (a), (b), and (c). (d) The velocities and

currents due to electrons and holes. The current flow is in the same direction( je & jh), even though the electron and holes have opposite velocities. Theelectron effective mass in the valence band is negative, but the hole behavesas if it has a positive mass.



6. FREE CARRIERS IN SEMICONDUCTORS: INTRINSIC CARRIERS



Total mobile carrier density = n + p



FREE CARRIERS IN SEMICONDUCTORS: INTRINSIC CARRIERS



(a) A schematic showing allowed energy bands in electrons in a metal. The electronsoccupying the highest partially occupied band are capable of carrying current.(b) A schematic showing the valence band and conduction band in a typical semiconductor.

In semiconductors only electrons in the conduction band and holes in the valence bandcan carry current.

FREE CARRIERS IN SEMICONDUCTORS: INTRINSIC CARRIERS



np is independent of the position of the Fermi level and is dependent only on the temperature and intrinsicproperties of the semiconductor. This observation is called the Law of mass action: a general description othe equilibrium condition; it defines the equilibrium constant expression. If n increases, p must decrease andvice versa. For intrinsic, n=ni =p=pi,

3

* * 3/ 2

24 ( ) exp( / )

2

Be h g B

k T np m m E k T

In intrinsic semiconductors, the electron concentration is equal to the hole concentration since each electronin the conduction band leaves a hole in the valence band.

* *3ln( / )

2 4

C v

Fi B h e

E E E k T m m

Take the square root ofthe equatn ***

***



Effective densities and intrinsic carrier concentrations of

Si, Ge and GaAs. The numbers for intrinsic carrier densities are the accepted values even though they aresmaller than the values obtained by using the equationsderived in the text.

7. DOPING OF SEMICONDUCTORS: DONORS AND ACCEPTORS (EXTRINSIC CARRIERS)



DOPING OF SEMICONDUCTORS: DONORS AND ACCEPTORS



(a) The energy-band diagram showingthe discrete donor energy states(T = 0 K)

(b) The energy-band diagram showingthe effect of some donor states beingionized creating free electrons (T > 0 K)

(b) The energy-band diagram showingthe effect of some acceptor states beingionized creating free holes (T > 0 K)

(a) The energy-band diagram showingthe discrete acceptor energy states(T = 0 K)



Schematic energy band representation of extrinsic

semiconductors with:(a) donor ions and (b) acceptor ions.

FREE CARRIERS IN DOPED SEMICONDUCTORS

If electron (hole) density is measured as a function of temperature in a doped S/C, one observes three regimes:



Electron density as a function of temperature for a Si sample with donor impurity concentrationof 1015 cm –3. It is preferable to operate devices in the saturation region where the free carrier density is approximately equal to the dopant density.

It is not possible to operate devices in the intrinsicregime, since the devices always have a highcarrier density that cannot be controlled by E.Every semiconductor has an upper T beyondwhich it cannot be used in devices.The larger the bandgap, the higher the upper limit.

Recall: 1.3: Quantum Mechanics a) Electrons in Free Space Electrons inside semiconductor can be regarded as „free‟ electrons under proper

conditions, uniform potential, V 0.



Energy of the electron:

E = ħ2k 2/2m 0 +V 0 = p* 2/2m 0 +V 0 *p = momentum

Energy-momentum or E-k or dispersion relation(Means, the allowed energies for the free e- from continuous “band”)

Shrödinger Equation:

1.5 Density of States (density of allowed solutions) = N (E)(is the number of available electronic states per unit volume per unit energy around an

energy E )

&N (E) = √2 m0

3/2 (E – V 0 ) ½

2 ħ3

compulsoryE = ħ2k 2/2m 0*+V 0



& 2 ħ3

1. E vs k (ħk is the momentum ) 2. N (E) vs E

Example 1The free electrons moving in the constant background potential of 2.0 eV.a) Find the N(E).b) Plot and describe the N(E) vs E diagram.c) Plot the N(E) for electrons that have E-k relation.



c) Plot the N(E) for electrons that have E k relation.

a)

N (E) = √2 m0 3/2

(E – V 0 ) ½

2 ħ3 3/2 1/2 30 3/2 1/2

0 0 0

2 3 2 34 3

56 1/2 1 3

0

1 1/ 219 19 3

56 1/2

0 1 3 6 3

21

2( ) ( ) 2(0.91 10 ) ( )) ( )

(1.05 10 )

1.07 10 ( ) .

1 1.6 10 1.6 10 11.07 10 ( )

. 1.0 1.0 10

6.8 10 (

m E V kg E V i N E

Js

E V J m

J J m E V

J m eV eV cm

E

1/ 2 1 3

0

21 1/ 2 1 3

) .

6.8 10 ( 2)) .

V eV cm

E eV cm

N (E) vs E

N(E) (eV -1.cm -3 ) x 10 21



E (eV)

0.0

6.0

4.0

2.0

0

4.02.0

N(E) = 0 when E < 2.0

C) Plot the N(E) for electrons that have E-k relation.

E = ħ2k 2/2m 0 + 2.0 eV



note: N(E) = 6.8 x 10 21 (E - 2.0 ) ½ eV -1 . cm -3 , E > 2.0 eV

E (eV) E = ħ2k 2/2m 0

k

Backgroundpotential = V0= 2.0 eV

Example 2Calculate the N(E) for a particle that has an E-k relation given by

E = - ħ2k 2/2m 0 with zero potentialAns:In this problem the allowed energies are present only for E < 0. For E > 0 no



p g p yenergies are allowed because the negative sign. The N(E) is

N (E) = √2 m0 3/2 (-E ) ½ , E < 0

2 ħ3

Example 3Calculate the N(E) at 0.1 eV electrons having a dispersion relationwith zero potential.

E = ħ2k 2/2m * ; m * = 0.1 m 0

Ans: (fr. Example 1)

N(E) = 6.8 x 10 21 x (0.1 ) 3/2 (0.1) ½

= 6.8 x 10 19 eV -1 . cm -3

N (E) = √2 m0 3/2 (E – V 0 )

½

2 ħ3

Example 4In a particular periodic potential electrons are found to have potential background of-5 eV, given the dispersion relation of;

E = ħ2k 2/2m * + (- 5.0) eV ; m * = 0.1 m 0



( ) ; 0

(note: the normal one is E = ħ2k 2/2m * + 5.0 eV,

Plot the N(E) vs E graph for the electrons between -6.0 eV and -8.0 eV

Ans: Refer example 1

N(E) = 6.8 x 10 21 eV -1 . cm -3 (E - (- 5) ) ½ eV -1. cm -3

N(E) = 6.8 x 10 21 eV -1 . cm -3 (E + 5) ) ½ eV -1. cm -3

(Means: E > - 5.0 eV = -4, -3, - 2,……)

Conclusion: There are no allowed states. Can’t plot N(E) vs E graph

N (E) = √2 m0

3/2 ((E –V 0 )

½ ) 2 ħ3

Example 5 (JFYI)Calculate the N(E) per unit volume with energies between 0 to 1.0 eV. TheN(E) per unit volume for free electron is given by

N (E) = 4 (2m )3/2 (E )½ )



Ans:

n = N(E) dE

n = 4 (2m 0 ) 3/2 (E ) ½ dE

h 3

n = 4 (2 x 9.1 x 10 -31 ) 3/2 2/3 E

(6.625 x 10 -34 ) 3

n = 1.06 x 10-56 . 2/3 (1 – 0) eV. x 1.60 x 10-19 J

n = 4.52 x 1027 m-3

N (E) = 4 (2m 0 ) 3/ (E ) ½

h 3

1

0

)

)1

0

3/2 1

0) [ ]

[1.0 eV

]3/2

Example 7(Continue from example 3)Plot the N(E) for electrons that have E-k relation, given by

a) E = - ħ2k2/2m0 with zero potential



a) E ħ k /2m 0 with zero potentialb) E = ħ2k 2/2m * + (- 5.0) eV

E (eV)

E = -ħ2k 2/2m 0

k

(a)

E (eV)

k

- 5.0 eV (b)

E = ħ2k 2/2m 0

b) Electrons in an Atom Free States : These states represent solutions where electron isnot confined near the atom and has so much kinetic energy thatthe attractive potential of the atom cannot hold it. All energies areallowed in this regime.



g

Bound states : These are of greatest interest in atomic physics.Not all energies are allowed for the bound states, where theelectron wavefunction is confined to 1 Å around the nucleus.Only certain discrete energies are allowed. The allowed energiesare separated by regions that are forbidden for the electron.

E n = - m 0 e 4 = -13.6 / n 2 (eV)

Where, n = 1, 2, 3…

2(4 0 ) 2 ħ2 n 2

c) Electrons in Crystalline Solids: Energy Bands

Within each allowed band the electron behaves as it were in free space, except it responds as it had a different mass known as effective mass, m*.



, How do the electrons distribute themselves among the various

allowed electronic states? Electrons have the property that at the most one electron can occupy an allowed states (“Pauli exclusion principle”) The distribution function, f (E) = Fermi-Dirac distribution function =

T k

E E E f

B

F exp1

1



For the case when (E – E F

>> k B

T )

T k

E E E f

B

f exp Maxwell – Boltzmann distribution

The electron densityor carrier density,

T k

E E N n

B

f c

0exp

Where N C = effective density of statesE 0 = the initial energy band/bandedge

N C = 2 (m0 /(2 ħ2))3/2(k B T) 3/2

Fermi Energy (E F ) (for semiconductor and T > 0)

1.Boltzmann approximation



E F - E 0 = k B T ln (n / N c )

2.Joyce-Dixon approximation

E F - E 0 = k B T [ln (n / N c ) + (1/√8) (n / N c)]

3. E F - E 0 = ( ħ2 /2 m0 ) (3 2 n)2/3 (T = 0 K)

Example 1Calculate the fermi level (E F ) at 77 K and 300 K for a case where theelectron density, n is 1019 cm-3. Use the Boltzmann and the Joyce-Dixonapproximation. Assume E 0 = Initial Energy = 0

A



Ans:

N c (300 K) = 2 (m0 /(2 ħ2

))3/2

(k B T) 3/2

= 2 (0.91 x 10 -30 kg / (2 (1.05 x 10 -34 Js) 2 ) 3/2

. [(1.380 x 10-23 J/K) x (300K)] 3/2

= (9.522 x 1055) . (2.66 x 10-31) = 2.53 x 1025 m-3

N c (77 K) = 2.53 x 1025 cm-3 x 77 3/2

300= 3.3 x 1024 cm-3

)

At 77 K, E F (k B T = 0.0067eV = [(1.380 x 10 -23

J/K) (77K) x (1 eV/(1.6 x 10 -19

J))] E F (Boltzmann) = 0.0067 ln 10 19 = -85 meV

E F (Joyce-Dixon) = 0.0067 ln 10 19 + 10 19

3.3 x 1024 ][ )



F ( y )√8( = -85 meV

At 300 K, E F (k B T = 0.026eV)

E F (Boltzmann) = 0.026 ln 10 19 = - 383 meV

E F (Joyce-Dixon) = 0.026 ln 10 19

+ 10 19

√8( ) = -383 meV

[ ) 2.53 x 1025

]3.3 x 1024)

[

2.53 x 1025

)

3.3 x 1024

)

]

[ ]2.53 x 1025

Example 2If the mass of the electron in the example 1 changes to 0.1 m0, calculate theN c.

Ans:N c (300 K) = 2.53 x 1019 (m* / m0)3/2 = 8.00 x 1017 cm-3

N c (77 K) = 3.3 x 1018 (m* / m0)3/2 = 1.04 x 1017 cm-3 note: m*= 0.1m0

Example 3In GaAs, electrons behave as if mass is m* = 0.067 m0. If 1018 electrons percm3 are placed in the conduction band of GaAs, what is the position of Fermilevel at 0 K? Using the Joyce-Dixon approximation, calculate the position ofthe Fermi level at 300 K.



Ans:The position of the Fermi level at 0 K is given in terms of the carrier densityby

E F = E 0 + ( ħ2 /2 m* ) (3 2 n)2/3

=(1.05 x 10-34 Js)2 x 3 x 2 (1024 m-3)2/3

2(0.067 x 0.91 x 10-30

kg)= 8.3 x 10-21 J = 5.18 x 10-2 eV

Using the Joyce-Dixon approximation withNC = 4.45 x 1017 cm-3 (from the table)

We find EF (300K) = 0.04 meV

Recall on the semiconductor:1. At 0 K, the valence band is completely occupied (ie, filled with electron),

while the conduction band is completely empty. How does the currentflow in semiconductor?

2 Every semiconductor has its own unique bandstructure (bandedge) (We



2.Every semiconductor has its own unique bandstructure (bandedge). (We

only interested in what E-k relation is near the top of the valence bandand near the bottom of the conduction band.

3. For conduction bandedge is at k= 0

E (k) = E C + ( ħ2 k 2 /2 m* )

WhereE C = the conduction bandedgem* = effective mass

4. For valence band

a) Heavy hole band E = E v – ( ħ2 k 2 /2 m hh * )

b) Light hole band E = E v – ( ħ2 k 2 /2 m lh * )

Example 4Find the k-value for an electron in the conduction band of GaAs having anenergy of 0.1 eV (measured from bandedge). The m* of electrons in GaAs is0.067 m 0 .



E (k) = E C + ( ħ2

k 2

/2 m* )

k = (√2m* E) / ħ

k = 2 (0.067 x 9.1 x 10-31 kg) (0.1 x 1.6 x 10-19 J)1/2 = 4.2 x 108 m-1

1.05 x 10-34

Js

Fermi Energy (E F )

1. Boltzmann approximation

E F = k B T ln (n / N c )

Where,

N c = effective density ofstates

= 2 (m /(2 ħ2))3/2(k T) 3/2



2. Joyce-Dixon approximation

E F = E 0 + k B T [ln (n / N c ) + (1/√8) (n / N c)]

3. E F = E 0 + ( ħ2 /2 m0 ) (3 2 n)2/3 (for metal and T = 0 K)

4.The ”Donor Energy”

E d = Ec - ( e4 / m*e ) / 2(4 )2ħ2 =

= 2 (m0 /(2 ħ2))3/2(k B T) 3/2

n = electron density = (carrier density) = N c exp (E F – E 0 )/ (k B T)

Example 6A donor atom (i.e indirect gap materials like Si)in a semiconductor has a donor energy of0.045 eV below the conduction band.Assuming the simple hydrogenic model for



Assuming the simple hydrogenic model for

donors, estimate the conduction bandedgemass. (Given: static dielectric constant forSi Si = S = 11.9 0)

Ans:

Ed

=

Ed – Ec = = - 0.045

= 0.045 (11.9 0 / 0 )2 = 0.47

13.6

Ed – Ec

Example 7Calculate the donor level energy (gap) for InAsGiven: static dielectric constant for InAs

= InAs = S = 15.15 0 : effective mass for InAs = m* = 0 027m



: effective mass for InAs = m = 0.027m0

Ans:

Ed – Ec =

= -13.6 (0.027) eV = -1.6 meV(15.15)2

Note: The donor level is very close to the conduction bandedge due to smalleffective mass of the electrons.

Energy-band diagrams for degenerately doped



(a) n-type (The Fermi energy level is inthe conduction band)

(b) p-type (The Fermi energy level is inthe valence band)

Energy-band diagrams for complete ionization

(a) Donor states (b) Acceptor states

Energy-band diagrams for position of the Fermi level at T = 0 K



(a) n-type semiconductors (b) p-type semiconductors

1. The intrinsic carrier concentration, n i is a very strong function of temperature. As thetemperature increases, additional electron-hole pairs are thermally generated so that n i

2 termmay begin to dominate.

2. The semiconductor will eventually lose its extrinsic characteristics.3. Partial ionization or freeze-out at low temperature.



Figure shows the electron concentration vs temperature in Si doped with 5 x 1014 donorsper cm3 (n-type)



(a) Energy-band diagrams showing position ofFermi level relative to intrinsic Fermi levelfor n-type (N d > N a ) semiconductor

(b) Energy-band diagrams showing position ofFermi level relative to intrinsic Fermi levelfor p-type (N a > N d ) semiconductor

Position of Fermi level as a function of donor concentration (n-type) andacceptor concentration (p-type)

Test Question

For GaAs, the ratio m*/m0 = mh*/me = 0.52 at temp,T = 300K, kBT = 0.026 eV, kB = 8.6 x 10-5 eV/K



a) Determine the position of the Fermi level of GaAsbelow EC if its band gap is 1.42 eV (In terms of EV)

b) Draw the positions of EC, EFi and EV.

c) How big is the shift of the Fermi level if the

temperature is 450K

Use : EFi

= ½(EV

+ EC

) + ¾k

BT ln (m

h*/m

e)

Ans:a) EC + EV = (EC - EV ) + 2 EV Recall: Eg = EC - EV = Eg + 2EV

EFi = ½(EV + EC ) + ¾ kBT ln (mh*/me)



EFi

= ½(Eg

+ 2EV

) + ¾kBT ln (m

h*/m

e)

= 1.42 + EV + ¾ (0.026) ln (0.52) = 0.71 + EV + (-0.0128)2

= EV + 0.6972 eV

b)

c) kBT = (8.6 x 10-5 )(450) = 0.0388 eV

EFi = 1.42 + EV + ¾ (0.0388) ln (0.52) = 0.6910 + EV2

∆E = 0.6972 – 0.6910 = 0.0062 eV

EC

EFi

EV

1.42 eV

0.6972 eV



Energy-band diagrams showing the redistribution of electrons whendonors are added

Carriers in doped semiconductors:

•n = total free electrons in the conduction band

•nd = electrons bound to the donors



•p = total free electrons in the valence band•pa = holes bound to the acceptors

• The fraction of electrons tied to the donor levels in a n-doped materialwith doping density Nd :

• The fraction of electron tied to the acceptors levels in a p-dopedmaterial with doping density N

a:

nd = 1 + (Nc / 2Nd) exp [- (Ec- Ed) / (k BT) ]-1 (n + nd )

pa = 1 + (Nv / 4Na) exp [- (Ea- Ev) / (k BT) ]-1 (p + pa )

Quiz QuestionThe fraction of total electrons still in the donor states at temperatureT = 450 K is given by

nd = 1 + (Nc / 2Nd) exp [- (Ec- Ed) / (k BT) ]-1 (n + nd )



Consider As doping in Si at a concentration, Nd = 1016cm-3 withNc = 2.8 x 1019cm-3 and Ed for As is 0.054 eV below Ec.

Ans:nd = 1 + Nc exp [- (Ec- Ed)] -1

(n + nd ) 2Nd k BT

= 1 + (2.8 x 1019 ) exp [ - 0.054] -1

2(1016) (8.6 x 10-5) (450)

= 1 + 1400 exp (- 1.3953) -1

= 2.8747 x 10 -3 0.3 % (Means 99.7 % electrons have transferred toconduction band)

Note:Ec- Ed = Eg = 0.054 eV

EC

Ed

EV

0.054 eV

Quiz /Test QuestionsDetermine the % of fraction of total electrons still in the donor states atT = 300 K (around room temperature).Assume Si is doped with phosphorus to a concentration of Nd = 5 x 1015 cm-3, Nc = 2.8x 1019cm-3, k BT = 0.0259.Ec- Ed = The ionization energy of the donor electrons (Impurity (donor = phosphorus)with Si) = 0.045 eV.



with Si) = 0.045 eV.

Use nd = 1 + (Nc / 2Nd) exp [- (Ec- Ed) / (k BT) ]-1 (n + nd )

Ans: 0.203%

Comment:

1. Shows that vast/majority of the donor electrons are in the conduction band.2. Only ~ 0.2 % of the donor electrons are still in the donor states.3. At room temp., we can say that the donor states are completely ionized.4. nd = The density of electrons occupying the donor level.5. Nd = The concentration of ionized donors

Quiz/Test Questions

Determine the phosphorus doping concentration in Si at 300K such that 1 % ofthe donor electrons are still in the donor states.Ans: Nd = 2.49 x 1016 cm-3

ExampleCalculate the density of electrons, n in a Si conduction band if theFermi level is 0.2 eV below the conduction band at 300K. Compare theresults by using the Boltzmann and the Joyce-Dixon approximation.Given: The effective density of the conduction band states in Si at 300K

is NC = 2 78 x 1019 cm-3



is NC = 2.78 x 10 cm

kBT (Si) = 0.026 eVAns:

Boltzmann approximationn = NC exp - (Ec- EF)

k BT

= (2.78 x 1019) exp - 0.20.026

= 1.27 x 1016 cm-3

Joyce-Dixon approximationE F - E C = k B T [ln (n / N c ) + (1/√8) (n / N c)]

- 0.2 = 0.026 ln n + 1 n n = 1.27 x 1016 cm-3 NC √8 NC

EC

EF = 0.2

EV

Example (Final Sem 1 09/10)

Given that the ionization energy of an incorporated particular impurity is 0.206 eVaway from the minimum of the conduction band of silicon.

a) What can be said about the type of this impurity. [1 mark]



b) Why do we need the impurities in the semiconductor devices? [3 marks]

c) What are the majority and minority carriers upon ionization of the impurity. [2 marks]

d) The resulting Fermi level is shifted by 0.356 eV away from its originalintrinsic location. Calculate the impurity concentration (i.e the majorityand minority carriers density). [4 marks]

e) Calculate the impurity intrinsic Fermi level for the majority and minoritycarriers (in terms of Ei) [2 Marks]

a) n-type (Donor-type) impurity.b) The addition of impurities to a semiconductor devices, known as

doping, has the effect of shifting the Fermi level within the

material. This results in a material with predominantly negative

(n type) or positive (p type) charge carriers depending on the

http://en.wikipedia.org/wiki/Semiconductor

http://en.wikipedia.org/wiki/Doping_(semiconductor)

http://en.wikipedia.org/wiki/Fermi_level

http://en.wikipedia.org/wiki/N-type_semiconductor

http://en.wikipedia.org/wiki/P-type_semiconductor

http://en.wikipedia.org/wiki/Charge_carrier















http://en.wikipedia.org/wiki/Semiconductor



10 16 30.356

exp (1.5 10 )exp 1.32 100.026

F ii

B

E E

n n x x cmk T

2 10 24 3

16

(1.5 10 )1.7 10

1.32 10

in x p x cm

n x

(n type) or positive (p type) charge carriers depending on the

dopant species.

c) Majority carriers = Electrons,

Minority carriers = Holes

d)

e)

16

10

1.32 10ln (0.026) ln 0.357

1.5 10Fn i B i i

i

n x E E k T E E eV

n x

4

10

1.7 10ln (0.026) ln 0.357

1.5 10Fp i B i i

i

p x E E k T E E eV

n x

Note: E F - E 0 = k B T ln (n / N c )

Figure 1: Periodic Table

Example (Final Sem 2 08/09)















A Silicon ingot is doped with 1016 Aluminium (p-type) atoms/cm3. Let T=300 K.(i) Find the majority and minority carrier concentrations.

State your assumption. [7 marks]

Assume complete ionization. (1) Majority carrier concentration/hole concentration,

p = Na =1016 cm-3 (2)

Minority carrier concentration/electron concentration,n = ni

2/p (2)

= (1.5x1010)2/1016 cm-3 (1)= 2.25 x104 cm-3 (1)

(ii) Find the positions of the Fermi level with reference to the top of the

valence band, using the Joyce-Dixon approximation.[3 marks]

16 16

1ln (1)

8F v B

v v

p p E E k T

N N



(iii) Find the positions of the Fermi level with reference to the intrinsicFermi level. [3 marks]

16 16

18 1810 1 100.025852 ln (1)

9.84 10 9.84 108

0.179 (1)

F v

F v

E E

E E eV

( ) /

16

10

( ) ln (1)

10( ) 0.025852 ln (1)

1.5 10

( ) 0.3487 (1)

F F i B E E k T

i

F F i B

i

F F i

F F i

pe

n

p E E k T

n

E E

E E eV

(iv) Draw the flat band diagram, indicating clearly the positionsof Ec, Ev, EF and EFi. [4 marks]



0.3487 eV

0.179 eV

Example (Test 1 Sem 1 09/10) [20 marks]

(a) Where is EF located (sketch the energy-band diagram) in the energy band of silicon at 300K with: [7 marks]

i) n = 1017 cm-3 from Ec17

19

10ln (0.026) ln 0.146

C F B

n E E k T eV



ii) p = 1014 cm-3 from Ev [7 marks]

Ec

EF

Ev

0.146 eV

2.78 10C N x (0.146)F C E E eV

14

18

10ln (0.026) ln 0.2989

9.84 10v F B

v

p E E k T eV

N x

(0.2989)F v E E eV

Ec

EF

Ev

0.2989 eV

19 16 30.2exp (2.78 10 )exp 1.27 10

C F E E

n N x x cm

ExampleCalculate the density of electrons, n in a Si conduction band if the Fermi level is 0.2eV below the conduction band at 300 K. [4 marks]

Revision



exp (2.78 10 )exp 1.27 100.025852C

B

n N x x cmk T

2 10 2

4 316

(1.5 10 ) 1.77 101.27 10

in x p x cmn x

Then, determine the density of holes, p in a Si conduction band. [2 marks]

2 2 3/ 2 1/ 2

1a). Plot the Energy - momentum, E-k relation. (2 marks)b). Plot the Density of states – Energy, N(E) – E relation for the electron

between -3.0 eV and -5.0 eV. (2 marks)c). Plot the Density of states – Energy, N(E) – E relation for the electron

between -1.0 eV and 1.0 eV. (2+1 marks)

Revision (almost like Final Sem 1 1213)



2 2

0 / 2 2.0 E k m eV 3/ 2 1/ 2

0 0

2 32 ( )( ) m E V N E

b) Can’t plot since:

c) Can plot since:

-2.0 eV

E(eV)

k

Given: ,

a)

N(E)

E(eV) -2.0

Allowed state, when E > -2 (-1, 0, 1, 2, 3, 4, …) Allowed state, when E > -2 (-1, 0, 1, 2, 3, 4, …)

d

d

n

n n

FINAL SEM 1 2010/2011 + Quiza) Determine the doping concentration of phosphorus (electrons) in Si at 300 Kwhere the percentage of is equal to 5.0 %.

Consider phosphorus doping in Si with Ed is 0.045 eV below Ec. [3 marks]

Egap Si = 1.124 eV



1

119

119

17 3

( )1 exp

2

2.78 10 0.0450.05 1 exp2 0.025852

2.78 100.05 1 0.175

2

1.2832 10

d C C d

d d B

d

d

d

n N E E

n n N k T

x N

x

N

N x cm

2 10 23

17

(1.5 10 )1753

1.2832 10

in x

p cmn x

b). Determine the doping concentration of phosphorus (holes) in Si at 300 K. [2 marks]



c). What is the percentage of the dopants (phosphorus) in the conduction level. [1 mark] (100 – 5)% = 95 %

d). Sketch the flat energy band diagram and label the values of the gap between Ec – Ed, andEd - EFi. [3 marks]

EFi

1.124 /2 = 0.562 eV

0.045 eV

0.517 eV

EC

EFn = Ed

SEM II 2010/2011 (10 marks)

As your first assignment as an electronicsengineer, you are to design a laser byrecommending a good material. You have topresent the following to your superior, with justifications, diagrams, calculations and



background information on the material chosen.

*If your answers is GaAs.(a)Crystal structure [3 marks]GaAs (1), Zinc blende (1), tetrahedral structures (1)

(b) Atom density, taking into account that thelattice constant is 5.65 Å. [4 marks]N(a3) = 8/8 +6/2 = 4 (2) Ga atom density = As atom density= 4/a3 = 4/(5.65x10-8)3=5x1022 atoms/cm3 (2)

(c) Band structure [3 marks]Graph on the left (1) direct bandgap (1)efficient light emission (1)

Sign for different E level

EC

EFn= Ed E = EFn – EC = -kBT ln (n/ni)(-)



EFi

EV

EFp= Ea

E = Egap= EC - EV

E = EV –

EFp = -kBT ln (p/ni)

E = EFn –

EFi = kBT ln (n/ni)

E = EFi - EFp = kBT ln (p/ni)

E = Enew gap = Vbi

(+)

(-)

(+)

EC

EFn= Ed EFn – EC = - 0.223 eV

FYI on the dopant in Si – an example

(-)

(+)



EFi

EV

EFp= Ea

Egap= 1.124 eV

EV –

EFp = - 0.041 eV

EFn –

EFi = (1.124/2)-0.223=0.339 eV

EFi - EFp = (1.124/2)-0.041=0.521 eV

Enew gap= Vbi = 0.339+0.521= 0.860 eV

(+)

(-)

chapter 1 - energy bands and

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