Important Chap3 1 SemiImportant Chap3-1-Semiconductor-1.pdf conductor 1

Chapter 3: semiconductor science and light emitting diodes

Of the 18 atoms shown in the figure, only 8 belong to the volume ao3.

Because the 8 corner atoms are each shared by 8 cubes, they contribute a total of 1 atom; the 6 face atoms are each shared by 2 cubes and thus contribute 3 atoms, and there are 4 atoms inside the cube. The atomic density is therefore 8/ao

3, which corresponds to 17.7, 5.00, and 4.43 X 1022 cm-3, respectively.

Semiconductor Lattice Structures

Diamond Lattices

The diamond-crystal lattice characterized byfour covalently bonded atoms.

The lattice constant, denoted by ao, is 0.356, 0.543 and 0.565 nm for diamond, silicon, and germanium, respectively.

Nearest neighbors are spaced units apart.( )4/3 oa

(After W. Shockley: Electrons and Holes in Semiconductors, Van Nostrand, Princeton, N.J., 1950.)

Semiconductor Lattice Structures

Diamond and Zincblende Lattices

Diamond latticeSi, Ge

Zincblende latticeGaAs, InP, ZnSe

Diamond lattice can be though of as an FCC structures with an extra atoms placed at a/4+b/4+c/4 from each of the FCC atoms

The Zincblende lattice consist of a face centered cubic Bravais point lattice which contains two different atoms per lattice point. The distance between the two atoms equals one quarter of the body diagonal of the cube.

How Many Silicon Atoms per cm-3?• Number of atoms in a unit cell:

• 4 atoms completely inside cell• Each of the 8 atoms on corners are shared among cells

count as 1 atom inside cell• Each of the 6 atoms on the faces are shared among 2

cells count as 3 atoms inside cell⇒ Total number inside the cell = 4 + 1 + 3 = 8

• Cell volume:(.543 nm)3 = 1.6 x 10-22 cm3

• Density of silicon atoms = (8 atoms) / (cell volume) = 5 x 1022 atoms/cm3

“diamond cubic” lattice

The Si Crystal

• Each Si atom has 4 nearest neighbors

• lattice constant= 5.431Å

Semiconductor Materials for Optoelectronic Devices

400~450 450~470 470~557 557~567 567~572 572~585 585~605 605~630 630~700 Pure Blue Blue Pure Green Green Yellow Green Yellow Amber Orange Red

Semiconductor Optical Sources

0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6

Blue

Gre

en

Ora

nge

Yello

w

Red

λ1.7

I n f r ar ed (um)Viol

et

GaA

s

GaA

s 0.55

P 0.45

GaAs1-yPy

InP

In0.

14G

a 0.8

6As

In1-xGaxAs1-yPyAlxGa1-xAs

x = 0.43

GaP

(N)

GaS

b

InG

aNS

iC(A

l )

In0.

7Ga 0.

3As 0.

66P 0.

34

In0.

57G

a 0.43

As 0.

95P 0.

05

In0.49AlxGa0.51-xP

Semiconductor Materials for Optoelectronic Devices

Quantization Concept

plank constant

Core electrons

Valence electrons

Periodic Table of the Elements

Group**

Period 1 IA 1A

18

VIIIA8A

1 1 H

1.008

2 IIA 2A

13

IIIA3A

14 IVA4A

15 VA5A

16 VIA6A

17 VIIA7A

2 He4.003

2 3

Li 6.941

4 Be 9.012

5 B

10.81

6 C

12.01

7 N

14.01

8 O

16.00

9 F

19.00

10 Ne20.18

8 9 10 3 11

Na 22.99

12 Mg 24.31

3 IIIB3B

4 IVB4B

5 VB5B

6 VIB6B

7 VIIB7B ------- VIII -------

------- 8 -------

11 IB 1B

12 IIB2B

13 Al26.98

14 Si

28.09

15 P

30.97

16 S

32.07

17 Cl

35.45

18 Ar39.95

4 19 K

39.10

20 Ca 40.08

21 Sc44.96

22 Ti

47.88

23 V

50.94

24 Cr52.00

25 Mn54.94

26 Fe55.85

27 Co58.47

28 Ni58.69

29 Cu63.55

30 Zn65.39

31 Ga69.72

32 Ge72.59

33 As74.92

34 Se78.96

35 Br79.90

36 Kr83.80

5 37

Rb 85.47

38 Sr

87.62

39 Y

88.91

40 Zr

91.22

41 Nb92.91

42 Mo95.94

43 Tc(98)

44 Ru101.1

45 Rh102.9

46 Pd106.4

47 Ag107.9

48 Cd112.4

49 In

114.8

50 Sn118.7

51 Sb121.8

52 Te127.6

53 I

126.9

54 Xe131.3

6 55 Cs 132.9

56 Ba 137.3

57 La*138.9

72 Hf178.5

73 Ta180.9

74 W

183.9

75 Re186.2

76 Os190.2

77 Ir

190.2

78 Pt

195.1

79 Au197.0

80 Hg200.5

81 Tl

204.4

82 Pb207.2

83 Bi

209.0

84 Po(210)

85 At(210)

86 Rn(222)

7 87 Fr

(223)

88 Ra (226)

89 Ac~(227)

104

Rf(257)

105

Db(260)

106

Sg(263)

107

Bh(262)

108

Hs(265)

109

Mt(266)

110 ---

()

111

---()

112

---()

114

---()

116

---()

118 ---

()

IV Compounds SiC, SiGe

III-V Binary CompoundsAlP, AlAs, AlSb, GaN, GaP, GaAs, GaSb, InP, InAs, InSb

III-V Ternary CompoundsAlGaAs, InGaAs, AlGaP

III-V Quternary CompoundsAlGaAsP, InGaAsP

II-VI Binary CompoundsZnS, ZnSe, ZnTe, CdS, CdSe, CdTe

II-VI Ternary CompoundsHgCdTe

Semiconductor Materials Semiconductor Materials

Atomic Bonding

a. Ionic bonding (such as NaCl)b. Metallic bonding (all metals)c. Covalent bonding (typical Si)d. Van der Waals bonding (water…)e. Mixed bonding (GaAs, ZnSe…, ionic & covalent)

Bonding forces in Solids

Covalent Bonding


plank constant

Core electrons

Valence electrons

2s2p

1sK

L


The shell model of the atom in which the electrons are confined to live within certain shells and in sub shells within shells.

The Shell Model

1s22s22p2 or [He]2s22p2

L shell with two sub shells

Nucleus

Band theory of solids

Two atoms brought together to form molecule

“splitting” of energy levels for outer electron shells

Energy Band Formation (I)

→ ← =

Splitting of energy states into allowed bandsseparated by a forbidden energy gap as the atomic spacing decreases.The electrical properties of a crystalline material correspond to specific allowed and forbidden energies associated with an atomic separation related to the lattice constant of the crystal.

Allowed energy levels of an electron acted on by the Coulomb potential of an atomic nucleus.

Energy Band Formation (I)

Energy Band Formation (II)

Strongly bonded materials: small interatomic distances. Thus, the strongly bonded materials can have larger energy bandgaps than do weakly bonded materials.

Energy Bandgapwhere ‘no’ states exist

As atoms are brought closer towardsone another and begin to bond together, their energy levels mustsplit into bands of discrete levelsso closely spaced in energy, theycan be considered a continuum ofallowed energy.

Pauli Exclusion Principle

Only 2 electrons, of spin ± 1/2, can occupy the same energy state at

the same point in space.

Energy Band Formation (III)

Conceptual development of the energy band model.

Elec

tron

ene

rgy

Ele

ctro

n en

ergy

isolatedSi atoms

Si latticespacing

Decreasing atom spacing

s

p

sp n = 3

N isolated Si-atoms

6N p-states total2N s-states total

(4N electrons total)

Elec

tron

ene

rgy

Crystalline Si N -atoms

4N allowed-states (Conduction Band)

4N allowed-states (Valance Band)

No states

4N empty states

2N+2N filled states

Elec

tron

ene

rgy Mostly

empty

Mostlyfilled

Etop

EcEg

Ev

Ebottom

Broadening of allowed energy levels into allowed energy bandsseparated by forbidden-energy gaps as more atoms influence each electron in a solid.

Energy Band Formation (IV)

One-dimensional representation

Two-dimensional diagram in which energy is plotted versus distance.

N electrons filling half of the 2Nallowed states, as can occur in a metal.

Energy Band

Energy band diagrams.

A completely empty band separated by an energy gap Eg from a band whose 2N states are completely filled by 2N electrons, representative of an insulator.

2 s Band

Overlapping energy bands

Electrons2 s2 p

3 s3 p

1 s 1sSOLIDATOM

E = 0

Free electronElectron Energy, E

2 p Band3s Band

Vacuumlevel

In a metal the various energy bands overlap to give a single band of energies that is only partially full of electrons. There are states with energies up to the vacuum level where the electron is free.

Typical band structures of Metal

Metals, Semiconductors, and Insulators

Electron energy, E

ConductionBand(CB)Empty ofelectrons at 0 K.

ValenceBand(VB)Full of electrons at 0 K.

Ec

Ev

0

Ec+χ

Covalent bond Si ion core (+4e)

A simplified two dimensional view of a region of the Si crystal showing covalent bonds.

The energy band diagram of electrons in the Si crystal at absolute zero of temperature.

Typical band structures of Semiconductor


Band gap = Eg

Carrier Flow for Metal


Carrier Flow for Semiconductors.mov

Carrier Flow for Semiconductor

Carrier Flow for Metals.mov


Insulator Semiconductor Metal

Typical band structures at 0 K.

10610310010-310-610-910-1210-1510-18 109

Semiconductors Conductors

1012

AgGraphite NiCrTeIntrinsic Si

Degeneratelydoped Si

Insulators

Diamond

SiO2

Superconductors

PETPVDF

AmorphousAs2Se3

Mica

Alumina

Borosilicate Pure SnO2

Inorganic Glasses

Alloys

Intrinsic GaAs

Soda silica glass

Manyceramics

MetalsPolypropylene


Range of conductivities exhibited by various materials.

Conductivity (Ωm)-1

Energy Band Diagram

Electrons within an infinite potential energy well of spatial width L, its energy is quantized.

e

nn m

kE2

)( 2h=

Lnknπ

= ...3,2,1=n

Energy increases parabolically with the wavevector kn.

nkh : electron momentum

This description can be used to the behavior of electron in a Metal within which their average potential energy is V(x) ≈ 0 inside, and very large outside.

x0 +a/2-a/2

V(x)

∞∞

V=0

m

infinite square potential

0 +a/2-a/2

∞∞

0 +a/2-a/2

∞∞

xϕ

x2ϕ

n=1

n=2

n=3

E1

E2

E3

Energy state Wavefunction Probability density

r

PE(r)

x

V(x)

x = Lx = 0 a 2a 3a

0aa

Surface SurfaceCrystal

Potential Energy of the electron around an isolated atom

When N atoms are arranged to form the crystal then there is an overlap of individual electron PEfunctions.

PE of the electron, V(x), inside the crystal is periodic with a period a.

The electron potential energy [PE, V(x)], inside the crystal is periodic with the same periodicity as that of the crystal, a. Far away outside the crystal, by choice, V = 0 (the electron is free and PE = 0).

3.1 Energy Band Diagram

E-k diagram, Bloch function.

Moving through Lattice.mov

within the Crystal!


[ ] 0)(222

2

=Ψ⋅−+Ψ xVEm

dxd e

h

Schrödinger equation

)()( maxVxV +=

Periodic Potential

xkikk exUx )()( =Ψ

Periodic Wave functionBloch Wavefunction

There are many Bloch wavefunction solutions to the one-dimensional crystal each identified with a particular k value, say kn which act as a kind of quantum number.

Each ψk (x) solution corresponds to a particular kn and represents a state with an energy Ek.

E-k diagram, Bloch function.

...3,2,1=m

Ek

kπ/a–π /a

Ec

Ev

ConductionBand (CB)

Ec

Ev

CB

The Energy BandDiagram

Emptyψk

Occupiedψkh+

e-

Eg

e-

h+

hυ

VB

hυ

ValenceBand (VB)

The E-k curve consists of many discrete points with each point corresponding to a possible state, wavefunction Ψk (x), that is allowed to exist in the crystal.

The points are so close that we normally draw the E-k relationship as a continuous curve. In the energy range Ev to Ec there are no points [no Ψk (x) solutions].

3.1 Band Diagram

E-k diagram of a direct bandgap semiconductor

E-k diagram


The bottom axis describe different directions of the crystal.

Si Ge GaAs

The energy is plotted as a function of the wave number, k, along the main crystallographic directions in the crystal.

E

CB

k–k

Direct Bandgap

GaAs

E

CB

VB

Indirect Bandgap, Eg

k–k

kcb

Si

E

k–k

Phonon

Si with a recombination center

Eg

Ec

Ev

Ec

Ev

kvb VB

CB

ErEc

Ev

Photon

VB

In GaAs the minimum of the CB is directly above the maximum of the VB. direct bandgapsemiconductor.

In Si, the minimum of the CB is displaced from the maximum of the VB.indirect bandgap semiconductor

Recombination of an electron and a hole in Si involves a recombination center.


E-k diagram

3.1 Energy Band

A simplified energy band diagram with the highest almost-filled band and the lowest almost-empty band.

valence band edge

conduction band edge

vacuum level

χ : electron affinity

e–hole

CB

VB

Ec

Ev

0

Ec+χ

Eg

Free e–hυ > Eg

Hole h+

Electron energy, E

hυ

3. 1 Electrons and Holes

A photon with an energy greater then Eg can excitation an electron from the VB to the CB.

Each line between Si-Si atoms is a valence electron in a bond.When a photon breaks a Si-Si bond, a free electron and a hole in the Si-Sibond is created.

Generation of Electrons and Holes Electrons: Electrons in the conduction band that are free to move throughout the crystal.

Holes: Missing electrons normally found in the valence band(or empty states in the valence band that would normally be filled).

Electrons and Holes

These “particles” carry electricity. Thus, we call these “carriers”

3.1 Effective Mass (I)

An electron moving in respond to an applied electric field.

E

E

within a Vacuum within a Semiconductor crystal

dtdmEqF v

0=−= dtdmEqF n

v∗=−=

It allow us to conceive of electron and holes as quasi-classical particles and to employ classical particle relationships in semiconductor crystals or in most device analysis.

3.1 Carrier Movement Within the Crystal

Density of States Effective Masses at 300 K

Ge and GaAs have “lighter electrons” than Si which results in faster devices

3.1 Effective Mass (II)

Electrons are not free but interact with periodic potential of the lattice.Wave-particle motion is not as same as in free space.

Curvature of the band determine m*.m* is positive in CB min., negative in VB max.

Moving through Lattice.mov


The bottom axis describe different directions of the crystal.

Si Ge GaAs

The energy is plotted as a function of the wave number, k, along the main crystallographic directions in the crystal.

The motion of electrons in a crystal can be visualized and described in a quasi-classical manner.

In most instancesThe electron can be thought of as a particle.The electronic motion can be modeled using Newtonian mechanics.

The effect of crystalline forces and quantum mechanical properties are incorporated into the effective mass factor.

m* > 0 : near the bottoms of all bandsm* < 0 : near the tops of all bands

Carriers in a crystal with energies near the top or bottom of an energy band typically exhibit a constant (energy-independent) effective mass.

` : near band edge

3.1 Mass Approximation

constant2

2

=

dk

Ed

Covalent Bonding

Covalent Bonding Band Occupation at Low Temperature

Band Occupation at High Temperature Band Occupation at High Temperature

Band Occupation at High Temperature Band Occupation at High Temperature

Band Occupation at High Temperature

Without “help” the total number of “carriers” (electrons and holes) is limited to 2ni.

For most materials, this is not that much, and leads to very high resistance and few useful applications.

We need to add carriers by modifying the crystal.

This process is known as “doping the crystal”.

Impurity Doping

The need for more control over carrier concentration

Regarding Doping, ...

Concept of a Donor “Adding extra” Electrons Concept of a Donor “Adding extra” Electrons

Concept of a Donor “Adding extra” Electrons Concept of a Donor “Adding extra” Electrons

Band diagram equivalent view

e–As+

x

As+ As+ As+ As+

Ec

Ed

CB

Ev

~0.05 eV

As atom sites every 106 Si atoms

Distance intocrystal

Electron Energy

The four valence electrons of As allow it to bond just like Si but the 5th electron is left orbiting the As site. The energy required to release to free fifth- electron into the CB is very small.

Energy band diagram for an n-type Si dopedwith 1 ppm As. There are donor energy levels

below Ecaround As+ sites.

Concept of a Donor “Adding extra” Electrons

n-type Impurity Doping of Si

just

Energy band diagram of an n-type semiconductor connected to a voltage supply of V volts.

The whole energy diagram tiltsbecause the electron now has an electrostatic potential energy as well.

Current flowingV

n-Type Semiconductor

Ec

EF − eV

A

B

V(x), PE (x)

x

PE (x) = – eV

EElectron Energy

Ec − eV

Ev− eV

V(x)

EF

Ev

Concept of a Donor “Adding extra” Electrons

Energy Band Diagram in an Applied Field

Concept of a Acceptor “Adding extra” Holes

All regions of

materialare neutrally

charged

One less bondmeans

the acceptor is electrically satisfied.

One less bondmeans

the neighboring Silicon is left with

an empty state.

Hole Movement

Empty state is located next to the Acceptor

Hole Movement

Another valence electron can fill the empty state located next tothe Acceptor leaving behind a positively charged “hole”.

Hole Movement

The positively charged “hole” can move throughout the crystal.(Really it is the valance electrons jumping from atom to atom that creates the hole motion)

Hole Movement


Hole Movement


Regionaround the“hole” hasone lesselectronand thus ispositivelycharged.

Hole Movement


Regionaround the“acceptor”hasone extraelectronand thus isnegativelycharged.


Band diagram equivalent view

B–h+

x

B–

Ev

Ea

B atom sites every 106 Si atoms

Distanceinto crystal

~0.05 eV

B– B– B–

h+

VB

Ec

Electron energy

p-type Impurity Doping of Si


Boron doped Si crystal. B has only three valence electrons. When it substitute for a Si atom one of its bond has an electronmissing and therefore a hole.

Energy band diagram for a p-type Si crystal doped with 1 ppm B. There are acceptor energy levels just above Ev around B- site. These acceptor levels accept electrons from the VB and therefore create holes in the VB.

Ec

Ev

EFi

CB

EFp

EFn

Ec

Ev

Ec

Ev

VB

Intrinsic semiconductors

In all cases, np=ni2

Note that donor and acceptor energy levels are not shown.

Intrinsic, n-Type, p-Type Semiconductors

Energy band diagrams

n-type semiconductors

p-type semiconductors

Impurity Doping Impurity Doping

Valence Band

Valence Band

Impurity Doping

Position of energy levels within the bandgap of Si for common dopants.

Energy-band diagram for a semiconductor showing the lower edge of the conduction band Ec, a donor level Ed within the forbidden band gap, and Fermi level Ef, an acceptor level Ea, and the top edge of the valence band Ev.

Energy Band

Energy band diagrams.

3.2B Semiconductor Statistics

dEEgc )(The number of conduction band states/cm3 lying in the energy range between E and E + dE (if E ≥ Ec).

The number of valence band states/cm3 lying in the energy range between E and E + dE (if E ≤ Ev).

dEEgv )(

Density of States Concept

General energy dependence of gc (E) and gv (E) near the band edges.


Density of States Concept

Quantum Mechanics tells us that the number of available states in a cm3 per unit of energy, the density of states, is given by:

Density of Statesin Conduction Band

Density of States in Valence Band

3.2B Fermi- Dirac function

Probability of Occupation (Fermi Function) Concept

Now that we know the number of available states at each energy, then how do the electrons occupy these states?

We need to know how the electrons are “distributed in energy”.

Again, Quantum Mechanics tells us that the electrons follow the “Fermi-distribution function”.

Ef ≡ Fermi energy (average energy in the crystal)k ≡ Boltzmann constant (k=8.617×10-5eV/K)T ≡Temperature in Kelvin (K)

f(E) is the probability that a state at energy E is occupied.1-f(E) is the probability that a state at energy E is unoccupied.

kTEE feEf /)(1

1)( −+=

Fermi function applies only under equilibrium conditions, however, is universal in the sense that it applies with all materials-insulators, semiconductors, and metals.

“The Fermi function f (E) is a probability distribution function that tells one the ratio of filled to total allowed states at a given energy E”

How do electrons and holes populate the bands?

Probability of Occupation (Fermi Function) Concept

Fermi-Dirac Distribution


Ef

Fermi Function• Probability that an available state at energy E is occupied:

• EF is called the Fermi energy or the Fermi level

There is only one Fermi level in a system at equilibrium.If E >> EF : If E << EF : If E = EF :

kTEE FeEf /)(1

1)( −+=


Probability of Occupation (Fermi function) Concept

Maxwell Boltzmann Distribution Function

Boltzmann Approximation

Probability that a state is empty (occupied by a hole):

kTEEF

FeEfkTEE /)()( ,3 If −−≅>−

kTEEF

FeEfkTEE /)(1)( ,3 If −−≅>−

kTEEkTEE FF eeEf /)(/)()(1 −−− =≅−

TYU

• Assume the Fermi level is 0.30eV below the conduction band energy (a) determine the probability of a state being occupied by an electron at E=Ec+KT at room temperature (300K).

TYU

• Determine the probability that an allowed energy state is empty of electron if the state is below the fermi level by (i) kT (ii) 3KT (iii) 6 KT


Example 2.2

The probability that a state is filled at the conduction band edge (Ec) is precisely equal to the probability that a state is empty at the valence band edge (Ev).Where is the Fermi energy locate?

SolutionThe Fermi function, f(E), specifies the probability of electron occupying states at a given energy E.The probability that a state is empty (not filled) at a given energy E is equal to 1- f(E).

( ) ( )VC EfEf −= 1

( ) ( ) kTEEC FCeEf /1

1−+

= ( ) ( ) ( ) kTEEkTEEV VFFV eeEf // 1

11

111 −− +=

+−=−

kT

EE

kT

EE FVFC −=

−

2VC

FEEE +

=

The density of electrons (or holes) occupying the statesin energy between E and E + dE is:


Probability of Occupation Concept

0 Otherwise

dEEfEgc )()(Electrons/cm3 in the conduction band between E and E + dE (if E ≥ Ec).

Holes/cm3 in the conduction band between E and E + dE (if E ≤ Ev).

dEEfEgv )()(


Probability of Occupation Concept

Typical band structures of Semiconductor

Ev

Ec

0

Ec+χ

EF

VB

CB

E

g(E)

g(E) ∝ (E–Ec)1/2

fE)

EF

E

Forelectrons

For holes

[1–f(E)]

Energy band diagram

Density of states Fermi-Diracprobability

function

probability of occupancy of a state

nE(E) or pE(E)

E

nE(E)

pE(E)

Area = p

Area

Ec

Ev

ndEEn E == ∫ )(

g(E) X f(E)Energy density of electrons in

the CB

number of electrons per unit energy per unit volumeThe area under nE(E) vs. E is the electron concentration.

number of states per unit energy per unit volume


The Density of Electrons is:

Probability the state is filled

Number of states per cm-3 in energy range dE

Probability the state is empty

Number of states per cm-3 in energy range dE

units of n and p are [ #/cm3]

The Density of Hole is:

Developing the Mathematical Model for Electrons and Holes concentrations

Electron Concentration (no)

TYU

Calculate the thermal equilibrium electron concentration in Si at T=300K for the case when the Fermi level is 0.25eV below the conduction band.

EC

EV

EF0.25eV

Hole Concentration (no)

TYU

• Calculate thermal equilibrium hole concentration in Si at T=300k for the case when the Fermi level is 0.20eV above the valance band energy Ev.

EC

EV 0.20eVEF

Degenerate and Nondegenerate Semiconductors

Nondegenerate Case

Useful approximations to the Fermi-Dirac integral:

( ) kTEEC

CfeNn −=

( ) kTEEV

fVeNp −=

Developing the Mathematical Model for Electrons and Holes

( ) kTEECi

CieNn −=

When n = ni, Ef = Ei (the intrinsic energy), then

orand

( ) kTEEVi

iVeNn −=( ) kTEE

iVVienN −=or

( ) kTEEiC

iCenN −=

The intrinsic carrier concentration

( ) kTEECo

CfeNn −= ( ) kTEEV

fVo eNp −=

Other useful relationships: n⋅p product:

( ) kTEECi

CieNn −= and ( ) kTEEVi

iVeNn −=

( ) kTEVC

kTEEVCi

gVC eNNeNNn −−− ==2

kTEVCi

geNNn 2−=

Semiconductor Statistics

TYU

Determine the intrinsic carrier concentration in GaAs (a) at T=200k and (b) T=400K

2ioo npn =

Law of mass Action

( ) kTEEio

ifenn −=( ) kTEE

iofiepp −=andSince

It is one of the fundamental principles of semiconductorsin thermal equilibrium

Example

Law of mass action

An intrinsic Silicon wafer has 1x1010 cm-3 holes. When 1x1018

cm-3 donors are added, what is the new hole concentration?

2ioo npn =

DNn ≅D

i

Nnp

2

≅and

AD NN ⟩⟩ iD nN ⟩⟩andif

TYU

• Find the hole concentration at 300K, if the electron concentration is no=1 x 1015 cm-3, which carrier is majority carrier and which carrier is minority carrier?

TYU: The concentration of majority carrier

electron is no=1 x 1015 cm-3 at 300K. Determine the concentration of phosphorus that are to be added and determine the concentration minority carriers holes.

Partial Ionization, Intrinsic Energy and Parameter Relationships.

Energy band diagram showing negative charges

Energy band diagram showing positive charges

If excess charge existed within the semiconductor, random motion of charge would imply net (AC) current flow.

Not possible!

Thus, all charges within the semiconductor must cancel.

Charge Neutrality:

( ) ( )[ ]( ) ( )[ ] 0=−+−⋅

+=++−

−+

nNNpq

nNNp

dA

ad

Mob

ile +

cha

rge

Imm

obile

-ch

arge

Imm

obile

+ c

harg

e

Mob

ile -

char

ge

3.5 Carrier concentration-effects of doping

NA¯ = Concentration of “ionized” acceptors = ~ NA

ND+ = Concentration of “ionized” Donors = ~ ND

Charge Neutrality: Total Ionization case

( ) ( ) 0=−+− +− nNNp dA

3.5 Developing the Mathematical Model for Electrons and Holes

The intrinsic carrier concentration as a function of temperature.

Electron concentration versus temperature for n-typeSemiconductor.

Carrier Concentration vs. Temperature

position of Fermi Energy level

( ) kTEEco

fceNn ][ −−=)/ln( occ nNkTEE F =−

)/ln( dcFc NNkTEE =−

Nd >> ni

Note: If we have a compensated semiconductor , then the Nd termin the above equation is simply replaced by Nd-Na.

( ) kTEEVo

fVeNp −=

)/ln( ovvF pNkTEE =−

)/ln( avvF NNkTEE =−

Na >> ni


Note: If we have a compensated semiconductor , then the Na termin the above equation is simply replaced by Na-Nd.

position of Fermi level as a function of carrier concentration

Where is Ei ?

Extrinsic Material:

Note: The Fermi-level is pictured here for 2 separate cases: acceptor and donor doped.

TYU

• Determine the position of the Fermi level with respect to the valence band energy in p-type GaAs at T=300K. The doping concentration are Na=5 x 1016 cm-3 and Na=4 x 1015 cm-3.


Extrinsic Material:

( ) kTEEio

fifenn −=( ) kTEE

ioffienp −=

Solving for (Ef - Efi)

−=

=−

iifif n

pkTnnkTEE lnln

=−

i

Dfif n

NkTEE ln

−=−

i

Afif n

NkTEE ln

AD NN ⟩⟩ iD nN ⟩⟩andfor DA NN ⟩⟩ iA nN ⟩⟩andfor

TYU 3.8

• Calculate the position of the Fermi level in n-type Si at T=300K with respect to the intrinsic Fermi energy level. The doping concentration are Nd=2 x 1017 cm-3 and Na=3 x 1016 cm-3

.

EC

EV

EFi

EF

Mobile Charge Carriers in Smiconductor devices

• Three primary types of carrier action occur inside a semiconductor:

– Drift: charged particle motion under the influence of an electric field.

– Diffusion: particle motion due to concentration gradient or temperature gradient.

– Recombination-generation (R-G)

Carrier Motion

Carrier Dynamics

Electron Drift

Hole Drift

Electron Diffusion

Hole Diffusion

Carrier DriftDirection of motion

Holes move in the direction of the electric field. (⊕ )

Electrons move in the opposite direction of the electric field. ( ⊕)

Motion is highly non-directional on a local scale, but has a net direction on a macroscopic scale.

Average net motion is described by the drift velocity, vd [cm/sec].

Net motion of charged particles gives rise to a current.

Instantaneous velocity is extremely fast

Describe the mechanism of the carrier drift and drift current due to an applied electric field.

Drift

Drift of Carriers

Electric Field

Drift of electron in a solid

The ball rolling down the smooth hill speeds up continuously, but the ball rolling down the stairs moves with a constant average velocity.

µ [cm2/Vsec] : mobility

Random thermal motion. Combined motion due to random thermal motion and an applied electric field.

Drift

Schematic path of an electron in a semiconductor.

EE

Drift

Random thermal motion. Combined motion due to random thermal motion and an applied electric field.

Drift

Conduction process in an n-type semiconductor

Thermal equilibrium Under a biasing condition

Drift

Given current density J ( I = J x Area ) flowing in a semiconductor block with face area A under the influence of electric field E, ρ is volume density, the component of J due to drift of carriers is:

Hole Drift Current Density

dp vpqJvJdrf

drf

d

⋅⋅==

Electron Drift Current Density

dn vneJ drf ⋅⋅−=and

dp

drf

vpeJ

vJ

drf

d

⋅⋅=

= ρ

Drift

At Low Electric Field Values,

EpeJ pDriftp ⋅⋅⋅= µ EneJ nDriftn ⋅⋅⋅= µand

µ [cm2/V·sec] is the “mobility” of the semiconductor and measures the ease with which carriers can move through the crystal.

The drift velocity increases with increasing applied electric field.:

EnpqJJJ npDriftnDriftpdrf ⋅+=+= )( µµ

Electron and hole mobilities of selected intrinsic semiconductors (T=300K)

Si Ge GaAs InAsµn (cm

2/V·s) 1350 3900 8500 30000µp (cm

2/V·s) 480 1900 400 500

sV

cmV/cmcm/s 2

⋅

=µ has the dimensions of v/ :

Electron and Hole Mobilities EX 4.1

• Consider a GaAs sample at 300K with doping concentration of Na=0 and Nd=1016 cm-3. Assume electron and hole mobitities given in table 4.1. Calculate the drift current density if the applied electric filed is E=10V/cm.

µ [cm2/Vsec] is the “mobility” of the semiconductor and measures the ease with which carriers can move through the crystal.

Mobility

µn ~ 1360 cm2/Vsec for Silicon @ 300Kµp ~ 460 cm2/Vsec for Silicon @ 300Kµn ~ 8000 cm2/Vsec for GaAs @ 300Kµp ~ 400 cm2/Vsec for GaAs @ 300K

[ ]sec2*

,, Vcm

mq

pnpn

τµ =

<τ > is the average time between “particle” collisions in the semiconductor.

Collisions can occur with lattice atoms, charged dopant atoms, or with other carriers.

Drift velocity vs. Electric field in Si.

Saturation velocity Saturation velocity

Drift velocity vs. Electric fieldDesigning devices to work at the peak results in faster operation

1/2mvth2=3/2kT=3/2(0.0259)=0.03885eV

Ohm’s law is valid only in the low-field region where drift velocity is independent of the applied electric field strength.Saturation velocity is approximately equal to the thermal velocity (107 cm/s).

[ ]sec2*

,, Vcm

mq

pnpn

τµ =

Drift

Drift velocity vs. Electric field in Si and GaAs.

Note that for n-type GaAs, there is a region of negative differential mobility.

[ se2*

,, Vcm

mq

pnpn

τµ =

Negative differential mobility

Electron distributions under various conditions of electric fields for a two-valley semiconductor.

m*n=0.067mo

m*n=0.55mo

Figure 3.24.

Velocity-Field characteristic of a Two-valley semiconductor.

Negative differential mobilityTYU

• Silicon at T=300K is doped with impurity concentration of Na=5 X 1016 cm-3 and Nd=2 x 1016 cm-3. (a) what are the electron and hole mobilities? (b) Determine the resistivity and conductivity of the material.

Mean Free Path

• Average distance traveled between collisions

mpthvl τ=

EX 4.2Using figure 4.3 determine electron and hole nobilities.

EX 4.2Using figure 4.3 determine electron and hole mobilities in (a) Si for Nd=1017 cm-3,Na=5 x 1016 cm-3 and (b) GaAs for Na=Nd=1017cm-3 Ex 4.2

Effect of Temperature on Mobility

Temperature dependence of mobility with both lattice and impurity scattering.

A carrier moving through the lattice encounters atoms which are out of their normal lattice positions due to the thermal vibrations.

The frequency of such scattering increases as temperature increases.

At low temp. lattice scattering is less important.

At low temperature, thermal motion of the carriers is slower, and ionized impurity scattering becomes dominant.

Since the slowing moving carrier is likely to be scattered more strongly by an interaction with charged ion.

Impurity scattering events cause a decrease in mobility with decreasing temperature.

As doping concentration increase, impurity scattering increase, then mobility decrease.

Mobility versus temperature Mobility versus temperature

Effect of Temperature on Mobility

Electron mobility in silicon versus temperature for various donor concentrations. Insert shows the theoretical temperature dependence of electron mobility.

Electron and hole mobilities in Silicon as functions of the total dopant concentration.

Effect of Doping concentration on Mobility

300 K

Resistivity and Conductivity

Ohms’ Law

Ohms Law [ ]2cmAEEJρ

σ =⋅=

Conductivityσ [ ]cmohm ⋅1

Resistivityρ [ ]cmohm ⋅

semiconductor conductivity and resistivity

Adding the Electron and Hole Drift Currents (at low electric fields)

Drift CurrentEnpeJJJ npDriftnDriftpdrf ⋅+=+= )( µµ

Conductivity)( npe np µµσ +=

Resistivity[ ])(11 pne pn µµσ

ρ +==

But since µn and µp change very little and n and p change several orders of magnitude:

for n-type with n>>p

pene

p

n

µσµσ

≅≅

for p-type with p>>n

[ ]sec2*

,, Vcm

mq

pnpn

τµ =

Particles diffuse from regions of higher concentration to regions of lower concentration region, due to random thermal motion.

Diffusion Diffusion

Nature attempts to reduce concentration gradients to zero.Example: a bad odor in a room, a drop of ink in a cup of water.

In semiconductors, this “flow of carriers” from one region of higher concentration to lower concentration results in a “Diffusion Current”.

Visualization of electron and hole diffusion on a macroscopic scale.

DiffusionpJDiffusionnJ

Diffuse Diffuse

dxdneDJ N=diffN, dx

dpeDJ P−=diffP,

D is the diffusion constant, or diffusivity.

x x

Diffusion Current

Diffusion current density

Fick’s law

Diffusion as the flux, F, (of particles in our case) is proportional to the gradient in concentration.

η∇−= DFη : Concentration

D : Diffusion Coefficient

For electrons and holes, the diffusion current density( Flux of particles times ± q )

nDqJ

pDqJ

nDiffusionn

pDiffusionp

∇⋅=

∇⋅−=

The opposite sign for electrons and holes

JN = JN,drift + JN,diff = qnµnε + dxdnqDN

JP = JP,drift + JP,diff = qpµpε –dxdpqDP

J = JN + JP

Total Current

Total Current

Total Current = Drift Current + Diffusion Current

nDqEnqJJJ

pDqEpqJJJ

nnDiffusionnDriftnn

ppDiffusionpDriftpp

∇⋅+⋅=+=

∇⋅−⋅=+=

µ

µ

np JJJ +=

TYU

• Consider a sample of Si at T=300K. Assume that electron concentration varies linearly with distance, as shown in figure.The diffusion current density is found to be Jn=0.19 A/ cm2. If the electron diffusion coefficient is Dn=25cm2/sec, determine the electron concentration at x=0.

dxdneDJ N=diffN,

dxdpeDJ P−=diffP,Jp=0.270 A/cm2

Dp=12 cm2/secFind the hole concentration at x=50um

Graded impurity distribution

Energy band diagram of a semiconductor in thermal equilibriumwith a nonuniform donor impurity concentration

Carrier Generation

Generation and Recombination

Band-to-band generation

Generation Mechanism

Band-to-Band Generation

Thermal Energyor

Light

Band-to-Band or “direct” (directly across the band) generation.

Does not have to be a “direct bandgap” material.

Mechanism that results in ni.

Basis for light absorption devices such as semiconductor photodetectors, solar cells, etc…

Gno=Gpo

Band-to-Band Recombination

Recombination Mechanism

Photon(single particle of light)

or

multiple phonons(single quantum of lattice vibration - equivalent tosaying thermal energy)

Band to Band or “direct” (directly across the band) recombination.

Does not have to be a “direct bandgap” material, but is typically very slow in “indirect bandgap” materials.

Basis for light emission devices such as semiconductor Lasers, LEDs, etc…

Rno=Rpo

In thermal equilibrium: Gno=Gpo=Rno=Rpo

Low-Level-Injection implies

00 , nnnp ≈<<∆

00 , pppn ≈<<∆

in a n-type material

in a p-type material

00 pppnnn +∆=+∆=

In Non-equilibrium, n·p does not equal ni2

Excess carrier Recombination and Generation

low level injection caseNd=1014/cm3 doped Si at 300K subject to a perturbation where ∆p =∆n =109/cm3.

n0 ≅ Nd =1014/cm3 and p0 ≅ ni2/Nd = 106/cm3

n = n0 + ∆n ≅ n0 and ∆p ≅ = 109/cm3 << n0 ≅ 1014/cm3

Although the majority carrier concentration remains essentially unperturbed under low-level injection, the minority carrier concentration can, and routinely does, increase by many orders of magnitude.

and

Example

Excess minority carrier lifetime

Carrier Lifetime

notentn τ−∆=∆ 0)(poteptp τ−∆=∆ 0)(

Light Pulses

Semiconductor

Rs

VARLI VL

+

_

Oscilloscope

Schematic diagram of photoconductivity decay measurement.

The minority carrier lifetime is the average time a minority carrier can survive in a large ensemble of majority carriers.If ∆p is negative Generation or an increase in carriers with time.If ∆p is positive Recombination or a decrease in carriers with time.Either way the system “tries to reach equilibrium”The rate of relaxation depends on how far away from equilibrium we are.

Material Response to “Non-Equilibrium”

Relaxation Concept

Consider a case when the hole concentration in an n-type sample is not in equilibrium, i.e., n·p ≠ ni

2

τpo is the minority carrier lifetime

po

ntp

tpR

τ)(' ∆

−=∂∂

=

Material Response to “Non-Equilibrium”

Relaxation Concept

Likewise when the electron concentration in an p-type sample is not in equilibrium, i.e., n·p does NOT equal ni

2

τn is the minority carrier lifetimenGRThermal

ntn

τ∆

−=∂∂

−

Indirect recombination-generation processes at thermal equilibrium.

Recombination-Generation center recombination

Generation and Recombination process

Recombination-Generation (R-G) Center Recombination

Energy loss can result in a Photonbut is more often multiple phonons

Also known as Shockley-Read-Hall (SRH) recombination.

Two steps:1 1st carrier is “trapped” (localized) at an defect/impurity (unintentional/intentional ).2 2nd carrier (opposite type) is attracted and annihilates the 1st carrier.

Useful for creating “fast switching” devices by quickly “killing off” EHP’s.

Recombination Mechanism Generation and Recombination process

Generation Mechanism

Recombination-Generation (R-G) Center Generation

Thermal Energy

Two steps:1 A bonding electron is “trapped” (localized) at an unintentional

defect/impurity generating a hole in the valence band.2 This trapped electron is then promoted to the conduction band

resulting in a new EHP.

Almost always detrimental to electronic devices. AVOID IF POSSIBLE!

Effects of recombination centers on solar cell

performance

EC

EV Light

a

b

c

dThe light-generated minoritycarrier can return to the groundstate through recombination

center before beingcollected by the junction:i) through path (a) ii) through path (c)

Without recombinationcenters paths (b) and (d)are dominated

Auger Recombination

Auger RecombinationAuger – “pronounced O-jay”

Requires 3 particles.

Two steps:

1 1st carrier and 2nd carrier of same type collide instantly annihilating the electron hole pair (1st and 3rd carrier). The energy lost in theannihilation process is given to the 2nd carrier.

2 2nd carrier gives off a series of phonons until it’s energy returns to equilibrium energy (E~=Ec) This process is known as thermalization.

3.3 p-n Junction Diode

p-n Junction

p-Type Material n-Type Material

p-n Junction principles

p-n Junction

p-Type Material


n-Type Material

A p-n junction diode is made by forming a p-type region of material directly next to a n-type region.

p-n Junction

But when the device has no external applied forces, no current can flow. Thus, the Fermi-level must be flat!

We can then fill in the junction region of the band diagram as:

p-n Junction Diode


EC

EV

EFEi

EC

EV

EF

Ei

But when the device has no external applied forces, no current can flow. Thus, the Fermi-level must be flat!

We can then fill in the junction region of the band diagram as:

p-Type Material

n-Type Material

p-n Junction Diode

EC

EV

EF

Ei EC

EV

EFEi

Built-in-potential

p-n Junction Diode

EC

EV

EF

Ei EC

EV

EFEi

- qVbi


x

Electrostatic Potential)(1

refC EEq

V −=

Built-in-potentialBIV

Built-in-potential

p-n Junction Diode

x

Electrostatic Potential)(1

refC EEq

V −=

Built-in-potentialBIV

x

Electric Field

dxdVE −=

Electric Field

x

Electric Field

dxdVE −=

Built-in-potential

p-n Junction Diode

Charge Density

x

dxdEK S 0ερ ⋅=

qND

qNA

Charge Density

- --

+ ++

Built-In Potential Vbi

=

=− −

i

A

isidepFi

nNkT

npkTEE

ln

ln)(

sideniFsidepFisiden Ssidep Sbi )()()( −−−− −+−=Φ+Φ= EEEEqqV

=

=− −

i

D

isideniF

nNkT

nnkTEE

ln

ln)(

For non-degenerately doped material:

TYU 5.1

• Calculate the built-in-potential barrier in a Si pm junction at T=300K for (a) Na=5 x 1017cm-3, Nd=1016cm-3 (b)Na=1015cm-3

x

Electric Field

dxdVE −=

Built-in-potential

p-n Junction Diode

Charge Density

x

dxdEK S 0ερ ⋅=

qND

qNA

Charge Density

- --

+ ++

n n o

xx = 0

p n o

p po

n po

l o g ( n ) , l o g ( p )

-e N a

e N d

M

x

E ( x )

B -

h +

p n

M

As +

e –

Wp Wn

V o

V ( x )

x

P E (x )

x

– Wp

Wn

0

e V o

x

– e V o

Hole Potential Energy PE (x)o

– E o

E0

ρ net

M

n i

p-n Junction Principles

Electron Potential Energy PE (x)

Space charge regionM

-Wp -WnNeutral n-regionNeutral p-region

Metallurgical Junction

Charge Density (NOT Resistivity)

Poisson’s Equation

p-n Junction

0ερ⋅

−=⋅∇SK

E0

/

0 εερ

⋅−

=⋅

−=S

DA

S KqN

KdxdE

in 1-dimension

Electric Field

Permittivity of free spaceRelative Permittivity of Semiconductor

(εr)

D

A

AD

qNxqNx

NNnpq

=−=

−+−=

)()(

)(

ρρρ

Movement of Electrons and Holes when Forming the Junction

Depletion Region Approximation: Step Junction Solution

Number of negative charges per unit area in the p region is equal to the number of positive charges per unit area in the n-region

Electric potential V(x) 0r φ(x)


Space charge width(depletion layer width)


n-region space charge width p-region space charge width

TYU5.2

• A silicon pn junction at T=300k with zero applied bias has doping concentration of Nd= 5 x 1016cm-3 and Na=5 x 1015 cm-3. Determine xn, xp, W, and |Ex(max)|, Vbi 0.718V

pn junction reverse applied bias


Diode under Forward Bias.mov Diode under no Bias.mov Diode under Reverse Bias.mov

Schematic representation of depletion layer width and energy band diagrams of a p-n junction under various biasing conditions.

Reverse-bias condition

Forward-bias condition

Thermal-equilibrium

pn junction reverse/forward applied bias

Depletion Region

Thus, only the boundary conditions change resulting indirect replacement of Vbi with (Vbi-VA) with VA ≠ 0.

pn junction forward bias applied bias


pn junction reverse/forward applied

Depletion Region Approximation: Step Junction Solution with VA ≠ 0Consider a p+n junction (heavily doped p-side, lightly doped n side)


Forward bias condition


Reverse bias condition

Space charge width and Electric field

2/1)(2

++

=+=NaNd

NdNe

VVxxW ARbispn

ε

2/11)(2

+

+

=da

Rbis

NNNaNd

eVVxp ε

2/11)(2

+

+=

dad

aRbis

NNNN

eVVxn ε

EX 5.3

• A Si pn junction at 300K is reverse bias at VR=8V, the doping concentration are Na=5 x 1015cm-3 and Nd= 5 x 1016 cm-3. Determine xn, xp and W, repeat for VR=12V.

p-n Junction I-V Characteristics

In Equilibrium (no bias)Total current balances due to the sum of the individual components

Electron Drift Current

Electron DiffusionCurrent

Hole Drift CurrentHole Diffusion

Current

Diode under no Bias.mov

no net current!

EC

EV

EFEi

p-Type Material n-Type Materialq VBI

+ + ++++ +++ ++ ++ ++++++

0=∇⋅+⋅=+= nDqnEqJJJ nnDiffusionnDriftnn µ

no net current!


EC

EV

EF

Ei

n vs. E

p vs. E

In Equilibrium (no bias)Total current balances due to the sum of the individual components

0=∇⋅+⋅=+= pDqpEqJJJ ppDiffusionpDriftpp µ


Forward Bias (VA > 0)

I

Hole Drift Current


Electron DiffusionCurrent

Hole Diffusion Current IP

IN

Diode under Forward Bias m

Current flow is dominated by majority carriers flowingacross the junction and becoming minority carriers

VA

Current flow is proportional to e(Va/Vref) due to the exponential decay of carriers into the majority carrier bands

Lowering of potential hill

by VA

surmount potential barrier

PN III +=

Hole Diffusion Current negligible due to large energy barrier

Hole Drift Current


Electron Diffusion Current negligible due to large energy barrier

Reverse Bias (VA < 0)

Diode under Reverse Bias.m


Current flow is constant due to thermally generated carriers swept out by E fields in the depletion region

Current flow is dominated by minority carriers flowing across the junction and becoming majority carriers

Increase of potential hill

by VA

Where does the Reverse Bias Current come from? Generation near the depletion region edges “replenishes” the current source.


P-N Junction Diodes Current Flowing through a Diode

I-V CharacteristicsQuantitative Analysis

(Math, math and more math)

Putting it all together


for Ideal diodeVref = kT/q

-I0

−

= 10

kT

qVexpII

η

η : Diode Ideality Factor


Diode Equation

Quantitative p-n Diode Solution

Assumptions:1) Steady state conditions2) Non- degenerate doping3) One- dimensional analysis4) Low- level injection5) No light (GL = 0)

Current equations:

)x(J)x(JJ npp +=

−=

dx

dpqDpEqJ ppp µ

−=

dxdnqDnEqJ nnn µ

Continuity Equations

Steady state : n(x) is time invariant.

Transient state : n(x) is time dependent.

xJ

qxF

tn

∂∂

=∂∂

−=∂∂ 1

F: Particle FluxJ: Current Density

Continuity Equation

)( xxJ p ∆−

)(xJ p

xx ∆+x

x∆

2Area, cmA

Ways Carrier Concentrations can be Altered


...

...

light as suchprocesses other All

GR ThermalDiffusionDrift


GR ThermalDiffusionDrift

tp

tp

tp

tp

tp

tn

tn

tn

tn

tn

∂∂

+∂∂

+∂∂

+∂∂

=∂∂

∂∂

+∂∂

+∂∂

+∂∂

=∂∂

−

−

There must be spatial and time continuity in the carrier concentrations.


PPzPyPx

DiffusionDrift

NNzNyNx

DiffusionDrift

Jqz

Jy

Jx

Jqt

ptp

Jqz

Jy

Jx

Jqt

ntn

⋅∇−=

∂∂

+∂

∂+

∂∂

−=∂∂

+∂∂

⋅∇=

∂∂

+∂

∂+

∂∂

=∂∂

+∂∂

11

11

...

...

1

1


GR ThermalP


GR ThermalN

tp

tpJ

qtp

tn

tnJ

qtn

∂∂

+∂∂

+⋅∇−=∂∂

∂∂

+∂∂

+⋅∇=∂∂

−

−

Ways Carrier Concentrations can be Altered

Continuity Equations: Special Case known as “Minority Carrier Diffusion Equation”

Simplifying Assumptions:

1) One dimensional case. We will use “x”.

2) We will only consider minority carriers.

3) Electric field is approximately zero in regions subject to analysis.

4) The minority carrier concentrations IN EQUILIBRIUM are not a function of position.

5) Low-level injection conditions apply.

6) SRH recombination-generation is the main recombination-generation mechanism.

7) The only “other” mechanism is photogeneration.


Minority CarrierDiffusion Equation

Because of (3) no electric field E = 0

Because of (5) - low level injection Because of (7) - Photogeneration

L


Gtn

=∂∂

...nGenerationombinationRe

ntn

τ∆

−=∂∂

−

2

2

20

2

2

2 )()(11x

nDx

nnDxnD

xJ

qJ

q nnnN

N ∂∆∂

=∂

∆+∂=

∂∂

=∂∂

=⋅∇

nDqnEqJJJ nnnnn ∇⋅+⋅=+= µDiffusionDrift

nDqJ nn ∇⋅==Diffusion


0 0

Finally

tn

tnn

tn

∂∆∂

=∂∆+∂

=∂∂ )()( 0


...light as suchprocesses other All

GenerationombinationReDiffusionDrift tn

tn

tn

tn

tn

∂∂

+∂∂

+∂∂

+∂∂

=∂∂

−

Ln

ppN

p Gn

xn

Dtn

+∆

−∂

∆∂=

∂

∆∂

τ)()()(

2

2

Lp

nnP

n Gpx

pDtp

+∆

−∂∆∂

=∂∆∂

τ)()()(

2

2

0

Minority Carrier Diffusion Equations



Further simplifications (as needed):

Steady State …

No minority carrier diffusion gradient …

No SRH recombination-generation …

00 =∆

−=∆

−pn

pandnττ

No light …

0→LG

0)(0)(

→∂∆∂

→∂

∆∂

tpand

tn np

0)(0)(

2

2

2

2

→∂∆∂

→∂∆∂

xpDand

xn

D nP

pN

Solutions to the“Minority Carrier Diffusion Equation”

SemiconductorSemiconductorLight x

Light absorbed in a thin skin.

Consider a semi-infinite p-type silicon sample with NA=1015 cm-3 constantly illuminated by light absorbed in a very thin region of the material creating a steady state excess of 1013 cm-3 minority carriers (x=0).

What is the minority carrier distribution in the region x> 0 ?

Ln

ppN

p Gn

xn

Dtn

+∆

−∂

∆∂=

∂

∆∂

τ)()()(

2

2

n

ppN

nx

nD

τ)()(

2

2 ∆=

∂

∆∂

0 0Steady state No excess carrier

generation


Steady-state carrier injection from one side.

Semi-infinite sample

Sample with thickness W

Direct generation and recombination of electron-hole pairs:

at thermal equilibrium under illumination. Surface recombination at x = 0. The minority carrier distribution near the surface is affected by the surface recombination velocity.


General Solution

NNNLxLx

p DL whereeeAxn NN τ⋅≡⋅+⋅=∆ +− )()( B)(

LN is the “Diffusion length” the average distance a minority carrier can move before recombining with a majority carrier.

Boundary Condition …

( )

0B

BeAxn

BAcmxn

NLp

p

=→

+==∞=∆

+===∆∞+

−

)0(0)(

10)0( 313

( ) 31310)( −−=∆ cmexn NLxp

0

Continue


SemiconductorSemiconductorx

Consider a p-type silicon sample with NA=1015 cm-3 and minority carrier lifetime τ=10 µsec constantly illuminated by light absorbed uniformly throughout the material creating an excess 1013 cm-3 minority carriers per second. The light has been on for a very long time. At time t=0, the light is shut off.

What is the minority carrier distribution in for t < 0 ?

Ln

ppN

p Gn

xn

Dtn

+∆

−∂

∆∂=

∂

∆∂

τ)()()(

2

2

3710)0,( −=⋅=<∆ cmGtx alln nLp τ

0

Light Light absorbed uniformly

0

SemiconductorSemiconductor

Uniform distribution


SemiconductorSemiconductor Light x

In the previous example: What is the minority carrier distribution in for t > 0 ?

Ln

ppN

p Gn

xn

Dtn

+∆

−∂

∆∂=

∂

∆∂

τ)()()(

2

2

)51(7

)(

10)(

)]0([)(−−

−

=∆

⋅=∆=∆et

p

tpp

etn

etntn nτ

Light absorbed uniformly

0 0

ContinueApplication of the Minority Carrier Diffusion Equation

Since electric fieldsexist in the depletion region, the minority

carrier diffusion equation does not

apply here.

Quisineutral Region Quisineutral Region

00 0 0


minority carrier diffusion eq. minority carrier diffusion eq.

kTEEi

kTEEi

fi

if

enp

enn)(

0

)(0

−

−

=

=

Equilibrium

kTFEi

kTEFi

Pi

iN

enp

enn)(

)(

−

−

=

=

Non-Equilibrium

The Fermi level is meaningful only when the system is in thermal equilibrium.

The non-equilibrium carrier concentration can be expressed by defining Quasi-Fermi levels Fn and Fp .

Equilibrium Non-Equilibrium

Quasi - Fermi Levels


Quantitative p-n Diode Solution (At the depletion regions edge)

kT)FF(i

PNennp −= 2quasi-Fermi levels formalism

?



+=

dxdnDnEqJ nnn µ

( )dx

nndqD p

n

∆+=

0

dx

ndqD p

n

∆=

+=

dx

dpDpEqJ ppp µ

( )dx

ppdqD n

p

∆+= 0

dx

pdqD n

p

∆=

0 0

Approach:Solve minority carrier diffusion equation in quasineutral regions.

Determine minority carrier currents from continuity equation.

Evaluate currents at the depletion region edges.

Add these together and multiply by area to determine the total current through the device.

Use translated axes, x x’ and -x x’’ in our solution.


x”=0 x’=0



x”=0 x’=0


Holes on the n-side



x”=0 x’=0 Holes on the n-side


Similarly for electrons on the p-side…


x”=0 x’=0


Thus, evaluating the current components at the depletion region edges, we have…

Note: Vref from our previous qualitative analysis equation is the thermal voltage, kT/q

J = Jn (x”=0) +Jp (x’=0) = Jn (x’=0) +Jn (x”=0) = Jn (x’=0) +Jp (x’=0)

Ideal Diode Equation Shockley Equation


x”=0 x’=0


Total on current is constant throughout the device. Thus, we can characterize the current flow components as…

-xp xn

J

A silicon pn junction at T=305K has the following parameters:NA = 5x1016 cm-3, ND = 1x1016 cm-3

, Dn = 25 cm2/sec, Dp = 10 cm2/sec, τn0 = 5x10-7 sec, and τp0 = 1x10-7 sec, ni305K = 1.5x1010 cm-3,. The cross-sectional area is A=10-3 cm2, and the forward-bias voltage is Va = 0.625 V.Calculate the (a) minority electron diffusion current at the space charge region.(b) minority hole diffusion current at the space charge edge.(c) total current in the pn junction diode.

Example

SolutionMinority electron diffusion current density nnn DL τ=

A silicon pn junction at T=305K has the following parameters:NA = 5x1016 cm-3, ND = 1x1016 cm-3

, Dn = 25 cm2/sec, Dp = 10 cm2/sec, τn0 = 5x10-7 sec, and τp0 = 1x10-7 sec, ni305K = 1.5x1010 cm-3,. The cross-sectional area is A=10-3 cm2, and the forward-bias voltage is Va = 0.625 V.Calculate the (a) minority electron diffusion current at the space charge region.(b) minority hole diffusion current at the space charge edge.(c) total current in the pn junction diode.

Example

SolutionMinority electron diffusion current density

Minority hole diffusion current density

nnn DL τ=

( ) 216

210

719

2

0

20

/154.010259.0625.0exp

105105.1

10525)106.1(

1exp1exp

cmmA

kTqV

NnDq

kTqV

LnqD

J a

A

i

n

na

n

pnn

=

−

××

××=

−

=

−

=

−−

τ

( ) 216

210

719

2

0

20

/09.110259.0625.0exp

101105.1

10110)106.1(

1exp1exp

cmmA

kTqV

NnD

qkTqV

LnqD

J a

D

i

p

pa

p

npp

=

−

××

××=

−

=

−

=

−−

τ

Total current density= 1.24 mA/cm2 Total current = AxJ=10-3x1.24 =1.24 µA

x

1017

108

103

-xp

2.1x10-2cm

1015

1010

xn

1.6x10-2cm

n or p(log scale)

105

nn

pp

nppn

The diode is forward biased.There is pile-up or minority carrier excess (∆np>0 and ∆pn>0 ) at the edges of the depletion region.

Given figure is a dimensioned plot of the steady state carrier concentrations inside a pn junction diode maintained at room temperature.(a) Is the diode forward or reverse biased? Explain how you arrived at your answer.(b) Do low-level injection conditions prevail in the quasineutral regions of diode?

Explain how you arrived at your answer.(c) Determine the applied voltage, VA.

Example

pn-junction diode structure used in the discussion of currents. The sketch shows the dimensions and the bias convention. The cross-sectional area A is assumed to be uniform.

Hole current (solid line) and recombining electron current (dashed line) in the quasi-neutral n-region of the long-base diode of Figure 5.5. The sum of the two currents J (dot-dash line) is constant.

Hole density in the quasi-neutral n-region of an ideal short-base diode under forward bias of Va volts.

The current components in the quasi-neutral regions of a long-base diode under moderate forward bias: J(1) injected minority-carrier current, J(2) majority-carrier current recombining with J(1), J(3) majority-carrier current injected across the junction. J(4) space-charge-region recombination current.

(a) Transient increase of excess stored holes in a long-base ideal diode for a constant current drive applied at time zero with the diode initially unbiased. Note the constant gradient at x = xn as time increases from (1) through (5), which indicates a constant injected hole current. (Circuit shown in inset.) (b) Diode voltage VD versus time.

J elec

x

n-region

J = J elec + J h ole

SCL

Minority carrier diffusioncurrent

M ajority carrier diffusionand drift current

Total current

J h ole

Wn–Wp

p-region

J

The total current anywhere in the device is constant.

Just outside the depletion region it is due to the diffusion of minority carriers.


Current-Voltage Characteristics

of a Typical Silicon p-n Junction Quantitative p-n Diode Solution

ExamplesDiode in a circuit


Current flow in a pn junction diode

(b) under reverse bias, only a small number of carriers are available todiffuse across the junction (once within the junction they drift to the other side). With increasing reverse bias the reverse current increases due to tunneling and carrier multiplication.

(a) under equilibrium, both diffusion currents arecancelled by opposing drift currents.

(c) under forward bias, the drift current is slightly reduced but the diffusion current is greatly increased.

(d) current-voltage characteristic

Summary

Built-in-Potential

−

= 1exp0 kT

qVJJ a

η

+=

p

np

n

pn

LpqD

LnqD

J 000

DiodeEquation

))(()(2 0Abi

DA

DAsnp VV

NNNN

qKxxW −

+=+=

ε

Width of Depletion Region

0ερ⋅

=⋅∇SK

E Poisson’s Equation

EX 5.7

• A silicon pn junction dide at T=300K is forward biased. The reverse saturation current is IS=4x 10-14A. Determine the required dide voltage to induce a diode current of ID=4.25mA.

Important Chap3 1 SemiImportant Chap3-1-Semiconductor-1.pdf conductor 1

Documents

Transcript of Important Chap3 1 SemiImportant Chap3-1-Semiconductor-1.pdf conductor 1