Chap 1. Introduction to Solid-State Electronics

Advanced Semiconductor Physics Chap. 1

Instructor: Pei-Wen LiDept. of E. E. NCU

1

Chap 1. Introduction to Solid-State Electronics

uPrinciples of Quantum MechanicsuSchrödinger’s Wave EquationuApplication of Schrödinger’s Wave Equation



2

Introduction

u In solids, there are about 1023 electrons and ions packed in a volume of 1 cm3. The consequences of this highly packing density :– Interparticle distance is very small: ~2x10-8 cm.⇒the instantaneous position and velocity of the particle are no longer

deterministic. Thus, the electrons motion in solids must be analyzed by a probability theory. Quantum mechanics ⇔Newtonian mechanicsSchrodinger’s equation: to describe the position probability of a particle.



3

Introduction

– The force acting on the j-th particle comes from all the other 1023-1 particles.

– The rate of collision between particles is very high, 1013

collisions/sec⇒average electron motion instead of the motion of each electron at a

given instance of time are interested. (Statistical Mechanics)equilibrium statistical mechanics:

Fermi-Dirac quantum-distribution ⇔Boltzmann classical distribution



4

Principles of Quantum Mechanics

u Principle of energy quantau Wave-Particle duality principleu Uncertainty principle



5

Energy Quanta

u Consider a light incident on a surface of a material as shown below:

u Classical theory: as long as the intensity of light is strong enough ⇒photoelectrons will be emitted from the material.

u Photoelectric Effect: experimental results shows “NOT”.u Observation:

– as the frequency of incident light ν < νo: no electron emitted.– as ν > νo:at const. frequency, intensity↑, emission rate↑, K.E. unchanged.

at const. intensity, the max. K. E. ∝ the frequency of incident light.



6

Quanta and Photon

u Planck postulated that thermal radiation is emitted from a heated surface in discrete energy called quanta. The energy of these quanta is given by

E = hν, h = 6.625 x 10-34 J-sec (Planck’s constant) u According to the photoelectric results, Einstein suggested that

the energy in a light wave is also contained in discrete packetscalled photon whose energy is also given by E = hν.The maximum K.E. of the photoelectron is

Tmax = ½mv2 = hν - hνo

u The momentum of a photon, p = h/λ



7

Wave-Particle Duality

u de Broglie postulated the existence of matter waves. He suggested that since waves exhibit particle-like behavior, then particles should be expected to show wave-like properties.

u de Broglie suggested that the wavelength of a particle is expressed as

λ = h /p, where p is the momentum of a particle u Davisson-Germer experimentally proved de Broglie postulation

of “Wave Nature of Electrons”.



8

Davisson-Germer Experiment

u Consider the experimental setup below:

u Observation: – the existence of a peak in the density of scattered electrons can be

explained as a constructive interference of waves scattered by the periodic atoms.

– the angular distribution of the deflected electrons is very similar to an interference pattern produced by light diffracted from a grating.



9

Conclusion

u In some cases, EM wave behaves like particles (photons) and sometimes particles behave as if they are waves.⇒Wave-particle duality principle applies primarily to SMALLparticles, e.g., electrons, protons, neutrons.For large particles, classical mechanics still apply.



10

Uncertainty Principle

u Heisenberg states that we cannot describe with absolute accuracy the behavior of the subatomic particles.

1. It is impossible to simultaneously describe with the absolute accuracy the position and momentum of a particle.

∆p ∆x ≥ ħ. (ħ = h/2π = 1.054x10-34 J-sec)2. It is impossible to simultaneously describe with the absolute

accuracy the energy of a particle and the instant of time the particle has this energy. ∆E ∆t ≥ ħ

u The uncertainty principle implies that these simultaneous measurements are in error to a certain extent. However, ħ is very small, the uncertainty principle is only significant for small particles.



11

Schrodinger’s Wave Equation

u Based on the principle of quanta and the wave-particle duality principle, Schrodinger’s equation describes the motion of electrons in a crystal.

u 1-D Schrodinger’s equation,

u Where Ψ(x,t) is the wave function, which is used to describe the behavior of the system, and mathematically can be a complex quantity.

u V(x) is the potential function.u Assume the wave function Ψ(x,t) = ψ(x)φ(t), then the

Schrodinger eq. becomes

ttxjtxxV

xtx

m ∂Ψ∂

=Ψ+∂Ψ∂

⋅− ),(),()(),(2 2

22

hh

ttxjtxxV

xxt

m ∂∂

=+∂

∂− )()()()()()()(2 2

22 φψφψ

ψφ h

h



12

Schrodinger’s Wave Equation

u where E is the total energy, and the solution of the eq. isand the time-indep. Schrodinger equation can be written as

u The physical meaning of wave function: – Ψ(x,t) is a complex function, so it can not by itself represent a real

physical quantity.– |Ψ2(x,t)| is the probability of finding the particle between x and x+dx

at a given time, or is a probability density function.– |Ψ2(x,t)|= Ψ(x,t) Ψ*(x,t) =ψ(x)* ψ(x) = |ψ(x)|2 -- indep. of time

Ett

tjxV

xx

xm=

∂∂

=+∂

∂− )()(

1)()()(

12 2

22 φφ

ψψ

hh

tEjet )/()( h−=φ

0)())((2)(22

2

=−+∂

∂ xxVEmx

xψ

ψh



13

Boundary Conditions

1. since |ψ(x)|2 represents the probability density function, then for a single particle, the probability of finding the particle somewhere is certain.If the total energy E and the potential V(x) are finite everywhere,

2. ψ(x) must be finite, single-valued, and continuous.3. ∂ψ(x)/∂x must be finite, single-valued, and continuous.

1)(2

=∫∞

∞−dxxψ



14

Applications of Schrodinger’s Eq.

u The infinite Potential Well

u In region I, III, ψ(x) = 0, since E is finite and a particle cannot penetrate the infinite potential barriers.

u In region II, the particle is contained within a finite region of space and V = 0. 1-D time-indep. Schrodinger’s eq. becomes

u the solution is given by

0)(2)(22

2

=+∂

∂ xmEx

xψ

ψh

2212 where,sincos)(h

mEKKxAKxAx =+=ψ



15

Infinite Potential Well

u Boundary conditions:1. ψ(x) must be continuous, so that ψ(x = 0) = ψ(x = a) = 0

⇒A1 = A2sinKa ≡ 0 ⇒ K = nπ/a, where n is a positive integer.2.

So the time-indep. Wave equation is given by

u The solution represents the electron in the infinite potential well is in a standing waveform. The parameter K is related to the total energy E, therefore,

1)(2

=∫∞

∞−dxxψ

aAKxdxA

a 21sin 22

0

22 =⇒=⇒ ∫

...3,2,1 where)sin(2)( == na

xna

x πψ

integer positive a isn where2 2

222

manEE n

πh==



16

Infinite Potential Well

u That means that the energy of the particle in the infinite potential well is “quantized”. That is, the energy of the particle can only have particular discrete values.



17

The Potential Barrier

u Consider the potential barrier function as shown:u Assume the total energy of an incident particle

E < Vo, as before, we could solve the Schrodinger’s equations in each region, and obtain

u We can solve B1, A2, B2, and A3 in terms of A1 from boundary conditions: – B3 = 0 , once a particle enters in region III, there is no potential

changes to cause a reflection, therefore, B3 must be zero.– At x = 0 and x = a, the corresponding wave function and its first

derivative must be continuous.

xjKxjK eBeAx 11111 )( −+=ψ

2221)(2 and 2 where

hh

EVmKmEK o −==

)( 22 222xKxK eBeAx −+=ψxjKxjK eBeAx 11

333 )( −+=ψ



18

The Potential Barrier

u The results implies that there is a finite probability that a particle will penetrate the barrier, that is so called “tunneling”.

u The transmission coefficient is defined byu If E<<Vo,

u This phenomenon is called “tunneling” and it violates classical mechanics.

*11

*33

AAAAT

⋅⋅

=

( )aKVE

VET

oo22exp116 −

−

≅



19

One-Electron Atom

u Consider the one-electron atom potential function due to the coulomb attraction between the proton and electron:

u Then we can generalize the Schrodinger’s eq. to 3-D in spherical coordinates:

u Assume the solution to the equation can be written as

u Then the solution Φ is of the form, Φ = ejmφ, where m is an integer.

rerV

oπε4)(

2−=

0))((2)(sinsin1

sin1)(1

2222

2

222

2 =−+∂∂

⋅∂∂

⋅+∂∂

⋅+∂∂

∂∂

⋅ ψθψ

θθθφ

ψθ

ψ rVEmrrr

rrr

o

h

)()()(),,( φθφθψ Φ⋅Θ⋅= rRr



20

One-Electron Atom

u Similarly, we can generate two additional constants n and l for the variables θ and r. n, l, and m are known as quantum numbers (integers)

u , each set of quantum numbers corresponds to a quantum state which the electron may occupy.

u The solution of the wave equation is designated by ψnlm. For the lowest energy state (n=1, l=0, m=0),

u The electron energy E is quantized,

0,...,1,0,...,3,2,1

,...3,2,1

−=

−−−==

llmnnnl

n

angstrom 529.0 where11 /2/3

100 =

⋅= −

oar

o

aea

o

πψ

( ) 222

4

24 nemE

o

on

hπε−

=



21

One Electron Atom

u The probability density function, or the probability of finding the electron at a particular distance form the nucleus, is proportional to ψ100ψ*100 and also to the differential volume of the shell around the nucleus.

u The electron is not localized at a given radius.



22

Two atoms

u The energy of a bound electron is quantized. And for the one-electron atom, the probability of finding the electron at a particular distance from the nucleus is not localized at a givenradius.

u Consider two atoms that are in close proximity to each other. The wave functions of the two atom electrons overlap, which means that the two electrons will interact. This interaction results in the discrete quantized energy level splitting into two discrete energy levels.



23

Energy Band Formation in Solids

u Consider a regular periodic arrangement of atoms in which each atoms contains more than one electron. If the atoms are initially far apart, the electrons in adjacent atoms will not interact andwill occupy the discrete energy levels.

u If the atoms are brought closer enough, the outmost electrons will interact and the energy levels will split into a band of allowed energies.



24

Formation of Energy Bands

u the Bond Model

uEnergy Band



25

Metal, Semiconductor, and Insulator

Insulator Semiconductor Metal



26

E-k Diagram of a semiconductor

T = 0 KAll electrons are confined within valence bands

T > 0 KSome electrons jump to conduction bands by gaining thermal energy



27

Concept of Hole



28

Hole in the Energy Bands



29

Energy Band

u Valence Band: – In all cases the valence-band maximum occurs at the zone center,

at k = 0– is actually composed of three subbands. Two are degenerate at k =

0, while the third band maximizes at a slightly reduced energy. The k = 0 degenerate band with the smaller curvature about k = 0 is called “heavy-hole” band, and the k = 0 degenerate band with the larger curvature is called “light-hole” band. The subbandmaximizing at a slightly reduced energy is the “split-off” band.

– Near k = 0 the shape and the curvature of the subbands is essentially orientation independent.

u Conduction band:– is composed of a number of subbands. The various subbands

exhibit localized and absolute minima at the zone center or along one of the high-symmetry directions.



30

E-k diagram of Si, Ge, GaAs



31

E-k diagram of GaAs



32

Conduction Band of Si, Ge, and GaAs

u In Ge the conduction-band minimum occurs right at the zone boundary along <111> direction. ( there are 8 equivalent conduction-band minima.)

u The Si conduction-band minimum occurs at k~0.9(2π/a) from the zone center along <100> direction. (6 equivalent conduction-band minima)

u GaAs has the conduction-band minimum at the zone center directly over the valence-band maximum. Morever, the L-valley at the zone boundary <111> direction lies only 0.29 eV above the conduction-band minimum. Even under equilibrium, the L-valley contains a non-negligible electron population at elevated temp. The intervalley transition should be taken into account.

u Direct bandgap: the valence band maximum and the conduction band minimum both occur at k = 0. Therefore, the transition between the two allowed bands can take place without change in crystal momentum.



33

Effective Mass

u In 3-D crystals the electron acceleration arising from an applied force is analogously by

where

uFor GaAs, , so mij = 0 if i≠j, and

therefore, we can define mii=me*, that is the the effective mass

tensor reduces to a scalar, giving rise to an orientation-indep. equation of motion like that of a classical particle.

Fmdt

dv⋅= *

1

=−−−

−−−

−−−

111

111

111

*1

zzzyzx

yzyyyx

xzxyxx

mmmmmmmmm

mzyxji

kkEm

jiij ,,,.. 1 2

21 =

∂∂∂

=−

h

)( 222zyxc kkkAEE ++=−

2111 2

h

Ammm zzyyxx === −−−

)(2

2222

2

zyxe

c kkkm

EE ++=−⇒h



34

Effective Mass

u For Si and Ge: E-Ec = Ak12+B(k2

2+k32)

so mij = 0 if i≠j, and

u Because m11 is associated with the k-space direction lying along the axis of revolution, it is called the longitudinal effective mass ml*. Similarly, m22 = m33, being associated with a direction perpendicular to the axis of revolution, is called the transverseeffective mass mt*.

211 2

h

Bmm zzyy == −−,22

1

h

Amxx =−

)(22

23

222

2212

2

kkm

km

EEtl

c ++=−⇒hh



35

Effective Mass of Si, Ge, and GaAs



36

Density of State Function

u To calculate the electron and hole concentrations in a material,we must determine the density of these allowed energy states as a function of energy.

u Electrons are allowed to move relatively freely in the conduction band of a semiconductor but are confined to the crystal.

u To simulate the density of allowed states, consider an appropriate model: A free electron confined to a 3-D infinite potential well, where the potential well represents the crystal.

u The potential of the well is defined as V(x,y,z) = 0 for 0<x<a, 0<y<a, 0<z<a, and V(x,y,z) = ∞elsewhere

u Solving the Schrodinger’s equation, we can obtain

++=++==⇒ 2

22222222

2 )(2a

nnnkkkkmEzyxzyx

πh



37


u The volume of a single quantum state is Vk =(π/a)3, and the differential volume in k-space is 4πk2dk

u Therefore, we can determine the density of quantum states in k-space as

– The factor, 2, takes into account the two spin states allowed for each quantum state; the next factor, 1/8, takes into account that we are considering only the quantum states for positive values of kx, ky, and kz.

32

2

3

24812)( adkk

a

dkkdkkgT ⋅=

=

ππ

π



38


u The density of states in conduction band is modified as

and the density of states in valence band is modified as

cnc EEmh

Eg −⋅⋅= 23*

3 )2(4)( π

vpv EEmh

Eg −⋅⋅= 23*

3 )2(4)( π

mn* and mp

* are the electron and hole density of states effective masses. In general, the effective mass used in the density of states expression must be an average of the band-structure effective masses.



39

Density of States Effective Mass

u Conduction Band--GaAs: the GaAs conduction band structure is approximately spherical and the electronss within the band are characterized by a single isotropic effective mass, me

*, ⇒

u Conduction Band--Si, Ge: the conduction band structure in Siand Ge is characterized by ellipsoidal energy surfaces centered, respectively, at points along the <100> and <111> directions in k-space.

u Valence Band--Si, Ge, GaAs: the valence band structures are al characterized by approximately spherical constant-energy surfaces (degenerate).

( )( ) ...Ge 4

...Si 63

13

2

31

32

2***

2***

tln

tln

mmm

mmm

=

=

...GaAs**en mm =

( ) ( )[ ] 32

23

23 ***

lhhhp mmm +=



40

Density of States Effective Mass



41

Statistics Mechanics

u Fermi-Dirac distribution function gives the probability that a quantum state at the energy E will be occupied by an electron.

u the Fermi energy (EF) determine the statistical distribution of electrons and does not have to correspond to an allowed energy level.

u At T = 0K, f(E < EF) = 1 and f(E >EF ) = 0, electrons are in the lowest possible energy states so that all states below EF are filled and all states above EF are empty.

)exp(1

1)(

kTEEEf

F−+

=



42

Fermi-Dirac Distribution, at T=0K



43

Fermi-Dirac Distribution, as T > 0 K

u For T > 0K, electrons gain a certain amount of thermal energy so that some electrons can jump to higher energy levels, which means that the distribution of electrons among the available energy states will change.

u For T > 0K, f(E = EF) = ½



44

Boltzmann Approximation

u Consider T >> 0K, the Fermi-Dirac function could be approximated by

which is known as the Maxwell-Boltzmann approximation.

−−

≈−

+=

kTEE

kTEEEf F

F

)(exp)exp(1

1)(



45

Equilibrium Distribution of Electrons and Holes

u The distribution of electrons in the conduction band is given bythe density of allowed quantum states times the probability thata state will be occupied.

The thermal equilibrium conc. of electrons no is given by

u Similarly, the distribution of holes in the valence band is given by the density of allowed quantum states times the probability that a state will not be occupied by an electron.

u And the thermal equilibrium conc. Of holes po is given by

)()()( EfEgEn c=

)](1)[()( EfEgEp v −=

)()(∫∞

=cE co EfEgn

∫ ∞−−= vE

vo EfEgp )](1)[(



46

Equilibrium Distribution of Electrons and Holes



47

The no and po eqs.

u Recall the thermal equilibrium conc. of electrons

u Assume that the Fermi energy is within the bandgap. For electrons in the conduction band, if Ec-EF >>kT, then E-EF>>kT, so the Fermi probability function reduces to the Boltzmannapproximation,

u Then

u We may define , (at T =300K, Nc ~1019 cm-3), which is called the effective density of states function in the conduction band

)()(∫∞

=cE co EfEgn

kTEEEf F )]([exp)( −−

≅

( )

−−

=

−−

−= ∫∞

kTEE

hkTmdE

kTEEEE

hmn FcnF

cEn

oc

)(exp22)(exp2423

2

*

3

23* ππ

23

2

*22

=

hkTmN n

cπ



48

The no and po eqs.

u The thermal equilibrium conc. of holes in the valence band is given by

u For energy states in the valence band, E<Ev. If (EF-Ev)>>kT,

u Then,

u We may define , (at T =300K, Nv ~1019 cm-3), which is called the effective density of states function in the valence band

∫ ∞−−= vE

vo EfEgp )](1)[(

kTEEEf F )]([exp)(1 −−

≅−

( )

−−

=

−−

−= ∫ ∞− kTEE

hkTm

dEkT

EEEEhm

p vFpFv

E po

v )(exp2

2)(exp24

23

2

*

3

23* ππ

23

2

*22

=

hkTm

N pv

π



49

nopo product

u The product of the general expressions for no and po are given by

⇒ for a semiconductor in thermal equilibrium, the product of no and po is always a constant for a given material and at a given temp.

u Effective Density of States Function

−=

−−

⋅

−−

=kTE

NNkT

EENkT

EENpn gvc

vFv

Fccoo exp)(exp )(exp



50

Intrinsic Carrier Concentration

u For an intrinsic semiconductor, the conc. of electrons in the conduction band, ni, is equal to the conc. of holes in the valence band, pi.

u The Fermi energy level for the intrinsic semiconductor is calledthe intrinsic Fermi energy, EFi.

u For an intrinsic semiconductor,

u For an given semiconductor at a constant temperature, the value of ni is constant, and independent of the Fermi energy.

−−

=

−−

==

kTEEN

kTEE

hkTmnn Fic

cFicn

io)(exp)(exp22

23

2

*π

−−

=

−−

==

kTEEN

kTEE

hkTm

pp vFiv

vFipio

)(exp)(exp2

223

2

*π

energy bandgap theis where,exp2g

gvci E

kTE

NNn

−=⇒



51

Intrinsic Carrier Conc.

u Commonly accepted values of ni at T = 300 K

Silicon ni = 1.5x1010 cm-3

GaAs ni = 1.8x106 cm-3

Germanium ni = 1.4x1013 cm-3



52

Intrinsic Fermi-Level Position

u For an intrinsic semiconductor, ni = pi,

u Emidgap =(Ec+Ev)/2: is called the midgap energy. u If mp

* = mn*, then EFi = Emidgap (exactly in the center of the

bandgap)u If mp

* > mn*, then EFi > Emidgap (above the center of the bandgap)

u If mp* < mn

*, then EFi < Emidgap (below the center of the bandgap)

)ln(43)(

21)ln(

43)(

21

])(exp[])(exp[

*

*

n

pvc

c

vvcFi

vFiv

Ficc

mm

kTEENNkTEEE

kTEEN

kTEEN

++=++=⇒

−−=

−−⇒



53

Dopant and Energy Levels



54

Acceptors and Energy Levels



55

Ionization Energy

u Ionization energy is the energy required to elevate the donor electron into the conduction band.



56

Extrinsic Semiconductor

u Adding donor or acceptor impurity atoms to a semiconductor will change the distribution of electrons and holes in the material, and therefore, the Fermi energy position will change correspondingly.

u Recall

−−

=

−−

=kT

EENkT

EENn vFiv

Ficci

)(exp)(exp

−+−−

=

−−

=kT

EEEENkT

EENn FiFFicc

Fcco

)()(exp)(exp

−+−−

=

−−

=kT

EEEENkT

EENp FivFiFv

vFvo

)()(exp)(exp

−−

=

−

=⇒kT

EEnpkT

EEnn FiFio

FiFio

)(exp and exp



57

Extrinsic Semiconductor

u When the donor impurity atoms are added, the density of electrons is greater than the density of holes, (no > po) ⇒ n-type;EF > EFi

u When the acceptor impurity atoms are added, the density of electrons is less than the density of holes, (no < po) ⇒ p-type; EF < EFi



58

Degenerate and Nondegenerate

u If the conc. of dopant atoms added is small compared to the density of the host atoms, then the impurity are far apart so that there is no interaction between donor electrons, for example, inan n-material.⇒nondegenerate semiconductor

u If the conc. of dopant atoms added increases such that the distance between the impurity atoms decreases and the donor electrons begin to interact with each other, then the single discrete donor energy will split into a band of energies. ⇒EF move toward Ec

u The widen of the band of donor states may overlap the bottom of the conduction band. This occurs when the donor conc. becomes comparable with the effective density of states, EF ≥ Ec

⇒degenerate semiconductor



59

Degenerate and Nondegenerate



60

Statistics of Donors and Acceptors

u The probability of electrons occupying the donor energy state was given by

where Nd is the conc. of donor atoms, nd is the density of electrons occupying the donor level and Ed is the energy of the donor level. g =2 since each donor level has two spin orientation, thus each donor level has two quantum states.

u Therefore the conc. of ionized donors Nd+ = Nd –nd

u Similarly, the conc. of ionized acceptors Na- = Na –pa, where

factor degeneracy: ,)exp(11

g

kTEE

g

NnFd

dd −

+=

GaAs and Siin levelacceptor for the 4 ,)exp(11

=−

+= g

kTEE

g

NpaF

aa



61

Complete Ionization

u If we assume Ed-EF>> kT or EF-Ea >> kT (e.g. T= 300 K), then

that is, the donor/acceptor states are almost completely ionized and all the donor/acceptorimpurity atoms have donated an electron/holeto the conduction/valence band.

ddddFd

dd NnNNkT

EENn ≅−=⇒

−−

≈ +)(exp2

aaaaaF

aa NpNNkT

EENp ≅−=⇒

−−

≈ −)(exp4



62

Freeze-out

u At T = 0K, no electrons from the donor state are thermally elevated into the conduction band; this effect is called freeze-out.

u At T = 0K, all electrons are in their lowest possible energy state; that is for an n-type semiconductor, each donor state must contain an electron, therefore, nd = Nd or Nd

+ = 0, which means that the Fermi level must be above the donor level.



63

Charge Neutrality

u In thermal equilibrium, the semiconductor is electrically neutral. The electrons distributing among the various energy states creating negative and positive charges, but the net charge density is zero.

u Compensated Semiconductors: is one that contains both donor and acceptor impurity atoms in the same region. A n-type compensated semiconductor occurs when Nd > Na and a p-type semiconductor occurs when Na > Nd.

u The charge neutrality condition is expressed by

where no and po are the thermal equilibrium conc. of e- and h+

in the conduction band and valence band, respectively. Nd+ is

the conc. Of positively charged donor states and Na- is the conc.

of negatively charged acceptor states.

+− +=+ doao NpNn



64

Compensated Semiconductor



65


u If we assume complete ionization, Nd+ = Nd and Na

- = Na, then

u If Na = Nd = 0, (for the intrinsic case), ⇒no = po

u If Nd >> Na, ⇒no = Nd

u If Na > Nd, is used to

calculate the conc. of holes in valence band

o

iodoao n

npNpNn2

recall , =+=+

( ) 22

222

22

0)(

iadad

o

ioadodn

iao

nNNNNn

nnNNnNnnNn

+

−

+−

=⇒

=−−−⇒+=+

( ) 22

22 idada

o nNNNNp +

−

+−

=⇒



66




67

Position of Fermi Level

u The position of Fermi level is a function of the doping concentration and a function of temperature, EF(n, p, T).

u Assume Boltzmann approximation is valid, we have

( ) ( )

ln and lnor

ln and ln

exp and exp

=−

=−

=−

=−⇒

−−=

−−=

i

ovFi

i

oFiF

o

vvF

o

cFc

vFvo

Fcco

npkTEE

nnkTEE

pNkTEE

nNkTEE

kTEENpkT

EENn



68

EF(n, p, T)



69

EF(n, p, T)

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Chap 1. Introduction to Solid-State Electronics

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