1

Chapter Eight Semiconductor crystals

Elements : Group IVC(graphite) 1S22S22P2

Si 1S22S22P63S23P2

Ge 1S22S22P63S23P63d104S24P2

Compounds :

IV-IV : SiC

III-V : GaAs, InSb, GaP, ..Ga 1S22S22P63S23P63d104S24P1As 1S22S22P63S23P63d104S24P3

II-VI : ZnS, CdSe, …

2

TkE

B

gdetermines intrinsic conductivity and intrinsic carrier concentration

(Eg can be obtained by optical absorption)

3

Two types of semiconductors

Indirect gapDirect gap

k0valence band

ε

Eg

conduction band

ε

k0valence band

Eg

conduction band

Evertical

Band edges of valence and conduction bans

at different ks

( Ge[111], Si[100], … )

Band edges (extremes)

at same k

(most compounds)

4

5

Direct absorption process Indirect absorption process

hω=Eg+ hΩ

energy of emitted phononhωg=Eg

6

A sharp threshold

Optical absorption of InSb

Eg=0.23eV

7

Why is your computer chip made of Si, but the laser in your CD player is made of GaAs (GaN in the future)?

Comparison of absorption

Si

Abs

orpt

ion

Energy of light photon

Weak absorption and emission

1.1eV Abs

orpt

ion

Energy of light photon

GaAs

1.5eV

red light

Light emission is related

– very high efficiency in GaAs for excited electron to emit light

– very low efficiency in Si

8

Why is GaN interesting?

Abs

orpt

ion

Energy of light (photon)

GaN

1.5eV

red light

3.4eV

ultraviolet light

After decades of efforts, finally it is possible to make blue light emitter and laser.

Shorter wavelength light focuses to smaller spot implies higher density of information on a CD.

9

Calculated band structure for Ge

Octahedron

a4π

X

L

XL Γ

FCC lattice w/. lattice constant a

In reciprocal space

Origin point Γ (0,0,0)

Point X (2π/a, 0, 0)

Point L (π/a, π/a, π/a)

10

0.0 0.5 1.00

1

2

3

4

5

6

-1.0 -0.8 -0.6 -0.4 -0.2 0.00

1

2

3

4

5

6

7

8

(010) (111)

(111)

FCC Lattice along ΓL

[111]

kx ky kz(π/a)

(110)(101)

(101) (110)

(111)(001)

(111)

(010)

(011)

(011)

(101)

(100)

(001)

(000)

(110)

(010)

ε (2

π2h

2 /ma2

)

kx (2π/a)

(100)

FCC Lattice along ΓX

[100]

(001)

(100)

(101)(011)(110)

(000)

ε

(3π2h

2 /2m

a2)

(100)(010)(001)

(110)(011)

(101)

ΓL X

11

Free electron model(Sommerfeld)

Nearly free electron model(Bloch theorem)

n , k( n is the band index,

hk is the crystal momentum)

k(hk is the momentum)

)k(ε)Gk(ε nnrrr

=+

For a given band n, no simple explicit form.

General property :2m

22kε(k) h=Energy

The mean velocity for e in band n with k

)(1mk(k)v k k

r

h

rhr

rε∇==Velocity)k(ε1(k)v nknr

h

rr∇=

)r(u)Tr(u)er(u)r(ψ

knkn

rkiknkn rrr

rr

rr

rr

rr

=+= •

Ve(r)ψ

rki

k

rr

r

•

=Wave function

12

General remarks about Bloch’s theorem

1. Bloch’s theorem introduces a wavevector k which turns out to play thesame fundamental role in the general problem of motion in a periodic potential as that in free electron model.But k is not proportional to the electronic momentum.

( ))r(u

iekψ

)er(ui

ψi

ψp̂

knrki

kn

rkiknknkn

rhh

rhh

rrr

r

rr

rrr

∇+=

∇=∇=

•

•

In general, ψnk is not an eigenstateof momentum operator.

Dynamical significance of k can be acquired when we consider themotions of Bloch electrons to externally electromagnetic fields.

2. k in Bloch’s theorem can always be confined to the first Brillouin zone.Any k’ out of zone can be back to zone by shifting a displacement G (reciprocal lattice vector). k’=k+G

13

3. For a given n, the eigenstates and eigenvalues are periodic functions of k in reciprocal lattice.

kn,Gkn,

kn,Gkn,

εε )r(ψ)r(ψ

rrr

rrrrr

==

+

+

4. An electron in a level ( band index n and wave number k) has anon-vanishing mean velocity.

)k(ε1)k(v nknr

h

rrr∇= Remarkable!

It asserts that there are stationary levels for Bloch electrons, in spite of the electron with fixed lattice of ions.

Against Drude’s naive picture of electron-ion scattering.

14

)kε(1)k(v kr

h

rrr∇=

In an external electric field E,

The energy gained by the electron in a time interval ∆t

dtEekdh

rr −

= FEedtkd rrr

h =−=

( )k)kε( )∆∆kε(1)Ee(t)v()Ee(Fε

k

krr

r

h

rrrlrr

r

r

∆•∇=

∇•−=∆•−=∆•=∆

setting ∆t→0

In general, equation of motion for an Bloch electron under Lorentz forces

×

∇+== B)kε(1EqF

dtkd

kext

rr

h

rr

h r SI unit

Weak external forces such that band structure still holds.

15

Real momentum

1C k k

eC)(k

G

2Gk

G

rG)i(kGk

==

==

∑

∑

+

•++rkψ

2Gk

G

2Gk

G

CGk

CG)i(ki

ki

kkp̂k

+

+

∑+=

∑ +=∇=r

hrh

hh

where

Under a weak external force F,

latticeelectronBtotal P∆P∆P∆dtFJrrrrr

+∫ === −

Impulse = the change of momentum of the crystal

( )kCGkk∆)P(∆P 2GkkGelectronBkelectronBrr

hr

hrr

rrrr ∆•∇∑+∆=•∇= +−−

( )kCGP∆ 2GkkGlatticerr

hr

rrr ∆•∇∑−= +

Fdtkd rr

h =k∆Jr

hr

=same as for free electronsTherefore,

16

Holes in semiconductorsIn a completely filled band (valence band), no current can flow since electrons are Fermions and obey the Pauli exclusion principle.

The empty states in the valence band are called “holes”.The electrons can “move” if there is an empty state (a hole) available.

A hole acts under the external forces as if it has a positive charge +e.Missing electron = producing hole

ε

17

In a full band : all pairs of states are filled and .

If an electron of wavevector ke is missing, . Alternatively speaking,

a hole of wavevector is produced and .

( )k ,k rr − ∑ = 0kr

∑ −= ekkrr

hkr

eh kkrr

−=

k

εSetting the energy of the top of valence band is zero,

the lower in the band the missing electron lies, the higher the energy of the system.

The band is symmetric :

)k()k()k( hheeeerrr

εεε −=−=

( )( ) eekhkh v11v eh =−∇−=∇= εε rr hhThe group velocity of the hole is the same as that of the electron.

18

How does a hole move?t1 t2 t3

×∇+−= B)k(ε1Ee)(

dtkd

eeke

e

rr

h

rr

h r the equation of a motion for an electron

Applying to a missing electron : creation of a hole

( )

×∇+−=

− B)k(ε1Ee)(dt

kdhhk

hh

rr

h

rr

h r

×∇++= B)k(ε1Ee)(

dtkd

hhkh

h

rr

h

rr

h r

the equation of a motion for a hole

exactly the equation of motion for a particle of positive charge

19

2

2

2 dkε(k)d1

m1

h=∗Effective mass (band mass)

◙ For a free electron ε(k) = h2k2/2m → m*=m

◙ For electrons in a band, their masses depend on band curvature.

( )

±+±≈

U2λ1

2mK~Uλε

22

K~h

2GkK~ −≡

free e

lectro

n

ε

k0 π/a=G/2

1st band

U>0

+

-

λ+U

λ-U

2nd band

λ

near the lower edge of the 2nd band

2

e

2

c K~

2mε)K~ε( h+= where εc=λ+U

2

h

2

v K~

2mε)K~ε( h−=

near the top edge of the 1st band

where εv=λ-U

distance to the zone boundary

20

+=

U2λ1K~

2mK~

2m2

22

e

2 hh

1 /U2λ1

mme

+=

−−=

U2λ1K~

2mK~

2m2

22

h

2 hh

1 /U2λ1

mmh

−=

)kε(1)k(v kr

h

rrr∇=

==

∇=

dtkd

dkε(k)d1

dtkd

dkε(k)d1

dt)kε(d1

dt(k)vd

2

2

22

2k

r

hh

r

h

r

h

rFr

2

2

2 dkε(k)d1

m1

h=∗

Definition of the effective massFrom Newton’s 2nd law

Considering an anisotropic energy surface

νµµν dkdkε(k)d1

m1 2

2h=

∗ where µ and ν are Cartesian coordinates.

reciprocal effective mass tensor (3x3)

21

In three (two) dimensions, constant energy surfaces (lines) are not necessarily spherical (circular), and the effective mass is a tensor:

νµµν dkdkε(k)d1

m1 2

2h=

∗

( )2y2xe

2

yx kk2m)k,(k += h

=

e

e*

m00mm µν

νµµν

δe

2

2 m1

dkdkε(k)d1

m1

==

∗ hQ

εfree electronsIn two dimensions,

The effective mass depends on the curvature of the bands;The flat bands have large effective massesThe curved bands have small effective masses

Near the bottom of a band, m* is positiveNear the top of a band, m* is negative

22

k

ε

k

ε22

2

dkε(k)dm

h=∗

m* 0

m* can be determined by cyclotron resonance measurements.

23

Effective mass in semiconductors

Cyclotron resonance energy surfaces of the conduction and valence bands near the band edgeeB

∗= mcω

where m* is the cyclotron effective mass

0.58--0.99Cu2O

0.0250.390.026InAs

0.0820.50.066GaAs

Light hole (mlh/m)

Heavy hole (mhh/m)

Electron (me/m)

Crystal

2

22

dkε(k)dm

h=∗

k

ε

Eg

direct conduction band

Heavy hole bandLight hole band

e

22

g m2kE h+=ε

h

22

m2kh

−=ε

24

Let B in the z direction and E in the x-y plane

( )

−+×+=

+

=

τvmBvEq

dtPd

dtPd

dtPd

scatteringfield rrrr

rrr

( )

( )BvEmqv

dtdv

BvEmqv

dtdv

xyyy

yxxx

−=+

+=+

τ

τ yxyx

EE

vv

iΕiu

+=

+= ( )Bmq

dtd iuΕuu −=+

τ

( ) ( )( ) ( )ŷ tsinx̂ tcos

ŷ tsinx̂ tcost

t

ωω

ωωω

ω

ooi

o

ooi

o

uueuuEEeEΕ

+==

+==

ẑB

E

Boltzmann transport,

where τ is the relaxation time of momentum

circularly polarized

( )

( )τωω

ωτ

/m q

Bmq1

c iΕiu

iuΕui

oo

ooo

−+−

=

−=

+ The maximum absorption of

electromagnetic energy by semiconductor occurs at the cyclotron frequency .

∗= mqB

cω

1c >>τω

25G.Dresselhaus, A.F. Kip, and C. Kittel, Phys. Rev. 98, 368 (1955)

ω=2.4×1010 Hz

T=4K

Silicon Germanium

26

Energy contours in k-space near the conduction band minimum.

There are six equivalent pockets.

Si

ellipsoids of revolutionConsidering the electrons situated close to (0,0, ko)dispersion relation can be written as,

( )

−+

+=

l

h

mkk

mkk

2

2oz

t

2y

2x

2

εBvedtkd rrr

h ×−=

( )

θ

θθ

θ

sinmBke

dtdk

sinm

kkBecosmBke

dtdk

cosmBke

dtdk

t

yz

oz

t

xy

t

yx

hh

hhh

hh

l

=

−−=

−=

kx

kz

ko

B

θ

Let kx=kxoeiωt, ky=kxoeiωt, kz=ko+kzoeiωt, c

2/1

2t

2

t

2

c meB

mcos

mmsineB =

+=

θθωl

depending on the orientation of Bwhen θ=0, mc=mtwhen θ=900, mc=(mlmt)1/2

t : transversel : longitudinal

27

( ) 1g

e

eV 6.0~5.0Emm

−≈

The perturbation theory of band edges suggests that for a direct gap crystal, the electron effective mass should be proportional to the bandgap Eg

The smaller the bandgap, the smaller the effective mass.

28

Ge 1S22S22P63S23P63d104S24P2

sp3 hybrid : a mixture of the s- and p- levelsTetrahedral bonding

The valence band edge at k=0

P3/2 states : fourfold degeneratemJ : 3/2, 1/2, -1/2, -3/2

P1/2 states : doubly degeneratemJ : 1/2, -1/2

Energy difference = Δ

Δ: an energy corresponding tospin-orbit interaction

29

( )2x2z2z2y2y2x2422 kkkkkkCkBAk +++±=ε ]2/[ CB,A, 2 mh

L LLLL Γ ΓX XUK UK

1.59

0.92

ml(m)

6.2

13.2

4.87

C

0.341

0.29

0.044

Δ(eV)

-4.5-6.98GaAs

0.828.48-13.4Ge

0.190.68-4.29Si

mt(m)BAcrystal

[111] axis

[100] axis

[111] [100]

30

Intrinsic carrier concentrationIn contrast to metallic conductivity, the conductivity of semiconductor

is strongly temperature dependent. -- “free” charges must be thermally excited and overcome Eg

charge carrier concentrations n & p have a strong dependence on T.

Semiconductor are called “intrinsic”,when “free” electrons and holes can be created only by electronic

excitations from the valence band to the conduction band.

εεε

εεε

dT),()f(Dp

dT),()f(Dn

v

hv

ec

∫=

∫=

∞−

∞

E

EcElectron concentration in the conduction band

Hole concentration in the valence band

−≈

−−+=−=

−−≈

+−=

Tkexp

T]k/)(exp[11)(f1)(f

Tkexp

1T]k/)exp[(1)(f

BBeh

BBe

µεµε

εε

µεµε

εFermi-Dirac distribution

μ-ε>> kBT

31

In the parabolic approximation (for simplicity),

The energy of an electron in the conduction band,

Density of states,

The energy of a hole in the valence band,

Density of states,

e

22

ck m2kE h+=ε

c

3/2

2e

2 Em2

21)D( −

= ε

πε

h

h

22

Vk m2kE h−=ε

επ

ε −

= v

3/2

2h

2 Em2

21)D(

h

∫

−−

=

∫

−−

=∫=

∞

∞∞

c

cc

E Bc

B

3/2

2e

2

E Bc

3/2

2e

2E

ec

dTk

expETk

exp2m2

1

dTk

expE2m2

1dT),()f(Dn

εεεµπ

εεµεπ

εεε

h

h

−∫

−∞

TkEexp*d

Tkexp

B

c

0 B

εεε

−

=

TkEexp

2Tkm2n

B

c2/3

2Be µ

πh

32

∫

−

−

=

∫

−−

=∫=

∞−

∞−∞−

v

vv

E

Bv

B

3/2

2h

2

E

Bv

3/2

2h

2

E

hv

dTk

expETk

exp2m2

1

dTk

expE2m2

1dT),()f(Dp

εεεµπ

εµεεπ

εεε

h

h

∫

−∞

TkEexp*d

Tkexp

B

v

0 B

εεε

−

=

TkEexp

2Tkm2p

B

v2/3

2Bh µ

πh

( )

−

=

−

−

=

TkE

expmm2

Tk4

TkEexp

2Tkm2

TkEexp

2Tkm2np

B

g2/3he

3

2B

B

v2/3

2Bh

B

c2/3

2Be

h

hh

π

µπ

µπ

=constant depends on material and temperature

The law of mass action. Eg=Ec-Ev Independent of EF, (µ)

33

For an intrinsic semiconductor n=p,

( )

−

==

T2kE

expmm2

Tk2pnB

g4/3he

2/3

2B

iihπ

Taking this value back to n(µ) or p(µ), we can obtain µ

+=

e

hBg m

m Tk43E

21 nlµFermi level

At T=0, µ lies half-way between the conduction and valence bands.

As T increases, µ moves toward the band with smaller effective mass

µ does not go far from mid-gap when mh≈me

34

Equal densities of states in the conduction and valence bands

Different densities of states in the conduction and valence bands

Intrinsic semiconductor : # of holes = # of electrons (n=p)

35

When me=mh, then µ=0.5Eg Fermi level is in the middle of the gap.

Intrinsic mobility µ

Electrical conductivity

=Edv

h

h2

e

e2

mpe

mne ττ

+=he pene µµσ +=

18003600Ge

12001800Diamond

4801350Si

600550PbS

45030000InAs

3008000GaAs

µh(cm2/Vs)µe(cm2/Vs)crystal The hole mobilities are typically smaller than the electron mobilities because of the occurrence of band degeneracy at the valence band edge at the zone center, thereby making possible interband scattering processes that reduce the mobility.

T=300K

36

Impurity conductivityDoping : addition of impurities to the crystal

(1)Donors – Group of V such as P, As, Sb

substitutional impurity for semiconductor

each dopant atom contribute an electron

(2)Acceptors – Group of III such as Al, Ga, In

attract electrons from valence band of semiconductor

create a hole per atom

N-type

P-type

Where do electrons / holes of the dopants go? free or boundLow T : bound

High T : free kBT > Ed (electron), Ea (hole)Acceptor activated energy

Donor activated energy

37

Activated energy – From Bohr model

λππε

nr 2 and r

mvr4

e 22

2

===o

F

22222

4

n

2

2

n

n13.6eV

n1

32meE

nanme

4r

−=−=

==

h

h

o

oo

επ

πε

Ionization energy 13.6eV

Hydrogen atom

N-doped Silicon P-doped Silicon

38

The fifth valence electron of P atom is not required for bounding and is thus, only weakly bound. The binding energy can be estimated by treating the system as a hydrogen atom embedded in a dielectric.

λππε

nr 2 and rvm

r 4e 2e

2

2

===F

==

==

mm13.6eV

n1

32meE

mman

em 4a

e222222

e4

d

eo2

e

2

κεκπ

κπκε

h

h

o

od

oκεε =

Dielectric constant

m of instead me~ao(10)(10)

~ 13.6eV(10-2)(10-1)

Donor

Electron energy

Ec

Ev

EDEd

Donor levelSi 45 54 43

Ge 13 14 10

P As Sb

Ionization energies Ed [meV]

11.7

15.8

κ

39

The valence –three Boron (B) accepts an electron from the Si lattice. The hole that is thereby created in the valence band orbits around the negatively charged impurity.

The Bohr model applies qualitatively for holes just as for electrons, but the degeneracy at the top of the valence band complicates the effective mass problem.

Electron energy

Ec

EvEAEa

Acceptor level Si 45 57 65 157

Ge 10.4 10.2 10.8 11.2

B Al Ga In

Ionization energies Ea [meV]

40

In a doped semiconductor,

an electron in the conduction band can originate either fromthe valence band or from the ionization of a donor;

a hoe in a valence band may correspond either tothe electron in the conduction band or to the negatively charged acceptor.

Density of doped donor Nd = Ndo + Nd+(ionized) donor

Density of doped acceptor Na = Nao + Na-(ionized) acceptor

Electron energy E

n

p

Nd+

_ _

Na-

++

( )[ ]

( )[ ] 1Tk/EEexp1NN

1Tk/EEexp1NN

BAFa

oa

BFDd

od

+−=

+−=

(neutral)

(neutral)

Neutrality condition n + Na- = p + Nd+

41

For pure N-type semiconductor : only donors are available2/3

2hB

o

2/3

2eB

o 2Tmk2p and

2Tmk2n where

=

=

hh ππ++=

−=

dodd

B

cFo

NNN

TkEEexpnn

n = Nd+ + p

For the simple case Nd+>> ni therefore, n ~ Nd+ = Nd - Ndo

( )[ ] ( )[ ]

−−+

=

+−

−=−≈Tk/EEexp1

1N1Tk/EEexp

11NNNnBFD

dBFD

dodd

=

Tk

EexpTk

Eexpnn

B

F

B

c

o

[ ] [ ]

−+≈

Tk/EexpT/kEexpnn1

1NnBDBc

o

d

gB ETk

42

++−

−=

TkEexp

nN411

TkEexp

2n n

B

d

o

d

B

dosolution

At low temperatures, such that

T2k

EexpNnn B

ddo

−≈

1Tk

EexpnN4

B

d

o

d >>

Freeze-out range

A sufficiently large number of donors still retain their valence electrons, i.e. are not ionized.

At the intermediate temperatures, such that 1TkEexp

nN4

B

d

o

d

43

At high temperatures, such that gB ETk ≈

−+≈+≈

T2kE

expnNnNnB

godid Intrinsic range

The concentration of electrons excited from the valence band across Egincreases and eventually outweighs the electron density due to donors.

It behaves as an intrinsic semiconductor.

N-doped Ge with a P concentration Nd=1×1013cm-3 : Ed~0.012eVIntrinsic Ge : ni=2.4×1013cm-3 (300K) : Eg~0.67eV

1Tk

EexpnN4 *

B

d

o

d =

T*= K

g**

B ETk ≈ T**=7800K

“Law of mass action” is also valid for the doped semiconductor

( )

−=

−

=

TkE

exppnTk

Eexpmm

2Tk4np

B

goo

B

g2/3he

3

2B

hπ

44

A semiconductor doped with Nd donor electrons

kBT>kBT>Ed

2/3

2Be

o

B

ddo

2Tkm2n where

,T2k

EexpNn~n

=

−

hπ

All carriers are excited (Saturation)

−+≈+≈

T2kE

expnNnNnB

godid

Dopant carriers are thermally excited to conduction band

n~NdEg~kBTIntrinsic carriers are excited fromvalence band

45

A semiconductor doped with Na acceptor holesSame results

Low Temperatures, kBT>kBT>Ea

p=Na

High Temperatures, Eg~kBT

−+≈+≈

T2kE

exppNpNpB

goaia

46

Fermi levelAssume all Nd electrons are excited into conduction band

n - p = NdMass action law np = nipi = ni2 =pi2 p= ni2/n

N-type semiconductors in saturation regime, n~Nd and p=ni2/n

d

2i

d NnnNpn +=+= ( )2i2dd

2id

2

4nNN21n

0nnNn

++=

=−−

n, p

ni=pi p~1/Nd

n~Nd

Nd

intrinsic0Eoffset TkE

expn)n(N

v

B

god

=

−=

µ

+=

o

dBg n

)n(NTkE nlµ

−=

d

oBg N

nTkE nlµsaturation

47

Electron energy

Ec

Ev

Electron energy

Ec

Ev

ED ED

0.5Eg

Ed

at low temperatures at high temperatures

N-type semiconductor

µ

µ

Electron energy

Ec

Ev

Electron energy

Ec

EvEA EAµ

µ0.5Eg

Ea

P-type semiconductor

48

Saturation range (Ed, Ea < kBT < Eg)

N-type : n ≈ Nd >> p dominated by electrons

electrical conductivity >0

Hall coefficienthed epeN µµσ +≈

eN1

en1R

dH −≈−≈

0

P-type : p ≈ Na >> n dominated by holes

electrical conductivity >0

Hall coefficienteha eneN µµσ +≈

eN1

ep1R

aH ≈≈

0

49

intrinsic

E.M. Conwell, Proc. IRE 40, 1327 (1952).

phonon

Data was obtained using the Hall effect.

50

E.M. Conwell, Proc. IRE 40, 1327 (1952).