Chapter 8 Semiconductor crystals
Transcript of Chapter 8 Semiconductor crystals
1
Chapter Eight Semiconductor crystals
Elements : Group IVC(graphite) 1S22S22P2
Si 1S22S22P63S23P2
Ge 1S22S22P63S23P63d104S24P2
Compounds :
IV-IV : SiC
III-V : GaAs, InSb, GaP, ..Ga 1S22S22P63S23P63d104S24P1
As 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
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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
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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.
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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)
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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
π2 h2 /m
a2 )
kx (2π/a)
(100)
FCC Lattice along ΓX
[100]
(001)
(100)
(101)(011)(110)
(000)
ε
(3π2 h
2 /2m
a2 )
(100)(010)(001)
(110)(011)
(101)
ΓL X
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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 nn
rrr=+
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 nkn
r
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
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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
ψp
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
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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 nkn
r
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.
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)kε(1)k(v k
r
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.
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Real momentum
1C k k
eC)(k
G
2Gk
G
rG)i(kGk
==
==
∑
∑
+
•++rkψ
2Gk
G
2Gk
G
CGk
CG)i(ki
ki
kkpk
+
+
∑+=
∑ +=∇=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(∆P2
GkkGelectronBkelectronB
rrh
rh
rrrrrr ∆•∇∑+∆=•∇=
+−−
( )kCGP∆2
GkkGlattice
rrh
rrrr ∆•∇∑−=
+
Fdtkd rr
h =k∆Jr
hr
=same as for free electronsTherefore,
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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
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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 ,krr
− ∑ = 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( hheeee
rrrεεε −=−=
( )( ) eekhkh v11veh
=−∇−=∇= εε rr
hhThe group velocity of the hole is the same as that of the electron.
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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
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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~2m
ε)K~ε( h+= where εc=λ+U
2
h
2
v K~2m
ε)K~ε( h−=
near the top edge of the 1st band
where εv=λ-U
distance to the zone boundary
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+=
U2λ1K~
2mK~
2m2
22
e
2 hh
1 /U2λ1
mme
+=
−−=
U2λ1K~
2mK~
2m2
22
h
2 hh
1 /U2λ1
mmh
−=
)kε(1)k(v k
r
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)
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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=
∗
( )2y
2x
e
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
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k
ε
k
ε2
22
dkε(k)dm h=∗
m* <0
k
ε
k
ε
m* >0
m* can be determined by cyclotron resonance measurements.
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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
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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 yx
yx
EE
vv
iΕiu
+=
+= ( )Bmq
dtd iuΕuu
−=+τ
( ) ( )( ) ( )y tsinx tcos
y tsinx tcost
t
ωω
ωωω
ω
ooi
o
ooi
o
uueuuEEeEΕ
+==
+==
zB
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
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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
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( ) 1
g
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.
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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
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( )2x
2z
2z
2y
2y
2x
2422 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]
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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
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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 Em22
1)D( −
= ε
πε
h
h
22
Vk m2kE h
−=ε
επ
ε −
= v
3/2
2h
2 Em22
1)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 2
2
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 2
e2
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
Ev
EAEa
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 <<
And
0NnnTk
Eexpn1 N
TkEexp
nn1n d
2
B
d
od
B
d
o
=−+
→≈
+
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 1Tk
EexpnN4
B
d
o
d <<
constant Nn d =≈ Saturation range
The concentration of donor electrons in the conduction band has reached the maximum possible value, equal to the concentration of donor.All donors are ionized.
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<Ed
Eg>>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~Nd
Eg~kBTIntrinsic carriers are excited fromvalence band
45
A semiconductor doped with Na acceptor holesSame results
Low Temperatures, kBT<Ea
2/3
2Bh
o
B
aao
2Tkm2p where
,T2k
EexpNp~p
=
−
hπIntermediate temperatures, Eg>>kBT>Ea
p=Na
High Temperatures, Eg~kBT
−+≈+≈
T2kE
exppNpNpB
goaia
46
Fermi levelAssume all Nd electrons are excited into conduction band
n - p = Nd
Mass action law np = nipi = ni2 =pi
2 p= ni2/n
N-type semiconductors in saturation regime, n~Nd and p=ni2/n
d
2i
d NnnNpn +=+= ( )2
i2dd
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
Ev
EA 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).