Semiconductor Industry:
Increasing miniaturization (Moore’s Law) is leading to Nanoelectronics.
Semiconductors:
(a)
(b)
1. Crystal Properties
Fig. 1 Relation of the external form of
crystals to the form of the elementary
building blocks. The building blocks are
identical in (a) and (b), but different crystal
faces are developed. (c) Cleaving of a crystal
of rocksalt. (Introduction to Solid State Physics,
Kittel, 7th Ed., John Wiley & Sons, Inc., 1996)
(c)
Translation Vector: T = u1a1+u2a2+u3a3
Primitive translation vectors: a1, a2, a3
Basis: Composed of at least one atom
Rj=xja1+yja2+zja3
0  xj, yj, zj  1
Lattice + Basis  Crystal Structure
Fig. 2 2D Protein crystal, Ibid.
Choices for primitive translation vectors and primitive unit cells which have
equal area (cells 1-3). Note cell #4 is not primitive.
a1
a2
Primitive cell in 3D
Basis with 2 atoms.
The body-centered cubic (bcc) lattice. The
primitive translation vectors are
a1 = a/2(1,1,-1), a2 = a/2(-1,1,1) and
a3 = a/2(1,-1,1)
bcc
fcc
The face-centered cubic (fcc) lattice.
The primitive translation vectors are
a1 = a/2(1,1,0), a2 = a/2(0,1,1) and
a3 = a/2(1,0,1)
Crystal Structure of diamond (C, Si, Ge, Sn)
Atomic positions in the cubic cell projected on a face
The diamond structure can be described as two
interpenetrating fcc lattices whose second
lattice is displaced along the body diagonal by
¼ of its length. That is, the fcc space lattice
consists of a basis with two identical atoms at
(000) and (¼ ¼ ¼).
The positions of the atoms can easily be
understood from this picture. The white
spheres correspond to atomic positions on
one fcc lattice, while the black spheres
correspond to atoms on the second fcc
lattice.
The Zinc-Blende crystal structure. The structure is very similar to
the diamond structure except the second fcc lattice consists of a
different type of atom. This is a very common crystal structure for
binary semiconductor structures shown in the table below:
Material a (Å)
GaAs
5.65
ZnSe
5.65
AlAs
5.66
ZnS
5.41
AlP
5.45
GaP
5.45
InSb
6.46
IA
VIII
IIA
IIIB IVB VB VIB VIIB
Index system for crystal planes
The plane intercepts the a1, a2, a3 axes at
3a1, 2a2, 2a3. The reciprocals of these
numbers are 1/3, ½, ½. The smallest three
integers having the same ration are 2, 3, 3,
thus the indices of the plane are (233).
Indices of important planes
in a cubic crystal.
Reciprocal Lattice Vectors:
a2  a3
b1  2
a1  a 2  a 3
a 3  a1
b 2  2
a1  a 2  a 3
bi  aj  2ij
a1  a 2
b3  2
a1  a 2  a3
G = v1b1+v2b2+v3b3
v1, v2, v3 are integers.
We want to describe periodic properties of the crystal using
reciprocal lattice vectors and Fourier analysis. For example, take the
electron density function n(r).
3-D:
1-D:
n( x)   np exp( i 2px / a )
p
n *  p  np
n(r )   nG exp( iG  r )
G
n(r  T )   nG exp( iG  r ) exp( iG  T )  n(r )
G
exp( iG  T )  exp i (v1b1  v 2b 2  v3b3)  (u1a1  u 2 a 2  u 3a 3)
 exp i 2 (v1u1  v 2u 2  v3u 3)  1
Brillouin Zone Construction
1-D
1st Brillouin Zone: Wigner-Seitz
primitive cell in the reciprocal
lattice.
2-D
Important Brillouin Zones in 3D for fcc and bcc
fcc
The face-centered cubic (fcc) lattice. The primitive
translation vectors are
a1 = a/2(1,1,0), a2 = a/2(0,1,1) and a3 = a/2(1,0,1)
b1=2/a(-1,1,1), b2=2/a(1,-1,1), and b3=2/a(1,1,-1)
bcc
Special points in the fcc and bcc Brillouin zones
bcc
fcc
Special points: X=(0,1,0)2/a, L=(1/2,1/2,1/2) 2/a,
K=(3/4,3/4,0)2/a, W=(1/2,1,0) 2/a,
U=(1/4,1,1/4)2/a
Free Electron Gas in metals:
Nucleus
Example: Electronic
Configuration of Na (Sodium)
1s22s22p63s1
eZa
Core electrons
-e(Za-Z)
Core: 1s22s22p6
Valence: 3s1
-eZ
Valence electrons
Za = atomic number
Drude Assumption:
-neZ valence
electrons
eZa
eZa
eZa
-e(Za-Z)
-e(Za-Z)
-e(Za-Z)
eZa
eZa
eZa
-e(Za-Z)
-e(Za-Z)
-e(Za-Z)
Quantum Theory of free electron gas:
1D metal  1D Shrodinger Equation:
V(x)
Inside
metal
Outside
metal
0
 2 d 2

 V ( x) ( x)  E ( x)

2
 2m dx

L
(x) = Asin(knx); knL=n and
kn = 2/n


n =2
Define Fermi Energy as the Energy of the highest
filled level: nF = N/2.
n =1
L
2
Now suppose we have N electrons in a 1D metal.
Each k-state can have spin-up  and spin-down 
electrons; ms = ½.
So
0
 2 kn2
 2  n 
En 



2me 2me  L 
n =3
V(x)

Also
(1)
 2 k F2
 2  N 
EF 



2me 2me  2 L 
2
Another approach for 1D, 2D and 3D electron gas:
L=Na
Solution to Eq. (1) can also be written as (x) = Aexp(ikx)
a
Suppose we have a ring of N atoms with length L.
Use periodic boundary conditions:
(Also known as Born Von Karman boundary conditions.)
x
(0) = (L)  kL=2n
X=0,L
and k=2n/L
For 3D, the Hamiltonian is
2 2
H 
  V (r )
2m
H (r )  E (r );
2
2
2
  2 2 2
x
y
z
2
and
 (r ) 
1
exp( ik  r )
V
3
*
d
r

 1

 2k 2
E
;
2m
k 2  k x2  k y2  k z2
Suppose we have a box with sides a, b, and c (i.e., V=abc)
Then periodic boundary conditions  (a,0,0) = (0,0,0) , (b,0,0) = (0,0,0) , and
(c,0,0) = (0,0,0).
kxa=2l, kyb=2m, and kzc=2p where l, m, p are 0, 1, 2 ,3 …
It’s possible to construct a “state-space”:
p
So
pml = n
m
 ak x  bk y  ck z 

n  plm  


2

2

2





abc

k x k y k z
3
(2 )
Since the actual number of
V
3

k
states is twice that for a give
3
(2 )
spin, it follows that
Number of states
with a given spin
in volume n
l
n  2n 
kz
It is customary to assume a spherical
approximation in k-space:
ky
kx
V 3
k
3
4
The volume of
the shell is
d3k=4k2dk
Density of States: Important concept for metals, semiconductors 1D, 2D and 3D.
V 3
V
V 2
2
d
k

4

k
dk

k dk
3
3
2
4
4

dn  g~ (k )dk  g ( E )dE

Note : k  k
g~ (k )dk  Number of states in the range kk+dk (assuming spherical symmetry)
dn  2dn 
E 
 2k 2
2me
 k  2mE /  2
1
dk V 2  dE  dE
~
g ( E )  g (k )
 2k 
  2k / m
 ;
dE 
 dk  dk
V 2mE   2
g ( E )  2 2 
  m
Therefore
g (E)  c E
1
3/ 2
2mE 
V  2m 
  2 2 
E
2 
 
2   
Also, for metals we can express the density of states in terms of the Fermi Energy:
N
EF
 g ( E )dE 
0
 g (E) 
3 N
2 EF
2 3/ 2
3
cE F  c  NEF3 / 2
3
2
k y2
 2  k x2
k z2 
E  Eo 


2  mxx m yy mzz 
 k x2
k y2
2
k z2 

Therefore,


1
2( E  Eo )  mxx m yy mzz 
or
 k x2 k y2 k z2 
 2  2  2  1
a
b
c 

where
a
c
ky
E
EF
For Semiconductors, vary often the surfaces of constant
energy aren’t spherical, but rather ellipsoidal.
Typically,
kz
2m yy ( E  Eo )
2mxx ( E  Eo )
,
b

, and
2
2
2mzz ( E  Eo )
2
kx
kF
Fermi Surface at EF
a,b,and c are the axes of the
ellipsoid.
mxx,myy, and mzz are the
effective masses
Surface of
constant E
ky
kz
c
b
a
kx
In the parabolic approximation of bands, the effective mass depends
generally on direction as is defined as:
 1 
 *
 m  


1  E (k )
1
 2
 M (k ) 
 k  k
2


E
Eo
The volume of the ellipsoid in k-space is
4
4 8

Vk  abc    6 mxx m yy mzz ( E  Eo )3 
3
3 


8 2
3 3
Remember,
1/ 2
mxx m yy mzz ( E  Eo )3 / 2
V 3
V
dn  g ( E )dE  3 d k  3 Vk ( E  dE )  Vk ( E )
4
4
V dVk
 3
dE (Simple Taylor expansion)
4 dE
V 4 2
1/ 2


 g (E)  3
m
m
m
E  Eo
xx yy zz
3
4

kx
If we define m*=(mxxmyymzz)1/3, then
g ( E )  (m*)3 / 2 E
At this point, it’s necessary to introduce the Fermi-Dirac distribution, f(E). So
far we have considered a temperature at absolute zero (T=0).
f (E) 
1
e ( E   ) / k BT  1
   (T )
f(E) gives the probability that an orbital (or state) will be occupied at energy E for
a given temperature T.  is the chemical potential and is need for the conservation
of electrons in the system. kB is Boltzmann’s constant, at kBT1/40 eV at T=300K.
Example:
Consider plot of f(E) vs. E/kB
for EF=4.2 eV (at T=0) and
TF=EF/kB=50,000 K.
f(E)
T=0, =EF, ,f=1/2
f(E)
1
0
EF
E
E/kB
The number of electrons between E
and E+dE is
dn  f ( E ) g ( E )dE
N
E2
 f ( E ) g ( E )dE
 E
g(E)
T=0
E1

f (E)g (E)
N total   f ( E ) g ( E )dE
0
Also, note here that the density of
states depends on the dimension
of the system:
T>0
Ntotal
E1/2
EF
g(E)
Const
E-1/2
E
E
Understanding Bands from the molecular orbital (MO) point of view.
A
A
B
B
MO2=1sA-1sB
=1s*
HA
HB
A
Consider the H2 molecule
B
MO1=1sA+1sB
=1s
1s*
Antibonding level
Antibonding MO
E
1s
Bonding MO
V(r)
Bonding level
When two atoms are brought together, a higher energy anti-bonding orbital (*) and a
lower energy bonding orbital () form which are linear combinations of atomic orbitals
(LCAO), as shown above. H2 is the simplest example of this effect. If instead of two
atoms, we bring N atoms together, there will be N distinct LCAO and N closely spaced
energy levels that will form a band.
Consider two Na atoms in a crystal. The spatial extent of the radial wavefunctions, r(r),
is shown below. Notice that there is appreciable overlap only for the 3s valence electrons.
Core electrons have significantly reduced overlap.
1s
# of nodes
=n-l-1
2p
3s
NaA
NaB
3.7 Å
2s
Tight-binding method for determination of the band sructure:
We want to calculate the electronic band dispersion for crystals in which the
electron wavefunctions are not significantly different from the atomic case, i.e.,
they are still tightly bound to the atom. This method is referred to as the tightbinding approximation.
Energy levels
V(r)
r
Bands form each
with N values of
k; the bandwidth
depends on the
overlap of the
wavefunctions.
n=3
n=2
n=1
d
d
d
When d is large: Hatn=Enn,
(spacing) -1
d
N-fold degenerate
levels
where Hat is the Hamiltonian for an isolated atom, n are the
atomic wavefunctions, En are the energy eigenvalues.
Let H = Hat +U(r), where U(r) contains all corrections to the atomic potential
required to produce the full periodic crystal potential.
r(r)
U(r)
When U(r) is added to a single atomic potential localized at the origin, the full
periodic crystal potential, U(r), is recovered.
(r) is an atomic wavefunction, localized at the origin. When |r(r)| large, |U(r)|
is small, and vice versa.
So,
 2 2
H at 
  Vat (r ),
2m
with U(r) = Vat(r) + U(r), and U(r+T)= U(r),
An important property of wavefunctions n(r) in crystals is provided by Bloch’s Theorem:
 k (r )  ei k r uk (r ),
 k (r  T )  ei k T k (r )
uk (r  T )  uk (r )
For N atoms, we want to construct linear combinations of the form:
 n k (r )   ei k T n (r  T ) where the periodic boundary conditions (i.e. kxLx=2n) are satisfied.
T
Test Bloch conditions:
 n k (r  T )   ei k T ' n (r  T  T ' )  ei k T  ei k (T 'T ) n (r  (T 'T ))
T'
 ei k T n k (r )
T'
thereby satisfying the Bloch condition.
There is a problem with this assumption, however, since negligible overlap of adjacent atomic
wavefunctions from site to site would be found for most states. This would lead to very little or
no dispersion in the energy bands, inconsistent with the experimental evidence. We want to
maintain the general form of the Block solution, but introduce new atomic-like functions (r).
(1)
 (r )   ei k T  (r  T )
T
(2)
 (r )   bn n (r )
n
where (r) is called a Wannier function. This is
expected to be similar to atomic wave functions, so
we can expand in terms of the the orthonormal set
of bound atomic wavefunctions.
LCAO (Linear combination of atomic orbitals)
H (r )  ( H at  U (r )) (r )  E (k ) (r )
 H at (r )  E (k ) (r )  U (r )) (r )
from the Crystal Hamiltonian
(3)
Multiply by m*(r) and integrate:
(3) 
*
3
*
3
*
3

H

d
r

(
H

)

d
r

E


d
r
m
at
at
m
m
m



  m* ( E (k )  U )d 3r  Em  m*d 3r
 ( E (k )  Em )  m* d 3 r   m* (r )U (r ) (r )d 3r
(4)
Now insert Wannier functions, by putting (1) and (2) into (4)
( E (k )  Em )  d 3r m* (r ) ei k T  bn n (r  T )   d 3r m* (r )U (r ) ei k T  bn n (r  T )
T
n
T
Simplify by considering T=0 terms and
n
3
d
 r m n   mn .


( E (k )  Em )bm  ( E (k )  Em )   m* (r ) n (r  T )e i k T d 3 r bn
n T  0





   m* (r )U (r ) n (r )d 3 r bn     m* (r )U (r ) n (r  T )e i k T d 3 r bn
n
n T  0

(5)
It’s important to note that the overlap is usually
small in the tight binding approximation so that
For our localized atomic levels:
3
*
d
r

 m (r ) n (r  T )  1
 (r )   bn n (r )
n
Now, let n run over these levels with energies that are degenerate or very close to a
fixed atomic level: s, p, or d level.
s-level 1-equation with one unknown energy dispersion E(k).
p-level 3 equations, 3 energy dispersions Ei(k)
d-level 5 equations, 5 energy dispersions Ei(k)
As an example, consider single atomic s-level (simplest) case:
 (r )  bs s (r )
where bs=1
Then Eq. (5) becomes
( E (k )  Es )  ( E (k )  Es )  d 3r * (r ) (r  T )ei k T
T 0
2
  d 3rU (r )  (r )    d 3r * (r )U (r ) (r  T )ei k T
T 0
(6)
Make some definitions:
Let
    d rU (r )  (r )
3
2
 (T )   d 3r * (r ) (r  T )
 (T )    d 3r * (r )U (r ) (r  T )
We can rewrite (6) as
E (k )  Es  ( E (k )  Es ) (T )ei k T      (T )ei k T
T 0
 E ( k )  Es 
T 0
    (T )ei k T
T 0
1   (T )e
i k T
(7)
T 0
There are symmetry arguments to consider for an s-level:
i. (r) is real and (r)=(r)  (-T)= (T)
ii. Inversion symmetry  U(-r) = U(r) and (-T)= (T)
iii. Also,  <<1 (Small Overlap for adjacent atomic sites)
This simplification gives:
E ( k )  Es   
 (T ) cos(k  T )
Nearest
Neighbors
Consider fcc lattice with 12 nearest neighbors:
T = a/2(±1, ±1,0), a/2(0, ±1, ±1), a/2(±1,0, ±1)
Then
k T 
a
 k x  k y , a  k x  k z , a  k y  k z 
2
2
2
Also, U(r) = U(x,y,z) has the full cubic symmetry of the lattice
and is unchanged by permutations of its arguments or signs.
Therefore (T) = const. = 
Therefore, we can simplify further:
E (k )  Es    4
 ki a k j a 
 

cos


2
2
i, j


x, y, z
  k a   kya 
k a k a
 k a   k a 
  cos y  cos z   cos z  cos x 
 Es    4 cos x  cos
 2   2 
 2   2 
  2   2 
using cos(A+B)=cosAcosB-sinAsinB and where  is given by
   d 3r * (r )U (r ) (r  To )
To 
a
a
xˆ  yˆ
2
2
An important characteristic of the tight-binding energy bands:
Bandwidth (spread between the maximum and minimum energy) is proportional to
the small overlap integral .
As the overlap decreases also || decreases.
It is therefore clear as   0, E(k)  const., forming N-fold degenerate levels.
Consider now ka<<1, kx=ky=kz, E(k) =Es-  - 12 + k2a2 = Eo + k2a2
E(k)
Appears similar to free electron case, i.e.
 2k 2
E
 Vo
2m
Eo
kx
Note: kx=ky=kz  [111] direction
2
*
2 d E 
m    2 
 dk 
1
1
 2 
2
*
2  E
2
2 1
   (2a ) 
For this tight-binding example: (m ) xx   
2 
2
2

a

k
x


Consider plot for full zone along the X-direction [100]:
E
  k a   kya 
k a k a
 k a   k a 
  cos y  cos z   cos z  cos x 
E (k )  Es    4 cos x  cos
 2   2 
 2   2 
  2   2 
Eo+16

 k a 
 Es    4 1  2 cos x 
 2 

k y  kz  0
Eo
0

[100]
kx
2/a
X
Bandwidth = 16
Extend the tight-binding method for p-levels, i..e. i= xf(r), yf(r), and zf(r)
LCAO involves expressing the wavefunctions as
 (r )(i )  b1i xf (r )  b2i yf (r )  b3i zf (r )
Note: these levels are degenerate.
Spherical coordinates and spherical harmonics:
x  r sin  cos  , y  r sin  sin 
z  r cos  
4 0
rY1
3
3 x  iy
Y 
8 r
LzY11  1Y11
1
1
Note for l=1 (p-level), ml=-1, 0, 1


( E (k )  Em )bm  ( E (k )  Em )   m* (r ) n (r  T )e i k T d 3 r bn
n T  0


(5)



   m* (r )U (r ) n (r )d 3 r bn     m* (r )U (r ) n (r  T )e i k T d 3 r bn
n
n T  0

(E(k )  E )1    ˆ(k ) b  0
For our p-levels:
where:
 b1 
b  b2 
b3 
p
ˆij (k )   ei k T  ij (T )
and
T
1 0 0 
1  0 1 0
0 0 1
 ij (T )    d 3r i (r ) j (r  T )U (r )
*
 ij   ij (T  0)

This is a 3 x 3 secular eigenvalue problem for E(k),
which, involves 3 eigenvalues and 3 eignevectors from:

det ( E (k )  E p )1    ˆ (k )  0
d-bands involve a 5 x 5 problem, etc..
E1(k)
In general, we get 3 bands
E
For d-bands, we would
calculate 5 bands.
E2(k)
for our p-bands:
E 3(k)

X
kx
Hybridization: LCAO for chemical bonding.
For C, Si, Ge, and Sn, the sp3 hybridization describes the covalent chemical
bonding in the diamond structure.The mixing of the wavefunctions results in bond orbitals with the famous 109.5 degree bond angle.
The four hybridized
1
1
1
1


s

p

p

pz
x
y
1
wavefunctions are:
2
2
2
2
p
sp3
E
s
Hybridization
1
1
1
1
px  p y  pz
2
2
2
2
1
1
1
1
 3  s  px  p y  pz
2
2
2
2
1
1
1
1
 4  s  px  p y  pz
2
2
2
2
2  s 
Consider
Silicon:
Conduction Band
States
1s22s22p63s23p2
Valence Band
States
Actual crystal spacing
Energy levels in Si a function of the inter-atomic spacing. At the actual atomic spacing of the
crystal, the 2N electrons in the 3s sub-shell and the 2N electrons in the 3p sub-shell undergo sp3
hybridization, and at T=0 all 4N electrons end up in the lower 4N states forming the valence
bands. The higher lying 4N states form the conduction band, and are empty at T=0. Note that there
is a band gap separating the valence and conduction bands.
Energy Bands:
T=0
E
Eg
EF
Semiconductor,
Insulator
Metal
E
Eg
T>0

Semiconductor,
Insulator
No sharp distinction between
semiconductors and insulators; the
energy gap for semiconductors is
usually less than ~2eV.
Metal
The bands of a metal are always
partially filled. Semiconductors
require thermal excitation across
the gap to populate conduction
levels.
Semiconductor or Insulator:
Metal:
T=0
E
Unoccupied states
EC Conduction band minimum (CBM)
 or EF Chemical potential or Fermi level
Eg
EV Valence Band maximum (VBM)
Occupied states
g(E)
Eg is the magnitude of the Band Gap.
E
Eg
T >0
Unoccupied states
Electrons near CBM
EC Conduction band minimum (CBM)
 or EF
EV Valence Band maximum (VBM)
Holes near VBM
Occupied states
g(E)
E
 or EF
Occupied
states
g(E)
For T > 0, electrons will be thermally
excited from the VBM to the CBM.
Conduction can only take place when
there are carriers (electrons or holes) in
partially filled bands.
We shall see that the fraction of
electrons excited across the band gap
(Eg) is proportional exp(- Eg/2kT).
For Eg=4 eV the fraction is ~10-35 at T =
300 K (kT=1/40 eV).
For Eg=0.25 eV the fraction is ~10-2 at
T = 300 K (kT=1/40 eV).
Metals vs semiconductors:
Transport properties:
F = ma =mdv/dt = -eE  v = -eEt/m if v(0) = 0.
If there is scattering with phonons or impurities,
then there is a characteristic collision time ,
whereby the velocity saturates at vd = -eE/m, where
vd is the carrier drift velocity.
j = nqvd =ne2E/m=neE (Ohm’s Law)
=E = E/   = ne2 /m and  = 1/
For semiconductors, the carrier
excitation across the band gap is
the dominant effect in determining
the resistivity () vs T behavior.
The basic result is that  increases
dramatically with T, as we shall see
quantitatively.
Also, vd = E by definition. So =e /m (mobility).
For metals the scattering rate increases with Temp.
(T) so that
Log()
Typical Metal
a
b

NDa < NDb
= 1/
Doping
1/T
1/T
Consider briefly optical properties of semiconductors:
Indirect Bandgap
Direct Bandgap

EC
EC

E
kc
EV
 = photon energy
 = phonon energy
Consider the emission
of a phonon:
  E g (For a direct band gap semiconductor)
  E g  

k
EV
(For an indirect band gap)
k photon  k c  k phonon  0
 k c  k phonon
(kc = electron momentum)
  0.01eV = phonon energy
o
1
k phonon  kc ~ 1 A
k photon 
2


2
3
o
~ 10 A
5000 A
  2.5eV = photon energy
Indirect transitions:
Almost all the energy is provided by the photon, while almost all the momentum
transfer is provided by the phonon in an indirect transition process.
o
1
The hole behaves like a
particle with positive
mass and positive charge.
Holes in semiconductors:
EC
EV

E
p   k , k h  ke

k
Xk k
h
e
ke=-kh
ke
Also, ideally in the
valence band at T=0,
d kh
 eE
dt
k
When one electron is
missing at ke, we have
k
Eh(kh) = -Ee(ke)
ke
k
The charge of a
hole is +e.
qh = -qe= +e
0
i
d ke

 e E
dt
Construct diagram to illustrate
hole dynamics:
A hole is an alternate
description of a band with one
missing electron.
kh
i
i
 k e  k h
i
The diagram construction gives
1
 k E (k )  v h  v e

1 d 2E
*1
m 
 mh  me
2
 dk
v
The motion of electrons in the conduction
band and holes in the valence band under the
influence of an electric field is summarized
by this sketch.
Equations of motion:

ve
e
je
j=qv
E
vh
h
jh
dkh
1


 e E  v h  B 
dt
c


Symmetry of the crystal  M is
real and symmetric, and we can
always find a set of orthogonal
principal axes such that
dke
1



 e E  v e  B 
dt
c


Details of the band structure:
2

Electrons: Ec (k )  Ec 
2
2

Holes: E (k )  E 
v
v
2
k (M


1
)  k
,
1
k
(
M
  )  k
 ,
mxx 1

1
M  0
 0

0
m yy
0
1
0 

0 
1
mzz 
Consider band structure of Si, Ge, GaAs:
me
For GaAs, CBM is at k=0 (i.e. direct bandgap)
E
Consider spin-orbit interactions for the p-like
hole bands. We need to consider addition of
angular momentum for orbit and spin:
s-like
Eg
J = L + S with L = 1 S = 1/2

k
Heavy Holes (hh)
mhh mlh
Light Holes
j = 3/2 and 1/2
For j=3/2,
mj = 3/2 and 1/2

p-like
mhh mlh
msoh
Split-off
holes
s-like
For j=1/2 mj = 1/2 (split-off)
 2k 2
Ec 
 Eg
2me
 2k 2
Ev (lh )  
2mlh
 2k 2
Ev (hh)  
2mhh
 2k 2
Ev ( soh )  

2mso
2
*
2 d E 
m    2 
 dk 
1
Curvature
For the conduction band, must consider the dispersion near
CBM in k-space:
2
2
See surfaces of constant energy
2
2
E (k )  E (0,0, ko ) 
(k x  k y ) 
(k z  ko ) 2
for the conduction bands.
2mT
2mL
Transverse (T) and longitudnal (L) effective masses (see also the table).
For Si, ko 0.85 (--X) where X is at (0,1,0)2/a,
For Ge, ko is at the L point which is (1,1,1)/a
Note again  defines the point (0,0,0).
Determine the number of carriers in thermal equilibrium:
Essential to know the carrier density as a function of temperature:
Ec
(T)
Ev
Definitions:
nc(T) = Number of electrons/vol. in conduction band
pv(T) = Number of holes/vol. in the valence band
gc(E) = Density of states (DOS) in the conduction band
gv(E)=DOS in the valence band.

nc (T )   dEgc ( E )
Ec
Using Fermi-Dirac statistics:
1
e ( E   ) / k BT  1
1
1

 v


pv (T )   dEgv ( E ) 1  ( E   ) / k T    dEgv ( E )  (   E ) / k T 
B
B
 1  
 1
 e
e

Ev
Note that the probability of
finding a hole is 1-f(E)
E
For most cases of interest we can approximate using Ec->>kBT and -Ev>>kBT, i.e.,  is
near the middle of the gap, or midgap. Therefore, for E > Ec and E< Ev
1
e
( E   ) / k BT
1
1
e (   E ) / k BT  1
 e  ( E   ) / k BT
(E > Ec)
 e  (   E ) / k BT
(E< Ev) This is the Boltzmann limit.
Therefore, the above expressions can be written as

nc (T )   dEgc ( E )
Ec
e ( E   ) / k BT  1
Ev
pv (T )   dEgv ( E )


1
1
e
(   E ) / k BT
1
  dEgc ( E )e
 ( E   ) / k BT
e
 ( Ec   ) / k B T
Ec
e
 (   Ev ) / k B T

 ( E  Ec ) / k B T
dEg
(
E
)
e
(1)
 c
Ec
Ev
 dEg ( E )e
v

 ( Ev  E ) / k B T
(2)

 ( E  Ec ) / k BT
N
(
T
)

dEg
(
E
)
e
c
c

Further define:
Ec
Ev
The DOS for electrons in
the conduction band and
holes in the valence band is
After integration:
Pv (T )   dEg v ( E )e ( Ev  E ) / k BT

g c ,v ( E ) 
mc3,/v2
 3 2
2 E  Ec , v
1  2m k T 
N C (T )   c 2B 
4   
1  2m k T 
PV (T )   v 2B 
4   
3/ 2
 nc  N C (T )e ( Ec   ) / k BT
(3)
3/ 2
 pv  PV (T )e (   EV ) / k BT
(4)
mc  (mxx m yy mzz )1/ 3
We can also reduce to a numerically convenient form:
m 
N C (T )  2.5 c 
 me 
m 
PV (T )  2.5 v 
 me 
3/ 2
3/ 2
 T 


 300 K 
 T 


 300 K 
3/ 2
1019 cm 3
3/ 2
1019 cm 3
Therefore, for mc,vme, the upper limit to the
carrier concentration is 1018 to 1019 cm-3 for
a non-degenerate semiconductor.
We can eliminate  altogether multiplying Eqs. (3) and (4):
nc pv  N c (T ) Pv (T )e  ( Ec Ev) / k BT
 N C PV e
This is a very useful formula known as
“the law of mass action.”
 E g / k BT
Consider now the intrinsic case in which all electrons in the conduction band and holes in the
valence band come from thermal excitation. In the intrinsic case, impurities don’t contribute to
the conductivity.
In the intrinsic case, nc=pv=ni
 ni  nc pv  N c (T ) Pv (T )e
1  2k T 
 ni   B2 
4   
Also nc=pv=ni  ni  pv  Pv e
i =  (intrinsic case)
3/ 2
(mc mv ) 3 / 4 e
 E g / 2 k BT
 E g / 2 k BT
 (  i  Ev ) / k B T
 N c (T ) Pv (T )e
 E g / 2 k BT
 Pv e ( i  Ev ) / k BT
 Eg
  Ev 
Pv
 
 exp  
 i
k BT 
Nc
 2 k BT
Eg
  Ev 1 Pv

 i
 ln
2 k BT
k BT
2 Nc
  i  Ev 
P
1
1
E g  k BT ln v
2
2
Nc
We can express i in terms of the masses:
1  2m k T 
N C (T )   c 2B 
4   
1  2m k T 
PV (T )   v 2B 
4   
3/ 2
3/ 2
  i  Ev 
Eg
m
3
 k BT ln v
2 4
mc
So, as T0, i  Eg/2
Even at room temperature kBT  1/40 eV and for Si, Eg ~1.1 eV so i  Eg/2
Notes for calculations: (mc)3/2  Rc(mde)3/2 where Rc is the number of equivalent
minima and mde is the DOS effective mass for electrons:
mde  (mxx myy mzz )1/ 3  (ml mt2 )1/ 3
Also (mv)3/2  (mdh)3/2 = mhh3/2 + mlh3/2, where mdh is the DOS effective mass for holes.
These masses can be found in standard tables.
For semiconductors, the addition of impurities to the crystal will significantly alter the carrier
density and conductivity.
Donors will add electrons to the conduction band.
Acceptors will add holes to the valence band.
Consider a group V impurity donor, e.g. As or Sb (s2p3) in Si or Ge (group IV with sp3)
Si
Si
Si
Si
Sb
Si
Si
Si
In order to preserve the tetrahedral
bonding and symmetry, the extra
electron is weakly bound to the
impurity donor.
Sb
Behaves like
hydrogen atom
Si
We have solved this problem in
quantum mechanics:
e2
1
 
c 137
+
Four sp3 hybrid bonds are formed, leaving one electron
in a hydrogen-like orbit around the Sb. Complete
removal of the electron will ionize the Sb donor. Thus,
we have a Coulomb force binding the electron as in the
hydrogen atom.
*
e 4 m* 1
2
1
0.51106 eV m* 1 1
2 m
ET   2 2 2   2 (me c )

2
2  n
2
me n
2(137) 2 me  2 n 2
m* 1 1
 13.6eV
me  2 n 2
ee 2 Coulomb
 potential
r
n is the principal quantum number, 1, 2, 3, ..
(fine structure constant)
m* 1
 EB 
13.6eV
me  2
The binding energy is given by EB = ET() – ET(n=1)
ro 
The Bohr radius ro for the impurity can also be easily written:
Where ao is the Bohr radius for atomic hydrogen:
m
ao
*
m
o
ao   / me  0.53 A
2
Example, typically m*0.1me and  10, so EB = 13.6eV(0.1/100) = 13.6 meV
Also ro = 0.53Å(10/0.1) = 53Å
A similar argument can be applied to group III acceptors (s2p):
_
h+
Check the table
which list the binding
energies for III
acceptors and V
donors.
Again, this is a hydrogen-like
problem except that the fixed charge
is negative and the orbiting charge is
a positively charged hole.
EC
ED

EA
EV
III
IV
V B
C
N Al
Si
P Ga
Ge
As
In
Sn
Sb
It is much easier to excite electrons and
holes from impurity levels than to excite
electrons across the bandgap (intrinsic
case).
i) Electrons are excited from ED to EC.
ii) Holes are excited from EA to EV.
(Also equivalent to filling bound hole
with e- promoted from the VBM.)
The statistics for the donors and acceptors are




1


N D  N D 1 


1
E


 1  exp D


g
k BT 
NA
N A 
E  

1  g exp  A
 k BT 
where ND+ is the number of ionized donors, g = 2
(ground state degeneracy of the donor which can
have spin up, down, or no electron).
where NA- is the number of ionized acceptors, g = 4
since we have spin up/down and two degenerate
valence bands which lead to mj = 3/2 and 1/2
(T) will adjust to maintain charge neutrality: Negative charge = Positive charge
For an n-type material (i.e., mostly donors) n = ND+ + p
For a p-type material (i.e., mostly acceptors) p = NA- + n,
where n = # of electrons in the conduction band and p = # of holes in the valence band.
For the partially compensated case:
present.
ND+ + p = NA- + n, where both dopants are
Consider an n-type case at T=300K, ND+ ND, NA=0.
1/ 2
N D  2 N D2 

  ni 
n
2
+
The Law of Mass Action becomes ni = np = n(n- ND )
4 
2 
Also,
ni2
 ni
p
Note that ni = 1.45 x 1010 cm-3 for Si at
n
 n  ni
T=300K
If ND>>ni, n  N D
Consider a determination of (T):
At T=0, Ec<  < ED since all electrons
are attached to donors. For very high
temperatures, there is thermal
excitation across the bandgap.To
understand what happens between
these limits consider again charge
neutrality conditions: n = ND+ + p
(a)
(b)
and
ni2
p
ND
usually very small.
(c)
 E  
 E  
1
  N D

 NC (T ) exp   C
 NV (T ) exp  V
k
T
1

2
exp[(


E
)
/
k
T
]
k
T
B
D
B


 B 
It is easy to solve for (T) using computational and iterative methods.
There are important features that can be understood by inspecting the
magnitude of terms (a), (b) and (c).
(1)
At intermediate and low temperatures (c) is negligible. This is called the saturation
range; n is controlled by ND.
(2)
In the intrinsic range at high temperatures (b)<<(c) and  is near the center of the gap.
(3)
At a given temperature,   Ec as ND increases.
- i (eV)
T (K)
(4) Note that Eg decreases as T increases. This is described by Varshni’s law:
4.73 10 4 T 2
E g  1.17 
T  656
where T is in K and Eg in eV.
The carrier concentration is described by the following graph:
Slope = -Eg/2
Saturation
Range
ln nc
Freeze-out Range
Slope=-(EC-ED)/2
Intrinsic
Range
Slope=-(EC-ED)
(very low-temp. range)
1/T
In the low-temperature case, term (c)
involving excitation of electrons from
the VBM is negligible, so that
 E  
E  
1
  N D

 N C (T ) exp   C
 NV (T ) exp  V
k
T
1

2
exp[(


E
)
/
k
T
]
k
T
B
D
B


 B 
 E    ND
0
 
 N C (T ) exp   C
exp[ (   ED ) / k BT ] Also at low T,
Ec<  < ED
2
 k BT 
 exp(
 E  ED 
2
N

)  D exp  C
k BT
2 NC
 k BT 
We can now solve for n, using the result for 
 E  
 E  ND
 E  ED 
  N C exp   C 

n  N C exp   C
exp  C
k BT 

 k BT  2 N C
 2 k BT 

 E  ED 
ND

exp   C
2NC
2 k BT 

and gives the slope illustrated in the
above graph in the freeze-out range.
There is always a finite amount of acceptor states even in an n-type
semiconductor. It is possible to show using a similar procedure that
 N  NA 
 E  ED 
 N C exp   C

n   D
k BT 
 2N A 

at even lower temperatures in the freeze-out range. Note the factor of two
is missing in the slope of the previous graph.
Example: A Si wafer is doped with 1016 As atoms/cm3. Find the carrier
concentrations and the Fermi level (i.e., the Chemical Potential) at T = 300
K.
Solution: (a) At room temperature, we assume complete ionization of impurity atoms.
So n  ND = 1016 cm-3
(b) From the law of mass action, p = ni2/n = ni2/ND = (1.45 x 1010)2/1016 =
2.1 x 104 cm-3.
(c) The Fermi level position is given by n= NCexp[-(EC - )/kBT]
EC -  = kBT ln (NC/ND) = 0.0259 ln (2.8 x 1019/1016) = 0.206 eV
Also, the Fermi level measured from the intrinsic Fermi level (i) is given by
nc  N C e  ( Ec   ) / k BT  N C e  ( Ec  i ) / k BT e (   i ) / k BT  ni e (   i ) / k BT
    i  k BT ln( nc / ni )  k BT ln( N D / ni )
 1016 
  0.354eV
 0.0259 ln 
10 
1
.
45

10


0.054 eV (Donor ionization energy)
(d) Examine the graph of the Fermi level:
Eg = 1.12 eV
(e) Check the assumption of almost complete
ionization at room temperature (T = 300 K).
0.206 eV
0.354 eV
EC
ED

i
EV








1
1
  1016 1 
N D  N D 1 
 0.994 1016 cm 3

1
0.206  0.054
 1  1 exp ED    
 1  exp

Therefore, 99.4% are


g
k BT 
2
0.0259 

indeed ionized and the
assumption is a good one.
Energy Gaps and Lattice Constants for several III-V Semiconductors