Metals, Semiconductors, and Insulators
Every solid has its own characteristic energy band structure.
In order for a material to be conductive, both free electrons and empty states must be
available.
Metals have free electrons and partially filled valence bands, therefore they are highly
conductive (a).
Semimetals have their highest band filled. This filled band, however, overlaps with the next
higher band, therefore they are conductive but with slightly higher resistivity than normal
metals (b). Examples: arsenic, bismuth, and antimony.
Insulators have filled valence bands and empty conduction bands, separated by a large
band gap Eg(typically >4eV), they have high resistivity (c ).
Semiconductors have similar band structure as insulators but with a much smaller band
gap. Some electrons can jump to the empty conduction band by thermal or optical excitation
(d). Eg=1.1 eV for Si, 0.67 eV for Ge and 1.43 eV for GaAs
Conduction in Terms of Band
Metals
An energy band is a range of allowed electron energies.
The energy band in a metal is only partially filled with electrons.
Metals have overlapping valence and conduction bands
Drude Model of Electrical Conduction in Metals
Conduction of electrons in metals – A Classical Approach:
In the absence of an applied electric field (ξ) the electrons move in random directions colliding
with random impurities and/or lattice imperfections in the crystal arising from thermal motion of
ions about their equilibrium positions.
The frequency of electron-lattice imperfection collisions can be described by a mean free path
λ -- the average distance an electron travels between collisions.
When an electric field is applied the electron drift (on average) in the direction opposite to that
v
of the field with drift velocity
The drift velocity is much less than the effective instantaneous speed (v) of the random
motion
−2
−1
In copper v ≈ 10 cm.s while
v ≈ 10 cm.s
8
−1
where
1
3
2
mev = k BT
2
2
The drift speed can be calculated in terms of the applied electric field ξ and of v and λ
When an electric field is applied to an electron in the metal it experiences a force qξ
resulting in acceleration (a) a = qξ
me
Then the electron collides with a lattice imperfection and changes its direction
randomly. The mean time between collisions is
q ⋅ ξ ⋅τ q ⋅ ξ ⋅ λ
=
v = a ⋅τ =
The drift velocity is
me
me ⋅ v
τ=
λ
v
If n is the number of conduction electrons per unit volume and J is the current density
Combining with the definition of resistivity gives
n ⋅q ⋅λ
σ=
me ⋅ v
2
q=1.6x10-19C
J = nq ν = σξ
q ⋅ λ q ⋅τ
μ=
=
me ⋅ v me
Electron Energy
For an electron to become free to conduct, it must be promoted into an
empty available energy state
For metals, these empty states are adjacent to the filled states
Generally, energy supplied by an electric field is enough to stimulate
electrons into an empty state
“Freedom”
Empty States
Energy Band
States Filled with Electrons
Distance
Band Diagram: Metal
T>0
EC
Fermi “filling”
function
Conduction band
(Partially Filled)
EF
Energy band to
be “filled”
E=0
At T = 0, all levels in conduction band below the Fermi energy EF are filled
with electrons, while all levels above EF are empty.
Electrons are free to move into “empty” states of conduction band with
only a small electric field E, leading to high electrical conductivity!
At T > 0, electrons have a probability to be thermally “excited” from below
the Fermi energy to above it.
Resistivity (ρ) in Metals
Resistivity typically increases linearly with temperature:
ρt = ρo + αT
Phonons scatter electrons. Where ρo and α are constants for an
specific material
Impurities tend to increase resistivity: Impurities scatter electrons in
metals
Plastic Deformation tends to raise resistivity dislocations scatter
electrons
σ=
1
ρ
= nqμ
The electrical conductivity is controlled by controlling the number of
charge carriers in the material (n) and the mobility or “ease of
movement” of the charge carriers (μ)
Temperature Dependence, Metals
There are three contributions to ρ:
ρt due to phonons (thermal)
ρi due to impurities
ρd due to deformation
ρ = ρt + ρi+ ρd
The number of electrons in the
conduction band does not vary
with temperature.
All the observed temperature
dependence of σ in metals arise
from changes in μ
Scattering by Impurities and Phonons
Thermal: Phonon scattering
Proportional to temperature
ρ t = ρ o + aT
Impurity or Composition scattering
Independent of temperature
Proportional to impurity concentration
Solid Solution
Two Phase
ρ i = Aci (1 − ci )
ρ t = ραVα + ρ β Vβ
Deformation
ρ d = must be experimentally determined
Insulator
The valence band and conduction band are separated by a large (> 4eV)
energy gap, which is a “forbidden” range of energies.
Electrons must be promoted across the energy gap to conduct, but the
energy gap is large. Energy gap º Eg
Electron
Energy
“Conduction Band”
Empty
“Forbidden”
“Valence Band”
Filled with Electrons
Distance
Energy
Gap
Band Diagram: Insulator
T>0
Conduction band
(Empty)
Egap
EC
EF
Valence band
(Filled)
EV
At T = 0, lower valence band is filled with electrons and upper conduction
band is empty, leading to zero conductivity.
Fermi energy EF is at midpoint of large energy gap (2-10 eV) between
conduction and valence bands.
At T > 0, electrons are usually NOT thermally “excited” from valence to
conduction band, leading to zero conductivity.
Conduction in Ionic Materials (Insulators)
Conduction by electrons (Electronic Conduction): In a ceramic, all
the outer (valence) electrons are involved in ionic or covalent bonds and thus
−Eg
they are restricted to an ambit of one or two atoms.
2 k BT
If Eg is the energy gap, the fraction of electrons in the conduction band is:
e
A good insulator will have a band gap >>5eV and kBT~0.025eV at room temperature
As a result of thermal excitation, the fraction of electrons in the conduction band is
~e-200 or 10-80.
There are other ways of changing the electrical conductivity in the ceramic which
have a far greater effect than temperature.
•Doping with an element whose valence is different from the atom it replaces. The
doping levels in an insulator are generally greater than the ones used in
semiconductors. Turning it around, material purity is important in making a good
insulator.
•If the valence of an ion can be variable (like iron), “hoping” of conduction can
occur, also known as “polaron” conduction. Transition elements.
•Transition elements: Empty or partially filled d or f orbitals can overlap providing a
conduction network throughout the solid.
Conduction by Ions: ionic conduction
It often occurs by movement of entire ions, since the energy gap is too large
for electrons to enter the conduction band.
Z .q.D
The mobility of the ions (charge carriers) is given by:
μ=
k B .T
Where q is the electronic charge ; D is the diffusion coefficient ; kB is
Boltzmann’s constant, T is the absolute temperature and Z is the valence of
the ion.
The mobility of the ions is many orders of magnitude lower than the mobility of
the electrons, hence the conductivity is very small:
σ = n.Z.q.μ
Example:
Suppose that the electrical conductivity of MgO is determined primarily by the
diffusion of Mg2+ ions. Estimate the mobility of Mg2+ ions and calculate the
electrical conductivity of MgO at 1800oC.
Data: Diffusion Constant of Mg in MgO = 0.0249cm2/s ; lattice parameter of
MgO a=0.396x10-7cm ; Activation Energy for the Diffusion of Mg2+ in MgO =
79,000cal/mol ; kB=1.987cal/K=k-mol; For MgO Z=2/ion; q=1.6x10-19C;
kB=1.38x10-23J/K-mol
First, we need to calculate the diffusion coefficient D
⎛ − QD
D = Do exp⎜
⎝ kT
⎛
⎞
− 79000cal / mol
cm 2
⎞
⎜
⎟⎟
=
0
.
0239
exp
⎟
⎜
s
⎠
⎝ 1.987cal / mol − Kx (1800 + 273)K ⎠
D=1.119x10-10cm2/s
Next, we need to find the mobility
2
Z .q .D ( 2carriers / ion )(1.6 × 10 −19 C )(1.1× 10 −10 )
C
.
cm
−9
=
.
μ=
=
1
12
×
10
(1.38 × 10 − 23 )(1800 + 273)
k B .T
J .s
C ~ Amp . sec ; J ~ Amp . sec .Volt
μ=1.12x10-9 cm2/V.s
MgO has the NaCl structure (with 4 Mg2+
and 4O2- per cell)
Thus, the Mg2+ ions per cubic cm is:
4Mg 2+ ions / cell
22
3
n=
=
6
.
4
×
10
ions
/
cm
(0.396 ×10 −7 cm )3
σ = nZqμ = (6.4 ×10 22 )( 2)(1.6 ×10 −19 )(1.12 ×10 −9 )
2
C
.
cm
σ = 22.94 ×10 −6 3
cm .V .s
C ~ Amp.sec ; V ~ Amp.Ω
σ = 2.294 x 10-5 (Ω.cm)-1
Example:
The soda silicate glass of composition 20%Na2O-80%SiO2 and a density of
approximately 2.4g.cm-3 has a conductivity of 8.25x10-6 (Ω-m)-1 at 150oC. If the
conduction occurs by the diffusion of Na+ ions, what is their drift mobility?
Data: Atomic masses of Na, O and Si are 23, 16 and 28.1 respectively
Solution:
We can calculate the drift mobility (μ) of the Na+ ions from the conductivity
expression
σ = ni × q × μ i
Where ni is the concentration of Na+ ions in the structure.
20%Na2O-80%SiO2 can be written as M At = 0.2 × ( 2( 23) + 1(16)) + 0.8 × (1( 281.1) + 2(16))
(Na2O)0.2-(SiO2)0.8 . Its mass can be
M At = 60.48g .mol −1
calculated as:
The number of (Na2O)0.2-(SiO2)0.8 units per unit volume can be found from the
density
n=
ρ × NA
M At
( 2.4g .cm −3 ) × (6.023x10 23 mol −1 )
=
60.48g .mol −1
n = 2.39 ×10 22 (Na 2O ) 0.2 (SiO 2 ) 0.8 units − cm −3
The concentration of Na+ ions (ni) can be obtained from the concentration of
(Na2O)0.2-(SiO2)0.8 units
⎡
⎤
0.2 × 2
22
21
−3
ni = ⎢
×
2
.
39
×
10
=
3
.
18
×
10
cm
⎥
0
.
2
×
(
2
+
1
)
+
0
.
8
×
(
1
+
2
)
⎣
⎦
And μi
σ
(8.25 × 10 −6 Ω −1m −1 )
μi =
=
q × ni (1.60 ×10 −19 C ) × (3.186 × 10 21 × 106 m −3 )
μ i = 1.62 ×10 −14 m 2V −1s −1
This is a very small mobility compared to semiconductors and metals
Electrical Breakdown
At a certain voltage gradient (field) an insulator will break down.
There is a catastrophic flow of electrons and the insulator is fragmented.
Breakdown is microstructure controlled rather than bonding controlled.
The presence of heterogeneities in an insulator reduces its breakdown field
strength from its theoretical maximum of ~109Vm-1 to practical values of 107V.m-1
Energy Bands in Semiconductors
Electron Energy
Energy Levels and Energy Gap in a Pure Semiconductor.
The energy gap is < 2 eV.
Energy gap º Eg
“Conduction Band” (Nearly)
Empty – Free electrons
“Forbidden”
Energy Gap
“Valence Band” (Nearly) Filled with
Electrons – Bonding electrons
Semiconductors have resistivities in between those of metals and insulators.
Elemental semiconductors (Si, Ge) are perfectly covalent; by symmetry electrons
shared between two atoms are to be found with equal probability in each atom.
Compound semiconductors (GaAs, CdSe) always have some degree of ionicity. In III-V
compounds, eg. Ga+3As+5, the five-valent As atoms retains slightly more charge than is
necessary to compensate for the positive As+5 charge of the ion core, while the charge of
Ga+3 is not entirely compensated. Sharing of electrons occurs still less fairly between the
ions Cd+2 and Se+6 in the II-VI compund CdSe.
Semiconductor Materials
Semiconductor
Carbon (Diamond)
Silicon
Germanium
Tin
Gallium Arsenide
Indium Phosphide
Silicon Carbide
Cadmium Selenide
Boron Nitride
Aluminum Nitride
Gallium Nitride
Indium Nitride
Bandgap Energy EG (eV)
5.47
1.12
0.66
0.082
IIIA
1.42
10.811
1.35
5
3.00
B
Bo ro n
1.70
7.50
6.20
3.40
1.90
13
IIB
30
65.37
Zn
Zinc
48
Portion of the Periodic Table Including the Most
Important Semiconductor Elements
112.40
Cd
C a d m ium
80
M e rc ury
Al
A lum inum
31
69.72
Ga
G a llium
49
114.82
In
6
12.01115
C
C a rb o n
14
81
204.37
Ti
Tha llium
28.086
Si
Silic o n
32
72.59
Ge
G e rm a nium
50
Ind ium
200.59
Hg
26.9815
IV A
118.69
Sn
Tin
82
207.19
Pb
Le a d
VA
14.0067
7
N
N itro g e n
15
30.9738
P
V IA
15.9994
8
O
O xy g e n
16
Pho sp ho rus
33
74.922
As
A rse nic
51
121.75
Sb
A ntim o ny
83
208.980
Bi
Bism uth
32.064
S
Sulfur
34
78.96
Se
Se le nium
52
127.60
Te
Te llurium
84
(210)
Po
Po lo nium
Band Diagram: Semiconductor with No Doping
T>0
Conduction band
(Partially Filled)
EF
Valence band
(Partially Empty)
EC
EV
At T = 0, lower valence band is filled with electrons and upper conduction
band is empty, leading to zero conductivity.
Fermi energy EF is at midpoint of small energy gap (<1 eV) between
conduction and valence bands.
At T > 0, electrons thermally “excited” from valence to conduction band,
leading to measurable conductivity.
Semi-conductors (intrinsic - ideal)
Perfectly crystalline (no perturbations in the periodic lattice).
Perfectly pure – no foreign atoms and no surface effects
At higher temperatures, e.g., room temperature (T @ 300 K), some electrons are
thermally excited from the valence band into the conduction band where they
are free to move.
“Holes” are left behind in the valence band. These holes behave like mobile
positive charges.
CB electrons and VB holes can
move around (carriers).
At edges of band the kinetic energy
of the carriers is nearly zero. The
electron energy increases upwards.
The hole energy increases
downwards.
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
positive
ion core
valence
electron
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
free
electron
free
hole
Si
Si
Si
Si
Si
Si
Si
Semiconductors in Group IV
Carbon
Silicon
Germanium
Tin
Each has 4 valence Electrons.
Covalent bond
Generation of Free Electrons and Holes
In an intrinsic semiconductor, the number of free electrons equals the
number of holes.
Thermal : The concentration of free electrons and holes increases with
increasing temperature.
Thermal : At a fixed temperature, an intrinsic semiconductor with a large
energy gap has smaller free electron and hole concentrations than a
semiconductor with a small energy gap.
Optical: Light can also generate free electrons and holes in a
semiconductor.
Optical: The energy of the photons (hν) must equal or exceed the energy
gap of the semiconductor (Eg) . If hν > Eg , a photon can be absorbed,
creating a free electron and a free hole.
This absorption process underlies the operation of photoconductive light
detectors, photodiodes, photovoltaic (solar) cells, and solid state camera
“chips”.
UV
Violet
Blue
Green
Yellow
Orange
Red
Near IR
100-400 nm 12.4-3.10 eV
400-425 nm 3.10-2.92 eV
425-492 nm 2.92-2.52 eV
492-575 nm 2.52-2.15 eV
575-585 nm 2.15-2.12 eV
585-647 nm 2.12-1.92 eV
647-700 nm 1.92-1.77 eV
10,000-700 nm 1.77-0.12 eV
Red
Orange
Yellow
Violet
Blue
Green
Photoconductivity
Eg
Conductivity is dependent on
the intensity of the incident
electromagnetic radiation
hων ≥EEgg
E = hν = hc/λ, c = λ(m)ν(sec -1)
Band Gaps:
Si Ge
GaAs
ZnSe
SiC
1.11 eV (Infra red)
0.66 eV (Infra red)
1.42 eV (Visible red)
2.70 eV (Visible yellow)
2.86 eV (Visible blue)
GaN
AlN
BN
3.40eV (Blue)
6.20eV (Blue-UV)
7.50eV (UV)
Total conductivity
For intrinsic semiconductors:
σ = σe + σh = nqμe + pqμh
n = p & σ = nq(μe + μh)
Question: How many electrons and holes are there in an intrinsic semiconductor in
thermal equilibrium?
Define:
no equilibrium (free) electron concentration in conduction band [cm-3]
po equilibrium hole concentration in valence band [cm-3]
Certainly in intrinsic semiconductor: no = po = ni
O
O
i
-3
ni intrinsic carrier concentration [cm ]
As T ↑ then ni ↑
As Eg ↑ then ni ↓
What is the detailed form of these dependencies?
We will use analogies to chemical reactions. The electron-hole formation can be
though of as a chemical reaction……..
bond ⇔ e − + h +
Similar to the chemical reaction………
H 2O ⇔ H + + (OH ) −
n =p =n
The Law-of-Mass-Action relates concentration of
reactants and reaction products. For water……
Where E is the energy released or consumed
during the reaction………….
[H + ][OH − ]
⎛ E ⎞
≈ exp⎜ − ⎟
K=
[H 2O ]
⎝ kT ⎠
This is a thermally activated process, where the rate of the reaction is limited by the
need to overcome an energy barrier (activation energy).
By analogy, for electron-hole formation:
Where [bonds] is the concentration of unbroken bonds
and Eg is the activation energy
⎛ Eg ⎞
[ no ][ po ]
≈ exp⎜⎜ − ⎟⎟
K=
[ bonds ]
⎝ kT ⎠
In general, relative few bonds are broken to form an
electron-hole and therefore the number of bonds are
approximately constant.
Two important results:
1)……………………..
⎛ Eg ⎞
⎟⎟
ni ≈ exp⎜⎜ −
⎝ 2kT ⎠
2)………………………………………….……..
[bonds] >> no ,po
[bonds] = cons tan t
⎛ Eg ⎞
no po ≈ exp⎜⎜ − ⎟⎟
⎝ kT ⎠
nO × pO = ni2
The equilibrium np product in a semiconductor at a certain temperature is a
constant specific to the semiconductor.
Effect of Temperature on Intrinsic Semiconductivity
The concentration of electrons with sufficient thermal energy to enter the
conduction band (and thus creating the same concentration of holes in the
valence band) ni is given by
⎛ − ΔE ⎞
⎟⎟
n i ≈ exp ⎜⎜
⎝ k BT ⎠
For intrinsic semiconductor, the energy is half way across the gap, so that
⎛ − Eg
n i ≈ exp ⎜⎜
⎝ 2 k BT
⎞
⎟⎟
⎠
Since the electrical conductivity σ is proportional to the concentration of
electrical charge carriers, then
⎛ − Eg
σ = σ O exp ⎜⎜
⎝ 2 k BT
⎞
⎟⎟
⎠
Example
Calculate the number of Si atoms per cubic meter. The density of silicon is 2.33g.cm-3
and its atomic mass is 28.03g.mol-1.
Then, calculate the electrical resistivity of intrinsic silicon at 300K. For Si at 300K
ni=1.5x1016carriers.m-3, q=1.60x10-19C, μe=0.135m2(V.s)-1 and μh=0.048m2.(V.s)-1
Solution
N A × ρSi
nSi =
= 5.00 ×10 28 Si − atoms.m −3
ASi
σ = ni × q × (μe + μ h ) = 0.4385 ×10 −3 (Ω − m ) −1
ρ = resistivity = 2.28 ×10 Ω − m
3
Example
The electrical resistivity of pure silicon is 2.3x103Ω-m at room temperature (27oC ~
300K). Calculate its electrical conductivity at 200oC (473K).
Assume that the Eg of Si is 1.1eV ; kB =8.62x10-5eV/K
⎛ − Eg ⎞
⎟⎟
σ = C . exp⎜⎜
⎝ 2k BT ⎠
σ 473
⎛ − Eg ⎞
⎟⎟
= C . exp⎜⎜
⎝ 2k B ( 473) ⎠
σ 300
⎛ − Eg ⎞
⎟⎟
= C . exp⎜⎜
⎝ 2k B (300) ⎠
− Eg ⎞
⎛ − Eg
σ 473
⎟⎟
= exp⎜⎜
−
σ 300
⎝ 2k B ( 473) 2k B (300) ⎠
− Eg
Eg
Eg ⎛ 1
⎛ σ 473 ⎞
1 ⎞
1.1eV
1 ⎞
⎛ 1
⎟⎟ =
+
=
−
−
ln⎜⎜
⎜
⎟=
⎜
⎟
−5
⎝ σ 300 ⎠ 2k B ( 473) 2k B (300) 2k B ⎝ 300 473 ⎠ 2(8.62 × 10 ) ⎝ 300 473 ⎠
⎛ σ 473 ⎞
⎟⎟ = 7.777
ln⎜⎜
⎝ σ 300 ⎠
1
−1
m
(
)
.
(
.
)
σ 473 = σ 300 ( 2385) =
2385
=
1
04
Ω
2.3 ×103
Example: For germanium at 25oC estimate
(a) the number of charge carriers,
(b) the fraction of total electrons in the valence band that are excited into the conduction
band and
⎛ − Eg ⎞
(c) the constant A in the expression n = A exp ⎜
⎟ when E=Eg/2
⎜ 2k T ⎟
⎝ B ⎠
Data: Ge has a diamond cubic structure with 8 atoms per cell and valence of 4 ;
a=0.56575nm ; Eg for Ge = 0.67eV ; μe = 3900cm2/V.s ; μh = 1900cm2/V.s
ρ = 43Ω-cm ; kB=8.63x10-5eV/K
(a) Number of carriers
T = 25o C
2k BT = ( 2)(8.63 × 10 −5 eV / K )( 273 + 25) = 0.0514eV
σ
0.023
13 electrons
n=
=
= 2.5 × 10
−19
q ( μe + μ h ) 1.6 ×10 (3900 + 1900)
cm 3
There are 2.5x1013 electrons/cm3 and 2.5x1013 holes/cm3 helping to conduct a charge
in germanium at room temperature.
;
b) the fraction of total electrons in the valence band that
are excited into the conduction band
The total number of electrons in the valence band of
germanium is :
Valence − electrons
=
( 8atoms / cell )( 4valence −electrons / atoms )
( 0.56575 x 10 −7 cm ) 3
Total − valence − electrons = 1.77 ×10 23 electrons / cm 3
number − excited − electrons / cm 3
2.5 ×1013
−10
Fraction − excited =
=
=
1
.
41
×
10
Total − valence − electrons / cm 3
1.77 × 10 23
(c) the constant A
n
A=
e
⎛ −E g
⎜⎜
⎝ 2 k BT
⎞
⎟⎟
⎠
=
2.5 × 1013
e
⎛ − 0.67 ⎞
⎜
⎟
⎝ 0.0514 ⎠
= 1.14 × 10 −19 carriers / cm 3
Direct and Indirect Semiconductors
The real band structure in 3D is calculated with various numerical methods,
plotted as E vs k. k is called wave vector
p = k p is momentum
For electron transition, both E and p (k) must be conserved.
A semiconductor is
direct if the maximum
of the conduction band
and the minimum of
the valence band has
the same k value
A semiconductor is indirect if the …do not have the same k value
Direct semiconductors are suitable for making light-emitting devices,
whereas the indirect semiconductors are not.