CHAPTER TWO: ELECTRONIC PROPERTIES OF MATERIALS
2.1
Introduction to Semiconductors
Materials can be grouped under three broad categories on the basis of their electric
conductivity: Conductors, semiconductors and Insulators. Semiconductors have conductivity
lying between that of conductors (mainly metals) and insulators (mainly non-metals). Mono
element semiconductors like carbon, silicon and Germanium are found in group IV of the
periodic table. Compound semiconductors are obtained from alloys of Group III and V as
well as Group II and VI for example Gallium Arsenide (GaAS) and Zinc Oxide (ZnO)
respectively. Ternary (such as Aluminium Gallium Nitride, AlGaN), quaternary and quinary
semiconductor alloys are also possible. We will concentrate on silicon, because of its
commercial usage in most of today’s electronics applications like transistors, diodes,
controlled rectifiers, and digital and analog integrated circuits.
The periodic Table
For a list of semiconductor materials, visit the URL:
http://en.wikipedia.org/wiki/List_of_semiconductor_materials.
Cosmas Mwikirize
CMP 1101: ELECTRONICS I-CHAPTER II
In a metallic conductor, current is carried by the flow of electrons. In semiconductors,
current can be carried either by the flow of electrons or by the flow of positively-charged
"holes" in the electron structure of the material.
Pure semiconductors are also known as intrinsic semiconductors. The conductivity of a
semiconductor material can be drastically changed by adding other elements, called
“impurities” to the melted intrinsic material and then allowing the melt to solidify into a
new and different crystal. This process is called doping and leads to extrinsic
semiconductors.
2.2
The Structure of Silicon
Silicon is in Group IV of the periodic table, with an atomic number of 14, and 4 electrons in
the outermost (bonding) shell.
The atomic density of silicon is 5 × 1022 𝑐𝑚−3. The
behavior of silicon can be explained using two theories:
The Bond Theory
A silicon atom has ten core electrons (tightly bound) and four valence electrons (loosely
bound) and responsible for its chemical properties. The atoms are tetrahedrally bonded
with each atom forming 4 strong covalent bonds (8 shared electrons) with the nearest
neighboring atoms in a regular diamond lattice.
At 0K, all valence electrons are involved in bonding and there are no free electrons. At finite
temperatures, electrons gain thermal energy and some bonds are broken. The breakage of a
bond yields a free mobile electron and a free mobile positive charge, the hole. A hole is a
missing electron, and behaves as a particle with the same properties as an electron, albeit
with a positive charge. This process where a covalent bond is broken to yield a free electron
Cosmas Mwikirize
CMP 1101: ELECTRONICS I-CHAPTER II
and a hole is known as Generation and it requires energy. Free electrons and holes are
known as carriers. Let n and p be electron and hole concentration respectively (𝑐𝑚−3). The
atomic density >> n and p so the number of breakable bonds is inexhaustible.
The reverse process to generation is recombination where a free electron and a free hole
get together to form a bond, with the release of energy. Both generation and recombination
occur continuously near the surface of the material where periodic crystalline structure is
broken. If G and R are the generation and recombination rates respectively, then,
𝐺 = 𝑓(𝑇) i.e. a function of temperature/ thermal, optical or any other energy supplied.
𝑅 𝛼 𝑛. 𝑝 i.e. dependent on the number of electrons and holes.
At thermal equilibrium (steady state and absence of external energy sources), the
recombination rate is equal to the generation rate. We write:
𝐺𝑜 = 𝑅𝑜 → 𝑛. 𝑝 = 𝑓(𝑇) = 𝑛𝑖2 (𝑇)
(1)
We use (1) to determine the number of carriers in an intrinsic pure semiconductor at a finite
temperature. We know that when a bond breaks, an electron and a hole are formed, hence
𝑛𝑜 = 𝑝𝑜 . Hence from (1), 𝑛 = 𝑝 = 𝑛𝑖 where 𝑛𝑖 is the intrinsic carrier concentration (𝑐𝑚−3).
For silicon at room temperature, 𝑛𝑖 ≈ 1010 𝑐𝑚−3. Hence for a given semiconductor in
thermal equilibrium, the 𝑛. 𝑝 product is a constant dependent on only temperature.
The Band Theory
Electron energy levels in materials expand into bands; closely spaced levels. In
semiconductors, the low energy levels are full while the upper energy levels are empty. The
highest filled band is the valence band while the lowest filled band is the conduction band.
The valence band and the conduction band in semiconductors are separated by a small
energy (band) gap. For an electron to be able to conduct current, it must be in the
conduction band.
Cosmas Mwikirize
CMP 1101: ELECTRONICS I-CHAPTER II
Conduction Band
𝐸𝑐
Band Gap
𝐸𝑓
𝐸𝑔
𝐸𝑣
Valence Band
Energy Band Diagram
𝐸𝑐 is the energy at the bottom of the conduction band while 𝐸𝑣 is the energy at the top of
the valence band. 𝐸𝑔 is the energy gap between the valence and the conduction bands.
Food for thought: How does the energy gap of semiconductors compare with that of
conductors and insulators?
Band Gap Energies of Selected Semiconductors
Material PbTe
Ge
Si
GaAs
GaP
Diamond
𝐸𝑔 (𝑒𝑉)
0.67
1.12
1.42
2.25
6.0
0.31
At 0K, the valence band is totally full while the conduction band is empty. There is no
current flow. With increase in temperature, electrons gain thermal energy and are able to
surmount the energy gap, and cross from the valence band to the conduction band. An
electron jump to the conduction band is accompanied by a hole left in the valence band.
Recombination of carriers occurs when an electron in the conduction band falls back into
the valence band. At thermal equilibrium, the generation and recombination rates are still
equal and the carrier concentration similar to that in the bond theory.
𝐸𝑓 is the Fermi level. It is the highest energy attainable by least held electrons in a solid at
absolute zero temperature. Note that for an intrinsic semiconductor, the Fermi level is in
the middle of the energy gap. Recall that 𝑛 = 𝑝 = 𝑛𝑖 .
Cosmas Mwikirize
CMP 1101: ELECTRONICS I-CHAPTER II
2.3
Material Statistics and the Fermi Function
Let me introduce some material statistics. Electrons cannot exist in similar states because of
the Pauli Exclusion Principle. At absolute zero temperature, they pack into the lowest
energy states, below the Fermi level. At higher temperatures, a certain fraction of electrons
will exist above the Fermi level, characterised by the Fermi function, which expresses the
probability that an available state at energy E is occupied.
𝑓(𝐸 ) =
1
(2)
𝐸−𝐸𝐹
1+𝑒 𝑘𝑇
1
𝑘 is the Boltzmann constant, = 1.380 × 10−23 𝐽𝐾 −1 . Note that at 𝐸 = 𝐸𝑓 , 𝑓(𝐸) = .
2
For your consideration: Prove that 𝑘𝑇 ≈ 26𝑚𝑉 at room temperature.
Note that at high energy levels, 𝐸 − 𝐸𝑓 ≫ 𝑘𝑇, 𝑓(𝐸) → 0 while at low energy levels, 𝐸 −
𝐸𝑓 ≪ 𝑘𝑇, 𝑓(𝐸) → 1. Because you are new to this domain, I will give you formulae for
calculating the equilibrium carrier concentrations. These will become clearer as we move
along.
𝐸𝑐 −𝐸𝐹
𝑘𝑇
𝑛 = 𝑁𝑐 𝑒 −
and 𝑝 = 𝑁𝑣 𝑒
𝐸 −𝐸
− 𝐹 𝑣
(3)
𝑘𝑇
𝑁𝑐 is the effective density of sates of the conduction band while 𝑁𝑣 is the effective density
of states of the valence band. For Silicon, 𝑁𝑐 = 2.8 × 1019 𝑐𝑚−3 while 𝑁𝑣 = 1.04 ×
1019 𝑐𝑚−3. Note that the closer 𝐸𝑓 moves towards 𝐸𝑓 E c , the larger n is. The closer E f
moves towards E v , the larger p is. This concept is handy in the next section.
Example 1
Calculate E c @E f for Silicon if n 10 cm@3 at room temperatureA
17
Solution:
Cosmas Mwikirize
CMP 1101: ELECTRONICS I-CHAPTER II
From
Efffffffffffffffffffffffffffffffff
c @E f
@
kT
n Nce
f
g
E c @E f
n
ln fffffffff  @ fffffffffffffffffffffffff
Nc
kT
f
h
g
17
i
n
10
k eV  0.1465 eV .
E c @E f  @kT ln fffffffff  @0.026 lnj fffffffffffffffffffffffffffff
19
Nc
2.8 B10
We know the energy gap for Silicon to be 1.12eV so we can locate the Fermi level:
0.1486 eV
Ec
EF
EV
Relation between np and Intrinsic Carrier Concentration
From (3), multiplying n and p yields
np  N c N v e
Effffffffffffffffffffffffffffff
v @E c
kT .
E c @E v  E g
so
Since
np  ni2 ,
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ni 
r
NcNve
Effffffffffffffffffffffffffffff
v @E c
kT
. We
know that
we
can write
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ni 
r
NcNve
@E
g
fffffffffffffffffffff
kT
(4)
It is important to note here that the relationship np  ni2 is obeyed by any semiconductor,
irrespective of its purity, but for a pure semiconductor, n  p  ni .
2.4
Doping
Pure semiconductors have low conductivity at ordinary temperatures. However, their
properties can be improved by introducing calculated amounts of other materials
(impurities) into them. The process of introduction of foreign atoms to engineer the
electrical properties of a semiconductor is known as doping.
Cosmas Mwikirize
CMP 1101: ELECTRONICS I-CHAPTER II
Doping is done using atoms from group III or V. We have two cases to consider:
Case I:
Consider a Group V atom such as Phosphorus, with 5 valence electrons, introduced in a
silicon crystal. 4 electrons participate in bonding with the silicon atoms. The fifth electron is
free and easy to release such that at room temperature, the atom releases one electron
that is free for conduction. We say the Phosphorous atom has donated a mobile electron to
the silicon crystal, and hence call it a donor atom. The donor site becomes positively
charged, but this is a fixed charge (immobile).
Let
N d be the number of phosphorous atoms introduced per unit volume (donor
concentration,
cm@3 ), if
N d << ni , doping is irrelevant and n  p  ni (intrinsic
semiconductor) . However, if N d >>>ni , the doping controls the carrier concentrations and
2
nffffffffff
we have an extrinsic semiconductor: n  N d , p  i and n>>p . Since the number of
Nd
electrons outweighs the number of holes, we call this an N-type semiconductor.
Example 2
15
What is the hole concentration in N-type silicon with 10 cm@3 of donor atoms?
Solution:
20
ni2 10
5
10
p  ffffffffff
 ffffffffffff
10 cm@3 . (Recall that for silicon at room temperature, ni 10 cm@3 ).
N d 1015
Case II
Consider a Group III atom such as Boron, with 3 valence electrons, introduced in a silicon
crystal. 3 electrons participate in bonding with the silicon atoms. One bonding site on the
Boron atom is unsatisfied and is easy to accept one neighboring bonding electron to
complete all bonds. This creates a hole (missing electron) on the silicon atom while the
Boron atom becomes negatively charged. We say the Boron atom has accepted an electron
from silicon, and hence call it an acceptor atom. The acceptor site becomes an immobile
negatively charged ion.
Let
N a be the number of phosphorous atoms introduced per unit volume (acceptor
concentration,
cm@3 ), if
N a << ni , doping is irrelevant and n  p  ni (intrinsic
semiconductor) . However, if N a >>>ni , the doping controls the carrier concentrations and
Cosmas Mwikirize
CMP 1101: ELECTRONICS I-CHAPTER II
ni2
we have an extrinsic semiconductor: p  N a , n  ffffffffff
and p>>n . Since the number of
Na
holes outweighs the number of electrons, we call this a P-type semiconductor.
Example 3
16
What is the electron concentration in P-type silicon with 10 cm@3 of acceptor atoms?
Solution:
20
ni2 10
4
n  ffffffffff
 ffffffffffff
10 cm@3 .
N a 1016
General Effects of Doping
What if a semiconductor is doped with both n and p type materials? We need to establish a
generic formula to use for such cases. We first introduce the principle of charge neutrality,
that the net charge on the semiconductor material is zero. Some clarity here for all the
species present and their charges:

Electrons, n (-)

Holes, p (+)

Donor atoms, N d (+)

Acceptor atoms, N a (-)
The total positive charge is equal to the total negative charge such that:
n Na p Nd
or n  N a @p @N d  0
(5)
We have assumed total ionization of all donor and acceptor atoms. The resultant type will
depend on which charge constitutes the majority carrier.
For
N
type
semiconductor,
N d @N a >>ni ,
and
2
nfffffff
i
n  N d @N a , p  .
n
ni2
N d >> N a , n  N d , p  ffffffffff
Nd
For
P
type
semiconductor,
N a >> N d , p  N a , n 
Cosmas Mwikirize
If
(6)
N a @N d >>ni ,
2
nffffffffff
i
Na
CMP 1101: ELECTRONICS I-CHAPTER II
p  N a @N d , n 
2
nfffffff
i
.
p
(7)
If
Making the latter approximation just saves computation time but a correct answer would
still be obtained using the first formulae in (6) and (7).
Example 4
Calculate the electron and hole concentrations for silicon with N d  6B10 cm@3 and
16
N a  2 B10 cm@3 .
16
Solution:
Clearly,
N d @N a >>ni .
n  N d @N a  4 B10 cm
16
@3
We
have
an
N
type
material
with
20
2
nfffffff
10
3
ffffffffffffffffffffffff
i
and p 

 2.5 B10 cm@3 .
16
n 4 B10
Effect of Doping on the Fermi Level
Let us now analyze the effect of doping from the angle of the band theory. Doping
introduces additional electron energy levels into the band gap that may or may not be
populated by electrons, dependent on circumstances and temperature, and causes the
Fermi level EF to shift from the intrinsic level.
Addition of donor atoms introduces the donor levels just below the conduction band and
the Fermi level is raised, with the donor states full. At room temperature, all donor
electrons have enough energy to cross from the donor states to the conduction band.
Ec
Donor Level
ED
Ev
T= 0 K
Increasing T
Room Temperature
The introduction of acceptor atoms has an opposite effect, acceptor levels are created in
the energy gap just above the valence band. The Fermi level is lowered. At 0K, the acceptor
levels are unpopulated by electrons. As temperature is increased, electrons cross from the
Cosmas Mwikirize
CMP 1101: ELECTRONICS I-CHAPTER II
valence band into the acceptor levels, leaving holes. At room temperature, all acceptor
atoms will have been ionized.
Ec
Ea
Acceptor Level
Ev
T= 0 K
2.5
Increasing T
Room Temperature
Current in Semiconductors
As we are by now aware, current in semiconductors is due to motion of electrons and holes,
what we call carriers. Let us look at the various modes of carrier transport:
2.5.1 Thermal motion
In thermal equilibrium, carriers undergo collisions with vibrating silicon atoms and
electrostatically interact with charged dopants and with one another. What results is zigzag
(random) motion.
The characteristic time constant of thermal motion is the mean free time between
collisions,  c . In between collisions, carriers acquire very high velocity, v th , travelling a
distance known as the mean free path   v th  c . This motion is hence not important for
electric current.
Cosmas Mwikirize
CMP 1101: ELECTRONICS I-CHAPTER II
You will recall from particle dynamics that the energy of a particle under thermal agitation is
given by
3fff
kT . To calculate the thermal velocity, we equate the kinetic energy of the
2
particle to the above energy:
1fff
3
m v th2  fffkT
2
2
(8)
2.5.2 Carrier Drift
When an electric field of intensity E (V/cm) is applied to a semiconductor, the net
electrostatic force on a carrier is F  F qE . Note that an electron will be urged in the
opposite direction to the field while a hole will be urged in the same direction as the field.
Consider the electron:
E
We have motion due to the electric force impinged on thermal motion. At every collision,
the carriers lose momentum. Between collisions, the carriers accelerate. From Newton’s
` a
` a Ffffffffffffffffffffff
` a @qE
dV
t
qE t
t
ffffffffffffffffffff
ffffffffffffffffff
 F qE . Integrating yields V t 
second law of motion, m
i.e. V t 
m
m
dt
` a
for electrons and V t 
qE
t
ffffffffffffff
for holes. Since velocity is randomised every  c due to
m
collisions, the net velocity, called the average drift velocity is given by
vd 
Ffffffffffffffffffffffffff
q E c
(cm/s)
m n,p
(9)
q
ffffffffffff
Let us now define the quantity  n,p  c . Substitution in (9) yields
mn,p
v d  = F  n,p E
(10)
The quantity  n,p is known as mobility of a carrier. (Units: cm 2 / V.s )
Hence, for electrons, v dn  @ n E and for holes: v dp   p E
(11)
Mobility is the measure of ease of carrier drift. A longer mean free time implies fewer
collisions and hence mobility of the carrier increases. Effective mass of the carrier also
Cosmas Mwikirize
CMP 1101: ELECTRONICS I-CHAPTER II
influences its mobility; the lighter the particle, the more its mobility. Since holes are heavier
than electrons,  n > p under similar conditions. Mobility is affected by doping. For a low
doping level, it is limited by collisions with the lattice. As the doping level increases, there is
collision with ionised impurities and this impedes carrier motion greatly, reducing mobility.
Selected Hole and Electron Effective Masses
mfffffffff
n
mo
mfffffffff
p
mo
Si
Ge
GaAS
GaP
0.26
0.12
0.068
0.82
0.39
0.30
0.5
0.6
Electron and Hole Mobilities of Selected Semiconductors
Si
Ge
GaAS
GaP
 n ( cm 2 /V.s )
1400
3900
8500
30,000
 p ( cm 2 /V.s )
470
1900
400
500
For consideration: Based on the above table alone, which carriers and which semiconductor
are attractive for use in high speed electronic devices?
Example 5
Given  p = 470 cm 2 / V·s, calculate the hole drift velocity at E=10 V/cm. Evaluate the mean
3
free time and the mean free path.
Solution:
Drift velocity v dp =  p E  470 B10  4.7 B10 cm/s =  4.7B10 m/s
3
Since
q c
P  ffffffffffff
,
mp
3
5
 p m p 470 B10@4 B 0.39 B 9.1 B10@31
@13
 c  ffffffffffffffffff  ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
 1 B10 s  0.1 ps , as
@19
q
1.602 B10
the mean free time. (Note the use of standard units)
Mean free path  p  v th  c . Note that v th is the thermal velocity, different from the drift
velocity.
3
1
From (8), fffkT  fffmv th2 [
2
2
Cosmas Mwikirize
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u
@23
u
3kT
3 B 1.38 B10 B 300
5
v th  s fffffffffffff  t ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
 1.87 B10
@31
m
0.39 B 9.1 B10
CMP 1101: ELECTRONICS I-CHAPTER II
m/s
 p  1.87 B10 B 1 B10
@13
5
 1.87 B10
@8
b
m  19 nm 187 A
c
0
Drift Current
Net velocity of carriers leads to flow of current which we call drift current, whose
magnitude depends on:

Carrier drift velocity

Carrier concentration

Carrier charge
b
Drift current density due to electrons
b
c
J n drift  @qn v dn  q n un E while that due to
c
holes is given by J p drift  q p v dp  q p u p E
(12)
Note that current due to electron drift is in the opposite direction to the motion of electrons
while current due to holes is in the same direction. Both currents are therefore in the
direction of the electric field.
b
c
b
c
b
c
Total drift current = J p drift  J n drift  q n n E  q p  p E  q nn  p  p E
(13)
Equation (13) has the shape of Ohm’s law : J  E
Derivation of J =  E
V
Recall that Ohm’s law is V= IR as we know it from High School! We are right to write I  fffff
R
I V
Dividing both sides by the cross sectional area A of a conductor yields fffff fffffffff
A RA
(14)
The LHS of (14), the current per unit area is the current density, J. We know that electric
field intensity E is equal to the potential gradient, thus E 
Vfffff
from which V  El , l being
l
El E
the length of a conductor. Substitution in (14) yields J  fffffffff fffffffff
l
RA fffffffffff
(15).
RA
The denominator of (15) looks familiar, it is indeed resistivity of the conductor,  whose
reciprocal is the conductivity,  .
E
We conclude that Ohm’s law V  IR  fffff  E

Cosmas Mwikirize
CMP 1101: ELECTRONICS I-CHAPTER II
We are now in position to conclude that by comparison,
b
c
  q n n  p p ( A
@1
cm@1 )
(16)
1
b
c
While   fffffffffffffffffffffffffffffffffffffffffff
q n n  p p
(17)
1
Resistivity is normally used to specify doping level. In an N type semiconductor,   ffffffffffffffffffffffff
q N d n
1
while in a P type semiconductor,   ffffffffffffffffffffffff
q N a p
(18)
Note that in both cases, resistivity is inversely related to the doping level.
Example 6
16
A silicon wafer is doped with phosphorous atoms with density of 3B10 cm@3 at 300K. If
the mobility of electrons is 1000 cm 2 /V.s, calculate the resistivity of the wafer. If an electric
field of 1 kV /cm is applied, calculate the drift velocity and the drift current density that
results. What is the time taken to drift through 0.1 m
Solution
1
1
 0.21 cm
Resistivity   fffffffffffffffffffffff fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
qN d  n 1.602 B10@19 B 3 B1016 B 1000
Drift velocity v dn  @n E  @1000 B 1000  @10 cm /s .The negative indicates that the
6
motion is opposite to the electric field, and can be neglected in subsequent computation.
E 1000
3
Drift current J dn  fffff  fffffffffffffff 4.76 B10 A/ cm 2
 0.21
@4
Time taken to drift =
2.5.3
L 0.1
B10 cm
fffffffff
ffffffffffffffffffffffffffffffffffffffffffff
v dn

10
6
 10 ps
Carrier Diffusion
This is the movement of carriers in response to a concentration gradient.
Consider
semiconductor material with carriers more concentrated in one region than another. The
particles tend to move from the region of high concentration to the region of low
concentration. Collisions and scattering still occur but overall particle movement is down
the concentration gradient.
Cosmas Mwikirize
CMP 1101: ELECTRONICS I-CHAPTER II
Diffusion of carriers follows Fick’s law: The diffusion flux is directly proportional to the
concentration gradient. Flux here refers to the number of particles diffusing per unit area
per unit time ( cm 2 s@1 ).
dp
dn
For electrons, the flux F n  @Dn ffffffff while for holes, F p  @D p ffffffff with Dn and D p as the
dx
dx
diffusion coefficients for electrons and holes respectively ( cm 2 /s).
(19)
Note the negative sign on both expressions because of the negative gradient.
n
x
The diffusion co-efficient measures the ease with which carriers can diffuse in response to a
concentration gradient. It is limited by vibrating lattice atoms and ionised dopant atoms.
The diffusion current density is got by multiplying the carrier charge by the diffusion flux.
b
c
b
c
dp
dn
ffffffff
Hence, J n diffusion = qDn ffffffff and J n diffusion = @qD p .
dx
dx
(20)
Please take note of the signs once again. The electron current is in the direction of
increasing concentration while the hole current is down the concentration gradient.
Einstein Relation
There is a relationship between diffusion and mobility and this is given by Einstein’s relation,
D
kT
D kT
p
fffffffff D
fffffffff
which we will derive for now: ffffff  ffffffffff such that n 
 ffffffffff
n  p
q

q
kT
Recall ffffffffff as the thermal voltage, approximately 25mV at room temperature. Given
q
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CMP 1101: ELECTRONICS I-CHAPTER II
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mobility, we can be able to calculate the diffusion coefficient for a particular carrier.
Total Current
From (13) and (20), the total current is given by the expressions:
dn
J n  J n drift  J n diffusion  q n  n E  q Dn ffffffff
dx
dp
J p  J p drift  J p diffusion  q p  p E @q D p ffffffff
dx
Total Current = J n  J p
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