Chapter 4
The semiconductor in
equilibrium
W.K. Chen
„
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Equilibrium (Thermal equilibrium)
‰
‰
‰
W.K. Chen
No external forces such as voltages, electric fields, magnetic fields, or
temperature gradients are acting on the semiconductor
All properties of the semiconductor will be independent of time at
equilibrium
Equilibrium is our starting point for developing the physics of the
semiconductor. We will then be able to determine the characteristics that
result when deviations from equilibrium occur
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Outline
„
„
„
„
„
„
Charge carriers in semiconductor
Dopant atoms and energy levels
The extrinsic semiconductor
Statistics of donars and acceptors
Charge neutrality
Position Fermi energy level
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4.1 Charge carriers in semiconductors
„
„
Current is the rate at which charge flow
Two types of carriers can contribute the current flow
‰
‰
„
Electrons in conduction band
Holes in valence band
The density of electrons and holes is related to the density of state function
and Fermi-Dirac distribution function
n( E )dE = g c ( E ) f F ( E )dE
n(E)dE: density of electrons in CB at energy levels between E and E+dE
p( E )dE = gυ ( E )[1 − f F ( E )]dE
p(E)dE: density of holes in VB at energy levels between E and E+dE
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Carriers concentration in intrinsic semiconductor
at equilibrium
n( E )dE = g c ( E ) f F ( E )dE
p( E )dE = gυ ( E )[1 − f F ( E )]dE
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no equation
The thermal equilibrium concentration of electrons in CB
top
top
Ec
Ec
no = ∫ n( E )dE = ∫ g c ( E ) f F ( E )dE
Q f F (E) =
1 + exp
1
(E − E f )
kT
4π (2mn* ) 3 / 2
gc (E) =
h3
E − Ec
For electrons in CB,
E > EC ⇒ ( E − E f ) >> kT
⇒ Q f F (E) =
W.K. Chen
1 + exp
− (E − E f )
1
≈ exp
(Boltzmann aproximation)
(E − E f )
kT
kT
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no = ∫
top
Ec
4π (2mn* ) 3 / 2
h3
E − Ec ⋅ exp
⎛ 2πmn* kT ⎞
⎟⎟
no = 2⎜⎜
2
h
⎝
⎠
3/ 2
exp
− (E − E f )
kT
dE
− ( EC − E f )
kT
We define effective density of states in CB,
⎛ 2πmn* kT ⎞
⎟⎟
N C = 2⎜⎜
2
h
⎝
⎠
3/ 2
The thermal-equilibrium electron concentration in CB
no = N C exp
W.K. Chen
− ( EC − E f )
kT
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Example 4.1 electron concentration
N c (300 K ) = 2.8 ×1019 cm −3
E f is 0.25 eV below Ec
⇒ Find the electron concentration in CB at 300K
Solution
no ≈ N c exp
− ( Ec − E f )
kT
= (2.8 ×1019 ) exp(
W.K. Chen
− 0.25
) = 1.8 ×1015 cm −3
0.0259
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po equation
The thermal equilibrium concentration of holes in VB
po = ∫
top
Ec
top
p ( E )dE = ∫ gυ ( E )(1 − f F ( E ))dE
Ec
Q1 − f F ( E ) = 1 −
gυ ( E ) =
1 + exp
1
=
(E − E f )
4π (2m*p ) 3 / 2
kT
1 + exp
1
(E f − E)
kT
Eυ − E
h3
For holes in VB,
E < Eυ ⇒ ( E f − E ) >> kT
Q1 − f F ( E ) =
⇒
1 + exp
− (E f − E)
1
≈ exp
(Boltzmann aproximation)
(E f − E)
kT
kT
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po = ∫
Eυ
0
4π (2m*p ) 3 / 2
h3
Eυ − E ⋅ exp
⎛ 2πm*p kT ⎞
⎟
po = 2⎜
⎜ h2 ⎟
⎠
⎝
3/ 2
exp
− (E f − E)
kT
dE
− ( E f − Eυ )
kT
We define effective density of states in CB,
⎛ 2πm*p kT ⎞
⎟
Nυ = 2⎜
⎜ h2 ⎟
⎠
⎝
3/ 2
The thermal-equilibrium electron concentration in CB
po = Nυ exp
W.K. Chen
− ( E f − Eυ )
kT
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Nc & Nυ values
⎛ 2πmn* kT ⎞
⎟⎟
N C = 2⎜⎜
2
h
⎝
⎠
„
„
3/ 2
⎛ 2πm*p kT ⎞
⎟
Nυ = 2⎜
⎜ h2 ⎟
⎝
⎠
3/ 2
Nc & Nυ are determined by the parameters of effective masses and
temperature. The effective mass is nearly a constant value ( a slight function
of temperature) for a semiconductor, which implies the Nc & Nυ will increase
in values with power of 3/2 with increasing temperature
Assume mn*=mo, then the value of the density of the effective density of
state at 300K is
N C = 2.5 ×1019 cm −3 if mn* = mo ,
„
which is lower the density of atoms in semiconductor
The effective mass of the electron in semiconductor is larger or smaller than
mo, but still of the same order of magnitude
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Example 4.2 hole concentration
Nυ (300 K ) = 1.04 ×1019 cm −3
E f is 0.27 eV above Eυ
⇒ Find the hole concentration in VB at 400K
Solution:
Nυ (400 K ) ⎛ 400 ⎞
=⎜
⎟
Nυ (300 K ) ⎝ 300 ⎠
3/ 2
po = Nυ exp
⎛ 400 ⎞
⇒ Nυ (400 K ) = (1.04 ×10 )⎜
⎟
⎝ 300 ⎠
− ( E f − Eυ )
19
= 1.60 ×1019 cm −3
kT
= (1.60 ×1019 ) exp(
W.K. Chen
3/ 2
− 0.27
) = 6.43 ×1015 cm −3
0.03453
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4.1.3 Intrinsic carrier concentration
„
For intrinsic semiconductor
‰
The concentration of electrons in CB is equal to the concentration of holes in VB
W.K. Chen
no = ni = N C exp
− ( EC − E f i )
kT
ni = pi
ni2 = N C exp
− ( EC − E f i )
ni2 = N C Nυ exp
po = pi = Nυ exp
− ( E f i − Eυ )
kT
for intrinsic semiconductor
kT
− ( EC − Eυ )
= N C Nυ exp
kT
„
13
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⋅ Nυ exp
− ( E f i − Eυ )
kT
− Eg
kT
The intrinsic carrier concentration is a function of bandgap, independent
of Fermi level
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Example 4.3 intrinsic carrier concentration
For GaAs, N c (300 K ) = 4.7 × 1017 cm −3
Nυ (300 K ) = 7.0 ×1018 cm −3
E g = 1.42 eV
⇒ Find the intrinsic carrier concentration at 300K and 400K
Solution:
ni2 = N C Nυ exp
− Eg
kT
ni2 (300 K ) = N c (300 K ) Nυ (300 K ) exp(
− 1.42
) = 5.09 × 1012
0.0259
ni (300 K ) = 2.26 × 106 cm −3
ni2 (400 K ) = N c (400 K ) Nυ (400 K ) exp(
− 1.42
) = 1.48 ×10 21
0.03885
ni (400 K ) = 3.85 ×1010 cm −3
W.K. Chen
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ni2 = N C Nυ exp
„
15
− Eg
kT
The intrinsic carrier concentration is
nearly increase exponentially with
temperature
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4.1.4 The intrinsic Fermi-level position
ni = pi
N C exp
− ( EC − E f i )
E fi =
kT
for intrinsic semiconductor
= Nυ exp
− ( E f i − Eυ )
kT
⎛N
1
1
( Ec + Eυ ) + kT ln⎜⎜ υ
2
2
⎝ Nc
⎛ m*p ⎞
1
1
E f i = ( Ec + Eυ ) + kT ln⎜ * ⎟
⎜m ⎟
2
2
⎝ p⎠
⎛ m*p ⎞
3
E f i = Emidgap + kT ln⎜ * ⎟
⎜m ⎟
4
⎝ p⎠
W.K. Chen
⎞
⎟⎟
⎠
Ec
E fi
3/ 2
Emidgap
Eυ
3/ 2
Emidgap =
1
( Ec + Eυ )
2
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Example 4.3 Intrinsic Fermi level
mn* = 1.08mo
m*p = 0.56mo
⇒ Find the intrinsic Fermi level
Solution:
⎛ m*p ⎞
3
E f i = Emidgap + kT ln⎜ * ⎟
⎜m ⎟
4
⎝ p⎠
3
⎛ 0.56 ⎞
E f i = Emidgap + (0.0259) ln⎜
⎟
4
⎝ 1.08 ⎠
W.K. Chen
3/ 2
3/ 2
= −0.0128eV = −12.8meV
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4.2 Dopant atoms and energy levels
The intrinsic semiconductor may be an interesting material, but the real
power of semiconductor is extrinsic semiconductor, realized by adding small,
controlled amounts of specific dopant, or impurity atom.
n-type semiconductor
„ A group V element, such as P atom is added into Si. 5 valence electrons, 4
of them contribute to the covalent bonding, leaving the 5th electron loosely
bound to P atom, referred as a donor electron
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n-type semiconductor
„ The energy level Ed is the energy state of donor impurity
„ If a small amount of energy, such as thermal energy, is added to the donor
electron, it can be elevated into CB, leaving behind a positively charged P
ion.
„ This type of impurity donates an electron to CB and so is called donor
impurity atoms, which add electrons to contribute the CB current, without
creating holes in VB
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p-type semiconductor
„ For silicon, a group III element, such as B atom is added. 3 valence
electrons are all taken up in covalent bonding, one covalent bonding position
appears to be empty. (Fig. a)
„ The valence electrons in the VB (Fig. b) may gain a small amount of thermal
energy and move to the empty state of group III element, forming empty
state in VB
„ The energy level Ea is the energy state of group III element in Si..
Acceptor energy level
VB
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p-type semiconductor
„ The empty positions in the VB are thought of as holes.
„ The group III atom accepts an electron from the VB and so is referred to as
an acceptor
„ The holes in the VB, formed due to the adding of acceptor impurity atoms
without creating electrons in the CB, can move through the crystal
generating a current
acceptor
B
hole
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4.2.2 Ionization energy
„
Ionization energy
The energy required to elevate the donor electron into the conduction band
The coulomb force of attraction between the electron and ion equal to the
centripetal force of the orbiting electron
e2
m*υ 2
=
Fc =
4πεrn2
rn
+
Due to the quantization of angular momentum
2
⎛ nh ⎞
⎜⎜
⎟ =υ2
* ⎟
⎝ rn ⋅ m ⎠
l = nh = rn ⋅ (m υ )
*
2
e2
m* ⎛ nh ⎞
n 2h 2
⎜
⎟ = 3 *
=
4πεrn2 rn ⎜⎝ rn ⋅ m* ⎟⎠
rn ⋅ m
4πεn 2 h 2
⎛m ⎞
= n 2ε r ⎜ o* ⎟ao
rn =
* 2
me
⎝m ⎠
W.K. Chen
rn ∝ n 2
o
4πεh 2
a
=
=
0
.
53
A
ao: Bohr radius o
mo e 2
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Consider the lowest energy state of donor impurity in silicon material, in
which n=1
4πεn 2 h 2
⎛ mo ⎞
2
ε
=
n
⎜ * ⎟ ao
r
m *e 2
⎝m ⎠
m
n = 1, ε r (Si) = 11.7, o* = 0.26
m
rn =
o
⇒ r1 = 45ao = 23.9 A >> ao
.
„
„
Because of the difference in the dielectric constant and effective electron
mass, the electron orbit in a semiconductor is much larger than that in a free
atom and the ionization energy of a donor state is considerably smaller than
that of hydrogen atom
The radius of lowest energy state in Si is about 4 lattice constant of Si, so
the radius of the orbiting donor electron encompasses many silicon atoms
The energy required to elevate the donor electron into the conduction band
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E = T +V
1 * 2
m *e 4
T = mυ =
2
2(nh) 2 (4πε ) 2
− e2
− m *e 4
V=
=
4πεrn (nh) 2 (4πε ) 2
− m *e 4
Ionization Energy E = T + V =
2(nh) 2 (4πε ) 2
Ionization energy of hydrigen : E (H) = −13.6 eV
Ionization energy of silicon :
W.K. Chen
E (Si) = −25.8m eV
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− m *e 4
E = T +V =
2(nh) 2 (4πε ) 2
„
Ordinary impurity
For ordinary impurities, such as P, As, and Sb in Si and Ge, the hydrogen
model works quite well.
„
Amphoteric impurity
For III-V semiconductor, the group IV such as Si and Ge is amphoteric
impurity. If a silicon atom replace a Ga atom, The Si impurity will act as a
donar, but if the Si atom replaces an As atom, then the Si impurity will act as
an acceptor
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4.3 Extrinsic semiconductor
„
„
Intrinsic semiconductor
A semiconductor with no impurity atoms present in the crystal
Extrinsic semiconductor
A semiconductor in which controlled amounts of specific dopant or impurity
atoms have been added so that the thermal-equilibrium electron and hole
concentrations are different the intrinsic carrier concentrations.
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4.3.1 Equilibrium distribution of electrons & holes
„
„
Adding donor or acceptor will change the distribution of electrons and holes in
semiconductor
Since the Fermi level is related to the distribution function, the Fermi level will change
as dopant atoms are added
no = N C exp
po = Nυ exp
− ( EC − E f )
kT
− ( E f − Eυ )
W.K. Chen
kT
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„
„
n-type semiconductor
When the density of electrons in CB is greater than the density of holes in VB
p-type semiconductor
When the density of holes in VB is greater than the density of electrons in CB
− ( EC − E f )
no = N C exp
kT
− ( E f − Eυ )
po = Nυ exp
kT
n - type no > po ⇒ E f > E fi
p - type po > no ⇒ E f < E fi
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29
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Example 4.5 thermal equilibrium concentrations
For Si, bandgap is 1.12 eV
Fermi energy is 0.25 eV below conduction band
N C = 2.8 ×1019 cm −3 Nυ = 1.04 ×1019 cm −3
Find the thermal equlibrium electron & hole concentrations
Solution:
no = N C exp
− ( EC − E f )
kT
po = Nυ exp
− ( E f − Eυ )
kT
− 0.25
= 1.8 ×1015 cm -3
0.0259
− 0.87
= 2.7 ×10 4 cm -3
po = (1.04 ×1019 ) exp
0.0259
no = (2.8 ×1019 ) exp
The electron and hole concentrations change by order of magnitude as the
Fermi energy changes by a few tenths of an electron-volt
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Another form of equations for no and po
n - type no = N C exp
− ( EC − E f )
intrinsic no = ni = N C exp
kT
− ( EC − E f i )
kT
⎡ − ( EC − E fi ) + ( E F − E fi ) ⎤
⇒ no = N C exp ⎢
⎥
kT
⎣
⎦
⎡ − ( EC − E fi ) ⎤
⎡ ( EF − E fi ) ⎤
no = N C exp ⎢
exp
⎥
⎢
⎥
kT
kT
⎣
⎦
⎣
⎦
⎡ ( EF − E fi ) ⎤
no = ni exp ⎢
⎥
kT
⎣
⎦
(n - type no > ni , E f > E fi )
⎡ − ( EF − E fi ) ⎤
po = ni exp ⎢
⎥
kT
⎣
⎦
(p - type no > ni , E f > E fi )
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4.3.2 The nopo product
no = N C exp
− ( EC − E f )
no po = N C exp
po = Nυ exp
kT
− ( EC − E f )
kT
⋅ Nυ exp
no po = N C ⋅ Nυ exp
„
„
„
− Eg
kT
− ( E f − Eυ )
kT
− ( E f − Eυ )
kT
= ni2
The product of no and po is always a constant for a given semiconductor material at a
given temperature, no matter it is intrinsic or extrinsic semiconductors,
The above equation is valid only when Boltzmann approximation is valid
The constant nopo product is one of the fundamental principles of the semiconductor
in thermal equilibrium
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4.3.3 The Fermi-Dirac Integral
„
If the Boltzmann approximation does not hold, the thermal equilibrium carrier
concentrations is expressed as Fermi-Dirac distribution function
n − type EC − E f > 3kT ⎫⎪
⎬ ⇒ Boltzmann approximation holds
p − type E f − Eυ > 3kT ⎪⎭
⇒ Fermi - Dirac distribution
elsewhere
Fermi-Dirac
Ec
Boltzmann
approximation
Ef
Eυ
Fermi-Dirac
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4π
no = 3 (2mn* ) 3 / 2 ∫
Ec
h
let η =
3kT
3kT
33
( E − Ec )1/ 2
dE
E − Ef
1 + exp(
)
kT
E − Ec
E − Ec
and η F = F
kT
kT
η 1/ 2
2mn* kT 3 / 2 ∞
no = 4π (
dη
) ∫
0 1 + exp(η − η )
h2
F
F1/ 2 (η F ) = ∫
∞
0
η 1/ 2
dη
1 + exp(η − η F )
3/ 2
⎛ 2πmn* kT ⎞
⎟⎟
N C = 2⎜⎜
2
h
⎝
⎠
3/ 2
*
⎛ 2πm p kT ⎞
⎟
Nυ = 2⎜
⎜ h2 ⎟
⎠
⎝
W.K. Chen
no =
2
po =
2
π
π
N c F1/ 2 (η F )
Nυ F1/ 2 (η F )
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4.3.4 Degenerate and nondegenerate
semiconductors
„
Nondegenerate semiconductor
If the doping is low, the impurity atoms are spread far enough apart so that
there is no interaction between donor (acceptor) electrons (holes), the
impurities introduced discrete, noninteracting energy states
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„
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Degenerate semiconductor
If the doping is high, the impurity atoms are close enough so that
donor/acceptor electrons/holes will begin to interact with each other. When
this occur, the single discrete energy state will split into a band of energy.
When the concentration of donor (acceptor) electrons (holes) exceeds the
effective density of states (Nc or Nυ), the Fermi level lies with the band. This
type of semiconductors are called degenerate semiconductors.
n - type degenerate semiconductor no > N c , E f > Ec
p - type degenerate semiconductor po > Nυ , Eυ > E f
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4.4 Statistics of donors and acceptors
Suppose we have Ni electrons and gi quantum states at ith impurity energy
level
Each donor level at least has two possible spin orientations for donor
electron, thus each donor has at least two quantum states
The distribution function of donor electrons in the donor energy states is
slightly different than the Fermi-Dirac function
f F (E) =
1
for allowed bands
E − Ef
1 + exp(
)
kT
f F (E) =
nd
=
Nd
1
E − Ef
1
1 + exp(
)
kT
2
for donor states
nd = N d − N d+
nd: the density of electrons occupying the donor level
Nd: the concentration of donor Nd+ : the concentration of ionized donor
Ed: the energy level of the donor
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f F (E) =
pa
=
Na
1
E − Ea
1
1 + exp( f
)
g
kT
37
for acceptor states
pa = N a − N a+
g is normally taken as four for acceptor level in Si and Ge because of the
detailed band structure
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4.4.2 Complete ionization and free-out
For non-degenerate semiconductor,
If Ed − E f >> kT
nd = N d f F ( E ) =
⎡ − ( Ed − E f ) ⎤
Nd
Nd
≈
= 2 N d exp ⎢
⎥
E − Ef
Ed − E f
1
1
kT
⎣
⎦
1 + exp(
)
exp(
)
kT
2
2
kT
no = N C exp
− ( EC − E f )
kT
The electron concentration in the conduction band is
no ≈ ni + ( N d − nd ) for n - type
The total number of electrons in and near the conduction band is
no + nd
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The ratio of electrons in the donor states to the total number of electrons
in and near the conduction band is
⎡ − ( Ed − E f ) ⎤
2 N d exp ⎢
⎥
kT
nd
⎣
⎦
=
− ( EC − E f )
nd + no
⎡ − ( Ed − E f ) ⎤
+
2 N d exp ⎢
exp
N
C
⎥
kT
kT
⎣
⎦
nd
1
=
nd + no 1 + N C exp − ( EC − Ed )
kT
2Nd
Ef
Ec
Ed + + + + +
3kT
Eυ
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Example 4.7
Si semiconductor, T = 300K, phosphorous doping
N d = 1016 cm -3
Solutions:
nd
=
nd + no
„
1
2.8 ×1019
− (0.045)
1+
exp
16
2(10 )
0.0259
= 0.0041 = 0.41%
Nondegenerate semiconductor
For shallow doped semiconductor, only a few electrons are still in the donor
states, essentially all of the electrons from the donor states are in the
conduction band. In this case, the donor states are said to be completely
ionized
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At T=300K
nd
1
=
nd + no 1 + N C exp − ( EC − Ed )
kT
2Nd
W.K. Chen
pa
1
=
pa + po 1 + Nυ exp − ( Ea − Eυ )
4Na
kT
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At T=0 K
All electrons are in their lowest possible energy states
f F (E) =
„
„
nd
=
Nd
Ed − E f
1
= 1 ⇒ exp(
) = 0 = exp(−∞) ⇒ E f > Ed
Ed − E f
1
kT
)
1 + exp(
2
kT
At T= 0 K, all the energy states below the Fermi level are full, and all the
states above the Fermi level are empty. Since the donor states is fully
occupied by donor electrons, the Fermi level must be well above the donor
energy state.
At T= 0K, no electrons from the donor state are thermally elevated into
conduction band ; this effect is called free-out.
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Example 4.8 90% ionization temperature
p - type Si semiconductor dpoed with boron (B)
N a = 1016 cm -3
Find the temperature at which 90% of acceptors are ionized
Solution:
pa
1
=
pa + po 1 + Nυ exp − ( Ea − Eυ )
4Na
kT
pa
= 0.1 =
pa + po
1
1+
T 3/ 2
)
− 0.045
300
exp
16
T
4(10 ) a
0.0259(
)
300
(1.04 ×1019 )(
⇒ T = 193 K
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4.5 Charge neutrality
„
„
„
Charge neutrality
In thermal equilibrium, the semiconductor crystal is electrically neutral
The charge-neutrality condition is used to determine the thermal equilibrium
electron and hole concentration as a function of impurity doping
concentration
Compensated semiconductor
A semiconductor contains both donor and acceptor impurity atoms in the
same region
‰
‰
‰
N-type compensated semiconductor ( Nd>Na)
P-type compensated semiconductor (Na>Nd)
Completely compensated semiconductor (Na=Nd)
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4.5.2 Equilibrium electron and hole concentration
Charge neutrality:
no + N a− = po + N d+
no + ( N a − pa ) = po + ( N d − nd )
Under complete ionization
nd = 0, pa = 0
no + N a = po + N d
ni2
no + N a =
+ Nd
no
no2 − ( N d − N a )no − ni2 = 0
(N − Na )
⎛ N − Na ⎞
2
n - type no = d
+ ⎜ d
⎟ + ni
2
2
⎝
⎠
2
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Electrophysics, NCTU
46
(N − Na )
⎛ N − Na ⎞
2
+ ⎜ d
n - type semiconductor no = d
⎟ + ni
2
2
⎝
⎠
2
ni2
po =
no
(N − Nd )
⎛ N − Nd ⎞
2
p - type semiconductor po = a
+ ⎜ a
⎟ + ni
2
2
⎝
⎠
2
ni2
no =
po
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47
Example 4.9 thermal equilibrium carrier conc.
n - type Si semiconductor, T = 300K, phosphorous doping
N d = 1016 cm -3 , N a = 0 and ni = 1.5 ×1010 cm -3
Solution:
majority carrier
2
⎛ 1016 ⎞
1016
⎟⎟ + (1.5 ×1010 ) 2 ≈ 1016 cm −3
+ ⎜⎜
no =
2
⎝ 2 ⎠
minority carrier
10 2
ni2 (1.5 ×10 )
=
= 2.25 ×10 4 cm −3
po =
16
10
no
The thermal equilibrium majority carrier concentration is essentially equal to
impurity concentration
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48
Redistribution of electrons when donors are
added
When add shallow donors
„
‰
‰
W.K. Chen
Most of donor electrons will gain
thermal energy and jump into the CB
A few of the donor electrons will fall
into the empty states in the VB and will
annihilate some of the intrinsic holes
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49
Electron concentration vs. temperature
„
Freeze-out (0 K)
nd = N d , no = 0
„
Partial ionization (0- ∼100K)
nd
1
=
nd + no 1 + N C exp − ( EC − Ed )
2Nd
kT
„
Extrinsic region (∼100K-400K)
no ≈ ni + ( N d − nd ) for n - type
„
Intrinsic region (>∼400 K)
− Eg
ni2 = N C ⋅ Nυ exp
kT
W.K. Chen
The above regions are determined by
type of semiconductor and doping
concentration
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50
Example 4.11 Compensated p-semiconductor.
p - type Si semiconductor, T = 300K,
N a = 1016 cm -3 , N d = 1.5 ×1015 cm -3 and ni = 1.5 ×1010 cm -3
Solution:
(N − Nd )
⎛ N − Nd ⎞
2
po = a
+ ⎜ a
⎟ + ni
2
2
⎠
⎝
2
majority carrier
(1016 − 3 × 1015 )
po =
+
2
2
⎛ 1016 − 3 ×1015 ⎞
⎜⎜
⎟⎟ + (1.5 ×1010 ) 2 ≈ 7 ×1015 cm −3
2
⎝
⎠
≈ (Na − Nd )
minority carrier
10 2
ni2 (1.5 ×10 )
no =
=
= 3.21×10 4 cm −3
15
po
7 ×10
Because of complete ionization at 300K, the majority carrier hole concentration
is just the difference between the acceptor and donor concentrations
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Electrophysics, NCTU
4.6 Position of Fermi energy level
no = N C exp
„
− ( EC − E f )
po = Nυ exp
kT
− ( E f − Eυ )
kT
When Boltzmann approximation holds,
The position of Fermi energy level moves through the bandgap energy,
depending on the electron and hole concentrations and temperature
⎛N
n - type EC − E f = kT ln⎜⎜ C
⎝ no
W.K. Chen
⎞
⎟⎟
⎠
⎛N
p - type EF − Eυ = kT ln⎜⎜ υ
⎝ po
Electrophysics, NCTU
⎞
⎟⎟
⎠
52
⎛N ⎞
n - type EC − E f = kT ln⎜⎜ C ⎟⎟
when N d >> N a , N d >> ni
⎝ Nd ⎠
⎛ NC ⎞
⎟⎟ when N d − N a = no , N d >> ni
= kT ln⎜⎜
N
−
N
a ⎠
⎝ d
⎛N ⎞
p - type E f − Eυ = kT ln⎜⎜ υ ⎟⎟
⎝ Na ⎠
⎛ Nυ
= kT ln⎜⎜
⎝ Na − Nd
W.K. Chen
when N a >> N d , N a >> ni
⎞
⎟⎟ when N a − N d = po , N a >> ni
⎠
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Example 4.13 Fermi energy
n - type Si semiconductor, T = 300K
N a = 1016 cm -3 and ni = 1.5 ×1010 cm -3
Fermi energy is 0.20 eV below the conduction band edge
Solution:
no = N C exp
− ( EC − E f )
kT
majority carrier
− 0.20
) = 1.24 ×1016 cm −3
0.0259
Q no = N d − N a
no = 2.8 ×1019 exp(
N d = 1.24 ×1016 + 1016 = 2.24 ×1016 cm −3
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Electrophysics, NCTU
54
Variation of Fermi level with doping conc at 300K
no = N C exp
po = Nυ exp
„
„
− ( EC − E f )
kT
− ( E f − Eυ )
kT
For silicon, at 300K, Nd, Na>>ni = (1.5x1010cm-3 )
The carrier concentrations are mainly determined by the impurity
concentration.
As the doping levels increase, the Fermi level moves closer to the
conduction band for n-type material and closer to valence band for the
p-type material
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Electrophysics, NCTU
55
Ef versus temperature
ni2 = N C ⋅ Nυ exp
„
„
„
− Eg
kT
The intrinsic carrier concentration ni is a strong function of temperature,
so the Ef is a function of temperature also
As high temperature, the semiconductor material begins to lose its
extrinsic characteristics and begins to behave more like an intrinsic
semiconductor
At the very low temperature, free-out occurs; the Fermi level goes
above Ed for the n-type material and below Ea for the p-type material
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Free-out
Ec
Extrinsic
Ef
Ed
Ec
E fi
E fi
Eυ
Eυ
W.K. Chen
Intrinsic
Ec
Ed
Ef
Ed
Ef
E fi
Eυ
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4.6.3 Relevance of the Fermi energy
„
If the two matetrials are brought into intimate contact, what would happen to
the carriers and Fermi level in these material?
The electrons will
tend to seek the
lowest possible
energy
In this example, the
electrons in
material A will flow
into the lower
energy states of
material B
„
Thermal equilibrium occurs when the distribution of electrons, as a
function of energy, is the same in the two materials.
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