Download Lecture 14: Intrinsic Semiconductors

Lecture 14: Intrinsic Semiconductors:
Intrinsic Semiconductors:
 They are pure semiconductors (no impurities). The
conductivity is dominated by the properties of the
pure crystal. It is controlled only by the
temperature.
 Because it is pure, concentration of electrons
equals concentration of holes.
 To obtain the conductivity: we need to obtain the n
umber of electrons in the conduction band.
First let’s calculate the number of electrons within a
band,
I’ll go through these steps:
1. Density of states
2. Population density
3. Number of electrons in the conduction band
Density of states within a band:
How are the energy levels distributed within a band?
We can assume similar to an electron in a potential well
of size a.
π2 ℏ2
2
2
2
En =
(n
+
n
+
n
)
x
y
z
2
2ma
n2 = nx 2 + ny 2 + nz 2
Equal energy of 𝐸𝑛 lie in the surface of sphere with
radius n.
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The number of quantum states with energy equal to or
smaller than 𝐸𝑛 is the proportional to the volume of the
sphere (assume this number=η)
Since n are positive integers, the number of energy
states, which is equal to or smaller than En (η) should be
given as:
14 3
η=
πn volume
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where the factor come from?
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3
2
2
2
n = (nx + ny + nz )
2
2ma
= ( 2 2)
π ℏ
3⁄
2
E
3⁄
2
3⁄
2
We are interested in calculating the density of states
(the number of energy states per unit energy).
dη
(
)
D E =
dE
2ma2
π
η= (
)
6 π2 ℏ2
3⁄
2
dη
D=
dE
2
3⁄
E 2
2
=
3 π 2ma
∙ v( 2 2)
2 6
π ℏ
2
=
3⁄
2
E
3⁄
2
v 2ma
( 2 2)
2
4π π ℏ
E
1⁄
2
1⁄
2
where v = a3
This parabolic relation between E versus D(E) is for free
electron case. In crystals, the density of states is
modified by the energy conditions within the first
Brillouin Zone.
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The largest number of states is found near the center of
a band.
The number of states that have energy equal or below
Eη
dη = D(E)dE
To obtain the number of electrons per unit energy
(population density)
N(E) = 2 D(E)F(E)
Where:
N(E): population density
2D(E) comes from Pauli Principle
F(E): Fermi Distribution
The number of electrons in the conduction band:
The number of electrons that have energy equal or
smaller than 𝐸𝑛
dN ∗ = N(E)dE
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N(E) = 2 D(E) ∙ F(E)
Where 𝐷 (𝐸 ) is the number of possible energy levels
1
(
)
𝐹 𝐸 =
(𝐸 − 𝐸𝑓 )
exp
+1
𝑘𝑇
For high energies 𝐸 ⋙ 𝐸𝑓 , the exp. factor is larger than
1
1
∴ 𝐹 (𝐸 ) =
(𝐸 − 𝐸𝑓 )
exp
𝑘𝑇
−(𝐸 − 𝐸𝑓 )
= (exp
)
𝑘𝑇
3⁄
2
v 2m
dN ∗ = 2 ∙ 2 ( 2 )
4π ℏ
∞
3⁄
2
v
2m
∗
N = 2∫ ( 2)
2π
ℏ
E
1
E ⁄2
0
=
3⁄
2
v 2𝑚
( )
2π2 ℏ2
1⁄
2
∞
E − Ef
exp [− (
)] dE
kBT
E − Ef
exp [− (
)] dE
kBT
3⁄
2
𝐸𝑓
2𝑚
exp (
)∫ ( 2 )
𝑘𝐵 𝑇
ℏ
0
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𝐸
1⁄
2
𝐸
𝑒𝑥𝑝 [− (
)] 𝑑𝐸
𝑘𝐵 𝑇
∞
∫
1
z ⁄2 e−nz dz
0
3⁄
2
v 2m
∗
N = 2( 2)
2π ℏ
1 π
√
=
2n n
Ef k B T
1
(πk B T) ⁄2
exp [
]
KBT 2
3⁄
2
v 2mk B
= (
)
4 πℏ2
Ef
3
exp (
) T ⁄2
kBT
Divide by v to get the number of electrons in the
conduction band per unit volume
∗
𝐸𝑔
𝑚
3
𝑒
𝑁𝑒 = 4.84 ∗ 1015 (
) 𝑇 ⁄2 exp [−
]
𝑚0
2𝑘𝐵 𝑇
𝐸𝑓 = −
𝐸𝑔
⁄
2
me ∗
We multiplied the equation by ( m ) where me ∗ is the
0
effective mass of the electron
Inspection of last equation shows that the number of
electrons in the conduction band depends on m:
 Energy gap
 Temperature
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Note: The number of holes in the valence band must
equal the number of electrons in the conduction band.
Conductivity (𝜎): We are in position now that we are
able to evaluate the conductivity-
𝑁𝑣𝑒
j=
= Neμ
ℇ
j = σℇ
j: current density ℇ: electric field
𝑁𝑣𝑒
=
= Neμ
ℇ
where μ = 𝑣⁄ℇ is the mobility of the current carriers
(drift velocity per unit electric field)
σ = 𝑁𝜇𝑒
The conductivity is due to electrons and holes.
σ = Ne eμe + Nh eμh
= Ne(μe + μh )
3⁄
2
∗
𝑚
= 4.84 ∗ 1015 ( )
𝑚0
Effect of Temperature:
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assume Ne = Nh
𝑒(μe + μh )𝑇
3⁄
2 exp [−
𝐸𝑔
]
𝑘𝐵 𝑇
Effect of impurity and lattice defects:
Leads to electron scattering at low temperatures.
Temperature dependence of the energy band gap:
The energy gap is slightly temperature dependent.
Why?
Empirical equations:
𝛼𝑇 2
𝐸𝑔 (𝑇) = 𝐸𝑔0 −
𝑇 + 𝜃𝐷
Eg : band gap at T = 0
α: constant
θD : Debye temperature
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HW: Calculate the number of electrons in the
𝑚∗
conduction band for silicon at T=300K. Assume 𝑚 = 1
0
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Extrinsic semiconductors
Extrinsic Semiconductor:
 They are doped materials
 n-type or p-type
 Conductivity depends on the impurity
concentration
n-type Semiconductor:
 Impurity has 1 extra valence electron → give donor
level just below the conduction band
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 Majority carriers: electrons
p-type Semiconductor:
 Impurity has less 1 valence electron→ acceptor
level just above the valence band.
 Majority carriers: holes
Let us study Si as an example.
Si doped with a group V impurity (n-type)
If we dope Si with an element of group V in the periodic
table (group after Si) such as Sb, As, or P
Each impurity at m (of group V) will replace as Si atom
and use 4 of its valence electrons to form covalent
bonding with Si atoms. There will be 1 extra electron
(not very tight to the atom). This electron needs less
energy to move to conduction band.
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Eg − ED ≈ 10−2 eV
(measured values)
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