Semiconductors
Semiconductors are materials with a (relatively) small band
gap (typically 1eV) between a filled valence band and an
empty conduction band.
Chemical potential μ (often called Fermi energy) lies in the
band gap.
Insulators at T=0, with a small density of electrons excited
at finite temperatures.
Typical semiconductors are Silicon and Germanium
or III-V compounds such as GaAs
2 atoms in primitive basis have 4 electrons each (or 3 + 5); 8
electrons fill 4 bands made of s and p orbitals
Band Structure of Semiconductors
Plot Energy versus k. EF separates
filled and empty states
Electrons are only
excited in a very
limited region of
energy
EF
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Energy levels of electrons and holes
Close to the band edge minima and maxima we can write:
conduction band ε = EG
valence band
= 2k 2
+
2me
= 2k 2
ε = −
2mh
Band gap determines the optical
properties - strong absorption
when hν > EG
Optical absorption
Excitation promotes an
electron from the valence
band to conduction band.
An empty state left in valence
band is known as a hole.
⎛ = 2k 2 ⎞
= 2k 2
⎟⎟
Δε = EG +
− ⎜⎜ −
2me
⎝ 2mh ⎠
⎛
= 2k 2 ⎞
= 2k 2
⎟⎟ +
Δε = ⎜⎜ EG +
2me ⎠
2mh
⎝
Electron energy
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Absorption above
the band gap is
strong ~ 106 m-1
Hole energy
2
If band minima and
maxima are at different
points then we have an
indirect semiconductor.
The classic example is
silicon.
This affects the optical
properties such as
absorption where Δk ≈ 0
Hole picture
Remove one electron from a filled band and electricity can be
conducted by the movement of all of the electrons present.
The sum of this motion is equivalent to one positively
charged particle: a hole
filled band
crystal
0
momentum
Energy
0
RJ Nicholas – Semicond. HT10
Electron state
removed
part-filled
band
hole
ke
-ke
kh = -ke
εe(ke)
-εe(ke)
εh(kh) = - εe(ke)
3
Holes and their properties
velocity
vh =
1 ∂ε
= ve
= ∂k
∂ε
F ∂ε
= F vg =
∂t
= ∂k
∂ε ∂k
=
∂k ∂t
∂k
∂k
∴ = e = F = −= h
∂t
∂t
Rate of change of momentum is opposite sign for holes
Work done
by a force F
=
∂ke
= F = − e (E + v × B )
∂t
=
∂k h
= + e (E + v × B )
∂t
Effective masses
Force accelerates the electron or hole:
∂v g
1 ∂ 2ε
1 ∂ 2ε ∂k
=
=
∂t
= ∂k ∂t
= ∂k 2 ∂t
∂v g
1 ∂ 2ε
= 2 2 F =
∂t
= ∂k
Effective mass:
Negative for holes unless
we say charge is positive
RJ Nicholas – Semicond. HT10
but F = =
∂k
∂t
F
m*
1
1 ∂ 2ε
= 2
= ∂k 2
m*
4
Meaning of the effective mass
2
2⎛ ∂ ε
m * = = ⎜⎜ 2
⎝ ∂k
⎞
⎟⎟
⎠
−1
Effective mass changes as we
move through a band in k-space
positive (electron-like) at the
bottom, becoming negative
(hole-like) at the top
Typical values in semiconductors are in the range 0.01 to 0.5 me
Concentrations of Electrons and Holes
Calculate carrier density from density of states and
distribution function:
∞
= 2k 2
n = ∫ g (ε ) f (ε ) dε , where ε = EG +
2me
EG
For most semiconductors the chemical potential, μ, (often
also called the Fermi Energy (EF)), lies in the band gap so:
e
F −D
f
e
F −D
≈ exp (
μ −ε
k BT
)
1 ⎛ 2me ⎞
g (ε ) =
⎜
⎟
2π 2 ⎝ = 2 ⎠
and
3/ 2
(ε − EG )1/ 2
giving :
1 ⎛ 2me ⎞
n =
⎜
⎟
2π 2 ⎝ = 2 ⎠
RJ Nicholas – Semicond. HT10
3/ 2
e
⎛ μ
⎜⎜
⎝ k BT
⎞ ∞
⎟⎟
⎠
∫
(ε − EG )
1/ 2
−
e
ε
k BT
dε
EG
5
1 ⎛ 2me k BT ⎞
n =
⎟
⎜
2π 2 ⎝ = 2 ⎠
⎛m k T ⎞
n = 2 ⎜⎜ e B 2 ⎟⎟
⎝ 2π = ⎠
3/ 2
e
3/ 2
e
⎛ μ − EG
⎜⎜
⎝ k BT
⎞ ∞
⎟⎟
⎠
∫
x1/ 2 e − x dx ,
with x =
0
⎛ μ − EG
⎜⎜
⎝ k BT
⎞
⎟⎟
⎠
= NC e
⎛ μ − EG
⎜⎜
⎝ k BT
ε − EG
k BT
⎞
⎟⎟
⎠
Effective number of conduction band states at ε = Eg
Now use same procedure for holes
0
= 2k 2
p = ∫ g (ε ) f (ε ) dε , where ε = −
2mh
−∞
1
1
f Fh− D = 1 − f Fe− D = 1 − (ε − μ ) / k BT
= ( μ − ε ) / k BT
≈ e (ε − μ ) / k BT
e
+1
e
+1
h
F −D
1 ⎛ 2mh ⎞
p =
⎜
⎟
2π 2 ⎝ = 2 ⎠
3/ 2
e
⎞ 0
⎟⎟
⎠
∫
ε
(−ε )
1/ 2
e
k BT
dε
−∞
1 ⎛ 2mh k BT ⎞
p =
⎜
⎟
2π 2 ⎝ = 2 ⎠
⎛m k T ⎞
p = 2 ⎜⎜ h B 2 ⎟⎟
⎝ 2π = ⎠
⎛ −μ
⎜⎜
⎝ k BT
3/ 2
e
3/ 2
⎛ −μ
⎜⎜
⎝ k BT
e
⎛ −μ
⎜⎜
⎝ k BT
⎞ ∞
⎟⎟
⎠
∫
x1/ 2 e − x dx ,
with x =
0
⎞
⎟⎟
⎠
= NV e
⎛ −μ
⎜⎜
⎝ k BT
− εe
ε
= h
k BT
k BT
⎞
⎟⎟
⎠
Effective number of valence band states at ε = 0
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Law of Mass Action
⎛ μ − EG
k BT
⎜⎜
Densities of electrons
n = N C e⎝
and holes depend on μ
⎞
⎟⎟
⎠
p = NV e
,
⎛ −μ
⎜⎜
⎝ k BT
⎞
⎟⎟
⎠
μ is determined by density of charge introduced by impurities
(doping)
Product np is independent of μ: Law of Mass Action
⎛ − EG
k BT
3
⎜
⎛ k BT ⎞
3 / 2 ⎜⎝
⎟ (me mh ) e
np = 4 ⎜⎜
2 ⎟
2
π
=
⎝
⎠
⎞
⎟⎟
⎠
= N C NV e
⎛ − EG
⎜⎜
⎝ k BT
⎞
⎟⎟
⎠
Intrinsic Semiconductors
A semiconductor is said to be intrinsic if it is undoped, and
the only source of electrons and holes is by thermal
excitation from the valence band to the conduction band.
In this case:
n = p = ni =
N C NV e
⎛ − EG
⎜⎜
⎝ 2 k BT
⎞
⎟⎟
⎠
⎛ k T ⎞
= 2 ⎜⎜ B 2 ⎟⎟
⎝ 2π = ⎠
3/ 2
(me mh )
3/ 4
e
⎛ − EG
⎜⎜
⎝ 2 k BT
⎞
⎟⎟
⎠
We can use this relation to measure the Band Gap, by
measuring the carrier densities from the Hall Effect
At low temperatures ni → 0 and impurities are important
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Doping Semiconductors
We can control the numbers of electrons and holes in a
semiconductor by adding impurities which dope the
material.
Donors donate an electron to make the material more n-type.
A typical example is by adding a group V element (such as
As or P) to a group IV semiconductor such as silicon.
Four of the electrons participate in the sp3 bonds as if they
were from silicon, but the fifth electron is left over. Extra
charge on P nucleus creates a +ive core, and the fifth
electron can be bound to this, but the binding is weak.
Shallow Donors
Impurity binding looks like a hydrogen atom
Binding energy
e 4 me*
me* / me
*
R =
2(4πε r ε 0= ) 2
=
εr2
R0
Binding energy is very small because:
(i) the effective mass is small (typically m* = 0.1 me)
(ii) The wavefunction is large (much more than the crystal
unit cell), so we must include the relative dielectric constant
of the medium εr, - typically ≈10.
∴ R * ≈ 10 −3 R0 = 13.6 meV (155 K )
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Acceptors
Dope semiconductors with holes by adding group III elements to
a group IV material. e.g. put Ga into silicon.
One valence electron is missing. This creates a vacant state, a
hole, which binds to the ion core of the Ga which is negatively
charged.
e 4 mh*
mh* / me
*
Binding energy is:
=
R =
R0
ε2
2(4πεε 0 = ) 2
Where is the energy level?
Ionized Acceptor or Donor is a free hole or electron at the
top or the bottom of the band
∴ Acceptor is R* above the valence band edge
Donor is R* below the conduction band edge
Extrinsic Carrier Densities
Density of impurities (e.g. Donors) usually much less than NC,
NV. Impurities can be ionized Nd+, or neutral Nd0 so:
Nd = Nd+ + Nd0
Using charge neutrality we have:
n = p + Nd+
Simple argument:
At high (e.g. Room) Temperature kT > R* therefore all
donors will be ionized but the density of holes created by
excitation across the band gap is still small
∴ n ≈ Nd+ ≈ Nd
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Chemical Potential μ
At high (room) temperature most impurities are ionized
∴ n ≈ Nd+ ≈ Nd
μ lies below the Donor level so that most impurities are
empty (ionized), but is still close to conduction band.
Minority Carriers
By Law of Mass action the (very small) density of holes is:
p = ni(T)2/Nd
Temperature
dependent density and
chemical potential
Density is constant in
region around room
temperature
High temperature gives
intrinsic behaviour
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Conductivity of Semiconductors
σ = neμ e + peμ h , μ =
eτ
m*
μ is the mobility, which is defined by v = μE, the dift
velocity per unit electric field.
Conductivity is dominated by variation in densities.
Mobility is determined by scattering rate:
Low T: impurity scattering gives μ ~ T3/2
High T: phonon scattering gives μ ~ T-3/2
p-n junction
• Unbiased junction
n ∼ Ncexp -(Eg/kT)
n ∼ Ncexp -(Eg/kT)
-------
• Forward bias of V
n ∼ Ncexp -(Eg - eV/kT)
n ∼ Ncexp -(Eg /kT)
-------
n ∼ Nc
n ∼ Nc
eV
p ∼ Nv
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p-n junctions
j = nev
What is current flow across junction?
What is v? - due to diffusion of carriers with
diffusion coefficient D and lifetime τ.
L
Diffusion length L = Dτ
so v =
=
τ
D
τ
Electron current/area from:
E
−
D
k T
e
(i) p → n
j p→n = e
NC e
τ
G
B
(ii) n → p
j
(iii) Total
e
n→ p
e
Total
j
= e
D
= e
D
−
EG
k BT
−
EG
k BT
NC e
τ
NC e
τ
(1 − 1) = 0
Add bias voltage V - This only affects electron current
from n → p
Electron current/area from:
(i) p → n
(ii) n → p
j
e
p →n
j
e
n→ p
e
Total
j
(iii) Total
= e
D
= e
D
= e
D
τ
τ
−
NC e
NC e
− EG + eV
k BT
−
NC e
EG
k BT
EG
k BT
(e
eV
k BT
− 1) = I 0 (e
eV
k BT
τ
An equivalent pair of currents flows for holes with the
− 1)
same formulae, except for masses and lifetimes of holes
(D
= μk BT / e, the Einstein relation )
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Low Dimensional Structures and Materials
• Anisotropic Materials
• Artificial layered structures Quantum Wells and
Superlattices
• Electric or Magnetic Fields
applied in one direction.
Layers may be only a few
atoms thick
Heterojunctions
Energy levels for 2 different
semiconductors
RJ Nicholas – Semicond. HT10
Energy line up at junction of
two (undoped) materials
13
Reduced Dimensionality
Quantum Well removes 1 Dimension by quantization
Electron is bound in well and can only move in plane
2-D system - motion in x, y plane
Quantum Well - Type I
Typical Materials:
1: GaAs
2: (Al0.35Ga0.65)As
(Eg = 1.5 eV)
(Eg = 2.0 eV)
Energy levels are quantized in
z-direction with values En for
both electrons and holes ∴
E = En + =2k⊥2/2m*
↑
1-D
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↑
2-D
14
Infinite well - Particle in a box
L
• 1-D Motion in z-direction
n=3
nπz
ψ n = A sin
L
= 2 d 2ψ
−
=Eψ
2m * dz 2
n=2
2
= 2 ⎛ nπ ⎞
= 2 k ⊥2
En =
⎜
⎟ +
2m *⎝ L ⎠
2m *
Typical values L = 10 nm,
n=1
me* = 0.07 me → En = 54 n2 meV
∴System is Two-Dimensional when:
E2 - E1
>
kT
162 meV
25 meV at 300 K
Density of States
g(k)dk
g(ε)dε
3-D
Travelling waves
eikx (eik.r)
4πk 2 dk
(2π / L )3
Periodic boundary condit.
ψ(x) = ψ(x + L)
2-D
∴ eikL = 1 → k = ±2nπ/L
→ δk = 2π/L
ε = =2k2/2m*,
dε = (=2/2m*) 2k dk
RJ Nicholas – Semicond. HT10
2πk dk
(2π / L )2
3/ 2
V ⎛ 2m * ⎞ 1/ 2
⎟ ε dε
⎜
(2π )2 ⎝ = 2 ⎠
A ⎛ 2m * ⎞
⎜
⎟ dε
4π ⎝ = 2 ⎠
1-D
2dk
2π / L
1/ 2
L ⎛ 2m * ⎞ −1/ 2
⎜
⎟ ε dε
2π ⎝ = 2 ⎠
15
Optical Properties
3-D
Absorption coefficient is
proportional to the
density of states:
∴ α ~ ε1/2
Modified close to the band
gap due to ‘excitons’
2-D - Big Changes
Multiple Band gaps Band gap shift Sharper edge
For wide wells the sum of
many 2-D absorptions becomes
equivalent to the 3-D
absorption shape (ε1/2)
Correspondence principle.
RJ Nicholas – Semicond. HT10
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GaAs/Al0.35Ga0.65As
Quantum Well
absorption
• Sharp peaks due to
excitons
• peaks doubled due to
heavy and light holes
Semiconductor
lasers
Forward biased p-n
junction
Quantum Well laser
Fibre Optic Communications,
CD players, laser pointers
RJ Nicholas – Semicond. HT10
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How do we achieve low dimensionality?
Naturally anisotropic crystals
Controlled growth of layers and/or apply external potential
Deposit thin layers of single crystals to create
‘heterostructures’
Two Main techniques:
I) Molecular Beam Epitaxy (MBE)
II) Metal Organic Vapour Phase Epitaxy (MOVPE)
Molecular Beam Epitaxy
(MBE)
• Ultra High Vacuum
evaporation of molecular
species of elements
(Molecular Beam)
• Epitaxy - maintaining
crystal structure of the
‘substrate’ - which is a
single crystal
RJ Nicholas – Semicond. HT10
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