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Semiconductor Lattice Structures
Diamond Lattices
Chapter 3: semiconductor
science and light emitting diodes
¾ The diamond-crystal lattice characterized by
four covalently bonded atoms.
¾ The lattice constant, denoted by ao, is 0.356,
0.543 and 0.565 nm for diamond, silicon, and
germanium, respectively.
¾ Nearest neighbors are spaced ( 3ao / 4) units apart.
¾ Of the 18 atoms shown in the figure, only 8 belong to the volume ao3.
Because the 8 corner atoms are each shared by 8 cubes, they contribute a
total of 1 atom; the 6 face atoms are each shared by 2 cubes and thus
contribute 3 atoms, and there are 4 atoms inside the cube.
The atomic density is therefore 8/ao3, which corresponds to 17.7, 5.00, and
4.43 X 1022 cm-3, respectively.
(After W. Shockley: Electrons and Holes in Semiconductors, Van Nostrand, Princeton, N.J., 1950.)
Semiconductor Lattice Structures
How Many Silicon Atoms per cm-3?
Diamond and Zincblende Lattices
Diamond lattice can be though of as an FCC structures with an
extra atoms placed at a/4+b/4+c/4 from each of the FCC atoms
• Number of atoms in a unit cell:
• 4 atoms completely inside cell
• Each of the 8 atoms on corners are shared among cells
Æ count as 1 atom inside cell
• Each of the 6 atoms on the faces are shared among 2
cells Æ count as 3 atoms inside cell
⇒ Total number inside the cell = 4 + 1 + 3 = 8
• Cell volume:
(.543 nm)3 = 1.6 x 10-22 cm3
Diamond lattice
Si, Ge
Zincblende lattice
GaAs, InP, ZnSe
The Zincblende lattice consist of a face centered cubic Bravais point lattice which contains
two different atoms per lattice point. The distance between the two atoms equals one quarter
of the body diagonal of the cube.
• Density of silicon atoms
= (8 atoms) / (cell volume) = 5 x 1022 atoms/cm3
Semiconductor Materials for
Optoelectronic Devices
The Si Crystal
• Each Si atom has 4 nea
rest neighbors
• lattice constant
= 5.431Å
“diamond cubic” lattice
400~450
450~470
470~557
557~567
567~572
572~585
585~605
605~630
630~700
Pure Blue
Blue
Pure Green
Green
Yellow Green
Yellow
Amber
Orange
Red
GaSb
In0.7Ga0.3As0.66P0.34
In0.14Ga0.86As
InP
GaAs
P
0.55 0.45
GaP(N)
GaAs
InGaN
SiC(Al)
In0.57Ga0.43As0.95P0.05
Semiconductor Materials for
Optoelectronic Devices
Semiconductor Optical Sources
GaAs1-yPy
x = 0.43
In1-xGaxAs1-yPy
AlxGa1-xAs
In0.49AlxGa0.51-xP
0.6
Red
Green
Yellow
Orange
0.5
Blue
Violet
0.4
0.7
0.8
λ
0.9
I n f r ar e d (um)
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
Quantization Concept
Periodic Table of the Elements
Group**
Period
plank constant
1
IA
1A
1
1
H
1.008
3
2
Semiconductor Materials
IV Compounds
SiC, SiGe
III-V Binary Compounds
AlP, AlAs, AlSb, GaN, GaP, GaAs,
GaSb, InP, InAs, InSb
III-V Ternary Compounds
AlGaAs, InGaAs, AlGaP
III-V Quternary Compounds
AlGaAsP, InGaAsP
II-VI Binary Compounds
ZnS, ZnSe, ZnTe, CdS, CdSe, CdTe
II-VI Ternary Compounds
HgCdTe
7
6
7
16
17
VIA VIIA
6A
7A
8
9
2
He
4.003
10
B
C
N
O
F
Ne
12.01
14.01
16.00
19.00
20.18
12
Na Mg
24.31
20
3
IIIB
3B
21
4
IVB
4B
22
5
VB
5B
23
K
Ca
Sc
Ti
V
39.10
40.08
44.96
47.88
50.94
38
39
40
41
6
VIB
6B
24
7
VIIB
7B
25
8
9
10
------- VIII ------------- 8 ------26
27
28
11
IB
1B
29
12
IIB
2B
30
34
35
36
Se
Br
Kr
78.96
79.90
83.80
42
43
44
45
46
47
48
Ru Rh
Pd
Ag Cd
101.1
106.4
107.9
95.94
74
(98)
75
76
102.9
77
78
Cs
Ba La* Hf
Ta
W
Re
Os
Ir
Pt
132.9
137.3
180.9
183.9
186.2
190.2
190.2
195.1
108
200.5
50
51
52
53
54
In
Sn
Sb
Te
I
Xe
114.8
118.7
121.8
127.6
126.9
131.3
81
82
83
84
85
86
Tl
Pb
Bi
Po
At
Rn
204.4
207.2
209.0
(210)
(210)
(222)
109
110
111
112
114
116
118
Ra Ac~ Rf
Db Sg
Bh
Hs
Mt
---
---
---
---
---
---
(226)
(260)
(262)
(265)
(266)
()
()
()
()
()
()
(263)
107
80
Au Hg
197.0
49
74.92
Fr
(257)
106
79
112.4
72.59
(223)
(227)
105
33
Ga Ge As
69.72
Nb Mo Tc
104
32
Zn
65.39
92.91
89
31
Cu
Zr
88
Ar
39.95
63.55
91.22
73
18
Cl
35.45
Ni
Y
72
17
S
32.07
58.69
88.91
178.5
16
P
Co
55.85
Sr
57
15
30.97
58.47
54.94
87.62
138.9
14
Si
28.09
Cr Mn Fe
Rb
56
13
Al
26.98
52.00
85.47
87
Valence electrons
5
15
VA
5A
10.81
55
6
4
14
IVA
4A
Be
37
5
13
IIIA
3A
9.012
19
4
2
IIA
2A
Li
22.99
Core electrons
8A
6.941
11
3
18
VIIIA
Semiconductor Materials
Atomic Bonding
Covalent Bonding
Bonding forces in Solids
a.
b.
c.
d.
e.
Ionic bonding (such as NaCl)
Metallic bonding (all metals)
Covalent bonding (typical Si)
Van der Waals bonding (water…)
Mixed bonding (GaAs, ZnSe…, ionic & covalent)
Quantization Concept
Quantization Concept
The Shell Model
plank constant
L shell with
two sub shells
Nucleus
1s
Core electrons
K
L
2s
2p
1s22s22p2 or [He]2s22p2
Valence electrons
¾ The shell model of the atom in which the electrons are confined
to live within certain shells and in sub shells within shells.
Energy Band Formation (I)
Energy Band Formation (I)
Band theory of solids
Two atoms brought together to form molecule
¾“splitting” of energy levels for outer electron shells
=
Allowed energy levels of
an electron acted on by
the Coulomb potential of
an atomic nucleus.
Energy Band Formation (II)
Pauli Exclusion Principle
Energy Band Formation (III)
Conceptual development of the energy band model.
where ‘no’ states exist
As atoms are brought closer towards
one another and begin to bond
together, their energy levels must
split into bands of discrete levels
so closely spaced in energy, they
can be considered a continuum of
allowed energy.
Electron energy
Electron energy
N isolated Si-atoms
Only 2 electrons, of spin ± 1/2, can
occupy the same energy state at
the same point in space.
¾ Strongly bonded materials: small
interatomic distances.
¾ Thus, the strongly bonded materials can
have larger energy bandgaps than do
weakly bonded materials.
The electrical properties of a crystalline material
correspond to specific allowed and forbidden energies
associated with an atomic separation related to the
lattice constant of the crystal.
p
s n=3
6N p-states total
2N s-states total
(4N electrons total)
p
Crystalline Si N -atoms
Electron energy
Energy Bandgap
Splitting of energy states into allowed bands
separated by a forbidden energy gap
as the atomic spacing decreases.
4N allowed-states
(Conduction Band)
No states
4N allowed-states
(Valance Band)
4N empty states
s
2N+2N filled states
isolated
Si atoms
Decreasing atom spacing
Si lattice
spacing
Electron energy
→ ←
Mostly
empty
Eg
Mostly
filled
Etop
Ec
Ev
Ebottom
Energy Band Formation (IV)
Energy Band
¾ Broadening of allowed energy levels into allowed energy bands
separated by forbidden-energy gaps as more atoms influence each
electron in a solid.
Energy band diagrams.
N electrons filling half of the 2N
allowed states, as can occur in a metal.
One-dimensional
representation
Two-dimensional diagram in which
energy is plotted versus distance.
Metals, Semiconductors, and Insulators
Metals, Semiconductors, and Insulators
Typical band structures of Metal
Electron Energy,
E
Typical band structures of Semiconductor
Covalent bond
Free electron
Vacuum
level
A completely empty band separated
by an energy gap Eg from a band
whose 2N states are completely filled
by 2N electrons, representative of an
insulator.
3s Band
Si ion core (+4e)
Electron energy, E
Ec+χ
E =0
2 p Band
3p
3s
2p
2s
2 s Band
ConductionBand(CB)
Empty of electrons at 0 K.
Ec
Overlapping
energy bands
Band gap = Eg
Ev
Electrons
Valence Band (VB)
Full of electrons at 0 K.
1s
ATOM
0
1s
SOLID
¾ In a metal the various energy bands overlap to give a single band of energies
that is only partially full of electrons.
¾ There are states with energies up to the vacuum level where the electron is free.
A simplified two dimensional
view of a region of the Si
crystal showing covalent
bonds.
The energy band diagram of
electrons in the Si crystal at
absolute zero of temperature.
Metals, Semiconductors, and Insulators
Metals, Semiconductors, and Insulators
Typical band structures at 0 K.
Carrier Flow for Metal
Carrier Flow for Metals.mov
Carrier Flow for Semiconductor
Carrier Flow for Semiconductors.mov
Insulator
Metals, Semiconductors, and Insulators
Semiconductors
Electrons within an infinite potential energy well of spatial width L,
its energy is quantized.
Conductors
En =
Many ceramics
Superconductors
Alumina
Diamond Inorganic Glasses
Polypropylene
PVDF Soda silica glass
Borosilicate Pure SnO2
PET
SiO2
10-18
10-15
Degenerately
doped Si
Alloys
Intrinsic Si
10-9
10-6
10-3
kn =
nπ
L
∞
n = 1,2,3...
h k n : electron momentum
E3
infinite square potential
∞
∞
Conductivity (Ωm)-1
106
109
1012
-a/2
0
ϕ2
x
E2
∞
n=3
x
n=2
-a/2
V=0
V(x)
103
∞
E1 n=1
Te Graphite NiCrAg
100
∞
ϕ
m
Amorphous
Intrinsic GaAs
As2Se3
10-12
( h k n2 )
2me
Energy increases
parabolically with the
wavevector kn.
Metals
Mica
Metal
Energy Band Diagram
Range of conductivities exhibited by various materials.
Insulators
Semiconductor
+a/2
x
Energy state
0 +a/2 -a/2
Wavefunction
0 +a/2
Probability density
¾ This description can be used to the behavior of electron in a Metal
within which their average potential energy is V(x) ≈ 0 inside, and very
large outside.
3.1 Energy Band Diagram
3.1 Energy Band Diagram
within the Crystal!
E-k diagram, Bloch function.
E-k diagram, Bloch function.
PE(r)
Potential Energy of the electron
around an isolated atom
r
When N atoms are arranged to
form the crystal then there is an
overlap of individual electron PE
functions.
V(x)
a
a
0
a
2a
Surface
Moving through Lattice.mov
¾ The electron potential energy [PE, V(x)], inside the crystal is periodic with the
same periodicity as that of the crystal, a.
¾ Far away outside the crystal, by choice, V = 0 (the electron is free and PE = 0).
3.1 Band Diagram
m = 1,2,3...
E-k diagram
There are many Bloch wavefunction solutions to the one-dimensional crystal each
identified with a particular k value, say kn which act as a kind of quantum number.
Each ψk (x) solution corresponds to a particular kn and
represents a state with an energy Ek.
The energy is plotted as a function of the wave number, k,
along the main crystallographic directions in the crystal.
The Energy Band
Diagram
E
k
¾
3.1 Energy Band Diagram
E-k diagram of a direct bandgap semiconductor
Si
Ge
GaAs
CB
Conduction
Band (CB)
eE
g
Valence
Band (VB)
Periodic Potential
Periodic Wave function
Surface
Crystal
Schrödinger equation
Bloch Wavefunction
x=L
3a
V ( x ) = V ( x + ma )
Ψk ( x ) = U k ( x ) e i k x
PE of the electron, V(x), inside the
crystal is periodic with a period a.
x
x=0
d 2 Ψ 2m e
+ 2 [ E − V ( x )] ⋅ Ψ = 0
dx 2
h
h+
Empty ψ
E
c
E
v
k
E
hυ
E
Occupied ψ
k
ec
v
hυ
h+
VB
k
– π /a
π/a
¾ The E-k curve consists of many discrete points with each point corresponding to a
possible state, wavefunction Ψk (x), that is allowed to exist in the crystal.
¾ The points are so close that we normally draw the E-k relationship as a continuous
curve. In the energy range Ev to Ec there are no points [no Ψk (x) solutions].
The bottom axis describe different directions of the crystal.
3.1 Energy Band Diagram
3.1 Energy Band
E-k diagram
A simplified energy band diagram with the highest almost-filled
band and the lowest almost-empty band.
E
E
E
CB
Direct Bandgap
Ec
Eg
Indirect Bandgap, Eg
Photon
CB
Ev
kcb
VB
k
–k
VB kvb
–k
GaAs
Er
Ev
k
vacuum level
CB
Ec
Ec
Phonon
Ev
VB
–k
χ : electron affinity
k
Si with a recombination center
Si
conduction band edge
In GaAs the minimum of
the CB is directly above
the maximum of the VB.
direct bandgap
semiconductor.
Recombination of an electron
In Si, the minimum of the CB is
displaced from the maximum of and a hole in Si involves a
recombination center.
the VB.
indirect bandgap semiconductor
valence band edge
3. 1 Electrons and Holes
Electrons and Holes
Electrons: Electrons in the conduction band that are free to move throughout the crystal.
Generation of Electrons and Holes
Holes:
Missing electrons normally found in the valence band
(or empty states in the valence band that would normally be filled).
Electron energy, E
Ec+χ
CB
hυ > Eg
Ec
Free e–
hυ
Eg
Ev
Hole h+
hole
e–
VB
0
A photon with an energy greater
then Eg can excitation an electron
from the VB to the CB.
Each line between Si-Si atoms is a
valence electron in a bond.
When a photon breaks a Si-Si bond, a
free electron and a hole in the Si-Si
bond is created.
These “particles” carry electricity.
Thus, we call these “carriers”
3.1 Effective Mass (I)
3.1 Carrier Movement Within the Crystal
An electron moving in respond to an applied electric field.
E
E
Density of States Effective Masses at 300 K
within a Vacuum
F = − q E = m0
dv
dt
within a Semiconductor crystal
F = − q E = m n∗
dv
dt
¾ It allow us to conceive of electron and holes as quasi-classical particles
and to employ classical particle relationships in semiconductor crystals or
in most device analysis.
3.1 Effective Mass (II)
Electrons are not free but interact with periodic potential of the lattice.
Wave-particle motion is not as same as in free space.
Ge and GaAs have “lighter electrons” than Si which results in faster devices
3.1 Energy Band Diagram
The energy is plotted as a function of the wave number, k,
along the main crystallographic directions in the crystal.
Si
Ge
GaAs
Moving through Lattice.mov
The bottom axis describe different directions of the crystal.
Curvature of the band determine m*.
m* is positive in CB min., negative in VB max.
3.1 Mass Approximation
Covalent Bonding
The motion of electrons in a crystal can be visualized and described
in a quasi-classical manner.
In most instances
¾ The electron can be thought of as a particle.
¾ The electronic motion can be modeled using Newtonian
mechanics.
The effect of crystalline forces and quantum mechanical properties
are incorporated into the effective mass factor.
¾ m* > 0 : near the bottoms of all bands
¾ m* < 0 : near the tops of all bands
Carriers in a crystal with energies near the top or bottom of an
energy band typically exhibit a constant (energy-independent)
effective mass.
`
 d 2E 
 2  = constant : near band edge
 dk 
Covalent Bonding
Band Occupation at Low Temperature
Band Occupation at High Temperature
Band Occupation at High Temperature
Band Occupation at High Temperature
Band Occupation at High Temperature
Band Occupation at High Temperature
Impurity Doping
The need for more control over carrier concentration
¾ Without “help” the total number of “carriers” (electrons and
holes) is limited to 2ni.
¾ For most materials, this is not that much, and leads to very
high resistance and few useful applications.
¾ We need to add carriers by modifying the crystal.
¾ This process is known as “doping the crystal”.
Regarding Doping, ...
Concept of a Donor “Adding extra” Electrons
Concept of a Donor “Adding extra” Electrons
Concept of a Donor “Adding extra” Electrons
Concept of a Donor “Adding extra” Electrons
Band diagram equivalent view
Concept of a Donor “Adding extra” Electrons
Concept of a Donor “Adding extra” Electrons
V(x), PE (x)
V(x)
n-type Impurity Doping of Si
x
Electron Energy
Energy Band Diagram in an
Applied Field
PE (x) = – eV
Electron Energy
E
CB
e–
As+
Ec
~0.05 eV E
d
As+
As+
Ec
EF
As+
As+
E c − eV
E F − eV
Ev
Ev
x
Distance into
crystal
E v − eV
A
6
As atom sites every 10 Si atoms
The four valence electrons of As allow
it to bond just like Si but the 5th
electron is left orbiting the As site.
The energy required to release to free
fifth- electron into the CB is very
small.
Energy band diagram for an n-type Si doped
with 1 ppm As. There are donor energy levels
just below Ec around As+ sites.
n-Type Semiconductor
B
¾ Energy band diagram of an n-type
semiconductor connected to a
voltage supply of V volts.
¾ The whole energy diagram tilts
because the electron now has an
electrostatic potential energy as
well.
¾ Current flowing
V
Concept of a Acceptor “Adding extra” Holes
Hole Movement
All regions
of
material
are neutrally
charged
One less bond
means
the acceptor is
electrically
satisfied.
One less bond
means
the neighboring
Silicon is left with
an empty state.
Hole Movement
Another valence electron can fill the empty state located next to
the Acceptor leaving behind a positively charged “hole”.
Empty state is located next to the Acceptor
Hole Movement
The positively charged “hole” can move throughout the crystal.
(Really it is the valance electrons jumping from atom to atom that creates the hole motion)
Hole Movement
Hole Movement
The positively charged “hole” can move throughout the crystal.
The positively charged “hole” can move throughout the crystal.
(Really it is the valance electrons jumping from atom to atom that creates the hole motion)
(Really it is the valance electrons jumping from atom to atom that creates the hole motion)
Hole Movement
Concept of a Acceptor “Adding extra” Holes
Band diagram equivalent view
Region
around the
“acceptor”
has
one extra
electron
and thus is
negatively
charged.
Region
around the
“hole” has
one less
electron
and thus is
positively
charged.
The positively charged “hole” can move throughout the crystal.
(Really it is the valance electrons jumping from atom to atom that creates the hole motion)
Concept of a Acceptor “Adding extra” Holes
p-type Impurity Doping of Si
Intrinsic, n-Type, p-Type Semiconductors
Energy band diagrams
Electron energy
B atom sites every 106 Si atoms
Ec
x Distance
into crystal
h+
CB
Ec
EFi
B–
Ea
Ev
Boron doped Si crystal. B has
only three valence electrons.
When it substitute for a Si atom
one of its bond has an electron
missing and therefore a hole.
B–
B–
h+
B–
B–
~0.05 eV
Ec
Ev
EFp
Ev
VB
VB
Energy band diagram for a p-type Si crystal
doped with 1 ppm B. There are acceptor
energy levels just above Ev around B- site.
These acceptor levels accept electrons
from the VB and therefore create holes in
the VB.
Impurity Doping
Ev
Ec
EFn
n-type
semiconductors
Intrinsic
semiconductors
¾ In all cases,
p-type
semiconductors
np=ni2
¾ Note that donor and acceptor energy levels are not shown.
Impurity Doping
Valence Band
Valence Band
Impurity Doping
Position of energy levels within the bandgap of Si for
Energy Band
Energy band diagrams.
common dopants.
¾ Energy-band diagram for a semiconductor showing the lower edge of the
conduction band Ec, a donor level Ed within the forbidden band gap,
and Fermi level Ef, an acceptor level Ea, and the top edge of the valence
band Ev.
3.2B Semiconductor Statistics
Density of States Concept
3.2B Semiconductor Statistics
Density of States Concept
Quantum Mechanics tells us that the number of available states in a
cm3 per unit of energy, the density of states, is given by:
gc ( E ) dE
gv ( E ) dE
General energy dependence of
gc (E) and gv (E) near the band edges.
The number of conduction band
states/cm3 lying in the energy
range between E and E + dE
(if E ≥ Ec).
The number of valence band
states/cm3 lying in the energy
range between E and E + dE
(if E ≤ Ev).
Density of States
in Conduction Band
Density of States
in Valence Band
3.2B Fermi- Dirac function
How do electrons and holes populate the bands?
Probability of Occupation (Fermi Function) Concept
Probability of Occupation (Fermi Function) Concept
¾ Now that we know the number of available states at each energy,
how do the electrons occupy these states?
then
¾ We need to know how the electrons are “distributed in energy”.
“Fer
¾ Again, Quantum Mechanics tells us that the electrons follow the
mi-distribution function”.
f (E) =
1
( E−E f ) / kT
1+ e
Ef ≡ Fermi energy (average energy in the crystal)
k ≡ Boltzmann constant (k=8.617×10-5eV/K)
T ≡Temperature in Kelvin (K)
“The Fermi function f (E) is a probability
distribution function that tells one the ratio of
filled to total allowed states at a given energy
E”
f(E) is the probability that a state at energy E is occupied.
1-f(E) is the probability that a state at energy E is unoccupied.
¾ Fermi function applies only under equilibrium conditions, however, is
universal in the sense that it applies with all materials-insulators,
semiconductors, and metals.
3.2B Semiconductor Statistics
Fermi Function
Fermi-Dirac Distribution
• Probability that an available state at energy E is occupied:
f ( E) =
1
1 + e( E − EF ) / kT
• EF is called the Fermi energy or the Fermi level
Ef
There is only one Fermi level in a system at equilibrium.
If E >> EF :
If E << EF :
If E = EF :
3.2B Semiconductor Statistics
Probability of Occupation (Fermi function) Concept
Maxwell Boltzmann Distribution Function
Boltzmann Approximation
TYU
If EF − E > 3kT , f ( E ) ≅ 1 − e( E − EF ) / kT
• Assume the Fermi level is 0.30eV below the
conduction band energy (a) determine the pro
bability of a state being occupied by an electr
on at E=Ec+KT at room temperature (300K).
If E − EF > 3kT , f ( E) ≅ e
− ( E − EF ) / kT
Probability that a state is empty (occupied by a hole):
( E − EF ) / kT
− ( EF − E ) / kT
1 − f ( E) ≅ e
=e
TYU
• Determine the probability that an allowed ene
rgy state is empty of electron if the state is be
low the fermi level by (i) kT (ii) 3KT (iii)
6 KT
How do electrons and holes populate the bands?
Example 2.2
The probability that a state is filled at the conduction band edge (Ec) is
precisely equal to the probability that a state is empty at the valence band
edge (Ev).
Where is the Fermi energy locate?
Solution
The Fermi function, f(E), specifies the probability of electron occupying
states at a given energy E.
The probability that a state is empty (not filled) at a given energy E is equal
to 1- f(E).
f ( EC ) = 1 − f ( EV )
f (EC ) =
EC − EF
kT
How do electrons and holes populate the bands?
Probability of Occupation Concept
The density of electrons (or holes) occupying the states
in energy between E and E + dE is:
gc ( E ) f ( E ) dE
Electrons/cm3 in the conduction
band between E and E + dE
(if E ≥ Ec).
gv ( E ) f ( E ) dE
Holes/cm3 in the conduction
band between E and E + dE
(if E ≤ Ev).
0
Otherwise
1
1 + e ( EC − E F ) / kT
=
EV − E F
kT
1 − f ( EV ) = 1 −
EF =
1
1
=
1 + e ( EV − E F ) / kT 1 + e ( E F − EV ) / kT
EC + EV
2
How do electrons and holes populate the bands?
Probability of Occupation Concept
How do electrons and holes populate the bands?
Developing the Mathematical Model
for Electrons and Holes concentrations
units of n and p are [ #/cm3]
Typical band structures of Semiconductor
E
The Density of Electrons is:
g (E) ∝ (E–Ec)1/2
E
Ec+χ
E
Probability the state is filled
[1–f(E)]
CB
For
electrons
Area = ∫ n E ( E ) dE = n
Ec
number of
states per unit
energy per unit
volume
EF
Ec
probability of
occupancy of
a state
EF
Ev
Ev
For holes
nE(E)
number of electrons per unit
energy per unit volume
The area under nE(E) vs. E is the
electron concentration.
pE(E)
Area = p
Number of states per cm-3 in energy range dE
The Density of Hole is:
VB
Probability the state is empty
0
fE)
nE(E) or pE(E)
Fermi-Dirac
probability
function
g(E) X f(E)
Energy density of electrons in
the CB
g(E)
Energy band
diagram
Density of states
Number of states per cm-3 in energy range dE
Electron Concentration (no)
TYU
Calculate the thermal equilibrium electron concen
tration in Si at T=300K for the case when the F
ermi level is 0.25eV below the conduction band
.
EC
EF
EV
0.25eV
Hole Concentration (no)
TYU
• Calculate thermal equilibrium hole concentrati
on in Si at T=300k for the case when the Fermi
level is 0.20eV above the valance band energy
Ev.
EC
EF
EV
0.20eV
Degenerate and Nondegenerate Semiconductors
Developing the Mathematical Model
for Electrons and Holes
Useful approximations to the Fermi-Dirac integral:
Nondegenerate Case
n = NCe
( E f − EC ) kT
p = NV e
Semiconductor Statistics
The intrinsic carrier concentration
no = N C e
(E f − EC ) kT
po = NV e
(EV − E f ) kT
When n = ni, Ef = Ei (the intrinsic energy), then
ni = N C e ( E i − EC ) kT
or
N C = ni e ( EC − E i ) kT
and
( EV − Ei ) kT
N = n e ( E i − EV ) kT
or
ni = NV e
V
i
(EV − E f ) kT
Other useful relationships: n⋅p product:
ni = N C e ( E i − EC ) kT
and
ni = N V e ( EV − Ei ) kT
ni = N C NV e − ( EC − EV ) kT = N C NV e
2
ni =
N C NV e
− E g 2 kT
− E g kT
TYU
Determine the intrinsic carrier concentration in
GaAs (a) at T=200k and (b) T=400K
Law of mass action
Law of mass Action
Since
(Ei − E f ) kT
(E − E ) kT
no = ni e f i and po = pi e
nopo = ni
Example
An intrinsic Silicon wafer has 1x1010 cm-3 holes. When 1x1018
cm-3 donors are added, what is the new hole concentration?
2
It is one of the fundamental principles of semiconductors
in thermal equilibrium
nopo = ni
2
TYU
if
N D 〉〉 N A
n ≅ ND
and
and
N D 〉〉 ni
p≅
ni2
ND
TYU
: The concentration of majority
carrier
electron is no=1 x 1015 cm-3 at 300K.
D
etermine the concentration of phosphorus th
at are to be added and determine the concentr
ation minority carriers holes.
• Find the hole concentration at 300K, if the
electron concentration is no=1 x 1015 cm-3,
which carrier is majority carrier and which
carrier is minority carrier?
Energy band diagram
showing negative
charges
Partial Ionization,
Intrinsic Energy
and Parameter Rel
ationships.
3.5 Carrier concentration-effects of doping
Charge Neutrality:
3.5 Developing the Mathematical Model
for Electrons and Holes
Charge Neutrality: Total Ionization case
¾ If excess charge existed within the semiconductor, random motion
of charge would imply net (AC) current flow.
¨ Not possible!
¾ Thus, all charges within the semiconductor must cancel.
[( p + N ) = (N + n )]
q ⋅ [( p − N ) + (N − n )]
+
d
Mobile - charge ¼
Immobile - charge ¼
−
A
¾ NA¯ = Concentration of “ionized” acceptors = ~ NA
¾ ND+ = Concentration of “ionized” Donors
( p − N )+ (N
−
A
−
a
Immobile + charge ¼
+
d
Mobile + charge ¼
Energy band diagram
showing positive
charges
=0
+
d
= ~ ND
)
−n =0
Electron concentration versus temperature for n-type
Semiconductor.
The intrinsic carrier concent
ration as a function of
temperature.
Carrier Concentration vs. Temper
ature
position of Fermi Energy level
no = N c e
(
)
[ − Ec − E f ] kT
Ec − EF = kT ln( Nc / no)
Nd >> ni
Ec − EF = kT ln( Nc / Nd )
Note: If we have a compensated semiconductor , then the Nd term
in the above equation is simply replaced by Nd-Na.
position of Fermi Energy level
po = N V e
(EV − E f ) kT
position of Fermi level as a function of carrier concentration
EF − Ev = kT ln( Nv / po)
Na >> ni
EF − Ev = kT ln( Nv / Na )
Note: If we have a compensated semiconductor , then the Na term
in the above equation is simply replaced by Na-Nd.
Where is Ei ?
Extrinsic Material:
TYU
• Determine the position of the Fermi level with res
pect to the valence band energy in p-type GaAs at
T=300K. The doping concentration are Na=5 x 1
016 cm-3 and Na=4 x 1015 cm-3.
Note: The Fermi-level is pictured here for 2 separate cases: acceptor and donor doped.
position of Fermi Energy level
Extrinsic Material:
no = ni e
(E f − E fi ) kT
po = ni e
(E fi − E f ) kT
Solving for (Ef - Efi)
 p
n
E f − E fi = kT ln  = − kT ln 
n
 ni 
 i
for N D 〉〉 N A
and
N D 〉〉 ni
N 
E f − E fi = kT ln D 
 ni 
for N A 〉〉 N D
and
N A 〉〉 ni
N 
E f − E fi = −kT ln A 
 ni 
TYU 3.8
• Calculate the position of the Fermi level in ntype Si at T=300K with respect to the intrinsi
c Fermi energy level. The doping concentrati
on are Nd=2 x 1017 cm-3 and Na=3 x 1016 cm-3
.
Mobile Charge Carriers in
Smiconductor devices
• Three primary types of carrier action occur inside a
semiconductor:
– Drift: charged particle motion under the influence of an
electric field.
– Diffusion: particle motion due to concentration gradient
or temperature gradient.
EC
EF
– Recombination-generation (R-G)
EFi
EV
Carrier Dynamics
Direction of motion
Carrier Drift
Describe the mechanism of the carrier drift and drift current
due to an applied electric field.
¾ Holes move in the direction of the electric field. (⊕F\)
Carrier Motion
Electron Drift
Electron Diffusion
Hole Drift
Hole Diffusion
¾ Electrons move in the opposite direction of the electric field. (\F⊕)
¾ Motion is highly non-directional on a local scale, but has a net direction
on a macroscopic scale.
¾ Average net motion is described by the drift velocity, vd [cm/sec].
¾ Net motion of charged particles gives rise to a current.
Instantaneous velocity is extremely fast
Drift
Drift of Carriers
Drift
Schematic path of an electron in a semiconductor.
E
Electric Field
Drift of electron in a solid
E
The ball rolling down the smooth hill speeds up
continuously, but the ball rolling down the
stairs moves with a constant average velocity.
Random thermal motion.
Combined motion due to random thermal
motion and an applied electric field.
µ [cm2/Vsec] : mobility
Drift
Drift
Conduction process in an n-type semiconductor
Random thermal motion.
Combined motion due to
random thermal motion and an
applied electric field.
Thermal equilibrium
Under a biasing condition
Drift
Drift
Jdrf = vd
J p drf = q ⋅ p ⋅ vd
At Low Electric Field Values,
Jp
Drift
= e⋅ p⋅µp ⋅ E
and
Jn
Drift
= e ⋅ n ⋅ µn ⋅ E
Given current density J ( I = J x Area ) flowing in a semiconductor block
with face area A under the influence of electric field E, ρ is volume
density, the component of J due to drift of carriers is:
J drf = ρvd
J n drf = −e ⋅ n ⋅ vd
and
J p drf = e ⋅ p ⋅ vd
Hole Drift Current Density
Electron Drift Current Density
Electron and Hole Mobilities
µ has the dimensions of v/ :
 cm/s cm2 
=


 V/cm V ⋅ s 
Electron and hole mobilities of selected
intrinsic semiconductors (T=300K)
µ n (cm /V·s)
µ p (cm2/V·s)
2
Si
Ge
GaAs
InAs
1350
3900
8500
30000
480
1900
400
500
¾ µ [cm2/V·sec] is the “mobility” of the semiconductor and measures the
ease with which carriers can move through the crystal.
¾ The drift velocity increases with increasing applied electric field.:
Jdrf = J p
Drift
+ Jn
Drift
= q ( µ p p + µ n n) ⋅ E
EX 4.1
• Consider a GaAs sample at 300K with dopin
g concentration of Na=0 and Nd=1016 cm-3.
Assume electron and hole mobitities given in
table 4.1. Calculate the drift current density if
the applied electric filed is E=10V/cm.
Mobility
¾ µ [cm2/Vsec] is the “mobility” of the semiconductor and measures the
ease with which carriers can move through the crystal.

µn ~ 1360 cm2/Vsec
µp ~ 460 cm2/Vsec
µn ~ 8000 cm2/Vsec
µp ~ 400 cm2/Vsec
µn, p =
qτ
m n* , p
for Silicon @ 300K
for Silicon @ 300K
for GaAs @ 300K
for GaAs @ 300K
[ cm
2
V sec]
¾ <τ > is the average time between “particle” collisions in the
semiconductor.
¾ Collisions can occur with lattice atoms, charged dopant atoms, or with
other carriers.
Saturation velocity
Drift velocity vs. Electric field in Si.
Saturation velocity
Drift velocity vs. Electric field
Designing devices to work at
the peak results in faster
operation
1/2mvth2=3/2kT=3/2(0.0259)
=0.03885eV
µn, p =
qτ
m n* , p
[ cm
2
V sec]
¾ Ohm’s law is valid only in the low-field region where drift velocity is independent
of the applied electric field strength.
¾ Saturation velocity is approximately equal to the thermal velocity (107 cm/s).
Negative differential mobility
Drift
Electron distributions under various conditions of electric
Drift velocity vs. Electric field in Si and GaAs.
fields for a two-valley semiconductor.
Note that for n-type GaAs,
there is a region of negative
differential mobility.
µn, p =
qτ
m n* , p
[ cm
Negative differential mobility
Velocity-Field characteristic of a Two-valley semiconductor.
2
V se
m*n=0.55mo
m*n=0.067mo
TYU
• Silicon at T=300K is doped with impurity
concentration of Na=5 X 1016 cm-3 and Nd=2
x 1016 cm-3. (a) what are the electron and hole
mobilities? (b) Determine the resistivity and
conductivity of the material.
Figure 3.24.
Mean Free Path
l = vthτ mp
• Average distance traveled between collisions
EX 4.2
Using figure 4.3 determine electron and hole nobilities.
EX 4.2
Using figure 4.3 determine electron and hole mobilities in (a) Si for Nd=1017 cm-3,
Na=5 x 1016 cm-3 and (b) GaAs for Na=Nd=1017cm-3
Ex 4.2
Mobility versus temperature
Mobility versus temperature
Effect of Temperature on Mobility
Effect of Temperature on Mobility
¾ Electron mobility in silicon
versus temperature for
various donor concentrations.
¾ Since the slowing moving carrier is
likely to be scattered more strongly by
an interaction with charged ion.
¾ Impurity scattering events cause a
decrease in mobility with decreasing
temperature.
¾ Insert shows the theoretical
temperature dependence of
electron mobility.
¾ A carrier moving through the lattice
encounters atoms which are out of
their normal lattice positions due to
the thermal vibrations.
¾ The frequency of such scattering
increases as temperature increases.
¾ At low temperature, thermal
motion of the carriers is
slower, and ionized impurity
scattering becomes dominant.
At low temp. lattice scattering is less important.
As doping concentration increase, impurity
scattering increase, then mobility decrease.
Temperature dependence of mobility with both lattice and impurity scattering.
Effect of Doping concentration on Mobility
Resistivity and Conductivity
Ohms’ Law
300 K
J =σ ⋅E =
E
[A cm ]
2
Ohms Law
σ [1 ohm ⋅ cm ]
Conductivity
ρ
ρ [ohm ⋅ cm ]
¾ Electron and hole mobilities in Silicon as functions of the total
dopant concentration.
Resistivity
semiconductor conductivity and resistivity
Adding the Electron and Hole Drift Currents (at low electric fields)
Jdrf = J p
Drift
+ Jn
Drift
= e( µ p p + µ n n ) ⋅ E
Drift Current
σ = e( µ p p + µ n n )
ρ=
1
σ
[
= 1 e( µ n n + µ p p )
Conductivity
]
Resistivity
¾ But since µn and µp change very little and n and p change several
orders of magnitude:
σ ≅ e µn n
σ ≅ eµ p p
for n-type with n>>p
for p-type with p>>n
µn, p =
qτ
m n* , p
2
V sec]
Diffusion
Diffusion
Particles diffuse from regions of higher concentration
to regions of lower concentration region, due to
random thermal motion.
[ cm
Nature attempts to reduce concentration gradients to zero.
Example: a bad odor in a room, a drop of ink in a cup of water.
¾ In semiconductors, this “flow of carriers” from one region of higher
concentration to lower concentration results in a “Diffusion Current”.
Jp
Diffuse
Diffusion
Jn
Diffusion
Diffuse
Visualization of electron and hole diffusion on a macroscopic scale.
Diffusion Current
J N,diff = eDN
dn
dx
J P,diff = −eDP
dp
dx
x
x
D is the diffusion constant, or diffusivity.
Diffusion current density
Total Current
Fick’s law
¾ Diffusion as the flux, F, (of particles in our case) is proportional to
the gradient in concentration.
F = − D∇η
η : Concentration
D : Diffusion Coefficient
¾ For electrons and holes, the diffusion current density
( Flux of particles times ± q )
Jp
Diffusion
= − q ⋅ D p∇p
Jn
Diffusion
= q ⋅ Dn∇n
The opposite sign for electrons and holes
J = JN + JP
ε + qD
JN = JN,drift + JN,diff = qnµn
N
ε – qD
JP = JP,drift + JP,diff = qpµp
P
dn
dx
dp
dx
Total Current
TYU
Total Current = Drift Current + Diffusion Current
Jp = Jp
Drift
+ Jp
Diffusion
= q ⋅ µ p p E − q ⋅ D p∇p
Jn = Jn
Drift
+ Jn
Diffusion
= q ⋅ µ n n E + q ⋅ Dn∇n
• Consider a sample of Si at T=300K. Assume that
electron concentration varies linearly with distance,
as shown in figure.The diffusion current density is
found to be Jn=0.19 A/ cm2. If the electron diffusio
n coefficient is Dn=25cm2/sec, determine the electr
on concentration at x=0.
J = J p + Jn
J N,diff = eDN
dn
dx
Jp=0.270 A/cm2
Dp=12 cm2/sec
Find the hole concentration at x=50um
J P,diff = −eDP
dp
dx
Graded impurity distribution
Energy band diagram of a semiconductor in thermal equilibrium
with a nonuniform donor impurity concentration
Generation and Recombination
Carrier Generation
Generation Mechanism
Band-to-Band Generation
Thermal Energy
or
Light
Band-to-band generation
Gno=Gpo
¾ Band-to-Band or “direct” (directly across the band) generation.
¾ Does not have to be a “direct bandgap” material.
¾ Mechanism that results in ni.
¾ Basis for light absorption devices such as semiconductor
photodetectors, solar cells, etc…
Recombination Mechanism
Band-to-Band Recombination
n = ∆n + n0
Photon
(single particle of light)
or
Rno=Rpo
multiple phonons
(single quantum of lattice
vibration - equivalent to
saying thermal energy)
¾ Band to Band or “direct” (directly across the band) recombination.
¾ Does not have to be a “direct bandgap” material, but is typically
very slow in “indirect bandgap” materials.
¾ Basis for light emission devices such as semiconductor Lasers,
LEDs, etc…
In thermal equilibrium: Gno=Gpo=Rno=Rpo
Excess minority carrier lifetime
Carrier Lifetime
∆p (t ) = ∆p0 e
− t τ po
∆n(t ) = ∆n0 e − t τ no
Schematic diagram of photoconductivity decay measurement.
Light Pulses
Rs
VA
Semiconductor
Excess carrier Recombination and Generation
+
I
RL
VL
_
Oscilloscope
and
p = ∆p + p0
In Non-equilibrium, n·p does not equal ni2
Low-Level-Injection implies
∆p << n0 ,
n ≈ n0
in a n-type material
∆n << p0 ,
p ≈ p0
in a p-type material
Example low level injection case
Nd=1014/cm3 doped Si at 300K subject to a perturbation where ∆p =∆n =109/cm3.
Ä n0 ≅ Nd =1014/cm3 and p0 ≅ ni2/Nd = 106/cm3
Ä n = n0 + ∆n ≅ n0 and ∆p ≅ = 109/cm3 << n0 ≅ 1014/cm3
Although the majority carrier concentration remains essentially
unperturbed under low-level injection, the minority carrier concentration
can, and routinely does, increase by many orders of magnitude.
Material Response to “Non-Equilibrium”
Relaxation Concept
¾ Consider a case when the hole concentration in an n-type sample is
not in equilibrium, i.e., n·p ≠ ni2
R' n =
∂p
∆p (t )
=−
τ po
∂t
τpo is the minority carrier lifetime
¾ The minority carrier lifetime is the average time a minority carrier can
survive in a large ensemble of majority carriers.
¾ If ∆p is negative ¼ Generation or an increase in carriers with time.
¾ If ∆p is positive ¼ Recombination or a decrease in carriers with time.
¾ Either way the system “tries to reach equilibrium”
¾ The rate of relaxation depends on how far away from equilibrium we are.
Material Response to “Non-Equilibrium”
Relaxation Concept
Generation and Recombination process
Recombination-Generation center recombination
¾ Likewise when the electron concentration in an p-type sample is not in
equilibrium, i.e., n·p does NOT equal ni2
∂n
∂t Thermal
=−
R −G
∆n
τn
τn is the minority carrier lifetime
Indirect recombinationgeneration processes at
thermal equilibrium.
Recombination Mechanism
Generation and Recombination process
Recombination-Generation (R-G) Center Recombination
Energy loss can result in
a Photon
but is more often
multiple phonons
¾ Also known as Shockley-Read-Hall (SRH) recombination.
¾ Two steps:
1 1st carrier is “trapped” (localized) at an defect/impurity (unintentional/intentional ).
2 2nd carrier (opposite type) is attracted and annihilates the 1st carrier.
¾ Useful for creating “fast switching” devices by quickly “killing off” EHP’s.
Generation Mechanism
Recombination-Generation (R-G) Center Generation
Effects of recombination
centers on solar cell
performance
b
Thermal Energy
¾ Two steps:
1 A bonding electron is “trapped” (localized) at an unintentional
defect/impurity generating a hole in the valence band.
2 This trapped electron is then promoted to the conduction band
resulting in a new EHP.
¾ Almost always detrimental to electronic devices. AVOID IF POSSIBLE!
The light-generated minority
carrier can return to the ground
state through recombination
center
before being
collected by the junction:
i) through path (a)
ii) through path (c)
Without recombination
centers paths (b) and (d)
are dominated
Light
d
EC
a
c
EV
Auger Recombination
3.3 p-n Junction Diode
Auger Recombination
Auger – “pronounced O-jay”
p-Type Material
n-Type Material
Requires 3 particles.
p-n Junction
¾ Two steps:
1 1st carrier and 2nd carrier of same type collide instantly annihilating
the electron hole pair (1st and 3rd carrier). The energy lost in the
annihilation process is given to the 2nd carrier.
2 2nd carrier gives off a series of phonons until it’s energy returns to
equilibrium energy (E~=Ec) This process is known as thermalization.
p-n Junction principles
p-n Junction
p-Type Material
n-Type Material
p-n Junction
p-Type Material
n-Type Material
¾ A p-n junction diode is made by forming a p-type region of
material directly next to a n-type region.
p-n Junction Diode
But when the device has no external applied forces, no current can
flow. Thus, the Fermi-level must be flat!
We can then fill in the junction region of the band diagram as:
EC
EC
EF
Ei
Ei
EF
EV
EV
p-Type Material
n-Type Material
p-n Junction Diode
p-n Junction Diode
Built-in-potential
But when the device has no external applied forces, no current can
flow. Thus, the Fermi-level must be flat!
p-Type Material
We can then fill in the junction region of the band diagram as:
n-Type Material
EC
- qVbi
Ei
EF
EC
EC
Ei
EF
EF
Ei
EV
EF
Ei
EV
EV
EV
Electrostatic Potential
p-Type Material
V=
1
( EC − E ref )
q
n-Type Material
VBI Built-in-potential
p-n Junction Diode
Built-in-potential
Electrostatic Potential
Electric Field
dV
dx
x
x
Charge Density
Electric Field
E=−
Electric Field
E=−
1
( EC − E ref )
q
VBI Built-in-potential
x
p-n Junction Diode
Built-in-potential
V=
EC
dV
dx
Charge Density
ρ = KS ⋅ε0
x
qND
+
-
dE
dx
+
-
qNA
+
x
Built-In Potential Vbi
TYU 5.1
qVbi = q(ΦS p−side + ΦS n −side ) = ( Ei − EF ) p−side + ( EF − Ei ) n −side
• Calculate the built-in-potential barrier in a Si
pm junction at T=300K for (a) Na=5 x 1017c
m-3, Nd=1016cm-3 (b)Na=1015cm-3
For non-degenerately doped material:
 p
( Ei − EF ) p−side = kT ln 
 ni 
N 
= kT ln A 
 ni 
n
( EF − Ei )n−side = kT ln 
 ni 
N 
= kT ln D 
 ni 
p
p-n Junction Diode
n
As +
Bh+
e–
Built-in-potential
M
Metallurgical Junction
Electric Field
dV
E=−
dx
Neutral p-region
M
E (x )
Neutral n-region
E0
-Wp
0
-Wn
x
–Eo
x
V (x )
M
Wp
lo g (n ), lo g (p )
Vo
Space charge region
Wn
p po
Charge Density
ρ = KS ⋅ε0
+
-
+
-
qNA
P E (x )
ni
dE
dx
eV o
pno
n po
Hole
o Potential Energy PE (x)
x
x = 0
ρ net
qND
x
nno
Charge Density
x
M
+
x
eN d
Electron Potential Energy PE (x)
– Wp
x
Wn
-e N a
–eV o
p-n Junction Principles
Movement of Electrons and Holes when
Forming the Junction
p-n Junction
Depletion Region Approximation: Step Junction Solution
Poisson’s Equation
Charge Density
(NOT Resistivity)
Electric Field
∇⋅E = −
ρ = q( p − n + N D − N A )
ρ ( x) = −qNA
ρ ( x) = qND
ρ
dE
ρ
− qNA / D
=−
=
K S ⋅ ε 0 dx
KS ⋅ε0
KS ⋅ε0
in 1-dimension
Relative Permittivity
of Semiconductor
(εr)
Permittivity of free space
Number of negative charges per unit area in the p region is
equal to the number of positive charges per unit area in the n-region
Electric potential V(x) 0r φ(x)
Depletion Region Approximation: Step Junction Solution
Space charge width(depletion layer
width)
Depletion Region Approximation: Step Junction Solution
n-region space charge width
p-region space charge width
TYU5.2
• A silicon pn junction at T=300k with zero ap
plied bias has doping concentration of Nd= 5
x 1016cm-3 and Na=5 x 1015 cm-3. Determine
xn, xp, W, and |Ex(max)|, Vbi 0.718V
pn junction reverse applied bias
Depletion Region Approximation: Step Junction Solution
pn junction reverse/forward applied
bias
Depletion Region
Schematic representation of
depletion layer width and
energy band diagrams of a
p-n junction under various
biasing conditions.
Thermal-equilibrium
Forward-bias condition
Diode under Forward Bias.mov
Diode under no Bias.mov
Reverse-bias condition
Diode under Reverse Bias.mov
pn junction forward bias applied bias
Depletion Region Approximation: Step Junction Solution
Thus, only the boundary conditions change resulting in
direct replacement of Vbi with (Vbi-VA) with VA ≠ 0.
pn junction reverse/forward applied
Depletion Region Approximation:
Step Junction Solution with VA ≠ 0
Consider a p+n junction (heavily doped p-side, lightly doped n side)
Movement of Electrons and Holes when
Forming the Junction
Forward bias condition
Movement of Electrons and Holes when
Forming the Junction
Reverse bias condition
Space charge width and Electric field
 2εs (Vbi + VR )  Nd 

1
xp = 



e
Na
N
N
+

 a
d 

 2εs (Vbi + VR )  N a

xn = 
e
 Nd

1/ 2


1


 Na + Nd 
1/ 2
 2εs (Vbi + VR )  N A + Nd 
W = xn + xp = 


e
 NaNd 

1/ 2
p-n
EX 5.3
Junction I-V Characteristics
In Equilibrium (no bias)
Total current balances due to the sum of the individual components
no net current!
• A Si pn junction at 300K is reverse bias at V
R=8V, the doping concentration are Na=5 x
1015cm-3 and Nd= 5 x 1016 cm-3. Determin
e xn, xp and W, repeat for VR=12V.
Electron Diffusion
Current
Electron Drift
Current
Hole Drift
Current
Hole Diffusion
Current
Diode under no Bias.mov
p-n
p-n
Junction I-V Characteristics
Forward Bias (VA > 0)
In Equilibrium (no bias)
IN
Total current balances due to the sum of the individual components
p-Type Material
q VBI
EC
++
EV
+ + ++ + + + + + + + + + + + + + EC
EF
Ei
EV
p vs. E
Jn = Jn
Drift
+ Jn
Diffusion
= q ⋅ µ n nE + q ⋅ Dn∇n = 0
Jp = Jp
+ Jp
Drift
Diffusion
= q ⋅ µ p pE + q ⋅ D p ∇p = 0
no net current!
Electron Diffusion
Current
Electron Drift
surmount potential barrier
Current
n vs. E
n-Type Material
E
EF i
Junction I-V Characteristics
Lowering of
potential hill
by VA
VA
Hole Diffusion
Current
Current flow is dominated
by majority carriers flowing
across the junction and
becoming minority carriers
IP
Current flow is
proportional to
e(Va/Vref) due to
the exponential
decay of carriers
into the majority
carrier bands
Hole Drift
Current
I = IN + IP
I
Diode under Forward Bias m
p-n
p-n
Junction I-V Characteristics
Reverse Bias (VA < 0)
Electron Drift
Current
Current flow is constant due
to thermally generated
carriers swept out by E fields
in the depletion region
Increase of
potential hill
by VA
Junction I-V Characteristics
Where does the Reverse Bias Current come from?
¾ Generation near the depletion region edges “replenishes” the
current source.
Electron Diffusion Current negligible due
to large energy barrier
Hole Diffusion Current negligible
due to large energy barrier
Current flow is dominated by
minority carriers flowing across
the junction and becoming
majority carriers
Hole Drift
Current
Diode under Reverse Bias.m
p-n
µ P-N Junction Diodes ¸
Junction I-V Characteristics
Putting it all together
Current Flowing through a Diode
I-V Characteristics
Quantitative Analysis
(Math, math and more math)
-I0
for Ideal diode
Vref = kT/q
p-n
Quantitative
Junction I-V Characteristics
Diode Equation
p-n Diode Solution
Assumptions:

 qV
I = I 0  exp 
 ηkT


1)
2)
3)
4)
5)
 
 − 1
 
 
Steady state conditions
Non- degenerate doping
One- dimensional analysis
Low- level injection
No light (GL = 0)
Current equations:
J p = J p ( x )+ J n ( x )
η : Diode Ideality Factor
Continuity Equations
Steady state
: n(x) is time invariant.
Transient state : n(x) is time dependent.
∆x
J p ( x − ∆x )
J p ( x)
Area, A cm 2
x
∂n
∂F 1 ∂J
=−
=
∂t
∂ x q ∂x
x + ∆x
Continuity Equation
F: Particle Flux
J: Current Density
 dn 

J n = qµ n nE − qD n 
 dx 
 dp 

J p = qµ p pE − qD p 
 dx 
Ways Carrier Concentrations can be Altered
Continuity Equations
Ways Carrier Concentrations can be Altered
Continuity Equations
∂n
∂t
∂p
∂t
∂n ∂n
=
∂t
∂t
∂p ∂p
=
∂t
∂t
+
Drift
+
Drift
∂n
∂t
∂p
∂t
+
∂n
∂n
+
All other processes
∂t Thermal R −G ∂t such
as light ...
+
∂p
∂p
+
∂t Thermal R −G ∂t
Diffusion
Diffusion
+
Drift
+
Drift
∂n
∂t
∂p
∂t
1  ∂J Nx ∂J Ny ∂J Nz
+
+
∂y
∂z
q  ∂x
=
Diffusion
Diffusion
 1
 = ∇ ⋅ JN
 q

∂J Py ∂J Pz 
1  ∂J
 = − 1 ∇ ⋅ JP
= −  Px +
+
∂y
∂z 
q  ∂x
q
∂n 1
∂n
∂n
= ∇ ⋅ JN +
+
All other processes
q
∂t
∂t Thermal R −G ∂t such
as light ...
∂p
∂p
∂p
1
= − ∇ ⋅ JP +
+
All other processes
∂t
∂t Thermal R −G ∂t such
q
as light ...
All other processes
such as light ...
¾ There must be spatial and time continuity in the carrier concentrations.
Continuity Equations: Special Case known as
“Minority Carrier Diffusion Equation”
Simplifying Assumptions:
Continuity Equations: Special Case known as
“Minority Carrier Diffusion Equation”
¾ Because of (3) no electric field E = 0
2) We will only consider minority carriers.
5) Low-level injection conditions apply.
6) SRH recombination-generation is the main recombination-generation
mechanism.
Minority Carrier
Diffusion Equation
+ Jn
Diffusion
0
Diffusion
= q ⋅ µ n nE + q ⋅ D n ∇n
= q ⋅ D n ∇n
∂ 2 ( n0 + ∆n)
∂ 2n
∂ 2 ( ∆n )
1
1 ∂J N
∇ ⋅ JN =
= Dn 2 = Dn
= Dn
2
q
q ∂x
∂x
∂x
∂x 2
¾ Because of (5) - low level injection
7) The only “other” mechanism is photogeneration.
Continuity Equations
Drift
= Jn
3) Electric field is approximately zero in regions subject to analysis.
4) The minority carrier concentrations IN EQUILIBRIUM are not a
function of position.
0
Jn = Jn
1) One dimensional case. We will use “x”.
∂n
∂t
=−
Reombination − Generation
∆n
τn
¾ Finally
∂n ∂ ( n0 + ∆n) ∂ ( ∆n)
=
=
∂t
∂t
∂t
¾ Because of (7) - Photogeneration
∂n
∂t
All other processes
such as light ...
= GL
Continuity Equations: Special Case known as
“Minority Carrier Diffusion Equation”
Further simplifications (as needed):
Continuity Equations
∂n ∂n
=
∂t
∂t
0
+
Drift
Continuity Equations: Special Case known as
“Minority Carrier Diffusion Equation”
∂n
∂t
+
Diffusion
∂n
∂t
+
Reombination − Generation
∂n
∂t
¾ Steady State …
All other processes
such as light ...
∂ ( ∆n p )
∂t
→0
∂ ( ∆ pn )
→0
∂t
and
¾ No minority carrier diffusion gradient …
DN
∂ ( ∆n p )
∂t
= DN
∂ 2 ( ∆n p )
∂x 2
−
( ∆n p )
τn
+ GL
∂ 2 ( ∆n p )
∂x
2
→0
∂ ( ∆ pn )
∂ ( ∆ pn ) ( ∆ pn )
= DP
−
+ GL
∂t
∂x 2
τp
2
∆n
τn
=0
and
¾ Consider a semi-infinite p-type silicon sample with NA=1015 cm-3 constantly
illuminated by light absorbed in a very thin region of the material creating a
steady state excess of 1013 cm-3 minority carriers (x=0).
Semi-infinite sample
Light absorbed in a thin skin.
∂t
No excess carrier
= DN
DN
∂x 2
∂ 2 ( ∆n p )
∂x
2
=
−
( ∆n p )
τn
+ GL
0generation
( ∆n p )
τn
=0
Steady-state carrier injection from one side.
Semiconductor
x
∂ 2 ( ∆n p )
∆p
τp
Solutions to the
“Minority Carrier Diffusion Equation”
¾ What is the minority carrier distribution in the region x> 0 ?
0
−
GL → 0
Solutions to the
“Minority Carrier Diffusion Equation”
∂ ( ∆n p )
∂ 2 ( ∆pn )
→0
∂x 2
¾ No light …
Minority Carrier Diffusion Equations
Steady state
DP
¾ No SRH recombination-generation …
−
Light
and
Sample with thickness W
Direct generation and recombination of electron-hole pairs:
at thermal equilibrium
under illumination.
Solutions to the
“Minority Carrier Diffusion Equation”
Continue
General Solution
0
( − x LN )
∆n p ( x ) = A ⋅ e
+ B ⋅ e ( + x LN )
where LN ≡ DN ⋅ τ N
Surface recombination at x = 0. The minority carrier distribution near the surface i
s affected by the surface recombination velocity.
Solutions to the
“Minority Carrier Diffusion Equation”
¾ Consider a p-type silicon sample with NA=1015 cm-3 and minority carrier lifetime τ
=10 µsec constantly illuminated by light absorbed uniformly throughout the
material creating an excess 1013 cm-3 minority carriers per second. The light
has been on for a very long time. At time t=0, the light is shut off.
¾ What is the minority carrier distribution in for t < 0 ?
¾ LN is the “Diffusion length” the average distance a minority carrier can
move before recombining with a majority carrier.
¾ Boundary Condition …
∆n p ( x = 0) = 1013 cm −3 = A + B
∆n p ( x = ∞ ) = 0 = A( 0) + Be (+ ∞ LN )
→ B=0
∆n p ( x ) = 1013 e (− x LN )cm − 3
Semiconductor
Light absorbed uniformly
Semiconductor
x
∂ ( ∆n p )
∂t
0
= DN
Light
Uniform
0 distribution
∂ 2 ( ∆n p )
∂x
2
−
( ∆n p )
τn
+ GL
∆n p (all x , t < 0) = G L ⋅ τ n = 10 7 cm −3
Solutions to the
“Minority Carrier Diffusion Equation”
Quantitative
Continue
p-n Diode Solution
Application of the Minority Carrier Diffusion Equation
¾ In the previous example: What is the minority carrier distribution in for t > 0 ?
Quisineutral Region
Semiconductor
Light absorbed uniformly
x
Quisineutral Region
Light
minority carrier diffusion eq.
∂ ( ∆n p )
∂t
= DN
∂ 2 ( ∆n p )
∂x
2
0
−
( ∆n p )
τn
+ GL
0
0
0
minority carrier diffusion eq.
Since electric fields
exist in the depletion
region, the minority
carrier diffusion
equation does not
apply here.
0
0
∆n p ( t ) = [∆n p ( t = 0)] ⋅ e ( − t τ n )
∆n p ( t ) = 10 7 e ( − t 1e − 5 )
Quasi - Fermi Levels
Equilibrium
n0 = ni e
Non-Equilibrium
( E f − E i ) kT
n = ni e ( FN − E i ) kT
( E i − E f ) kT
p = ni e ( E i − FP ) kT
p0 = ni e
¾ The Fermi level is meaningful only when the system is in thermal equilibrium.
¾ The non-equilibrium carrier concentration can be expressed by defining QuasiFermi levels Fn and Fp .
Equilibrium
Non-Equilibrium
Quantitative p-n Diode Solution
(At the depletion regions edge)
Quisineutral Region
Quisineutral Region
quasi-Fermi levels formalism
np = n i e ( F N − F P
2
) kT
Quantitative
p-n Diode Solution
Quisineutral Region
Quantitative
Quisineutral Region
Quisineutral Region
Quisineutral Region
x”=0

dn 

J n = q  µ n nE + D n
dx 
 0
= qD n
= qD n
(
d n 0 + ∆n p
)
dx
dp 


J p = q  µ p pE + D p
dx 
 0
?
= qD p
d∆n p
= qD p
dx
d ( p 0 + ∆p n
dx
d∆p n
dx
)
p-n Diode Solution
x’=0
Approach:
¾ Solve minority carrier diffusion equation in quasineutral regions.
¾ Determine minority carrier currents from continuity equation.
¾ Evaluate currents at the depletion region edges.
¾ Add these together and multiply by area to determine the total current
through the device.
¾ Use translated axes, x t x’ and -x t x’’ in our solution.
Quantitative
p-n Diode Solution
Quisineutral Region
Quisineutral Region
x”=0
x’=0
Holes on the n-side
Quantitative
Quisineutral Region
p-n Diode Solution
Quisineutral Region
x”=0
x’=0 Holes on the n-side
Quantitative
p-n Diode Solution
Quisineutral Region
Quantitative
p-n Diode Solution
Quisineutral Region
Thus, evaluating the current components at the depletion region edges,
we have…
x”=0
J = Jn (x”=0) +Jp (x’=0) = Jn (x’=0) +Jn (x”=0) = Jn (x’=0) +Jp (x’=0)
x’=0
Similarly for electrons on the p-side…
Ideal Diode Equation
Shockley Equation
Note: Vref from our previous qualitative analysis equation is the thermal voltage, kT/q
Quantitative
p-n Diode Solution
Quisineutral Region
Quisineutral Region
x”=0
x’=0
Total on current is constant throughout the device.
Thus, we can characterize the current flow components as…
J
Example
A silicon pn junction at T=305K has the following parameters:
NA = 5x1016 cm-3, ND = 1x1016 cm-3 , Dn = 25 cm2/sec, Dp = 10 cm2/sec, τn0 = 5x10-7 sec,
and τp0 = 1x10-7 sec, ni305K = 1.5x1010 cm-3,.
The cross-sectional area is A=10-3 cm2, and the forward-bias voltage is Va = 0.625 V.
Calculate the
(a) minority electron diffusion current at the space charge region.
(b) minority hole diffusion current at the space charge edge.
(c) total current in the pn junction diode.
Solution
Minority electron diffusion current density
-xp
xn
Ln =
Dnτ n
Example
Example
A silicon pn junction at T=305K has the following parameters:
NA = 5x1016 cm-3, ND = 1x1016 cm-3 , Dn = 25 cm2/sec, Dp = 10 cm2/sec, τn0 = 5x10-7 sec,
and τp0 = 1x10-7 sec, ni305K = 1.5x1010 cm-3,.
The cross-sectional area is A=10-3 cm2, and the forward-bias voltage is Va = 0.625 V.
Calculate the
(a) minority electron diffusion current at the space charge region.
(b) minority hole diffusion current at the space charge edge.
(c) total current in the pn junction diode.
Solution
Ln =
Minority electron diffusion current density
2
Jn =
Dnτ n
Given figure is a dimensioned plot of the steady state carrier concentrations inside a
pn junction diode maintained at room temperature.
(a) Is the diode forward or reverse biased? Explain how you arrived at your answer.
(b) Do low-level injection conditions prevail in the quasineutral regions of diode?
Explain how you arrived at your answer.
(c) Determine the applied voltage, VA.
n or p
(log scale)
pp
25 (1.5 × 1010 )   0.625  
2
exp
 − 1 = 0.154 mA / cm
5 × 10 − 7 5 × 1016   0.0259  
2
= (1.6 × 10 −19 )
1015
1010
np
2
qD p nn 0   qVa  
D p ni 2   qVa  
exp
exp
−1 = q
 −1
L p   kT  
τ p 0 N D   kT  
pn
108
Minority hole diffusion current density
Jp =
nn
1017
2
qDn n p 0   qVa  
Dn ni   qVa  
exp
exp
−1 = q
 −1
Ln   kT  
τ n 0 N A   kT  
The diode is forward biased.
There is pile-up or minority carrier
excess (∆np>0 and ∆pn>0 ) at the
edges of the depletion region.
103
-xp
2.1x10-2cm
105
x
xn
1.6x10-2cm
10 (1.5 × 1010 )   0.625  
2
)
exp
 − 1 = 1.09 mA / cm
1 × 10 − 7 1 × 1016   0.0259  
2
= (1.6 × 10
− 19
Total current density= 1.24 mA/cm2
Total current = AxJ=10-3x1.24 =1.24 µA
pn-junction diode structure used in the discussion of currents. The sketch
shows the dimensions and the bias convention. The cross-sectional area
A is assumed to be uniform.
Hole current (solid line) and recombining electron current (dashed line) in the quasi-neutr
al n-region of the long-base diode of Figure 5.5. The sum of the two currents J (dot-dash l
ine) is constant.
Hole density in the quasi-neutral n-region of an ideal short-base
diode under forward bias of Va volts.
The current components in the quasi-neutral regions of a long-base diode
under moderate forward bias: J(1) injected minority-carrier current, J(2)
majority-carrier current recombining with J(1), J(3) majority-carrier current
injected across the junction. J(4) space-charge-region recombination current.
Quantitative
p-n Diode Solution
J
p-region
SCL
n-region
J = J elec + J h ole
T otal current
M ajority carrier diffusion
and drift current
J h ole
J elec
(a) Transient increase of excess stored holes in a long-base ideal diode for a
constant current drive applied at time zero with the diode initially unbiased. Note
the constant gradient at x = xn as time increases from (1) through (5), which
indicates a constant injected hole current. (Circuit shown in inset.) (b) Diode
voltage VD versus time.
M inority carrier diffusion
current
–W p
Wn
x
¾ The total current anywhere in the device is constant.
¾ Just outside the depletion region it is due to the diffusion of minority
carriers.
Current-Voltage Characteristics
of a Typical Silicon
Quantitative
p-n Junction
p-n Diode Solution
Examples
Diode in a circuit
Quantitative
Summary
p-n Diode Solution
Current flow in a pn junction diode
∇⋅E =
ρ
KS ⋅ε0
Poisson’s Equation
Built-in-Potential
(a) under equilibrium, both diffusion currents are
cancelled by opposing drift currents.
(b) under reverse bias, only a small number of carriers are available to
diffuse across the junction (once within the junction they drift to the
other side). With increasing reverse bias the reverse current increases
due to tunneling and carrier multiplication.
W = x p + xn =
2 K sε 0 ( N A + N D )
(Vbi − (V A ))
q
N AND
Width of Depletion Region

 qVa  
J = J 0 exp 
−1
 η kT  

(c) under forward bias, the drift current is slightly reduced
but the diffusion current is greatly increased.
(d) current-voltage characteristic
 qDn n p 0 qD p pn 0  Diode
+
J0 = 
L p  Equation
 Ln
EX 5.7
• A silicon pn junction dide at T=300K is forw
ard biased. The reverse saturation current is
IS=4x 10-14A. Determine the required dide v
oltage to induce a diode current of ID=4.25m
A.