important chap3 1 semiimportant chap3-1-semiconductor-1.pdf conductor 1
DESCRIPTION
Important Chap3 1 SemiImportant Chap3-1-Semiconductor-1.pdf conductor 1TRANSCRIPT
Chapter 3: semiconductor science and light emitting diodes
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/ao
3, which corresponds to 17.7, 5.00, and 4.43 X 1022 cm-3, respectively.
Semiconductor Lattice Structures
Diamond Lattices
The diamond-crystal lattice characterized byfour covalently bonded atoms.
The lattice constant, denoted by ao, is 0.356, 0.543 and 0.565 nm for diamond, silicon, and germanium, respectively.
Nearest neighbors are spaced units apart.( )4/3 oa
(After W. Shockley: Electrons and Holes in Semiconductors, Van Nostrand, Princeton, N.J., 1950.)
Semiconductor Lattice Structures
Diamond and Zincblende Lattices
Diamond latticeSi, Ge
Zincblende latticeGaAs, InP, ZnSe
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
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.
How Many Silicon Atoms per cm-3?• Number of atoms in a unit cell:
• 4 atoms completely inside cell• Each of the 8 atoms on corners are shared among cells
count as 1 atom inside cell• Each of the 6 atoms on the faces are shared among 2
cells 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
• Density of silicon atoms = (8 atoms) / (cell volume) = 5 x 1022 atoms/cm3
“diamond cubic” lattice
The Si Crystal
• Each Si atom has 4 nearest neighbors
• lattice constant= 5.431Å
Semiconductor Materials for Optoelectronic Devices
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
Semiconductor Optical Sources
0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6
Blue
Gre
en
Ora
nge
Yello
w
Red
λ1.7
I n f r ar ed (um)Viol
et
GaA
s
GaA
s 0.55
P 0.45
GaAs1-yPy
InP
In0.
14G
a 0.8
6As
In1-xGaxAs1-yPyAlxGa1-xAs
x = 0.43
GaP
(N)
GaS
b
InG
aNS
iC(A
l )
In0.
7Ga 0.
3As 0.
66P 0.
34
In0.
57G
a 0.43
As 0.
95P 0.
05
In0.49AlxGa0.51-xP
Semiconductor Materials for Optoelectronic Devices
Quantization Concept
plank constant
Core electrons
Valence electrons
Periodic Table of the Elements
Group**
Period 1 IA 1A
18
VIIIA8A
1 1 H
1.008
2 IIA 2A
13
IIIA3A
14 IVA4A
15 VA5A
16 VIA6A
17 VIIA7A
2 He4.003
2 3
Li 6.941
4 Be 9.012
5 B
10.81
6 C
12.01
7 N
14.01
8 O
16.00
9 F
19.00
10 Ne20.18
8 9 10 3 11
Na 22.99
12 Mg 24.31
3 IIIB3B
4 IVB4B
5 VB5B
6 VIB6B
7 VIIB7B ------- VIII -------
------- 8 -------
11 IB 1B
12 IIB2B
13 Al26.98
14 Si
28.09
15 P
30.97
16 S
32.07
17 Cl
35.45
18 Ar39.95
4 19 K
39.10
20 Ca 40.08
21 Sc44.96
22 Ti
47.88
23 V
50.94
24 Cr52.00
25 Mn54.94
26 Fe55.85
27 Co58.47
28 Ni58.69
29 Cu63.55
30 Zn65.39
31 Ga69.72
32 Ge72.59
33 As74.92
34 Se78.96
35 Br79.90
36 Kr83.80
5 37
Rb 85.47
38 Sr
87.62
39 Y
88.91
40 Zr
91.22
41 Nb92.91
42 Mo95.94
43 Tc(98)
44 Ru101.1
45 Rh102.9
46 Pd106.4
47 Ag107.9
48 Cd112.4
49 In
114.8
50 Sn118.7
51 Sb121.8
52 Te127.6
53 I
126.9
54 Xe131.3
6 55 Cs 132.9
56 Ba 137.3
57 La*138.9
72 Hf178.5
73 Ta180.9
74 W
183.9
75 Re186.2
76 Os190.2
77 Ir
190.2
78 Pt
195.1
79 Au197.0
80 Hg200.5
81 Tl
204.4
82 Pb207.2
83 Bi
209.0
84 Po(210)
85 At(210)
86 Rn(222)
7 87 Fr
(223)
88 Ra (226)
89 Ac~(227)
104
Rf(257)
105
Db(260)
106
Sg(263)
107
Bh(262)
108
Hs(265)
109
Mt(266)
110 ---
()
111
---()
112
---()
114
---()
116
---()
118 ---
()
IV Compounds SiC, SiGe
III-V Binary CompoundsAlP, AlAs, AlSb, GaN, GaP, GaAs, GaSb, InP, InAs, InSb
III-V Ternary CompoundsAlGaAs, InGaAs, AlGaP
III-V Quternary CompoundsAlGaAsP, InGaAsP
II-VI Binary CompoundsZnS, ZnSe, ZnTe, CdS, CdSe, CdTe
II-VI Ternary CompoundsHgCdTe
Semiconductor Materials Semiconductor Materials
Atomic Bonding
a. Ionic bonding (such as NaCl)b. Metallic bonding (all metals)c. Covalent bonding (typical Si)d. Van der Waals bonding (water…)e. Mixed bonding (GaAs, ZnSe…, ionic & covalent)
Bonding forces in Solids
Covalent Bonding
Quantization Concept
plank constant
Core electrons
Valence electrons
2s2p
1sK
L
Quantization Concept
The shell model of the atom in which the electrons are confined to live within certain shells and in sub shells within shells.
The Shell Model
1s22s22p2 or [He]2s22p2
L shell with two sub shells
Nucleus
Band theory of solids
Two atoms brought together to form molecule
“splitting” of energy levels for outer electron shells
Energy Band Formation (I)
→ ← =
Splitting of energy states into allowed bandsseparated by a forbidden energy gap as the atomic spacing decreases.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.
Allowed energy levels of an electron acted on by the Coulomb potential of an atomic nucleus.
Energy Band Formation (I)
Energy Band Formation (II)
Strongly bonded materials: small interatomic distances. Thus, the strongly bonded materials can have larger energy bandgaps than do weakly bonded materials.
Energy Bandgapwhere ‘no’ states exist
As atoms are brought closer towardsone another and begin to bond together, their energy levels mustsplit into bands of discrete levelsso closely spaced in energy, theycan be considered a continuum ofallowed energy.
Pauli Exclusion Principle
Only 2 electrons, of spin ± 1/2, can occupy the same energy state at
the same point in space.
Energy Band Formation (III)
Conceptual development of the energy band model.
Elec
tron
ene
rgy
Ele
ctro
n en
ergy
isolatedSi atoms
Si latticespacing
Decreasing atom spacing
s
p
sp n = 3
N isolated Si-atoms
6N p-states total2N s-states total
(4N electrons total)
Elec
tron
ene
rgy
Crystalline Si N -atoms
4N allowed-states (Conduction Band)
4N allowed-states (Valance Band)
No states
4N empty states
2N+2N filled states
Elec
tron
ene
rgy Mostly
empty
Mostlyfilled
Etop
EcEg
Ev
Ebottom
Broadening of allowed energy levels into allowed energy bandsseparated by forbidden-energy gaps as more atoms influence each electron in a solid.
Energy Band Formation (IV)
One-dimensional representation
Two-dimensional diagram in which energy is plotted versus distance.
N electrons filling half of the 2Nallowed states, as can occur in a metal.
Energy Band
Energy band diagrams.
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.
2 s Band
Overlapping energy bands
Electrons2 s2 p
3 s3 p
1 s 1sSOLIDATOM
E = 0
Free electronElectron Energy, E
2 p Band3s Band
Vacuumlevel
In a metal the various energy bands overlap to give a single band of energies that is only partially full of electrons. There are states with energies up to the vacuum level where the electron is free.
Typical band structures of Metal
Metals, Semiconductors, and Insulators
Electron energy, E
ConductionBand(CB)Empty ofelectrons at 0 K.
ValenceBand(VB)Full of electrons at 0 K.
Ec
Ev
0
Ec+χ
Covalent bond Si ion core (+4e)
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.
Typical band structures of Semiconductor
Metals, Semiconductors, and Insulators
Band gap = Eg
Carrier Flow for Metal
Metals, Semiconductors, and Insulators
Carrier Flow for Semiconductors.mov
Carrier Flow for Semiconductor
Carrier Flow for Metals.mov
Metals, Semiconductors, and Insulators
Insulator Semiconductor Metal
Typical band structures at 0 K.
10610310010-310-610-910-1210-1510-18 109
Semiconductors Conductors
1012
AgGraphite NiCrTeIntrinsic Si
Degeneratelydoped Si
Insulators
Diamond
SiO2
Superconductors
PETPVDF
AmorphousAs2Se3
Mica
Alumina
Borosilicate Pure SnO2
Inorganic Glasses
Alloys
Intrinsic GaAs
Soda silica glass
Manyceramics
MetalsPolypropylene
Metals, Semiconductors, and Insulators
Range of conductivities exhibited by various materials.
Conductivity (Ωm)-1
Energy Band Diagram
Electrons within an infinite potential energy well of spatial width L, its energy is quantized.
e
nn m
kE2
)( 2h=
Lnknπ
= ...3,2,1=n
Energy increases parabolically with the wavevector kn.
nkh : electron momentum
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.
x0 +a/2-a/2
V(x)
∞∞
V=0
m
infinite square potential
0 +a/2-a/2
∞∞
0 +a/2-a/2
∞∞
xϕ
x2ϕ
n=1
n=2
n=3
E1
E2
E3
Energy state Wavefunction Probability density
r
PE(r)
x
V(x)
x = Lx = 0 a 2a 3a
0aa
Surface SurfaceCrystal
Potential Energy of the electron around an isolated atom
When N atoms are arranged to form the crystal then there is an overlap of individual electron PEfunctions.
PE of the electron, V(x), inside the crystal is periodic with a period a.
The electron potential energy [PE, V(x)], inside the crystal is periodic with the same periodicity as that of the crystal, a. Far away outside the crystal, by choice, V = 0 (the electron is free and PE = 0).
3.1 Energy Band Diagram
E-k diagram, Bloch function.
Moving through Lattice.mov
within the Crystal!
3.1 Energy Band Diagram
[ ] 0)(222
2
=Ψ⋅−+Ψ xVEm
dxd e
h
Schrödinger equation
)()( maxVxV +=
Periodic Potential
xkikk exUx )()( =Ψ
Periodic Wave functionBloch Wavefunction
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.
E-k diagram, Bloch function.
...3,2,1=m
Ek
kπ/a–π /a
Ec
Ev
ConductionBand (CB)
Ec
Ev
CB
The Energy BandDiagram
Emptyψk
Occupiedψkh+
e-
Eg
e-
h+
hυ
VB
hυ
ValenceBand (VB)
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.
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].
3.1 Band Diagram
E-k diagram of a direct bandgap semiconductor
E-k diagram
3.1 Energy Band Diagram
The bottom axis describe different directions of the crystal.
Si Ge GaAs
The energy is plotted as a function of the wave number, k, along the main crystallographic directions in the crystal.
E
CB
k–k
Direct Bandgap
GaAs
E
CB
VB
Indirect Bandgap, Eg
k–k
kcb
Si
E
k–k
Phonon
Si with a recombination center
Eg
Ec
Ev
Ec
Ev
kvb VB
CB
ErEc
Ev
Photon
VB
In GaAs the minimum of the CB is directly above the maximum of the VB. direct bandgapsemiconductor.
In Si, the minimum of the CB is displaced from the maximum of the VB.indirect bandgap semiconductor
Recombination of an electron and a hole in Si involves a recombination center.
3.1 Energy Band Diagram
E-k diagram
3.1 Energy Band
A simplified energy band diagram with the highest almost-filled band and the lowest almost-empty band.
valence band edge
conduction band edge
vacuum level
χ : electron affinity
e–hole
CB
VB
Ec
Ev
0
Ec+χ
Eg
Free e–hυ > Eg
Hole h+
Electron energy, E
hυ
3. 1 Electrons and Holes
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-Sibond is created.
Generation of Electrons and Holes Electrons: Electrons in the conduction band that are free to move throughout the crystal.
Holes: Missing electrons normally found in the valence band(or empty states in the valence band that would normally be filled).
Electrons and Holes
These “particles” carry electricity. Thus, we call these “carriers”
3.1 Effective Mass (I)
An electron moving in respond to an applied electric field.
E
E
within a Vacuum within a Semiconductor crystal
dtdmEqF v
0=−= dtdmEqF n
v∗=−=
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 Carrier Movement Within the Crystal
Density of States Effective Masses at 300 K
Ge and GaAs have “lighter electrons” than Si which results in faster devices
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.
Curvature of the band determine m*.m* is positive in CB min., negative in VB max.
Moving through Lattice.mov
3.1 Energy Band Diagram
The bottom axis describe different directions of the crystal.
Si Ge GaAs
The energy is plotted as a function of the wave number, k, along the main crystallographic directions in the crystal.
The motion of electrons in a crystal can be visualized and described in a quasi-classical manner.
In most instancesThe electron can be thought of as a particle.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.
m* > 0 : near the bottoms of all bandsm* < 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.
` : near band edge
3.1 Mass Approximation
constant2
2
=
dk
Ed
Covalent Bonding
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
Without “help” the total number of “carriers” (electrons and holes) is limited to 2ni.
For most materials, this is not that much, and leads to very high resistance and few useful applications.
We need to add carriers by modifying the crystal.
This process is known as “doping the crystal”.
Impurity Doping
The need for more control over carrier concentration
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
e–As+
x
As+ As+ As+ As+
Ec
Ed
CB
Ev
~0.05 eV
As atom sites every 106 Si atoms
Distance intocrystal
Electron Energy
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 dopedwith 1 ppm As. There are donor energy levels
below Ecaround As+ sites.
Concept of a Donor “Adding extra” Electrons
n-type Impurity Doping of Si
just
Energy band diagram of an n-type semiconductor connected to a voltage supply of V volts.
The whole energy diagram tiltsbecause the electron now has an electrostatic potential energy as well.
Current flowingV
n-Type Semiconductor
Ec
EF − eV
A
B
V(x), PE (x)
x
PE (x) = – eV
EElectron Energy
Ec − eV
Ev− eV
V(x)
EF
Ev
Concept of a Donor “Adding extra” Electrons
Energy Band Diagram in an Applied Field
Concept of a Acceptor “Adding extra” Holes
All regions of
materialare neutrally
charged
One less bondmeans
the acceptor is electrically satisfied.
One less bondmeans
the neighboring Silicon is left with
an empty state.
Hole Movement
Empty state is located next to the Acceptor
Hole Movement
Another valence electron can fill the empty state located next tothe Acceptor leaving behind a positively charged “hole”.
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
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
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)
Regionaround the“hole” hasone lesselectronand thus ispositivelycharged.
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)
Regionaround the“acceptor”hasone extraelectronand thus isnegativelycharged.
Concept of a Acceptor “Adding extra” Holes
Band diagram equivalent view
B–h+
x
B–
Ev
Ea
B atom sites every 106 Si atoms
Distanceinto crystal
~0.05 eV
B– B– B–
h+
VB
Ec
Electron energy
p-type Impurity Doping of Si
Concept of a Acceptor “Adding extra” Holes
Boron doped Si crystal. B has only three valence electrons. When it substitute for a Si atom one of its bond has an electronmissing and therefore a hole.
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.
Ec
Ev
EFi
CB
EFp
EFn
Ec
Ev
Ec
Ev
VB
Intrinsic semiconductors
In all cases, np=ni2
Note that donor and acceptor energy levels are not shown.
Intrinsic, n-Type, p-Type Semiconductors
Energy band diagrams
n-type semiconductors
p-type semiconductors
Impurity Doping Impurity Doping
Valence Band
Valence Band
Impurity Doping
Position of energy levels within the bandgap of Si for common dopants.
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.
Energy Band
Energy band diagrams.
3.2B Semiconductor Statistics
dEEgc )(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).
dEEgv )(
Density of States Concept
General energy dependence of gc (E) and gv (E) near the band edges.
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:
Density of Statesin Conduction Band
Density of States in Valence Band
3.2B Fermi- Dirac function
Probability of Occupation (Fermi Function) Concept
Now that we know the number of available states at each energy, then how do the electrons occupy these states?
We need to know how the electrons are “distributed in energy”.
Again, Quantum Mechanics tells us that the electrons follow the “Fermi-distribution function”.
Ef ≡ Fermi energy (average energy in the crystal)k ≡ Boltzmann constant (k=8.617×10-5eV/K)T ≡Temperature in Kelvin (K)
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.
kTEE feEf /)(1
1)( −+=
Fermi function applies only under equilibrium conditions, however, is universal in the sense that it applies with all materials-insulators, semiconductors, and metals.
“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”
How do electrons and holes populate the bands?
Probability of Occupation (Fermi Function) Concept
Fermi-Dirac Distribution
3.2B Semiconductor Statistics
Ef
Fermi Function• Probability that an available state at energy E is occupied:
• EF is called the Fermi energy or the Fermi level
There is only one Fermi level in a system at equilibrium.If E >> EF : If E << EF : If E = EF :
kTEE FeEf /)(1
1)( −+=
3.2B Semiconductor Statistics
Probability of Occupation (Fermi function) Concept
Maxwell Boltzmann Distribution Function
Boltzmann Approximation
Probability that a state is empty (occupied by a hole):
kTEEF
FeEfkTEE /)()( ,3 If −−≅>−
kTEEF
FeEfkTEE /)(1)( ,3 If −−≅>−
kTEEkTEE FF eeEf /)(/)()(1 −−− =≅−
TYU
• Assume the Fermi level is 0.30eV below the conduction band energy (a) determine the probability of a state being occupied by an electron at E=Ec+KT at room temperature (300K).
TYU
• Determine the probability that an allowed energy state is empty of electron if the state is below 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?
SolutionThe 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).
( ) ( )VC EfEf −= 1
( ) ( ) kTEEC FCeEf /1
1−+
= ( ) ( ) ( ) kTEEkTEEV VFFV eeEf // 1
11
111 −− +=
+−=−
kT
EE
kT
EE FVFC −=
−
2VC
FEEE +
=
The density of electrons (or holes) occupying the statesin energy between E and E + dE is:
How do electrons and holes populate the bands?
Probability of Occupation Concept
0 Otherwise
dEEfEgc )()(Electrons/cm3 in the conduction band between E and E + dE (if E ≥ Ec).
Holes/cm3 in the conduction band between E and E + dE (if E ≤ Ev).
dEEfEgv )()(
How do electrons and holes populate the bands?
Probability of Occupation Concept
Typical band structures of Semiconductor
Ev
Ec
0
Ec+χ
EF
VB
CB
E
g(E)
g(E) ∝ (E–Ec)1/2
fE)
EF
E
Forelectrons
For holes
[1–f(E)]
Energy band diagram
Density of states Fermi-Diracprobability
function
probability of occupancy of a state
nE(E) or pE(E)
E
nE(E)
pE(E)
Area = p
Area
Ec
Ev
ndEEn E == ∫ )(
g(E) X f(E)Energy density of electrons in
the CB
number of electrons per unit energy per unit volumeThe area under nE(E) vs. E is the electron concentration.
number of states per unit energy per unit volume
How do electrons and holes populate the bands?
The Density of Electrons is:
Probability the state is filled
Number of states per cm-3 in energy range dE
Probability the state is empty
Number of states per cm-3 in energy range dE
units of n and p are [ #/cm3]
The Density of Hole is:
Developing the Mathematical Model for Electrons and Holes concentrations
Electron Concentration (no)
TYU
Calculate the thermal equilibrium electron concentration in Si at T=300K for the case when the Fermi level is 0.25eV below the conduction band.
EC
EV
EF0.25eV
Hole Concentration (no)
TYU
• Calculate thermal equilibrium hole concentration in Si at T=300k for the case when the Fermi level is 0.20eV above the valance band energy Ev.
EC
EV 0.20eVEF
Degenerate and Nondegenerate Semiconductors
Nondegenerate Case
Useful approximations to the Fermi-Dirac integral:
( ) kTEEC
CfeNn −=
( ) kTEEV
fVeNp −=
Developing the Mathematical Model for Electrons and Holes
( ) kTEECi
CieNn −=
When n = ni, Ef = Ei (the intrinsic energy), then
orand
( ) kTEEVi
iVeNn −=( ) kTEE
iVVienN −=or
( ) kTEEiC
iCenN −=
The intrinsic carrier concentration
( ) kTEECo
CfeNn −= ( ) kTEEV
fVo eNp −=
Other useful relationships: n⋅p product:
( ) kTEECi
CieNn −= and ( ) kTEEVi
iVeNn −=
( ) kTEVC
kTEEVCi
gVC eNNeNNn −−− ==2
kTEVCi
geNNn 2−=
Semiconductor Statistics
TYU
Determine the intrinsic carrier concentration in GaAs (a) at T=200k and (b) T=400K
2ioo npn =
Law of mass Action
( ) kTEEio
ifenn −=( ) kTEE
iofiepp −=andSince
It is one of the fundamental principles of semiconductorsin thermal equilibrium
Example
Law of mass action
An intrinsic Silicon wafer has 1x1010 cm-3 holes. When 1x1018
cm-3 donors are added, what is the new hole concentration?
2ioo npn =
DNn ≅D
i
Nnp
2
≅and
AD NN ⟩⟩ iD nN ⟩⟩andif
TYU
• 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?
TYU: The concentration of majority carrier
electron is no=1 x 1015 cm-3 at 300K. Determine the concentration of phosphorus that are to be added and determine the concentration minority carriers holes.
Partial Ionization, Intrinsic Energy and Parameter Relationships.
Energy band diagram showing negative charges
Energy band diagram showing positive charges
If excess charge existed within the semiconductor, random motion of charge would imply net (AC) current flow.
Not possible!
Thus, all charges within the semiconductor must cancel.
Charge Neutrality:
( ) ( )[ ]( ) ( )[ ] 0=−+−⋅
+=++−
−+
nNNpq
nNNp
dA
ad
Mob
ile +
cha
rge
Imm
obile
-ch
arge
Imm
obile
+ c
harg
e
Mob
ile -
char
ge
3.5 Carrier concentration-effects of doping
NA¯ = Concentration of “ionized” acceptors = ~ NA
ND+ = Concentration of “ionized” Donors = ~ ND
Charge Neutrality: Total Ionization case
( ) ( ) 0=−+− +− nNNp dA
3.5 Developing the Mathematical Model for Electrons and Holes
The intrinsic carrier concentration as a function of temperature.
Electron concentration versus temperature for n-typeSemiconductor.
Carrier Concentration vs. Temperature
position of Fermi Energy level
( ) kTEEco
fceNn ][ −−=)/ln( occ nNkTEE F =−
)/ln( dcFc NNkTEE =−
Nd >> ni
Note: If we have a compensated semiconductor , then the Nd termin the above equation is simply replaced by Nd-Na.
( ) kTEEVo
fVeNp −=
)/ln( ovvF pNkTEE =−
)/ln( avvF NNkTEE =−
Na >> ni
position of Fermi Energy level
Note: If we have a compensated semiconductor , then the Na termin the above equation is simply replaced by Na-Nd.
position of Fermi level as a function of carrier concentration
Where is Ei ?
Extrinsic Material:
Note: The Fermi-level is pictured here for 2 separate cases: acceptor and donor doped.
TYU
• Determine the position of the Fermi level with respect to the valence band energy in p-type GaAs at T=300K. The doping concentration are Na=5 x 1016 cm-3 and Na=4 x 1015 cm-3.
position of Fermi Energy level
Extrinsic Material:
( ) kTEEio
fifenn −=( ) kTEE
ioffienp −=
Solving for (Ef - Efi)
−=
=−
iifif n
pkTnnkTEE lnln
=−
i
Dfif n
NkTEE ln
−=−
i
Afif n
NkTEE ln
AD NN ⟩⟩ iD nN ⟩⟩andfor DA NN ⟩⟩ iA nN ⟩⟩andfor
TYU 3.8
• Calculate the position of the Fermi level in n-type Si at T=300K with respect to the intrinsic Fermi energy level. The doping concentration are Nd=2 x 1017 cm-3 and Na=3 x 1016 cm-3
.
EC
EV
EFi
EF
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.
– Recombination-generation (R-G)
Carrier Motion
Carrier Dynamics
Electron Drift
Hole Drift
Electron Diffusion
Hole Diffusion
Carrier DriftDirection of motion
Holes move in the direction of the electric field. (⊕ )
Electrons move in the opposite direction of the electric field. ( ⊕)
Motion is highly non-directional on a local scale, but has a net direction on a macroscopic scale.
Average net motion is described by the drift velocity, vd [cm/sec].
Net motion of charged particles gives rise to a current.
Instantaneous velocity is extremely fast
Describe the mechanism of the carrier drift and drift current due to an applied electric field.
Drift
Drift of Carriers
Electric Field
Drift of electron in a solid
The ball rolling down the smooth hill speeds up continuously, but the ball rolling down the stairs moves with a constant average velocity.
µ [cm2/Vsec] : mobility
Random thermal motion. Combined motion due to random thermal motion and an applied electric field.
Drift
Schematic path of an electron in a semiconductor.
EE
Drift
Random thermal motion. Combined motion due to random thermal motion and an applied electric field.
Drift
Conduction process in an n-type semiconductor
Thermal equilibrium Under a biasing condition
Drift
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:
Hole Drift Current Density
dp vpqJvJdrf
drf
d
⋅⋅==
Electron Drift Current Density
dn vneJ drf ⋅⋅−=and
dp
drf
vpeJ
vJ
drf
d
⋅⋅=
= ρ
Drift
At Low Electric Field Values,
EpeJ pDriftp ⋅⋅⋅= µ EneJ nDriftn ⋅⋅⋅= µand
µ [cm2/V·sec] is the “mobility” of the semiconductor and measures the ease with which carriers can move through the crystal.
The drift velocity increases with increasing applied electric field.:
EnpqJJJ npDriftnDriftpdrf ⋅+=+= )( µµ
Electron and hole mobilities of selected intrinsic semiconductors (T=300K)
Si Ge GaAs InAsµn (cm
2/V·s) 1350 3900 8500 30000µp (cm
2/V·s) 480 1900 400 500
sV
cmV/cmcm/s 2
⋅
=µ has the dimensions of v/ :
Electron and Hole Mobilities EX 4.1
• Consider a GaAs sample at 300K with doping 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.
µ [cm2/Vsec] is the “mobility” of the semiconductor and measures the ease with which carriers can move through the crystal.
Mobility
µn ~ 1360 cm2/Vsec for Silicon @ 300Kµp ~ 460 cm2/Vsec for Silicon @ 300Kµn ~ 8000 cm2/Vsec for GaAs @ 300Kµp ~ 400 cm2/Vsec for GaAs @ 300K
[ ]sec2*
,, Vcm
mq
pnpn
τµ =
<τ > is the average time between “particle” collisions in the semiconductor.
Collisions can occur with lattice atoms, charged dopant atoms, or with other carriers.
Drift velocity vs. Electric field in Si.
Saturation velocity Saturation velocity
Drift velocity vs. Electric fieldDesigning devices to work at the peak results in faster operation
1/2mvth2=3/2kT=3/2(0.0259)=0.03885eV
Ohm’s law is valid only in the low-field region where drift velocity is independent of the applied electric field strength.Saturation velocity is approximately equal to the thermal velocity (107 cm/s).
[ ]sec2*
,, Vcm
mq
pnpn
τµ =
Drift
Drift velocity vs. Electric field in Si and GaAs.
Note that for n-type GaAs, there is a region of negative differential mobility.
[ se2*
,, Vcm
mq
pnpn
τµ =
Negative differential mobility
Electron distributions under various conditions of electric fields for a two-valley semiconductor.
m*n=0.067mo
m*n=0.55mo
Figure 3.24.
Velocity-Field characteristic of a Two-valley semiconductor.
Negative differential mobilityTYU
• 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.
Mean Free Path
• Average distance traveled between collisions
mpthvl τ=
EX 4.2Using figure 4.3 determine electron and hole nobilities.
EX 4.2Using 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
Effect of Temperature on Mobility
Temperature dependence of mobility with both lattice and impurity scattering.
A carrier moving through the lattice encounters atoms which are out of their normal lattice positions due to the thermal vibrations.
The frequency of such scattering increases as temperature increases.
At low temp. lattice scattering is less important.
At low temperature, thermal motion of the carriers is slower, and ionized impurity scattering becomes dominant.
Since the slowing moving carrier is likely to be scattered more strongly by an interaction with charged ion.
Impurity scattering events cause a decrease in mobility with decreasing temperature.
As doping concentration increase, impurity scattering increase, then mobility decrease.
Mobility versus temperature Mobility versus temperature
Effect of Temperature on Mobility
Electron mobility in silicon versus temperature for various donor concentrations. Insert shows the theoretical temperature dependence of electron mobility.
Electron and hole mobilities in Silicon as functions of the total dopant concentration.
Effect of Doping concentration on Mobility
300 K
Resistivity and Conductivity
Ohms’ Law
Ohms Law [ ]2cmAEEJρ
σ =⋅=
Conductivityσ [ ]cmohm ⋅1
Resistivityρ [ ]cmohm ⋅
semiconductor conductivity and resistivity
Adding the Electron and Hole Drift Currents (at low electric fields)
Drift CurrentEnpeJJJ npDriftnDriftpdrf ⋅+=+= )( µµ
Conductivity)( npe np µµσ +=
Resistivity[ ])(11 pne pn µµσ
ρ +==
But since µn and µp change very little and n and p change several orders of magnitude:
for n-type with n>>p
pene
p
n
µσµσ
≅≅
for p-type with p>>n
[ ]sec2*
,, Vcm
mq
pnpn
τµ =
Particles diffuse from regions of higher concentration to regions of lower concentration region, due to random thermal motion.
Diffusion Diffusion
Nature attempts to reduce concentration gradients to zero.Example: a bad odor in a room, a drop of ink in a cup of water.
In semiconductors, this “flow of carriers” from one region of higher concentration to lower concentration results in a “Diffusion Current”.
Visualization of electron and hole diffusion on a macroscopic scale.
DiffusionpJDiffusionnJ
Diffuse Diffuse
dxdneDJ N=diffN, dx
dpeDJ P−=diffP,
D is the diffusion constant, or diffusivity.
x x
Diffusion Current
Diffusion current density
Fick’s law
Diffusion as the flux, F, (of particles in our case) is proportional to the gradient in concentration.
η∇−= DFη : Concentration
D : Diffusion Coefficient
For electrons and holes, the diffusion current density( Flux of particles times ± q )
nDqJ
pDqJ
nDiffusionn
pDiffusionp
∇⋅=
∇⋅−=
The opposite sign for electrons and holes
JN = JN,drift + JN,diff = qnµnε + dxdnqDN
JP = JP,drift + JP,diff = qpµpε –dxdpqDP
J = JN + JP
Total Current
Total Current
Total Current = Drift Current + Diffusion Current
nDqEnqJJJ
pDqEpqJJJ
nnDiffusionnDriftnn
ppDiffusionpDriftpp
∇⋅+⋅=+=
∇⋅−⋅=+=
µ
µ
np JJJ +=
TYU
• 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 diffusion coefficient is Dn=25cm2/sec, determine the electron concentration at x=0.
dxdneDJ N=diffN,
dxdpeDJ P−=diffP,Jp=0.270 A/cm2
Dp=12 cm2/secFind the hole concentration at x=50um
Graded impurity distribution
Energy band diagram of a semiconductor in thermal equilibriumwith a nonuniform donor impurity concentration
Carrier Generation
Generation and Recombination
Band-to-band generation
Generation Mechanism
Band-to-Band Generation
Thermal Energyor
Light
Band-to-Band or “direct” (directly across the band) generation.
Does not have to be a “direct bandgap” material.
Mechanism that results in ni.
Basis for light absorption devices such as semiconductor photodetectors, solar cells, etc…
Gno=Gpo
Band-to-Band Recombination
Recombination Mechanism
Photon(single particle of light)
or
multiple phonons(single quantum of lattice vibration - equivalent tosaying thermal energy)
Band to Band or “direct” (directly across the band) recombination.
Does not have to be a “direct bandgap” material, but is typically very slow in “indirect bandgap” materials.
Basis for light emission devices such as semiconductor Lasers, LEDs, etc…
Rno=Rpo
In thermal equilibrium: Gno=Gpo=Rno=Rpo
Low-Level-Injection implies
00 , nnnp ≈<<∆
00 , pppn ≈<<∆
in a n-type material
in a p-type material
00 pppnnn +∆=+∆=
In Non-equilibrium, n·p does not equal ni2
Excess carrier Recombination and Generation
low level injection caseNd=1014/cm3 doped Si at 300K subject to a perturbation where ∆p =∆n =109/cm3.
n0 ≅ Nd =1014/cm3 and p0 ≅ ni2/Nd = 106/cm3
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.
and
Example
Excess minority carrier lifetime
Carrier Lifetime
notentn τ−∆=∆ 0)(poteptp τ−∆=∆ 0)(
Light Pulses
Semiconductor
Rs
VARLI VL
+
_
Oscilloscope
Schematic diagram of photoconductivity decay measurement.
The minority carrier lifetime is the average time a minority carrier can survive in a large ensemble of majority carriers.If ∆p is negative Generation or an increase in carriers with time.If ∆p is positive Recombination or a decrease in carriers with time.Either way the system “tries to reach equilibrium”The rate of relaxation depends on how far away from equilibrium we are.
Material Response to “Non-Equilibrium”
Relaxation Concept
Consider a case when the hole concentration in an n-type sample is not in equilibrium, i.e., n·p ≠ ni
2
τpo is the minority carrier lifetime
po
ntp
tpR
τ)(' ∆
−=∂∂
=
Material Response to “Non-Equilibrium”
Relaxation Concept
Likewise when the electron concentration in an p-type sample is not in equilibrium, i.e., n·p does NOT equal ni
2
τn is the minority carrier lifetimenGRThermal
ntn
τ∆
−=∂∂
−
Indirect recombination-generation processes at thermal equilibrium.
Recombination-Generation center recombination
Generation and Recombination process
Recombination-Generation (R-G) Center Recombination
Energy loss can result in a Photonbut is more often multiple phonons
Also known as Shockley-Read-Hall (SRH) recombination.
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.
Useful for creating “fast switching” devices by quickly “killing off” EHP’s.
Recombination Mechanism Generation and Recombination process
Generation Mechanism
Recombination-Generation (R-G) Center Generation
Thermal Energy
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.
Almost always detrimental to electronic devices. AVOID IF POSSIBLE!
Effects of recombination centers on solar cell
performance
EC
EV Light
a
b
c
dThe light-generated minoritycarrier can return to the groundstate through recombination
center before beingcollected by the junction:i) through path (a) ii) through path (c)
Without recombinationcenters paths (b) and (d)are dominated
Auger Recombination
Auger RecombinationAuger – “pronounced O-jay”
Requires 3 particles.
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 theannihilation 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.
3.3 p-n Junction Diode
p-n Junction
p-Type Material n-Type Material
p-n Junction principles
p-n Junction
p-Type Material
p-Type Material n-Type Material
n-Type Material
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
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:
p-n Junction Diode
p-Type Material n-Type Material
EC
EV
EFEi
EC
EV
EF
Ei
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:
p-Type Material
n-Type Material
p-n Junction Diode
EC
EV
EF
Ei EC
EV
EFEi
Built-in-potential
p-n Junction Diode
EC
EV
EF
Ei EC
EV
EFEi
- qVbi
p-Type Material n-Type Material
x
Electrostatic Potential)(1
refC EEq
V −=
Built-in-potentialBIV
Built-in-potential
p-n Junction Diode
x
Electrostatic Potential)(1
refC EEq
V −=
Built-in-potentialBIV
x
Electric Field
dxdVE −=
Electric Field
x
Electric Field
dxdVE −=
Built-in-potential
p-n Junction Diode
Charge Density
x
dxdEK S 0ερ ⋅=
qND
qNA
Charge Density
- --
+ ++
Built-In Potential Vbi
=
=− −
i
A
isidepFi
nNkT
npkTEE
ln
ln)(
sideniFsidepFisiden Ssidep Sbi )()()( −−−− −+−=Φ+Φ= EEEEqqV
=
=− −
i
D
isideniF
nNkT
nnkTEE
ln
ln)(
For non-degenerately doped material:
TYU 5.1
• Calculate the built-in-potential barrier in a Si pm junction at T=300K for (a) Na=5 x 1017cm-3, Nd=1016cm-3 (b)Na=1015cm-3
x
Electric Field
dxdVE −=
Built-in-potential
p-n Junction Diode
Charge Density
x
dxdEK S 0ερ ⋅=
qND
qNA
Charge Density
- --
+ ++
n n o
xx = 0
p n o
p po
n po
l o g ( n ) , l o g ( p )
-e N a
e N d
M
x
E ( x )
B -
h +
p n
M
As +
e –
Wp Wn
V o
V ( x )
x
P E (x )
x
– Wp
Wn
0
e V o
x
– e V o
Hole Potential Energy PE (x)o
– E o
E0
ρ net
M
n i
p-n Junction Principles
Electron Potential Energy PE (x)
Space charge regionM
-Wp -WnNeutral n-regionNeutral p-region
Metallurgical Junction
Charge Density (NOT Resistivity)
Poisson’s Equation
p-n Junction
0ερ⋅
−=⋅∇SK
E0
/
0 εερ
⋅−
=⋅
−=S
DA
S KqN
KdxdE
in 1-dimension
Electric Field
Permittivity of free spaceRelative Permittivity of Semiconductor
(εr)
D
A
AD
qNxqNx
NNnpq
=−=
−+−=
)()(
)(
ρρρ
Movement of Electrons and Holes when Forming the Junction
Depletion Region Approximation: Step Junction Solution
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 applied 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
Diode under Forward Bias.mov Diode under no Bias.mov Diode under Reverse Bias.mov
Schematic representation of depletion layer width and energy band diagrams of a p-n junction under various biasing conditions.
Reverse-bias condition
Forward-bias condition
Thermal-equilibrium
pn junction reverse/forward applied bias
Depletion Region
Thus, only the boundary conditions change resulting indirect replacement of Vbi with (Vbi-VA) with VA ≠ 0.
pn junction forward bias applied bias
Depletion Region Approximation: Step Junction Solution
pn junction reverse/forward applied
Depletion Region Approximation: Step Junction Solution with VA ≠ 0Consider 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
2/1)(2
++
=+=NaNd
NdNe
VVxxW ARbispn
ε
2/11)(2
+
+
=da
Rbis
NNNaNd
eVVxp ε
2/11)(2
+
+=
dad
aRbis
NNNN
eVVxn ε
EX 5.3
• A Si pn junction at 300K is reverse bias at VR=8V, the doping concentration are Na=5 x 1015cm-3 and Nd= 5 x 1016 cm-3. Determine xn, xp and W, repeat for VR=12V.
p-n Junction I-V Characteristics
In Equilibrium (no bias)Total current balances due to the sum of the individual components
Electron Drift Current
Electron DiffusionCurrent
Hole Drift CurrentHole Diffusion
Current
Diode under no Bias.mov
no net current!
EC
EV
EFEi
p-Type Material n-Type Materialq VBI
+ + ++++ +++ ++ ++ ++++++
0=∇⋅+⋅=+= nDqnEqJJJ nnDiffusionnDriftnn µ
no net current!
p-n Junction I-V Characteristics
EC
EV
EF
Ei
n vs. E
p vs. E
In Equilibrium (no bias)Total current balances due to the sum of the individual components
0=∇⋅+⋅=+= pDqpEqJJJ ppDiffusionpDriftpp µ
p-n Junction I-V Characteristics
Forward Bias (VA > 0)
I
Hole Drift Current
Electron Drift Current
Electron DiffusionCurrent
Hole Diffusion Current IP
IN
Diode under Forward Bias m
Current flow is dominated by majority carriers flowingacross the junction and becoming minority carriers
VA
Current flow is proportional to e(Va/Vref) due to the exponential decay of carriers into the majority carrier bands
Lowering of potential hill
by VA
surmount potential barrier
PN III +=
Hole Diffusion Current negligible due to large energy barrier
Hole Drift Current
Electron Drift Current
Electron Diffusion Current negligible due to large energy barrier
Reverse Bias (VA < 0)
Diode under Reverse Bias.m
p-n Junction I-V Characteristics
Current flow is constant due to thermally generated carriers swept out by E fields in the depletion region
Current flow is dominated by minority carriers flowing across the junction and becoming majority carriers
Increase of potential hill
by VA
Where does the Reverse Bias Current come from? Generation near the depletion region edges “replenishes” the current source.
p-n Junction I-V Characteristics
P-N Junction Diodes Current Flowing through a Diode
I-V CharacteristicsQuantitative Analysis
(Math, math and more math)
Putting it all together
p-n Junction I-V Characteristics
for Ideal diodeVref = kT/q
-I0
−
= 10
kT
qVexpII
η
η : Diode Ideality Factor
p-n Junction I-V Characteristics
Diode Equation
Quantitative p-n Diode Solution
Assumptions:1) Steady state conditions2) Non- degenerate doping3) One- dimensional analysis4) Low- level injection5) No light (GL = 0)
Current equations:
)x(J)x(JJ npp +=
−=
dx
dpqDpEqJ ppp µ
−=
dxdnqDnEqJ nnn µ
Continuity Equations
Steady state : n(x) is time invariant.
Transient state : n(x) is time dependent.
xJ
qxF
tn
∂∂
=∂∂
−=∂∂ 1
F: Particle FluxJ: Current Density
Continuity Equation
)( xxJ p ∆−
)(xJ p
xx ∆+x
x∆
2Area, cmA
Ways Carrier Concentrations can be Altered
Continuity Equations
...
...
light as suchprocesses other All
GR ThermalDiffusionDrift
light as suchprocesses other All
GR ThermalDiffusionDrift
tp
tp
tp
tp
tp
tn
tn
tn
tn
tn
∂∂
+∂∂
+∂∂
+∂∂
=∂∂
∂∂
+∂∂
+∂∂
+∂∂
=∂∂
−
−
There must be spatial and time continuity in the carrier concentrations.
Continuity Equations
PPzPyPx
DiffusionDrift
NNzNyNx
DiffusionDrift
Jqz
Jy
Jx
Jqt
ptp
Jqz
Jy
Jx
Jqt
ntn
⋅∇−=
∂∂
+∂
∂+
∂∂
−=∂∂
+∂∂
⋅∇=
∂∂
+∂
∂+
∂∂
=∂∂
+∂∂
11
11
...
...
1
1
light as suchprocesses other All
GR ThermalP
light as suchprocesses other All
GR ThermalN
tp
tpJ
qtp
tn
tnJ
qtn
∂∂
+∂∂
+⋅∇−=∂∂
∂∂
+∂∂
+⋅∇=∂∂
−
−
Ways Carrier Concentrations can be Altered
Continuity Equations: Special Case known as “Minority Carrier Diffusion Equation”
Simplifying Assumptions:
1) One dimensional case. We will use “x”.
2) We will only consider minority carriers.
3) Electric field is approximately zero in regions subject to analysis.
4) The minority carrier concentrations IN EQUILIBRIUM are not a function of position.
5) Low-level injection conditions apply.
6) SRH recombination-generation is the main recombination-generation mechanism.
7) The only “other” mechanism is photogeneration.
Continuity Equations
Minority CarrierDiffusion Equation
Because of (3) no electric field E = 0
Because of (5) - low level injection Because of (7) - Photogeneration
L
light as suchprocesses other All
Gtn
=∂∂
...nGenerationombinationRe
ntn
τ∆
−=∂∂
−
2
2
20
2
2
2 )()(11x
nDx
nnDxnD
xJ
qJ
q nnnN
N ∂∆∂
=∂
∆+∂=
∂∂
=∂∂
=⋅∇
nDqnEqJJJ nnnnn ∇⋅+⋅=+= µDiffusionDrift
nDqJ nn ∇⋅==Diffusion
Continuity Equations: Special Case known as “Minority Carrier Diffusion Equation”
0 0
Finally
tn
tnn
tn
∂∆∂
=∂∆+∂
=∂∂ )()( 0
Continuity Equations: Special Case known as “Minority Carrier Diffusion Equation”
...light as suchprocesses other All
GenerationombinationReDiffusionDrift tn
tn
tn
tn
tn
∂∂
+∂∂
+∂∂
+∂∂
=∂∂
−
Ln
ppN
p Gn
xn
Dtn
+∆
−∂
∆∂=
∂
∆∂
τ)()()(
2
2
Lp
nnP
n Gpx
pDtp
+∆
−∂∆∂
=∂∆∂
τ)()()(
2
2
0
Minority Carrier Diffusion Equations
Continuity Equations
Continuity Equations: Special Case known as “Minority Carrier Diffusion Equation”
Further simplifications (as needed):
Steady State …
No minority carrier diffusion gradient …
No SRH recombination-generation …
00 =∆
−=∆
−pn
pandnττ
No light …
0→LG
0)(0)(
→∂∆∂
→∂
∆∂
tpand
tn np
0)(0)(
2
2
2
2
→∂∆∂
→∂∆∂
xpDand
xn
D nP
pN
Solutions to the“Minority Carrier Diffusion Equation”
SemiconductorSemiconductorLight x
Light absorbed in a thin skin.
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).
What is the minority carrier distribution in the region x> 0 ?
Ln
ppN
p Gn
xn
Dtn
+∆
−∂
∆∂=
∂
∆∂
τ)()()(
2
2
n
ppN
nx
nD
τ)()(
2
2 ∆=
∂
∆∂
0 0Steady state No excess carrier
generation
Solutions to the“Minority Carrier Diffusion Equation”
Steady-state carrier injection from one side.
Semi-infinite sample
Sample with thickness W
Direct generation and recombination of electron-hole pairs:
at thermal equilibrium under illumination. Surface recombination at x = 0. The minority carrier distribution near the surface is affected by the surface recombination velocity.
Solutions to the“Minority Carrier Diffusion Equation”
General Solution
NNNLxLx
p DL whereeeAxn NN τ⋅≡⋅+⋅=∆ +− )()( B)(
LN is the “Diffusion length” the average distance a minority carrier can move before recombining with a majority carrier.
Boundary Condition …
( )
0B
BeAxn
BAcmxn
NLp
p
=→
+==∞=∆
+===∆∞+
−
)0(0)(
10)0( 313
( ) 31310)( −−=∆ cmexn NLxp
0
Continue
Solutions to the“Minority Carrier Diffusion Equation”
SemiconductorSemiconductorx
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.
What is the minority carrier distribution in for t < 0 ?
Ln
ppN
p Gn
xn
Dtn
+∆
−∂
∆∂=
∂
∆∂
τ)()()(
2
2
3710)0,( −=⋅=<∆ cmGtx alln nLp τ
0
Light Light absorbed uniformly
0
SemiconductorSemiconductor
Uniform distribution
Solutions to the“Minority Carrier Diffusion Equation”
SemiconductorSemiconductor Light x
In the previous example: What is the minority carrier distribution in for t > 0 ?
Ln
ppN
p Gn
xn
Dtn
+∆
−∂
∆∂=
∂
∆∂
τ)()()(
2
2
)51(7
)(
10)(
)]0([)(−−
−
=∆
⋅=∆=∆et
p
tpp
etn
etntn nτ
Light absorbed uniformly
0 0
ContinueApplication of the Minority Carrier Diffusion Equation
Since electric fieldsexist in the depletion region, the minority
carrier diffusion equation does not
apply here.
Quisineutral Region Quisineutral Region
00 0 0
Quantitative p-n Diode Solution
minority carrier diffusion eq. minority carrier diffusion eq.
kTEEi
kTEEi
fi
if
enp
enn)(
0
)(0
−
−
=
=
Equilibrium
kTFEi
kTEFi
Pi
iN
enp
enn)(
)(
−
−
=
=
Non-Equilibrium
The Fermi level is meaningful only when the system is in thermal equilibrium.
The non-equilibrium carrier concentration can be expressed by defining Quasi-Fermi levels Fn and Fp .
Equilibrium Non-Equilibrium
Quasi - Fermi Levels
Quisineutral Region Quisineutral Region
Quantitative p-n Diode Solution (At the depletion regions edge)
kT)FF(i
PNennp −= 2quasi-Fermi levels formalism
?
Quisineutral Region Quisineutral Region
Quantitative p-n Diode Solution
+=
dxdnDnEqJ nnn µ
( )dx
nndqD p
n
∆+=
0
dx
ndqD p
n
∆=
+=
dx
dpDpEqJ ppp µ
( )dx
ppdqD n
p
∆+= 0
dx
pdqD n
p
∆=
0 0
Approach:Solve minority carrier diffusion equation in quasineutral regions.
Determine minority carrier currents from continuity equation.
Evaluate currents at the depletion region edges.
Add these together and multiply by area to determine the total current through the device.
Use translated axes, x x’ and -x x’’ in our solution.
Quisineutral Region Quisineutral Region
x”=0 x’=0
Quantitative p-n Diode Solution
Quisineutral Region Quisineutral Region
x”=0 x’=0
Quantitative p-n Diode Solution
Holes on the n-side
Quantitative p-n Diode Solution
Quisineutral Region Quisineutral Region
x”=0 x’=0 Holes on the n-side
Quantitative p-n Diode Solution
Similarly for electrons on the p-side…
Quisineutral Region Quisineutral Region
x”=0 x’=0
Quantitative p-n Diode Solution
Thus, evaluating the current components at the depletion region edges, we have…
Note: Vref from our previous qualitative analysis equation is the thermal voltage, kT/q
J = Jn (x”=0) +Jp (x’=0) = Jn (x’=0) +Jn (x”=0) = Jn (x’=0) +Jp (x’=0)
Ideal Diode Equation Shockley Equation
Quisineutral Region Quisineutral Region
x”=0 x’=0
Quantitative p-n Diode Solution
Total on current is constant throughout the device. Thus, we can characterize the current flow components as…
-xp xn
J
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.
Example
SolutionMinority electron diffusion current density nnn DL τ=
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.
Example
SolutionMinority electron diffusion current density
Minority hole diffusion current density
nnn DL τ=
( ) 216
210
719
2
0
20
/154.010259.0625.0exp
105105.1
10525)106.1(
1exp1exp
cmmA
kTqV
NnDq
kTqV
LnqD
J a
A
i
n
na
n
pnn
=
−
××
××=
−
=
−
=
−−
τ
( ) 216
210
719
2
0
20
/09.110259.0625.0exp
101105.1
10110)106.1(
1exp1exp
cmmA
kTqV
NnD
qkTqV
LnqD
J a
D
i
p
pa
p
npp
=
−
××
××=
−
=
−
=
−−
τ
Total current density= 1.24 mA/cm2 Total current = AxJ=10-3x1.24 =1.24 µA
x
1017
108
103
-xp
2.1x10-2cm
1015
1010
xn
1.6x10-2cm
n or p(log scale)
105
nn
pp
nppn
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.
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.
Example
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-neutral n-region of the long-base diode of Figure 5.5. The sum of the two currents J (dot-dash line) 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.
(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.
J elec
x
n-region
J = J elec + J h ole
SCL
Minority carrier diffusioncurrent
M ajority carrier diffusionand drift current
Total current
J h ole
Wn–Wp
p-region
J
The total current anywhere in the device is constant.
Just outside the depletion region it is due to the diffusion of minority carriers.
Quantitative p-n Diode Solution
Current-Voltage Characteristics
of a Typical Silicon p-n Junction Quantitative p-n Diode Solution
ExamplesDiode in a circuit
Quantitative p-n Diode Solution
Current flow in a pn junction diode
(b) under reverse bias, only a small number of carriers are available todiffuse 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.
(a) under equilibrium, both diffusion currents arecancelled by opposing drift currents.
(c) under forward bias, the drift current is slightly reduced but the diffusion current is greatly increased.
(d) current-voltage characteristic
Summary
Built-in-Potential
−
= 1exp0 kT
qVJJ a
η
+=
p
np
n
pn
LpqD
LnqD
J 000
DiodeEquation
))(()(2 0Abi
DA
DAsnp VV
NNNN
qKxxW −
+=+=
ε
Width of Depletion Region
0ερ⋅
=⋅∇SK
E Poisson’s Equation
EX 5.7
• A silicon pn junction dide at T=300K is forward biased. The reverse saturation current is IS=4x 10-14A. Determine the required dide voltage to induce a diode current of ID=4.25mA.