Chapter 1 Electrons and Holes
in Semiconductors
1.1 Silicon Crystal Structure
•
Unit cell of silicon crystal is
cubic.
•
Each Si atom has 4 nearest
neighbors.
5.43 Å
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Slide 1-1
Silicon Wafers and Crystal Planes
z
z
z
y
(100) x
y
(011)
y
x
(111)
x
(100)
plane
(011)
flat
Si (111) plane
• The standard notation
for crystal planes is
based on the cubic
unit cell.
• Silicon wafers are
usually cut along the
(100) plane with a flat
or notch to help orient
the wafer during IC
fabrication.
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Slide 1-2
1.2 Bond Model of Electrons and Holes
Si
Si
Si
Si
Si
Si
Si
Si
Si
• Silicon crystal in
a two-dimensional
representation.
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Dopants in Silicon
Si
Si
Si
Si
Si
Si
Si
As
Si
Si
B
Si
Si
Si
Si
Si
Si
Si
• As, a Group V element, introduces conduction electrons and creates
N-type silicon, and is called a donor.
• B, a Group III element, introduces holes and creates P-type silicon,
and is called an acceptor.
• Donors and acceptors are known
as dopants. Dopant ionization
energy ~50meV (very low).
Hydrogen:
E ion =
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m0 q4
8ε02h2
= 13.6 eV
Slide 1-4
GaAs, III-V Compound Semiconductors, and Their Dopants
Ga
As
Ga As
Ga
As Ga
As
Ga As
Ga
• GaAs has the same crystal structure as Si.
• GaAs, GaP, GaN are III-V compound semiconductors, important for
optoelectronics.
• Which group of elements are candidates for donors? acceptors?
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1.3 Energy Band Model
}
2s2s
Empty
upper bands
Lowest empty band
( conduction band)
2p2p
Highest filled band
(valence band)
}
Filled lower bands
(a)
(b)
• Energy states of Si atom (a) expand into energy bands of Si crystal (b).
• The lower bands are filled and higher bands are empty in a semiconductor.
• The highest filled band is the valence band.
• The lowest empty band is the conduction band .
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1.3.1 Energy Band Diagram
Conduction band
Ec
Band gap
Eg
Ev
Valence band
• Energy band diagram shows the bottom edge of conduction band,
Ec , and top edge of valence band, Ev .
• Ec and Ev are separated by the band gap energy, Eg .
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Slide 1-7
Measuring the Band Gap Energy by Light Absorption
electron
Ec
photons
Eg
photon energy: h v > E g
Ev
hole
• Eg can be determined from the minimum energy (hν) of
photons that are absorbed by the semiconductor.
Bandgap energies of selected semiconductors
Semiconductor PbTe Ge
Si GaAs GaP Diamond
E g (eV)
0.31
0.67 1.12
1.42
2.25
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6.0
Slide 1-8
1.3.2 Donor and Acceptor Levels in the Band Model
Conduction Band
Donor Level
Acceptor Level
Ed
Ec
Donor ionization energy
Acceptor ionization energy
Ea
Valence Band
Ev
Ionization energy of selected donors and acceptors in silicon
Donors
Dopant
Sb
Ionization energy, E c –E d or E a –E v (meV) 39
P
44
Acceptors
As
54
B
45
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Al
57
In
160
Slide 1-9
1.4 Semiconductors, Insulators, and Conductors
Ec
E g = 1.1 eV
Ec
E g= 9 eV
empty
Ev
Ev
Si (Semiconductor)
Top of
conduction band
SiO (Insulator)
2
filled
Ec
Conductor
• Totally filled bands and totally empty bands do not allow
current flow. (Just as there is no motion of liquid in a
. totally empty bottle.)
totally filled or
• Metal conduction band is half-filled.
• Semiconductors have lower Eg 's than insulators and can be
doped.
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Slide 1-10
increasing hole energy
increasing electron energy
1.5 Electrons and Holes
electron kinetic energy
Ec
Ev
hole kinetic energy
• Both electrons and holes tend to seek their lowest
energy positions.
• Electrons tend to fall in the energy band diagram.
• Holes float up like bubbles in water.
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Slide 1-11
1.5.1 Effective Mass
In an electric field, , an electron or a hole accelerates.
electrons
holes
Electron and hole effective masses
m n /m 0
m p /m 0
Si
0.26
0.39
Ge
0.12
0.30
GaAs
0.068
0.50
GaP
0.82
0.60
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1.5.2 How to Measure the Effective Mass
Cyclotron Resonance
Technique
B
-
Centripetal force =
Lorentzian force
-
-
-
E
mn v 2
= qvB
r
qBr
v=
mn
v
qB
f cr =
=
2πr 2πmn
Microwave
fcr is the Cyclotron resonance frequency.
It is independent of v and r. Electrons
strongly absorb microwaves of that frequency.
By measuring fcr, mn can be found.
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1.6 Density of States
E
∆Ε
Ec
Dc
Ec
D
Ev
Ev
Dv
Dc ( E ) ≡
number of states in ∆E 
1


3 
∆E ⋅ volume
 eV ⋅ cm 
Dc ( E ) ≡
8πmn 2mn (E − Ec )
h3
Dv ( E ) ≡
8πm p 2m p (Ev − E )
h3
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1.7 Thermal Equilibrium
1.7.1 An Analogy for Thermal Equilibrium
Sand particles
Dish
Vibrating Table
• There is a certain probability for the electrons in the
conduction band to occupy high-energy states under
the agitation of thermal energy (vibrating atoms, etc.)
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Slide 1-15
1.7.2 Fermi Function–The Probability of an Energy State
Being Occupied by an Electron
f (E) =
1
1+ e
Ef is called the Fermi energy or
the Fermi level.
( E − E f ) / kT
Boltzmann approximation:
f (E) ≈ e
E
Ef + 3kT
f (E) ≈ e
(
− E−E f
f (E) ≈ 1 − e
Ef
Ef + kT
Ef
Ef – kT
Ef – 2kT
f (E) ≈ 1 − e
f(E)
0.5
1
) kT
E − E f >> kT
) kT
Ef + 2kT
Ef – 3kT
(
− E−E f
(
)
− E f − E kT
(
)
− E f − E kT
E − E f << − kT
Remember: there is only
one Fermi-level in a system
at equilibrium.
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1.8 Electron and Hole Concentrations
1.8.1 Derivation of n and p from D(E) and f(E)
n=∫
top of conduction band
Ec
8πmn 2mn
n=
h3
∫
∞
Ec
f ( E ) Dc ( E ) dE
E − Ec e
(
− E−E f
) kT
dE
8πmn 2mn −(Ec − E f ) kT ∞
− ( E − Ec ) kT
=
e
E
−
E
e
dE
c
3
∫
0
h
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Slide 1-17
Electron and Hole Concentrations
n = Nce
− ( Ec − E f ) / kT
 2πmn kT 
N c ≡ 2
2

 h

p = Nve
32
Nc is called the effective
density of states (of the
conduction band) .
− ( E f − Ev ) / kT
32
Nv is called the effective
density of states of the
valence band.
 2πm p kT 
N v ≡ 2

2
h


Remember: the closer Ef moves up to N c , the larger n is;
the closer Ef moves down to Nv , the larger p is.
For Si, Nc = 2.8 ×1019 cm-3 and Nv = 1.04×1019 cm-3 .
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Slide 1-18
1.8.2 The Fermi Level and Carrier Concentrations
Where is Ef for n =1017 cm-3? And for p = 1014 cm-3?
Solution: (a) n = N c e
− ( Ec − E f ) / kT
(
)
Ec − E f = kT ln( N c n ) = 0.026 ln 2.8 ×1019 / 1017 = 0.146 eV
(b) For p = 1014cm-3, from Eq.(1.8.8),
(
)
E f − Ev = kT ln( N v p ) = 0.026 ln 1.04 ×1019 / 1014 = 0.31 eV
0.146 eV
Ec
Ec
Ef
Ef
Ev
(a)
0.31 eV
Ev
(b)
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Slide 1-19
1.8.2 The Fermi Level and Carrier Concentrations
Ec
300K
ped
o
d
r
o
n
Ef, Do
Ef , Acce
400K
n = Nce
− ( Ec − E f ) / kT
E f = Ec − kT ln ( N c n )
ptor-dop
ed
400K
300K
Ev
1013
1014
1015
1016
1017
1018
1019
1020
N a or N d (cm-3)
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1.8.3 The np Product and the Intrinsic Carrier Concentration
Multiply n = N c e
− ( Ec − E f ) / kT
np = N c N v e
and
− ( Ec − Ev ) / kT
np = ni
p = Nve
= Nc Nve
− ( E f − Ev ) / kT
− E g / kT
2
ni = N c N v e
− E g / 2 kT
• In an intrinsic (undoped) semiconductor, n = p = ni .
• ni is the intrinsic carrier concentration, ~1010 cm-3 for Si.
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EXAMPLE: Carrier Concentrations
Question: What is the hole concentration in an N-type semiconductor
with 1015 cm-3 of donors?
Solution: n = 1015 cm-3.
2
ni
10 20 cm -3
≈ 15 −3 = 105 cm -3
p=
n 10 cm
After increasing T by 60°C, n remains the same at 1015 cm-3 while p
− E / kT
increases by about a factor of 2300 because ni 2 ∝ e g .
Question: What is n if p = 1017cm-3 in a P-type silicon wafer?
Solution:
2
ni
10 20 cm -3
≈ 17 −3 = 103 cm -3
n=
p 10 cm
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EXAMPLE: Complete ionization of the dopant atoms
Nd = 1017 cm-3. What fraction of the donors are not ionized?
Solution: First assume that all the donors are ionized.
n = N d = 1017 cm −3 ⇒ E f = Ec − 146meV
45meV
146 meV
Ed Ec
Ef
Ev
Probability of non-ionization ≈
1
1+ e
( E d − E f ) / kT
=
1
1+ e
((146 − 45 ) meV ) / 26 meV
= 0.02
Therefore, it is reasonable to assume complete ionization, i.e., n = Nd .
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1.9 General Effects of Doping on n and p
Charge neutrality:
n + Na = p + Nd
np = ni
2

N a − N d  N a − N d 
2
p=
+ 
 + ni 
2
2



2
1/ 2

N d − N a  N d − N a 
2
n=
+ 
 + ni 
2
2



2
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1/ 2
Slide 1-24
1.9 General Effects of Doping on n and p
I. N d − N a >> ni (i.e., N-type)
n = Nd − Na
2
p = ni n
If N d >> N a ,
n = Nd
II. N a − N d >> ni (i.e., P-type)
If N a >> N d ,
p = Na
p = ni
and
2
Nd
p = Na − Nd
n = ni
and
2
p
2
n = ni N a
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EXAMPLE: Dopant Compensation
What are n and p in Si with (a) Nd = 6×1016 cm-3 and Na = 2×1016 cm-3
and (b) additional 6×1016 cm-3 of Na?
(a) n = N d − N a = 4 ×1016 cm −3
2
p = ni / n = 10 20 / 4 × 1016 = 2.5 × 103 cm −3
(b)
Na = 2×1016 + 6×1016 = 8×1016 cm-3 > Nd!
p = N a − N d = 8 × 1016 − 6 × 1016 = 2 × 1016 cm −3
2
n = ni / p = 10 20 / 2 × 1016 = 5 × 103 cm −3
n = 4×1016 cm-3
++++++
......
++++++
Nd = 6×1016 cm-3
. . . .N.a .=.2×10
. . . . cm
16
-3
......
Nd = 6×1016 cm-3
Na = 8×1016 cm-3
- - - - - - - -. . . . . .
p = 2×1016 cm-3
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Infrared Detector Based on Freeze-out
To image the black-body radiation emitted by tumors requires
a photodetector that responds to hν’s around 0.1 eV. In doped
Si operating in the freeze-out mode, conduction electrons are
created when the infrared photons provide the energy to
ionized the donor atoms.
electron
photon
Ec
Ed
Ev
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1.10 Carrier Concentrations at Extremely High
and Low Temperatures
ln n
intrinsic regime
n = Nd
freeze-out regime
1/T
high
temp.
room
temperature
cryogenic
temperature
high T: n = p = ni =
Nc Nv e
− E g / 2 kT
1/ 2
N c N d  −( Ec − Ed ) / 2 kT

low T: n =
 2  e


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1.11 Chapter Summary
Energy band diagram. Acceptor. Donor. mn, mp. Fermi function. Ef.
n = Nce
− ( Ec − E f ) / kT
p = Nve
− ( E f − Ev ) / kT
n = Nd − Na
p = Na − Nd
np = ni
2
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