Transcript Fundamentals of Semiconductors

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.
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Slide 1-1
Silicon Wafers and Crystal Planes
z
z
z
y
(100) x
y
y
x
(011)
(111)
x
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.
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
Si
(a)
 When an electron
breaks loose and becomes(b)a conduction
electron, a hole is also created.
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Slide 1-3
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
8e0
= 13.6 eV
2h 2
Slide 1-4
GaAs, III-V Compound Semiconductors, and Their Dopants
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.
Wich group of elements are candidates for donors? acceptors?
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Slide 1-5
1.3 Energy Band Model
}
Empty upper bands
(conduction band)
2p
2s
(valence band)
}
(a)
Filled lower bands
(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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Slide 1-6
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 .
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
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 (hn) of
photons that are absorbed by the semiconductor.
Bandgap energies of selected semiconductors
Semiconductor
InSb
Ge
Si
GaAs
GaP
ZnSe
Diamond
Eg (eV)
0.18
0.67
1.12
1.42
2.25
2.7
6
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Slide 1-8
1.3.2 Donor and Acceptor in the Band Model
Conduction Band
Ed
Donor Level
Ec
Donor ionization energy
Acceptor ionization energy
Acceptor Level
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
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B
45
Al
57
In
160
Slide 1-9
1.4 Semiconductors, Insulators, and Conductors
Ec
Top of
conduction band
Ec
E g= 9 eV
empty
E g = 1.1 eV
Ev
Ev
Si (Semiconductor)
SiO (Insulator)
filled
Ec
Conductor
2
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.
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Slide 1-11
1.5.1 Effective Mass
The electron wave function is the solution of the three
dimensional Schrodinger wave equation
2

2  V ( r )   
2m0
The solution is of the form exp(  k  r)
k = wave vector = 2π/electron wavelength
For each k, there is a corresponding E.
qe d 2 E F
accelerati on   2

2
 dk
m
2
effective mass  2
d E / dk 2
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Slide 1-12
1.5.1 Effective Mass
In an electric field, E, an electron or a hole accelerates.
electrons
holes
Electron and hole effective masses
mn/m0
mp/m0
Si
Ge
GaAs
InAs
AlAs
0.26
0.39
0.12
0.3
0.068
0.5
0.023
0.3
2
0.3
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Slide 1-13
1.5.2 How to Measure the Effective Mass
B
Cyclotron Resonance Technique
-
-
Centripetal force = Lorentzian force
mn v 2
 qvB
r
qBr
mn
v
qB
f cr 

2r 2mn
v
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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Slide 1-14
1.6 Density of States
E
DE
Ec
Dc
Ec
D
Ev
Ev
Dv
Dc ( E ) 
number of states in DE 
1


3 
DE  volume
eV

cm


8mn 2mn E  Ec 
Dc ( E ) 
h3
Dv ( E ) 
8m p 2m p Ev  E 
h
Derived in Appendix I
3
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Slide 1-15
1.7 Thermal Equilibrium and the Fermi Function
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.
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Slide 1-16
Appendix II. Probability of a State at E being Occupied
•There are g1 states at E1, g2 states at
E2… There are N electrons, which
constantly shift among all the states
but the average electron energy is
fixed at 3kT/2.
•There are many ways to distribute
N among n1, n2, n3….and satisfy the
3kT/2 condition.
•The equilibrium distribution is the distribution that
maximizes the number of combinations of placing n1 in g1
slots, n2 in g2 slots…. :
ni/gi =
EF is a constant determined by the condition
n
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i
N
1.7.2 Fermi Function–The Probability of an Energy State
Being Occupied by an Electron
f (E) 
Ef is called the Fermi energy or
the Fermi level.
1
1 e
( E  E f ) / kT
Boltzmann approximation:
E
Ef + 3kT
f (E)  e

 EE f
f (E)  e
 kT
Ef + 2kT
E
f
f (E)  1  e
Ef + kT
Ef

 EE f

 kT

 E f  E kT
E  E f  kT
E  E f  kT
Ef – kT
Ef – 2kT
Ef – 3kT
f (E)  1  e
f(E)
0.5


 E f  E kT
Remember: there is only
one Fermi-level in a system
at equilibrium.
1
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Slide 1-18
1.8 Electron and Hole Concentrations
1.8.1 Derivation of n and p from D(E) and f(E)
n
top of conductionband
Ec
8mn 2mn
n
h3


Ec
f ( E ) Dc ( E )dE
E  Ec e

 EE f
 kT
dE
8mn 2mn Ec  E f  kT 
  E  Ec  kT

e
E

E
e
d ( E  Ec)
c
3

0
h
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Slide 1-19
Electron and Hole Concentrations
n  Nc e
 ( Ec  E f ) / kT
 2mn kT 
N c  2
2

 h

p  Nv e
32
Nc is called the effective
density of states (of the
conduction band) .
 ( E f  Ev ) / kT
Nv is called the effective
density of states of the
valence band.
 2m 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 Ev , the larger p is.
For Si, Nc = 2.8 ´ 1019 cm-3 and Nv = 1.04 ´ 1019 cm-3 .
32
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Slide 1-20
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  Nc e
 ( Ec  E f ) / kT


Ec  E f  kT ln Nc 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 Nv 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 1-21
1.8.2 The Fermi Level and Carrier Concentrations
Ec
n  Nc e
 ( Ec  E f ) / kT
E f  Ec  kT ln  N c n 
Ev
1013
1014
1015
1016
1017
1018
1019
1020
N a or N d (cm-3)
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Slide 1-22
1.8.3 The np Product and the Intrinsic Carrier Concentration
Multiply n  Nc e
 ( Ec  E f ) / kT
and
p  Nv e
np  Nc Nv e( Ec Ev ) / kT  Nc Nv e
np  ni
 ( E f  Ev ) / kT
 Eg / 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.
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Slide 1-23
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
p
 15 3  105 cm -3
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
n
 17 3  103 cm -3
p 10 cm
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Slide 1-24
Ionized concentration for donors and acceptors
From neutrality equation:
n  N A,Shallow  N A,Deep  p  N D ,Shallow  N D ,Deep
Shallow dopant: close to EC or EV , Deep dopant: far away from EC or EV
We usually use NA- to represent NA,Shallow- , and use ND+ to represent ND,Shallow+
ND
N 
1  g D exp[( E F  ED ) / kT ]

D
NA
N 
1  g A exp[( E A  EF ) / kT ]

A
gD = 2: e- spin up and spin down
gA = 4: e- spin up and spin down
Two valence bands (hh and lh)
At very low temperature Nd+ is 0 and Na- is 0
At very low temperature Ef is between Ec and Ed : Ef = (Ec+Ed)/2
1.9 General Theory of n and p
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  Nd  1017 cm3  E f  Ec 146meV
45meV
146 meV
Ed Ec
Ef
Ev
Probability of not 
being ionized
1
1 ( E  E ) / kT
1 e d f
2

1
1
1  e((14645) meV ) / 26meV
2
 0.04
Therefore, it is reasonable to assume complete ionization, i.e., n = Nd .
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
(what happens when Ef is close to Ed?)
Slide 1-26
1.9 General Theory of 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
1/ 2
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Slide 1-27
1.9 General Theory of on n and p
I. N d  N a  ni (i.e., N-type)
n  Nd  Na
p  ni n
2
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 Nd
2
and
p  Na  Nd
n  ni
and
2
p
n  ni Na
2
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Slide 1-28
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
n = 41016 cm-3
......
++++++
Nd = 61016 cm-3
p  ni / n  1020 / 4 1016  2.5 103 cm3
2
16
-3
. . . .N.a.=. .210
. . . cm
(b) Na = 21016 + 61016 = 81016 cm-3 > Nd
p  N a  N d  8 10  6 10  2 10 cm
16
16
16
3
......
++++++
Nd = 61016 cm-3
n  ni / p  1020 / 2 1016  5 103 cm3
2
Na = 81016 cm-3
-------- ......
p = 21016 cm-3
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Slide 1-29
1.10 Carrier Concentrations at Extremely High
and Low Temperatures
ln n
intrinsic regime
n = Nd
freeze-out regime
1/T
High
temp
Room temp
Cryogenic temp
high T: n  p  ni  N c N v e
 E g / 2 kT
1/ 2
N c N d  ( Ec  Ed ) / 2 kT

low T: n  
e

 2 
Modern Semiconductor Devices for Integrated Circuits (C. Hu)
Slide 1-30
Ec
Ed
Efn
E
m*p > m*n
Ei
Ea
Ev
Efp
Temperature
Infrared Detector Based on Freeze-out
•To image the black-body radiation emitted by tumors
requires a photodetector that responds to hn’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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Slide 1-32
1.11 Chapter Summary
Energy band diagram. Acceptor. Donor. mn, mp.
Fermi function. Ef .
n  Nc e
 ( Ec  E f ) / kT
p  Nv e
 ( E f  Ev ) / kT
n  Nd  Na
p  Na  Nd
np  ni
2
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Slide 1-33