Transcript 1.1 Silicon Crystal Structure

Integrated Circuit Devices
Professor Ali Javey
Summer 2009
Semiconductor Fundamentals
Evolution of Devices
Yesterday’s Transistor (1947)
Today’s Transistor (2006)
Why “Semiconductors”?
• Conductors – e.g Metals
• Insulators – e.g. Sand (SiO2)
• Semiconductors
– conductivity between conductors and insulators
– Generally crystalline in structure
• In recent years, non-crystalline semiconductors have
become commercially very important
Polycrystalline amorphous crystalline
What are semiconductors
Elements: Si, Ge, C
Binary: GaAs, InSb, SiC, CdSe, etc.
Ternary+: AlGaAs, InGaAs, etc.
Electrons and Holes in Semiconductors
Å Structure
Silicon Crystal
• Unit cell of silicon crystal is
cubic.
• Each Si atom has 4 nearest
neighbors.
5.43 Å
Silicon Wafers and Crystal Planes
z
z
z
y
(100) x
y
y
x
(011)
(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.
Bond Model of Electrons and Holes (Intrinsic Si)
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
 When an electron breaks loose and becomes a conduction
electron, a hole is also created.
Dopants in Silicon
Si
Si
Si
Si
Si
Si
Si
As
Si
Si
B
Si
Si
Si
Si
Si
Si
Si
N-type Si
P-type Si
 As (Arsenic), a Group V element, introduces conduction electrons and creates
N-type silicon, and is called a donor.
 B (Boron), a Group III element, introduces holes and creates P-type silicon,
and is called an acceptor.
 Donors and acceptors are known as dopants.
Types of charges in semiconductors
Hole
Electron
Ionized
Donor
Ionized
Acceptor
Mobile Charge Carriers
they contribute to current flow
with electric field is applied.
Immobile Charges
they DO NOT
contribute to current flow
with electric field is applied.
However, they affect the
local electric field
EE143 – Vivek Subramanian
Slide 1-9
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?
From Atoms to Crystals
conduction band
Energy
p
Pauli exclusion
principle
s
valence band
isolated atoms
lattice spacing
Decreasing atomic separation
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 .
Energy Band Diagram
Conduction band
Ec
Eg Band gap
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 .
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
Material
PbTe Ge
Si
GaAs GaP
E g (eV)
0.31
0.67
1.12
1.42
2.25
Diamond
6.0
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 E 's than insulators and can be
g
doped.
Donor and Acceptor Levels 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
Hydrogen:
E
ion =
P
44
m0 q 4
8e0
2h2
Acceptors
As
54
B
45
= 13.6 eV
Al
57
In
160
Dopants and Free Carriers
Donors
n-type
Acceptors
p-type
Dopant ionization
energy ~50meV (very low).
General Effects of Doping on n and p
_
Charge neutrality:
+
+
n Na p Nd = 0
_
Na
Nd
: number of ionized acceptors /cm3
+
: number of ionized donors /cm3
Assuming total ionization of acceptors and donors:
n + Na - p - Nd
Na
: number of acceptors /cm3
Nd
: number of donors /cm3
=0
Density of States
E
gc
DE
Ec
Ec
g(E)
Ev
Ev
gv
gc ( E ) 
number of statesin DE 
1


3
DE  volume
 eV  cm 
mn* 2mn* E - Ec 
gc ( E) 
 2 h3
gv ( E ) 
m*p 2m*p Ev - E 
 2 h3
Thermal Equilibrium
Thermal Equilibrium
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.)
At E=EF, f(E)=1/2
Question
• If f(E) is the probability of a state being occupied
by an electron, what is the probability of a state
being occupied by a hole?
Nc is called the effective density of states
(of the conduction band) .
Nv is called the effective density of states
of the valence band.
Intrinsic Semiconductor
• Extremely pure semiconductor sample containing an insignificant
amount of impurity atoms.
n = p = ni
Ef lies in the middle of the band gap
Material
Ge
Si
GaAs
Eg (eV)
0.67
1.12
1.42
ni (1/cm3)
2 x 1013
1 x 1010
2 x 106
Remember: the closer Ef moves up to E 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 .
Ec
Ev
Ef
Ec
Ev
Ef
Example: The Fermi Level and Carrier Concentrations
Where is Ef for n =1017 cm-3? Solution:
n  Nce
-( Ec - E f ) / kT


Ec - E f  kT lnNc n  0.026ln 2.81019 / 1017  0.146 eV
0.146 eV
E
E
E
c
f
v
The np Product and the Intrinsic Carrier Concentration
Multiply n  Nce
-( Ec - E f ) / kT
and
p  Nv e
np  Nc Nve-( Ec -Ev ) / kT  Nc Nve
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 .
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
1020 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
1020 cm-3
n
 17 -3  103 cm-3
p 10 cm
General Effects of Doping 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
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  Nd - Na  4 1016 cm-3
p  ni / n  1020 / 4 1016  2.5 103 cm-3
2
(b) Na = 21016 + 61016 = 81016 cm-3 > Nd!
p  Na - Nd  8 1016 - 6 1016  2 1016 cm-3
n  ni / p  1020 / 2 1016  5 103 cm-3
2
n = 41016 cm-3
++++++
......
++++++
......
Nd = 61016 cm-3
Nd = 61016 cm-3
Na = 21016 cm-3
Na = 81016 cm-3
...........
- - - - - - - -. . . . . .
p = 21016 cm-3
Chapter Summary
Energy band diagram. Acceptor. Donor. mn, mp. Fermi function. Ef.
n  Nce
-( Ec - E f ) / kT
p  Nv e
-( E f - Ev ) / kT
n  Nd - Na
p  Na - Nd
np  ni
2
Thermal Motion
• Zig-zag motion is due to collisions or scattering
with imperfections in the crystal.
• Net thermal velocity is zero.
• Mean time between collisions (mean free time) is m ~ 0.1ps
Thermal Energy and Thermal Velocity
3
2
1
2
electron or hole kinetic energy  kT  mv th2
vth 
3kT

meff
3  1.38 10-23 JK -1  300K
0.26  9.1 10-31 kg
 2.3 105 m/s  2.3 107 cm/s
~8.3 X 105 km/hr
Drift
Electron and Hole Mobilities
• Drift is the motion caused by an electric field.
Effective Mass
In an electric field, E, an electron or a hole accelerates.
electrons
Remember :
F=ma=-qE
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
Remember :
F=ma=mV/t = -qE
Electron and Hole Mobilities
mp v  qE  mp
qE  mp
v
mp
v  pE
q mp
p 
mp
v  - nE
q
 n  mn
mn
• p is the hole mobility and n is the electron mobility
Electron and Hole Mobilities
v = E ;  has the dimensions of v/E  cm/s
cm 2 


.
 V/cm V  s 
Electron and hole mobilities of selected
semiconductors
 n (cm2/V∙s)
 p (cm2/V∙s)
Si
1400
Ge
3900
GaAs
8500
InAs
30000
470
1900
400
500
Based on the above table alone, which semiconductor and which carriers
(electrons or holes) are attractive for applications in high-speed devices?
Drift Velocity, Mean Free Time, Mean Free Path
EXAMPLE: Given p = 470 cm2/V·s, what is the hole drift velocity at
E = 103 V/cm? What is mp and what is the distance traveled between
collisions (called the mean free path)? Hint: When in doubt, use the
MKS system of units.
Solution: n = pE = 470 cm2/V·s  103 V/cm = 4.7 105 cm/s
mp = pmp/q =470 cm2/V ·s  0.39  9.110-31 kg/1.610-19 C
= 0.047 m2/V ·s  2.210-12 kg/C = 110-13s = 0.1 ps
mean free path = mhnth ~ 1 10-13 s  2.2107 cm/s
= 2.210-6 cm = 220 Å = 22 nm
This is smaller than the typical dimensions of devices, but getting close.
Mechanisms of Carrier Scattering
There are two main causes of carrier scattering:
1. Phonon Scattering
2. Impurity (Dopant) Ion Scattering
Phonon scattering mobility decreases when temperature rises:
 phonon   phonon 
1
1
-3 / 2


T
phonon density carrier therm al velocity T  T 1/ 2
 = q/m
T
vth  T1/2
Impurity (Dopant)-Ion Scattering or Coulombic Scattering
Boron
_ Ion
-
Electron
-
+
Electron
Arsenic
Ion
There is less change in the direction of travel if the electron zips by
the ion at a higher speed.
impurity
3
th
3/ 2
v
T


Na + Nd
Na + Nd
Total Mobility
1600
1
1400

Electrons
1
1000

2
-1
-1
Mobility (cm V s )
1200


1
 phonon
1
 phonon
+
1
 impurity
+
1
 impurity
800
600
400
Holes
200
0
1E14
1E15
1E16
1E17
1E18
1E19
Na +Concenration
Nd (cm-3)
Total Impurity
(atoms cm )
-3
1E20
Temperature Effect on Mobility
1015
Question:
What Nd will make
dn/dT = 0 at room
temperature?
Drift Current and Conductivity
E
Jp
unit
area
+
+
n
Current density
EXAMPLE:
Jp = qpv
A/cm2 or C/cm2·sec
If p = 1015cm-3 and v = 104 cm/s, then
Jp= 1.610-19C  1015cm-3  104cm/s
= A/cm2
1.6 C/s  cm2  1.6
Drift Current and Conductivity
E
Jp
unit
area
+
+
n
+
E
-
Remember:
• Holes travel in the direction of the Electric field
• Electrons travel in the direction opposite to that of the E-field
Drift Current and Conductivity
Jp,drift = qpv = qppE
Jn,drift = –qnv = qnnE
Jdrift = Jn,drift + Jp,drift = (qnn+qpp)E =  E


conductivity of a semiconductor is  = qnn + qpp
resistivity of a semiconductor is  = 1/
DOPANT DENSITY cm-3
Relationship between Resistivity and Dopant Density
P-type
N-type
RESISTIVITY (cm)
 = 1/
Ec and Ev vary in the opposite
direction from the voltage. That
is, Ec and Ev are higher where
the voltage is lower.
V
x
Ec
E
Ec-Ereference = -qV
Ev
x
Variation in Ec with position is called band bending
Diffusion Current
Particles diffuse from a higher-concentration location
to a lower-concentration location.
Diffusion Current
dn
J n ,diffusion  qDn
dx
J p ,diffusion
dp
 -qD p
dx
D is called the diffusion constant. Signs explained:
p
n
x
x
Total Current – Review of Four Current Components
JTOTAL = Jn + Jp
dn
Jn = Jn,drift + Jn,diff = qnnE + qDn
dx
dp
Jp = Jp,drift + Jp,diff = qppE – qD p
dx
JTOTAL = Jn,drift + Jn,diff + Jp,drift + Jp,diff
Chapter Summary
vp   pE
vn  -  nE
J p ,drift qp pE
Jn ,drift qn nE
dn
J n ,diffusion  qDn
dx
J p ,diffusion
dp
 -qD p
dx
kT
Dn 
n
q
kT
Dp 
p
q