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EEE 3394
Electronic Materials
Chris Ferekides
Fall 2014
Week 8
Metal to Metal Contacts
Electrons
Electrons
-
e transfer due to difference in energy
net e-transfer leaves behind a
positive charge, while making the
other metal more negative
the result is charge separation … Efield and V! known as the Contact
Potential
= 4.20 eV
Vacuum
Fermi level
Mo
Electrons
Electrons
F (Pt) F (Mo) = 1.16 eV = eDV
Vacuum
5.36 eV
-
Fermi level
F (Mo)
= 5.36 eV
Fermi level
Vacuum
Pt
Vacuum
Fermi level
4.20 eV
Fermi level
Vacuum
Mo
F(Pt)
Vacuum
= 4.20 eV
F(Pt)
= 5.36 eV
Pt
F (Mo)
What happens when we bring two metals together ??
Seebeck Effect
Net diffusion of electrons from the “hot” to the “cold” region of a metal …
Temperature, DT
-
-
Fermi function at a higher T
“spreads” more toward higher
energies
High energy e’s move to fill in lower
energy states ...
This process leaves behind a net
positive charge … therefore E-field
… Voltage!
The Seebeck Coefficient or
Thermoelectric Power is:
dV
S
dT
Cold
Hot
Conductor
E
EFC
EFH
0
Hot
+
+
+
+
+
E
f(E)
1
Voltage, DV
0
1
Cold
f(E)
Seebeck Effect
The electron movement is NOT always from Hot to Cold because the diffusion
process depends on several parameters including the mean free path …
which also depends on T!
The Seebeck coefficient can be –ve or +ve depending on which mechanism
dominates …
dV
S
dT
Temperature, DT
Conductor
E
T2
ΔV   SdT
0
Hot
+
+
+
+
+
E
EFC
EFH
T1
π 2k 2 T
S
x
3eEFO
Cold
Hot
f(E)
1
Voltage, DV
0
1
Cold
f(E)
Thermocouple
What is a thermocouple ??
Can we measure the voltage generated by the temperature difference ??
Metal
Hot
Cold
0 °C
100 °C
0
Metal
mV
Metal
So how can we utilize the Thermoelectric Power ??
Metal
type A
Hot
Cold
0 °C
100 °C
Metal
type B
0
mV
Metal
type B
Thermocouple
The difference in Seebeck coefficients will result in a net Voltage across the two
wires …
Metal
type A
Hot
Cold
0 °C
100 °C
0
Metal
type B
mV
T2
Metal
type B
VAB   (SA - SB )dT
T1
Thermionic Emission
• Heated filament in a vacuum will emit electrons !
I
Plate or Anode
Vacuum
I
Saturation current
E
Cathode
V
Filament
(a )
E
Probability
(b)
(a) Thermionic electron emission in a vacuum tube.
(b) Current-voltage characteristics of a vacuum diode.
Free Electron
T3
T2
T1
EF + 
Electron concentration
per unit energy
T3
T2
T1
EF
f(E)
0
0
1.0
0
n(E) = g(E)f(E)
Thermionic Emission - Rectifier
• Why does the current saturate ?
• What happens when the voltage is reversed ?
I
Plate or Anode
Vacuum
I
Saturation current
Cathode
V
Filament
(a )
(b)
(a) Thermionic electron emission in a vacuum tube.
(b) Current-voltage characteristics of a vacuum diode.
Thermionic Emission Equation
• Where Bo is a constant called the RichardsonDushman constant
I
 
J  BoT exp  

 kT 
Plate or Anode
2
Vacuum
I
Saturation current
Cathode
V
Filament
(a )
(b)
(a) Thermionic electron emission in a vacuum tube.
(b) Current-voltage characteristics of a vacuum diode.
Silicon (Si) Bonding Model
Silicon:
- 14 electrons;
- 10 are tightly bound to nucleus; (core
electrons);
- 4 weakly bound; valence electrons - participate
in chemical reactions.
(Ge similar to Si with 28 core electrons)
Bonding Model:
- Rem: four nearest neighbors;
- covalent bonding - sharing of electrons between neighboring atoms;
- each atom contributes four “shared” electrons;
- each atom accepts four shared electrons from its neighbors;
Note: at room temperature some bonds are “broken” i.e. free electrons.
Semiconductor Terminology
CHARGE CARRIERS:
 In conductors:
electrons
 In semiconductors?:
 At 0ºK no broken bonds i.e. no free electrons.
 At 0ºK no electrons in conduction band, valence band completely filled- energy band
model
 electrons in valence band can still move but net momentum (quantized) is zero;
therefore no NET current flow.
 At room temperature “some” bonds are broken and there exist electrons in the
conduction band: conduction electrons.
 Breaking of a bond also creates a “void”; vacancy in the valence band: a HOLE is
also a charge carrier. (see figs)
TERMINOLOGY:
 Dopants: certain impurity atoms added to semiconductors in order to control the number
of holes/electrons.
 Intrinsic semiconductor: pure (undoped) semiconductor.
 Extrinsic semiconductor: doped semiconductor; properties determined by added
impurities.
What Happens @ 0 K ?
S i c r y s ta l in 2 -D
E le c tr o n e n e r g y
E c+ c
C o n d u c t io n B a n d
(C B )
E m p t y o f e le c t r o n s a t 0 K .
Ec
yB
Bandgap = E g
Ev
V a le n c e B a n d ( V B )
F u ll o f e le c t r o n s a t 0 K .
0
(b )
(c)
Intrinsic means … n=p
g(E) (E Ec)1/2
Ec+c
E
E
E
[1-f(E)]
CB
Area = nE (E )dE  n
For
electrons
Ec
Ec
nE(E)
Ev
pE(E)
EF
EF
Ev
For holes
Area = p
VB
0
g(E)
f(E)
nE(E) or pE(E)
Ex
Hole energy
(a)
Electron Energy
V(x)
Electrostatic PE(x)
Ex
CB
CB
(b)
VB
VB
x
x=0
x=L
When an electric field is applied, electrons in the CB and holes in the
VB can drift and contribute to the conductivity. (a) A simplified
illustration of drift in Ex. (b) Applied field bends the energy bands
since the electrostatic PE of the electron is -eV(x) and V(x) decreases in
the direction of Ex whereas PE increases.
From Principles of Electronic Materials and Devices, Third Edition, S.O. Kasap (© McGraw-Hill, 2005)
Bonding Model
REM – We are considering Si for most examples
a donor atom has 5 valence electrons;
when a donor atom replaces a Si atom 4 of its 5 valence electrons will participate in the
formation of the four covalent bonds;
the 5th electron is weakly bound to the donor atom;
what does weakly bound means?
It takes about 1 eV to break a Si-Si bond (i.e. to free an electron in pure Si.)
it takes about 0.1 eV or less to remove the extra electron from the donor atom.
Most donor and acceptor binding energies are about 1/20 Eg (Si).
(Same can be described for an acceptor atom with one less electron!).
Note:
when a donor atom gives up its extra electron the net charge of the donor is +1. This
charged donor is FIXED.
n-type Doping
Electron Energy
CB
Ec
As+
e-
~0.03 eV
Ed
As+
As+
As+
As+
Ev
As atom sites every 106 Si atoms
Arsenic doped Si crystal. The four valence electrons of As allow it to
bond just like Si but the fifth electron is left orbiting the As site. The
energy required to release to free fifth-electron into the CB is very
small.
From Principles of Electronic Materials and
Devices, Third Edition, S.O. Kasap (©
McGraw-Hill, 2005)
x Distance
into
crystal
p-type Doping
h+
Electron energy
B-
B-
B atom sites every 106 Si atoms
x Distance
Ec
into crystal
Free
B-
B-
B-
B-
Ea
(a)
(b)
h+
~ 0.05eV
Ev
Boron doped Si crystal. B has only three valence electrons. When it
substitutes for a Si atom one of its bonds has an electron missing and
therefore a hole as shown in (a). The hole orbits around the B- site by
the tunneling of electrons from neighboring bonds as shown in (b).
Eventually, thermally vibrating Si atoms provides enough energy to
free the hole from the B- site into the VB as shown.
From Principles of Electronic Materials and
Devices, Third Edition, S.O. Kasap (©
McGraw-Hill, 2005)
VB
Light Effects
Electron energy
Ec+c
CB
Ec
Free e-
h > Eg
h
Eg
Ev
hole
e-
Hole
VB
0
(a)
(b)
(a) A photon with an energy greater than Eg can excite an electron from the VB to the
CB. (b) When a photon breaks a Si-Si bond, a free electron and a hole in the Si-Si bond is
created.
From Principles of Electronic Materials and
Devices, Third Edition, S.O. Kasap (©
McGraw-Hill, 2005)
Semiconductors – Bonding/Energy Band Model
the binding energy for a donor
electron is about 1/20 Eg.
Donor
if energy equal to binding energy is
supplied to the crystal the extra
electron will leave the donor and end
up in the conduction band.
Note:
at OºK no thermal energy therefore
no donor electrons can be excited to the
conduction band.
The creation of a free electron from a
donor atom does not result in the
creation of a hole (rem. Intrinsic
semiconductor).
Acceptor
Semiconductors n & p
EQUILIBRIUM CARRIER CONCETRATIONS
REM:
 gC(E)d(E) represents the number of available states (cm-3) in the energy
interval E+dE.
 f(E)
is the probability a state is occupied by an electron; (1-f(E)
holes);
 gC(E)f(E)d(E)
gives the number of electrons (cm-3) in the interval E+dE;
Therefore the TOTAL number of electrons n (and holes p) in the conduction band
(and in the valence band) can be obtained by integrating the relationships:
no 
ETO P
g
C
(E)f(E)dE
EC
gc (E) 
po 
EV
g
V
(E)[1  f(E)]dE
EBotom
m
*
n
2m (E  Ec )
*
n
2 3
 
gv (E) 
mp* 2mp* (Ev  E)
 2 3