Physical Properties
of Materials
1. Electrical Properties
2. Optical Properties
3. Magnetic Properties
1
1. electrical properties
2
(1) electrical conduction in metals
classical Model
• metallic bonds make free movement of
valence electrons possible
• outer valence electrons are completely free
to move between positive ion cores
• positive ion cores vibrate with greater
amplitude with increasing temperature
• the motion of electrons are random and
restricted in absence of electric field
no net electron flow
• in presence of electric field, electrons attain
directed drift velocity that is proportional to
the applied field but in the opposite direction
3
Ohm’s law – electric current flow i is directly
proportional to the applied voltage V and
inversely proportional to resistance of the
wire
V
i = ──
R
i: electric current (A)
V: potential difference (V)
R: resistance of wire (Ω)
electric resistivity
A
ρ = R ──
l
l: length of the conductor
A: cross-sectional area of the conductor
electric conductivity
σ = 1/ ρ
4
conductor σ > 107 (Ω·m)-1 metals
insulator
σ ~ 10-14 (Ω·m)-1 polymer
semiconductor Si, Ge
ex. a wire whose diameter is 0.20 cm carries
20 A current, the maximum power is 4 W/m,
calculate the minimum allowable conductivity
of the wire
P = iV =i2R
R=ρl/A
σ = 1/ρ
i2l
(20 A)2 × (1 m)
σ = ── = ────────2= 3.18 × 107 (Ω·m)-1
PA
(4 W) × π(0.001 m)
ex. Cu wire conducts 10A of current with
maximum voltage drop of 0.4 V/m. what is
the minimum diameter?
R = V/i
ρ = RA/ l = AV / i l
σ = i l / AV
5.8 × 107 (Ω·m)-1 = (10 A)(1 m) / A (0.4 V)
(10 A)(1 m)
-7 m2
A = ──────────
=
4.31
×
10
(0.4 V) 5.8 × 107 (Ω·m)-1
d = √4 (4.31 × 10-7)/π = 7.37 × 10-4 m
microscopic Ohm's law
J=E/ρ
or J = σ E
J: current density A/m2
E: electric field V/m
5
drift velocity of electrons
electrons are accelerated when electric field E
is applied and collide with ion cores
after a collision, they accelerate again
electron velocity varies in a saw tooth manner
drift velocity
vd = μ E
μ: electron mobility m2/(V·s)
the electron flow in metal wire depends on the
number of electrons per unit volume, the
electronic charge (-1.60 × 10-19 C), and the
drift velocity of electrons
the current density
J = nevd
direction of current flow
is opposite to that of electrons
6
electrical resistivity
ρtotal = ρT + ρr
ρT thermal component – elastic waves
(phonons) generated due to vibration of
electron core scatter electrons
• resistivity increases with temperature
ρT = ρ0oC (1+ αTT)
ρ0oC: resistivity at 0oC
αT: coefficient of resistivity
T: temperature of the metal
7
ex. calculate the electrical resistivity of Cu at
132oC
ρT = ρ0oC (1+ αTT) = 1.6 × 10-6 (1 + 0.0039 × 132)
= 2.42 × 10-6 Ω·m
ρr residual component – due to structural
imperfections like dislocations, grain
boundaries, impurity atoms
• ρr is almost independent of temperature
• alloying increases resistivity
addition of various elements to Cu
8
(2) energy-band model of electrical conduction
valence electrons are delocalized, interact and
interpenetrate each other
sharply defined energy levels are broadened
into wider regions called energy bands
ex. Na has 1 valence electron (3s1).
there are N Na atoms, there are N distinct
3s1 energy levels in 3s energy band
energy bands
Na is a good conductor since it has half filled
outer 3s band, very little energy is required to
9
produce electron flow
energy-band diagrams for Na, Mg, and Al
in insulators, electrons are tightly bound to
the bonding atoms
large energy gap Eg separates lower filled
valence band and upper empty conduction
band
to be available for conduction, the electron
should jump the energy gap, which may as
much as 6~7 eV (for diamond)
10
(3) intrinsic semiconductors
semiconductors – electrical conductivities are
between good conductors and insulators.
intrinsic semiconductors – pure semiconductors
and conductivities are determined by their
inherent properties ex. Si and Ge
sufficient energy is needed
to excite valence electrons
away from their bonding
position
bonding electron becomes
a free conduction electron
and leaves a hole behind
both electrons and holes are charge carriers
hole is attracted to negative terminal, electron
to positive terminal
11
the motion of a hole in an electric field
energy-band diagram for an intrinsic
semiconductor such as pure Si (Eg = 1.1 eV)
quantitative relationship of electrical conduction
current density J = nqvn* + pqvp*
n: number of conduction electrons per unit
volume
p: number of conduction holes per unit
volume
q: absolute value of electron or hole charge
(1.6 x 10-19C)
12
vn, vp: drift velocities of electrons and holes
J=σE
pqvp*
J
nqvn*
σ = ─ = ─── + ───
E
E
E
vn/E: electron mobility μn
vp/E: hole mobility
μp
σ = nqμn + pqμp
for intrinsic semiconductor n = p = ni
σ = niq(μn + μp)
ex. calculate the number of Si atoms per m3
(d = 2.33 Mg/m3, atomic mass: 28.08)
6.023 × 1023 atoms
1
2.33 × 106 g
(────────)(─────)(──────)
mol
28.08 g/mol
m3
= 5.00 × 1028 atoms/m3
ex. calculate the electrical resistivity of Si at
300 K
σ = niq(μn + μp) = (1.5 × 1016 m-3) × (1.6 × 10-19 C) ×
(0.135 + 0.048 m2/V·s) = 4.39 × 10-4 Ω-1·m-113
ρ = 1/σ = 2.28 × 103 Ω·m
effect of temperature on intrinsic semiconductors
• the conduction band is completely empty at
0oK
• at higher temperatures, valence electrons are
excited to conduction bands
• conductivities increase with increasing
temperature
ni ∞ e-(Eg – Eav)/kT
ni: concentrations of electrons having
energy to enter conduction band.
Eg: energy gap
Eav: average energy across gap
k: Boltzmann's constant
T: temperature
Eav = Eg/2
ni ∞ e-(Eg)/2kT
σ = σ0 e-(Eg)/2kT
σ0: constant depending on the mobility
Eg
ln σ = ln σ0 - ──
2kT
Eg can be determined from the slope of the
plot of ln σ vs. 1/T for intrinsic semiconductor
14
ex. plot of ln σ vs. 1/T for Si
ex. electrical resistivity of pure Si is 2.3 × 103
Ω·m at 27oC, calculate its conductivity at
200oC (Eg = 1.1 eV, k =8.62 × 10-5 eV/K)
Eg
ln σ = ln σ0 - ──
2kT
σ473
1.1 eV
1
1
ln ──
= ─────────
(──
──)
σ300 2 (8.62 × 10-5 eV/K) 300 473
= 7.777
σ473 = e7.777σ300 = 2385(1/2.3 × 103)
= 1.04 Ω-1·m-1
15
(4) extrinsic semiconductors
have impurity atoms (100-1000 ppm) that have
different valance characteristics
(i) n – type extrinsic semiconductors – impurities
donate electrons for conduction
ex. group VA atoms (P, As, Sb) added to Si
only 0.044 eV of energy is required to remove
the excess electron from its parent nucleus
16
(ii) p-type extrinsic semiconductors – group IIIA
atoms when added to silicon, a hole is created
since one of the bonding electrons is missing
when electric field is applied, electrons from
the neighboring bond move to the hole
boron atom gets ionized and hole moves
towards negative terminal
B, Al, Ga provide acceptor level energy and
are hence called acceptor atoms
17
doping – impurity atoms (dopants) are deposited
into silicon by diffusion at 11000C
(iii) effect of doping on carrier concentration
the mass action law at constant temperature
np = ni2
ni : intrinsic concentration of carriers in a
semiconductor and a constant at a given
temperature
n, p : concentration of electrons or holes
since the semiconductor must be electrically
neutral
Na + n = Nd + p
Na and Nd are concentrations of negative
acceptor ions and positive donor ions
in a n-type semiconductor, Na = 0 and n >> p
hence nn ≈ Nd
pn = ni2/nn ≈ ni2/Nd
18
for p-type semiconductor np = ni2/pp ≈ ni2/Na
typical carrier concentrations in intrinsic and
extrinsic semiconductors
Si at 300 K ni = 1.5 × 1016 carriers/m3
for extrinsic Si doped with As at a typical
concentration of 1021 impurity atoms/m3
nn = 1021 electrons/m3
pn = 2.25 × 1011 holes/m3
ex. Si wafer doped with 1021 P atoms/m3
calculate (a) the majority-carrier
concentration, (b) the minority-carrier
concentration, (c) the electrical resistivity
of this n-type semiconductor at room
temperature (assume complete
ionization of the dopant atoms; and
ni = 1.5 × 1016 m-3, μn = 0.135 m2/V·s,
μp = 0.048 m2/V·s)
(a) nn = 1021 electrons/m3
(b) pn = (1.5 × 1016 m-3)2 /1021 m-3
= 2.25 × 1011 holes/m3
(c)
1
1
ρ = —— = ———————————————
(1.6 × 10-19 C)(0.135 m2/V·s)(1021 m-3)
qnnμn
= 0.0463 Ω·m
19
ex. P-doped Si wafer has an electrical
resistivity of 8.33 × 10-5 Ω·m at 27oC.
assume μn = 0.135 m2/V·s, μp = 0.048
m2/V·s, (a) what is its majority-carrier
concentration if complete ionization is
assumed? (b) what is the ratio of P to Si
atoms in this semiconductor?
(a) this is an n-type semiconductor, the majoritycarrier is electron and the resistivity
1
1
ρ = ———
hence nn = ———
qnnμn
ρ q μn
1
nn = ——————————————————
-5
-19
2
(8.33 × 10 Ω·m)(1.6 × 10
C)(0.135 m /V·s)
= 5.56 × 1023 electrons/m3
(b) pure Si contains 5 × 1028 atoms/m3
P/Si ratio = 5.56 × 1023/ 5 × 1028 = 1.11 × 10-5
(iv) effect of total ionized impurity concentration
as the concentration of impurities increase,
mobility of carriers decrease
20
ex. Si is doped with 1.4 × 1016 B atoms/cm3 +
1.0 × 1016 P atoms/cm3 at 27oC. Calculate
(a) the equilibrium electron and hole
concentrations, (b) the mobilities of
electrons and holes, and (c) the electrical
resistivity (ni = 1.5 × 1010 cm-3)
(a) majority-carrier concentration
pp ≈ Na – Nd = 1.4 × 1016 - 1.0 × 1016
= 4 × 1015 holes/cm3
minority-carrier concentration
np = ni2/pp = (1.5 × 1010)2 / 4 × 1015
= 5.6 × 104 electrons/cm3
(b) total impurity concentration = 2.4 × 1016 atoms/cm3
check figure 14.26 find
μn = 900 cm2/V·s
μp = 300 cm2/V·s
(c) electrical resistivity
1
1
ρ = —— = ———————————————
qppμp (1.6 × 10-19 C)(300 cm2/V·s)(4 × 1015 cm-3)
= 5.2 Ω·cm
21
(v) effect of temperature on electrical conductivity
extrinsic range – electrical conductivity
increases with temperature as more and more
impurity atoms are ionized
exhaustion range – temperature at which donor
atom becomes completely ionized for n-type
semiconductor
saturation range – acceptor atoms become
completely ionized for p-type semiconductor
beyond these ranges, temperature does not
change conductivity substantially
further increase in temperature results in
intrinsic conduction becoming dominant 22
and is called intrinsic range
to provide an exhaustion (or saturation) range
at about room temperature is important since
it provides temperature range that have
essentially constant electrical conductivity
ex. (a) As-doped Si
(b) B-doped Si
23
(5) semiconductor devices
(a) pn junction – n-type and p-type silicon
semiconductors are joined together
• formed by doping a single crystal of silicon
first by n-type and then by p-type material
• majority carriers cross over the junction and
recombine but the process stops later as
electrons repelled by negative ions giving
rise to depletion region
• under equilibrium conditions, there exists a
barrier to majority carrier flow
there is no net current flow
24
(b) biased – an external voltage is applied to a pn
junction
reverse biased – n-type is connected to the
positive terminal of battery, electrons of ntype material and holes of p-type material
move away from junction
current resulting from majority carriers does
not flow
leakage current (of the order of μA) flows
due to minority carriers
25
forward biased – n-type is connected to the
negative terminal and p-type to positive
majority carriers are repelled to the junction
and recombine and considerable current flows
26
(c) application of pn junction diode
rectifier diodes – converts alternating voltage
into direct voltage (rectification)
when AC signal is applied
to diode, current flows
only when p-region is
positive and hence half
way rectification is
achieved
signal can be further smoothed out by using
other electronic devices
breakdown diodes (zener diodes)
in reverse bias, small
leakage currents flow
upon breakdown voltage
is reached, reverse
current increases rapidly
electrons gain sufficient energy to knock
more electrons from covalent bonds
these are available for conduction in reverse
27
bias
(d) bipolar junction transistor (BJT)
• BJT consists of two pn junctions occurring
sequentially on a single crystal
• can serve as current amplifier
ex. npn BJT
emitter – emits electrons
base – controls flow of
charge (0.1mm thick)
collector – collects
electrons
emitter base junction is forward biased and
collector base junction is reverse biased
small base current can be used to control
large collector current
28
(6) microelectronics
thousands of transistors
on a “chip” of Si about
5 mm2 and 0.2 mm thick
(a) microelectronic planar bipolar transistors
ex. planar bipolar npn transistor
• relatively large island of n-type silicon is
formed first in a p-type Si substrate
• smaller island of p and n type silicon are
created on larger n type island
29
(b) microelectronic planar field-effect transistor
MOSFET (metal oxide semiconductor field
effect transistor)
NMOS – n-type MOSFET
PMOS – p-type MOSFET
ex. NMOS
overall structure
cross-sectional view
• two islands of n-type silicon are created in a
substrate of p-type silicon
• source – the contact where electrons enter
drain – the contact where electrons leave 30
• gate – third contact, a layer of polysilicon
• no voltage is applied to the gate, only a few
electrons are attracted to the drain
• when a positive voltage is applied to the
gate, electrons will flow between source and
drain if there is positive voltage difference
between them
• MOSFET is capable of current amplification
• MOSFET technology is the basis for most
LSI digital memory circuits
31
(c) fabrication of microelectric integrated circuits
laying out IC network
photolithographic mask
chrome mask emulsion mask
32
(i) photolithography – the process by which
microscopic pattern is transferred from a
photomask to the silicon wafer surface
step 2 wafer is coated with photoresist
step 3 then exposed to UV light
through photomask
step 4 pattern of photoresist is left
where mask is transparent
step 5 wafer is immersed in
hydrofluoric acid
step 6 photoresist pattern is removed
by other chemicals
33
(ii) diffusion technique
• impurity atoms are diffused into Si wafers
at about 1000 ~ 1100oC
• thin silicon dioxide patterns serve as
masks to prevent dopant atoms from
penetrating into silicon
• high concentration of dopant is deposited
near surface in predeposit step
• in drive-in diffusion step, the wafers are
placed in high temperature furnace and
necessary concentration of dopant atoms at
particular depth is attained.
34
(iii) ion implantation technique
• carried out at room temperature
• dopant atoms are ionized and accelerated to
high energies through a high potential
difference of 50-100 KV
• on striking, ions embed in Si
• photoresist or SiO2 is used to mask desired
regions
• damage to Si lattice is caused but can be
healed by annealing
35
(iv) MOS integrated circuit fabrication
step 1 deposition of SiN4 by CVD
boron ions are implanted to suppress
unwanted conduction
SiO2 layer is grown in inactive regions
step 2 SiN4 is removed by etchants
step 3 insulating layer is deposited on wafer
by CVD
step 4 Al is deposited on wafer
step 5 protective layer is deposited on entire
surface
36
(v) NMOS integrated circuits fabrication
37
(vi) complimentary metal oxide semiconductor
(CMOS) devices
circuits containing both types of MOSFETs
– NMOS and PMOS
• made by isolating all NMOS devices with
islands of p-type material.
• used in LSI circuits in microprocessors
and computer memories.
38
(7) compound semiconductors
MX semiconductors are major compound
semiconductors IIIA-VA and IIA-VIA
electrical properties of intrinsic semiconductors
at room temperature
Eg
μn
μp
Group Material eV m2/(V·s) m2/(V·s)
IV A Si
1.10 0.135
0.048
Ge
0.67 0.390
0.190
IIIA - GaP
2.25 0.030
0.015
V A GaAs
1.47 0.720
0.020
GaSb
0.68 0.500
0.100
InP
1.27 0.460
0.010
InAs
0.36 3.300
0.045
InSb
0.17 8.000
0.045
IIA - ZnSe
2.67 0.053
0.002
VI A ZnTe
2.26 0.053
0.090
CdSe
2.59 0.034
0.002
CdTe
1.50 0.070
0.007
lattice
ni
constant carrier/m3
5.4307 1.5 × 1016
5.257 2.4 × 1019
5.450
5.653 1.4 × 1012
6.096
5.869
6.058
6.479 1.35 × 1022
5.669
6.104
5.820
6.481
39
• as molecular mass increases
energy band gap decreases and electron
mobility increases
• increasing ionic bonding character
energy band gap increases and electron
mobility decreases
gallium arsenide (GaAs) is the most important
of all compound semiconductors
advantages: higher mobility
better radiation resistance
ex. (a) calculate the intrinsic electrical
conductivity of GaAs at room temperature
(b) what fraction of the current is carried by
the electrons in GaAs at 27oC?
(a) σ = niq(μn + μp)
= (1.4 × 1012)(1.60 × 1019)(0.072 + 0.002)
= 1.66 × 10-7 (Ω·m)-1
(b) σn
niqμn
0.072
── = ───── = ───── = 0.973
σ
niq(μn + μp)
0.072 + 0.002
40
(8) electrical properties of ceramics
• used for electrical insulators for low- and
high- voltage electric currents
• application in various types of capacitors
• piezoelectrics
(a) basic properties of dielectrics
(i) dielectric constant
parallel-plate capacitor
q = CV or C = q/V
C: capacitance (farad = coulomb/volt)
area dimension A >> distance d
A
C = ε0──
d
permittivity of free space
ε0 = 8.854 × 10-12 F/m
a dielectric fills the space between plates
κε0 A
C = ─── κ : dielectric constant41
d
the energy stored in a capacity of a given
volume at given voltage is increased by the
factor of dielectric constant
(ii) dielectric strength
• a measure of the ability of the material to
hold energy at high voltage
• defined as voltage per unit length
• the maximum electric field that the
dielectric can maintain without electrical
breakdown
• commonly measured in volts/mil (1 mil =
0.001 in) or kilovolts/mm
(iii) dielectric loss factor κ tan δ
dielectric loss angle δ
ex. a parallel-plate capacitor made to store
5.0 × 10-6 C at a potential of 8000 V
separation distance = 0.30 mm, calculate
the area of the plates if dielectric is (a)42a
vacuum (κ = 1) and (b) aluminum (κ = 9)
κε0 A
Cd
(5.0 × 10-6/8000)(3 × 10-4)
C = ───
A = ── = ──────────
d
κε0
k (8.85 × 10-12)
(a) κ = 1 A = 0.021 m2
(b) κ = 9 A = 2.35 × 10-3 m2
(b) ceramic insulator materials
electrical and mechanical properties make
ceramic materials suitable for insulator
applications
(i) electrical porcelain
• 50% clay (Al2O3．2SiO2．2H2O) +
25% silica (SiO2) +
25% feldspar (K2O．Al2O3．6SiO2)
• good green-body plasticity and a wide
firing temperature
• disadvantage: high power-loss factor
(ii) steatite porcelain
• 90% talc (3MgO．4SiO2．H2O) +
10% clay
• good electrical insulator due to low powerloss factor, low moisture absorption, and
good impact strength
43
(iii) fosterite
• chemical formula: Mg2SiO4
• has high resistivity and low electrical loss
with increasing temperature
• has lower-loss dielectrica properties at high
frequencies
(iv) alumina ceramic
• Al2O3 as the crystalline phase bonded with
a glassy matrix
• has relatively high dielectric strength and
low dielectric loss and relatively high
strength
44
(c) ceramic materials for capacitors
• most common type: disk ceramic
capacitors, consist mainly of BaTiO3 and
other additives
dielectric constant
325
2100
6500
formulation
BaTiO3 + CaTiO3 + low% Bi2Sn3O9
BaTiO3 + low% CaZrO3 and Nb2O5
BaTiO3 + low% CaZrO3 or CaTiO3
+ BaZrO3
steps in manufacture:
(i) after firing ceramic disk
(ii) after applying Ag electrodes
(iii) after soldering Pb
(iv) after applying dipped phenolic coating
45
(d) ceramic semiconductors
• thermistor (thermally sensitive resistor) is
used for temperature measurement and
control
• negative temperature coefficient (NTC) –
resistance decreases with increasing
temperature, as the temperature increases,
the termistor becomes more conductive
• most commonly used for NTC thermistors:
oxides of Mn, Ni, Fe, Co, Cu
ex. Fe3O4 mixed with MgCr2O4 to adjust the
resistivity
46
(e) ferroelectric ceramics
ferroelectric – ionic crystalline materials have
unit cells that do not have a center of
symmetry and contain small electric dipole
industrially important material: BaTiO3
• > 120oC regular cubic symmetrical
perovskite structure
• < 120oC (the Curie temperature) Ti4+ and
O2- ions shift slightly in opposite directions
the crystal structure changes from cubic to
slightly tetragonal
formation of ferroelectric domain
47
the piezoelectric (PZT) effect
an effect by which mechanical forces can
produce an electrical response, or electrical
forces a mechanical response
• an excess of positive and negative charges
at two ends in piezoelectric material
• when compressive stresses are applied, the
length of the sample is reduced and the
distance between the unit dipoles are
reduced
the overall dipole moment is reduced and
the voltage difference between the ends is
changed
• an electric field is applied across the ends,
the charge density at each end will be
changed
it causes the change in dimensions in the48
direction of the applied field
examples of industrial applications
• compression accelerator
mechanical → electrical
• ultrasonic cleaning
transducer
electrical → mechanical
49