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Electrical Conduction in Solids
What is Electric Current?
What is current density?
What is Drift? … motion of charges under the influence of an Electric Field
What is drift velocity ?… average velocity of charges under an applied E-field
NOTE: many of these quantities are VECTORS !!
tq
dtdq
i
tAq
Adtdq
J
Electrical Conduction in Solids
• Which way do negative charges move with respect to the applied E-field?
• Which way do positive charges move with respect to the applied E-field?
• What type of charges are responsible for current flow in conductors (metals)?
• What type of charges are responsible for current flow in semiconductors?
• What is mobility μ ?
A
Jx
Dx
vd x
Ex
Quick Derivation of Current Density
ΔtvΔx dxN/Vn
xNx3x2x1dx v...vvvN1v
ΔxAneΔq ΔtvAneΔq dx
(t)vne(t)J dxx
A
Jx
Dx
vd x
Ex
Important Current Density Relationships!
(t)vne(t)J dxx
μEvdx
Eμne(t)J x
Eσ(t)J x
• Mobility … expresses how “easy” electrons can move (drift) under the influence of an electric field; it is directly related to relaxation (or mean scattering) time
• Conductivity … the ability of a solid to conduct electricity (cousin of mobility!)
• ‘Mean Free Time’
pn μpeμneσor μneσ
em
e μ
Temperature Dependence of Resistivity
• Electrons are scattered by the vibrating metal ions
• Therefore the time between collisions (scattering time) can be affected by the vibration frequency … which implies that the mobility and therefore the conductivity and resistivity are also affected by the vibrating atoms (ions in metals)
• What should be the relationship between frequency of vibration and time between collisions ?– The higher the freq. the shorter the time because
the metal ions are vibrating faster (i.e. higher KE) and therefore colliding more often with moving electrons
• Shorter scattering times imply lower mobility … lower conductivity and therefore higher resistivity!
• (the math/derivation is in the book! Let’s look at the expressions…)
Temperature Dependence of Mobility and Conductivity
TmCe
μe
d
nCeTm
μne1
σ1
ρ 2e
d
• IMPORTANT to remember the dependence of conductivity … mobility … etc. on T! … i.e. directly/inversely proportional
Alloys … and Matthiessen’s Rule
ILd μ1
μ1
μ1
• Previous discussion was based on the ASSUMPTION that the material was a pure metal & perfect crystal! … i.e. no impurities no point defects etc.
• When impurities are present then one needs to consider two scattering mechanisms: (1) host atoms (2) impurity atoms
• The result is:
– where subscript L is for Lattice and I is for Impurity• Therefore the effective mobility is lower (than that of a
pure metal)
• The “net” resistivity is:
which is known as Matthiessen’s Rule
ITIL
ρρμne
1μne
1ρ
tI
Strained region by impurity exerts ascattering force F = d(PE) /dx
tT
Temperature Dependence of Resistivity
BATρρρ RT
• All non-intrinsic effects (impurities, crystal defects etc.) on the reisstivity can be “summed” up in a “residual” reisistivity term and the resistivity can be re-written as:
• NOTE that the thermal vibrations term is temperature dependent!! but the extrinsic component is not.
• TEMPERATURE COEFFICIENT OF RESISTIVITY α is the fractional change in resistivity per unit temperature at the reference temperature TO
• Valid over a narrow temperature range… not a bad approximation BUT be cautious !
)T-(Tα1ρρ to leads which δTδρ
ρ1
α ooooTo
O
r µ T
Tungsten
Silver
Copper
Iron
Nickel
Platinum
NiCr Heating Wire
Tin
Monel-400
Inconel-825
10
100
1000
2000
100 1000 10000Temperature (K)
Res
i sti v
it y (
n W m
)
0.00001
0.0001
0.001
0.01
0.1
1
10
100
1 10 100 1000 10000Temperature (K)
00.5
11.5
2
2.53
3.5
0 20 40 60 80 100T(K)
r µ T
r µ T5
r = rRr = rR
r µ T5
r µ T
r (nW m)
Res is
ti vit y
(nW
m)
Temperature Dependence of Resistivity
Solid Solutions
• What is a solid solution ?
• The second term in the above equation is NOT temperature dependent
• Therefore when forming a solid solution of two metals … i.e. one metal is the “host” (and the other the “impurity”, then the addition of the “impurity” metal will cause an increase in the resistivity and make the total resistivity less and less temperature dependent !
• NOTE: at large “impurity” amounts the material will become an ALLOY! And we need to consider “alloy effects”
etc. defects impurities ofeffect includes ρ where ρρρ RRT
1 0 0 % C u a t . % N i
C u - N i A l l o y s
0
1 0 0
2 0 0
3 0 0
4 0 0
5 0 0
6 0 0
0 2 0 4 0 6 0 8 0 1 0 0
1 0 0 % N i
Re
sis
tiv
ity
(nW
m)
Nordheim’s Rule of Solid Solutions
• A simple way to determine the effect on resistivity component – the one due to “alloying impurities”
• … and combining Matthiessen’s and Nordheim’s rules then we get
tcoefficien Nordheim the -constant a isC and solute the of fraction the is Xwhere
X)-CX(1ρ I
X)-CX(1ρρ MATRIX
Solid Solutions vs. Mixtures
bbaaeff ρχρχρ
• Solid solutions are homogenous and mixing takes place at the atomic level … Nordheim’s rule applies.
• What happens when the mixture is not homogenous? i.e. 2 phases
bbaaeff σχσχσ
Jx
a b
A
L
L
A
Jy
– First case: series mixture
– Second case: parallel mixture
Mixtures – Continuous Phase w Dispersed 2nd Phase
cdd
d
ceff ρ 10 ρfor χ1
χ21
1ρρ
cd
d
dceff ρ 0.1 ρfor
2χ1χ1
ρρ
• If we have a mixture where the “host” material is continuous (c) and the added material is dispersed (d) therein … then the following two empirical expressions apply:
L
A
Dispersed phaseContinuous phase
y
Jx
Cd
Cd
C
C
2σσσσ
χ2σσσσ
:expression general theOr
Fig 2.15
Rule ???
X 1X 2
1 0 0 % B1 0 0 % A X (% B )
L iqu id , L+L +L
T AT B
T 1
Tw o ph a se reg io n
O n e p h a s ere g io n : o n ly
T E
(a )
Tem
per
atur
e
BA
C o m p o s i t io n X (% B )0 X 1 X 2 1 0 0 % B
N o rd h e im 's R u le
M ix tu re R u le
(b )
Resi
stiv
ity
(a) The phase diagram for a binary, eutectic forming alloy. (b) Theresistivity vs composition for the binary alloy.
V ib ra tin g C u + io n sE lec tro n G as
H O T C O L DH E A T
Thermal conduction in a metal involves transferring energy from the hot region to the cold region by conduction electrons. More energetic electrons (shown with longer velocity vectors) from the hotter regions arrive at cooler regions and collide there with lattice vibrations and transfer their energy. Lengths of arrowed lines on atoms represent the magnitudes of atomic vibrations.
Thermal Conduction
Thermal Conductivity
• Good Electrical Conductors (like metals) are also good Thermal Conductors …
• Why ??
• … electrons are responsible for the electrical and thermal conductivity …
• … they pick up the thermal energy from vibrating atoms and transfer to atoms elsewhere …
• It is easy to think of Thermal Conductivity the same way we think of Electrical Conductivity …
AL
k1
θ ... AL
σ1
AL
ρR
δxδT
-AkdtdQ
Q' similarly ... δxδV
-Aσdtdq
Ior ... σEJ
Thermal Conductivity
• σ is the electrical conductivity and k the thermal conductivity
• Since electrons are responsible for both … the two are related i.e.
• … where CWFL is the Weidemann-Franz-Lorenz coefficient
WFLCσTk
Conduction of heat through a component in (a) can be modeled as athermal resistance shown in (b) where Q= T/.
H o t
L
T
(a)
Q C o ld
AQ
T
Q
(b)
Q = T /
Thermal Conductivity
• σ is related to temperature (in a certain range!) as 1/T
… the above relationship suggests that k is temperature
INDEPENDENT
• Since there are no free electrons in non-metals … how does the heat transfer take place ?
• VIBRATIONS … vibrations will increase at the hot end of a material and the vibrational energy will be transferred along the material depending on the type of bonding
WFLCσTk
0
100
200
300
400
450
0 10 20 30 40 50 60 70Electrical conductivity, s, 106 W-1 m-1
Al
Ag
Au
Cu
Brass (Cu-30Zn)
Be
Pd-40Ag
Bronze (95Cu-5Sn)
Ag-3Cu
Ag-20Cu
WMo
Hg
Mg
Steel (1080)
Ni
ks = TCWFL
The
r mal
con
duct
ivity
, k (
W K
-1 m
-1)
Electrical Conductivity
10610310010-310-610-910-1210-1510-18 109
Semiconductors Conductors
1012
Conductivity (Wm)-1
AgGraphite NiCrTeIntrinsic Si
DegeneratelyDoped Si
Insulators
Diamond
SiO2
Superconductors
PETPVDF
AmorphousAs2Se3
Mica
Alumina
Borosilicate Pure SnO2
Inorganic Glasses
Alloys
Intrinsic GaAs
Soda silica glass
Many ceramics
MetalsPolypropylene
Charges under the influence !
• What happens when a charge finds itself under the influence of an electric field ?
E
Charges under the influence !
• What happens when a charge finds itself under the influence of a magnetic field ?
B
Right vs. Left-Handed Oriented xyz System…. And the Cross product!
yxz
xzy
zyx
ˆˆˆ
ˆˆˆ
ˆˆˆ
z y x i.e.
.... order In
yzx
xyz
zxy
ˆˆˆ
ˆˆˆ
ˆˆˆ
xyz i.e.
order reverse In
The Hall EffectExample with p-type semiconductor;
i.e. holes are the majority charge carriers;
apply voltage in x direction i.e. current Ix.
apply a B-field in the z-direction. Total Force on the charge carriers
due to E and B fields is)BvEq(F
)Bvq(EF zxyy The y-component of the force is
As the holes flow in the x-direction they experience a force in the y-direction due to the B-field.
Holes will accumulate in the -y end of the bar setting up an electric field, i.e. a voltage VAB in the y direction. The net force in the y-direction becomes zero when the two components of the force i.e. due to the electric and due to the magnetic field are equal.
The Hall Effect The “setting up” of the E-field in the
y-direction is known as the Hall Effect.
The voltage VAB is known as the Hall Voltage.
This experiment is used to measure the mobility of the charge carriers as explained below:
wEV yAB
qpJνqpνJ X
XXX
zxHzx
zxyy BJRBqpJ
BvE0F
)Bvq(EF zxyy
qp1
RH AB
zx
AB
zx
y
zx
qtVBI
/w)q(Vwt)BI(
qEBJ
p
The Hall Effect
Note: the current and magnetic fields are known quantities since they are externally applied; the hall voltage can be measured.
The resistivity of the sample can also be calculated by measuring the resistance R of the bar.
wtρL
R
ρR
qR1q
ρ1
qpσ
μqpμσ H
H
pp
Hall Effect
Key Points• Direction of electric field is also the direction of conventional
current.• Hall electric field is set by the movement charge due to Lorentz
force in the presence of a magnetic field.
Right hand rule• Index finger -> Current / charge velocity• Thumb -> Lorentz force• Rest of the fingers -> Magnetic filed
Example #1
The resistivity of an alloy of Cu is 10-7 Ω-m; the dimensions of a rectangular bar of this material are w=4 mm, t=10 mm, and l=50 mm; the thermal conductivity of this alloy is:
Example #2• Points A, B, and C
– What phases are present– What is the composition of each phase– What is the fraction of each phase present– Max solubility of C in Fe to maintain the Ferrite
phase
Example #3
The resistivity of aluminum at 25 °C has been measured to be 2.72x10-8 Ω-m. The thermal coefficient of resistivity of aluminum at 0 °C is 4.29 x 10-3 K-1. Aluminum has a valency of 3, a density of 2.70 g cm-3, and an atomic mass of 27.
(a) what is the resistivity @ -40 °C
Example #4
The resistivity of aluminum at 25 °C has been measured to be 2.72x10-8 Ω-m. The thermal coefficient of resistivity of aluminum at 0 °C is 4.29 x 10-3 K-1. Aluminum has a valency of 3, a density of 2.70 g cm-3, and an atomic mass of 27.
(b) what is the thermal coefficient of resistivity @ -40 °C
Example #4The resistivity of aluminum at 25 °C has been measured to be 2.72x10-8 Ω-m. The thermal coefficient of resistivity of aluminum at 0 °C is 4.29 x 10-3 K-1. Aluminum has a valency of 3, a density of 2.70 g cm-3, and an atomic mass of 27.
(c) Estimate the mean free time between collisions for the conduction electrons in aluminum at 25 °C, and hence estimate their drift mobility.
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