Transcript e - Help-A-Bull

Electrical Conduction in Solids
What is Electric Current?
What is current density?
i
dq q

dt t
J
dq
q

Adt At
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 !!
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?
•
E
A
 x responsible
What type of charges are
for current flow in
conductors (metals)?
v
x
dx
•
What type of charges are responsible
for current flow in
J
x
semiconductors?
•
What is mobility μ ?
Voltage Gradient and E-Field
Ex
u
x
+
Vibrating Cu ions
(a)
V
(b)
Quick Derivation of Current Density
vdx  1 vx1  vx2  vx3  ...  vxN 
N
E
x
n  N/V
x
Δx  vdx  Δt
v
A
dx
Δq  e  n  A  Δx
Δq  e  n  A  vdx  Δt
Jx (t)  e  n  vdx (t)
J
x
Important Current Density Relationships!
•
Jx (t)  e  n  vdx (t)
e
vdx  μE μ 
me •
Jx (t)  e  n  μ  E
Jx (t)  σ  E
σ  e n μ
•
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’
or σ  e  n  μn  e  p  μp
J = σE, Ohm’s Law
•
OHM’S LAW (V=IR!)
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
eC
μd 
me T
me  T
1
1
ρ 
 2
σ e  n  μd e nC
•
IMPORTANT to remember the dependence of conductivity …
mobility … etc. on T! … i.e. directly/inversely proportional
Alloys … and Matthiessen’s Rule
•
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
–
Strained 1
region by1
impurity1exerts a
 F = d(PE) /dx
is: scattering force
μd
μL
μI
where subscript L is for Lattice and I is for Impurity
I
•
Therefore the effective mobility is lower (than that of a pure
metal)
•
1
1

 ρT  ρI
The “net” resistivity is: ρ 
e  n  μL e  n  μI
T
which is known as Matthiessen’s Rule
Resistivity versus temperature for pure iron and 4%
C steel.
Nichrome ??
Temperature Dependence of Resistivity
•
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:
ρ  ρT  ρR  AT  B
•
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
1
αO 
ρo
•
 δρ 
 δT 
 To
which leads to
ρ  ρo 1  αo (T - To )
Valid over a narrow temperature range… not a bad
approximation BUT be cautious !
Temperature Dependence of Resistivity
100
2000
rT
Inconel-825
NiCr Heating Wire
1000
10
Iron
Tungsten
r T
Resistivity(nWm)
Resistivity (nWm)
Monel-400
Tin
100 Platinum
1
0.1
r  T5
0.01
Copper
Nickel
0.001
Silver
r  rR
0.0001
10
100
r (nW m)
3.5
rT
3
2.5
2
1.5
r  T5
1
0.5 r = rR
0
0 20 40 60 80 100
T (K)
1000
Temperature (K)
10000
0.00001
1
10
100
Temperature (K)
1000
10000
Solid Solutions
•
What is a solid solution ?
ρ  ρT  ρR
•
where ρR includes effect of impurities defects etc.
600
The second
term in the above equation is NOT
temperature Cdependent
u -N i A llo y s
•
•
R e s is tiv ity ( n W m )
500
Therefore
4 0 0 when forming a solid solution of two metals …
i.e. one metal is the “host” (and the other the “impurity”,
3 0 0 addition of the “impurity” metal will cause an
then the
increase in the resistivity and make the total resistivity
200
less and
less temperature dependent !
100
NOTE: at large “impurity” amounts the material will
0
become an
ALLOY! And we need to consider “alloy
0
20
40
60
80
100
effects”
100% C u
a t.% N i
100% N i
Nordheim’s Rule of Solid Solutions
•
A simple way to determine the effect on resistivity
component – the one due to “alloying impurities”
ρI  CX(1- X)
where X is the fraction of the solute and C is a constant - the Nordheim coefficien t
•
… and combining Matthiessen’s and Nordheim’s rules then
we get
ρ  ρMATRIX  CX(1- X)
Solid Solutions vs. Mixtures
•
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
L
–
First case: series mixture
ρeff  χa ρa  χb ρb
–
Second case: parallel
mixture
A
Jx
a b
Jy
σ eff  χa σa  χb σb
L
A
Mixtures – Continuous Phase w Dispersed 2nd
Phase
•
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:
ρeff
1 

 1  χd 
2 
 ρc 
1  χd 

1  χd 
ρeff  ρc
1  2χd 
for ρd  10 ρc
Continuous phase
Dispersed phase y
A
Jx
for ρd  0.1 ρc
σd  σC
σ  σC
Or the general expression :
χ
σ  2σ C
σ d  2σ C
L
Temperature
Rule ???
TA
(a)
Liquid, L
TE a
T1
100%A
X1
a+L
TB
b +L
b
Two phase region
ab
X (% B)
X2
One phase
region: b
only
100%B
Resistivity
(b)
Mixture Rule
Nordheim's Rule
rA
0 X1
Fig 2.15
Composition X (% B)
rB
X 2 100%B
(a) The phase diagram for a binary, eutectic forming alloy. (b) The
resistivity vs composition for the binary alloy.
Thermal Conduction
HOT
COLD
HEAT
Electron Gas
+
Vibrating Cu ions
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 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 …
dq
δV
dQ
δT
J  σE ... or I 
 -Aσ
... similarly Q'
 -Ak
dt
δx
dt
δx
L 1L
1L
Rρ 
... θ 
A σA
kA
Thermal Conductivity
Q = T/
T
Hot
Cold
T
Q
Q
Q
A

L
(b)
(a)
Conduction of heat through a component in (a) can be modeled as a
thermal resistance  shown in (b) where Q = T/.
•
•
•
σ is the electrical conductivity and k the thermal
conductivity
Since electrons are responsible for both … the two are
related i.e.
k
 CWFL
σT
… where CWFL is the Weidemann-Franz-Lorenz coefficient
Thermal Conductivity
•
k
 CWFL
σT
σ is related to temperature
(in a certain range!) as 1/T
… the above relationship
suggests that k is
temperature
INDEPENDENT
450
•
Since there are no free
electrons in non-metals …
how does the heat transfer
take place ?
Thermal conductivity, k (WK-1 m-1)
•
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
Ag
400
Ag-3Cu
Cu
Ag-20Cu
k
s = TCWFL
300
Au
Al
200
W
Be
Mg
Mo
Brass (Cu-30Zn)
Ni
Bronze (95Cu-5Sn)
Steel (1080)
Pd-40Ag
Hg
100
0
0
10
20
30
40 6 50
60
-1
-1
Electrical conductivity, s, 10 W m
70
Electrical Conductivity
Insulators
Semiconductors
Conductors
Many ceramics
Superconductors
Alumina
Diamond Inorganic Glasses
Mica
Polypropylene
PVDF Soda silica glass
Borosilicate Pure SnO2
PET
SiO2
10-18
Metals
Degenerately
Doped Si
Alloys
Intrinsic Si
Amorphous Intrinsic GaAs
As2Se3
10-15 10-12
10-9
10-6
10-3
100
Conductivity(Wm)-1
Te Graphite NiCr Ag
103
106
109
1012
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!
In order ....
i.e. x  y  z
xˆ  yˆ  zˆ
yˆ  zˆ  xˆ
zˆ  xˆ  yˆ
In reverse order
i.e. z  y  x
yˆ  xˆ   zˆ
zˆ  yˆ   xˆ
xˆ  zˆ   yˆ
The Hall Effect
Example 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

  
F  q(E  v  B)
 The y-component of the force is
Fy  q(Ey  v x  Bz )
 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 ydirection 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:
VAB  Ey w
JX  qpνX  νX 
JX
qp
Fy  q(Ey  v x  Bz )
Fy  0  Ey  v xBz 
1
RH 

qp
Jx
Bz  RH JxBz
qp
( Ix wt)Bz
JxBz
IxBz
p 


qEy
q(VAB /w)
qtVAB
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 r of the sample can also
be calculated by measuring the resistance
R of the bar.
ρL
R 
wt
σ  qpμp  μp 
σ

qp
1
ρ

q 1

 qRH 

RH
ρ
Hall Effect
Right hand rule
• Index finger -> Current / charge velocity
• Thumb -> Lorentz force
• Rest of the fingers -> Magnetic filed
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.
B
Bz
Jy = 0
+
+
x
Ey
eE y
Jx
Ex
v hx
eE y
VAB>0
>0
+
+
A
Jx
z
vex
evhxB z
+
y
evexB z
+
A
Bz
V
VAB>0
<0
Application of Hall Effect
V12<0,
Charge?
Charge (-)ve
V12 ?
Charge (-)ve, V12 <0
B?
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:
k
= CWFL = 2.44x10 -8 éëWWK -2 ùû
sT
(300)(2.44x10 -8 )(K )(WWK -2 )
-1
-1
k=
=
73.2Wm
K
-7
(10 )(W - m)
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
r (@25°C) = ro (@0°C)[1+ a o (@0°C)(25 - 0)]
r (@- 40°C) = ro (@0°C)[1+ a o (@0°C)(-40 - 0)]
r (@25°C)
[1+ a o (@0°C)(25 - 0)]
=
r (@- 40°C) [1+ a o (@0°C)(-40 - 0)]
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 #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.
(c) Estimate the mean free time between collisions for the conduction
electrons in aluminum at 25 °C, and hence estimate their drift mobility.
s  nq n 
n 
1
r
q
me
q nq 2
 (nq )

r
me
me
1
....we need n ... electron density
 kg 
6.022 x10 23 (atoms ) 1[mol ] 2.7[ g ] 103 g 
 6.022 x10 28 atoms / m 3
3
3
mol
0.027[kg ] [cm ]  m 
10 2 cm 
3electrons
.
atom
3 x6.022 x10 28 electrons / m 3  18.066 x10 28 electrons / m 3
1


 nq 2 r 
 18.066 x10 29 (1 / 602 x10 19 ) 2 (2.72 x10 8 ) 
 
 

m
(9.109 x10 31 )


 e 
1