Conductors: Conductivity & Resistance
EE 141 Lecture Notes
Topic 16
Professor K. E. Oughstun
School of Engineering
College of Engineering & Mathematical Sciences
University of Vermont
2014
Motivation
Conductivity
The action of an applied electric field E(r) in a conducting medium is
to accelerate both electrons & holes through the Coulombic force
F(r) = qE(r). Due to collisions with the molecules of the material,
this motion is reduced to an average drift velocity which is
proportional to the applied field strength E.
The average electron drift velocity is then given by
ūe = −µe E
(1)
where µe is the electron mobility [in (m/s)/(V /m) = m2 /Vs], and
the average hole drift velocity is given by
ūh = +µh E
where µh is the hole mobility.
(2)
Conductivity
With the number densities defined as
Ne ≡ # free electrons per unit volume,
Nh ≡ # free holes per unit volume,
the electron & hole charge densities are given by
̺e = −Ne qe
& ̺h = +Nh qe
(3)
respectively, where qe ≃ 1.602 × 10−19 C is the magnitude of the
electronic charge. The conduction current density (in A/m2 ) is then
given by
Jc (r) = Je (r) + Jh (r) = ̺e ue + ̺h uh
=
− ̺e µe + ̺h µh E(r).
(4)
Conductivity
The static electric conductivity σ0 is then defined through the point
form of Ohm’s law
Jc (r) = σ0 E(r)
(5)
where
σ0 = −̺e µe + ̺h µh
= Ne µe + Nh µh qe
(6)
in units ✵/m where ✵ ≡ A/(Vm). The conductivity is also expressed
in Siemens/m or S/m after Ernst Werner von Siemens (1816–1892).
For a good conductor, Nh µh ≪ Ne µe and σ0 ≈ −̺e µe = Ne µe qe .
Two special limiting cases:
Perfect Dielectric: σ0 = 0 =⇒ Jc = 0 regardless of E.
Perfect Conductor σ0 = ∞ =⇒ E = σJc0 = 0 regardless of Jc .
Historical Perspective
Ernst Werner von Siemens
(1816 - 1892)
Comic Relief
Resistance & Conductance
Consider a length ℓ of a conductor with static conductivity σ0 and
uniform cross-section surface S.
Application of a voltage V = V12 = V1 − V2 between cross-sectional
surfaces 1 and 2 of the conductor results in an electrostatic field E
and a conduction current density Jc = σ0 E, where
Z S2
V = V1 − V2 = −
E · d ~ℓ.
(7)
S1
Resistance & Conductance
The total current I flowing through the conductor is given by
ZZ
ZZ
I =
Jc · d s =
σ0 E · d s
S
(8)
S
in Amperes (A).
The resistance of the conductor is then defined as
R
− C E · d ~ℓ
V
= RR
R≡
(Ω)
I
σ E · ds
S 0
(9)
and the conductance is
RR
σ0 E · d s
1
G ≡ = SR
R
− C E · d ~ℓ
(✵ = S)
(10)
For example, for a uniform wire of cross-sectional area A and length
ℓ, V = E ℓ and I = σ0 EA so that R = VI = σ0ℓA .
Resistance & Capacitance
From Eqs. (15.8) and (16.9) for a pair of conductors embedded in a
material with uniform static dielectric permittivity ǫs and conductivity
σ0 , the capacitance and resistance are given by
H
E · ds
,
(11)
C = −ǫs R SS11
E · d ~ℓ
S2
R S1
1 S2 E · d ~ℓ
R =− H
,
σ0 S1 E · d s
(12)
from which it then follows that
RC =
ǫs
σ0
(13)
Conductor-Conductor Boundary Conditions
The boundary conditions on the conduction current density Jc across
the interface S separating two conductors with static conductivities
σ1 and σ2 are directly obtained from the boundary conditions on the
electric field intensity through the static form of Ohm’s law Jc = σ0 E
as follows:
Tangential Boundary Conditions
E1t = E2t =⇒
J1t
J2t
=
σ1
σ2
(14)
Normal Boundary Conditions
D1n − D2n = ̺s =⇒
ǫ2
ǫ1
J1n − J2n = ̺s
σ1
σ2
However, by the conservation of charge, J1n = J2n , so that
ǫ1
ǫ2
J1n
= ̺s .
(15)
−
σ1 σ2
Example: Conductance of a Coaxial Cable
Consider a length ℓ of a coaxial cable with inner radius a and outer
radius b filled with an insulating material with static conductivity σ0 .
σ0
l
Let I denote the total current flowing from the inner conductor to the
outer conductor through the insulating material. The area through
which the current is flowing is then given by A = 2πr ℓ with a < r < b
so that the current density in the insulating material is given by
I
Jc (r ) = 1̂r
, a < r < b.
2πr ℓ
Example: Conductance of a Coaxial Cable
By Ohm’s law Jc (r) = σ0 E(r), the electrostatic field between the
inner & outer conductors is given by
E(r ) = 1̂r
I
,
2πσ0 r ℓ
a < r < b.
In a resistor, the current flows from higher to lower electric potential.
Because Jc is in the +1r -direction, the inner conductor must be at a
higher potential than the outer conductor. Accordingly, the potential
difference between the two conductors is given by
Z b
Z b
dr
I
I
=
ln (b/a).
V =−
E(r ) · 1̂r dr = −
2πσ0 ℓ a r
2πσ0 ℓ
a
The conductance is then given by G = I /V and the conductance per
unit length is
G
2πσ0
G′ ≡
(✵/m)
(16)
=
ℓ
ln (b/a)
Joule’s Law
Consider a differential volume element ∆V containing the electron
charge qe = ̺e ∆V and hole charge qh = ̺h ∆V. Under the action of
an applied electrostatic field E, these charges experience the forces
Fe = qe E = ̺e E∆V,
Fh = qh E = ̺h E∆V,
respectively. The work expended in moving qe the differential
distance ∆~ℓe and qh the differential distance ∆~ℓh is then given by
∆W = Fe · ∆~ℓe + Fh · ∆~ℓh
with associated power expended ∆P ≡ ∆W /∆t given by
∆P = Fe ·
∆~ℓe
∆~ℓh
+ Fh ·
.
∆t
∆t
Joule’s Law
With ue ≡ ∆~ℓe /dt denoting the electron drift velocity and
uh ≡ ∆~ℓh /dt denoting the hole drift velocity, this expression for the
expended power becomes
∆P = Fe · ue + Fh · uh
= ̺e ue + ̺h uh · E ∆V
= Jc · E ∆V,
where Jc denotes the conduction current density [see Eq. (4)]. The
total dissipated power in a volume V is then given by Joule’s Law
ZZZ
P=
Jc (r) · E(r) d 3r (W )
(17)
V
With the point form of Ohm’s Law Jc (r) = σ0 (r)E(r), Joule’s law
becomes
ZZZ
(18)
P=
σ0 (r)|E(r)|2 d 3 r (W )
V
Joule’s Law
For a uniform wire with cross-sectional area A and length ℓ and with
constant static conductivity σ0 , Joule’s law yields
ZZ
Z
2
P =
σ0 |E|d r |E| d ℓ
ℓ
A
= (σ0 |E|A) (|E|ℓ)
| {z } | {z }
I
V
yielding the familiar result
P = IV = I 2 R.
(19)