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 ¯ e = −µe E u

(1)

where µe is the electron mobility [in (m/s)/(V /m) = m2 /Vs], and the average hole drift velocity is given by ¯ h = +µh E u 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 ˆ r I , a < r < b. Jc (r ) = 1 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 ˆr E(r ) = 1

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 ˆ r dr = − = ln (b/a). V =− E(r ) · 1 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)