Lesson 10Steady Electric Currents

楊尚達 Shang-Da YangInstitute of Photonics TechnologiesDepartment of Electrical EngineeringNational Tsing Hua University, Taiwan

Outline

Current densityCurrent lawsBoundary conditionsEvaluation of resistance

Sec. 10-1Current Density

1. Definition2. Convection currents3. Conduction currents

Definition-1

( )3mC qN=ρVolume charge density:

Within a time interval , the amount of charge enclosed by a volume will pass through the elemental surface

Definition-2

tΔ

( )[ ] satuv

n Δ⋅Δ=Δ

vv

( ) ( ) ( )sautsatuvQ nn Δ⋅Δ=Δ⋅Δ=Δ=Δ vvvv ρρρ

QΔ vΔ

Definition-3

( ),sautQ nΔ⋅Δ=Δ vvρ ( )sautQI nΔ⋅=

ΔΔ

=⇒ vvρ

( ),saJI nΔ⋅= vv

uJ vvρ=⇒

Current I can be regarded as the “flux” of a volume current density , i.e.,J

v

(A/m2)

Convection currents

Convection currents result from motion of charged particles (e.g. electrons, ions) in vacuum (e.g. cathode ray tube), involving with mass transport but without collision.

q

Example 10-1: Vacuum tube diode (1)

1.Electrons are boiled away from the incandescent cathode with zero initial velocity (space-charge limited condition).

2.Anode has positive potential, attracting the free electrons.

?)( 0 =VJv

0V

Example 10-1: Vacuum tube diode (2)

By planar symmetry, ⇒

1. Potential: , BCs:

2. Volume charge density:

3. Charge velocity:

)( yV 0)( ,0)0( VdVV ==

)( yρ

)(yuau yvv =

Example 10-1: Vacuum tube diode (3)

By space-charge limited condition, ⇒

1. Zero initial charge velocity:

2. Zero boundary E-field:

0)0( =u

,0)0( =yE 0)0( =′⇒ V

0,0

=

=

Euv

v

Example 10-1: Vacuum tube diode (4)

(1) In steady state, JaJ yvv

−=

,uJ vvQ ρ= constant,)()( =−=⇒ yuyJ ρ

)()(

yuJy −=ρ

(2) By energy conservation, )(21)( 2 ymuyeV =

,)(2)(m

yeVyu =⇒)(2

)(yeV

mJy −=ρ

Example 10-1: Vacuum tube diode (5)

(3) Since free charges exist inside the tube:

ερ

−=∇ V2 …Poisson’s equation

)(2)(

002

2

yeVmJy

dyVd

εερ

=−=⇒…Nonlinear

ODE

dydV2

(4) Shortcut to get :

Multiply both sides by

)( 0VJ

)(222

02

2

yeVm

dydVJ

dyVd

dydV

ε=

Example 10-1: Vacuum tube diode (6)

Integrate with respect to y:

)(222

02

2

yeVm

dydVJ

dyVd

dydV

ε=

( )∫∫ −=′′⋅′ dVVe

mJdyyVyV 2/1

0 22)()(2ε

ce

ymVJdydV

+=⎟⎟⎠

⎞⎜⎜⎝

⎛2

)(4

0

2

ε

Example 10-1: Vacuum tube diode (7)

By BCs of the potential: 0)0( ,0)0( =′= VV4/1

0 2)(2 ,0 ⎥⎦⎤

⎢⎣⎡==⇒

eymVJ

dydVc

ε

dye

mJdVV4/1

0

4/1

22 ⎟

⎠⎞

⎜⎝⎛=⇒ −

ε

⎟⎠⎞⎜

⎝⎛⎟

⎠⎞

⎜⎝⎛=⇒ ∫∫ − dV

dye

mJdVV

0

4/1

0

0

4/1

220

ε

Example 10-1: Vacuum tube diode (8)

Child-Langmuir law:

2

2/300 2

94

dV

meJ ε

=

Differ from Ohm’s law ( ) of conduction current.

Differ from forward-biased semiconductor diode ( ).

VI ∝

VeI α∝

Example 10-1: Vacuum tube diode (9)

Triode (transistor)

Conduction (drift) currents-1

Conduction currents result from drift motion of free electrons due to applied E-field in conductors, involving with frequent collisions with immobile ions.

Ev

kTvm thn 23

21 2 =

E.g. At T =300K,vth ~ 105 m/sec

-eEffective mass of e-, depending on the lattice.

mean scattering time between collisions

Conduction (drift) currents-2

In steady state, two forces are balanced with each other (Drude model):

1. Electric force:

2. Frictional force:

Eev

−

τdnum v

−

EEmeu e

nd

vvv μτ−=−=⇒

average momentum change in each collision

…drift (average) velocity

For typical fields, ud ~ mm/sec

Conduction (drift) currents-3

Electron mobility describes

how easy an external E-field can influence the motion of conduction electrons.

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=secV

m 2

ne m

eτμ

E.g.

Silicon:

1-D carbon nanotube:

Copper:

,1035.1 1−×=eμ 2108.4 −×=hμ

10=eμ3103.2 −×=eμ

Conduction (drift) currents-4

,EuJ eede

vvvμρρ −==⇒

Eu ed

vv μ−=

EJvv

σ=⇒

eeμρσ −=

(A/m2) …conduction current density

(S/m) …electric conductivity

hhee μρμρσ +−=

For semiconductors:

,uJ vvρ=By ,

holes

<0

Sec. 10-2Current Laws

1. Ohm’s law2. Electromotive force & KVL3. Equation of continuity & KCL4. Joule’s law

Ohm’s law-1

Consider a piece of homogeneous, imperfectconductor (σ < ∞) with arbitrary shape:

, 2112 ∫ ⋅=−=L

ldEVVVvv

∫∫ ⋅=⋅=AA

sdEσsdJI vvvv

=⋅

⋅==

∫∫

A

L

sdE

ldE

IVR

12

vv

vv

σ⇒ constant

EJvv

σ=

microscopic law:

Since the spatial distribution of is independent of V12

Ev

Ohm’s law-2

Consider a piece of homogeneous, imperfect conductor (σ < ∞) with uniform cross-section:

SL

ESEL

sdE

ldER

A

L

σσσ==

⋅

⋅=

∫∫

vv

vv

⇒

Electromotive force (emf)-1

If there is only conservative electric field (created by charges):

⇒

Ev

,EJvv

σ= ( ) 0

=⋅=⋅ ∫∫ CCldJldEvvvv

σ

⇒ no steady loop current!

⇒ Non-conservative field is required to drive charges in a closed loop

Electromotive force (emf)-2

Consider an open-circuited battery:

chemical reaction

impressed field

∫ ⋅≡1

2 ldEi

vvV

emf: the strength of non-conservative force

Electromotive force (emf)-3

By :,inside iEEvv

−= 0

=⋅∫CldEvv

0 1

2

2

1 outside

1

2 inside

2

1 outside =⋅−⋅=⋅+⋅=⋅ ∫∫∫∫∫ ldEldEldEldEldE iC

vvvvvvvvvvV

21

2

1 outside VVldE −=⋅= ∫vv

V

Kirchhoff’s voltage law (KVL)

If the two terminals are connected by a uniform conducting wire of resistance:

SLR

σ=

,outsideEJvv

σ=

IRS

ILldJ

ldE

C==⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛=

⋅=

∫

∫

σσ

2

1 outside

v

v

vvV

,SIJ =

∑∑ =k

kkj

j IRVMulti-sources: …KVL

Equation of continuity-1

S

VtΔ

),( trvρ ),( ttr Δ+vρ

∫=V

dvtrQ

),(vρ ∫ Δ+=′V

dvttrQ

),(vρ

Principle of conservation of charge, ⇒ a net current I flowing out of V must be due to the decrease of the enclosed charge:

Equation of continuity-2

tQQI

t Δ′−

=→Δ 0

lim

∫∫∫ ∂∂−

=Δ

Δ+−=⋅

VVSdv

ttrdv

tttrtrsdJ

),(),(),( vvvvv ρρρ

( ) ∫∫ ∂∂

−=⋅∇VV

dvt

dvJ

ρv

tJ

∂∂

−=⋅∇ρv

⇒

decrease of change within V during Δt

Divergence theorem

For steady currents,

Kirchhoff’s current law (KCL)

tJ

∂∂

−=⋅∇ρv

0=∂∂

tρ

0=⋅∇ Jv …no current

source/sink

( ) 0

=⋅∇=⋅= ∫∫∑ VSj

j dvJsdJIvvv

S

VI1

I2

I3

Example 10-2: Dynamics of charge density (1)

Ev

?),( =trvρ

0=ρ0=E

v

0=t

tJ

∂∂

−=⋅∇ρv

EJvv

σ=( ) ( )

tEEJ

∂∂

−=⋅∇=⋅∇=⋅∇ρσσ

vvv

constnat=σconstnat=ε

Example 10-2: Dynamics of charge density (2)

ρ=⋅∇ Dr

EDvv

ε=( ) ( ) ρεε =⋅∇=⋅∇=⋅∇ EED

vvr

ερ

=⋅∇ Ev

tE

∂∂

−=⋅∇ρ

σ1v

0=+∂∂ ρ

εσρ

t

,)( 0τρρ tet −=⇒

σετ = …Lifetime of free

charge in a conductor

Joule’s law-1

With , collisions among free electrons (with drift velocity ) & immobile atoms transfer energy from to thermal vibration.

Ev

Ev

Evd

uv

Joule’s law-2

Work done by to move charge Q for is:Ev

Ev

dlv

ΔDifferential drift displacement within Δt:

,dlEQwvv

Δ⋅=Δdlv

Δ

corresponding to a power dissipation:

dtuEQ

twp vv

⋅=ΔΔ

=→Δ 0

lim

Q

Joule’s law-3

Power dissipated in a differential volume dv is:

)(rE vv

dvrdQ )(vρ=

⇒ (W/m3) is power density,

( ) ( ) ( )dvJEdvuEuEdvp dd

vvvvvv⋅=⋅=⋅=Δ ρρ

JEvv

⋅

( )∫ ⋅=V

dvJEP

vv⇒ … Joule’s law

Joule’s law-4

Power dissipated in a homogeneous conductor of uniform cross section:

( ) ( ) IVlIdESdlJEdvJEPLLV 12 =⋅=⋅=⋅= ∫∫∫

vvvvvv

Sec. 10-3Boundary Conditions

1. BCs of current density2. Examples: Two lossy media

connected in series

1. Divergence & curl relations of :(i) For steady currents:(ii)

Derivation of BCs-1

Jv

0=⋅∇ Jv

EJvv

σ=

0=×∇ Ev ( ) 0=×∇=×∇ σJE

vv

2. Integral forms:(i)

(ii)

0

=⋅∫SsdJ vv

0

=⋅∫CldJ v

v

σ

3. Apply the integral relations to a(i) differential pillbox:

(ii) differential loop:

Derivation of BCs-2

0

=⋅∫SsdJ vv

0

=⋅∫CldJ v

v

σ

nn JJ 21 =

2

2

1

1

σσtt JJ

=

Example 10-3: BCs between two lossy media (1)

Represent the surface charge density ρs by either D1n or D2n

Example 10-3: BCs between two lossy media (2)

( ) sn DDa ρ=−⋅ 212

vvv

EDvv

ε=

nn

nns

EEDD

2211

21

εερ

−=−=

EJvv

σ=

nn JJ 21 =

⇒

nn EE 2211 σσ =⇒nn EE 2

1

21 σ

σ=

nn EE 12

12 σ

σ=

Example 10-3: BCs between two lossy media (3)

⇒ ;1 221

12ns D⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

εσεσρ

ρs=0 only if:

(i) , rarely happens;

(ii) …both media are lossless2

2

1

1

εσ

εσ

=

021 == σσ

,2

22

1

21222

1

212211 ⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−=−⎟⎟

⎠

⎞⎜⎜⎝

⎛=−=

εε

σσεε

σσεεερ n

nnnnsDEEEE

⇒ ns D112

211 ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

εσεσρ

similarly,

Example 10-4: Electromagnetostatic field (1)

Find , , and surface charge densities ρsJv

Ev

By planar symmetry and ,nn JJ 21 = JaJ yvv

−=⇒

By ,EJvv

σ= ,iyi

i EaJE vv

v−==⇒

σ ii

JEσ

=

y

Example 10-4: Electromagnetostatic field (2)

,2

2

1

122110 ⎟⎟

⎠

⎞⎜⎜⎝

⎛+=+=

σσddJdEdEVBy

( ) ( )2211

0

σσ ddVJ+

=⇒

,2112

02

11 dd

VJEσσ

σσ +

==2112

01

22 dd

VJEσσ

σσ +

==

Example 10-4: Electromagnetostatic field (3)

2112

021

1111

ddV

EDs

σσσε

ερ

+=

==

2112

012

2222

ddV

EDs

σσσε

ερ

+−=

−=−=

ns D112

211 ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

εσεσρBy

( )2112

0211211

12

21 )(1dd

VEsi σσσεσεε

εσεσρ

+−

=−⎟⎟⎠

⎞⎜⎜⎝

⎛−=

Comments

but,21 ss ρρ ≠

021 =++ siss ρρρ

Gauss’s law, ⇒ no E-field outside the capacitor.

Static charge and steady current (i.e., static magnetic field) coexist, ⇒ one of rare examples of electromagnetostatic field.

Sec. 10-4Evaluation of Resistance

1. Standard procedures2. Example3. Relation between R and C

Evaluation of single-(imperfect)conductor resistance

1. Assume between 2 selected end faces

2. Find by solving with BCs

Ev

3. Find by

5. Find R by , independent of

0V

)(rV v02 =∇ V

)(rVE vv−∇=

4. Find total current by

0V

∫∫ ⋅=⋅=SS

sdEσsdJI vvvv

IVR 0=

Example 10-5: Resistance of a washer (1)

Find R of a quarter-circular washer of rectangular cross section and σ < ∞, if the two electrodes are located at φ = 0, π/2, respectively

,0)0( ==φVAssume: (1) 0)2( VV == πφ

V=0

V=V0

Example 10-5: Resistance of a washer (2)

,0)0( ==φV

(2) Current flow and are in

0)2( VV == πφ

Ev

φav−

)(),,( φφ VzrV =⇒

,02 =∇ V 0)( =′′⇒ φV

21)( ccV += φφ

φπ

φ 02)( VV =

Example 10-5: Resistance of a washer (3)

rVaVE 12 0

πφvv

−=−∇= ⎟⎠⎞

⎜⎝⎛∝

r1

rVaEJ 12 0

πσσ φ

vvv−==

⎟⎠⎞

⎜⎝⎛=⋅

⋅=⋅= ∫∫ a

bhVhdrr

VsdJIb

aSln22 0

0

πσ

πσvv

( )abhIVR

ln20

σπ

==

Evaluation of resistance between two perfect conductors

IQ

VQ

IVRC =⋅=

σε

σ

ε=

⋅

⋅=

⋅

⋅=

∫∫

∫∫

S

S

S

S

sdE

sdE

sdJ

sdD

vv

vv

vv

vv

⇒ one can evaluate R by finding C first!

Example 10-6: Resistance of a coaxial cable

Find the leakage resistance between inner and outer conductors of a coaxial cable of length L, where a lossy medium of conductivity σ is filled

( )abLC

ln2πε

=

( )Lab

CR

πσσε

2ln

==⇒