Lesson Steady lectric Currents 楊尚達 Shang-Da Yang Institute of Photonics Technologies Department of lectrical ngineering National Tsing Hua Uniersity, Taiwan
Outline Current density Current laws Boundary conditions aluation of resistance
Sec. - Current Density. Definition. Conection currents 3. Conduction currents
Definition- olume charge density: ρ qn ( 3 C m )
Within a time interal, the amount of charge enclosed by a olume will pass through the elemental surface Definition- Δt ( ) [ ] s a t u n Δ Δ Δ ( ) ( ) ( ) s a u t s a t u Q n n Δ Δ Δ Δ Δ Δ ρ ρ ρ Q Δ Δ
Definition-3 Δ Q Δ t( u a s), ρ I ρ( u a s) n Δ ΔQ Δt nδ Current I can be regarded as the flux of a I J olume current density J, i.e., ( a s), J ρu (A/m ) n Δ
Conection currents Conection currents result from motion of charged particles (e.g. electrons, ions) in acuum (e.g. cathode ray tube), inoling with mass transport but without collision. q
xample -: acuum tube diode () J ( )?. lectrons are boiled away from the incandescent cathode with zero initial elocity (space-charge limited condition).. Anode has positie potential, attracting the free electrons.
xample -: acuum tube diode () By planar symmetry,. Potential: ( y), BCs: ( ), ( d). olume charge density: ρ( y) 3. Charge elocity: u a u(y) y
xample -: acuum tube diode (3) By space-charge limited condition,. Zero initial charge elocity:. Zero boundary -field: u( ) ( ), ( ) y u,
xample -: acuum tube diode (4) () In steady state, J ayj Q J ρu, J ρ( y) u( y) constant, J ρ( y) u( y) () By energy conseration, e y mu ( y) e ( y) m u ( y), ρ( y) J m e ( y) ( )
xample -: acuum tube diode (5) (3) Since free charges exist inside the tube: ρ Poisson s equation d dy ρ( y) J J ( ) (4) Shortcut to get : Multiply both sides by d dy d dy J d dy m e ( y) d dy m e ( y) Nonlinear OD
xample -: acuum tube diode (6) Integrate with respect to y: d dy d dy J d dy m e ( y) J m ( y) ( y) dy d e ( ) / d dy 4J m ( y) e + c
xample -: acuum tube diode (7) By BCs of the potential: ( ), () c, d dy J m ( y) e / 4 / 4 J m d e / 4 dy / 4 d J m e / 4 d dy
xample -: acuum tube diode (8) Child-Langmuir law: J 4 9 e m d 3 / Differ from Ohm s law ( I ) of conduction current. Differ from forward-biased semiconductor α diode ( I e ).
xample -: acuum tube diode (9) Triode (transistor)
Conduction (drift) currents- Conduction currents result from drift motion of free electrons due to applied -field in conductors, inoling with frequent collisions with immobile ions. -e m th 3 n kt ffectie mass of e -, depending on the lattice..g. At T 3K, th ~ 5 m/sec
Conduction (drift) currents- In steady state, two forces are balanced with each other (Drude model):. lectric force: e. Frictional force: m u n d τ aerage momentum change in each collision mean scattering time between collisions u d eτ m n μ e drift (aerage) elocity For typical fields, u d ~ mm/sec
Conduction (drift) currents-3 lectron mobility μ e eτ m describes how easy an external -field can influence the motion of conduction electrons..g. Silicon: -D carbon nanotube: Copper: μ e μ e.35.3 3, n μ e μ h m sec 4.8
Conduction (drift) currents-4 u By J ρu, d μ, e J ρ u ρ μ, e d J e < e (A/m ) conduction current density ρ e μ e (S/m) electric conductiity For semiconductors: ρ μ + ρ e e h μ holes h
Sec. - Current Laws. Ohm s law. lectromotie force & KL 3. quation of continuity & KCL 4. Joule s law
Ohm s law- Consider a piece of homogeneous, imperfect conductor ( < ) with arbitrary shape: microscopic law: J dl, Since the spatial distribution of is independent of L R I I A J ds L A dl ds A ds constant
Ohm s law- Consider a piece of homogeneous, imperfect conductor ( < ) with uniform cross-section: R dl L ds A L S L S
lectromotie force (emf)- If there is only conseratie electric field (created by charges): J, dl ( J ) dl C C no steady loop current! Non-conseratie field is required to drie charges in a closed loop
lectromotie force (emf)- Consider an open-circuited battery: chemical reaction impressed field i dl emf: the strength of non-conseratie force
lectromotie force (emf)-3 By :, inside i C l d outside inside outside + l d dl dl dl dl i C outside dl
Kirchhoff s oltage law (KL) If the two terminals are connected by a uniform conducting wire of resistance: R Multi-sources: j j k J R L S outside, J I S, outside dl J IL dl IR C S k I k KL
quation of continuity- S ( r ρ, t) ρ( r, t + Δt) Δt Q ρ( r, t) d Q ρ( r, t + Δt) d
quation of continuity- Principle of conseration of charge, a net current I flowing out of must be due to the decrease of the enclosed charge: Q Q I lim Δt Δt ρ( r, t) J ds S Diergence theorem ( J ) d decrease of change within during Δt ρ( r, t Δt ρ d t + Δt) d J ρ t ρ( r, t) t d
Kirchhoff s current law (KCL) I For steady currents, J ρ t S I 3 j I j ρ t J J ds S no current source/sink ( J ) d I
xample -: Dynamics of charge density () t constnat constnat ρ( r, t)? ρ J J ρ t J ( ) ( ) ρ t
xample -: Dynamics of charge density () D r D ρ r D ( ) ( ) ρ ρ t ρ ρ( t) ρe t τ, ρ + ρ t τ Lifetime of free charge in a conductor
Joule s law- With, collisions among free electrons (with drift elocity u ) & immobile atoms transfer energy from to thermal ibration. d
Joule s law- Differential drift displacement within Δt: Δl d Q Work done by to moe charge Q for Δl is: d Δ w Q Δ, corresponding to a power dissipation: p lim Δt Δw Δt l d Q u d
Joule s law-3 (r) dq ρ( r ) d Power dissipated in a differential olume d is: Δp d u u d J ρ J P d ( ) ( ρ ) ( )d d (W/m 3 ) is power density, ( J ) d Joule s law
Joule s law-4 Power dissipated in a homogeneous conductor of uniform cross section: P ( J ) d ( J ) Sdl Idl I L L
Sec. -3 Boundary Conditions. BCs of current density. xamples: Two lossy media connected in series
Deriation of BCs-. Diergence & curl relations of J : (i) For steady currents: J (ii) J ( ) J. Integral forms: (i) J ds S J (ii) dl C
Deriation of BCs- 3. Apply the integral relations to a (i) differential pillbox: J ds S J n J n (ii) differential loop: J dl C J t J t
xample -3: BCs between two lossy media () Represent the surface charge density ρ s by either D n or D n
xample -3: BCs between two lossy media () ( ) s n D D a ρ D n n n n s D D ρ J n n J J n n n n n n
xample -3: BCs between two lossy media (3) ; s D n ρ ρ s only if: (i), rarely happens; (ii) both media are lossless, ρ n n n n n s D n s D ρ similarly,
xample -4: lectromagnetostatic field () Find J,, and surface charge densities ρ s y By planar symmetry and n n, J By J, a, i i J J J ayj J y i i i
xample -4: lectromagnetostatic field (), + + d d J d d By ( ) ( ) d d J +, d d J + d d J +
xample -4: lectromagnetostatic field (3) d d D s ρ + d d D s ρ + s D n ρ By ( ) ) ( d d si ρ +
Comments ρ ρ s s, but ρ s + ρs + ρsi Gauss s law, no -field outside the capacitor. Static charge and steady current (i.e., static magnetic field) coexist, one of rare examples of electromagnetostatic field.
Sec. -4 aluation of Resistance. Standard procedures. xample 3. Relation between R and C
aluation of single-(imperfect)conductor resistance. Assume between selected end faces. Find (r ) by soling with BCs 3. Find by (r ) 4. Find total current by I R 5. Find R by, independent of I S J ds S ds
xample -5: Resistance of a washer () Find R of a quarter-circular washer of rectangular cross section and <, if the two electrodes are located at φ, π/, respectiely Assume: () ( φ ), ( φ π )
xample -5: Resistance of a washer () () Current flow and are in ( r, φ, z) ( φ) a φ, ( φ) ( φ) c φ + c ( ) ( φ ), ( φ π ) π φ φ
xample -5: Resistance of a washer (3) r a π φ r r a J π φ a b h hdr r ds J I b a S ln π π ( ) a b h I R ln π
aluation of resistance between two perfect conductors I Q Q I RC S S S S ds ds ds J ds D one can ealuate R by finding C first!
xample -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 conductiity is filled πl C ln R ( b a) C ( b a) ln πl