Lecture 14 Current Density Ohm s Law in Differential Form

Transcription

1 Lecture 14 Current Density Ohm s Law in Differential Form Sections: 5.1, 5.2, 5.3 Homework: See homework file

2 Direct Electric Current Review DC is the flow of charge under electrostatic forces in conductors Georg Simon Ohm was the 1 st to observe and explain the lack of charge acceleration in metals electrons move with uniform averaged speed (drift velocity) the electrostatic force is provided by external sources: battery (chemical), charged capacitor (storage), dynamo (mechanical) LECTURE 14 slide 2

3 Current Density the current flowing through the cross-section Δs of a conductor is the amount of transferred charge ΔQ per unit time Q 1 dq I( s) =, A=C s I = t dt Q= ρv V = ρv s L= ρv sv t, C Q I( s) = = ρvv s, A t V L I( s) = Jn s where J = ρ v, A/m2 n v current density, normal component J n s e e e e e e e e L Q = v t I V the current density is a vector J = n Ja n J = ρ, A/m 2 v v LECTURE 14 slide 3

4 Current and Current Density the current I is the flux of the current density J I = I = J s = J s di = J ds ( s) n I = d s, A J S Two cylindrical wires are connected in series. Current I = 10 A flows through the junction. The radii of the wires are: r 1 = 1 mm, r 2 = 2 mm. Find the current densities J 1 and J 2 in the two wires. LECTURE 14 slide 4

5 Charge Mobility charge velocity in a conductor depends on the charge mobility v = µ E, v = µ E, v = µ E, m/s e e h h i i metals support electron current drift electron velocity in metals: v d = μ e E semiconductors support both electron and hole currents most electrolytes support both electron and ion currents in general plasmas support both electron and ion currents mobility may in general depend on E (nonlinear conductors) LECTURE 14 slide 5

6 Specific Conductivity 1 J = ρeve + ρpvp = ( ρµ e e + ρ pµ p) E, p hi, σ Ohm s law note: ρ e < 0 the specific conductivity σ depends on the free-charge density and its mobility σ= ρµ e e + ρ pµ p, S/m=( Ω m) the charge density depends on the number of charge carriers per unit volume (number density), e.g., ρ e = en 19 e e , C 1 σ semi = ( Neµ e + Nhµ p) e σ = N µ e metal e e in pure semiconductors N e = N h LECTURE 14 slide 6

7 Specific Conductivity 2 typical carrier number densities, mobilities, conductivities (below THz) μ e μ h N e (m 3 ) N h (m 3 ) σ (S/m) pure Ge x x pure Si x x x10 4 Cu x x10 7 Al x x10 7 Ag x x10 7 σ = Au S/m Voltage V = 1 V is applied to a piece of copper wire of length L = 10 cm. What is the electron drift velocity v d? LECTURE 14 slide 7

8 Ohm s Law in Point (Differential) Form J = σ E Ohm s law in circuits I = GV = V / R, A assume uniform current distribution in the cross-section of the conductor between points A and B V I = J s V = EL AB E =, l = LAB l use Ohm s law in point form to arrive at Ohm s law for resistors s I = Js= σ Es = σ V l G= R 1 s 1 l G = σ, R= l σ s this is the conductance/resistance of a conductor of length l, constant crosssection s, and constant current density distribution in s LECTURE 14 slide 8

9 Technology Brief: Resistive Sensors [from Ulaby&Ravaioli, Fundamentals of Applied Electromagnetics] piezoresistivity deformation of a wire changes its resistance R 1 l = σ s α F R= R0 1+ s0 F = applied force R 0 = resistance when F = 0 s 0 = cross-section when F = 0 α = piezoresistive coefficient of material LECTURE 14 slide 9

10 Technology Brief: Resistive Sensors, cont. [from Ulaby&Ravaioli, Fundamentals of Applied Electromagnetics] longer wires have better sensitivity LECTURE 14 slide 10

11 Technology Brief: Resistive Sensors, cont. [from Ulaby&Ravaioli, Fundamentals of Applied Electromagnetics] Wheatstone bridge converts small changes in resistance to voltage LECTURE 14 slide 11

12 LECTURE 14 slide 12 General Expression for Resistance, B A s d V R I d = = Ω E l J s, B A s d R d σ = Ω E l E s in homogeneous medium 1, B A s d R d σ = Ω E l E s 1, S s B A d G R d σ = = E s E l

13 General Expression for Resistance: Example Current I = 1 A flows between two concentric spherical electrodes of radii a = 1 cm and b = 5 cm. The direction of the flow is from the small electrode toward the large one. (a) Find J(r), J(a), and J(b). (b) Find the resistance R if the conductivity of the medium between the electrodes is σ = 2 S/m. (c) Find the dissipated power P. LECTURE 14 slide 13

14 DC Resistance per Unit Length twin-lead line 1 L I R = 2, Ω A σ A R 1 R = = 2, Ω/m I L σ A A L coaxial line 1 L 1 L R = +, Ω σπa2 σπ( c2 b2) R = +, Ω/m σπ a2 c2 b2 I c I b a LECTURE 14 slide 14

15 Conservation of Charge (Continuity of Current Law) 1 consider the current flowing through a closed surface I = ds J positive flux corresponds to outflow of charge (charge inside volume decreases) s [ v] I = d s = J s [ v] dq dt conservation of charge (continuity of current) law in integral form encl NOTE THE NEGATIVE SIGN! LECTURE 14 slide 15

16 KCL and Conservation of Charge in circuits we assume that nodes do not accumulate or lose charge (charge goes in and out of the node at the same rate) I = J ds= J ds+ J ds+ J ds= [ v] s s s s I I I dq dt encl = 0 I 3 Kirchhoff s current law follows from conservation of charge I 1 s 3 I n = n 0 s 1 s 2 s I 2 KCL does NOT hold in high-frequency electronics LECTURE 14 slide 16

17 Conservation of Charge in Differential (Point) Form apply Gauss (divergence) theorem to conservation-of-charge law dqencl d I = J d s = J dv = = ρvdv dt dt s v v [ v] ρv J = t continuity of current (conservation of charge) in point form LECTURE 14 slide 17

18 equation of charge relaxation Charge Relaxation hm ( εe) = ρv E= D also ( σ E) = J ρ t v hm ρ ε v E= 1 ρv σ t ρv σ + ρ v = 0 t ε charge relaxation constant solution of the charge-relaxation equation σ t ε 0 0 t / τ ρ( t) = ρ e = ρ e, τ = ε / σ LECTURE 14 slide 18

19 Charge Relaxation in Plain Words consider an isolated conductor into which some charge Q 0 is injected initially Coulomb forces push the charge carriers apart until they redistribute and settle on the surface the process continues until no free charge is left inside the conductor at which stage the field inside is zero E = 0 the time for this to happen is about 3τ where τ = ε / σ this is also the time required to discharge a charged capacitor through a shorting conductor (resistor) LECTURE 14 slide 19

20 ρ Charge Relaxation Illustrated τ = 3 s ρ = 0 1 tangent at t = 0 intersects time axis at t = τ exp(-t/τ) 1/e α 0 0 τ = time (s) ( t/ τ ) 0 ρ0e = Example: Calculate the time required to restore charge neutrality in Cu where ε = ε 0 and σ = 5.8x10 7 S/m T 3τ = 3 ε 19 0 / σ = , s d dt t= 0 ρ τ = tanα LECTURE 14 slide 20

21 Power Density and Joule s Law in Differential Form consider sufficiently small volume Δv = Δs ΔL where the E-field and the charge density ρ v are constant since charge is moving with uniform drift velocity u d, the E-field does work on the charge (this work is converted into heat) We = Q E L= we v, J F power is work done per unit time We we v QE L P= = p v= = = QEu t t t power density P QEu d p = = = ρveu d = EJ, W/m 3 v v, W Joule s law in differential form: dissipated power per unit volume p = EJ = σ( EE ) = σ E, W/m 2 3 LECTURE 14 slide 21 d

22 power dissipated in conductors Joule s Law in Integral Form P = pdv = EJ dv = σ dv v v v Joule s law in circuit theory P = EJ dsdl = EJdLds E 2, W o assume that in a piece of conductor, E does not depend on the cross-section, only J (or σ) does, while J (or σ) does not depend on the length o assume that E and J are collinear v P = EdL Jds = V I = RI 2 = V 2 / R, W L p dv S S L LECTURE 14 slide 22

23 Joule s Law in Integral Form: Example Current I = 1 A flows between two concentric spherical electrodes of radii a = 1 cm and b = 5 cm. The direction of the flow is from the small electrode toward the large one. (a) Find E(r). (b) Find the dissipated power P using E(r). LECTURE 14 slide 23

24 You have learned: what current density is and how it relates to the total current that drift velocity of charge in conductors is proportional to the strength of E and the coefficient of proportionality is the mobility what specific conductivity is and how it relates J to E through Ohm s law in differential (point) form how to compute the resistance/conductance of conducting bodies that charge is preserved and the rate of change of the charge density determines the divergence of the current density ρv J = t what charge relaxation is and how it depends on the permittivity and conductivity of the material how to find the dissipated power from the E field using Joule s law LECTURE 14 slide 24