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Lecture 17
March 11, 2005
Chapter 5 Steady Electric Current
5-1 Introduction
a) Conduction currents in conductors and semiconductor: drift motion of
conduction electrons and /or holes.
b) Electrolytic currents: migration of positive and negative ions.
c) Convection current: results from motion of electrons and /or ions in a
vacuum.
5-2 Current Density and Ohm’s Law
If N is the number of charge carriers per unit volume, then in time ∆t each
carrier moves a distance u∆t , the amount of charge passing through the
surface ∆s is
( C).
∆Q = Nqu ⋅ nˆ∆s∆t
Since current is the time rate of change of charge, we have
∆I =
∆Q
= Nqu ⋅ nˆ ∆s = Nqu ⋅ ∆s
∆t
( A ).
Define a vector point function, volume current density, J
J = Nqu
( A/m ) ,
2
The total current flowing through S
I = ∫ J ⋅ ds
S
We may rewrite the current density as
J = ρu
and ∆I =J ⋅ ∆s.
( A ).
( A/m ) ,
2
which is the relation between the convection current density and the velocity
of the charge.
Example 5-1 Vacuum tube diode
Solution
The net electric field at the cathode is zero(space charge limited condition).
ˆ y ( 0 ) = − yˆ
E ( 0 ) = yE
dV ( y )
dy
= 0.
y =0
In the steady state the current density is constant, independent of y:
J = − yˆ J = yˆ ρ ( y ) u ( y ) ,
where the charge density ρ ( y ) is a negative quantity. The velocity u = yu(y)
is related to the electric field intensity E(y) = yE(y) by Newton’s law of
motion:
m
du ( y )
= −eE ( y ) = e
dt
dV ( y )
dy
,
where m = 9.11 x 10-31 (kg) and – e = 1.60 x 10-19 (C) are the mass and
charge of electron. Noting that
m
du
du dy
du d  1

=m
= mu
=  mu 2  ,
dt
dy dt
dy dy  2

we can rewrite
d 1
dV
2
.
 mu  = e
dy  2
dy

Integration of this equation gives
1
mu 2 = eV ,
2
where the constant of integration = 0, at y = 0, u(0) = V(0) = 0.
1/ 2
 2e 
u= V
m 
.
In order to find V(y) in the interelectrode region we must solve Poisson’s
equation with ρ expressed in terms of V(y)
ρ =−
J
m −1/ 2
= −J
V .
u
2e
We have
ρ J
d 2V
=− =
2
ε0 ε0
dy
m −1/ 2
V .
2e
This equation can be integrated if both sides are first multiplied by 2dV/dy.
The result is
2
 dV 
4J

 =
ε0
 dy 
At y = 0, V = 0, and dV/dy = 0, so c = 0.
m 1/ 2
V + c.
2e
1/ 4
V
−1/ 4
J m
dV = 2
ε 0  2e 
dy .
Integrating the left side of from V = 0 to V0 and the right side from y = 0 to d,
we obtain
1/ 4
J m
4 3/ 4
V0 = 2
3
ε 0  2e 
d , or J =
4ε 0
9d 2
2e 3/ 2
V0
m
( A/m ) .
2
This equation states that the convection current density in a space-charge
limited vacuum diode is proportional to the three-halves power of the
potential difference between the anode and the cathode: Child Langmuir
law.
For most conducting materials the average drift velocity is directly
proportional to the electric field intensity.
u = − µe E
( m/s ) ,
where µe is the electron mobility measured in (m2/Vs). The electron mobility
for copper is 3.2 x 10-3 (m2/V.s). From eqs. J = ρu and u = µeE we obtain
J = ρeµeE,
where ρe = -Ne is the charge density of the drifting electrons and is negative
quantity.
J = σE (A/m2),
where the proportionality constant, σ = -ρeµe, is a macroscopic constitutive
parameter of the medium called conductivity.
Isotropic materials for which the linear relation J = σE are called
ohmic media. The unit for σ is (A/V.m).or (S/m). The reciprocal of
conductivity is called resistivity, in (Ω.m).
Ohm’s law:
V12 = RI .
The point form of Ohm’s law: J = σE.
The potential difference between terminals 1 and 2 is
V12 = EA or E =
E=
V12
.
A
V12
A
The total current is
I = ∫ J ⋅ ds = JS or J =
S
Using Eqs. E =
I
.
S
V12
and J = I/S in J = σE, we obtain
A
I
V
 A 
= σ 12 or V12 = 
 I = RI ,
S
A
σS 
which is the same as V12 = RI . The formula for the resistance of a straight
piece of homogeneous material of a uniform cross section for steady current:
R=
A
σS
(Ω).