### Re: What exactly is the Child Langmuir effect or the law of space charge ?

Date: Tue Jun 13 22:04:46 2000
Posted By: Yaxun Liu, Grad student, Electrical Engineering, National University of Singapore
Area of science: Physics
ID: 960591390.Ph
Message:

The Child-Langmuir Law describes the relation between
the current and the voltage between the cathode and
the anode.

Generally the equation for Child-Langmuir Law is

J=4/9*K/s*V^(3/2)=2.330*10^(-6)*V^(3/2)/s

while

J is the current (A)
V is the voltage between the cathode and the anode (V)
K = eps0*sqrt(2*e/m) is a constant, e is the charge of
electron, m is the mass of electron, eps0 is the
permittivity of vacuum
s is a constant depending on the shape of the anode
and cathode, its unit is cm^2.

For simplicity, consider a 1-D problem, i.e. the anode
and the cathode are two infinite parallel plates,
one at x=0, the other at x=d. In this case the shape
parameter s in the Child-Langmuir Law is

s=d^2

Now let's derive this equation.

Assume the potential is V(x), we'll have Possion's equation

d^2 V     rho
----- = - ----
d^2 x     eps

while rho(x) is the density of the charge, eps is the
permittivity of the medium.

Assume V(0)=0, electrons move from x=0 toward x=d,
at x=x' electron will gain energy e*V(x'), therefore
its speed is

v = sqrt(2*e*V/m)

while m is the mass of the electron

Assume the current distribution is J(x). It
should be noticed that in steady state J is
constant due to the conservation of electric
charges:

d rho      d J
----- = - ------ = 0
d t        d x

The relation between rho(x) and J is

J = rho * v

Therefore

rho = J/v = J/sqrt(2*e*V/m)

By substitution of the above equation into Possion's
equation, we can obtain

d^2 V     J/sqrt(2*e*V/m)
----- = - ----------------
d^2 x          eps

This ordinary differential equation can be easily solved
by assuming V(x)=a*x^b and substituting it into the
above equation
J/sqrt(2*e/m)
a*b*(b-1)*x^(b-2) = - ---------------- /sqrt(a*x^b)
eps
therefore

b-2=-b/2
a^(3/2)*b*(b-1)=-J/sqrt(2*e/m)/eps

From the above equations we obtain

b=4/3
a=[-J/sqrt(2*e/m)/eps*9/4]^(2/3)

Therefore

V(x)=[-J/sqrt(2*e/m)/eps*9/4]^(2/3)*x^(4/3)

which can be simplified as

J=-4/9/x^2*eps*sqrt(2*e/m)*V^(3/2)

If V(x)>0, J<0, which means the current flows from
higher potential to lower potential, just in inverse
direction of the motion of the electrons.

References

J. P. Barbour et al., Space-charge effects in field
emission, Phys. Rev. 92 (1): 45-51, 1953.

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