Space Charge Limitation

The charges at the cathode reduces the potential that is present in the absence of the electron so that further emission is prevented. The maxium current is said to be the space charge limitation. Child and Langmuir first studied the space charge limitation in 1910s[1,2]. It assumes that two infinite parallel plane electrodes at fixed voltage $V_0$ in vacuum seprarated by a distance D[3]. Both the traditional way and a new approach were presented in Ref[3]. However the key procedure to solve the nonlinear 2nd order ordinary differential equation was not presented in Ref[3,4]. Here it is presented.

Poisson’s Law applies and the equation for the potential V is

$\frac{d^2V}{dz^2}=-\frac{\rho}{\epsilon_0}$ (1)

Where V is the potential, $\rho$ is the charge density and $\epsilon_0$ is the vacuum permittivity. The charge density $\rho$ is related to the current density J and the electron velocity u(x). The current density is a constant according to the charge conservation. so, the current density is

$J=\rho(z)v(z)=-J_{CL}$ (2)

and from the conservation of energy,

$\frac{mv^2}{2}=eV$ (3)

where m an e are the electron’s mass and charge respectively. Substituting Eq.(3) and Eq.(2) into Eq.(1) we have a 2nd order nonlinear differential equation for the potential

$\frac{d^2V}{dz^2}=\frac{J_{CL}}{\epsilon_0\sqrt{2eV/m}}$ (4)

with the following boundary conditions

$V'(0)=0, V(0)=0$ (5)

It is special nonlinear ODE with z missing. Let y=V'(x), then

$v”(x)=\frac{dy}{dx}=\frac{dy}{dV}\cdot\frac{dV}{dx}=y\frac{dy}{dV}=\frac{J_{CL}}{\epsilon_0\sqrt{2eV/m}}$ (6)

It is a 1st order separable ODE. Integrate V on both side,

$\int y\frac{dy}{dV}dV=\int \frac{J_{CL}}{\epsilon_0\sqrt{2eV/m}} dV$ (7)
$\int ydy=\int \frac{J_{CL}}{\epsilon_0\sqrt{2eV/m}} dV$
$\frac{y^2}{2}=\frac{2J_{CL}}{\epsilon_0\sqrt{2e/m}}V^{1/2} +C$

The integral constant C=0 because y(0)=0 and V(0)=0, so

$\frac{y^2}{2}=\frac{2J_{CL}}{\epsilon_0\sqrt{2e/m}}V^{1/2}$
$y=V'(z)=\pm k V^{1/4}$

Where $k=\sqrt{\frac{4J_{CL}}{\epsilon_0\sqrt{2e/m}}}$ is a constant and the potential increases with z, the slope of the potential is positive. Then we get a 1st order separable ODE again

$y=V'(z)=kV^{1/4}$

Integrate z and simplify

$\int V^{-1/4}\frac{dV}{dz}dz=\int k dz$
$\int V^{-1/4}dV=\int k dz$
$\frac{V^{3/4}}{3/4}=kz+C$

Since V(0)=0, C=0 again.

$V(z)=V0\cdot z^{4/3}$

Usually, V(z) is in the following form

$V(z)=V0\cdot (\frac{z}{D})^{\frac{4}{3}}$ （8）

substituting Eq.(8) into Eq.(1), the charge density is

$\rho=-\frac{4 \text{V0} \epsilon _0}{9 D^2 \left(\frac{z}{D}\right)^{2/3}}$ (9)

Substituting Eq.(9) , Eq.(8) and Eq.(3) into Eq.(2), the charge limited current density is given by

$\rho=-\frac{4 \text{V0} \epsilon _0}{9 D^2 \left(\frac{z}{D}\right)^{2/3}}=-\frac{J_{CL}}{\sqrt{2eV0\cdot (\frac{z}{D})^{\frac{4}{3}}/m}}$

Simplify it to obtain the result

$J_{CL}=\frac{4\epsilon_0}{9D^2}\sqrt{\frac{2e}{m}}V_0^{\frac{3}{2}}$

It is well known as the Child-Langmuir law. It states that the behavior of the current density is proportional to the three-halves power of the potential and inversely proportional to the square of the gap distance between the electrodes.

Reference
[1] C.D. Child, “Discharge from hot CaO”, Phys. Rev. 32, 492-511 (1911)
[2] I. Langmuir, “The effect of space charge and residual gases on thermionic currents in high vacuum”, Phys. Rev. 2, 450-486 (1913)
[3] https://arxiv.org/pdf/1506.07417.pdf
[4] Gilmour A S. Klystrons, Traveling Wave Tubes, Magnetrons, Crossed-field Amplifiers, and Gyrotrons[M].  Artech House, 2011.