Child’s Law

Child’s Law (or the Child-Langmuir Law or three-halves-power law) gives the maximum space-charge-limited current in a planar diode of infinite radius (i.e. one-dimensional beam) as a function of the length and potential difference between anode and cathode.

J = K V_d^{3/2} / d^2

where

  • J = current density (mA mm-2)

  • V_d = potential difference between anode and cathode (V)

  • d = distance between anode and cathode (mm)

  • K is a constant given by K = (4/9) \epsilon_0 (-2q/m)^{1/2} with units mA V-3/2, where

    • \epsilon_0 = permittivity of free-space ~ 8.854187817*10-12 F m-1 [1]

    • q/m = charge (C) to mass (kg) ratio of particle (absolute value). For an electron, q/m = 1.758820150*1011 C kg-1, [2]

    • For an electron, K \approx 0.0023340 mA V-3/2.

Note units on left-side of constant:

(F m^{-1}) C^{1/2} kg^{-1/2} = (C^2 J^{-1}) m^{-1} C^{1/2} kg^{-1/2} = C^{5/2} J^{-1} (J^{-1/2} s^{-1}) = (C s^{-1}) (J/C)^{-3/2} = A V^{-3/2}

This can be applied near cathode surfaces in space-charge limited cathode emissions. A correction may be applied for non-zero temperature or non-planar surfaces.

Derivation

Child’s Law is a fairly direct result of the poission_equation:

\bnabla^2 V = -\rho / \epsilon_0

where charge density \rho = J/v, for velocity v, and where v can be expressed in terms of kinetic energy and related to potential energy: (1/2) m v^2 = -q V, for mass m and charge q. This gives

\bnabla^2 V = dV'/dx = A V^{-1/2}

where A = -J \epsilon_0^{-1} (-(1/2) m/q)^{1/2} is a constant and V' = dV/dx.

Substitute dx = dV/V' to obtain a separable differential equation and solve, while applying boundary conditions V(x=0) = 0, V(x=d) = Vd, and V'(x=0) = 0:

V' dV' = A V^{-1/2} dV

V' = dV/dx = 2 A^{1/2} V^{1/4}

V^{-1/4} dV = 2 A^{1/2} dx

(2/3)V^{3/4} = A^{1/2} x

where A = (4/9)V_d^{3/2} d^{-2} (due to V(x=d) boundary condition)

Now knowing A, we can solve for J in the expression for A:

(4/9)V^{3/2} d^{-2} = A = -J \epsilon_0^{-1} (-(1/2) m/q)^{1/2} \mathrm{ implies}

J = K V^{3/2} d^{-2} \textrm{ where } K = (4/9) \epsilon_0 (-2q/m)^{1/2}

which is the desired result.

Note also that at x=0, v=0 and \rho=\infty.

See Also