# Child-Langmuir space charge law for non-zero cathode potential (non-zero initial electron velocity)

I'm trying to reconcile some conflicting results that I've found in publications that address the idea of the current in a vacuum diode in the case where the cathode has a non-zero potential, in other words, the electrons are emitted with a non-zero initial velocity.

The traditional Child-Langmuir Space Charge Law, also known as the 3/2 Law is as follows:

$$J_{CL}= \frac{4\epsilon_{0}}{9}\sqrt{\frac{2e}{m_{e}}} \frac{V_{a}^{3/2}}{d^{2}}$$

This is pretty straightforward to me, however, I am interested in the case where the cathode also has a non-zero potential. According to the literature, there are at least two different results, and I have been unable to convert one to the other, even though they are supposed to describe the same situation.

In S.E Sampayan, Nuc. Inst. and Meth. in Phys. Res. A, (1994) A 340, pp. 90-95, Eqn. A-13 & A-14, in which the increased electron current is justified by the existence of a "virtual cathode". The derivation eventually leads to the result:

$$J= J_{CL}\frac{[1+(1+ \Psi_{0} )^{3/4} ]^{2} }{ \Psi_{0} ^{3/2} } = \frac{4 \epsilon_{0} }{9} \sqrt{ \frac{2e}{m_{e} } } \frac{\phi_{0}^{3/2} }{d^{2} } \frac{[1+(1+ \Psi_{0} )^{3/4} ]^{2} }{ \Psi_{0} ^{3/2} }$$

where $\Psi_{0}= \frac{e \phi_{0}}{E_{0} }$, $E_{0}= \text{initial electron energy}$, and $\phi_{0}= \text{anode potential}$.

However, as the anode potential, $\phi_{0}$ tends to 0, $\Psi_{0}$ also tends to 0, and the current goes to infinity, when it should reduce to the original Child-Langmuir Law.

In another publication, G. Jaffe, Phys. Rev. (1944) Vol. 65, No. 3 & 4 pp. 91-98 , Eqn. 28, the result is given as:

$$J_{CL}= \frac{4\epsilon_{0}}{9}\sqrt{\frac{2e}{m_{e}}} \frac{( \sqrt{V_{c}}+\sqrt{V_{c}+V_{a}})^{3}} {d^{2}}$$

where $V_{c}= \text{cathode voltage}$, and here, it is clear that as the cathode potential tends to 0, the Child-Langmuir Law is recovered.

The same results can also be seen, (with minor typographic errors to the equation) in H. Riege, Nuc. Inst. and Meth. in Phys. Res. A, (2000) A 451, pp. 394-405, Eqn. 3.

I'm trying to figure out if either or both are correct, and in the latter case, how one converts from one to another. As far as I can tell, they both describe the same conditions, with a virtual cathode being the means of increased electron emission.

Any ideas?

• These formulas might be attempts to combine two effects: 1) the actual anode-to-cathode accelerating voltage potential $V_{AK}$, which can only depend on the difference between the two potentials, 2) an "effective" (e.g. fake) cathode potential $V_{eff}$ that models the initial velocity of the electrons that are "boiling" out of the heated cathode. – Art Brown May 14 '14 at 3:12
• @ArtBrown What you propose is interesting and could be true. First, can these potentials be interchanged (can you recast one potential in the form of the other)? Is it possible to model both together? Regardless, the Sampayan eqn is odd in that it does not reduce to the Child-Langmuir Law. Another question is, is there a difference between electrons with an initial velocity, and the cathode being biased to a specific voltage? I'd really like to see this reconciled, because the author of the first paper has not responded to requests, and the second paper is 70+ yrs old. How do we combine them? – iwantmyphd Mar 7 '15 at 13:30

I don't have access to the Sampayan paper, but I think you're conflating two different things:

1. The total anode-to-cathode accelerating potential $V_{AK}$. (Note it's the difference between the two potentials that matters.) The cathode is assumed to be capable of emitting unlimited numbers of electrons, so it is solely the repelling electric field of said electrons that limits the total current. (The current is said to be space-charge limited.) If one assumes that the electrons emitted by the cathode have 0 initial velocity, the Child-Langmuir current $J_{CL}$ results.

2. The effect of a non-zero initial electron velocity $v_e$, which can be characterized by the initial electron kinetic energy $E_e=m_e v_e^2/2$ or by an effective potential $V_e = E_e/e$, with $m_e$ the mass of the electron and $e$ the absolute value of its charge. Here $V_e$ is the potential that would result in the same kinetic energy as possessed by the emitted electrons. The electrons start with this kinetic energy and are then accelerated further by $V_{AK}$. Intuitively, one would expect the increased energy to give a current exceeding $J_{CL}$.

In these terms, the Sampayan equation for the current density $J_S$ becomes:

\begin{align*} J_S & = \frac{4}{9} \epsilon_0 \sqrt{\frac{2e}{m_e}} \, \frac{V_{AK}^{3/2}}{d^2} \frac{\left[1+ \left( 1 + V_{AK}/V_e \right)^{3/4} \right]^2}{(V_{AK}/V_e)^{3/2}} \\ & = J_{CL} \left[ \left( 1+ V_e/V_{AK} \right)^{3/4} + \left(V_e/V_{AK}\right)^{3/4} \right]^2 \end{align*}

which does in fact reduce to the Child-Langmuir current $J_{CL}$ in the limit $V_e \rightarrow 0$.

By contrast, Jaffe's current density $J_J$ is (again using my notation):

\begin{equation*} J_J = J_{CL}\, \left[ \left(1+V_e/V_{AK} \right)^{1/2} + \left( V_e/V_{AK} \right)^{1/2} \right]^3 \end{equation*}

which, as you noted, also reduces to $J_{CL}$ when $V_e=0$.

The two results have different dependencies on $V_e$. Which is (more) correct? I don't know, but here are some considerations.

• Jaffe's is a purely theoretical calculation, which in particular assumes that all the electrons are emitted with the exact same velocity. He references work by Schottky and others that considers the presumably more realistic (and presumably more complicated) case of a Maxwellian velocity distribution.
• The extract to Sampayan's paper states that it is an experimental investigation of ferroelectric cathodes. Without the paper, I'll say no more.
• Thanks for your post. (1) I can provide the Sampayan paper if that would help. Please let me know how best to get it to you. (2) I'm confused by your first point: doesn't a non-zero (negative) cathode potential immediately imply a non-zero electron initial velocity? Perhaps you address this in your second point, but my read of it looked like they were different physical cases. (3) Given that the powers are equivalent (3/4)*2 = (1/2)*3, the units work out, would the choice of how those powers are broken down (3/4 vs. 1/2), have any physical meaning? It seems strange to break them down this way. – iwantmyphd Mar 8 '15 at 22:25
• (1) thanks but I'll pass, sorry. (2) In this formulation, the initial electron velocity is just that, an initial condition. Once emitted from the cathode, an electron is accelerated by the local electric field, which is determined by (a) the voltage difference between anode and cathode and (b) the electric field from other emitted electrons (the space charge). The value of the cathode voltage alone is insufficient. (3) Presumably there are different physical assumptions behind the two formulations, but I don't have anything intelligent to say about them. – Art Brown Mar 10 '15 at 4:59