Summary: Pierce type electron gun essentials and most basic geometry

Introduction

After I wrote some blog posts about thermionic emission cathodes as well as a simple improvised uncollimated electron source from scrape stuff I thought it’s time to write a blog post about one of the more simple types of electron guns - the Pierce electron gun design.

As one knows the main problem for an electron gun to provide a collimated non diverging beam of electrons is the divergence of the beam. This usually occurs because electrons are charged negative and are repelling each other. This leads to natural beam divergence in charged particle beams. The idea behind a Pierce type electron gun is to counter this effect by the electric field formed by the cathode-anode pair by choosing a clever geometry.

To recall the current limits for electron guns - there are two modes in which a thermionic source can operate (for more details see my blog post about thermionic emission cathodes):

First there’s thermally limited mode. In thermally limited mode the number of electrons increases as the temperature of the cathode increases. In this mode the Richardson-Dushman equation describes the current density limit - $A$ is a material constant, $\psi$ is the work function and $T$ the temperature of the cathode:

$$J_{th} = A * T^2 * e^{- \frac{\psi}{k_B * T}}$$

The constant $k_B$ is the Boltzmann constant, the work function should be specified in Joule ($1 eV \approx 1.602 * 10^{-19} J$).

$$k_B = 1.380 * 10^{-23} \frac{J}{K}$$

On the other hand there’s space charge limited mode that’s reached at higher temperatures. In this mode the repulsive force between electrons inside the electron cloud is limiting the release of electrons from the cathode. The current limit is described by Child’s law:

$$J_{sc} = \underbrace{\frac{4 \epsilon_0}{9} * \sqrt{\frac{2 e}{m_e}}}_{c_1} * \frac{V_a^{\frac{3}{2}}}{d^2} \\ J_{sc} = c_1 * \frac{V_a^{\frac{3}{2}}}{d^2}$$

The quantities $\epsilon_0$, $e$ and $m_e$ are the vacuum permittivity, the charge of an electron and the mass of an electron respectively - and thus constant:

$$\epsilon_0 = 8.854 * 10^{-12} \frac{As}{Vm} \\ e = 1.602*10^{-19} C \\ m_e = 9.109 * 10^{-31} kg \\ \frac{4 \epsilon_0}{9} * \sqrt{\frac{2 e}{m_e}} \approx 2.2338 * 10^{-6} m A V^{-\frac{3}{2}}$$

The quantity $V_a$ is the acceleration voltage between cathode and anode, $d$ is the distance between these two electrodes in meters.

Pierce type approach

Usually for Pierce type electron guns only space charge mode is assumed - the thermally emitted electrons are basically ignored. The idea is pretty simple - the beam emitted from an cathode should be emitted in a cylindrical form - the area around the beam should be free of charge carriers. Since one knows that

$$U = \int E ds \\ \to \nabla U = E \\ \nabla * E = \underbrace{\frac{\rho}{\epsilon_0}}_{0} \\ \to \nabla * E = 0 \\ \to \nabla \underbrace{\nabla U}_{E} = 0 \\ \Delta U = 0$$

As one can see the requirement for charge carrier free space is just that the potential fulfills the Laplace equation. This can be achieved with a huge number of geometric configurations - the most basic Pierce type gun usually uses a pretty simply construction.

Simple solution

One can show that the Laplace equation is satisfied by any function that depends on a complex number and that’s twice differentiable:

$$u = y + i * x \\ \implies \Delta f(u) = 0$$

This can easily be shown:

$$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} \\ = \frac{\partial}{\partial x} \left(\frac{\partial u}{\partial x} \frac{\partial f}{\partial u} \right) + \frac{\partial}{\partial y} \left(\frac{\partial u}{\partial y} \frac{\partial f}{\partial u}\right) \\ = \frac{\partial}{\partial x} \left(i * \frac{\partial f}{\partial u}\right) + \frac{\partial}{\partial y} \left(\frac{\partial f}{\partial u}\right) \\ = \frac{\partial u}{\partial x} \frac{\partial}{\partial u} \left(i \frac{\partial f}{\partial u}\right) + \frac{\partial u}{\partial y} \frac{\partial}{\partial u} \left(\frac{\partial f}{\partial u}\right) \\ = i^2 \frac{\partial^2 f}{\partial u^2} + \frac{\partial^2 f}{\partial u^2} \\ = -\frac{\partial^2 f}{\partial u^2} + \frac{\partial^2 f}{\partial u^2} = 0$$

This leads to a simple method to construct such a function. Simply define $V = \mathfrak{Re}(f(u)) \forall x > 0$. Note that I’ll be using cylindrical coordinates during the construction. The origin will be placed at the cathode, the beam will be traveling into positive y direction and the x axis will be parallel to the plain cathode surface / perpendicular to the beam and symmetry axis. $x=0$ will be located at a sharp edge of the beam. For $x < 0$ one will enter the zone that contains charge carriers, $x > 0$ will be charge carrier free. This will allow easy modeling.

Note that the graphically shown emitter surface might either be an electron emitting surface itself or the emitting surface from a Wehnelt cylinder assembly.

Since one can assume that the potential has to be continuous at $x = 0$ one can assume that - assuming space charge limited mode - at $x = 0, y = 0$ Child’s law provides the boundary condition for the potential:

$$V = V_a * \mathfrak{Re}\left(\left(\frac{u}{d}\right)^{\frac{4}{3}})\right)$$

Now moving to polar coordinates makes life easier:

$$u = r * e^{i * \theta} \\ V = V_a * \mathfrak{Re} \left( \left(\frac{r}{d} e^\theta\right)^{\frac{4}{3}} \right) \forall x > 0 \\ = V_a * \left(\frac{r}{d}\right)^\frac{4}{3} * \mathfrak{Re}\left(e^{\frac{4}{3}\theta}\right) \\ \to V = V_a \left(\frac{r}{d}\right)^{\frac{4}{3}} * \cos \left(\frac{4}{3} \theta\right)$$

Geometry of the cathode

Now first let’s calculate the geometry of the cathode. I’m going to assume arbitrary that $V=0$ for the cathode as this is a commonly chosen configuration in electron tubes - for particle accelerators and similar equipment the cathode might be placed at negative potential which would only add an total offset to the voltage and thus not change the geometry of the electron gun anyways.

$$V=0 \to 0 = V_a \left(\frac{r}{d}\right)^{\frac{4}{3}} * \cos \left(\frac{4}{3} \theta\right) \\ \to \cos \left(\frac{4}{3} \theta\right) = 0 \\ \to \frac{4}{3} \theta = \frac{\pi}{2} \\ \implies \theta = \frac{3 \pi}{8} \\ \theta = 67.5^{\circ}$$

As one can see the angle outside the cathode surface will be fixed - it’s simply a cone with fixed angle

Geometry of the anode

The anode will be a little bit more complex. We’ll be assuming that $V=V_a$:

$$V=V_a \to V_a = V_a \left(\frac{r}{d}\right)^{\frac{4}{3}} * \cos \left(\frac{4}{3} \theta\right) \\ 1 = \left(\frac{r}{d}\right)^{\frac{4}{3}} * \cos \left(\frac{4}{3} \theta\right) \\ 1 = \frac{r}{d} \cos^{\frac{4}{3}}\left(\frac{4}{3} \theta \right) \\ r = d * \left(\frac{1}{\cos(\frac{4}{3}\theta)}\right)^{\frac{3}{4}}$$

This article is tagged: Physics, Particle source