Fundamentals of Electrical Gas Discharges

Abstract

This chapter is a brief, and in certain passages qualitative, description of how an electrical gas discharge works. The main purpose of Part I of this textbook is not to present a detailed description of the operation of the various types of discharges, but instead to analyse the various aspects of the electron kinetics that are in the origin of the electrical discharges operation from a microscopic point of view. However, the present chapter is not in line with this perspective, since its purpose is to bring the reader to the physical system under study, which will be studied from Chap. 2 throughout based on the electron Boltzmann transport equation.

Appendices

Appendices

1.1 A.1.1 Particle Current Density Upon a Surface

The expression of the particle current density upon a surface due to the random movement used in equations (1.1) and (1.84) can be derived as follows. Let us consider a certain type of particles (electrons, ions, atoms) in which the number of particles, per volume unit, with the absolute value of velocity between v and v + dv is

$$\displaystyle{ dn = f(v)\;4\pi v^{2}\;dv, }$$

(1.98)

being f(v) the particle velocity distribution normalized such that

$$\displaystyle{ n =\int _{ 0}^{\infty }f(v)\;4\pi v^{2}\;dv, }$$

(1.99)

with n denoting the particle number density. The particles within the angle solid $d\Omega =\sin \theta \; d\theta \;d\phi$ will hit the plane xOy in the time interval dt if they lie at a distance $dz = v\;\cos \theta \;dt$ of the plane (see Fig. 1.14). Since the fractional number of particles within the angle solid is $d\Omega /4\pi$, the number of particles, per surface unit, with velocities between v and v + dv and within the angle solid $d\Omega$, that impinge the plane xOy in the time interval dt is

$$\displaystyle{ f(v)\;4\pi v^{2}\;dv\;\frac{d\Omega } {4\pi } \;v\;\cos \theta \;dt. }$$

(1.100)

Fig. 1.14

Particles within the angle solid $d\Omega =\sin \theta \; d\theta \;d\phi$ and at a distance $dz = v\;\cos \theta \;dt$ of the plane xOy, impinging on it during the time interval dt

Integrating first over all velocities

$$\displaystyle{ \cos \theta \;\frac{d\Omega } {4\pi } \;dt\int _{0}^{\infty }v\;f(v)\;4\pi v^{2}\;dv, }$$

(1.101)

we obtain

$$\displaystyle{ \cos \theta \;\frac{d\Omega } {4\pi } \;dt\;n <\!\! v\!\!>, }$$

(1.102)

being

$$\displaystyle{ <\!\! v\!\!>= \frac{1} {n}\int _{0}^{\infty }v\;f(v)\;4\pi v^{2}\;dv }$$

(1.103)

the average particle velocity. Integrating now over all directions in the semispace z ≥ 0, we obtain the number of collisions per surface unit in the time interval dt

$$\displaystyle{ \frac{n <\!\! v\!\!>} {4\pi } \;dt\int _{0}^{\pi /2}\cos \theta \;\sin \theta \;d\theta \int _{ 0}^{2\pi }d\phi = \frac{n <\!\! v\!\!>} {4} \;dt. }$$

(1.104)

Thus, the number of collisions per surface and time units, i.e. the particle current density is

$$\displaystyle{ \Gamma = \frac{n <\!\! v\!\!>} {4}. }$$

(1.105)

1.2 A.1.2 Solution of Equation (1.15)

Let us start by multiplying both members of equation (1.15) by 2 dV∕dx and write it under the form

$$\displaystyle{ \frac{d} {dx}\left [\left (\frac{dV } {dx} \right )^{\!\!2}\right ] = 2\;C\;V ^{-1/2}\;\frac{dV } {dx}. }$$

(1.106)

This equation is easily integrated to get

$$\displaystyle{ \left (\frac{dV } {dx} \right )^{\!\!2} =\; 4\;C\;V ^{1/2} +\; K, }$$

(1.107)

in which K = 0 due to the condition at the cathode $E_{0} = (dV/dx)_{0} = 0$. We have therefore

$$\displaystyle{ \frac{dV } {dx} = 2\;\sqrt{C}\;V ^{1/4} }$$

(1.108)

and now this equation may be integrated again

$$\displaystyle{ \int _{0}^{V }V ^{-1/4}dV = 2\;\sqrt{C}\int _{ 0}^{x}dx }$$

(1.109)

allowing to obtain

$$\displaystyle{ \frac{V ^{3/4}} {3/4} = 2\;\sqrt{C}\;x, }$$

(1.110)

which at the end takes the form of equation (1.17)

$$\displaystyle{ V ^{3/2} = \frac{9} {4}\;C\;x^{2}. }$$

(1.111)

1.3 A.1.3 Multiplication Factor (1.30)

Each primary electron originates a multiplication α and in this case they arrive $(e^{\alpha d} - 1)/\alpha d$ electrons at the anode as given by equation (1.26). Thus they are created $[(e^{\alpha d} - 1)/\alpha d] - 1$ electron-ion pairs and these ions extract from the cathode $\gamma \;\{[(e^{\alpha d} - 1)/\alpha d] - 1\}$ secondary electrons. After the secondary electrons create more $\gamma \;\{[(e^{\alpha d} - 1)/\alpha d] - 1\}\;(e^{\alpha d} - 1)$ ions during their travel to the anode and these extract more $\gamma ^{2}\;\{[(e^{\alpha d} - 1)/\alpha d] - 1\}\;(e^{\alpha d} - 1)$ electrons from the cathode. The sum of all electrons entering the anode per each primary electron is

$$\displaystyle\begin{array}{rcl} m& =& 1 + \left (\frac{e^{\alpha d} - 1} {\alpha d} - 1\right ) +\;\gamma \; \left (\frac{e^{\alpha d} - 1} {\alpha d} - 1\right ) \\ & & +\;\gamma \;\left (\frac{e^{\alpha d} - 1} {\alpha d} - 1\right )\left (e^{\alpha d} - 1\right ) +\;\gamma ^{2}\;\left (\frac{e^{\alpha d} - 1} {\alpha d} - 1\right )\left (e^{\alpha d} - 1\right ) \\ & & +\;\gamma ^{2}\;\left (\frac{e^{\alpha d} - 1} {\alpha d} - 1\right )\left (e^{\alpha d} - 1\right )^{2} +\;\ldots \ldots \ldots \ldots {}\end{array}$$

(1.112)

This series may hence be written under the form

$$\displaystyle\begin{array}{rcl} m& =& \frac{e^{\alpha d} - 1} {\alpha d} \; +\; \left (\frac{e^{\alpha d} - 1} {\alpha d} - 1\right )\;\gamma \;e^{\alpha d}\left (1 +\;\gamma \; (e^{\alpha d} - 1) +\ldots \ldots.\right ) \\ & =& \frac{e^{\alpha d} - 1} {\alpha d} \; +\; \left (\frac{e^{\alpha d} - 1} {\alpha d} - 1\right )\;\gamma \;e^{\alpha d}\; \frac{1} {1 -\gamma \; (e^{\alpha d} - 1)},{}\end{array}$$

(1.113)

alloing to obtain at the end the multiplication factor

$$\displaystyle{ m = \frac{(1+\gamma )(e^{\alpha d} - 1)/\alpha d -\gamma \; e^{\alpha d}} {1 -\gamma \; (e^{\alpha d} - 1)}. }$$

(1.114)

1.4 A.1.4 Demonstration of Equation (1.59)

Defining the variable x as

$$\displaystyle{ x =\ln \left ( \frac{r} {r_{c}}\right ), }$$

(1.115)

we have

$$\displaystyle{ \frac{dx} {dr} = \frac{1} {r} }$$

(1.116)

and the first member of equation (1.55) writes as

$$\displaystyle{ \frac{d} {dr}\left (r\;\frac{dV } {dr} \right ) =\; \frac{1} {r}\;\frac{d^{2}V } {dx^{2}}. }$$

(1.117)

Using now the variable β ² defined by equation (1.58), equation (1.55) takes the form

$$\displaystyle{ \frac{d^{2}V } {dx^{2}} = \frac{4} {9}\;\frac{V } {\beta ^{2}}. }$$

(1.118)

Differentiating $\ln \beta ^{2}$, we obtain

$$\displaystyle{ \frac{d} {dx}\left (\ln \beta ^{2}\right ) = \frac{3} {2}\; \frac{1} {V }\;\frac{dV } {dx} \; - 1 }$$

(1.119)

and therefore

$$\displaystyle{ \frac{dV } {dx} = \frac{2} {3}\;V \left (1 + \frac{2} {\beta } \; \frac{d\beta } {dx}\right ). }$$

(1.120)

Differentiating again with respect to x, and using equation (1.120) to eliminate dV∕dx, we obtain the following equation

$$\displaystyle{ \frac{d^{2}V } {dx^{2}} = \frac{4} {9}\;V \left (1 + \frac{4} {\beta } \; \frac{d\beta } {dx}\; + \frac{1} {\beta ^{2}} \left ( \frac{d\beta } {dx}\right )^{\!\!2} + \frac{3} {\beta } \; \frac{d^{2}\beta } {dx^{2}}\right ) }$$

(1.121)

and substituting it in equation (1.118), we obtain at the end equation (1.59)

$$\displaystyle{ 1 =\beta ^{2} +\; 4\beta \; \frac{d\beta } {dx}\; + \left ( \frac{d\beta } {dx}\right )^{\!\!2} +\; 3\beta \; \frac{d^{2}\beta } {dx^{2}}. }$$

(1.122)

Exercises

Exercise 1.1.

An electrical potential difference V is applied to a plane parallel gap of width 2d which contains a constant uniformly distributed space-charge of density ρ between the plane x = 0 and a plane at distance d from it, while the remaining space is entirely free of space-charge (see Fig. 1.15). Determine the expressions for the potential and the electric field in the gap of width 2d (von Engel 1965).

Fig. 1.15

Plane parallel electrodes limiting a gap with positive ions of density ρ uniformly distributed in one-half of the interspace (von Engel 1965)

Resolution:

Assuming $\mathbf{E} = -\;E\;\mathbf{e_{x}}$, the Poisson’s equation writes $dE/dx = -\rho /\epsilon _{0}$, so that we obtain at the regions 0 < x < d and d < x < 2d, respectively,

$$\displaystyle\begin{array}{rcl} & & E_{1}(x) = -\;\frac{\rho } {\epsilon _{0}}\;x\; +\; C_{1} {}\\ & & E_{2}(x) = C_{2}, {}\\ \end{array}$$

while for the electric potential, we obtain from $E = dV/dx$

$$\displaystyle\begin{array}{rcl} & & V _{1}(x) = -\; \frac{\rho } {2\epsilon _{0}}\;x^{2}\; +\; C_{ 1}\;x {}\\ & & V _{2}(x) = C_{2}\;x\; +\; C_{3}, {}\\ \end{array}$$

having assumed V ₁(0) = 0.

Because the separation surface x = d is not electrically charged, we have E ₁ = E ₂ at the boundary, which together with the continuity of the potential, allows to write

$$\displaystyle\begin{array}{rcl} & & -\;\frac{\rho } {\epsilon _{0}}\;d\; +\; C_{1} = C_{2} {}\\ & & -\; \frac{\rho } {2\epsilon _{0}}\;d^{2}\; +\; C_{ 1}\;d = C_{2}\;d\; +\; C_{3}. {}\\ \end{array}$$

Because at the plane x = 2d, we have $V _{2}(2d) = C_{2}\;2d + C_{3} = V$, the electrical potential is

$$\displaystyle\begin{array}{rcl} & & V _{1}(x) = -\; \frac{\rho } {2\epsilon _{0}}\;x^{2} + \left ( \frac{V } {2d} + \frac{3} {4}\;\frac{\rho d} {\epsilon _{0}} \right )x {}\\ & & V _{2}(x) = \left ( \frac{V } {2d}\; -\; \frac{\rho d} {4\epsilon _{0}}\right )x + \frac{\rho d^{2}} {2\epsilon _{0}}, {}\\ \end{array}$$

while for the electrical field, we find

$$\displaystyle\begin{array}{rcl} & & E_{1}(x) = -\;\frac{\rho } {\epsilon _{0}}\;x + \left ( \frac{V } {2d} + \frac{3} {4}\;\frac{\rho d} {\epsilon _{0}} \right ) {}\\ & & E_{2} = \left ( \frac{V } {2d}\; -\; \frac{\rho d} {4\epsilon _{0}}\right ) = \mbox{ const}. {}\\ \end{array}$$

We note that E ₂ < 0 for ρ > 2ε ₀ V∕d ².

Exercise 1.2.

Consider two infinite plane parallel in a gas at pressure such that the mean free path of ions is much smaller than the electrode separation, $\lambda _{i} \ll d$. One electrode at x = d emits an unlimited number of positive ions, while the other at x = 0 emits electrons at the same rate. If the back scattering and the charge multiplication in the interelectrode spacing are neglected, determine the spatial distributions of the electric field, potential, and ion and electron densities.

Resolution:

Since $\lambda _{i} \ll d$ we must use a collision model and find the same situation as in equations (1.10) to (1.14). Assuming $\mathbf{E} = -\;E\;\mathbf{e_{x}}$ and $\mathbf{J} = -\;J\;\mathbf{e_{x}}$, with $J = J_{e} + J_{i}$, in order E and J are positive, Poisson’s equation writes as

$$\displaystyle{\frac{dE} {dx} = \frac{1} {\epsilon _{0}E}\left (\frac{J_{e}} {\mu _{e}} -\frac{J_{i}} {\mu _{i}} \right ).}$$

Because of μ _e  ≫ μ _i and $J_{i} = J_{e} = J/2$, we have approximately

$$\displaystyle{E\;\frac{dE} {dx} \; \simeq \;-\; \frac{J} {2\epsilon _{0}\mu _{i}}.}$$

This equation can be integrated now from x = d (where $n_{i} = \infty$ and E = 0) allowing to obtain the following expression for the electric field

$$\displaystyle{E(x) = \sqrt{ \frac{J} {\epsilon _{0}\mu _{i}}}\;\;(d - x)^{1/2},}$$

while for the potential, with V (0) = 0, we find

$$\displaystyle{V (x) = \frac{2} {3}\;\sqrt{ \frac{J} {\epsilon _{0}\mu _{i}}}\;\left [d^{3/2} - (d - x)^{3/2}\right ].}$$

Then, the densities of charged species are

$$\displaystyle\begin{array}{rcl} & & n_{i}(x) = \frac{J} {2e\mu _{i}E(x)} = \frac{1} {2e}\;\sqrt{\frac{\epsilon _{0 } J} {\mu _{i}}} \;(d - x)^{-1/2} {}\\ & & n_{e}(x) = \frac{J} {2e\mu _{e}E(x)} = \frac{\mu _{i}} {\mu _{e}}\;n_{i}(x). {}\\ \end{array}$$

Exercise 1.3.

Consider a gas uniformly irradiated with X-rays contained in a long concentric cylindre of radius r ₀ and r ₁, with an applied potential (V ₁ − V ₀) of such a value that multiplication of charges does not occur. Find the relation between the current per unity length i∕l and the potential difference (V ₁ − V ₀), assuming the density of electrons and ions uniformly distributed, with n _e  = n _i , the pressure high enough to use a collisional model, and neglecting space-charge field effects.

Resolution:

Assuming that there is no multiplication of charges and because of the large difference between the mobilities of electrons of ions, we have

$$\displaystyle{i/l = 2\pi r\;e\Gamma _{e}(r) = \mbox{ const},}$$

with $\Gamma _{e}(r) = n_{e}\mu _{e}\;E(r)$ denoting the electron particle current density. Since in the absence of space-charge field effects equation (1.38) yields to write

$$\displaystyle{r\;E(r) = \frac{(V _{1} - V _{0})} {\ln (r_{1}/r_{0})},}$$

we obtain

$$\displaystyle{i/l = 2\pi \;en_{e}\mu _{e}\;\frac{(V _{1} - V _{0})} {\ln (r_{1}/r_{0})}.}$$