Townsend current growth with two coefficients

In Townsend's theory the current in a gas grows by two ionization processes, each described by a coefficient. In Townsend's original form these are:

• $\alpha$ = number of ionizing collisions made by an electron per cm of travel (first coefficient)
• $\beta$ = number of ionizing collisions made by a positive ion per cm of travel (second coefficient)

Derivation

Let $n_0$ electrons per second leave the cathode, $n_x$ = electrons crossing plane $x$, and $p_x$ = positive ions crossing plane $x$ (moving towards the cathode). In steady state the total current is the same at every plane. At the anode ($x = d$) there are no positive ions, so

$$n_x + p_x = n_d \;\Rightarrow\; p_x = n_d - n_x$$

In length $dx$, new electrons are created by electrons and by ions:

$$\begin{aligned} \frac{dn_x}{dx} &= \alpha n_x + \beta p_x = \alpha n_x + \beta(n_d - n_x) \\ &= (\alpha - \beta)\,n_x + \beta n_d \end{aligned}$$

Solving this linear equation with $n_x = n_0$ at $x = 0$:

$$n_x = \left[n_0 + \frac{\beta n_d}{\alpha - \beta}\right]e^{(\alpha - \beta)x} - \frac{\beta n_d}{\alpha - \beta}$$

Put $x = d$:

$$\begin{aligned} n_d\left[1 + \frac{\beta}{\alpha - \beta} - \frac{\beta\,e^{(\alpha - \beta)d}}{\alpha - \beta}\right] &= n_0\, e^{(\alpha - \beta)d} \\ n_d\,\frac{\alpha - \beta\,e^{(\alpha - \beta)d}}{\alpha - \beta} &= n_0\, e^{(\alpha - \beta)d} \end{aligned}$$ $$\boxed{I = I_0\,\frac{(\alpha - \beta)\,e^{(\alpha - \beta)d}}{\alpha - \beta\,e^{(\alpha - \beta)d}}}$$

Breakdown condition

The current becomes infinite (self-sustained discharge) when the denominator is zero:

$$\alpha = \beta\,e^{(\alpha - \beta)d}$$

Since $\beta \ll \alpha$, this is approximately $\beta\,e^{\alpha d} = \alpha$.

Modern form ($\alpha$ and $\gamma$)

Positive ions in fact have too little energy to ionize gas, so the second process is now taken as electron release from the cathode by ions, photons and metastables, with coefficient $\gamma$. The same steps give

$$I = \frac{I_0\, e^{\alpha d}}{1 - \gamma\left(e^{\alpha d} - 1\right)}, \qquad \text{breakdown: } \gamma\left(e^{\alpha d} - 1\right) = 1$$

Both forms show the current rising faster than $e^{\alpha d}$ as the gap or voltage increases, ending in breakdown.