---
title: "Townsend current growth with two coefficients"
subject: "High Voltage Engineering"
chapter: 6
key: "semester_8/high_voltage_engineering:townsend-two-coefficients"
source: https://www.bctengineer.com/q/semester_8/high_voltage_engineering/townsend-two-coefficients
site: BCT Engineering
---

# Townsend current growth with two coefficients

High Voltage Engineering · Chapter 6: Dielectric Breakdown

- Asked in: 2074 Bhadra, 6 marks
- Video: [Townsend's Current growth equation by considering Secondary Ionization Process| Derivation|HVE Lect](https://www.youtube.com/watch?v=PjB4uZT-k4o) — Simplified EEE Studies
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/high_voltage_engineering/townsend-two-coefficients
- Whole chapter: https://www.bctengineer.com/md/semester_8/high_voltage_engineering/chapter-6

## Question

Derive the Townsend's current growth equation based on two different ionization coefficients.

## Short answer

By considering both electron (α) and positive ion (β) ionization, and using n_x + p_x = n_d, the current growth equation is I = I0 (α-β) e^{(α-β)d} / [α - β e^{(α-β)d}]. Breakdown occurs when denominator zero, giving α = β e^{(α-β)d}.

## Answer

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What does beta represent in Townsend's original two-coefficient theory?
   A. Ionizing collisions per cm made by an electron
   B. Ionizing collisions per cm made by a positive ion
   C. Electrons released per ion at the cathode
   D. Attachment of electrons per cm
   Answer: B. alpha is for electrons, beta for positive ions. (Key point: Meaning of beta)

2. What is used to get p_x = n_d - n_x?
   A. Electrons are absorbed at the cathode
   B. Ions move faster than electrons
   C. There are no electrons at x = d
   D. Total current is the same at every plane and no ions exist at the anode
   Answer: D. At x = d only electrons cross, so n_x + p_x = n_d. (Key point: Boundary condition)

3. Which differential equation is solved?
   A. dn/dx = (alpha - beta) n + beta n_d
   B. dn/dx = alpha n only
   C. dn/dx = beta n - alpha n_d
   D. dn/dx = (alpha + beta) n
   Answer: A. New electrons come from electrons (alpha n_x) and ions (beta p_x). (Key point: Growth equation)

4. When does the current become infinite?
   A. When the numerator is zero
   B. When alpha = beta
   C. When the denominator alpha - beta e^((alpha - beta)d) = 0
   D. When d = 0
   Answer: C. This gives self-sustained discharge, roughly beta e^(alpha d) = alpha. (Key point: Breakdown condition)

## Related in High Voltage Engineering

- [Townsend current growth with secondary ionization](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Atownsend-current-growth-secondary) — High Voltage Engineering, chapter 6, asked 2×
- [Townsend's primary ionization coefficient (0.4 cm to 0.1 cm)](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Atownsend-alpha-0-1cm) — High Voltage Engineering, chapter 6, asked 6×
- [Townsend's primary ionization coefficient (0.4 cm to 0.1 mm)](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Atownsend-alpha-0-1mm) — High Voltage Engineering, chapter 6, asked 6×
- [Townsend breakdown condition with secondary ionization](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Atownsend-breakdown-condition) — High Voltage Engineering, chapter 6
- [Townsend's first and second coefficients from data](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Atownsend-first-second-coefficients) — High Voltage Engineering, chapter 6

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