Paschen's law; vacuum and high-pressure gas strength

Paschen's law states that the breakdown voltage of a uniform-field gap in a given gas depends only on the product of gas pressure $p$ and gap length $d$:

$$V_b = f(pd)$$

Derivation (outline)

Townsend's breakdown criterion is $\gamma(e^{\alpha d} - 1) = 1$. With $\alpha/p = A e^{-Bp/E}$ and $E = V_b/d$:

$$\begin{aligned} \alpha d &= \ln\left(1 + \frac{1}{\gamma}\right) \\ Apd\, e^{-Bpd/V_b} &= \ln\left(1 + \frac{1}{\gamma}\right) \\ V_b &= \frac{B\,pd}{\ln (A\,pd) - \ln\left[\ln\left(1 + \frac{1}{\gamma}\right)\right]} \end{aligned}$$

Since $A$, $B$ and $\gamma$ are gas constants, $V_b$ is a function of $pd$ only.

Paschen curve

 Vb
  |\                         /
  | \                      /
  |  \                  /
  |   \             /
  |    \_______ /
  |         min (about 327 V
  |          for air)
  +----------------------------> pd
     low pd    (pd)min    high pd

Setting $dV_b/d(pd) = 0$ gives a minimum: $(pd)_{min} = \frac{2.718}{A}\ln(1 + 1/\gamma)$. For air the minimum sparking voltage is about 327 V at $pd \approx 0.57$ torr·cm. No uniform-field air gap can break down below this voltage.

Air at normal pressure with practical gaps lies far to the right of this minimum. For example, a 1 cm gap at 760 torr has $pd = 760$ torr·cm, and breaks down at about 30 kV. If the pressure of such a gap is reduced, $V_b$ first falls until $(pd)_{min}$ is reached, and then rises steeply as the pressure falls further.

Why vacuum (low $pd$) is strong

On the left of the minimum the gas is very thin:

• The mean free path of electrons becomes comparable to or greater than the gap, so an electron crosses the gap making very few collisions.
• Few collisions means few ionizations ($\alpha d$ small), so the avalanche cannot build up.
• To get enough ionization a much higher voltage is needed, so $V_b$ rises steeply as $pd$ falls.

In high vacuum (below about $10^{-4}$ torr) there is almost no gas ionization at all; breakdown then depends on electrode effects (field emission, particle exchange, clumps), and strengths of the order of $10^5$–$10^6$ V/cm are obtained. This is why vacuum is used in vacuum circuit breakers.

Why high-pressure gas is strong

On the right of the minimum the gas is dense:

• The mean free path is very short, so an electron collides often but gains little energy between collisions (energy gained $= eE\lambda$).
• Most collisions are elastic or exciting, not ionizing, so $\alpha$ falls.
• A larger field is needed to give electrons ionizing energy over the short free path, so $V_b$ rises almost linearly with $pd$. Raising the pressure above 1 atm moves a gap further up this rising branch.

This is the basis of compressed-gas insulation (SF₆ at 3–6 bar in GIS, compressed nitrogen capacitors). At very high pressures (above about 10–20 bar) the rise becomes less than linear because electrode surface effects start to dominate.

Summary