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Limitations of Townsend theory; Paschen's law

Short answer

Townsend's theory fails at high pd: it predicts large time lags, cathode independence, diffuse glow, and neglects space charge, not explaining streamers. Paschen's law V_b = B pd / ln[Apd / ln(1 + 1/γ)] is derived by substituting α = Ap e^{-Bpd/Vb} into αd = ln(1+1/γ), giving breakdown voltage as a function of pd only.

What are the limitations of Townsend's theory? Derive the relation given by Paschen law.

Limitations of Townsend's theory

Townsend's theory (avalanche growth by α\alpha, sustained by cathode secondary emission γ\gamma, breakdown when γeαd=1\gamma e^{\alpha d} = 1) explains breakdown well at low pdpd, but fails in several respects at higher pdpd (above about 1000 torr·cm, e.g. long gaps in atmospheric air):

  1. Time lag: Townsend predicts formative times of the order of 10−510^{-5} s (time for ions to cross the gap many times). Observed times in long gaps are about 10−810^{-8}–10−710^{-7} s, far too short for ion transit.
  2. Cathode independence: At high pressures breakdown voltage is nearly independent of cathode material, which should not happen if γ\gamma at the cathode controls breakdown.
  3. Channel shape: Townsend predicts a diffuse glow-type discharge, but actual sparks at high pdpd are thin, bright, branched and zig-zag channels.
  4. Space charge ignored: The theory neglects the distortion of the field by the space charge of the avalanche, which becomes significant when the avalanche contains about 10810^8 electrons.
  5. Non-uniform fields: It does not explain corona and breakdown in non-uniform fields, or polarity effects.
  6. Breakdown with no cathode involvement: Discharges can start mid-gap and grow towards both electrodes, which Townsend's mechanism cannot explain.

These limitations led to the streamer (Raether–Meek) theory for high pdpd.

Derivation of Paschen's law

Townsend's breakdown criterion:

γ(eαd−1)=1  ⇒  αd=ln⁡(1+1γ)(1)\gamma\left(e^{\alpha d} - 1\right) = 1 \;\Rightarrow\; \alpha d = \ln\left(1 + \frac{1}{\gamma}\right) \qquad (1)

Experimentally, the first coefficient depends on E/pE/p:

αp=A e−Bp/E(2)\frac{\alpha}{p} = A\, e^{-Bp/E} \qquad (2)

where AA and BB are gas constants. In a uniform field E=V/dE = V/d, so at breakdown E=Vb/dE = V_b/d:

α=Ap e−Bpd/Vb(3)\alpha = Ap\, e^{-Bpd/V_b} \qquad (3)

Substitute (3) in (1):

Apd e−Bpd/Vb=ln⁡(1+1γ)=Ke−Bpd/Vb=KApdBpdVb=ln⁡ApdKVb=B pdln⁡(A pd)−ln⁡K\begin{aligned} Apd\, e^{-Bpd/V_b} &= \ln\left(1 + \frac{1}{\gamma}\right) = K \\ e^{-Bpd/V_b} &= \frac{K}{Apd} \\ \frac{Bpd}{V_b} &= \ln\frac{Apd}{K} \\ V_b &= \frac{B\,pd}{\ln (A\,pd) - \ln K} \end{aligned} Vb=B (pd)ln⁡[A (pd)ln⁡(1+1/γ)]=f(pd)\boxed{V_b = \frac{B\,(pd)}{\ln\left[\dfrac{A\,(pd)}{\ln(1 + 1/\gamma)}\right]} = f(pd)}

Since AA, BB and γ\gamma are constants of the gas (and electrode), the breakdown voltage of a uniform-field gap is a function of the product pdpd only. This is Paschen's law. For air at temperature other than 20 °C, pp is replaced by gas density (NdNd or δd\delta d).

Minimum sparking voltage

Setting dVbd(pd)=0\dfrac{dV_b}{d(pd)} = 0:

(pd)min=eAln⁡(1+1γ),Vb,min=eBAln⁡(1+1γ)(pd)_{min} = \frac{e}{A}\ln\left(1 + \frac{1}{\gamma}\right), \qquad V_{b,min} = \frac{eB}{A}\ln\left(1 + \frac{1}{\gamma}\right)

(e=2.718e = 2.718.) For air, Vb,min≈327V_{b,min} \approx 327 V at pd≈0.57pd \approx 0.57 torr·cm.

 Vb
  |\                      /
  | \                  /
  |  \             /
  |   \_______ /
  |      min ~327 V (air)
  +---------------------------> pd

Left of the minimum, too few collisions; right of the minimum, too little energy per collision; in both cases VbV_b rises.

✓ Answer checked by Claude Opus 5.5 (high effort)

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