Paschen's law

For air at standard conditions for temperature and pressure (STP), the voltage needed to arc a 1-metre gap is about 3.4 MV.[7] The intensity of the electric field for this gap is therefore 3.4 MV/m.

The electric field needed to arc across the minimal-voltage gap is much greater than what is necessary to arc a gap of one metre. For a 7.5 μm gap the arc voltage is 327 V, which is 43 MV/m. This is about 13 times greater than the field strength for the 1-metre gap. The phenomenon is well verified experimentally and is referred to as the Paschen minimum.

The equation loses accuracy for gaps under about 10 μm in air at one atmosphere[8] and incorrectly predicts an infinite arc voltage at a gap of about 2.7 micrometres. Breakdown voltage can also differ from the Paschen curve prediction for very small electrode gaps, when field emission from the cathode surface becomes important.

To calculate the breakthrough voltage, a homogeneous electrical field is assumed. This is the case in a parallel-plate capacitor setup. The electrodes may have the distance ${\displaystyle d}$. The cathode is located at the point ${\displaystyle x=0}$.

To get impact ionization, the electron energy ${\displaystyle E_{e}}$ must become greater than the ionization energy ${\displaystyle E_{\text{I}}}$ of the gas atoms between the plates. Per length of path ${\displaystyle x}$ a number of ${\displaystyle \alpha }$ ionizations will occur. ${\displaystyle \alpha }$ is known as the first Townsend coefficient as it was introduced by Townsend.[10] The increase of the electron current ${\displaystyle \Gamma _{e}}$, can be described for the assumed setup as

${\displaystyle \Gamma _{e}(x=d)=\Gamma _{e}(x=0)\,e^{\alpha d}.\qquad \qquad (1)}$

(So the number of free electrons at the anode is equal to the number of free electrons at the cathode that were multiplied by impact ionization. The larger ${\displaystyle d}$ and/or ${\displaystyle \alpha }$, the more free electrons are created.)

The number of created electrons is

${\displaystyle \Gamma _{e}(d)-\Gamma _{e}(0)=\Gamma _{e}(0)\left(e^{\alpha d}-1\right).\qquad \qquad (2)}$

Neglecting possible multiple ionizations of the same atom, the number of created ions is the same as the number of created electrons:

${\displaystyle \Gamma _{i}(0)-\Gamma _{i}(d)=\Gamma _{e}(0)\left(e^{\alpha d}-1\right).\qquad \qquad (3)}$

${\displaystyle \Gamma _{i}}$ is the ion current. To keep the discharge going on, free electrons must be created at the cathode surface. This is possible because the ions hitting the cathode release secondary electrons at the impact. (For very large applied voltages also field electron emission can occur.) Without field emission, we can write

${\displaystyle \Gamma _{e}(0)=\gamma \Gamma _{i}(0),\qquad \qquad (4)}$

where ${\displaystyle \gamma }$ is the mean number of generated secondary electrons per ion. This is also known as the second Townsend coefficient. Assuming that ${\displaystyle \Gamma _{i}(d)=0}$, one gets the relation between the Townsend coefficients by putting (4) into (3) and transforming:

${\displaystyle \alpha d=\ln \left(1+{\frac {1}{\gamma }}\right).\qquad \qquad (5)}$

What is the amount of ${\displaystyle \alpha }$? The number of ionization depends upon the probability that an electron hits a gas molecule. This probability ${\displaystyle P}$ is the relation of the cross-sectional area of a collision between electron and ion ${\displaystyle \sigma }$ in relation to the overall area ${\displaystyle A}$ that is available for the electron to fly through:

${\displaystyle P={\frac {N\sigma }{A}}={\frac {x}{\lambda }}\qquad \qquad (6)}$

As expressed by the second part of the equation, it is also possible to express the probability as relation of the path traveled by the electron ${\displaystyle x}$ to the mean free path ${\displaystyle \lambda }$ (distance at which another collision occurs).

Visualization of the cross-section ${\displaystyle \sigma }$: If the center of particle b penetrates the blue circle, a collision occurs with particle a. So the area of the circle is the cross-section and its radius ${\displaystyle r}$ is the sum of the radii of the particles.

${\displaystyle N}$ is the number of molecules which electrons can hit. It can be calculated using the equation of state of the ideal gas

${\displaystyle pV=Nk_{B}T\qquad \qquad (7)}$

(${\displaystyle p}$: pressure, ${\displaystyle V}$: volume, ${\displaystyle k_{B}}$: Boltzmann constant, ${\displaystyle T}$: temperature)

The adjoining sketch illustrates that ${\displaystyle \sigma =\pi (r_{a}+r_{b})^{2}}$. As the radius of an electron can be neglected compared to the radius of an ion ${\displaystyle r_{I}}$ it simplifies to ${\displaystyle \sigma =\pi r_{I}^{2}}$. Using this relation, putting (7) into (6) and transforming to ${\displaystyle \lambda }$ one gets

${\displaystyle \lambda ={\frac {k_{B}T}{p\pi r_{I}^{2}}}={\frac {1}{L\cdot p}}\qquad \qquad (8)}$

where the factor ${\displaystyle L}$ was only introduced for a better overview.

The alteration of the current of not yet collided electrons at every point in the path ${\displaystyle x}$ can be expressed as

${\displaystyle \mathrm {d} \Gamma _{e}(x)=-\Gamma _{e}(x)\,{\frac {\mathrm {d} x}{\lambda _{e}}}\qquad \qquad (9)}$

This differential equation can easily be solved:

${\displaystyle \Gamma _{e}(x)=\Gamma _{e}(0)\,\exp {\left(-{\frac {x}{\lambda _{e}}}\right)}\qquad \qquad (10)}$

The probability that ${\displaystyle \lambda >x}$ (that there was not yet a collision at the point ${\displaystyle x}$) is

${\displaystyle P(\lambda >x)={\frac {\Gamma _{e}(x)}{\Gamma _{e}(0)}}=\exp {\left(-{\frac {x}{\lambda _{e}}}\right)}\qquad \qquad (11)}$

According to its definition ${\displaystyle \alpha }$ is the number of ionizations per length of path and thus the relation of the probability that there was no collision in the mean free path of the ions, and the mean free path of the electrons:

${\displaystyle \alpha ={\frac {P(\lambda >\lambda _{I})}{\lambda _{e}}}={\frac {1}{\lambda _{e}}}\exp \left({\mbox{-}}{\frac {\lambda _{I}}{\lambda _{e}}}\right)={\frac {1}{\lambda _{e}}}\exp \left({\mbox{-}}{\frac {E_{I}}{E_{e}}}\right)\qquad \qquad (12)}$

It was hereby considered that the energy ${\displaystyle E}$ that a charged particle can get between a collision depends on the electric field strength ${\displaystyle {\mathcal {E}}}$ and the charge ${\displaystyle Q}$:

${\displaystyle E=\lambda Q{\mathcal {E}}\qquad \qquad (13)}$

For the parallel-plate capacitor we have ${\displaystyle {\mathcal {E}}={\frac {U}{d}}}$, where ${\displaystyle U}$ is the applied voltage. As a single ionization was assumed ${\displaystyle Q}$ is the elementary charge ${\displaystyle e}$. We can now put (13) and (8) into (12) and get

${\displaystyle \alpha =L\cdot p\,\exp \left({\mbox{-}}{\frac {L\cdot p\cdot d\cdot E_{I}}{eU}}\right)\qquad \qquad (14)}$

Putting this into (5) and transforming to ${\displaystyle U}$ we get the Paschen law for the breakdown voltage ${\displaystyle U_{\mathrm {breakdown} }}$ that was first investigated by Paschen in[4] and whose formula was first derived by Townsend in,[11] section 227:

${\displaystyle U_{\mathrm {breakdown} }={\frac {L\cdot p\cdot d\cdot E_{I}}{e\left(\ln(L\cdot p\cdot d)-\ln \left(\ln \left(1+\gamma ^{-1}\right)\right)\right)}}\qquad \qquad (15)}$

with ${\displaystyle \textstyle L={\frac {\pi r_{I}^{2}}{k_{B}T}}}$

Plasma ignition in definition of Townsend (Townsend discharge) is a self-sustaining discharge, independent of an external source of free electrons. This means that electrons from the cathode can reach the anode in the distance ${\displaystyle d}$ and ionize at least one atom on their way. So according to the definition of ${\displaystyle \alpha }$ this relation must be fulfilled:

${\displaystyle \alpha d\geq 1\qquad \qquad (16)}$

If ${\displaystyle \alpha d=1}$ is used instead of (5) one gets for the breakdown voltage

${\displaystyle U_{\mathrm {breakdown\,Townsend} }={\frac {L\cdot p\cdot d\cdot E_{I}}{e\cdot \ln(L\cdot p\cdot d)}}={\frac {d\cdot E_{I}}{e\cdot \lambda _{e}\,\ln \left({\frac {d}{\lambda _{e}}}\right)}}\qquad \qquad (17)}$