Basics 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 d. The cathode is located at the point x = 0. To get
impact ionization, the electron energy E_e must become greater than the ionization energy E_\text{I} of the gas atoms between the plates. Per length of path x a number of \alpha ionizations will occur. \alpha is known as the first Townsend coefficient as it was introduced by Townsend. The increase of the electron current \Gamma_e, can be described for the assumed setup as {{NumBlk|:|\Gamma_e(x = d) = \Gamma_e(x = 0)\,e^{\alpha d}. |}} (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 d and/or \alpha, the more free electrons are created.) The number of created electrons is {{NumBlk|:|\Gamma_e(d) - \Gamma_e(0) = \Gamma_e(0) \left(e^{\alpha d} - 1\right). |}} Neglecting possible multiple ionizations of the same atom, the number of created ions is the same as the number of created electrons: {{NumBlk|:|\Gamma_i(0) - \Gamma_i(d) = \Gamma_e(0) \left(e^{\alpha d} - 1\right). |}} \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 where \gamma is the mean number of generated secondary electrons per ion. This is also known as the second Townsend coefficient. Assuming that \Gamma_i(d) = 0, one gets the relation between the Townsend coefficients by putting () into () and transforming: {{NumBlk|:|\alpha d = \ln\left(1 + \frac{1}{\gamma}\right). |}}
Impact ionization What is the amount of \alpha? The number of ionization depends upon the probability that an electron hits a gas molecule. This probability P is the relation of the
cross-sectional area of a collision between electron and ion \sigma in relation to the overall area A that is available for the electron to fly through: {{NumBlk|:|P = \frac{N\sigma}{A} = \frac{x}{\lambda}|}} 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 x to the
mean free path \lambda (distance at which another collision occurs). N is the number of molecules which electrons can hit. It can be calculated using the equation of state of the
ideal gas {{NumBlk|:|p V = N k_{B} T |}} :(p: pressure, V: volume, k_B:
Boltzmann constant, T: temperature) The adjoining sketch illustrates that \sigma = \pi (r_a + r_b)^2. As the radius of an electron can be neglected compared to the radius of an ion r_I it simplifies to \sigma = \pi r_I^2. Using this relation, putting () into () and transforming to \lambda one gets {{NumBlk|:|\lambda = \frac{k_{B}T}{p\pi r_{I}^{2}}=\frac{1}{L\cdot p} |}} where the factor L was only introduced for a better overview. The alteration of the current of not yet collided electrons at every point in the path x can be expressed as {{NumBlk|:|\mathrm{d}\Gamma_e(x) = -\Gamma_e(x)\,\frac{\mathrm{d}x}{\lambda_e} |}} This
differential equation can easily be solved: {{NumBlk|:|\Gamma_e(x) = \Gamma_e(0)\,\exp{\left(-\frac{x}{\lambda_e}\right)} |}} The probability that \lambda > x (that there was not yet a collision at the point x) is {{NumBlk|:|P(\lambda > x) = \frac{\Gamma_e(x)}{\Gamma_e(0)} = \exp{\left(-\frac{x}{\lambda_e}\right)}|}} According to its definition \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: {{NumBlk|:|\alpha = \frac{P(\lambda > \lambda_I)}{\lambda_e} = \frac{1}{\lambda_e}\exp\left(-\frac{\lambda_{I}}{\lambda_{e}}\right) = \frac{1}{\lambda_e}\exp\left(-\frac{E_{I}}{E_{e}}\right)|}} It was hereby considered that the energy E that a
charged particle can get between a collision depends on the
electric field strength \mathcal{E} and the charge Q: {{NumBlk|:|E = \lambda Q\mathcal{E}|}}
Breakdown voltage For the parallel-plate capacitor we have \mathcal{E} = \frac{U}{d}, where U is the applied voltage. As a single ionization was assumed Q is the
elementary charge e. We can now put () and () into () and get {{NumBlk|:|\alpha=L\cdot p\,\exp\left(-\frac{L\cdot p\cdot d\cdot E_{I}}{eU}\right)|}} Putting this into (5) and transforming to U we get the Paschen law for the breakdown voltage U_{\mathrm{breakdown}} that was first investigated by Paschen in and whose formula was first derived by Townsend in {{NumBlk|:|U_{\text{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 L = \frac{\pi r_{I}^{2}}{k_{B}T}
Plasma ignition Plasma ignition in the 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 d and ionize at least one atom on their way. So according to the definition of \alpha this relation must be fulfilled: If \alpha d = 1 is used instead of () one gets for the breakdown voltage {{NumBlk|:|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)}|}} == Conclusions, validity ==