The ionization degree of a plasma is given by the Saha equation, which depends on the temperature and the particle specific ionization energy. In thermal equilibrium, the relation between ionization and its electric potential is given by the Boltzmann relation.

It is common practice to induce a plasma via electric voltage (for example in Neon lamps), and there are also configurations with periodically applied electric fields, such as the dielectric barrier discharge.

enter image description here

From what I understand, the idea is to induce polarization and consequently electrical breakdown of the gas (how does this work?), and raise the temperature via an electrical arc (is this already the plamsa?).

My question is:

In artificially generated plasmas what is the quantitative relation between the applied electric field/voltage and the ionization?

I'd also be interested in the time dependence of the plasma on the applied field frequency.

  • $\begingroup$ I don't know a much about plasma, but I have the feeling that the answers to your questions could fill several books. $\endgroup$ – user2963 Feb 28 '12 at 18:48

And indeed it fills books. Look Y.P. Raizer, Gas Discharge physics. It will answer everything you would ever need.

To make a short answer, the Saha equation works well for plasmas at thermal equilibrium. These plasmas you call "artificial plasmas" are gas discharge plasmas, and they are non-equilibrium. While the gas has almost the same temperature as it had before the discharge took place, the electrons are "very hot", tens of thousands of °K. This can be because the degree of ionization (density of electrons over density of neutrals) of such plasmas is very weak, say 10^-7 in Neon tubes, AND electrons lose only a very small fraction of their kinetic energy in each collision, due to their very small mass compared to the mass of ions and neutrals (in plasmas physics a lot of things are due to this assimetry). This means that, both the power lost to gas heating is low, and electrons need to be very fast before the loss due to each collision compensate for the energy they gain through the work of the external electric field, which depends only on the mean free path.

Electric breakdown works this way: there are seed electrons in any gas, even if their density is ridiculously small (like 1000 in a cubic meter, in air, due to cosmic rays). These electrons are accelerated by the external electric field. When the mean kinetic energy of the electrons starts to be comparable to the energy of atomic levels, the electrons will excite atoms, and if you rise the field further they will ionize atoms. This way, electrons will multiply, creating the electron density of the plasma. Then, stabilizing mechanisms take place (look in the book I mentionned above), and a stable "glow discharge" occurs, when the conditions are adequate (low reasonable currents, low density (compared to atmospheric pressure), and adequate voltage (see Paschen's curve)).

Now, if you increase the voltage, the current will rise a lot, then rise slowly (normal and abnormal glow discharge). When the metal plates inside the discharge tube (the cathode and anode) become too hot, they radiate electrons much more easily, too easily. This increases the current a lot while weakening the needed electric field. And at that moment, the concentration in electrons, and various phenomena, make possible the transfer of kinetic energy between electrons and the rest of the gas, heating it to a temperature comparable to the electron temperature: several thousands of °K. This is the "arc discharge", and this is what happens in sparks used to ignite motors: glow discharges, immediately transformed in an "arc", violently heating the gas inside it.

glow discharge ("stable, cold discharges") plasmas can indeed be sustained by DC or RF power. But dielectric barrier discharges are very particular types of plasmas. First, in their case, the current never closes in continuous regime (due to the dielectric barrier preventing the current from flowing): it is only a matter of displacement current, dE/dt. And second, their development are closely influenced by the presence, shape, and properties (dielectric constant, and time response), of the surface on which they develop.

You came to gas discharge plasmas from the "wrong" end (from the high temperature physics end), I hope this way you can understand it better :-)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.