# I Corona discharge current formula

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1. Aug 27, 2016

### G Cooke

I'm trying to find a formula for the current flowing through a wire that is only undergoing corona discharge into air. The wire is connected to a charged sphere, which provides the voltage.

So far, I've just found formulas for certain ionization coefficients and electric fields, but no current formulas. Is there such a formula? I imagine it would involve the voltage on the sphere, the resistance on the wire, and perhaps the thinness of the wire. Also, would it matter if the wire is bent rather than straight?

I am assuming that corona discharge will occur because the wire ends at a sharp point, but I have noticed that corona discharge experiments usually have the discharging point near an oppositely charged conductor. Is this necessary, and would the current be adversely affected if the discharging point were instead near a similarly charged conductor?

2. Aug 27, 2016

### Henryk

Corona discharge has been studied for year and still is poorly understood. Certainly, there is no universal formula that would give you the corona current.
Basically, corona discharge occurs when you have a point or small diameter wire at high voltage, around 3 kV. If you look at the field near a point you notice that it is very high near the point and drops off quickly with the distance from the point. If the field near the point is strong enough (around 30 kV/cm for dry air) you get dielectric breakdown that starts the corona discharge. Yes, the diameter of the point (or wire) does matter. The smaller diameter, the stronger the field near it and lower voltage is required to initiate the discharge. The distance to a counter electrode is less important because the field far away from the point is much weaker.
What is important though is to have a region of the electric field strong enough to cause the dielectric breakdown surrounded by a region of the field low enough to make sure that the breakdown does not occur. Otherwise, you get an arc, not corona discharge.

Another thing you need to start corona discharge is 'seed' electron, something supplied by, for instance, cosmic radiation in the region of the strong field.
The electron will gain energy and when colliding with a gas molecule will ionize it, that is produce another electron, then you have two electrons accelerating producing more and more electrons in an avalanche process.

The thing is, an average energy gained by electron in 30 kV/cm field is 0.15 eV and energy required to ionize air is around 15 eV, 100 times larger. The reality is the electron have to collide many times before it will actually ionize a neutral molecule. Therefore, the corona discharge is going to be strongly affected by even a small concentration of molecules that tend 'scavenge' free electrons.

Another thing is, the corona current is not constant. Once you start the avalanche process, you create a region of plasma which is conductive. The plasma will tend to screen the applied field and stop the discharge. Then the plasma neutralizes, the field is restored and the discharge current starts again. This is known as Tichler oscillations.
Then, there is a difference between negative and positive corona. It has to do with how you get more electrons.
Yes, the corona discharge is a very complex process and there is no simple formula.

3. Aug 28, 2016

### G Cooke

Thank you for your very informative reply, Henryk.
It looks like you thought I was referring to the discharge current. Of course the discharge current couldn't be constant because it is an avalanche (and apparently oscillates). I was actually referring to the current inside the wire, i.e., pre-avalanche discharge.

I found this recent lab report on corona discharge, which shows that it is possible to obtain linear I-V characteristics using tungsten-coated ZnO nanowires. See Fig. 5 for the setup and Fig. 8 for the results showing the linear I-V characteristics measured by the picoammeter:

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC4754239/
I am curious about positive corona discharge. It seems like if my setup with sphere and wire were to undergo positive corona discharge, it would mean that the wire is "sucking in" electrons from the air rather than actually discharging them. Would a positively charged sphere discharge* at the same rate as an equally negatively charged sphere via corona discharge?

*Note that I am again referring to the rate of "discharge" inside the wire, not the actual corona discharge, which occurs in the air.

4. Aug 28, 2016

### G Cooke

I feel that a few points of clarification are necessary:

1) My question does not imply that I am expecting the current through the wire to be constant. Here is some insight into the basis of my question:

I know that the voltage function for a discharging capacitor is V(t) = Vo*e^-t/RC, where Vo is the initial voltage, R is the resistance through which it is discharging, and C is the capacitance. For a sphere, the capacitance is r/k, where r is the radius and k is Coulomb's constant. Therefore, a charged sphere connected to the ground by a wire would discharge voltage according to the function V(t) = Vo*e^-kt/Rr. My question arises from the following curiosity: What if the sphere is not discharging into the ground, but rather into the air via corona discharge? What would the voltage function be then? I asked about current I(t) with the intent to then find this voltage function V(t) via Ohm's Law.

2) Note that the purpose of the experiment in the lab report I linked to in the above post was not to obtain linear I-V characteristics, but rather to minimize the voltage required to induce corona discharge, which the scientists accomplished by using tungsten-coated ZnO nanowires. I wonder, then, whether the I-V characteristics would have been linear regardless of the material of the discharger (but that is irrelevant to my question).

3) I am actually more curious now about Townsend discharge than corona discharge. It appears from Fig. 6 in the link the above post that Townsend discharge also exhibits a region of linear (ohmic) I-V characteristics, but starting from a much lower voltage (near zero from the looks of it). My present question is then: What would be the discharging voltage function of a sphere undergoing Townsend discharge into air, and would it actually undergo Townsend discharge even though it is merely a charged sphere and not, as in the experimental setup, a battery?

This is all assuming, of course, that the voltage function of a capacitor discharging by conventional means (into the ground, etc.) does not simply extend to a capacitor discharging via one of these gas discharge processes.

5. Sep 1, 2016

### Henryk

For all practical purpose, the corona discharge current is equal to the current in the wire.

No.
It has to do with the generation of more electrons to maintain the avalanche. You can read more about this in
https://en.wikipedia.org/wiki/Corona_discharge#Mechanism

This is correct, however, if you plug in numbers, a 1 mm sphere has a capacitance of 0.1 pF!!. This is very small and time constants are going to be very short. However, in the case of corona discharge you have a capacitance between the corona point and the conductive plasma surrounding the point. This capacitance is much larger and that's the one that matters.

Ohm's law is actually an approximation. It is valid only if the electric field is small enough that affects the motion of electrons (or other charge carriers) linearly with the applied field. It is valid in metals, where the intrinsic electron velocities are very high, several orders of magnitude higher that an electron can gain due to applied field before losing it all to collision. But the Ohm's is not obeyed in certain semiconductors in high field regime. It definitely does not apply to corona discharge.

6. Sep 6, 2016

### G Cooke

I see. I had a false understanding of the corona discharge process, but the Wikipedia article on it cleared things up.

I guess it makes sense that a capacitance that has to do with the charges in the air is more applicable than the sphere's capacitance since corona discharge is apparently more of an external (to the electrode) process that is merely enabled by the electrode rather than the electrode actually discharging like a capacitor (as I had previously thought). Could you point me towards getting an idea of the value of this plasma capacitance? Even just an estimate or a keyword that would return useful search results is fine ("corona discharge plasma capacitance" didn't return what I'm looking for).

Separately, it seems from your statement that the plasma capacitance is independent of the electrode's capacitance, so a large enough electrode capacitance would overpower the effects of the plasma capacitance.

So I suppose no one has done an experiment where corona or Townsend discharge was repeatedly induced while varying parameters such as electrode starting voltage and capacitance and V(t) was recorded for each case and then the data was consolidated into a multivariate function of V...

And I take it that Townsend discharge is just as poorly understood as corona discharge...