Current Path for Corona Discharge

[KyleGranger]

TL;DR

How does current flow with corona?

I am trying to understand the current path for corona. Assuming the corona appears on the high voltage electrode, how are the electrons actually flowing? I understand that with flashover a low resistance current path is created as the air is ionized but what is happening with corona?

 

[KyleGranger]

Like with this plasma globe, where is the current going?

 

[Baluncore]

Corona is where the electric field gradient near a tight radius, is so steep that it tears electrons from the air molecules. Recombination of the ions emit light. But the field gradient away from that point is less, so a flash-over arc is not supported. Air pressure is important.
https://en.wikipedia.org/wiki/Corona_discharge

 

[KyleGranger]

As that article states, "It represents a local region where the air (or other fluid) has undergone electrical breakdown and become conductive". My question is where does the current conduct to? It's easy enough to make general statements but understanding what's truly happing is tougher. So where does the current flow? If flashover is sort of an "amplified" corona that completely ionizes a channel between electrodes allowing current to flow between them, where is the current flowing without flashover? Because corona causes leakage in the system. It also can cause EMI issues so there is current flowing. I'm not understanding how. Maybe most others don't either and we just accept that it does?

 

[KyleGranger]

So with the image above, I believe the air is being ionized and emitting light in that particular region. It terminates at the glass so the channel stops there. Since we know that corona can cause EMI issues, what is the current path? If I had sensitive electronics on the other side of the glass, what would the coupling paths be?

 

[Baluncore]

The local voltage gradient is reduced by the radius of the corona discharge. The circuit is closed within the corona where the ion charges recombine to be neutral, with the emission of light and heat.
https://en.wikipedia.org/wiki/Paschen's_law

If the gas pressure is very low, and the available current very much limited, there may be sufficiently high resistance in the gas to maintain the threads of arc discharge.

 

[KyleGranger]

The local voltage gradient is reduced by the radius of the corona discharge. The circuit is closed within the corona where the ion charges recombine to be neutral, with the emission of light and heat.
https://en.wikipedia.org/wiki/Paschen's_law

If the gas pressure is very low, and the available current very much limited, there may be sufficiently high resistance in the gas to maintain the threads of arc discharge.

If I have two electrodes and start to see corona on one, there is current flow. So I'm getting from this the current isn't flowing back to ground, it's just "disappearing" in thin air as the electrons are combining with air molecules. First of all, if the molecules were originally neutral, combining with an electron would leave them charged. Maybe that doesn't matter for this discussion though.

Now let's say we have the same scenario where we have two electrodes and the field intensity is large enough to cause corona. If we increase it further, we'll eventually have flashover. But if we add an insulator, let's say glass, between, there will be no flashover and still only corona. That's because the glass has a much higher breakdown voltage than air and the localized breakdown stops at the glass. But I'm still having a hard time understanding what's actually happening. If I have sensitive electronics on the other side of the glass, are there the potential for EMI issues? if I still have the potential for EMI issues, what is the coupling path and how does the current flow from the corona in air to the device?

 

[KyleGranger]

The local voltage gradient is reduced by the radius of the corona discharge. The circuit is closed within the corona where the ion charges recombine to be neutral, with the emission of light and heat.
https://en.wikipedia.org/wiki/Paschen's_law

I understand Paschen's law. Basically, the probability of an electron being ejected, gaining enough velocity, and having enough probability of striking other molecules that can eject more electrons causing avalanche are a function of pressure, distance, and electric field. if the pressure is low enough, there may be enough velocity but the probability of having other molecules in the vicinity are lower so the critical electric field is higher. if the pressure is high but the distance is low, it may still gain enough velocity and have high enough probability of striking other molecules to cause avalanche. It's a function that's really dependent on 3 variables, e-field and pressure*distance. I get that but I don't think that's relevant in the discussion of the current path when corona does occur.

 

[Baluncore]

First of all, if the molecules were originally neutral, combining with an electron would leave them charged. Maybe that doesn't matter for this discussion though.

They were neutral. They were then ionised by the local voltage gradient. The ions then move to where they recombine, so they are neutral again.

To understand corona you must study the field gradient about a point source distant from a ground plate, not between parallel plates.

If I have sensitive electronics on the other side of the glass, are there the potential for EMI issues? if I still have the potential for EMI issues, what is the coupling path and how does the current flow from the corona in air to the device?

The current that flows generates a magnetic field. That is radiated as a dipole field about each and every filament of current. The fast rise and fall of the current in the filaments cause harmonics throughout the radio spectrum.

The fast rise and fall of the current in the corona filaments also cause expansion and contraction of air that is radiated as an ultrasonic sound wave. For corona on AC power lines you can hear the thermal effects at harmonics of the line frequency.

 

[KyleGranger]

They were neutral. They were then ionised by the local voltage gradient. The ions then move to where they recombine, so they are neutral again.
To understand corona you must study the field gradient about a point source distant from a ground plate, not between parallel plates.

The current that flows generates a magnetic field. That is radiated as a dipole field about each and every filament of current. The fast rise and fall of the current in the filaments causeharmonics throughout the radio spectrum.

The fast rise and fall of the current in the corona filaments also cause expansion and contraction of air that is radiated as an ultrasonic sound wave. For corona on AC power lines you can hear the thermal effects at harmonics of the line frequency.

So the coupling is purely radiated? That's part of what I was trying to understand. If I have corona on a wire in air and it's causing EMI issues to nearby electronics, I was wondering if placing a barrier would reduce the EMI (my thoughts were it would not). If it's purely radiated, the effect would be minimal. It would only be effected by the slight change in velocity while propagating through the different medium of the barrier (assuming nonconductive).

I know that corona occurs in the UV spectrum which has a very high frequency so figured the primary coupling source would be radiated. Is it reasonable to completely eliminate conducted (capacitive, inductive, conduction)?

 

[Baluncore]

If I have corona on a wire in air and it's causing EMI issues to nearby electronics, I was wondering if placing a barrier would reduce the EMI (my thoughts were it would not).

If corona happens at the end of a wire, attach a wire frame or conductive ball at the end of the wire.

If corona happens on the length of a wire, use a thicker wire, a thin wall tube, or a spaced cage of three or more parallel connected wires. That will give the electric field a greater radius of curvature, which should prevent corona.

I know that corona occurs in the UV spectrum which has a very high frequency so figured the primary coupling source would be radiated. Is it reasonable to completely eliminate conducted (capacitive, inductive, conduction)?

EMI due to a nearby corona will be capacitively and/or inductively coupled.

EMI due to a distant corona will be like any other radio interference.

Electronic devices should be enclosed to protect them from EMI and static discharge. A grounded conductive foil should be more than sufficient to stop conducted interference.

 

[KyleGranger]

If corona happens at the end of a wire, attach a wire frame or conductive ball at the end of the wire.

If corona happens on the length of a wire, use a thicker wire, a thin wall tube, or a spaced cage of three or more parallel connected wires. That will give the electric field a greater radius of curvature, which should prevent corona.EMI due to a nearby corona will be capacitively and/or inductively coupled.

EMI due to a distant corona will be like any other radio interference.

Electronic devices should be enclosed to protect them from EMI and static discharge. A grounded conductive foil should be more than sufficient to stop conducted interference.

Thanks for the tips. I understand how to get rid of corona (larger radius), but I'm just trying to understand what's actually happening when it does occur. I understand attaching conductive balls at the end of wires, or even using acorn nuts on screws, can help. But when it does occur, how would I draw that equivalent circuit?

You mentioned that nearby corona will be capacitively and/or inductively coupled and distant will be like any other radio interference. So I guess the distance depends on frequency corona (UV spectrum) and the distance before it can be considered a plane wave. With the frequency of corona being in the UV spectrum, that distance is very very small so that would mean practically that any coupling would be primarily radiated, right?

 

[Baluncore]

So I guess the distance depends on frequency corona (UV spectrum) and the distance before it can be considered a plane wave.

I was not talking about UV, which is like light and does not cause EMI to electronic devices.
UV is specified in wavelength.

I was referring to EMI = RF; where frequency is usually specified.
Antenna terms; Near-field is a confusion of waves that goes out to about 60 λ.
Beyond that is far-field, where a plane wave has formed.

But when it does occur, how would I draw that equivalent circuit?

Corona discharge is usually not on any RF schematic circuit. Maybe you could represent it with a broad band noise source, like a gas discharge tube.
For UV, you could write up the chemistry of the molecular ionisation and recombination, with the emission of photons.

 

[KyleGranger]

I was not talking about UV, which is like light and does not cause EMI to electronic devices.
UV is specified in wavelength.

I was referring to EMI = RF; where frequency is usually specified.
Antenna terms; Near-field is a confusion of waves that goes out to about 60 λ.
Beyond that is far-field, where a plane wave has formed.Corona discharge is usually not on any RF schematic circuit. Maybe you could represent it with a broad band noise source, like a gas discharge tube.
For UV, you could write up the chemistry of the molecular ionisation and recombination, with the emission of photons.

When corona occurs, the frequency is in the UV spectrum. Thinking about it, since we can also hear it, does that mean it's also in the lower frequency range as well?

WHen I'm referring to EMI, I'm questioning what all should be considered. I understand that corona isn't uslly represented on a schematic but I have a feeling that's because systems are usually designed to not have corona in the first place. Again, there is no denying that if corona is present in a system the system may behave different than intended. I was hoping to find out how to represent the circuit while modeling corona.

In other words, if I know I have high levels of corona, I don't think saying "reduce the voltage (or increase the radius, etc) until there is no corona" is necessarily acceptable when asked how the corona affects the performance of nearby circuits. I also don't think its acceptable to say something along the lines of "just disregard the corona because we don't usually model light for EMI. I'm not sure how you can not have EMI when there is clearly high frequency current. Also, arent there quite a few papers discussing EMI effects of corona? Here's a quick one.

Chen Shixiu, Sun Youlin and Me Hengkun, "Characteristics of electromagnetic wave radiated from corona discharge," 2001 IEEE EMC International Symposium. Symposium Record. International Symposium on Electromagnetic Compatibility (Cat. No.01CH37161), 2001, pp. 1279-1282 vol.2, doi: 10.1109/ISEMC.2001.950630.

 

[Baluncore]

When corona occurs, the frequency is in the UV spectrum.

There is no hope of understanding if you cannot separate the photons released by the chemistry, from the varying current that flows during the discharge and which radiates EMI at radio frequencies.

Thinking about it, since we can also hear it, does that mean it's also in the lower frequency range as well?

You hear thunder following a lightning strike because the air is rapidly heated by the discharge and so expands, radiating a pressure wave. The same thing happens during corona discharge, you hear the start-stop nature of the heating caused by the currents.

I am sorry that you are annoyed by the inability to draw a circuit diagram of corona discharge. Maybe you could simulate a gas discharge tube that would have the same characteristics.

Because there are far greater sources of hostile EMI that must be kept out of the electronics, it matters little what EMI is generated by corona discharge. Once you protect the electronics package from the average EMI, corona will no longer be a problem.

 

[tech99]

We all know that if we have a wire with a sharp point at its end then the potential gradient can be very high, and if above about 30kV/m the air starts to break down. If the wire is positive, then electrons from the ionised air will be attracted to the wire and positive ions will be repelled. Notice, however that these both constitute a conventional current flowing in the same direction, away from the wire. The current will be flowing into the capacitance of the wire to free space, so it looks as if AC or RF is required for the corona to be sustained. No doubt the corona will extend outwards until it forms a sphere large enough to reduce the potential gradient below the breakdown threshold.
As mentioned, the filaments of current which suddenly form are a source of electrical noise. From the radio wave perspective, this noise will create current fluctuations in the wire, which will form an antenna. Radiation directly from the small current filaments is not going to be very large due to their small dimensions. However, there will be strong electric and magnetic fields around the corona, which constitute part of the induction near field of the antenna. We expect the induction near field of an antenna to expent outwards for a relatively short distance, often quoted as lambda/2 pi for a dipole.

 

[DaveE]

So the coupling is purely radiated? That's part of what I was trying to understand.

Not necessarily, although radiated noise is often the most noticed. Corona discharge occurs in tiny pulses similar to a capacitor charging slowly and discharging quickly. That will perturb the circuit, and you will see small rapid current pulses in the circuit if you look very carefully. You can look up "Biddle Tester" to learn more. Also, this link: https://llis.nasa.gov/lesson/813

PS: Also "Megger" for corona (partial discharge) testers

 

[tech99]

I think that, if close to the corona, the coupling is mainly by the strong fluctuating fields, but if far away, the mechanism will be by radiation, which is mostly from the wire.