Author Topic: Impedance of a plasma arc  (Read 2266 times)

0 Members and 1 Guest are viewing this topic.

Offline EteslaTopic starter

  • Regular Contributor
  • *
  • Posts: 149
  • Country: us
Re: Impedance of a plasma arc
« Reply #25 on: March 20, 2024, 11:29:20 pm »
Ok this is interesting. So you're implying that I should model the arc as a series RC circuit? I don't really understand why that's a good approach, probably I'm misunderstanding.

Going by definitions, 1/Q = tan(delta), where delta is the angle between the vectors R and the reactance of the capacitor [1/(2*pi*frequencySwitchingHz*C)]? So if I assumed R was 1, and you're telling me Q is 30, then delta is like 2 degrees ish, and at 20 khz C would be like 500uF? I guess if I model it as a series RC thing, how would I get the R or the C?

I explained why R isn't ~1 on average.  An RF discharge is billions of microscopic events; it is not a continuous process.  This is simultaneously why so much wideband noise is emitted.

I don't know how you would get 500uF out of, say, a sphere a couple inches diameter at most; more like a couple pF.  The ESR will be in the kohms to Mohms range, depending on frequency and Q.  Connect this to the secondary coil, modeled as a parallel RLC circuit; and couple this to whatever the primary driving winding and amplifier is.

The methods won't make a whole lot of sense without some network theory, but I would suggest just putting together a basic circuit in SPICE and playing with values.

Tim


I guess I should be more clear about my starting point. I'm working in LTSpice to simulate all of this. I have a good model of my core that I'm testing as I go along, complete with saturation. My point in the previous post is that I don't understand what you're talking about with 'Q' and 'dissipation factor'. I've only used those in the context of capacitor characterization before, and when I tried to apply them in that way I got nonsensical values like you saw with the 500uF.



The goal of this topic is to get an idea of what that question mark should be assuming the load is an electric arc in air on the 1-2" order of magnitude that looks 'pretty hot'. Think about what the arc looks like in this dudes video for hottness: https://youtu.be/UMRvqS953AE?feature=shared although I wouldn't mind going quite a bit hotter and possibly longer.

Basically if I know that impedance roughly, then I can do a lot better job optimizing my secondary windings for that sort of load to dump as much power into the arc as I can.
 

Offline EteslaTopic starter

  • Regular Contributor
  • *
  • Posts: 149
  • Country: us
Re: Impedance of a plasma arc
« Reply #26 on: March 20, 2024, 11:31:26 pm »
Ok interesting. I'm hearing that the resistance of a 1-2" arc is going to be on the order of single digit ohms of resistance? Does anyone have evidence against or for that?

As I remember from science paper, it has fraction of Ohms and very close to a copper conductor of the same size. But it depends on conditions (air pressure, temperature, chemestry, etc), which also depends on the voltage and current. So, there is no exact value and it can be higher than copper, but in average you can assume it the same as a copper conductor.

This is what I am looking for. From you I'm hearing it's in the < 1 ohm territory. Do you remember what that paper was or any further details?
 

Offline IanB

  • Super Contributor
  • ***
  • Posts: 11891
  • Country: us
Re: Impedance of a plasma arc
« Reply #27 on: March 20, 2024, 11:41:23 pm »
Apparently an arc doesn't look like a short circuit, otherwise the breaker would cut out on overcurrent:

 

Offline T3sl4co1l

  • Super Contributor
  • ***
  • Posts: 21688
  • Country: us
  • Expert, Analog Electronics, PCB Layout, EMC
    • Seven Transistor Labs
Re: Impedance of a plasma arc
« Reply #28 on: March 21, 2024, 12:18:48 am »
At the simplest, all you need is an AC voltage source, some LC components representing the transformer's small-signal parameters, and R+C representing the load.

Core saturation isn't important as that is easily checked separately, but there's no harm in using it in a simulation.  Maybe the results generate a little slower (it adds nodes to the matrix, if nothing else).

The inverter square wave isn't even important, because, even if a pulsed waveform makes it to the output, the fundamental frequency will deliver most of the power, and all harmonics will be some percentage of that (typically under 10%).  There are some cases where this isn't true, and there are cases where square pulses are simpler anyway (square-pulse type SMPS analysis), but this is an assumption we can validate.

Note that a simple RC load won't reflect the whole situation, because for short time scales, most plasma filaments, present at a given instant in time, remain conductive; therefore there is less resistance at high frequencies.  But there is also less reactance at high frequencies.  I don't know what the actual balance is, but I would assume just for starters that the effect is constant Q.  Thus, we emphasize what the resistance is: an effective series resistance (ESR), not a real physical resistor in the system.  (Because, again, it's not a resistor at all, it's myriad very small-value resistors incessantly making and breaking connections to incremental capacitances; it's a time-averaged equivalent of the very intricate microscopic physics going on.)

At least, I'm assuming (or, have been) that you want a corona discharge, consistent with the first request: a plasma tweeter.

Plasma tweeters do NOT use continuous point-to-point arc discharges.  Arcs have smaller volume, and, I think something about the acoustics that makes them poorer besides?  Certainly, the convection of a fat thermal arc will produce subtle but persistent doppler distortion, on top of increased noise (the roiling boiling sound as the arc wanders and convects); corona discharges still suffer from this, which is why -- besides the ozone and NOx -- they're just not very good at this, but, they are the best case among related mechanisms, so that's the way it has been done.


Apparently an arc doesn't look like a short circuit, otherwise the breaker would cut out on overcurrent:

Right.  Several posters here have been less than helpful, posting factoids without relating them to actual in-circuit parameters.

A welding arc has low resistance because it is short length; there is still a built-in potential due to the voltage drop near the electrodes (typically 10-20V depending on gas, electrode, pressure, etc.), and there is a quasi-ohmic drop through the plasma itself.  Arcs don't simply have one voltage and that's it.  This is highly obvious when you measure the voltage of a welding arc, while moving it around; it grows with distance, which is why stick/TIG power supplies use a relatively high open-circuit voltage (50-80V; up to 300V for plasma cutters, though that may be only during ignition, I forget), and, at available current and ~1atm ambient pressure, this limits arc length to less than a cm.

Or why you see thousands of volts dropped across long glow or arc discharges, even rather hot ones like the above video.

Which, I have no idea why protection didn't fire in that case, maybe it wasn't equipped, maybe it wasn't operational, maybe the lines are designed for as much current (and thus the switchgear at either end as well) and there was enough line resistance (and reactance!) to limit current below the fusing threshold.  Would have to look up the particular case and see if there was any analysis or a report on it.  Such incidents are fairly rare, highly visible (for obvious reasons, lol) and tend to make the news, at least locally, so there's probably something out there.  I'm not interested enough to look it up myself but anyone who wants to give it a try is welcome to do it -- research is good exercise.

In any case, this is why I've been trying to emphasize the collective effect of myriad minute discharges.  A corona discharge isn't a static effect, it's highly dynamic, far more dynamic than you can see by eye, probably than you can see by any high-speed camera even.  It looks like a quasi-static or chaotic discharge by eye, but this is simply because your eyes can't resolve billions or trillions of discharges per second.

The power dissipated by those myriad discharge paths, must necessarily come from the energy spent charging and discharging the air surrounding the breakout.  There's no other energy flow, and current is limited by the reactance of that surrounding air.

And neither can your circuit [resolve these myriad paths]; nor does it care.  All it cares about is load current at the fundamental, and harmonics if relevant.  Each individual discharge lets off a microscopic EMP burst, contributing to radiation from MW to microwaves -- which is how corona and arc discharges are so noisy, they emit a ton of RF noise.  But averaged over a cycle of say 40kHz, it doesn't amount to much, and just looks like a resistor load with excess noise.

Air breaks down because the maximum field strength is exceeded, electrons are torn from molecules, and avalanche discharge ensues.  It doesn't matter where this happens, it can happen at random in free air, but it's most likely to occur near an electrode (at a point or corner, a sharp curvature, maximum field strength at the surface), or extending from a discharge which now pierces into the surrounding low-voltage field, making its own needle tip and so on.  Streamers shoot and branch as they fill into nearby space, charging it up, making it unipotential to the electrode, at least momentarily until the voltage reverses and the whole process repeats anew.

But none of this is material to the electronics, all that's needed is a representative RC load at the driven fundamental frequency.  Most likely, the transformer will have considerable leakage inductance and secondary stray capacitance, making its resonance quite strong, and a full-wave inverter driving that is quite effective, the leakage acting to filter harmonics -- which validates our earlier assumption of single-frequency analysis, and which without, we might want to choose a different analysis (a more detailed RCRC load equivalent, perhaps, and a square-wave source).

Tim
« Last Edit: March 21, 2024, 12:25:07 am by T3sl4co1l »
Seven Transistor Labs, LLC
Electronic design, from concept to prototype.
Bringing a project to life?  Send me a message!
 

Offline EteslaTopic starter

  • Regular Contributor
  • *
  • Posts: 149
  • Country: us
Re: Impedance of a plasma arc
« Reply #29 on: March 21, 2024, 12:28:59 am »
Ah. Ok I think I misspoke again. I don't want a plasma tweeter, I just thought that's what I want is called. For now I just want a hot solid arc like this:



don't worry about the audio part. I guess I didn't need to include that in my description, it was just context for my motivation for my initial question.

I am looking for a circuit that roughly approximates the impedance of a hot solid arc that looks like that and is 1-2" in length. I understand that whatever is going on 'under the hood' is not at all like an R L C thing, but I don't care for now. I just want a rough estimate to help me design my secondary winding. My goal for this thread is to find someone who indeed has REAL data or numbers. I do think the arc will mainly look like an 'R'. I want to know what that R is in ohms.
« Last Edit: March 21, 2024, 12:33:08 am by Etesla »
 

Online coppercone2

  • Super Contributor
  • ***
  • Posts: 9454
  • Country: us
  • $
Re: Impedance of a plasma arc
« Reply #30 on: March 21, 2024, 12:46:57 am »
I gave you numbers from a good text book written before simulations
 

Offline ejeffrey

  • Super Contributor
  • ***
  • Posts: 3719
  • Country: us
Re: Impedance of a plasma arc
« Reply #31 on: March 21, 2024, 12:59:59 am »

As I remember from science paper, it has fraction of Ohms and very close to a copper conductor of the same size. But it depends on conditions (air pressure, temperature, chemestry, etc), which also depends on the voltage and current. So, there is no exact value and it can be higher than copper, but in average you can assume it the same as a copper conductor.

Not at all. The resistivity of a plastma is orders of magnitude higher than copper under any reasonable conditions.  For mm scale arcs at atmospheric pressure and many amps you have ohms or fractions of an ohm but a copper conductor of the same size would be milliohms or less.

 

Offline IanB

  • Super Contributor
  • ***
  • Posts: 11891
  • Country: us
Re: Impedance of a plasma arc
« Reply #32 on: March 21, 2024, 01:06:10 am »
Which, I have no idea why protection didn't fire in that case, maybe it wasn't equipped, maybe it wasn't operational, maybe the lines are designed for as much current (and thus the switchgear at either end as well) and there was enough line resistance (and reactance!) to limit current below the fusing threshold.  Would have to look up the particular case and see if there was any analysis or a report on it.  Such incidents are fairly rare, highly visible (for obvious reasons, lol) and tend to make the news, at least locally, so there's probably something out there.  I'm not interested enough to look it up myself but anyone who wants to give it a try is welcome to do it -- research is good exercise.

That incident happened somewhere overseas (Turkey, maybe?) and from what I read, it seems like it was a maintenance issue. It wouldn't necessarily trip on overcurrent protection since the circuit is designed to deliver that kind of power to a normal load, but there should have been an arc fault detector upstream that would have interrupted the circuit. For whatever reason, the automatic circuit protection didn't operate and it had to wait for someone to manually switch it off.

The arc most likely started due to dirty insulators (they have to be cleaned regularly to prevent this kind of thing happening).
 

Offline EteslaTopic starter

  • Regular Contributor
  • *
  • Posts: 149
  • Country: us
Re: Impedance of a plasma arc
« Reply #33 on: March 21, 2024, 01:14:46 am »
nonlinear at low currents. At high currents between say 50 amps it should be around 0.5 ohms for a 1mm length welding arc (taken from old book). And at 250 amps something like 90 miliohms. (the voltage drop stays constant) *

At lower currents say 10 amps it comes out to a few ohms . (say in this case I got about 3 ohms but its hard to read these old tiny charts)


But this is for a welding arc

It changes with distance just like it does with a wire. Linearly (at high currents). The temperature of the gas is probobly the main factor for conductivity ? At low currents I don't know. I think its kinda unstable


and
At very low currents it looks like it tops out at about 30 ohms per mm (2-3 amper range)


*keep in mind its i2r, so even if you have decreased resistance, the power of the arc still goes up even if the resistance decreases. I.e. 1kW @ 50A and 4kW @ 250 amps So if that heat is being carried into the work piece you know how a welder works, its like a torch jet. Then you also have the electricity doing something on the work piece, but if you experiment, 50 amps on a piece of steel actually don't do too much ! nothing like the arc heat.

Indeed you did! I didn't notice you had a 2-3A range number. Was this from a book that had even lower currents documented? I'm not sure if I can achieve 2-3A with the wire diameters I'll be forced to use for the secondary coil regardless of anything else.
 

Offline themadhippy

  • Super Contributor
  • ***
  • Posts: 2583
  • Country: gb
Re: Impedance of a plasma arc
« Reply #34 on: March 21, 2024, 01:21:03 am »
Quote
but there should have been an arc fault detector upstream
maybe they'd fitted  a hager afd
 
The following users thanked this post: Gyro

Offline T3sl4co1l

  • Super Contributor
  • ***
  • Posts: 21688
  • Country: us
  • Expert, Analog Electronics, PCB Layout, EMC
    • Seven Transistor Labs
Re: Impedance of a plasma arc
« Reply #35 on: March 21, 2024, 03:15:08 am »
Ok.  For a HV arc, hot glow can be sustained for even mere 10s of mA.  The main thing is it's a continuous arc when it remains hot and ionized before the next pulse and so on.  It's hard to get this from a DC supply like a CRT flyback, with internal rectifier, because the internal capacitance discharges suddenly, a lot of nothing happens for a while, then another pulse comes along and so on; AC of enough frequency I suppose, or enough energy per pulse, can maintain a conductive channel and you get the hot appearance.

A continuous arc is mostly resistance; the impedance is low enough in general that capacitance is swamped.  Also because, if capacitance were dominant, it would spark as a relaxation oscillator instead.  The loaded Q of the secondary will most likely be low, unless you add reactance to it; when a resonant supply is used, it can be helpful to use a series capacitor for example (use low-loss C0G or SL type).  Or inductor, at low impedance -- whichever is more convenient for the system.

Tim
Seven Transistor Labs, LLC
Electronic design, from concept to prototype.
Bringing a project to life?  Send me a message!
 

Offline Berni

  • Super Contributor
  • ***
  • Posts: 4956
  • Country: si
Re: Impedance of a plasma arc
« Reply #36 on: March 21, 2024, 06:34:54 am »
I am looking for a circuit that roughly approximates the impedance of a hot solid arc that looks like that and is 1-2" in length. I understand that whatever is going on 'under the hood' is not at all like an R L C thing, but I don't care for now. I just want a rough estimate to help me design my secondary winding. My goal for this thread is to find someone who indeed has REAL data or numbers. I do think the arc will mainly look like an 'R'. I want to know what that R is in ohms.

There is no one single number of resistance for an arc.

This is like asking what is the resistance of a diode? You can apply certain amps and volts to a diode and use ohms law to calculate the equivalent resistance of the diode at that point. But that resistance is not useful since it will change as soon as you change the conditions the diode is operating in, heck even just changing the ambient temperature might change it a lot.

For this reason your power source can NOT be designed to operate only with a specific load resistance. Instead the feature that arc sustaining power supplies have is that they can maintain a large current flow over a wide range of equivalent load resistances( usually down to 0 Ohm). This way the power source adapts to the arc, so to give it however much power it needs to sustain itself. So in your simulation you need to put in different resistors and have your power source operate with all of them (otherwise your arc will extinguish when its resistance sways into that range)

A very hot short arc might be less than an Ohm, a long barely sustained corona arc might be > 100s of kOhm.

Usually what you design for is Amps you want to push into an arc. The more amps you make available to the arc the hotter it will get until it draws all the available amps. Lots of amps is what makes these white hot arcs (id say at a minimum >100mA). Then how high of a voltage you can sustain at such a load determines how far you can drag out that arc before it extinguishes (since drawing out the arc increases resistance by both making it a longer path and also making the plasma less conductive due to cooling it off)
 

Offline Marco

  • Super Contributor
  • ***
  • Posts: 6722
  • Country: nl
Re: Impedance of a plasma arc
« Reply #37 on: March 21, 2024, 08:06:04 am »
Arcs chew up electrodes ...

You can find old papers on glow discharge speakers such as "Modeling of a DC glow plasma loudspeaker".
 

Offline EteslaTopic starter

  • Regular Contributor
  • *
  • Posts: 149
  • Country: us
Re: Impedance of a plasma arc
« Reply #38 on: March 28, 2024, 04:25:37 am »
In case anyone is curious, I got an arc going, I built myself a high voltage probe, and I started taking measurements. After collection the data, I learned that the impedance of a plasma arc in the single digit to low 10's of mm length, in the 20mA to 100mA range, is going to look like a resistor in the single digit to 10s of kilo ohms. I was doing this in an AC situation at ~55kHz so the 'current' is the peak current. The actual data points I actually have are:

length (mm), peak current (mA), Arc resistance (ohms)
3, 20, 18000
3, 30, 13333
3, 40, 10000
3, 50, 8400
3, 60, 7333
3, 80, 6250
3, 100, 5500
4, 40, 10500
5, 40, 11000
6, 40, 11500
8, 40, 12500

If I do a best fit for a surface with the equation
Resistance = Current(Amps)^A * (B + C*length(mm))

then I get

A = -0.7923
B = 712.58
C = 31.515

I'm just going to use this equation to model the arcs resistance in spice. Hope someone else finds this useful.
« Last Edit: April 01, 2024, 01:25:42 am by Etesla »
 
The following users thanked this post: Someone, T3sl4co1l

Offline Vovk_Z

  • Super Contributor
  • ***
  • Posts: 1418
  • Country: ua
Re: Impedance of a plasma arc
« Reply #39 on: March 30, 2024, 04:42:18 am »
Apparently an arc doesn't look like a short circuit, otherwise the breaker would cut out on overcurrent:


The arc is quite similar to the short circuit but it is not a good metal short circuit but a bit higher impedance 'short'. If it is not turned off by protection then it means some problem exists with this line and protection. Possibly protecton has low sensitivity (line has both high impedance and high load) or mallfunctioning.
 

Online coppercone2

  • Super Contributor
  • ***
  • Posts: 9454
  • Country: us
  • $
Re: Impedance of a plasma arc
« Reply #40 on: March 30, 2024, 06:10:09 am »
wow that is alot lower resistance. I figured it would be around 40 ohms. Maybe it further increases at really low currents.
 

Offline johansen

  • Frequent Contributor
  • **
  • Posts: 996
Re: Impedance of a plasma arc
« Reply #41 on: March 30, 2024, 06:11:35 am »
The impedance of those big generator step up transformers  transformers is on the order of 20 %, yet 99% efficiency. The inductance of the transmission line further de magnetizes the generators  which have a regulation of 30%.

The reduced voltage to the customers, trips equipment off line, reducing the load.

So the arc is not the dead short that it appears.
 


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf