How much heat does it take to make a (neon tube) plasma?

[Nereid]

This thread is intended to address a question in the now-closed Q&A thread Does this qualify as an electrical current flow?

The question comes from neilzero, in response to a post by korjik:

Wait a minute, while a plasma can be created using heat, it can also be produced by using electricity.

Doesn't a plasma need to be constantly energised (ionised) to stay a plasma?

The electricity goes to heat the gas till it ionizes. The reason electricity is used so often is that it is a good way to the the energy density needed to make the plasma.

No a plasma dosent need to be energized. It just needs to stay hot enough. Not [losing] the energy is the tricky part.

The 1/25 watt neon light has largely been replaced by LEDs. In the dark you can see the light of ionization at one or two milliwatts, so I'm having trouble thinking much heat is involved in the 200 cubic millimeters of low pressure neon gas. Neil

I replied to this question as follows:

It is quite easy to get confused if one tries to work all the effects out, just with words.

As you have done, it is necessary to get quantitative to appreciate the time-scales.

[snip]

In the case of the neon light, do you know what the pressure is? If we have that, we could take a stab at doing an OOM (order of magnitude) calculation of the heat required to ionise it, and the time it would take for it to cool to the temperature of the glass vessel it is contained in. If it's a 1/25 watt light, even with a quite low conversion ratio (input energy to kinetic energy of the plasma + energy of ionisation), I suspect the heat required is indeed miniscule. Likewise the time it takes for the plasma to cool very short.

[snip]

So here are the questions this thread seeks answers to:

1) What is the pressure inside a 1/25 watt neon light?

2) How much energy does it take to turn the neutral gas in such a light into a plasma (in the light's normal working state)?

3) How long after the energy source is removed does the plasma turn back into a gas, near ambient temperature?

In all cases, an order of magnitude (OOM) answer is what I'm after.

For 1) and 2), standard SI units please.

I'd really like it if amateurs and high school or undergrad university students tried to answer these ... I'd be astonished if any professional physicist/space scientist/astrophysicist or grad student couldn't!

Note that one way to answer 2) avoids a need to know a great deal about neon.

And if anyone finds (or knows of) a site with both a nice piccie and specs for a neon light of the kind neilzero describes, I'm sure that would help readers who have may never have heard of, let alone seen, such a thing ...

[JohnD]

Neried,
I hope I'm not treading on anyone's toes if I suggest that anyone wanting to pursue your question tries earching for "fluorescent" as well as 'neon'? Lots of different hits about the same subject.

John

[galacsi]

It is a very strange question , as it is well known a neon tube is electrically powered.

Are you really seeking an anwser for yourself or are you giving us some homework ?

[trinitree88]

out of sequence...
you only need the first ionizationof energy of neon to separate the remains of the neon atom from it's electron...but that would produce a feeble few photons as it recombined.

to run the light at 25 watts you need 25 joules/sec provided the bulb is 100% efficient...(it's not unless the Supreme Being sent it to you for Christmas),so you need the heat losses from the glass and the ohmic losses from the conductors involved.

the recombination occurs at ~ the speed of sound in the plasma...it's called quenching time in particle detector spark chambers' or wire chambers' cells, and limits the counting rate, so a trace of chlorinated hydrocarbons greatly improves the response, as they photodisintegrate to release small quantities of "free" chlorine. For that reason they gold coat the wires so they don't corrode during the avalanche of electrons produced in the burst...(you happened to ask about the CEBAF OOPS detector I worked on one summer...in a roundabout sort of way....same principles)...pete.

[TrAI]

I assume its one of those neon indicator lamps that is talked about, not a neon tube.

If i recall correctly the gas is a low pressure mixture of about 99% neon and the rest is argon, since the mix ionize easier than the pure gas would, but the mixes do vary a bit, i think . The lamps usualy have a breakdown voltage of around 90-100 volts, though frequency and gas mixture will affect this. They are usualy limited by a resistor to less than 2-3 mA, since they exhibit a negative resistance, that is, when lighted the buildup of ions causes a drop in resistance, that would, unless limited lead to a higher current flow and more ions, and so on until the lamp is destroyed, after all, they are glow discharge lamps not arc lamps... This also means the holding voltage is lower than the voltage needed to start the discharge.

Usualy lamps like this is used as indicators, but sometimes they are used as voltage limiters. And, of course, those flickering flame bulbs are neon lamps too.

[neilzero]

I believe everything TrAI typed is correct. The 1/25 th watt = 40 milliwatt neon bulbs can work as a relaxation oscillator up to about 500 hertz, likely because the 90% de-ionization time is about one millisecond. Old design florescent tubes, and more powerful neon lamps have a deionization time of at least ten milliseconds, I think, as it is difficult to get them to function as a relaxation oscillator. A hydrogen thyratron can deionize in about 0.1 millisecond at power levels up to a megawatt. They were used to pulse the magnetron in early radar.
In a very dark room, you can see the ionization in a typical 4 foot florescent tube, with about 500 volts dc applied, with the current limited to about 100 microamps = 50 milliwatts input. Neil

[TrAI]

Are you asking how they work? or just the figures that make it work? e.g pressure, temperature, electrical input?

Cos I'm a little confused, are you Nereid eluding to, is it the heat or electrical energy doing the "work"?

Throw this into the mix

Hold a fluro tube next to (not touch) a toy plasma globe and it "glows"....How?

No wires, no heat..how?

I am not really very good at explaining stuff, and I am quite sleepy right now, so I may make a muddle of this, but here goes:

It is electricity that lights the tube, of course...

Actually, a fluorecent tube needs very little current to make a visible light, as long as the potential difference between two points along the tube is over a certain voltage. Now, there are a few ways that electricity can "jump" through a nonconducting material, like air, one is to ionize it, another is capacitive coupling, I expect it is the last type we are dealing with here.

You may know that capacitors tend to block DC currents, but conduct AC currents and the current in depends on frequency, capacitance and impedance/resistance of the load? well, the air around the globe acts as a dielectric in a capacitor, while the globe is one plate(this is simplified, of course) and the tube is the other(and, of course, you are holding the tube by the other side, and so closer to ground potential), and since the globe uses an alternating current of several kHz, you tend to get at least a slight current flow, and so the field around the globe is more than sufficient to light up the tube out to a certain distance...

[Nereid]

It is a very strange question , as it is well known a neon tube is electrically powered.

Are you really seeking an anwser for yourself or are you giving us some homework ?

Neither.

As I explained in the OP, neilzero asked a good question, in a thread that's now closed.

Those who are interested to discover for themselves - by finding the relevant formulae and concepts, and working through the calculations - will gain a deeper appreciation of the nature of plasmas, and learn - through a specific example - what korjik (and others) meant.

If you, or any other reader, have no interest in doing this, fine.

If you, or any other reader, are stuck on how to go about doing it (or on any part of the work), then just ask!

[TrAI]

By the way, I seem to recall that in glow discharge ionisation the electricity does not ionize the gas by directly heating it, rather that the electricity cause already existing ions in the gas to move towards the cathode, and it is these ions hitting neutral atoms that cause further ionisation. I expect you get more ionisation from heat in arc discharges...

[publius]

By the way, I seem to recall that in glow discharge ionisation the electricity does not ionize the gas by directly heating it, rather that the electricity cause already existing ions in the gas to move towards the cathode, and it is these ions hitting neutral atoms that cause further ionisation. I expect you get more ionisation from heat in arc discharges...

Yes, that's true. The "glow discharge" phase is distinct from the "full arc" phase. It gets very complex, lots of physics and chemistry afoot there. The "blue fire" was actually what got me interested in physics as a kid, and I've been playing with it every since. Arc welding is one of my favorite hobbies.

At one time, I got very interested in all this, but alas have forgotten most of the details. Makes me mad at myself.

But here's how a glow discharge, sometimes called corona (there may be a slight difference in usuage between the two, but they are basically the same process) works. Keep in mind this all depends on all sort of variables, including the shape of the electrodes, and the polarity of each plays a big role. Making one electrode a "sharp point" helps as the local electric field gets high locally, and the other electrode may need to be spread out more.

Now, take a positive sharp electrode with a spread out negative one, and put a low pressure gas between them and start increasing the voltage between electrodes, watching what happens.

At low voltages, nothing much will happen. You get some miniscule "leakage" current from various mechanisms, but nothing of any consequence. As you increase the voltage, you cross into the glow discharge region.

There will be always been some extraneous local ionization events from various sources, UV or higher photons, heck a cosmic ray or derivative, or even other local small background radiation. Anyway, a local atom gets an electron knocked off. Normally, it would just recombine quickly. But when the local electric field around the positive electrode gets high enough, it will accelerate that loose electron, which will slam into another neutral atom or molecule floating around. That's enough to knock off another electron from that electron.

The result is a "electron avalanche" or shower of free electrons from the initial ionization event. All those accelerated electrons flying around, falling into orbitals and out again produce the photons of the glow.

Some of those loose electrons fly back into anode, and that makes the small current. Now, positive ions have to migrate back to the negative electrode, and they are heavy and slow. The brunt of the ionization is concentrated around the anode, and those positive ions are the majority charge carriers throughout the rest of the gas.

Reverse the polarity, and things are very different. Most gases can support a "positive corona" as described above, but not all will support a negative one, at least easily. The gas atoms or molecules have to support negative ions, and be able to carry the extra electrons back to the positive anode. If the gas doesn't like to do that, you won't get much of a negative corona. Second, in the high field region near a sharp negative electrode, free electrons are accelerated away, not toward it and so the avalanche process is not as energetic, and it depends on be able to pull electrons off the electrode itself.

The behavior between the two cases is very different.

The above corona process is very different from a full blown arc. In corona, the electric field is not strong enough to cause ionization on its own, but requires an external source.

When you increase the voltage enough, the field does become strong enough, and you have a dielectric breakdown of the gas. Now, you're going to town. You have a full blown plasma, with free electrons all the way between electrodes.

More later.....

-Richard

[publius]

Now, in glow corona, we don't have much plasma, just a small "avalanche region" close to the "sharp" electrode, with a relative trickle of heavy ion carries completing the circuit through the bulk of the gas. Not very energetic.

Get into full blown arc mode, though, and it's Katie bar the door. That can get very energetic indeed.

There's an interesting "women in science" angle to arcs. A Brit woman named Hertha Ayrton did extensive experimentation with arcs back in the late 19th century, published in "The Electrician" (the dude in my avatar published in there a lot )and derived an empirical arc equation known as Ayrton's equation. The motivation was arc lighting, IIRC. I forget how it went, but her husband was a physicist and electrical engineer and that's how she got started with this. For her work, she became the first ever female member of the Brit IEE, the Brit national version of the IEEE at the time.

She compared arcs to mules (and maybe even men ), stubborn with a mind of their own.

Her equation relates the voltage across an arc to the length and current, amongst other things, and goes something like this. I'm quoting this from memory, so don't hold me to it exactly, but it looks right:

V = A + B*d + C/I

Where 'd' is the length of the arc. This applies to steady state, well behaved arcs. "Turbulent" arcs will deviate from this. There are other empircal formulas derived since, but Ayrton's works pretty well for most.

'A' is a constant, indepedent of length, which depends mainly on the properties (and shape, actually) of the electrodes, which would include the work function and all that good stuff. IIRC, this value is around 30V for copper rod electrodes.

B is a length constant, and it depends on the pressure of the gas primarily (as well as other gas properties). B *increases* with pressure. C is the inverse current, negative differential resistance coefficient (and I believe it includes a 'd' term itself, dependent on length, but I can't remember that part).

The C part looks like "constant power", but that tends to be misleading. This term is responsible for the runaway behavior of arcs.

So, to maintain an arc, we've got to have some minimum voltage, 'A' to sustain it. Drop our voltage below that and it dies no matter what the other variables are. We've got another term that increases linearly with length. Double the length, double the additional voltage above A. And the more pressure, the more voltage per length.

Now, what does this tell us. A high pressure arc, for a given current, takes more voltage drop to maintain, and by VI, is going to be more energetic that a low pressure arc.

The third term is the runaway term. As I mentioned, the functional form can be misleading at first blush, but it is indeed a runaway term. Get a current started, and without some series current limiting device, the current tends to increase without limit. Solve a simple DC circuit with a resistor in series with a term like that, then let R go to zero, and you'll see what I mean.

The result is basically that once we get an arc started, we've got a monster on our hands that tends to suck all available power from our source. Now, the higher the pressure, the worse that monster will be (if we've got the voltage available).

A low pressure arc can be a dainty, well behaved and civilized thing, not drawing too much power. Fluoresecent tubes are example of this. And yes, the are full blown arcs, but low pressure arcs. The gas inside is at relatively low pressure, the power is relatively low. Arc stability is a while 'nother complicated kettle of fish, but it depends on the size of the arc and the shape of the electrodes. When fluorescent electrodes are near the end of their life, the electrodes get pitted and worn, and stability is sometimes lost. That's that swirling, flickering, and jumping you see.

That's a "turbulent" arc. The mechanical stresses of that turbulence, along with little pieces of electrode flying off can break the seal, let air pressure in, and the tube goes out. THere's nowhere enough voltage and available power to run an arc in full atmosphere. And if there was, it would be too strong, that monster I was talking about. That happenned out in my shop just the other day. I noticed one of the 8' tubes started flickering, got a nasty swirl in it, and started making a popping noise. Then it went out with a "squeeeeeee" sound as air rushed in through a little pin hole made by the jerking on the electrodes.

The low pressure arc is not all that hot. Now, the free electrons in there would be at very high temperature indeed, but the majority of the mass, and thus the total heat capacity in the ions and rest isn't all that hot at all. You can touch the tubes are they're just warm to the touch most of the time. In hot weather, they'll feel a little hot, but it's nothing like an incandescent bulb.

HID (high intensity discharge) lightning of various types use higher pressure arcs to get more output, and they do get fairly hot. But fluorescents are very mild and civilized arcs.

-Richard

[trinitree88]

Now, in glow corona, we don't have much plasma, just a small "avalanche region" close to the "sharp" electrode, with a relative trickle of heavy ion carries completing the circuit through the bulk of the gas. Not very energetic.

Get into full blown arc mode, though, and it's Katie bar the door. That can get very energetic indeed.

There's an interesting "women in science" angle to arcs. A Brit woman named Hertha Ayrton did extensive experimentation with arcs back in the late 19th century, published in "The Electrician" (the dude in my avatar published in there a lot )and derived an empirical arc equation known as Ayrton's equation. The motivation was arc lighting, IIRC. I forget how it went, but her husband was a physicist and electrical engineer and that's how she got started with this. For her work, she became the first ever female member of the Brit IEE, the Brit national version of the IEEE at the time.

She compared arcs to mules (and maybe even men ), stubborn with a mind of their own.

Her equation relates the voltage across an arc to the length and current, amongst other things, and goes something like this. I'm quoting this from memory, so don't hold me to it exactly, but it looks right:

V = A + B*d + C/I

Where 'd' is the length of the arc. This applies to steady state, well behaved arcs. "Turbulent" arcs will deviate from this. There are other empircal formulas derived since, but Ayrton's works pretty well for most.

'A' is a constant, indepedent of length, which depends mainly on the properties (and shape, actually) of the electrodes, which would include the work function and all that good stuff. IIRC, this value is around 30V for copper rod electrodes.

B is a length constant, and it depends on the pressure of the gas primarily (as well as other gas properties). B *increases* with pressure. C is the inverse current, negative differential resistance coefficient (and I believe it includes a 'd' term itself, dependent on length, but I can't remember that part).

The C part looks like "constant power", but that tends to be misleading. This term is responsible for the runaway behavior of arcs.

So, to maintain an arc, we've got to have some minimum voltage, 'A' to sustain it. Drop our voltage below that and it dies no matter what the other variables are. We've got another term that increases linearly with length. Double the length, double the additional voltage above A. And the more pressure, the more voltage per length.

Now, what does this tell us. A high pressure arc, for a given current, takes more voltage drop to maintain, and by VI, is going to be more energetic that a low pressure arc.

The third term is the runaway term. As I mentioned, the functional form can be misleading at first blush, but it is indeed a runaway term. Get a current started, and without some series current limiting device, the current tends to increase without limit. Solve a simple DC circuit with a resistor in series with a term like that, then let R go to zero, and you'll see what I mean.

The result is basically that once we get an arc started, we've got a monster on our hands that tends to suck all available power from our source. Now, the higher the pressure, the worse that monster will be (if we've got the voltage available).

A low pressure arc can be a dainty, well behaved and civilized thing, not drawing too much power. Fluoresecent tubes are example of this. And yes, the are full blown arcs, but low pressure arcs. The gas inside is at relatively low pressure, the power is relatively low. Arc stability is a while 'nother complicated kettle of fish, but it depends on the size of the arc and the shape of the electrodes. When fluorescent electrodes are near the end of their life, the electrodes get pitted and worn, and stability is sometimes lost. That's that swirling, flickering, and jumping you see.

That's a "turbulent" arc. The mechanical stresses of that turbulence, along with little pieces of electrode flying off can break the seal, let air pressure in, and the tube goes out. THere's nowhere enough voltage and available power to run an arc in full atmosphere. And if there was, it would be too strong, that monster I was talking about. That happenned out in my shop just the other day. I noticed one of the 8' tubes started flickering, got a nasty swirl in it, and started making a popping noise. Then it went out with a "squeeeeeee" sound as air rushed in through a little pin hole made by the jerking on the electrodes.

The low pressure arc is not all that hot. Now, the free electrons in there would be at very high temperature indeed, but the majority of the mass, and thus the total heat capacity in the ions and rest isn't all that hot at all. You can touch the tubes are they're just warm to the touch most of the time. In hot weather, they'll feel a little hot, but it's nothing like an incandescent bulb.

HID (high intensity discharge) lightning of various types use higher pressure arcs to get more output, and they do get fairly hot. But fluorescents are very mild and civilized arcs.

-Richard

Good stuff, Richard. Then there's the fun of carrying a fluorescent tube while hiking the pole lines at night....you don't need a flashlight under the 400,000 volt lines from the nearby nuclear plant...just a working fluorescent tube, the induced currents from ~ 50 feet below light them up nicely....wonder how they bill you for that?....pete.

[publius]

Good stuff, Richard. Then there's the fun of carrying a fluorescent tube while hiking the pole lines at night....you don't need a flashlight under the 400,000 volt lines from the nearby nuclear plant...just a working fluorescent tube, the induced currents from ~ 50 feet below light them up nicely....wonder how they bill you for that?....pete.

Thanks, and that brings up something else. Many "anti power line" types (had to deal with a few around here recently) use that ol' fluorescent under the HV transmission line as evidence of the "radiation" of a power line. IOW, if you're near one, all that radiation is frying you.

Nothing could be further from the truth. If you define radiation as an EM radiation field, that is *nil* near a power line. What you're doing is (capacitively) coupling to the phase closest to you, and that arrangement can pull some power (and yes, you can actually pull some substantial power from a line like this if you know what you're doing. It has been done, and court cases or two have sprung from it). In field terms, that alters the field locally and pulls power off the line that wouldn't otherwise flow out sideways. And that is detectable from the altered V-I curves if the parasistic loss is large enough. If power lines were radiating a power density like that all the time, the power companies would be loosing money big time.

But that is used a lot to scare people who don't understand all this.

-Richard

[publius]

And, after checking some figures, to half-way answer Nereid's original question, the power for discharge lamps is usually expressed in watts per arc length. That's based on standard lamp diameters.

For fluorescents and low pressure arcs, that value is < 1W/cm. For high pressure lamps, it is >20W/cm, getting as high as 50W/cm in some designs.

The voltage drop across arcs is a complex and fascinating thing, too. You've got the electrode region, and differences between anode and cathode (which switches back and forth with AC, so averages out) generally, plus the bulk region. There is extensive literature on investigating what goes on in all these regions.

-Richard

[publius]

Now those are nice, controlled and well behaved arcs for lighting purposes. If you didn't have a nice ballast for current limiting there, the current even in fluorescent tube would try to runaway, probably hitting 100A or so before it blew itself up. The higher pressure arc tubes would be worse and make an even bigger boom.

Now, when I'm arc welding, I've got some higher power arcs at work there. Generally the arc voltage is 25 - 30V or so, with currents in the 100 - 200A range, so I'm putting around 2.5 - 6kW in a little roughly cylindrical region about 1/4" long by 1/4" diameter, roughly ball parking. It varies. Arc temperature is in the 5000F range (plasma cutters get a thin arc jet of about 7000F). With a smoothe DC source, it can nice and smooth. My first machine as a kid was a little AC buzz box. An AC arc has "BZZZZZZOOOOOT" sound to it, AC "vibration", which will become quite pronounced with longer arcs. A DC arc does not have the sound, it's just more of a "HSSSSSHHHHH". and when I first started playing with DC (much, much better) an arc of a given current didn't sound nearly as strong as the equivalent AC, which took some getting used to.

One of my favorite machines is an old 50's era Westinghouse shipyard welder I managed to acquire a few years ago. It is a 3-phase motor turning a DC generator -- built like a tank, weighs about 800lbs, and was made for some serious production stick welding. It is rated 500A, *continuous, 100% duty cycle*. The DC commutator output is smooth as a baby's bottom, compared to modern eeeee-lectronic rectified DC stuff. The modern stuff is fancy and will do amazing things, make no mistake, but there's nothing like welding with a commutator machine. And they don't make too many of those any more. Too big and expensive. It is dead smooth, and a pleasure to run. I've got some high tensile E9010 pipe electrodes (electrode numbering and application is a long, long story). Those things are very hard to start and maintain. That commuator DC machine will run those easy as pie and make some beautiful beads. And it has another nice advantage. The arc circuit is completely isolated from the power system, coupled only mechanically, and you don't put any noise on the line, which can tear up radios.

I've got an inverter based MIG machine that will absolutely tear up radios and put lines on the TV screens, even in the house, which is on a different supply transformer. That's not just arc noise, but the noise of the high frequency, high current switching of the inverter supply system.

Anyway, welding arcs are a little less civilized than your lighting arcs, but nothing like a monster arc fault on a power system can be. You cheat a little bit with stick welding, using a coating that among other things such a burning the O2 out of the air away before it contaminates your bead, is easier to ionize than air at 1 atm. A "bare arc" in atmosphere would be a bit too energetic, too much voltage required and too hot to handle, not to mention making a pile of slag, rather than a good bead. Plasma cutters do use a bare compressed air arc, arc voltage around 100 - 130V (current is 20 - 50A), and that's in a different VI region than stick welding, but that more powerful arc is safely shielded and controlled in a nozzle assembly.


-Richard