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Registered Member #59110
Joined: Mon Apr 11 2016, 04:35PM
Location: Camarillo, California
Posts: 74
A few friends and I where hiking last Sunday when the subject of current flow in a plasma tube came up. I've been telling them about my recent adventures making plasma stuff so it's not completely random.
I've known for some time that I have no detailed explanation as to how a plasma tube with an ungrounded end works.
Here is the set-up. A glass tube filled with low pressure gas has an electrode wrapped around the outside. This electrode is connected to a high voltage high frequency AC power source. This AC potential on the surface of the glass will induce an alternating layer of charge on the inside of the glass, like a capacitor with only one electrode. If the voltage gets high enough inside the glass tube the glass will break down and start to conduct.
This is the point at which things get foggy to me. Break down to what? I have read vague explanations about the environment around the glass acting like a virtual capacitor but would like to get more detailed than this. Does a distributed current flow through the entire glass surface out into the air?
It's experimentally obvious that placing a (sort of) grounded object (like a human hand) will cause an increase in the current flow through the glass. We have seen this many times with a commercial plasma ball.
Any plausible explanation is welcome. If it can be verified by an easy experiment, even better.
The electric field outside goes to some extent into the tube and will wiggle free electrons there. They will then hit neutral gas atoms, which will ionise, i.e. strip electrons from them, if the field is strong enough. These ions will then recombine with the electrons and emit light.
Registered Member #59110
Joined: Mon Apr 11 2016, 04:35PM
Location: Camarillo, California
Posts: 74
Sulaiman,
Thanks for the link. However I'm not sure that it answers the question I have ie What is the complete circuit of a plasma tube/ball?
The displacement current certainly explains how the current gets from the outside electrode to the inner section. And at that point is will create an electric field that, when it reaches a high enough level, will break down the gas inside the tube. But where does the current go when it breaks down?
Like I wrote earlier the only explanation I've read before is that there is a mysterious grounded capacitor around the tube that completes the circuit. If this is true why doesn't this virtual capacitor just steal the current from the outer electrode and keep the tube from lighting up at all?
Registered Member #2939
Joined: Fri Jun 25 2010, 04:25AM
Location:
Posts: 615
If this is true why doesn't this virtual capacitor just steal the current from the outer electrode and keep the tube from lighting up at all?
It does take current from the outer electrode. It just can't take all of it. Consider the situation as two parallel capacitors - they will share the current in proportion to their capacitance. You also have to remember that the voltage source has a very low impedance compared to these small capacitances - in other words its capable of supplying far more current than is required to light the tube.
Registered Member #60240
Joined: Mon May 16 2016, 07:01PM
Location:
Posts: 304
Hi alan sailer
I have seen a pulsed excimer laser beam (UV, invisible) focussed by a quartz lens, f was about 30 cm. Around the focus I could see real sparks in the air comparable to sparks generated by a small induction coil. But no electrodes! If the electric field is high enough it can produce a big number of ions and free charges in the air. As a consequence you get also discharges in the free air.
But where does the current go when it breaks down?
Currents are moving charges. There are charge carriers inside the gas, which will move back and forth due to the oscillating field. This is a current which does not require conductor connecting to it. This is similar to the arcs of Tesla coils, which can end in mid air.
In the case of a DC field positive charge carriers will pile up on side of the tube and negative ones on the other. This will eventually create a field opposite to the external one and cancel it. From there on there will be no more current.
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