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jock88
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Remember that the 'usual' domestic microwave oven is rated at about 900 watts but the duty cycle of the oven is low.
The manufacturer assumes that the domestic user will not simply run the oven for hours at a time like you would get in a commercial (hotel kitchen
say) situation.
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wish i had a kraken!!!
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Quote: Originally posted by jock88  |
Remember that the 'usual' domestic microwave oven is rated at about 900 watts but the duty cycle of the oven is low.
The manufacturer assumes that the domestic user will not simply run the oven for hours at a time like you would get in a commercial (hotel kitchen
say) situation. |
I know that , thats why I want to make a more powerful High Voltage Power supply .I think bombarder neon transformers are good for this purpose but:
1.I'm not sure
2.I don't know how to make a neon sign bombarder transformer
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jock88
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A book here gives some info. on HV transformers.
https://archive.org/details/neonsignsmanufac00millrich
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wish i had a kraken!!!
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Many thanks :-)
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gregxy
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What is the impedance of your plasma system? The same current flows in the transformer secondary as in the plasma load. If the load impedance is
too low, then most of the electrical power will be wasted heating the windings of the transformer.
Compare a neon sign with an arc welder. The sign has a very long low pressure plasma (an arc of several feet in length) so the impedance (Z) is high.
Therefore 1000s of volts be a few mA of current are used. Z=V/I = 1e6
For the welder, the arc is 1/8 inch in length, 20V at 100A are used. Z =0.2
It takes a high voltage to initiate the arc, but once started the impedance drops dramatically. Try measuring the voltage across the plasma load and
see if it remains high. If it drops sharply then try reducing the number of turns in the secondary. This will reduce the voltage and the secondary
resistance to better match the load, but will also reduce the plasma power.
Shorting the primary of the ballast makes it into a resistor which only wastes power.
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wish i had a kraken!!!
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I'll check the impedance , But I noticed when I draw arc from the PSU, after some few minutes the transformers get hot !Does it help ? I think the
impedance drop is not the case here !, by the way I want to know what type of solutions U might offer ? Shall I have a resistor in series with plasma
? How can I save the powerful plasma?
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gregxy
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When you draw an arc in air, its impedance drops to ~10 ohms.
The series resistance of the transformer coils for the MOT secondary is probably >10 ohms. Power=I^2*R. I is the same in the arc as the
transformer coil so most of the power goes into heating the transformer. Furthermore I is limited by the coil resistance. Here is a paper on arc
resistance. You need to work out the arc length, current and voltage
you want and then make sure the transformer resistance is low enough
to supply them.
http://link.springer.com/article/10.1007/s002020100074#page-...
You will probably need an external HV source to start the arc
or else start it and quickly draw the electrodes apart.
Try using your MOTs in parallel (however they need to be the same voltage
and you need to get the phasing correct). This should give 1/4 the series
resistance and more power to the load.
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jock88
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| Quote: Originally posted by gregxy | When you draw an arc in air, its impedance drops to ~10 ohms.
The series resistance of the transformer coils for the MOT secondary is probably >10 ohms.............
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It is the magnetic shunts in the microwave transformer core that do all the current limiting. The output windings (secondary) resistance has little
to do with it.
You can easily measure the current by adding a 'hot' ammeter. Just obtain a cheap moving coil meter that measures a few amps and 'hang' it on the
wire (in series of course). Treat it as hot and dangerous of course.
If you can get your hands on a variac you can turn down the output voltage / power as soon as the arc strikes. It will be an expensive option but a
good big variac is a good investment for electrical work.
[Edited on 13-1-2015 by jock88]
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gregxy
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The shunts probably do limit the current, but since the transformer is getting
hot excess power is being dissipated in the winding resistance. Anyway it indicates too much current.
If you put 2 identical 1000W MOTs in parallel, you should get 2000V at 1A,
(at least for a while, MOTs are not designed for continuous duty)
From the paper above: V = 29,000 L/I^0.4
Solving for L (the arc length) gives 8.6cm to match the arc to the MOTs.
The equation from the paper is questionable at this low current, the
lowest current they measured when fitting the eqn was 100A. There are
probably better suited papers out there.
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