Understanding current draw on a fusor

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Adrian Turner
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Understanding current draw on a fusor

Post by Adrian Turner »

So, without doing enough research on the subject, I thought a fusor didn't actually draw much electrical power and rather the light and heat produced inside of the jar was a result of the power produced by the fusion reaction occurring. With this in mind, since I already have a vacuum pump and high voltage test equipment on hand (I have access to either a 75 kV DC or 150 kV DC hipot through my job) I thought I could achieve this task relatively cheaply.

I work in the electric utility industry and in my experience, vacuum is a good insulator, it is better than air. Some low and medium voltage circuit breakers have the interrupting contacts in vacuum bottles because of this. With that knowledge, I intuitively thought there would be almost no load on the fusor. So I built my fusor and tested the resistance between the inner sphere and outer with no vacuum using a Megohmmeter (megger) and I had about 300,000 Megohms @ 15kV, very good! As soon as I applied a vacuum the resistance went down and the load increased! The needle indicating resistance pinged out low and the voltage dropped to 1000V. So, I decided to use a hipot instead this time, also used for testing purposes, but has a higher VA. The hipot it trips out at 7 microamperes, and I could only get about 2000V before I reached that level. At that voltage I had some arcing inside and small amount of purple corona?

I can't wrap my head around it. What is going on inside that chamber that makes it want to draw more power when it has a vacuum instead of less?

Any suggestions for a power supplies? Would using a 15kV 60mA NST and a voltage doubling circuit be a good choice? Furthermore, NSTs with ratings like that, does that mean it can produce a continuous 60mA at 15kV?

What VA rating should the transformer supplying a fusor have? And roughly how much current would you expect on the secondary (high voltage side) of the power supply at 30kV?
Andrew Haynes
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Re: Understanding current draw on a fusor

Post by Andrew Haynes »

Hi
The way I understand it is the high voltage ionizes the atoms make a plasma. Plasmas conductive formula is
k = e^2*Ne*t/u
k = conducive
e = electron charge
Me = electron number density
t = mean free path
u = electron mass

P = 1/k
P resistance constant

Ohms = p*(l/a)
L = length
A = area

With vacuum the mean free path gets A lot longer more than the number density, to a point.
A vacuum is like a short circuit, if the voltage is high enough to make it a cold plasma.
Andrew Haynes
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Richard Hull
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Re: Understanding current draw on a fusor

Post by Richard Hull »

The fusor will go back to near infinite resistance if the vacuum goes much below 10e-5 torr. The nature of the fusor and dong fusion demands a working pressure of about 4 to 20 microns. (4-20 millitorr - 4-20 X 10e-3 torr.) The voltage applied will force a bit of electron field emission to create ions which snowballs to appear as a virtual short circuit (actually a rather stiff load to any high voltage applied.)

The fusor operates on the edge of a full blown arc-over condition. The art is in adjusting the vacuum and voltage correctly so that you have a plasma glow and yet limiting the current to about a few milliamps to over 30 milliamps in range for any voltage. It is an art.

The neon transformer is a special case and will self-lower the voltage under load. Thus, no fusion can be done with a neon transformer as it doesn't pack the gear needed to have a sustained current due to its magneitcally shunted core self-limiting both its voltage and current under load. Neon signs work this way. They need 5 to 15 thousand volts to start, but then want 5000 amps to keep running. The neon transformer is designed to buckle under the load and drops back to a running, "stay lit", voltage of about 600 volts at 20 milliamps.

If a working fusor was heated due to fusion alone, you would not live more than a few hours after operating it. The radiation would kill you real quick.

A good fusor, operated by an old hand, will produce a few millionths of one watt of fusion energy, while you pour in about 500 or more watts from your wall outlet. The source of the heat is wasted electrical power in the device to get the one millionth of a watt out in fusion energy. (About a billion to one net operational loss!)

Remember, doing fusion is, technically, very easy to do. However, doing fusion for actual net energy output is currently absolutely impossible even if you have a billion dollars to spend on the effort. We know!....Billions have already been spent by world governments on fusion. Successful fusion has been done. Yet, not one watt of usable electrical energy has ever been produced by fusion in a controlled manner.

Richard Hull
Progress may have been a good thing once, but it just went on too long. - Yogi Berra
Fusion is the energy of the future....and it always will be
The more complex the idea put forward by the poor amateur, the more likely it will never see embodiment
Adrian Turner
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Joined: Sun Feb 19, 2017 5:24 pm
Real name: Adrian Turner

Re: Understanding current draw on a fusor

Post by Adrian Turner »

Very good answers, thank you!

Also, part of my question was already answered in the FAQs so I apologize for that, but thanks for answering a question you have seen probably 1000 times, Richard.
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