FAQ: FUSION VOLTAGE VS. CURRENT AND EFFICIENCY

It may be difficult to separate "theory" from "application," but let''s see if this helps facilitate the discussion.
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Richard Hull
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FAQ: FUSION VOLTAGE VS. CURRENT AND EFFICIENCY

Post by Richard Hull »

The following discussion is really what happens, but is assumptive that the full applied potential will be applied or transferred to every ion produced. This is patently absurd in the real world and in the fusor, specifically. It also assumes that all ions produced will arrive at the fusion zone, (central grid region). Again, also patently absurd in the real world and our fusors. Given these ideal conditions, the following is how potential and current affect the fusion process within any electrically accelerated fusion system.

The potential or VOLTAGE applied to a fusor does one thing......... At any given current from one billionth of one ampere to one billion amperes (makes no difference).... The potential applied determines the PROBABILITY that fusion will occur based on experimentally derived nuclear cross sectional curves. All fusion cross sectional curves are complicated and virtually all have an OPTIMUM ENERGY for fusion located somewhere on them. Energy for any reactant or particle is related to the potential used to accelerate it. SO... the applied potential perfectly transferred to an idea fuel ion only determines the probability that you will do fusion if any two of these ideal, full potential, fuel particles collide.

The CURRENT in a fusor only does one thing......... regardless of voltage from one hundredth of one volt to one billion volts.....The current determines the ABSOLUTE MAXIMUM NUMBER OF PARTICLES you have available to bang together. More particles, (more fusor current), means more fusion for any given potential applied

To sum up...If you have an ideal fusion voltage applied, right at the peak of the cross sectional energy curve, the current will determine the amount of fusion you will be doing, (the number of neutrons produced in the case of D-D fusion). If you now reduce the voltage applied, but keep the current the same, you have just reduced the number of probable fusions, but still are paying the same price in generated hopeful particles. (Fewer of the particles you are producing will participate in useful fusion.)

Thus, at some point in applied voltage, more voltage CAN MEAN LESS FUSION! (you are beyond the optimum energy). We, on this list, need never worry about this!

Increasing current to a fusor, as long as you are somewhere on the fusion cross sectional energy curve, will always increase the amount of fusion. You will not be at the most efficient point of operation, however, until you are at the optimum voltage level on the cross sectional curves, regardless of current.

The usefulness of a fusion device has nothing to do with the efficiency of operation related to current and voltage via the cross sectional curves.

We need to supply energy to any fusion device. In all cases, thus far studied, that energy has vastly exceeded the useful fusion output energy. Such sluggardly performance has been due to losses within the device. These losses are the result of useless heating of the device by such processes as non-reactant production and attendant losses, losses within the reactant itself, (slowing down due to collisions), and simple ohmic losses within the device and its support systems.

Nothing, even on the horizon, looks hopeful as a solution to these issues which keep fusion pushed way back from any realizable, effective, energy production scenario.

The term "ignition" is the ultimate goal in the minds of all fusioneers. This is where all input or "seed" energy to a fusion device is removed once started and the device will run forever as long as fuel reactants are kept in continuous supply. (much like a gasoline engine) Again, not even a hint of a clue as to how this might be done.

Most fusioneers would settle for a process of fusion whereby we have to keep the seed energy in place and flowing constantly, but we would get more energy out of the reactor than was put into it. This is actually what has been searched for all these years, but even this far less lofty goal eludes all comers.

Fusion, in short, is maddeningly easy to do, as is so brilliantly attested to on this list, being done by rather unskilled amateur hands in poorly funded efforts. It appears, however, virtually impossible for billion dollar efforts by the brightest minds, spaning one half a century to make it into a viable power prodicing entity.

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
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Re: FAQ: FUSION VOLTAGE VS. CURRENT AND EFFICIENCY

Post by Todd Massure »

Yes, it is the current that ionizes the particles. It's a partial "dielectric breakdown" and the atoms become ions as they are disassociated from their electrons. On the atomic level it may be more violent than the typical dielectric breakdown as high energy electrons which have been accelerated outward knock electrons loose from deuterium atoms creating positive ions which then will be accelerated towards the inner grid. This is the only way that a "typical" fusor can ionize deuterium, however some other methods have been discussed.
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Re: FAQ: FUSION VOLTAGE VS. CURRENT AND EFFICIENCY

Post by Frank Sanns »

Amps are the number of coulombs per second. The fundemental charge is 1.6 x 10-19 coulombs so the number of amps exactly tells you how many fundemental charges (ions in this case) will be moving between electrodes.

There is no way to know precisely which of these ions will strike another and fuse. It is like rolling dice. There is no way to know which numbers will come up on a given roll but over time you know that they will all come up.

There is also the matter of nuclear cross section. You may want to think of it as the APPARENT size of the ion. At higher voltage up to a point, the deuteron (or any other fusible ion) will apear to be larger that it is at rest. This has the effect of increasing the chance for a collision that will result in fusion.

Frank Sanns
Achiever's madness; when enough is still not enough. ---FS
We have to stop looking at the world through our physical eyes. The universe is NOT what we see. It is the quantum world that is real. The rest is just an electron illusion. ---FS
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Re: FAQ: FUSION VOLTAGE VS. CURRENT AND EFFICIENCY

Post by Steven Sesselmann »

Have any of you guys with working fusors produced any papers,
documenting the outcome of your experiments.

I am interested in seeing some curves, showing neutron and gamma
emission at various voltages with amps consumed.

Do you all agree on optimum pressure, voltage etc?

What is the highest voltage used by anyone in the group?

Other than a grid melt down is there any other reason why efficiency
would drop with higher voltage ?

Steven
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Re: FAQ: FUSION VOLTAGE VS. CURRENT AND EFFICIENCY

Post by Carl Willis »

I did write a paper and graphs from it should be in the Neutron and Radiation Detection forum from a couple years ago. There should be simultaneous graphs of neutron emission rate, cathode voltage, cathode current, and chamber pressure. If there's a problem finding this post I'll look for it and make a link.

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Re: FAQ: FUSION VOLTAGE VS. CURRENT AND EFFICIENCY

Post by Steven Sesselmann »

Carl, did a quick search and confirmed that you have been very busy.

I found some graphs on various materials that you have activated, but
couldn't find any on the fusor output itself. maybe you could remember
some search words that I could try.

Thanks..

Steven
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Re: FAQ: FUSION VOLTAGE VS. CURRENT AND EFFICIENCY

Post by 001userid »

Givers,

Here is my rambeling on efficiency,

Efficiency in systems of course is input vs. output.
Initial review of the spherical fusor lended it to be
inefficient for one simple reasons, HIGH VOLTAGE.

High voltage in most systems tend to create energy leakage. Un
less the leak is for obvious reasons, such as creating
signals. High energy is always finding ways to escape.

Voltage as related to probability of fusion of particles is
relative to the system it is in. Furthermore, how the input
energy is used to create the fusion relative the parameters involved.

If biota create fusion (at will), How can voltage be related to
probability of fusion in this aspect?

It comes back to a simple question. How can the electron(s) be
most efficiently used to create predictable fusion reation(s).

First things first, to input the electron, the work function
needs to be minimal.

Second, the electron doesn't need to be carrying excessive
large amounts of energy after release from the conductor. (as
it will give it away at will)

Third, the electron will need a direct task and a "obvious forced
route" to achieve that task (with just the right amount of
potential and little excess). High Energy Electrons released
at random rarely provide an efficient outcome.

I have seen only a few taking the low voltage direction. Los
Alamos with Tritium production from palladium at about
1500-2500 Volts (still rather high). Kenneth Shoulders also
has low voltage aspects.

And so continues the low voltage limbo. How low can we go?

Joe S.
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Re: FAQ: FUSION VOLTAGE VS. CURRENT AND EFFICIENCY

Post by DaveC »

I think there is quite a bit more to this topic than simply potential sets the fusion probability, and current sets the gross number of events at the probability set by the potential.

The reasons I have for this are that within the volume of the fusor, ions are created more or less everywhere outward from the inner negative potential grid electrode (aka the cathode). Since this device generally operates in the glow discharge regime, and since most fusors have a glow throughout the volume, the evidence is before us that most of the volume produces ions.

The ions, having a positive charge, are accelerated toward the inner grid. Most of the ions strike the grid, causing the grid heating problems all fusors experience. A well-designed accelerating grid, will intercept very few charge particles. This I know from personal experience, having built many electron guns and done more than a few hundred ion/electron trajectory models, in my daily work. Actually we ALL know well designed ion optics do not intercept more than miniscule amounts of the ions, because the CRTs we all watch, dissipate their input energies in two places only: the cathode heater power, and the beam input power which lights the screen.

Sooo. the inner grid electrode of the fusor, one of the two large energy consumers of the fusor, is a very inefficiently designed electrode. Much if not most of the input power... coulombs/sec at the various potentials goes into grid heat. Most of the rest goes into wall heating. The tiniest fraction goes into actually creating collisions between ions.or perhaps more accurately, causing ions to deflect. This IS what goes on in the present fusors. Their simplicity comes at a substantial price.... they are inherently inefficient.

My earlier post suggesting we consider carefully ways to increase the efficiency by a factor of say 10x was made from this standoint. What will it take to increase the neutron yield, WITHOUT increasing the kV and current.

That the deuterons are made at all locations in the fusor, means that they will have a large range of energies imparted to them.

The energy they can acquire is a complex blend of the inital velocity gained falling into the grid area the very first time, and any subsequent energy exchanges through collisions with other ions. This is an extremely complicated picture to unravel.

Simply measuring gross neutron output at a point in space outside the fusor, and then extrapolating on the presumption of isotropic generation to get an inferred neutron output, is really not very accurate, despite usually being all we can do....and may either underestimate or over estimate the true emission.

Finally, a short answer to Newfusy... High voltages do not intrinsically imply any form of inefficiency. Leakage currents in well designed high voltage devices, can be as low as in your 2 cell flashlight. Witness the 3rd generation image intensifiers... 1.5 volt AA cell powered 3 stages of 15 kv or so, operating life 100 hrs or more. Now thats high voltage and high efficiency.


Great discussions, and thanks Richard for kicking it off, with a fine review, of first principles.

Dave Cooper
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Re: FAQ: FUSION VOLTAGE VS. CURRENT AND EFFICIENCY

Post by Steven Sesselmann »

Richard, As you know, my fusor is still some time away from any
experimental results, so I have been looking for documented fusor runs
from you guys with existing fusors and plenty of neutrons beneath your
skin.

What are your experiments reading in regards to voltage, neutrons amps
etc.

X Axis / Y axis
time / pressure
time / volts
time / amps
time / neutrons/sec

Has any amateur gone to 50 or 100 KV ?

Steven
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Re: FAQ: FUSION VOLTAGE VS. CURRENT AND EFFICIENCY

Post by 001userid »

Self,
In 90 seconds name 10 devices that continuously supply
30,000-100,000 volts to two electrodes exposed to each other in a gas rich environment. The devices must run for 100 hours on a flashlight battery.

Self, the list is pretty short. Furthermore, of the devices on the
list how many can I acquire or reproduce economically?

High voltage is intrinsically a part of the seed energy being used
and directly affects efficiency.(in an apples to apples sort of way)

Even looking to gain a factor of 10X I have to see the 10,000 volt
beans traveling down the pipe and say humm.

Joe S.
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Re: FAQ: FUSION VOLTAGE VS. CURRENT AND EFFICIENCY

Post by Richard Hull »

This FAQ, if you will notice, goes back to 2003 when I created it. A post is brought forward in the que by anyone replying to it in a distant future. These replys are all good discussions Thanks to Dave for adding a large bit of more info.

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
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Re: FAQ: FUSION VOLTAGE VS. CURRENT AND EFFICIENCY

Post by Richard Hull »

Sorry for taking so long to respond. I have been out of town.

Your requests are unusual in that the time axis against the values such as voltage, current, etc., are not of any value except on a complex graph together and then all values would be dependant on each other and how sucessful the operator was in bringing the device to operational status. It is not that simple.
I wish it was.

Most amateur fusors are sloppy and contaminated. Most have leaks to greater or lesser degrees and outgas from the central grid as the device is started up . One hour is often needed to get a fusor fusing and another hour to get it to fusing well.

About the best one can hope for is a "sweet spot" where the device performs well within their ability to stabilize it.

My sweet spot tend to be about 27 kv applied with 12ma of current and ~11 to 16 microns of D2 pressure. This gives about 50,000-100,000n/sec. with fusor IV.

I have pushed my fusor to 30kv + but it starts to act up and go unstable. I think that only Jon Rosenstiel has used voltages in the 50kv range.

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
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Re: FAQ: FUSION VOLTAGE VS. CURRENT AND EFFICIENCY

Post by Steven Sesselmann »

Hi Richard, welcome back.
The whole forum went quiet while you were away :)
Thanks for the data, it was as I expected. I guess your grid comes under
heavy attack at voltages over 30 KV. I look forward to get my fusor
operational , I hope to be able to go to much higher voltages, as my fusor
will not have that problem, my voltage will be limited to the breakdown
voltage of about 2.5 times the permittivity of empty space over about 70
mm. I am still looking for a -100KV power supply. Last resort would be to
buy a new one for some $8,000, pretty expensive, but nothing compared
to the divorse.
By the way, while you were away, I found a paper written by Subramanian
of Wisconsin Uni on IEC, it is very extensive, and covers a lot of
questions that come up here (written Dec 2004).

Steven
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Re: FAQ: FUSION VOLTAGE VS. CURRENT AND EFFICIENCY

Post by Richard Hull »

The main thing is that is a REAL world out there and nothing ever works according to theory unless the most rigid hold is had on every input condition and variable. (this is almost never the case.) Good luck on your efforts. You will see all too soon the difficulties involved in the doing.

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
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Re: FAQ: FUSION VOLTAGE VS. CURRENT AND EFFICIENCY

Post by Javier Lopez »

Hello all.

I think that fusor has low efficiency due to it having too low a density of the plasma. Also, heat plasma too much, so it has losses.
Also, most of the energy is used to heat electrons, but electrons do not fuse but generate a lot of losses.
Also, electrostatic pressure is much lower than the magnetic one

A question: Fusor fuses ions using electrostatic pressure to break the electrostatic repulsion of ions or only heat the ions to accelerate to collide between them?

If the response is the second, then it is more efficient the magnetic fusion


I have placed a formula to obtain magnetic and electrostatic pressure in a plasma here (I am sorry because it is a Spanish forum):
http://www.cientificosaficionados.com/f ... hp?t=22691

I cant write latex formula here, but place it here:

Code: Select all

𝑃_𝑒=\π‘“π‘Ÿπ‘Žπ‘{𝐹}{𝐴}=\π‘“π‘Ÿπ‘Žπ‘{𝐸*𝑄}{𝐴} //
\𝑖𝑛𝑑 \π‘œπ‘£π‘’π‘Ÿπ‘™π‘–π‘›π‘’{𝐸}*\π‘œπ‘£π‘’π‘Ÿπ‘™π‘–π‘›π‘’{𝑑𝑠}=\π‘“π‘Ÿπ‘Žπ‘{𝑄}{\π‘£π‘Žπ‘Ÿπ‘’π‘π‘ π‘–π‘™π‘œπ‘› }  //
𝐸*𝐴=\π‘“π‘Ÿπ‘Žπ‘{𝑄}{\π‘£π‘Žπ‘Ÿπ‘’π‘π‘ π‘–π‘™π‘œπ‘› } //
𝑃_𝑒=\π‘“π‘Ÿπ‘Žπ‘{𝐸^2*𝐴*\π‘£π‘Žπ‘Ÿπ‘’π‘π‘ π‘–π‘™π‘œπ‘›}{𝐴}=𝐸^2*\π‘£π‘Žπ‘Ÿπ‘’π‘π‘ π‘–π‘™π‘œπ‘› //
𝑃=π‘ƒπ‘š+𝑃𝑒=\π‘“π‘Ÿπ‘Žπ‘{𝐡^{2}}{2*\π‘šπ‘’_0}+𝐸^2*\π‘£π‘Žπ‘Ÿπ‘’π‘π‘ π‘–π‘™π‘œπ‘›
For example, in a cylindrical system with only ten megaamps we can obtain 400 gigapascals of compressing the plasma magnetically, but with two megavolts we can obtain only 0.32 megapascals (3.2 bars)
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