How a fusor works! Detailed theory

[Richard Hull]

The simple fusor is the best example of how fusion can be done on the cheap.

Physical description........

In our example, we will take a stainless steel spherical fusor of the type so common to budding fusioneers found here.

The unit consists of an electrically conductive, vacuum tight, metallic outer shell.

Within the outer spheical shell casing is found a single, centrally located, electrically conductive spherical grid that is more or less transparent in that it is made up of a hollow, wire ball. This "central" or "inner" grid is supported by and electrically connected to the outside world, by an insulated, electrically conductive stalk or post which is run through a "feedthrough" insulator that is also vacuum tight.

There are other add-ons that facilitate operational control and observation, such as view ports, vacuum guage ports and the manadatory vacuum system connection which allows the air inside to be removed. There is also a mandatory gas line which allows the introduction of a fusionable gas. This gas is always deuterium in a simple amateur fusor.

What verbage describes the fusor best?..............

The fusor can be considered to be an " electrostatically focused,and accelerated, deuteron collider type of fusion device" relying on "inertial electrostatic confinement" to allow fusion to take place in "velocity space".

How is it viewed in terms of its fusion function?...................

The fusor, as a form of accelerator, is a closed electrical system, voltage gradiented device. It demands input energy to achieve fusion. It will not self sustain or achieve "ignition" as is classically sought in an energy producing fusion reactor. (none of this desirable breed have ever existed on earth). As such, the fusor is not a net energy producer.

What is the fusion reaction?..........

The reactions found in the deuterium-deuterium fusor (d-d) manifest themselves in two forms, each being roughly of equal probability. (50:50). These reactions are as follows:

d + d = He3 (.82mev) + n (2.45mev)
d + d = H3(1mev) + p (3mev)

Stated verbally, d-d can make a reaction occur that yields a Helium 3 atom, (stable), with a kinetic energy of .82mev and a Neutron of 2.45 mev kinetic energy. 50% of the time, d-d can also form a reaction that yields a tritium atom, (radioactive), with a kinetic energy of 1 mev and a proton of 3 mev kinetic energy.

All of these particles except the neutron will NEVER leave the fusor, but collide with other gas atoms in the device and or the metal outer shell wall. Here, their kinetic energy will be tranformed into HEAT and X-rays/gamma rays. NOTE** these X-rays/gamma rays can be of massive energy, (up to 3mev!), but result in a normal, external, net x-ray current in the sub atto-ampere range and effectively be undetectable due to their large penetrating power. The neutrons will pass right through the casing as if it were not there. Thus, we say that the fusor is a "neutron producing device".

There is a third reaction possible about every 10,000+ fusions that is not part of any real discussion of d-d "hot" fusion and it is:

d + d = He4 + gamma ray with about ~20 mev of energy distributed among the two particles.

What is the physical process and how does it happen mechanically?.........

The fusor device, first, has all the air extracted via a vacuum pump. This is much easier said than done. Much time, expense and effort is put forth in attaining this mandatory goal.

The required pressure for evacuation is a minimum of one micron or 10e-3 torr. It is far better if one can achieve lower pressures in the 10e-4 to 10e-6 range. Such higher vacuum levels would indicate a more professional job and represent very "clean" and well sealed fusor.

Ultimately, any vacuum achieved will be filled back up to a pressure of about 10 microns (10e-2 torr) with the reactant gas, deuterium, from which the fusion is actually derived. Getting the gas there and regulating it is another mission that must be accomplished for fusion to take place.

We now have a fusor device that is evacuated of all air and repressurized to only about 1/100,000th of an atmosphere of pure deuterium gas. There is still a vacuum in the vessel, obvisously, but all the gas in the vesssel is a fusion ready, deuterium gas.

To make fusion happen, we must apply energy externally to the device. This energy is electrical energy. This electrical energy is applied as a very high voltage gradient across the two fusor electrical components, the outer shell and the inner grid. This potential gradient can be as low as a few kilovolts to cause fusion to commence, though over 20 kilovolts is needed to make readily detectable fusion with normal instrumentation found in amateur hands.

This application of electrical energy does two very important things.

1. It supplies the energy necessary to strip the outershell electrons from deuterium gas atoms. This turns them into "ions" called DEUTERONS which are a naked hydrogen nucleus with one neutron and one proton in it.

2. The potential gradient established between the negative inner grid and the positive outer spherical shell forces the, now positive, deuterons created within the inter grid gas region to push away or be repelled from the positive outer shell and rush or be accelerated towards the highly negative inner grid. (Opposite charges attract, like charges repel)

It only takes a minimum gradient of 14 volts or so to ionize a deuterium atom, transforming it into a deuteron. With such a huge gradient as we apply, deuterons can be created over the entire fusion gas volume!! This process is called field ionization. Most of this ionization occurs near the inner grid due to the small radius wires. Deuterons created here are lost to fusion.

Due to the laws of physics and conservation of energy, the location of a particular deuteron's creation within the volume of chamber gas is of key importance in its successful fusion.

WHY?..................

In any accelerator, the bombarding particle is accelerated by falling through a potential gradient. The ideal is to have the particle fall through the entire gradient to allow it to rise in energy to the full gradient potential. Thus a deuteron falling from its creational point at zero energy, to a target of the opposite potential of say 100,000 volts would arrive at that target with a 0.1 mev energy or 0.1 million electron volts.

In the average fusor the field gradient is spherical. This is fantastic, for it allows not just a beam of deuterons to collide with the target (other deuterons) but deuterons from anywhere in the vessel to fall into a central point. This is ideally where they are at a maximum velocity and can collide with each other summing their velocities and quadrupling the collisonal kinetic energy (1/2mv^2).

In reality, and as deuterons are created all over the gaseous fill region, a 20kv fusor would have deuterons of 4kev colliding with deuterons of 10kev and of 1kev colliding with 18kev, etc, etc.

What's wrong with this?

CROSS SECTION..........................

There is a term called collisonal cross section and this relates to the probabilities of two identical particles actually being able to collide and do fusion. Without delving into the specifics, in general, higher energy particles colliding have more probability of fusing than lower energy particles. The graph of energy versus cross section are all non-linear and some are very bizarre. Most cross sectional data is gathered empirically from experiment.

There exists a well known chart for the cross section of the d-d collisonal fusion reaction. As the deuterons rise in energy there is an ever increasing probability of fusion up to nearly 3 mev. where it rolls off again towards zero. All such collisional fusion in the amateur fusor is accomplished via a quantum tunneling process not to be explained here.

It turns out that in a practical situation, for easy neutron detection, (the normal signature heralding fusion in a fusor), the fusor needs to have an applied gradient potential of about 20kv. The bulk of the fusions demand head on, near full energy collisions of deuterons. Inspite of currents of 10e14 to 10e16 deuterons/second, only about 30,000 d-d fusions occur per second. This makes the fusor a poor energy conversion device, (electrical energy to fusion energy). BUT, It is cheap and easy to fabricate judged by any other fusion energy standards around today.

What happened to all those other deuterons in the current?............

They just fall back to their creational point, for they can go no farther than their zero energy point. Thus, they turn around and are re-accelerated back for another go around. The thrifty fusor can reuse some of its old deuterons! Unfortunately, at the operational pressures in an amateur fusor, the very density of deuterons that make the device so attractive, also limits the "mean free path" of any particular deuteron to about 6 inches, though many do go farther. As such, there is some "recirculation" but even a very lucky deuteron would rarely get a third pass in an 6"-8" diameter amateur fusor. Some deuterons will not even be able to complete one pass!!

Those deuterons that do not fuse and do not recirculate (99.99999 percent of the total deuteron count), just recombine with electrons and become fast neutrals losing the energy they have slamming into the walls of the vessel, thereby, heating it a small amount.

There is also a degree of beam-on-target fusion due to fast deterium neutrals entering the chamber wall's metallic surface and fusing with other fast neutral deuterium atoms impacting them in the wall. The fusor has many fusion reactions occuring all ove the place.

Fusion process......................

All fusion relies on quantum tunneling as both deuterons are positively charged. They can never touch and fuse electrostatically! Quatum tunneling is a bit complicated to explain, but suffice it to say that the energy of the approaching particles often gets the two particles within range of the nuclear strong force which takes over from colombic repulsion and they fuse.

What about all those electrons we stripped off way back at the beginning of this FAQ?.....

These represent the greatest loss and maximal heating component of the fusor shell. These are accelerated just like the deuteron, but towards the shell wall!! They slam into it producing x-rays and accounting for the bulk of the heating of outer shell wall. NOTE**** at voltages above 30 kv applied these electron generated 'wall x-rays' will start to "shine" through the shell along with the neutrons, creating a new and very serious radiation hazard that must be shielded against.

So there you have the rough workings of the amateur fusor.

If you have gathered the fusor is a terribly ineffiecent fusion engine, then you have listened well. If you think it can be improved, then have at it in a hands on mode and keep us informed.

The one positve note is that there is no currently avaialble d-d fusion system or engine that can out perform, watt for watt out of the wall outlet, and dollar for dollar out of pocket, the simple Farnsworth fusor!

Given that we are limited to d-d fusion for the most part, as amateurs, and that materials limit other aspects of amateur fusion, the best you might hope to do is about 1 million fusions per second in a 6-8" fusor with 50kv applied to the device. Order of magnitude mechanically related improvements are just not possible beyond raising the potential gradient.

It would also be very instructive to look at the "mean free path" FAQ in the vacuum forum for more data on how the fusor works.

viewtopic.php?f=25&t=3557#p20965

Addendum:

This section added 12/14/04 relates to the rather high loss of created deuterons close to the inner grid.........RH

The fusor, as we build it, gets its ionization through field emission and the subsequent collisons of electrons with neutrals. So, where we find the most electrons is where most of the ionization takes place. The electrons and therefore the ions will be created at the highest density at points of high field gradient within the fusor. The highest gradient is just at the inner grid wires. (Tiny radius wire= high field zone)

This literally means that virtually zero ions created in the area of greatest ionization stand any chance of fusing. This is all lost energy. While ions are created throughout the volume of the fusor, the bulk of ionization occurs at a place within the volume where the deuterons can do little good.

The best fusor is a gunned fusor or a fusor with an ionizing grid. very close to the shell. While still grossly inefficient, a very much larger percentage of ions created per unit input energy will stand a better CHANCE at fusion.

With the inner grid still in place, we would still have the high losses of ionization near it. That would not go away. It might turn out that we would apply about double the energy to ionize, (one price at the inner grid and another creating them where we want them), but at least half would now be doing decent ion production in the proper region of the fusor.

Richard Hull

[3l]

Hi Richard:

A great post and timely too!
The group is getting large enough that personal attention of the old days is going out the window....just answering questions is turning into a full time job!

Thanx Richard.

Happy Fusoring!
Larry Leins
Fusor Tech

[Adam Szendrey]

Richard, i have noticed that in other posts aswell you say that deuterons are repelled away from the chamber walls. But the chamber wall is neutral (grounded, nor positively, nor negatively charged), it's only positive relative to the center cathode, but a deuteron would not see it as positive. Why would it repell deuterons? Ofcourse we can say that it seems like it's repelling deuterons, because the cathode in the center attracts them.
Is that what you mean?

Adam

[Richard Hull]

Wrongo Adam! Sorry. Never confuse a driven ground rod for neutral or zero potential. It is just a safety reference for us meat engines designed to avoid death. In an electrical circuit ground can be any voltage or polarity you want or that you design a system to accept.

Kinetic energy is vectoral fixed energy tagged to mass. charge is a scalar until you look at another charge. Then there is potential energy due to electrostatic forces this can induce or reduce kinetic energy in bound charge matter particles in motion.

The outer electrode is never neutral, to the deuteron! The deuteron is always created at near zero energy somewhere in the chamber, the potential of the outer shell is always, 100% of the time, positive and the inner grid is always, 100% of the time, negative so far as any deuteron is concerned. It will never change, it can never change.

Let us say we have a positive deuteron which is field created in a 6" fusor with 20kv applied. Let us say it is created at a distance of 1.5 inches from the poissor center. It sees the poissor as being 10kv negative and very attractive and inviting. It sees the shell wall as being a very repellant 10kv positive. Which way will it go? Towards the inner grid only. Barring any other happenings, if it totally misses anything in the poissor adventure and speeds out the other side, it now sees a 20kv totally repellant wall coming right at it. But hey, it now has 10kev of enegy to burn and plows onward against the electrostatic repuslive forces until it just slows to a stop at exactly 1.5 inches out the other side of the poissor and 1.5 inches away from the wall of the shell. it now is still positive and has zero energy, but that grid is lookin' awful pretty again and so ZIP off it heads back to the poissor for another pass. No matter how many trips it makes, it will never enter the poissor with more than 10kev of energy and it will never get closer to the shell wall than 1.5 inches.

The important thing is that inside the Fusor shell we have a CLOSED ELECTRICAL UNIVERSE! Outside of the shell, in our world, it is a grounded point and will not kill us. The inside of a fusor is a gardient driven, closed electrical system. All deuterons are permanently electrically tagged at birth and will always see the spherical wall as positive as few deuterons are created at the wall surface. Likewise, they will always view the poissor as negative. There is a nearly infinite number of deuteron shells orbiting at a nearly infinite number of distances all arriving at the center of the poissor with a finite range of energies from 14ev to just under 20kv. By the way, the electrons are of the same mindset, but they have NO chance of survival as they can never recirculate via the same mechanism as the deuterons for they are forever doomed to slam into a hard, attractive, shell that is always positive with almost 100% loss of all energy. The electrons would also have energies from a few ev to about 20kev.

If, at any time, a flying deuteron, for any reason becomes a neutral, it is doomed to hit the fusor shell wall unless reionized on the way there.

It is not charge that keeps deuterons from ever hitting the walls, it is where they were born in their electrical universe, within their sea of electrical energy gradients. You can't change where you were born.

Again, due to the nature of the geometry of the system, 99.9999999% of all d-d, d- neutral and neutral neutral collisions are DETREMENTAL to BOTH particles pre-existing vectoral energies. This translates, wonderfully, into more fusion opportunities (only at the time of impact) per unit collision and less individual particle energy gains due to the same collison in question.

I hope this clears things up.

Richard Hull

[Adam Szendrey]

Thanks Richard,

What i meant is, if i would have an outer shell, that has an absolute zero charge (not relatively), it would "look positive" to a deuteron inside the chamber, because there is a cathode in the center. What i meant by "a deuteron would not see it as positive", was for a situation when there is no external electrical field present, and i would "put" a deuteron near an absolutely neutral metal plate (just a thought experiment).
Essentially i said the right thing (and put it in wrong words), that because of the attraction of the cathode the deuteron sees the chamber wall as positive, but ofcourse this is a crude way to put it. But after all, potentials are relative. The chamber wall can be considered +50 kV and the cathode to be zero, ofcourse i understand this. It was just a bit confusing , saying that the chamber wall is positive, because it translated to me as if the chamber was positively charged to the outside world. My mistake! It should have been obvious to me what you mean.
Your explanation is much more accurate, thanks.

Adam

[Richard Hull]

Thanks for you clarification on what you meant. The key is to realize that inside a closed conductive sphere you have an isolated electrical universe. Charges and relative potentials created within that mini-universe have no communication or even relevance to the world outside. This all goes away if you have leadin wires from outside (like in a van de graff). It returns in a fusor even with lead in wires due to the creation of unit charges within the volume. At this point they break free of that communication. The leadin wires at this point just supply and maintain an internal potential gradient. The fusor is truly unique.

Obviously, the actual realized external fusor current in a fusor is a 50:50 balance of dueterons crashing into the inner grid and electrons carshing into the outer shell. Equally apparent is the fact that at any moment of time within the fusor, there is an unknown quantity of balanced electrons and deuterons involved in fusion, missed fusion, neutral production and ionization. These are part of the plasma free inside the device.

Based on a simple ammeter reading, you can calculate the quadrillions of dying deuterons and electrons every second which are the bulk of the losses within the device.

Some mental calcs shows about 10e16 particles per second are doing nothing but dying in an average fusor. Based on 10e5 fusions/second that would be a ratio of 1 fusion for every 100 biilion deuteron-electron pairs. Not very pretty. Equal in elegance to killing a nest of wasps with a tactical nuclear weapon.

Richard Hull

[DaveC]

Nice, helpful summary Richard. You have good way of explaining.

I had just recently calculated fields thorughout the normal fusor, and concluded that ionization is essentially spontaneous everywhere inside. So as you have said, deuterons are created literally everywhere. Since the potential falls off as 1/r inside of spherical electrodes, the higher voltages will allow a greater fraction of ions to reach a fusible collision energy.

Thank you.

Dave Cooper

[badflash]

Dave-
You say the field drops off as 1/r and not 1/r**2. Is that due to the spherical geometry?

[walter_b_marvin]

Im sure this has been asked before, but is there any way to frequency modulate the conductors so as to prevent the deuterons that miss from hitting the wall by reionization? I would envision an oscillation of the collision locality at one "sweet" frequency

[Adam Szendrey]

Though i think nobody could verify this (not even Richard Hull, who has talked to some key members of the Farnsworth team, so i'm a little sceptical about it) but i've read in a book that the original fusor team used a 100 MHz (well, around 100 MHz) HV RF power source. Indeed resonance has been discussed here before, but not too much conlcusion has been drawn. It's not pulsed, it's not DC, it's not just simple RF operation, it's a resonant RF operation.

Adam

[walter_b_marvin]

Yes duty cycle frequency and wave form are all questions on resonance. On thinking about it what one really wants is a ping-pong paddle effect. A high enough potential to ionize virtually all the duterium between the meshes. Then a resting period to allow them to fuse, or miss each other and recombine. Then hit them again before they leak out of the outer mesh. This assumes the containment vessle is not the outer mesh. For a particular geomentry it should not be too hard to calculate a resonate frequency. Initially a square wave would be used. I expect a slightly larger than 50% duty cycle to do the ionization and push them beyond the inner electrode would be required. If the ions are accelerated to saay 10% of the speed of light, then they travel about 4 inches per nanosecond. So I expect the frequency to be several hundred Mhz to 1 GHz. Doesnt sound that practical at Kilovolt potentials.

[Sanchez-Yamagishi]

Just to point this out because it confused me, but the product energies for the DD reaction should be Mev, not mev. This probably would seem obvious to anybody who read it but just for those who are new to these conventions..

great read nonetheless though.

~Javier D Sanchez-Yamagishi

[Richard Hull]

I think all involved in this site know we are talking millions of electron volts in nuclear reactions and the M vs. m formality is often dropped of left uncorrected in the hast of battle in hammering out posts.

A lot of strict scientific formalism is often abandoned here. I am most often guilty of it. I know that it is 3He for helium 3, but always post it here as it is spoken ,"He3". We play over such a narrow field in these forums on amateur fusion that the lock step formalism is not enforced. Most anyone here would pick through any form of jumbles associated with a post and understand. Still, good habits here might save embarrasement before more august associations in future.

I would certainly use the formalism in any scientific paper or communication with a publication or such as that. I would also proof read the material more carefully than I do here.

Richard Hull

[DaveC]

In calculation velocities remember to use the ion mass, which is about 3700 times that of an electron. So speaking approximately, ion velocities are about 60 times smaller than electrons for the same energies. This gives a few MHz for frequencies.

A particular excitation frequency should tend to pick out a particular energy of ion. Thus tending to "bunch" or group ions . Whether this is beneficial to fusing efficiencies is hard to anticipate. But it is an interesting concept.

To Marvin's question, the potentials fall off as 1/r, which makes the fields vary approximately as (1/r)^2.

Dave Cooper

[Sanchez-Yamagishi]

Less trivial comment this time, something that has been bothering me for a while.

What are the energy ranges to cause D-D reaction to happen? Shouldn't it just be the energy necessary to overcome the Coulomb Potential to come within range of the strong force? So we have:

W = 1/(4*pi*epsilon) * (q^2 / r )

with r = 10e-15 m, the range of the strong force you get energies on the order of 1.4 Mev.

even if you ease up on r, because there is the possibility in D that they will collide with the neutrons facing each other so you have r = 3*10e-15 m you still get 0.5 MeV.

Given that the only thing driving the ions is the potential, the highest KE energy they can reach is the magnitude of the bias so how does fusion happen at anything less then 500 kV, let alone at a few kV.

I'm assuming the answer here will probably be through tunneling, I just haven't found it mentioned anywhere else though

[Richard Hull]

Quatum tunneling is indeed the fusion enabler at the lower kev levels. Actually, at levels approaching 2mev or more the oppenhiemer-phillips reaction takes over and fusion diminishes as deuteron "stripping" occurs more frequently than fusion.

So D-D fusion starts at a few tens of ev and continues up to well over 2 mev with the best fusing region occuring just around 1 mev. Oddly, there is little difference between 200kev and 1 mev based on the cross sectional curves.

In a well designed device using a very good, plus ultra neutron detection device it is easy to detect fusion at 10kev. At 20kev, even a poor neutron counter will indicate fusion. Easy to detect neutron activation experiments can be done in a fusor at 30-35kev applied.

Richard Hull

[Alex Aitken]

In a pure deuterium plasma, fusion is a stripping reaction.

When a deuterium splits into a proton and a neutron at >2mev energies this is called 'break up'.

http://www.helsinki.fi/~fyl_www/common/ ... cla98.html

For an example of some other stripping reactions.

[MD1994]

So from what I read it seems that the d ions will tend to return to there orginal positions and if thats 1.5 in away from the center of the inner grid orginally then if will return to 1.5 in away from the inner grid if it doesn't fuse. So wouldn't it be possible to use a a positive inner grid to repel ions and attarct electrons and ionizing them, but then could one change the inner grids charge to negative allowing all the ions that were formed in the walls of the fusor chamber to rush to the center and any of them that don't fuse couldn't they keep going to a distance equilvalent positions and since all the d ions were formed at that distance there wouldn't have been any d ions formed in the center that weren't accelerated and so they weren't fused. Of course you would have to do this I believe every time you put new deutrium gas into the chamber. But I just want to know if this is where they are born thing still applies in this situation, and if this will allow a greater number of d ions to fuse. Correct if I'm wrong which I probably am, sorry if I wasted anyone's time with this.

[Richard Hull]

In theory, in the simple fusor, and barring any collisions, an ion is doomed to perpetually circulate through a more of less central point in the device in an orbit with its starting point being where it is created. Thus, its energy can never exceed that of its relative creational, positional, acceleratory, fractional energy related to the potential difference between the outer wall and inner grid.

The above sad tale was in theory. However, in fact, it is usually even far worse than this already bad situation! Thus, we need hundreds of billions of ions created each second to get just a few tens of thousands to fuse. There are about (20!+6) valid scenarios of what can happen in a simple fusor to any one specific ion given an acceleratory potential of 40kv and an operational current 11.8ma when operated at something both above and below STP.

** Note: The (20!+6) factor shall henceforth be known as The "Hull ion life scenario factor for velocity space". This factor was carefully derived, metaphysically, while musing upon th' throne.

This is why it is called a simple fusor. Adding grids and thermionic cathodes can, again, in theory, aid the process by some factor, yet to be determined by any amateur.

Try your ideas and get back to us on how it worked out with full data and details versus your results using a simple fusor. Ideas need to be tested. Don't expect me to applaud or poo-poo your idea. A bit of "hands-on" will separate theory from actual experience. Pumping fusion up by novel ideas is possible by inch-worming degrees, but energetically useful fusion outside of a multi-mega-ton H bomb blast is currently impossible.

Wanna' destroy tens of thousands of lives? We've got your fusion solution.
Wanna' keep warm in the winter and watch some TV? Burn some coal.

You will not improve the simple fusor, save in microscopic steps. The more steps the more improvement and then the fusor will not be simple anymore, but become an evermore bloated, power hungry and complex megatherium of its former self.

Richard Hull

[Richard Hull]

Pushed to the top of the pile as there is no FAQ banner or repository in this forum.

Richard Hull