Farnsworth Fusor

Fred's Fusor


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Date/Time: 2015-07-29 06:11:22 PDT

Name: Zorislav Casar

Feedback: You made a excellent work. Thanks for very useful information from your site.
I've made aluminizing chamber for telescope mirror and planing to modify it to fusor.

If you are interested there is some my photos on this sites:

It's in Croatian but pictures maters.

I hope you will help me in my task.
Thank you

Thank you for your kind feedback!
Looking at your impressive photographs I’m sure that a fusor is a very suitable new project for you.
Please let me know when I can be of any help to you.

Date/Time: 2015-07-29 21:10 GMT+1

Name: Zorislav Casar

Hi Fred,
Allow me to inform about my current fusor status, but first Sorry for my bad english.

Since D2O is not cheap (I bought 2,5cc Deuterium Oxide 99,90% purity from ebay for 9€)

I need something to produce minimum quantity of D2:

Option 1:
Making micro electrolyser capable to work with few drops of D20 (picture v0.1 below).

Option 2:
Built small PEM cell.

Option 3:
Buying this:
But I didn't find minimum quantity of D2O to soak PEM membrane for this device

To reduce D2 consumption (since My chamber jar is pretty big - ID 30cm, Height 50cm ) I'm thinking to replace it with smaller one.
Any suggestions for this my ideas.

Electrolyzer v0.1


You do not need to apologize for your English. It is very OK and I do understand very well your questions.

With 2.5 cc D2O you can in theory produce about 2.5 liters of D2 gas.

Option 1:
The drawing of your micro electrolyses is very ingenious. The disadvantage is however that you need to add baking soda (NaHCO3) to your D2O to convert it into an electrolyte. This will produce H2 gas impurity in your D2 gas, which is unwanted.
Please read my webpage about D2 electrolysis: http://www.fusor.eu/deuterium.html

Option 2:
The smallest PEM cell available can be found here: http://fuelcellstore.com/horizon-mini-pem-fuel-cell with dimensions outside of 3.2 x 3.2 x 3.2 cm

Option 3:
The standard Horizon PEM cell needs about 5 cc of D2O. You may need two bottles of 2.5 cc. Note: it is important that the PEM tissue always remains soaked or the cell will be destroyed. Therefore after use store the cell wet in a hermetically sealed plastic bag.

Fuel consumption by the Fusor in absolute volume per time unit is very little and rather independent of size: 1 SCCM (one standard cubic centimeter per minute). It is not the size of the chamber that determines the amount of D2 consumed but the absorption by the grid and the loss through the vacuum pump. The size of the chamber is only permitting the length of the way that the Deuterium ions travel and permit them to gain speed for fusion. You are not saturating the total volume of the chamber with ions but concentrating the ions with high speed due to high voltage potential into the grid sphere area where they are supposed to fuse. When they not fuse they will pass the grid (or hit the grid wire and be adsorbed and go lost) and travel through the chamber for gradually loosing speed and then reversing their direction and gaining speed again towards the grid again and so on.

Note: A Deuterium gas flow of 1 SCCM introduces in the vacuum chamber per minute a quantity of n=PV/RT = (1 atm)x(0.001 L)/(0.08206 L atm mol-1 K-1 x 273.15 K) =  4.46 x 10-5 mol. The gas constant R has a value of 8.314 J/mol K or 0.08206 L atm mol-1 K-1. Avogadro's number is used to convert from moles into molecules: 4.46 x 10-5 mol x 6.02 x1023 molecules/mol = 2.69 x 1019 molecules Deuterium per minute or (theoretically) 5.37 x 1019 Deuterium atoms per minute or 8.95 x 1017 Deuterium atoms per second, which may turn into ions when all Deuterium atoms will become ionized. Ionization of Deuterium atoms depends on the number of charges occurring in the vacuum chamber. A current of 1A is equivalent to 1 Coulomb per second which is 6.24 x 1018 electrons per second. In our example fusor we apply a current of 10 mA, which equals 6.24 x 1016 electrons/sec, potentially creating an equal amount of ions per second. Ions will be circulating a number of times, expressed as the recirculation factor which may vary between 10 and 100. The total amount of Deuterium ions present in the vacuum chamber at a given timepoint will therefore be higher than the number of ions per second formed by ionization. This calculation is, however, based on the flow of Deuterium gas into the vacuum chamber. However, we know that the vacuum in our example fusor is kept constant at a pressure of of 10-2 Torr (10 micron) and that our spherical fusor has a radius of 10 cm. The volume of the vacuum chamber as a function of the radius is:
volume bol
or 4.1888 x 103 = 4188.79 cm3 which equals 4.18879 L.
Since nV=P/RT and R = 62.3637 (L Torr)/(mol K), P = 0.01 Torr, T = 273.15 + 20 = 293.15 K and V = 4.18879 L we get n = (4.18879 L x 0.01 Torr)/(62.3637 L Torr K-1 mol-1 x 273.15 K)  = 2.459 x 10-6 mols in the vacuum chamber, which equals 2.459 x 10-6 mol x 6.02 x 1023 (Avogadro's number) molecules/mol = 1.48 x 1018 molecules in the vacuum chamber or 2.96 x 1018 atoms Deuterium, which (may) get ionized.
We notice that with a flow of 1 SCCM of Deuterium gas, a pressure of 10 micron and a current of 10 mA, one or the other way we obtain roughly 1018 Deuterium ions per second in our Fusor.

With a constant flow of 1 SCCM Deuterium gas, roughly per second 1018  Deuterium ions will enter the vacuum chamber and randomly occupy the volume of the vacuum chamber. The attracting forces of the cathode compete more or less with the "suction force" applied by the vacuum pump(s). This "suction force" is not to be taken litterally as the pump(s) only constantly remove(s) gas particles which causes the remaining particles inside the system to re-arrange themselves in order ideally to get an equal distribution. At the same time more Deuterium ions arrive into the system (over 1018 per second!) and an equilibrium will be the result. This equilibrium can be observed from the fact that the instream of Deuterium gas is regulated in such a way that the pressure (vacuum) of the system is kept at a constant pressure.

Date/Time: 2015-07-30 08:27 GMT+1

Name: Zorislav Casar

Dear Fred,
after some thinking I'll try to build micro electrolyser but with addition of
high voltage short pulse generator. This option will (drastically?) reduce amount of additional salt.
(Google-d "Water Electrolysis with Inductive Voltage Pulses",...)

It will take some time to make this piece of hardware and after that I'll inform about the result.

Thank you for informing me about the inductive pulse electrolysis process.
I was not yet aware of this technique, but i am an old school person!
Next weeks I shall update my website with this info.

Please keep me informed about your progress.

Good luck!

No update of the website has been made so far due to complexity of the suggested method. Constructing a high voltage DC nano pulse generator (HVNPG) requires a special static inductive thyristor (a SIThy), which is manufactured by NGK in Japan but appears not to be available to private persons.

Concerning this HVNPG electrolysis method a search on the internet yielded information with the following timeline:
  • Japanese team discovered/invented nano-pulse method, published article in 2005-2006 (reference 1).
  • Japanese team receives 1st August 2006 United States Patent 7,084,528 B2
  • Latvian team analyzed pulse behavior in various electrolyte concentrations, 2011 (reference 2).
  • Next, in 2012 an article was published in India with results (reference 3).

The researchers in india reported a power saving, compared to conventional electrolysis methods, of 98.6%!
Most amazing, however, is the fact that with the HVNPG method the so-called overunity world comes peeping around the corner. This fact and the fact that the SIThy cannot be obtained makes it impossible (or inconvenient) for me to publish it on this website.
For those who are interested to receive the concerned PDF files (literature as mentioned in the timeline) that I have collected, please ask me through the feedback submission modus on this page.

Date/Time: 2015-09-23 22:46:26 PDT

Name: Tom Clemmerson

Feedback: I'll be your first feedback :) I am the guy you got the small PMT and LiI crystal from. The two things you need to get set up and going first are the vacuum system and the HV system. Practice using air, nitrogen, helium, then pure hydrogen for your fusor before using your deuterium. Instead of the usual spherical geometry try a tetrahederal wire frame geometry for the cathode. It will reduce scattering and increase actual fusion. Oh btw, I have some green fast neutron detection fibers currently.

Hi Tom,

Thank you for your feedback (and the PMT and LiI crystal 😄). I’m sorry that it is the second feedback and not the first one, which was not yet published due to the fact that I needed to sort out the high voltage nanopuls electrolysis issue first.
Nevertheless your feedback is very welcome.
The vacuum system is almost complete, though a rotary vane pump is still in the process of being overhauled.
The HV system gives me more problems. I do have the X-ray transformer but it has a very low duty cycle. Using this transformer requires heavy cooling of the transformer oil in order to make it operate for a sufficiently long time. Therefore I shall need a cooling compressor with a liquid/liquid heat exchanger. This is a general problem is setting up the project: having realized one issue, another issue turns up. BTW Any suggestions for a better and cheaper HV power system?
It is indeed my intention to practice with air, helium and hydrogen prior to using deuterium.
I shall dive into your suggestion of a tetrahedral wire frame geometry for the grid. This geometry is new for me. Is there any reference to this geometry or is it your own experience?
I noticed the green fast neutron detection fibers on eBay and I’m interested in assembling a fast neutron detection system. Is one fiber sufficient for detecting neutrons? As you have a lot of stuff on eBay: what do I need more (which PIN photo diode assembly?)

Date/Time: 2015-09-25 02:15 GMT+1

Name: Tom Clemmerson

The tetrahederal geometry effectively creates an potential well gradient at the center of the device rather than an infinitesimally small zeropoint focus with a zone of spherical aberration around it. As you may be aware the main effect in these fusors is scattering with a standard spherical geometry. Simply it is very hard to make a truly perfect and uniform field with a non uniform wire cage. In a nutshell using a tetrahedron wire frame will make the nuclei more likely to fuse because the probability the nuclei hitting each other with the magnetic field vectors lining up correctly is greatly increased. A couple magnets from a magnetron aligned at the base and top of the tetrahedron will increase the acceleration even more, but you'll have to find the sweet spot as the increase also comes with deflection the stronger the field gets. What you will end up with is a hybrid z-pinch/fransworth fusor.

To make the x-ray transformer run at a better duty cycle, you should get a larger transformer say for a GE, Ritter or SS White x-ray head. Using a dynarad field x ray transformer would be the best though since it is designed to be actively cooled. The trick to cooling them is to use a manifold with jets that actually make the oil flow past the transformer. as for cooling the oil, a simple $30 transmission cooler from an auto parts store and a couple of server fans will do the job nicely. The one you have looks like a Orilix 70 dental machine transformer, and yeah it is a bit difficult to cool. You will also need to soak it in hot oil and vaccuum out the air and moisture or it will self destruct- violently.

The neutron detecting fibers were used in an array in a device (64 of them) that looked for atomic bomb material down to the low miligram range from a distance of at least 6 meters. One or two fibers should do fine for fusor neutron detection since it can yield a decent amount of neutrons when it is reacting strongly. They would be best used in combination with your LiI in a moderator as a confirmatory measure. Your moderated LiI will detect first, then when the fiber starts detecting, then you know you got fusion going in inside. They are best read with an APD detector. The one i recommend for you is the unit with the four wires, as they came from a device that used a laser to measure the weight of very small particles and can sense a single photon when cooled and well shielded. Another item I recommend is the CsI:Tl+ photodide low energy gamma detector to measure the x-ray output for safety and other reasons like detection of sputtering arcing before any real damage is done.

Anyway, I hope this information is helpful to you :)

Your information is very useful to me and it makes me think away from the current path.
I have a feeling that the tetrahedron grid concept comes near to Bussard’s patents for realizing standing waves of ions. Apparently it should also be beneficial to adapt the outer vacuum chamber into a similar tetrahedron or even octahedron shape instead of the traditional spherical shape? Anyway, I shall need to dive deeply into this matter as it is an interesting issue.

Your suggestion for the air cooled oil cooling device for the X-ray transformer is also a good idea. The funny thing is that a transmission cooler will be difficult to find in Europe. The majority of cars here have manual gear boxes! Nevertheless, an oil or water cooler as used on certain motorcycles can be found easily.

Concerning the fiber, the ADP and the CsI:Tl thin detector with photodiode: I hope to be in business with you coming weekend!

Date/Time: 2015-10-13 17:12:50 PDT

Name: Roberto Ferrari

Congratulations for all your effort and great presentation.
I have been around forums and web pages-fusor related and I have a question:
What would be wrong using heavy water vapor as fuel?
Would the negative ions bother deuteron interactions?


Thank you for your feedback.
Those fusioneers who use heavy water to produce deuterium gas for their Fusor always put a gas drier in the supply line to the Fusor to eliminate heavy water vapor. Apparently, heavy water vapor in the Fusor has a negative effect on Deuterium fusion.
Knowing this, it never occurred to me to try to find the scientific rationale for this phenomenon.

Nevertheless it is known that the impact of electrons on heavy water vapor will cause radiolysis of the heavy water and form a selection of different ions: D2O+, D+, OD+, D3O+, O+, D2+, O2+, OD-, D- and O-. The mentioned ions will not be present in equal quantities as this may differ depending on the energy applied. The relatively huge amount of different particles will obstruct fusion of D+ particles. In this perspective we should note that for proper fusion of Deuterium gas it is essential that the vacuum chamber of the Fusor is evacuated up to a pressure (vacuum) of 10E-3 Torr (1 micron) and preferably better, i.e. 10-4 to 10-6 Torr (0.1 to 0.001 micron), followed by filling with Deuterium gas up to a pressure of the vacuum chamber of 10-2 Torr (10 micron). This shows that a relatively high amount of D+ particles are required to obtain fusion.

Note: see above the earlier answer where has been calculated that roughly 1018 ions Deuterium per second flow into the vacuum chamber.


engineering works
control systems
Ion Source


Ref. 1: NAOHIRO SHIMIZU, SOUZABURO HOTTA, TAKAYUKI SEKIYA and OSAMU ODA: A novel method of hydrogen generation by water electrolysis usingan ultra-short-pulse power supply, Journal of Applied Electrochemistry (2006) 36:419–423

Ref 2.: Martins Vanags, Janis Kleperis, Gunars Bajars, Andrejs Lusis: ANALYSIS OF INDUCTIVE CURRENT PULSE DYNAMICS IN WATER ELECTROLYSES CEL,4th World Hydrogen Technologies Convention, 2011, Glasgow, U.K., Paper ID: 0103

Ref. 3: Dharmaraj C.H, AdishKumar S.: Economical hydrogen production by electrolysis using nano
pulsed DC, International Journal of Energy and Environment (IJEE), Volume 3, Issue 1, 2012, pp.129-136

Last Updated on: Wed Oct 21 11:02:54 2015

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