Mind twisting riddles of fusor operation
Posted: Tue Dec 10, 2019 4:58 pm
I've spent last two weeks digging through the forum archives and reading books and articles on gas discharges, ionization, mean free paths, cross sections and electric fields and trying to put together what is really happening inside a fusor. The more I know, the more complicated it becomes. It's really amazing that such a simple device can have such complexity inside.
I've found following threads very interesting:
FAQ - Fusor - detailed theory of operation
#7 FAQ- mean free path
Nevertheless I've some thoughts and ideas I would like to share and discuss.
The mean free path of deuteron at typical fusor pressure of 0,03torr (0,04hPa) is in order of 10cm (about 4") so it is comparable with the diameter of the chamber. For electrons it's about 5 times more. It means that most of the electrons will pass through the space between the cathode and the anode without any collision (so it will not ionize any dueter molecule). It is consistent with the glow discharge data - extrapolating form Steenbeck measurements the length of the cathode dark space for hydrogen and iron cathode at 0,03torr will be 30cm so self sustained glow discharge at this pressure is not possible (typical glow discharges are down to 1torr, so we are operating two orders of magnitude lower).
For other hand for my fusor with grid wire diameter of 0,4mm and grid shell distance of 33mm the maximum electric field at 20kV can be estimated as 10E5-10E6 V/cm. This is quite strong field. It's one order to weak for significant cold electron emission, but should be enough for some tunnel ionization of deuterium molecules. I'm still searching for some good papers on tunnel ionization and for the moment I've no means of estimating ionization rate. As the fusor is filled with deuterium molecules (D2) not atoms most of the formed ions would be dideuterium cations not deuterons. And most of those ions will be lost either hitting the grid (they are formed just around the grid in the strong electric field) or will immediately recombine with slow electrons. The trace of this process is faint glow around the grid wires, where recombined neutral atoms return to the ground state. The electrons which have not recombined with the ions around the grid are quickly accelerated towards the anode (fusor shell).
The maximum cross section for impact ionization of deuterium by electrons is at about 70eV. With -20kV at the cathode and taking in account very long mean free path of electrons at fusor operating pressures it's almost impossible for this kind of ionization to happen. The collision must occur in the very narrow band between anode and cathode, where electrons have just right energy to ionize. So it seems that 100% electrons hits the anode. But those electrons generate quite nice stream of X-rays focused towards the center of the fusor. And X-rays do ionize and dissociate deuterium into deuterons in the whole volume of the chamber. Quite much of them has chance to be accelerated to the fusion energies.
And there is another process happening on the grid. The "lost" ions created by strong field heat the grid up and the thermionic electron emission begins, leading to increased X-ray production, thus increased fusion rate.
So summing up, my model is:
- Almost all ion current is due to tunnel ionized dideuterium cations returning to the grid and heating it.
- Most of the current flowing in the fusor is the anode-cathode electron current generating X-rays.
- All fusable deuterons are ionized and dissociated by X-rays.
- Grid heating contributes to the electron current and thus to the ionization rate.
And if it is correct:
- The resistance heated thermionic cathode should increase fusion ratio and it should happen at lower voltages and currents.
- Very permeable thin wire grid may be not as good as it seams. If there is thermionic emission the tunneling ionization is not needed, as 100% ions created this way is lost and non-fusable. And high energy, accelerated deuterons will not hit the grid that easily.
- External source of ionizing radiation should increase fusion ratio.
That it is. I would appreciate any critique, alternative explanations and pointing out mistakes.
I've found following threads very interesting:
FAQ - Fusor - detailed theory of operation
#7 FAQ- mean free path
Nevertheless I've some thoughts and ideas I would like to share and discuss.
The mean free path of deuteron at typical fusor pressure of 0,03torr (0,04hPa) is in order of 10cm (about 4") so it is comparable with the diameter of the chamber. For electrons it's about 5 times more. It means that most of the electrons will pass through the space between the cathode and the anode without any collision (so it will not ionize any dueter molecule). It is consistent with the glow discharge data - extrapolating form Steenbeck measurements the length of the cathode dark space for hydrogen and iron cathode at 0,03torr will be 30cm so self sustained glow discharge at this pressure is not possible (typical glow discharges are down to 1torr, so we are operating two orders of magnitude lower).
For other hand for my fusor with grid wire diameter of 0,4mm and grid shell distance of 33mm the maximum electric field at 20kV can be estimated as 10E5-10E6 V/cm. This is quite strong field. It's one order to weak for significant cold electron emission, but should be enough for some tunnel ionization of deuterium molecules. I'm still searching for some good papers on tunnel ionization and for the moment I've no means of estimating ionization rate. As the fusor is filled with deuterium molecules (D2) not atoms most of the formed ions would be dideuterium cations not deuterons. And most of those ions will be lost either hitting the grid (they are formed just around the grid in the strong electric field) or will immediately recombine with slow electrons. The trace of this process is faint glow around the grid wires, where recombined neutral atoms return to the ground state. The electrons which have not recombined with the ions around the grid are quickly accelerated towards the anode (fusor shell).
The maximum cross section for impact ionization of deuterium by electrons is at about 70eV. With -20kV at the cathode and taking in account very long mean free path of electrons at fusor operating pressures it's almost impossible for this kind of ionization to happen. The collision must occur in the very narrow band between anode and cathode, where electrons have just right energy to ionize. So it seems that 100% electrons hits the anode. But those electrons generate quite nice stream of X-rays focused towards the center of the fusor. And X-rays do ionize and dissociate deuterium into deuterons in the whole volume of the chamber. Quite much of them has chance to be accelerated to the fusion energies.
And there is another process happening on the grid. The "lost" ions created by strong field heat the grid up and the thermionic electron emission begins, leading to increased X-ray production, thus increased fusion rate.
So summing up, my model is:
- Almost all ion current is due to tunnel ionized dideuterium cations returning to the grid and heating it.
- Most of the current flowing in the fusor is the anode-cathode electron current generating X-rays.
- All fusable deuterons are ionized and dissociated by X-rays.
- Grid heating contributes to the electron current and thus to the ionization rate.
And if it is correct:
- The resistance heated thermionic cathode should increase fusion ratio and it should happen at lower voltages and currents.
- Very permeable thin wire grid may be not as good as it seams. If there is thermionic emission the tunneling ionization is not needed, as 100% ions created this way is lost and non-fusable. And high energy, accelerated deuterons will not hit the grid that easily.
- External source of ionizing radiation should increase fusion ratio.
That it is. I would appreciate any critique, alternative explanations and pointing out mistakes.