Viewing 15 posts - 1 through 15 (of 41 total)
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  • #801
    vansig
    Participant

    my understanding of the onion is, that it consists of many layers of metallic foil, eg: aluminum, arranged such that an xray photon knocks an electron out of position from a section of foil, and into a neighbouring zone where it is collected. this raises some questions.

    this is sort-of like a photovoltaic cell, but tuned for xray spectrum. do we use semi-conductors for this?

    an xray may be scattered, absorbed, or transmitted. what wavelengths of xrays are we talking about?

    if i used ordinary household aluminum foil (0.6 mil thickness, or 15.2 µm), what is the expected penetration, per layer?

    do we alternate layers of xray-transparent and xray-absorbing material, both conductive, separated by an insulator, to build charges?
    what would be its breakdown voltage?

    #6475
    KeithPickering
    Participant

    I would assume that we engineer this something like a solar cell, i.e., separate the foil with diodes, forcing any current to flow only one way, doing work in the process.

    #6534
    vansig
    Participant

    Yes. It might look like this IR – UV PV cell, recently reported.
    http://www.physorg.com/news188637189.html

    I wonder if this works as well when x-rays scatter instead of being absorbed?

    #6537
    jamesr
    Participant

    vansig wrote: Yes. It might look like this IR – UV PV cell, recently reported.
    http://www.physorg.com/news188637189.html

    I wonder if this works as well when x-rays scatter instead of being absorbed?

    The x-rays will have a continuous spectrum of energies upto around 70keV (ie the electron Temperature). most of these will be photo-electrically absorbed creating an electron/ion pair (the electron gets pretty much all energy the photon had). At these energies a small proportion will be compton scattered one or more times before the photons will be absorbed. Note the level of PE absorption rises roughly with the 4th power of the matierial’s atomic number (& hence number of electrons as a possible target)

    The primary high energy electron created by the absorption rips through the material ionizing thousands of atoms in its path, until it has slowed and given up all its energy. So if the ionization energy of the material is say 1.5eV, a 70keV primary electron will create 70000/1.5=~47000 secondary electrons. These would normally just recombine with the ions after a short time. To tap them off as an electric current you need the secondary electrons to be created in an insulating layer separated by two electrodes in which you apply a bias voltage to cause the electrons to drift to the anode & ions/holes to the cathode. If the insulator layer is too thick then the electrons will still have a chance of recombining & hence lower the efficiency

    So you want thin foil electrodes made of a light metal, separated by insulator/semi-conductor layers with a high proportion of heavy elements to absorb the x-ray photons, and allow the drift of the secondary electrons to form a current.

    #8080
    vansig
    Participant

    Would a wide band gap semiconductor do, for this purpose? such as
    Aluminium gallium indium nitride (AlGaInN) ?

    #8082
    jamesr
    Participant

    vansig wrote: Would a wide band gap semiconductor do, for this purpose? such as
    Aluminium gallium indium nitride (AlGaInN) ?

    I just checked up the various methods for normal diagnostic X-ray detectors, such as in X-Ray Data Booklet. They use different materials are used for different energy ranges an fluxes. We are operating in a completely different regeme to these detectors, as the flux of x-rays will be much larger (and harder). Although the premise is the same – to produce a current from the energy deposited by the x-rays as efficiently as possible. It is just that in our case the total current should be substantial.

    A rough calculation of 15kJ of X-rays at average of 50keV emitted over 50ns onto a 50cm radius sphere puts the flux at ~10^21 photons/cm^2/s. Which is many orders of magnitude greater than a normal detector would saturate at.

    Googling for recent papers I found one using a different wide band gap semiconductor (Fast High-Flux Response of CdZnTe X-Ray Detectors by Optical Manipulation of Deep Level Defect Occupations), quoted as high flux levels a figure of 10^9photons/cm^2/s on a 2mm thick slice.

    So unlike detectors which are built to be as sensitive as possible and absorb as many x-ray photons in a small volume as they can. We need a material that as it saturates the rest of the photons pass through to the next layer & so on. Rather than one which absorbs more than it can cope with, and the excess ending up wasted as heat.

    So if we have a flux of 10^21photons/cm^2/s and an onion with 1000 layers, (ignoring the fact that the outer layers have a larger area), then each layer has to cope with absorbing 10^18 photons/cm^2/s – still 9 orders of magnitude higher that their wide band-gap semiconductor can cope with.

    Of course this assumes the materials don’t exhibit some weird non-linear response at such high fluxes & short pulse duration. I guess the only way to really find out is by experiment (or very detailed modeling).

    #8091
    KeithPickering
    Participant

    Thanks for the research on this, jamesr. Probably easier than finding a material that absorbs x-rays would be finding a material that is *almost* transparent to x-rays (but not quite!), so that it could handle the flux. With enough layers, you could get good absorption overall without saturating any one layer. In fact, I suspect that a thin enough foil layer might be just the ticket. Studies of x-ray absorption by various metals has been done, eg here:

    http://scripts.iucr.org/cgi-bin/paper?a04630

    #8126
    vansig
    Participant

    i’m guessing that instead of a 2mm thick slice, we have closer to 2µm thick slices, and that we have many thousands more of them. also, something tells me this will be thick enough that the greater surface area of outer layers does, indeed, matter.

    this is a job for transparent aluminum

    #8128
    vansig
    Participant

    jamesr wrote:
    A rough calculation of 15kJ of X-rays at average of 50keV emitted over 50ns onto a 50cm radius sphere puts the flux at ~10^21 photons/cm^2/s. Which is many orders of magnitude greater than a normal detector would saturate at.

    something’s not right. Google calculator tells me 50 keV x 10^21 = 8.0 MJ.

    if we were emitting 5MW of xrays, that’d be 160 W/cm² at a distance of 50 cm.

    ==Edit==
    . o Oh! you mean flux as the peak intensity. that makes much more sense.

    this really is new territory

    #8133
    jamesr
    Participant

    vansig wrote:
    . o Oh! you mean flux as the peak intensity. that makes much more sense.

    this really is new territory

    Yep – I assumed although you get some soft xrays during the few microseconds of each pulse as the plasma heats up & filaments, the bulk will come out in a very short flash of ~50ns while the plasmoid it at the 50-500keV temperature.

    The flux levels are not unheard of – it is just that getting the energy out as a usable current will not be trivial. I doubt early prototypes would get anything like the efficiency needed for a working power generator.

    This is the kind of research which really could be done alongside the main focus fusion experimental program. There are plenty of x-ray sources around the world that could be used to test and play with different laminations of material.

    #8141
    MTd2
    Participant

    What is the angular distribution of x-rays around the plasmoid?

    #8144
    jamesr
    Participant

    MTd2 wrote: What is the angular distribution of x-rays around the plasmoid?

    It will be isotropic. So any part of the device, such as the base of the electrodes, pump inlets, rogowski coil etc. which covers area the onion could be gathering energy from will also detract from the overall efficiency.

    #8148
    vansig
    Participant

    seems like we want all electrical conductors to be transparent to x-ray, then.

    every layer of the onion will be a sandwich, of: a thin semiconductor, that catches a small fraction of x-rays; and two plates of a thicker foil, that carries the current, but is otherwise transparent. there is a maximum thickness, due to skin effect, as well.

    if i’m looking at this correctly, then
    within each uniformly distributed pulse, upon each layer of the onion, there’d be ~100 nm lateral separation between collisions.

    each x-ray absorption imparts an electron with a lot of momentum, like striking a pool cue ball, during the break, causing maybe 10,000 secondary collisions, and lots of electrons rise up about 4 or 5V to the conduction band. the respective plates are thus charged, and we want this current to be collected.

    the onion is looking like a big capacitor. the capacitance calculator at http://deepfriedneon.com/tesla_f_calccap.html
    gives me ~500 µF, if we choose 1000 of each +ve and -ve plates, with 1mm spacing, assuming average 5317×5317 mm overlap.

    unlike an ordinary capacitor, every other layer of insulation is the semiconductor.

    will a greater charge than the breakdown voltage develop, from a single pinch?

    #8150
    vansig
    Participant

    i just realized that because each PV cell is itself a diode, it would be better to have ~12,500 of these 4 to 5V cells arranged in series, and ~.01 mm interlayer spacing, to bring it to the 50 kV range needed for the next pinch.

    with total thickness 25cm, and average overlap of 1570 x 2000 mm, total capacitance would drop to a few dozen pF.

    easiest way to manufacture is probably to grow the PN junction epitaxially onto one side of a continuous, 2 m wide by 20km length of foil, as you roll it up into a cylinder. a continuous process would yield an even thickness.

    you’d want to pull out gases, and bake, to anneal it nicely, and finally cut through and insulate along its width, on one side, changing it from a coil into the final cylindrical PV battery.

    a 15 kJ pulse would then raise 1.2 J on each cell

    #8155
    MTd2
    Participant

    Why not immersing a PFF device in a pool of mercury and make its vapor moving helices?

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