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Investigations into the Radient Spike

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  • #16
    Hi ZPDM --

    I like your video and the information I am seeing on the results. I see that the capacitor marked as 100 ufd @ 400 volts being used at far less voltages then the rating is one of the factors for the great results. When using a capacitor for keeping a smaller voltage longer you need to use a far greater rated capacitor. Now the 1 ufd Orange Drop type capacitor I believe is a capacitor rated at 400 volts to 600 volts but I may be wrong on my guess. You can let me know what that voltage rating is. The other variable which is unknown is the coil value. Could you tell me the wire gauge and measure the resistance of that coil? If you have a LCR meter I would like to know the exact measurement at 10KHz frequency or more depending on your LCR meter if you own one. It would be nice to see the results with this type of capacitor below to see if it holds a charge longer then a Orange Drop type Capacitor.

    Polypropylene capacitor applications

    PP capacitors find uses in many areas of electronics. Although only available in leaded versions they are still widely used.

    They are seen in many applications:
    • High power / High AC voltage circuit applications.
    • Circuits with high peak current levels.
    • High frequency resonant circuits.
    • Precision timing circuits.
    • Lighting ballast systems.
    • Switching power supplies.
    • Sample and hold circuits.
    • Premium audio applications where many enthusiast believe they offer better performance and hence a better sound quality.
    • High frequency pulse discharge circuits
    • Energy storage circuits - their high resistance enables low levels of self discharge.
    Dielectric: Polypropylene (PP) %3D

    Information PDF on this type of capacitor.

    -- James


    • #17
      HI James,

      Glad you like the video and are enjoying the results. You are correct the 100uF is a 400 volt cap, the 1 uF is a, would need to double check, but I believe a 640 volt film capacitor. I will measure the coil resistance and capacitance and get back to you with that. Was thinking of moving on but will spend some more time on this as you bring up some good questions. One thing I will tell you is I don't think this is a finicky finding at all, but fair enough, I am not certain and really should confirm that. What I would say is I am 99.99% certain you can generalize this. If you discharge 1 cap directly into another they, of course, both equalize to the same voltage. If you discharge one cap into another through a conducting diode and an inductor the two caps do not both end up at the same voltage once the discharge is complete. Take the 100uF cap at 20 volts discharge through conducting diode -> inductor -> 1 uF cap. The 1 uF cap goes to maybe 30 or 35 volts. The 100 uF cap also drops more to preserve conservation of charge as the small cap went to 35 volts. What I saw for the first time a week ago that surprised me is take the 1 uF cap at 20 volts discharge it into the empty 100 uF cap, the 100 UF cap goes higher than it should and the small cap goes to -14 volts! all to preserve conservation of charge.

      I will diagram out either from software or by hand the 4 arrangements I am talking about, but I know the great work you and others on the board have done, I am talking about Junior High level sort of stuff here, two caps, one diode, an inductor and a switch and varying the arrangement of the diode and caps, that's all, but you see interesting behavior! As I said in the video the conducting diode preserves conservation of charge, even if this means flipping one cap negative. The blocking diode just plain violates conservation of charge to the upside. Now one thing that occurred to me that I am still trying to wrap my head around and so I should just plain gather the data but yes the conducting diode doesn't violate conservation of charge you just end up with the two caps at different voltages after discharge. So consider the following if one just plain discharges a 1 uF cap at 20 volts into a 100 uF they both end up at ((1/101)*20) or 0.198 volts. If you instead discharge the 1 uF at 20 volts through a conducting diode and inductor the 100 uF cap ends up at something like 0.337 volts and the 1 uF goes to negative 14 volts to preserve conservation of charge. Now if you flip -14 volt cap over it is at positive 14 volts. Discharge this +14 volts now into the large cap that is at 0.337 volts, the large cap (haven't done it yet) but it will go to something like 0.57 volts and the small cap to -10 volts. The problem is we now have more charge then we started with, haha. Even though each discharge followed conservation of charge, it was just the flipping of the cap that led to this, so that is probably worth confirming as it is a bit of head scratcher. Will diagram out the four approaches shown in the video and check back in. Oh, I did automate the bridging diode set-up, from a source of 20 volts you can charge a cap to 18 volts while saving about 40% or more of the charge.