Display MoreSo I have this challenge to readers: forget for a moment everything about LENR, especially about Z-pinch condensed plasmoids.
Consider this simple circuit in plain air, room temperature:
(Flyback primary circuit not drawn up, it is a 1 transistor circuit)
And these results:
Using power-electronics and plasma physics principles, please explain:
1. What charges the capacitor bank back up in 2nd and 3rd waveform occasionally, with respect to 1st baseline waveform?
2. Why does C1 fails and becomes open circuit? C1 = C2 = Wima MKP10 1.5nF 2500Vdc. Pulse repetition rate varies, on average is in higher tens/ lower hundreds Hz range. Input DC power to flyback converter is 1-1.5W.
What is the point of the C1 and C2 being in series? Why not use a single capacitor of value (C1 + C2) / 2? It would be equivalent.
QuoteWhat charges the capacitor bank back up in 2nd and 3rd waveform occasionally, with respect to 1st baseline waveform?
Those look like transmission-line reflections. It seems that your load isn't impedance-matched to the impedance of your transmission-line. To understand transmission-lines consider that an EM wave with an electric field of 10 V travelling through a transmission-line that has an impedance of 10 Ohms will appear to travel with a current of 1 A. In fact, the impedance of a transmission line is usually simplified to sqrt(L / C). Capacitance is related to permittivity of a dielectric, that simply tells us how well it polarized electrically, while the inductance is related to permeability of a dielectric, which tells us how well that medium magnetizes in response to a magnetic field. So we can say that the impedance of a transmission line tells us what the ratio of the electric field is to the magnetic field.
If your load does not have the same impedance as your transmission-line, you will have reflections, to understand why you have to remember that the 10 V EM wave will travel with 1 A of current through an impedance of 10 Ohms. If this same EM wave were now to attempt to pass through a resistive load of 20 Ohms, it would be unable to push 1 A of current through it as it would require an electric field of 20 V. Logically at this junction the same amount of current must be passing through. Further since this junction is infinitely small, it should have the same amount of voltage. We can intuitively now sense that there will be a reflection since you cannot have 1 A of current coming in and 0.5 A of current coming out, you will also note that since the reflection reduces the incoming current, it will also reduce the outcoming current, so some equilibrium must be found.
So we can write V1(+) + V1(-) = V2, I1(+) + I1(-) = I2. Further we can rewrite the voltages as function of the impedance (Z), so V1(+) = I1(+) * Z1, V1(-) = I1(-) * Z1, V2 = I2 * Z2. From there you will be able to derive the following equations:
Our reflected current: I1(-) = (Z1 - Z2)/(Z1 + Z2) * I1(+)
Our passing current: I2 = 2 * Z1 / (Z1 - Z2) * I1(+)
I don't know what your transmission line's characteristic impedance is, so I can't calculate what your reflections should be, but if you find it out you can see if the reflections match what you're seeing. A spark gap should be pretty hard to control the impedance of, so that is likely your first source of reflections, and depending if there is enough time your load could also be reflecting.
QuoteWhy does C1 fails and becomes open circuit? C1 = C2 = Wima MKP10 1.5nF 2500Vdc. Pulse repetition rate varies, on average is in higher tens/ lower hundreds Hz range. Input DC power to flyback converter is 1-1.5W.
Not sure what you mean by it.