The secret to Overunity

[sm0ky2]

Void, try charging your capacitor from a high voltage source that is capable
of greatly exceeding the charge current of the equation (max value on the data sheet).


What is the charge time?

[Void]

Void, try charging your capacitor from a high voltage source that is capable
of greatly exceeding the charge current of the equation (max value on the data sheet).
What is the charge time?

Hi Sm0ky2. If you have a high voltage DC power source with very high current capability,
and you have very little resistance and component losses, then the capacitor initial charge
current would be very high. To have a high voltage DC power source that can supply that
kind of current would be quite dangerous to work around. I am not interested in trying it.

People have used microwave oven transformers as a fairly high voltage power source with
fairly high current capacity, and if you use something like that (careful, it's dangerous) with a suitable
high current rated HV diode to rectify the AC to DC, even a relatively large capacitance HV cap can be charged
to a high voltage very quickly, and the initial capacitor charge current would be relatively pretty high.
Doing this still consumes power from the DC power supply however in proportion to the voltage you are
charging up the capacitor to, regardless of the power supply voltage you are actually using. 

All the best...

[SolarLab]

F.Y.I.

Vyacheslav Gorchilin shows [proves] mathematically that excess (free) energy is achievable.
He also presents several methods/techniques and suggests some practical implementation
approaches. Animated Mathcad on-line graphs are made available to aid in the process as well.

His approach to the solution [proof] is rather brilliant IMHO!

From:  http://gorchilin.com/articles/
(use translation (flag in upper right hand corner) if required - unfortunately Google Translate will not properly
display the formulas, however Yandex - the flags - will work OK)

    Energy parametric RLC-circuits

Free energy in a parametric RLC-circuits of the first kind of the second order
http://gorchilin.com/articles/energy/RLC_5?lang=en

"In this work we consider the electric circuit containing the nonlinear reactive elements: the
capacitor and inductance. Their nonlinearity is determined by the parametric dependence:
the capacitance — voltage on it, and inductance from the current flowing through it. The
resistance is constant, but even if its value changed, for example, from the time the proof is
not affected . According to the classification here will be considered generators of the first kind
and second order. In real devices the parametric dependence can be only one element
capacitance or inductance, but here we show the General case and prove that these elements
are independent from each other to influence change in the efficiency of the second kind. "

Insights:  " Insights on circuits of the second order similar to those done on the circuits of the
first order. However, repeat them.
In this work we proved that it is impossible to gain energy in parametric circuits of the second
order in the full cycle (FCC) because the energy dissipated in the resistance is always equal to
the energy expended by the power formula (5.8).

But if the cycle is incomplete, the receiving gain becomes achievable task.

If reactive elements contain potential energy in the beginning of the
cycle (PCCIE), the gain can be found using the formula (5.12). If the energy in the reactive elements
is at the end of the cycle, then the conditions for receiving allowances, we can find from formula (5.14),
and the increment of efficiency by (5.16).

You need to understand that there is a mathematically strictly proved potentially achievable values of
the increment of energy, part of which, in the real reactance, can be spent inefficiently, for example, on
heating. However, on the basis of evidence about the energy increment in the fractional cycles, one
can obtain special cases for engineering calculations, which, in turn, will allow you to build a real device
with high efficiency.
Additional materials and some special cases with examples from real wireless components you can
see here. "

Application to the proof of parametric RLC-circuits of the first kind
http://gorchilin.com/articles/energy/RLC_51?lang=en

Parametric inductance in PCCFE

In this case, without knowledge of the schema and according to   of the power source,
the increase in efficiency we can not calculate, but we can search conditions for     from
the formula (4.15). Assume that our coil is described by the parametric dependence of
inductance from its current next next:
         {see text for formulas - unfortunately they can not be reproduced here}
For rows with a large value of the degree calculations are either very bulky, or, in principle,
not derived analytically. But for example, it will be enough and second degree.

Reference: {further background study}

SAINT-PETERSBURG   STATE  INSTITUTE   ACCURATE   MECHANICS   AND   OPTICS
(TECHNICAL   UNIVERSITY)
Department of Electrical Engineering and Precision Electromechanical Systems
Yu.M. Osipov
FREQUENCY   AND   TEMPORAL   ANALYSIS  STATIONARY   AND   TRANSITIONAL 
CHARACTERISTICS   LINEAR   ELECTRICAL   CIRCUITS
A manual on electrical engineering and TOE courses
Part two
St. Petersburg  2002

3.3 Transient processes in second-order circuits
http://ets.ifmo.ru/osipov/os1/3_3.htm
Translation to English (Google)
https://translate.google.com/translate?depth=1&hl=en&rurl=translate.google.com&sl=auto&tl=en&u=http://ets.ifmo.ru/osipov/os1/3_3.htm

*** Seasons Greetings and best wishes for the New Year ***

FIN

[sm0ky2]

Hi Sm0ky2. If you have a high voltage DC power source with very high current capability,
and you have very little resistance and component losses, then the capacitor initial charge
current would be very high. To have a high voltage DC power source that can supply that
kind of current would be quite dangerous to work around. I am not interested in trying it.

People have used microwave oven transformers as a fairly high voltage power source with
fairly high current capacity, and if you use something like that (careful, it's dangerous) with a suitable
high current rated HV diode to rectify the AC to DC, even a relatively large capacitance HV cap can be charged
to a high voltage very quickly, and the initial capacitor charge current would be relatively pretty high.
Doing this still consumes power from the DC power supply however in proportion to the voltage you are
charging up the capacitor to, regardless of the power supply voltage you are actually using. 

All the best...

Losses are negligible, and at high enough voltages there is no resistance in the circuit
External to the capicitor


The only impedance is then the internal resistance curve of the capacitor that defines charging current


Without adding extraneous components, and making assumptions outside of the situation.


What I am asking is, what is the time-frame over which the charging current exists?
And what portion of this time does the current hold a 'high-value'?


My HV source provides infinite instantaneous current.
And exists as only potential except the moment charge is drawn from it.
Current, defined in Amperes, is a time-derived quantity.


The magnitude of the time defines the order of magnitude of the current.
Here's a hint: 1 Amp is 1 coulomb per second.


If the total charge time is 1/82 of a second,
Current is only of appreciable value for 1/6000th of a second
after which internal resistance of the capacitor reaches such high value
that the rest of the charging time, current is calculably minuscule.


So what exactly is the charge current you keep arguing about?
And how does the power rating compare to the output of the capacitor discharge?

[Void]

Hi sm0ky2. The formula for the capacitor charge current makes it quite clear what will happen.
I am not going to keep repeating myself, and this is very basic and well established stuff anyway. 
If you doubt it, you can always experiment and see for yourself how things work.
A very simple setup to see what happens is if you charge one capacitor from another capacitor, whether
at high voltage or not, the source capacitor will lose charge (a loss of energy) to charge the other capacitor.

Be very careful if you mess with high voltages, especially at higher current capacity. Not recommended. It can certainly be lethal.

All the best...