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Overunity Machines Forum



OU/COP>1 switched cap PS cct like Tesla's 'charge siphoning'

Started by nul-points, April 04, 2008, 11:49:23 PM

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nul-points

hi all

i've now automated the capacitor charge/discharge sequence and run some further tests which use a different method to measure and calculate the energies involved in the switched-charge process

full result details, with test schematic, are posted in the 'Latest' section of my 'Doc Ringwood' website (linked below)

my first energy tests relied entirely on measurements of the initial and final voltage on input and output capacitors

these results have prompted some discussion, with some forum members and visitors to my website, about the energy needed to charge the output capacitor

the surprise finding is that the text-books appear to be wrong in stating that the amount of energy converted as work in storing energy into a capacitor is the same as the amount of energy which gets stored

for the example test shown on the website, the work required to transfer charge to the output capacitor was only 77% of the final energy actually stored (scope trace shown below)

this finding suggests that the energy calaculations can't assume a doubling of energy converted, based on the final value stored in the output capacitor - however, there is still sufficient energy being converted in the resistive load during charging which gets added to energy discharged through it from the output capacitor to bring the efficiency over 100%

the new measurement test shows a total circuit efficiency of 127% and a (resistive-load energy) / (input energy) of 106%, supporting the results of my previous 'energy anomaly' experiments with switching charge into a capacitor


next step for me, hopefully, will be heat tests

all the best
sandy

Doc Ringwood's Free Energy site  http://ringcomps.co.uk/doc
"To do is to be" ---  Descartes;
"To be is to do"  ---  Jean Paul Sarte;
"Do be do be do" ---  F. Sinatra

nul-points

...i should have identified the scope traces: Red is the voltage charging up on C2; Blue is the voltage across the 10 ohm resistor in series with C2

the burst of 24 pulses show that charging current is passing thro' the 10 ohm resistor to charge C2; the larger single negative pulse shows the discharge of the final stored energy from C2, again passing as current thro' the 10 ohm resistor

full details in the 'Latest' section at the link below

all the best
sandy

Doc Ringwood's Free Energy site  http://ringcomps.co.uk/doc
"To do is to be" ---  Descartes;
"To be is to do"  ---  Jean Paul Sarte;
"Do be do be do" ---  F. Sinatra

alan

Hi again,
you can try to do this experiment at different temperatures and see if it is temperature dependent.

oh, wait, that's your next step ;)

nul-points

hi all


as i understand it, the official line on fly-back energy is that you only get back what you put in

the measurements i'm making on this switched-charge circuit show something different!


the traces below show just the first and last pulses, as examples, in the switched-charge burst used to charge C2 - the applied pulse across the RLC branch (in Red) and the voltage across the 10 ohm resistor (in Blue)

fly-back energy is returned to the RLC branch, after the pulse, via the diode at the positive end of the coil (see the negative portion of each Red trace)

you can see that, at the point that the pulses switch off, the current through the RLC branch was still rising, so the coil field was still being 'charged'

the energy gain can be seen on the 'load' waveform (Blue trace) - the traces clearly show that energy input to the RLC branch during the pulse is a fraction of the energy input which continues after the pulse stops

the left-hand side of the triangle waveform (representing energy input pulse) is smaller than the right-hand side of the triangle waveform (representing additional energy gained from the flyback)

Energy input to the load for the first pulse:

57.6uJoules
  16.0uJ from the pulse
  41.6uJ from flyback

Energy input to the load for the last pulse:

20.2uJoules
  8.9uJ from the pulse
11.3uJ from flyback

the proportion of flyback energy to pulse energy reduced with each pulse between the first and last, as the stored voltage across the capacitor increased, but the ratio was always > 1

compared to the energy provided by the pulse to the load:
an additional 260% energy was returned to the load by flyback after the first pulse
an additional 127% energy was returned to the load by flyback after the last pulse

this gives further support to the overunity results i'm seeing on the switched-charge experiment - until now i couldn't see where the additional energy was entering the system - it appears now that it's being supplied from the coil field collapse as flyback energy, extra to that supplied from the input capacitor

-----------------------------------------------
some more details on test setup:-

the coil used for these tests was approx 2.5mH, 0.5 ohm DC resistance, so the I^2*R loss in the coil was only 1/20th of the power developed across the load

pulse width as shown here was approx 77us, but this is not a critical value** - it's just an example

the first pulse peak-voltage shown was 8.3V approx (the initial voltage on my input cap) and the final pulse peak-voltage was 7.8V approx


**the pulse width could be increased until the RLC current, as monitored by the load resistor, is limited by the supply
(this assumes the core doesn't get saturated before this point)

any more width on the pulse, after that point, will not increase the coil field, it will just result in increasing energy lost in the coil thro' I^2*R loss

to achieve most benefit from flyback, the pulse width should always be less than the value which limits the RLC current (or saturates the core)

it also appears that the geometry of the waveform is important: the gain of returned coil energy to pulse input energy is better when the rising and falling sides of the 'triangle' waveform approximate straight lines

if the coil charge current increases too much then the rising side curves 'positively', raising the area of energy under the left-hand side of the triangle, while the falling side curves 'negatively' (like a discharging capacitor), reducing the area of energy under the right-hand side of the triangle - the net result would be to reduce the energy gain and instead produce an energy loss

all the best
sandy

Doc Ringwood's Free Energy site  http://ringcomps.co.uk/doc
"To do is to be" ---  Descartes;
"To be is to do"  ---  Jean Paul Sarte;
"Do be do be do" ---  F. Sinatra

alan

You rock!
Which digital osc. are you using?

"you can see that, at the point that the pulses switch off, the current through the RLC branch was still rising, so the coil field was still being 'charged'"

I don't see this  :-[
Do you mean the red damped waveform on the right?

"the left-hand side of the triangle waveform (representing energy input pulse) is smaller than the right-hand side of the triangle waveform (representing additional energy gained from the flyback)"

You mean the area beneath the blue curve?

What does simulation show

(maybe I ask obvious stuff, but I am trying to understand)