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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 alan

thanks for the interest

@your previous comment re: the heat tests - my plan is to measure the heat output from the Rload & compare with heat output from same Rload with equivalent DC supply


> Which digital osc. are you using?

i'm using a PC to drive 2 optos which get shaped via 4093B Schmitt NAND gates & then feed the MOSFET gates

the optos are powered by the PC; the 4093 is powered from the input cap, so my Efficiencies are actually slightly greater than i show


>> "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 current through the RLC branch is proportional to the voltage across the 10 ohm load res (Blue trace) - just divide a particular voltage on the Blue trace by 10 to get the current

the applied pulses are shown on the Red traces - you can see where a pulse ends - if you look at a blue trace you'll see it increasing from 0V at the start of a pulse up to some voltage at the end of a pulse

the current through the coil (& load & into the cap) is increasing all thro' both pulses shown, hence the coil still being charged when the pulse ends

the pulse start edges are slightly obscured by the first of each pair of time markers

the time markers are set at +/- 77us wrt end of each pulse to show that the energy thro' the load due to the pulse is less than the energy thro' the load due to the flyback action


>> "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?

well, both the left- and right-hand sides of each blue triangle waveform have the same height, so they both cover the same voltage range and their average voltage is the same - hence the average current is the same - and therefore the average power is the same

but because they have different periods the energy of each side will differ - the load energy from the flyback, relating to the right-hand side of each triangle is greater than the load energy from the pulse, relating to the left-hand side of each triangle

so, the area isn't directly proportional to energy on that trace but you can see that one side represents a greater energy than the other

i've calculated the energy for each side of both triangle waveforms using Excel on the trace data files - the results are shown in my post above


> What does simulation show

i haven't been able to get simulation to give results which look very much like reality - the simulator in CircuitMaker seems to like making circuits greatly overunity!

since these are such simple circuits, i'm finding it quicker just to connect a real circuit and test it


> maybe I ask obvious stuff, but I am trying to understand

me too !  :)

but i think these overunity flyback energy gain results have started to provide an answer


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

Thanks for taking the time to explain!

With digital osc., I meant oscilloscope connected to the pc :)

If I understand correctly, the charge and discharge period should be the same, hence RLC constant is the same.

Are you familiar with applying Laplace Tansform and feeding it a stepresponse?
Maybe that could be a good basis to show that experiment doesnt comply to theory.

nul-points

hi alan

> With digital osc., I meant oscilloscope connected to the pc

apologies, i thought your 'osc.' meant oscillator

i'm using PicoScope 6 software supplied with PicoScope 2203 2-channel PC scope


> If I understand correctly, the charge and discharge period should be the same, hence RLC constant is the same.

my understanding is that the energy recovered from coil field-collapse (aka 'flyback' energy) 'should be' equal to the energy used to establish the field

also, that the energy stored by the coil field is proportional to the current squared - if the av. input current = av. returned current (as noted earlier) then equal energies should require the same periods (all else being equal)

the input & returned periods shown in each trace are different - my interpretation of this is that more energy is being returned than was stored


> Are you familiar with applying Laplace Tansform and feeding it a stepresponse?
   Maybe that could be a good basis to show that experiment doesnt comply to theory.

i think i could only just manage to spell Laplace Transform now - it is many decades since i saw one swimming across a 'whiteboard' at college!

if you feel you have the time, capability & interest to carry out such a procedure... go for it!  :)


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

AbbaRue

What is the value of the diode D1?
The diagram is that of a zener diode, or is this just a drawing mistake?
A regular diode has a straight line, a zener uses the   "Z".


nul-points

g'day AbbaRue

diode D1 is a 1A Schottky barrier rectifier 1N5817

the diagram is a generic Schottky device - the line has folded-back ends to denote an 'S'

a Zener has slanting ends to denote a 'Z', as shown below

hope this helps

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