KapaNoDrawBack circuit from a German friend

[magpwr]

hi NickZ,

The suspicion on high current draw which doesn't allow the meter to reveal the current spikes can be quelled easily by using a large capacitor around 1000uf...4700uf at the circuit input.
I have previously created a youtube "HV to current convertor 2.0" where current draw do not surge upon shorting 5 turns windings.There isn't any increase in current drawn via 1000uf capacitor at input.


The above experiment for this topic is valid.Quite similar concept to the "E-stress" device.

This is a good link with all the related replication videos-(The last video is always the best one capacitor in series with bulb)
http://realstrannik.ru/forum/19-svobodnaya-energiya/229-transformator-zaczarinina.html

[NoDrawBack]

Hi!

you wrote:
>The suspicion on high current draw which doesn't allow the meter to reveal the current spikes
>can be quelled easily by using a large capacitor around 1000uf...4700uf at the circuit input.

In fact i have such a big CAP (1000uF) between the circuit and the supply which is just not been shown in the drawings and
video to not complicate this thing.

The cap is connected to the power supply via two diodes (now feedback possible) and a 50 watt Shunt of 0.1 Ohm for measuring the current via scope.
On the scope one could see just very little ripple (< 100mV) on DC voltage / current over shunt.

I do HF measurement since 20 years now and do not cheat myself....

Anyway - thanks to point this out - even I may overlook something someday 

BTW: Yesterday i tried to use a cap on the output L1 and L2 to get DC from the "standing wave" mentioned - that thought was way to short - no DC as expected.
Afterwards obvious.....

BUT: When connecting a CAP (10uF) on the OUTPUT L1, the INPUT drops from  500mA to 1mA !!!! Efficiency increased from 80% to 99% !!! Circuit runs on 1000uF CAP alone for several seconds !!
There is no feedback path to the CAP yet (remember i am using a diode in front of FET)- do not know how power is feed back?

Interesting! I have to take a closer look on this - just had a crude quick and dirty test yesterday and ran out of time - weekend can come....

[gyulasun]

...
BUT: When connecting a CAP (10uF) on the OUTPUT L1, the INPUT drops from  500mA to 1mA !!!! Efficiency increased from 80% to 99% !!! Circuit runs on 1000uF CAP alone for several seconds !!
...

Hi NoDrawBack,

Please tell where exactly did you connect the 10uF cap?  Stefan shows a schematic on Page 1, first post of this thread, where the bifilar L1 is shown: with wire ends labeled as I and C and J and B. 
It is possible that the addition of the 10uF increases the AC impedance of the L1 by tuning it towards a parallel resonance, question is whether the brightness of the light bulb (indicated in the schematic as 220V/10W) changes by the 10uF addition or not?

Regarding the observation of member magpwr on the "masking" effect of using a large puffer capacitor:  it is okay that the sudden high current loads are taken from the puffer cap but once we take out charge from the puffer, the supply voltage feeding the cap should start to charge up the cap again, this would mean an increased current draw between the supply output and the cap for during the time the charge is restored in the cap, obviously this supply current would be much less than the sudden high current value. This could be seen across a (second) series current sense resistor placed between the output of the supply and the puffer cap. 
(This process would be analog to a flywheel when you take out mechanical energy from it suddenly and for a short time: the flywheel continues to spin but with a reduced speed, unless you supply the energy to the wheel to increase its rpm to the original one.)

Thanks for the great job you have been doing.
Gyula

[Vortex1]

The reason the output goes to zero when DC is applied to L3:

Applying DC from a low impedance source such as a battery or well regulated power supply is effectively the same as shorting L3 as far as high frequencies are concerned. A suitable isolating (blocking) RF choke (inductor) must be added between the source of current and the L3 winding to prevent the effective short circuit.

The effective shorted L3 soaks all the power from the primary coil and dissipates it as heat.

[NickZ]

   It would still be a good to idea to test this circuit using a battery, first without the ferrite/copper rod inserted, and also, with the copper rod inside the air core. And do run time tests this way. Also to show a similar bulb running on the grid source, placed next to the test device.
  We really need to determine if the extra bulb brightness is also causing a higher draw, or not. Using higher wattage bulbs, like a 100 watt bulb would also help to displace further doubts.  Big claims need big proof. (1mA )