Zero and Q device

[v8karlo]

If Karlos doesn't mind, i'd like to attach here one more circuit of the same way of thinking. It is for the experimenters. The waveform of the returned energy back to the source, was attached at the Kapanadje thread before few days. If A.G likes to bring it here it would be appreciated.

I don't recommend anyone to build it from scratch. It is just for showing you here that if potential become higher than input then a good portion of energy can return back to the source and raising this way the efficiency.
All coils 1:1

I don't mind at all, you are welcome for any idea.

[v8karlo]

Wow! Yes. It has to be something that simple. I'll test "J" topology asap. Nice to hear from wistiti that Zero circuit worked well. Keep on the good work Karlos!

Q-topology didn't work well with my push-pull setup. But feels like that with some refinements it can become better. The general idea looks valid.

ps. Don't have to be so deffensive. If your circuits are genuine then they will tell their story by themselves. 

Regards

edit: I meant deffensive..sorry

Hehe, ok, no problem.

Q device isn't finished. It is start.
I am searching for way to make Q returning,
feed-backing to source in both phases, not just one phase.

Yesterday I was looking at Tesla patent 577670.
He placed inductors on feedback line on both sides.
In J device we have pulses on feedback line to source.

If I place inductors on feedback line, the back EMF will add it's
energy to feedback line.

Because feedback line energy is not lost. It goes to source.
In that case back Emf from inductors will be extra energy on that line
and it will only add energy to that line.
That is something which I have to test. For now it is idea.

[lancaIV]

v8karlo, some " Studentenfutter" + espacenet " Baltzar von Platen" electro motor
http://rexresearch.com/platen/platen.htm
https://worldwide.espacenet.com/searchResults?submitted=true&locale=en_EP&DB=EPODOC&ST=advanced&TI=&AB=&PN=&AP=&PR=&PD=&PA=Baltzar+von+platen&IN=&CPC=&IC=&Submit=Search


Tesla,  later Otto von Traun ( co-op.  Hermann  Plauson) and from 2019 ten years back :

Project base
http://www.patent-de.com/20100114/DE102008032666A1.html

  some developments :https://worldwide.espacenet.com/searchResults?submitted=true&locale=en_EP&DB=EPODOC&ST=advanced&TI=&AB=&PN=&AP=&PR=&PD=&PA=Envez&IN=&CPC=&IC=&Submit=Search

[Void]

Regarding 'returning energy back to the source'.  You have to look at how much energy is being
returned.  Returning currents through a load back to the power source in no way means all
the energy consumed by the load is returned back to source. That is a misunderstanding.
You have to think in terms of energy, not current.

The energy dissipated in the load is 'consumed' energy. The current returning back to the source
through a load that is consuming energy from the power source will be at a reduced energy level.
This is because there is a voltage drop across the load.

You don't need fancy or complicated circuits to test this concept. You can just connect an efficient DC to DC boost
converter to a battery and boost the output voltage high enough so that you can connect a light bulb
(or resistor) between the boost converter output plus wire and the battery plus terminal. The light
bulb will light up and all the current passing through that light bulb returns to the battery plus terminal,
but your battery will still steadily run down. The energy consumed by the bulb is not returned to
the battery just because your setup returns all the current passing through your bulb back to the battery.

If anyone doubts this, you can easily test it if you have a DC to DC boost converter.
For example if you have a 12V battery and a suitable DC to DC boost converter, you can connect the
boost converter to the 12V battery and adjust the output voltage of the boost converter to 24V.
You can then connect a 12V light bulb between the boost converter output plus terminal and the battery
plus terminal, and all the current passing through the 12V light bulb is returning to the battery plus terminal.

Test how long it takes a fully charged 12V battery to run down to some voltage lighting a 12V light bulb directly,
compared to how log it takes to run down the same charged battery using the boost converter (at the exact same light bulb current)
with all the current from the bulb returning to the battery plus terminal. See which way is more efficient.

[lancaIV]

Void,  yes ! But how convert the electric cycle to an environmental energy converting device , charge pump