Kapanadze Cousin - DALLY FREE ENERGY

[magpwr]

Hi magpwr,

I think I understand your test circuit (I mean its schematic, I made a quick sketch for myself as per your description).
What I do not get is: why you say you have achieved OU in ANY sense? Would like to understand your thinking.

I think you would have to consider the input power your signal generator provides for your circuit. It is okay you use
very low duty cycle, it makes input power surely low but I believe it can be enough to maintain 3+Volts in the 4700uF
after disconnecting the 3.9V battery.  And if you use the diode bridge instead of the single diode, the forward voltage
loss doubles (the charge current has to go through two diodes in the bridge) and the remaining lower current cannot
maintain the charge in the cap any more.
So this can also happen in this test circuit. You (we) do not know that in the single good diode's case how much excess
current would remain over the charge-maintaining current, which would be already enough to feed a tiny 2.5MHz pulse
oscillator with a few uA consumption from the 3+V cap voltage (provided of course that a low duty 2.5MHz  oscillator
could be built with a 1-2uA current draw, this may be possible).   
Would you tell the 2.5MHz input amplitude to the gate-source what the signal generator gives at its output? Just curious.

Thanks,  Gyula

hi Gyula,

I have confirmed without feedback diode be it fast diode or simple 1n4148 as tested the capacitor voltage does keep dropping normally after providing  short connection from the battery.

Another thing i noted i can't get capacitor to sustain voltage if i use schottky diode eg:1n5819 for the feedback.Even my "ultra lowest 25mV joule thief" as found in my youtube did not perform well with this diode which i just recall.BAT85 diode performed little better than 1n4148.Maybe we should not even use these 1n5819 for the 3v ou light as well.
I got to rush off to work it's 2.5khz not mhz.But without a low power oscillator to self sustain or to create a close loop it's hard to prove anything.

The signal generator is connected via 1k resistor to mosfet.Not forgetting the capacitor is powering the 10 turns switched via mosfet.
The "green" toroid used is a iron powdered toroid.

Recalled MC34063 or any other circuit  used a normal step voltage booster only 1 diode is needed to recover bemf/spike to get higher voltage at the output.

[gyulasun]

Hi magpwr,

I drew the schematic as per your description, please tell if it is correct or needs some modification.

Your tests with the different diode types show that we have to find the best performing type for a specific job,  gotoluc, RomeroUK (and others) also wrote about this.  One would think that Schottky types should be superior but this is not always the case.
If you happen to have a Germanium diode in junkbox, it may work also well. Or if you have a Germanium transistor (any type), you could try it too, connecting it as a diode.

Ok on the 2.5 kHz frequency instead of the 2.5 MHz you wrote earlier.

Can you tell the voltage amplitude that drives the gate-source via the 1 kOhm from the signal generator? What type of generator you have I wonder.

Thanks, Gyula

[Farmhand]

Hi magpwr,

I think I understand your test circuit (I mean its schematic, I made a quick sketch for myself as per your description).
What I do not get is: why you say you have achieved OU in ANY sense? Would like to understand your thinking.

I think you would have to consider the input power your signal generator provides for your circuit. It is okay you use
very low duty cycle, it makes input power surely low but I believe it can be enough to maintain 3+Volts in the 4700uF
after disconnecting the 3.9V battery.  And if you use the diode bridge instead of the single diode, the forward voltage
loss doubles (the charge current has to go through two diodes in the bridge) and the remaining lower current cannot
maintain the charge in the cap any more.
So this can also happen in this test circuit. You (we) do not know that in the single good diode's case how much excess
current would remain over the charge-maintaining current, which would be already enough to feed a tiny 2.5MHz pulse
oscillator with a few uA consumption from the 3+V cap voltage (provided of course that a low duty 2.5MHz  oscillator
could be built with a 1-2uA current draw, this may be possible).   
Would you tell the 2.5MHz input amplitude to the gate-source what the signal generator gives at its output? Just curious.

Thanks,  Gyula

I hope no one takes this as criticism but in my humble opinion if a function generator is used then it's entire input power should be considered as input as it is part of the device, without it there is no working circuit. When using IC's to produce signals I can see that the input power to the IC is very small with low frequency output to a mosfet gate, but as the frequency increases the input power can become significant if thinking in mA x volts. If a mosfet just for example has an input capacitance of 2000 pF and the operating frequency is 100 kHz then the 2000 pF is charged to gate drive voltage and discharged 100 000 times per second, that energy is supplied by the FG and just considering the power required to drive the mosfet means we must consider this power and the input power to the device that provides it as part of the input, not even considering any energy that might be transferred to the output via the mosfet capacitance.

If we use an IC to provide signals then we consider/measure all of it's input power draw not just the signal output power as it is part of the circuit/device. Just my opinion.

..

[NickZ]

  Gyula:
   quote
   "EDIT 2: There is a full wave rectifier circuit variation with two (independent) diodes but this would double the output DC voltage. The drawback in this circuit for your MBR20100 types is that the output coil ought to give only 50V peak AC to match the 100V max rating. Nevertheless, see the second figure in this link http://www.daenotes.com/electronics/devices-circuits/voltage-multipler   Should you test it, maybe first reduce the number of turns of the output coils or use taps on it. Use one-one diode from the two MBRs"
                                          end quote.
  I am going try the MBR20100CT diodes first, to see what they can do.
The AC voltage of the output coil can be reduced, as needed, to suit the MBRs, which can also help to raise the current output level as well.
  So, can you, or Sea Monkey,  direct me as to how to connect two of those diodes together to form a full wave rectifier? Please...  I will place that bridge on my circuit's second yoke, before placing another lesser voltage diode inline. also.
  I'll also connect a 2200uf 16v cap across the input, as well as another higher uf cap on the DC output side, as you suggested.

[SeaMonkey]

Nick,

The MBR20100CT Dual Diode Common Cathode is
designed to be used as a full wave rectifier in
conjunction with a center-tapped transformer winding.

When it is used in such a fashion it achieves full wave
rectification with less loss than a bridge rectifier.  It
does mean though that the output winding of your
device would have to be wound as a center tapped
winding in order to feed the Common Cathode Diode
Pair properly.

In order to make a full wave bridge rectifier using the
MBR20100CT it would be necessary to obtain one of
its complements the MBR20100CTR.  The final R in that
designation connotes "Reverse Polarity" or Common Anode
configuration.

With one each of a Common Cathode and a Common Anode
the two can be easily connected to make the full wave
bridge rectifier.

It may be easiest to just add some additional copper wire
to your device to make a center-tapped output winding in
order to utilize a single MBR20100CT as a full wave rectifier.