An alternative Bedini circuit

[lanenal]

WIthout the rotor, sure this won't work, but when the rotor is taken into play, it could work. I only worry about that diverge of the flyback current.

@Gyulasum



Yes i understand your idea, but it will not work.
Simply say´d:
The Transistor first become not and Voltage to start.
If the circuit comes - with some triks- to oszillate,
and old this supply current over this CAP (that separate the DC to flow)
than THIS Cap charges up to an voltage, that the Transistor becomes no

more supply current to work. Pls try it...
I have done my life so mutch unusal transistor circuits,
testequipments and used unusal functions of semiconductors.
So most schematics, i must not more "try to see".
I have also "invented" that device that "Popular Electronics" -named-

NEGISTOR .
I have TISONE published that years before in an european electronic news.

- But now over 40 years ago... No more mutch value. It was used... but

nobody ave understand, te cause and function that this work.

Your ideas to find simply schematic transistors-circuits , instead of

IC´s . 
i find very good, also to understand the electronics.
Gustav Pese

[gyulasun]

I agree with Gyula, just one correction: the trigger coil closes the transistor when the magnet leaves the core.
See this youtube video: http://www.youtube.com/watch?v=yi7cmUpMdX8

Another thing that I am not quite sure with: when flyback pulse from L1 is directed to B2 battery via diodes D3 and D1, there is another possible path via D2 (but is that really possible? I don't know yet).

lanenal

Hi Lanenal,

No need for correction in your first sentence above: I already wrote it, see this line I quote from my text:
"When the magnet is moving away, transistor switches off, flyback pulse from L1 is directed to B2 battery via diodes D3 and D1,  ok?"  This has the same meaning, isn't it?   

Regarding the role of D2 when capturing the flyback pulse of L1, yes this is an interesting question.  From DC point of view D2 is connected parallel with D1 via L2 wires, so flyback current may divide through them in the function of their forward voltage difference, actually no problem with this in DC respect.  However, from AC point of view I see the followings:
The flyback pulse when appears across power coil L1, it does appear across L2 due to normal induction, right? And the polarity of this pulse is such that the upper wire end of L1 is negative, (the lower wire end of L1 is positive,) so the polarity of of the upper end of L2 is also negative (assuming bifilar windings), right?  And this negative pulse also appears diode D2's cathode, giving a reverse polarity to it hence D2 cannot really conduct while flyback pulse lasts.  This is how I see it. In fact some measurements with an oscilloscope is warranted on a built circuit... to say for certain.

rgds,  Gyula

[gyulasun]

Hi Pese,

I agree with you in that this circuit cannot work without the rotating rotor magnets and we must find some means to start the rotor rotating either by our hands or by another motor.

however I do not understand what capacitor you refer to here (if your word CAP means capacitor):

....snip
If the circuit comes - with some triks- to oszillate, and old this supply current over this CAP (that separate the DC to flow) than THIS Cap charges up to an voltage, that the Transistor becomes no more supply current to work. Pls try it...
....snip

Would you clarify it?

Thanks,  Gyula

[lanenal]

Hi Guyla: sorry mate, I got it wrong, you were right  , and I am glad we agree on this point too  Also thank for the analysis on the question I raised. According to your analysis, there should be no diverged current through R1 when the flyback is charging the battery, is that correct?

lanenal

Hi Lanenal,

No need for correction in your first sentence above: I already wrote it, see this line I quote from my text:
"When the magnet is moving away, transistor switches off, flyback pulse from L1 is directed to B2 battery via diodes D3 and D1,  ok?"  This has the same meaning, isn't it?   

Regarding the role of D2 when capturing the flyback pulse of L1, yes this is an interesting question.  From DC point of view D2 is connected parallel with D1 via L2 wires, so flyback current may divide through them in the function of their forward voltage difference, actually no problem with this in DC respect.  However, from AC point of view I see the followings:
The flyback pulse when appears across power coil L1, it does appear across L2 due to normal induction, right? And the polarity of this pulse is such that the upper wire end of L1 is negative, (the lower wire end of L1 is positive,) so the polarity of of the upper end of L2 is also negative (assuming bifilar windings), right?  And this negative pulse also appears diode D2's cathode, giving a reverse polarity to it hence D2 cannot really conduct while flyback pulse lasts.  This is how I see it. In fact some measurements with an oscilloscope is warranted on a built circuit... to say for certain.

rgds,  Gyula

[gyulasun]

....snip
According to your analysis, there should be no diverged current through R1 when the flyback is charging the battery, is that correct?
....snip

Hi Lanenal,

Yes, there should be not current from the flyback pulse  (induced in L2 from L1)  through R1 because the pulse voltage polarity is also negative at the common point of R1 and D2's cathode now with respect to Q1's emitter, so this is a reverse bias for the Base-Emitter junction.  Maybe the maximum reverse voltage limit for Q1's B-E should be considered from its data sheet, not to overload (and ruin) its junction.

rgds,  Gyula