HV (bemf) Spikes - What to do with them?

[mondrasek]

I've built several different system now that can be tapped for HV spikes from the collapse of the EM field in an inductor (bemf).  Interesting, but does it have a use other than conditioning or charging a battery?

For example, a system that generates these HV spikes can charge a cap to 85 V.  But that cap can only run a small DC motor for a split second.

But the reason for this post is this:  If I connect the HV spikes directly to the motor it will spin (pulsed) very slowly.  If I connect it to the cap and then discharge the cap into the motor it will spin very fast for a very brief period of time.  So 10 seconds of direct HV spikes to the motor may yield 1/4 turn of the motor.  But 10 seconds of HV spikes to the cap and then discharged to the motor can yield 100s of times more turns.

So why?  Do the direct to motor HV spikes get consumed by breaking the stiction of the stationary motor over and over again while the cap discharge only breaks it once?  Is the work done by the motor really the same in each case?

Thanks,

M.

[amigo]

I suppose one could charge/discharge the cap as fast as possible, hook up the cap to another transformer with primary/secondary, create a tank circuit from that cap and the primary coil so it's tuned, then perhaps get some useful energy on the secondary.

Just my theoretical rambling, it's probably nonsense...

[Paul-R]

But the reason for this post is this:  If I connect the HV spikes directly to the motor it will spin (pulsed) very slowly. 

I doubt that it would thank you for this. The spikes are high voltage; the Bedini SSG circuit
calls for 1000v diodes to route the spikes into the battery. It is not what the motor wants.
Paul.

[gyulasun]

I've built several different system now that can be tapped for HV spikes from the collapse of the EM field in an inductor (bemf).  Interesting, but does it have a use other than conditioning or charging a battery?

Yes.  The flyback pulse (created by the collapse of EM field in an inductor) is utilized in switch mode power supplies for instance.  See these for some more info:
http://www.maxim-ic.com/appnotes.cfm/appnote_number/2031
http://www.butlerwinding.com/elelectronic-transformer/flyback/index.html

For example, a system that generates these HV spikes can charge a cap to 85 V.  But that cap can only run a small DC motor for a split second.

But the reason for this post is this:  If I connect the HV spikes directly to the motor it will spin (pulsed) very slowly.  If I connect it to the cap and then discharge the cap into the motor it will spin very fast for a very brief period of time.  So 10 seconds of direct HV spikes to the motor may yield 1/4 turn of the motor.  But 10 seconds of HV spikes to the cap and then discharged to the motor can yield 100s of times more turns.

So why?  Do the direct to motor HV spikes get consumed by breaking the stiction of the stationary motor over and over again while the cap discharge only breaks it once?  Is the work done by the motor really the same in each case?

Whenever you connect a load directly into the circuit connectors between which your HV spike (i.e. the flyback pulse) is created  the reason your the motor (the load) will spin very slowly is that the flyback pulse's amplitude gets very much reduced just by the direct load and its average energy content will be small.
In case of a capacitor charged up by your flyback pulse:  this is surely a lighter load for the pulse wrt your motor, then a charged capacitor can behave much like a battery of a given capacity (while a flyback pulse's circuit that creates it cannot) so this is one reason the motor makes much more turns. But if the cap is like a battery, it is surely a HV battery for your motor because your motor is designed for  5 - 10 -15V max? and you blast on it 85V? No wonder Paul-R warned you on this..

Now please read these posts here because they will give further answers for your last Why? question:
http://www.overunity.com/index.php/topic,4728.msg128183.html#msg128183
http://www.overunity.com/index.php/topic,4728.msg128208.html#msg128208

rgds,  Gyula

[sparks]

     If the motor becomes the inductor in a resonant circuit then it's own collapsing magnetic field is utilized to charge the capacitor.  The input from the pulsed transformer then needs only to support the ohmic losses through the copper and not have to overcome the total impedance of the motor.