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

[mondrasek]

@sparks,

Easiest to understand (for me) analogy I've heard/read so far.  Thanks so much.

Maybe I need all these electrical concepts related to pneumatics/hydraulics? 

One of the concepts that is much clearer is that the current is analogous to a compressible fluid, ie. a gas and not a liquid.  The whole "tank circuit" analogy was confusing.  Your analogy is much better, and more accurate I believe.

M.

[mondrasek]

To use spark's analogy...

The HV spikes are analogous to high pressure pulses of air into an air compressor tank.  The tank is attached to a pump with an attached flywheel (the DC motor in this case).  Between the compressor tank and the pump is a valve.  If the valve is left open, the pump only moves a slight amount with each air pulse.  In 10 seconds it will rotate the flywheel only a few degrees.  But if the valve is closed for 10 seconds the pressure will build in the compressor tank.  Once opened the pump will spin to high RPM very quickly and then wind down under friction to stop again.  This time the pump has rotate what, 100 times?

So in the first case 10 seconds of pulsed air gives us several degrees of rotation.  The result of the case of 10 seconds of stored and then released air is magnitudes more rotation. 

Okay?

M.

[gyulasun]

The high pressure air pulse from the reservoir to the windmill (case where the cap is charged and then allowed to release into the motor) can create (impart?) inertia in the windmill.  The high pressure puffs of air (HV spikes alone) cannot. 

I'm not entirely clear on why.   Have I fallen or skipped off course?

Is this analogous to the stationary wave concept?

M

Hi,

I am not a "professor" in mechanical-electrical analogy so I would stay at electrical explanations if I can (sometimes I am rusty though...) 

So then your basic question is: Is the work done by the motor really the same in each case?   
My answer is no.
Although the flyback pulse's energy is the same in both cases, an inductive load (your motor) opposes any current the pulse voltage wishes to drive through it and being a very narrow pulse with fast rise and fall times it gets quasi 'choked' : the increase of current can be only gradual starting from zero and mainly governed by the value of your motor self inductance.  So in case this self inductance is high and the width of pulse is narrow there can be no time for the current to reach the maximum possible peak value. 
In case of a capacitive load the situation is totally different, a capacitor always behaves as a short circuit when uncharged and then any voltage is connected across it for charging. Hence the flyback pulse 'sees' first a short circuit as a load and all its energy content is able to get transferred into the capacitor (there is not any resistance, 'choking' effect in its way, except the series loss resistance of the cap which is very small and the inner impedance of the circuit that creates the flyback pulse). 
Now comes you connect the motor across the charged up capacitor (capacitor is separated from the flyback source). One thing is the 85V is many times higher than the 5V for your motor so a huge current is able to flow in the first moment too (pressure is very high).   Say your capacitor would have been charged up for only 5-6V from the pulse your motor would behave in a much modest way.
Now comes Erfinder and Grumpy posts I referred to (i.e. Tesla conversation with the Counsel)  on how quickly you take out the stored energy from a capacitor.

Your question on:  I'm still curious why my 8.4 V rated NiCds now charge to 9.7 V after being HV spike conditioned if anyone knows.     A battery can be compared to a high value capacitor (not a 100% analogy of course) so it can take up easily the flyback pulse.  However, be careful because NiCd or NiMh and probably Li do not like being pulse charged when they are otherwise in good shape. Pulse charging them is not recommended on the long run, though in case the first 2 types start already 'dying'  (they cannot take up the normal charge from their normal charger after many cyles of usage) then just treating them with pulsing current may make wonders. But after such treatment, the usage of the normal charger is recommended again.

rgds,  Gyula

[sparks]

   I kinda jumped the gun to try to demonstrate that a tuned resonant circuit is always better to use when working with electricity.   You just get more bang for the buck!    Without resonance in the pneumatic analogy we would open the valve and get one flow from tank to tank until the pressure is equalized.   With the inertia of the inductor tuned so that it empties one tank while it fills the other we get alot more net current flow for the same amount of voltage input.

[Paul-R]

I'm still curious why my 8.4 V rated NiCds now charge to 9.7 V after being HV spike conditioned if anyone knows. 

The Bedini people speak of conditioning the battery with numerous chargings of radiant
energy. In some way, this alters the battery. This may be what you are experiencing
with the new charge voltage. It might pay you to join the bedini_monopole3 Yahoo group.
Paul.