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.
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...
Quote from: mondrasek on September 25, 2008, 07:15:31 PM
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.
Quote from: mondrasek on September 25, 2008, 07:15:31 PM
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
Quote
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
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.
@Paul-R,
As Gyula concluded, the motor is a very small brushless (I think) pager motor from a toy helicopter. I believe it is runs on 5V in the original toy.
Just thought I'd add this fact that I omitted in the original post. I did not know it could be relavant.
Thanks to all for the replies. Looks like I have some more reading assignments. Much appreciated.
M.
PS. 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.
@ Gyula,
Ah, Erfinder and Grumpy's posts are always an enlightening read.
So to equate to a mechanical system: The HV spikes are like a short pulsing high pressure puff of air. The motor is a windmill. If they interact, the windmill sees only pulsing high pressure puffs of air. It will move very slightly with each puff, but be stopped by friction and thus never accelerate enough to spin freely.
The cap is like an air pressure resevoir, only of incredibly tiny volume. When the puffs of air are introduced to it the pressure builds very high rather quickly. But the volume is so small that there is still not alot of air in there. So when it is released into the windmill the windmill sees a slighly longer pulse of much higher pressure. The windmill spins up higher for a split second and then slows back to a stop due to friction.
Does this sound right so far?
yupp keep going :)
Quote from: mondrasek on September 26, 2008, 01:45:43 PM
@ Gyula,
Ah, Erfinder and Grumpy's posts are always an enlightening read.
So to equate to a mechanical system: The HV spikes are like a short pulsing high pressure puff of air. The motor is a windmill. If they interact, the windmill sees only pulsing high pressure puffs of air. It will move very slightly with each puff, but be stopped by friction and thus never accelerate enough to spin freely.
The cap is like an air pressure resevoir, only of incredibly tiny volume. When the puffs of air are introduced to it the pressure builds very high rather quickly. But the volume is so small that there is still not alot of air in there. So when it is released into the windmill the windmill sees a slighly longer pulse of much higher pressure. The windmill spins up higher for a split second and then slows back to a stop due to friction.
Does this sound right so far?
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
A resonant circuit can be analogous to the below:
Two air compressor tanks form the capacitor plates. An air motor/pump with an attached Flywheel the inductor. The motor is connected between the two tanks. One tank is full of compressed air the other empty. The capacitor is charged. We open the valve and the air pressure flows to the motor but the flow is resisted by the flywheel. The pressure slowly accelerates the flywheel with an increase in flow/current. Meanwhile the tanks are approaching equal pressure differential. The flow does not stop after the pressure equalizes because of the flywheel inertia. The flywheel energy turns the air motor into an air compressor which voids the first tank and pressures the 2nd. This action continues until the inertial energy of the flywheel can no longer overcome the pressure differential. The air flow/current now reverses and the process is repeated over and over until the frictional losses of the system bleed out the energy. An antennae attached to the system would take on the form of a balloon teed into the circuit which is alternately inflated and deflated as the circuit resonates.
@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.
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.
Quote from: mondrasek on September 26, 2008, 02:18:30 PM
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
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.
Quote from: mondrasek on September 26, 2008, 01:23:28 PM
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.
@sparks
Your post on resonance was great. I've been playing with the idea of tuning a simplified four battery current siphon (Tesla switch) in the thread http://www.overunity.com/index.php/topic,1645.0/topicseen.html . This thread here was a side topic question related to that, so your post on resonance was right on. I didn't realize until you mentioned it that is was slightly off topic because it is very relavant to the thread that spun this one off.
@Paul-R
Unfortunately the firewall at work does not allow the Yahoo groups. I miss a lot of references from there, as well as some other sites that Gyula and others have offered up. I do most of my reading at work, since at home I am usually trying to build or take care of the rest of "life".
@Gyula
Thanks again.
Also, I pulse "conditioned' brand new uncharged 9V NiCds from RadioShack on my Bedini fan one time. They climbed to 9.7V very quickly (rated at 8.4). Just like any new uncharged NiCd this was a fluff charge only, ie no capacity. But since them I have conventionally charged them, maybe 20 times. They always conventionally charge to ~9.7 volts. They run great but once decayed down as far as 8.4 they are nearly dead. I guess I should buy another and not condition it and try to compare the overall capacity and/or life. That behavior is just something weird that I thought maybe someone here already had the explanation for.
@modrasek;
Capacitors work on what is know as Q quantity of charge, litterally quantity of electrons.
One can look at the height of pulse as being created by a stack of electrons piled "end on end"
while the width of the pulse is created by a volume of electrons. So a pulse on the oscilloscope
can be looked on as an integration (summation) of the height of the pulse times the width of the
pulse.
A small capacitor will fill up fast, so a quantity of electrons will make it go to high voltage faster.
A larger (more uF) capacitor will go to a given voltage more slowly and require more quantity
of electron charge to get there.
Your motor winding has an inductance. It tends to reject (block) short (narrow) spikes. But
when you use those short spikes to charge a capacitor the voltage slowly rises to meet
an objective voltage. The DC voltage on the capacitor then runs the motor with DC for a
short time period which the inductor conducts more vigorously, but in the end then drains off.
=>In effect the capacitor has performed the function of down converting the high voltage spikes to
much lower voltage, but the quantity of electrons and therefore the energy, has been conserved
by the capacitor.
The impedance of the capacitor is a better match for the motor's impedance then the short
spikes circuit was. so you are using the charging cap as impedance matching.
---
photo flash caps are exactly the same as any capacitor except the high working
voltage (~300Vdc) and moderate capacitance (~250uF) allows them to store a
relatively high ( and relatively more hazardous) amount of electrical charge.
They are really the second rung on the power (rather than signal) capacitor scale.
---
Nicad batteries, of course, are infamous for the "memory effect" which happens when the cell's
of the battery are not discharged evenly. Voltage on some cells are still high while others
are low and there is no good way to "reach through" the charged cells to get to the
discharged ones. The spike voltage charging, I don't think is recommended for NiCad's,
and I believe LmNiH batteries are touchy to charge too. In acid/lead batteries you can
reach through with a "polishing charge". That leaves alkalines, which I don't know too
much about but would be worth a try.
---
Eventually, I think you should convert to small gel-cell acid leads that would be a better match
for the chemistry and pulse charging and a better match for your relay's power requirement
which is I suspect to large for the the 9Volt size batteries. I have not been able to locate
a good source for the 9Volt sized acid leads.
:S:MSCoffman
@MSCoffman,
Thanks for the explanations on capacitors and impedance matching. Makes good sense to me.
For what it is worth, my experiences with charging alkaline batteries off of HV has been somewhat successfull. AA, C, and D cells appear to take on a charge again, however very slowly, especially so for the large D cells. Smaller cells are more prone to burning up. One 9V became very hot after failing to reach 9V over several days. After disassembling it was clear that one of the 6 cells had burnt up, while the other 5 appeared fine. A 12V "remote controll" type battery from a DMM actually exploded! It is made of small button cells like used in a watch. One of them exploded, causing the case to split and the other cells to fly around my work bench. So I'm done with trying to charge alkalines.
I've always intended to switch to 12V SLAs on the current siphon if it became stable or the NiCds were limiting the testing. I've been holding off only because of the expense and additional danger of higher current sources for now. Since I'm at the point of winding my own solenoid coils now I might consider dropping down to 6 V since the rest of the set up is made of low current components.
M.