Inductive Circuits - The "Classical" Approach

[Inquorate]

LC tank circuit v veljko oscillator

I've yet to see an LC circuit that is given a little push like the veljko oscillator pendulum - at the apex of it's swing. I've bought a cheap scope on ebay with which I intend to investigate the possibility of an LC circuit that mirrors a veljko oscillator.

As for a capacitive analogue to the inductive OU circuit, point 99, I think it's fairly clear that we don't know what you're hinting at..

Throw a dog a bone

Love and light

[Hoppy]

Aaron wrote: -

Maybe not but this sidesteps the fact that there are mosfets specifically designed to self-oscillate as there are mosfets that are designed to specifically not oscillate. When it oscillates, it may encounter avalanche.

Aaron, you are getting very confused about this. Mosfets's are not designed to self-oscillate, this is nonsense! Repetative avalanche simply means that the physical build of the mosfet can withstand multiple high energy pulses to a certain rated level without breaking down.

Hoppy

[Aaron]

self-oscillation frequency

There is no purposely-designed avalanche circuit in MOSFETs.

Maybe not but this sidesteps the fact that there are mosfets specifically designed to self-oscillate as there are mosfets that are designed to specifically not oscillate. When it oscillates, it may encounter avalanche.

IRFPG50 allows for oscillation and repetitive avalanche IF it should occur.

Since you want to stick to facts and empirical evidence, it would be a good idea for you to stick to the facts. I don't have much time for this thread but I do have to point out - a fact.

A very specific part of the premise to your argument for one is this:

"Using a pulse width of 15.42us (3.7% of 2.4kHz) as shown above, equates to about 72% wasted energy"

Now read this from the actual Quantum article.




That says:

The required level of oscillation is achieved by setting the duty cycle at 3.7% 'on' at a frequency of 2.4 kHz.

Reducing the gate current of the mosfet results in an oscillation that
overrides the predetermined frequency and duty cycle. The frequency
oscillates between 143 kHz and 200 kHz and the duty cycle defaults
to approximately 1.3% on.

So here is a fact. Does the circuit claim of 17.0 COP come from the circuit
running at 3.7% duty cycle and 2.4 kHz frequency. The answer is NO; it does NOT.

That is what the timing signal was set at when it was triggered into
self-oscillation - the circuit ran in self-oscillation and the high gains came from self-oscillation.

And what was the circuit running at when in oscillation?

143 kHz to 200 kHz at about 1.3% duty cycle. There is a big difference.

Take that all into account first, then re-edit your explanation based on
the 3.7% and 2.4 kHz because the fact above contradicts the point
you make about 3.7% at 2.4 kHz frequency.

You don't even discuss what the circuit was running at for the high COP claim.

You started this thread, do what you want, you invited me and I'm just
pointing out this one very important fact that continues to be sidestepped
by everyone that is skeptical of these claims.

In any case, the above is all I have to say for now.

[quantumuppercut]

Okay, I will try to answer as many questions as possible with my point of view.

First of all, is Milhigh your buddy? Your name is David, right? Anyway, he got a good point. A wire is an inductor, even if it straight. That lead to your point of view about an inductor is the opposite a capacitor is somewhat not exactly correct. An inductor and a capacitor are like springs with different stiffness. Pure capacitor is spring with high stiffness. That means it has a very high vibrating frequency, the opposite is true for inductor. A mechanical view is that they're pipes with differnt diameter. High pressure, low volumetric flow (inductor) or low pressure, high volumetric flow(capacitor). This is the opposite you're thinking.

OK, What's wrong with conventional EM theory.

It states that P=IV
it also state that P=I^2R

This is the mistake I spot in conventional theory. Those are different P. IV is electrical current while I^2R are EM waves. They're two totally different things.

What is the equilvalent OU capacitor? All AC capacitors are OU with EM waves feed back. The reason is as stated above about conventional theory.

[poynt99]

Give it a shot

So far there are no responses to the original question in this thread.

Anyone?

There's an equivalent capacitive circuit that behaves much the same way (taking into account the main differences between capacitors and inductors), and is still a simple circuit.

Can anyone get their mind around how it would look and work?

Come on guys, I know there are a number of you watching this thread. Don't be shy

.99

[poynt99]

Keep in mind too, that as I understand Rosemary's claim, it has to do with conserved energy in the load resistor being stored at the point of manufacture, and that it would leak into the field and add to it resulting in a breakdown of the load resistor. Please correct me if I am wrong here.

What is leaking into what, and what is breaking down? Sorry, haven't a clue what you mean by all that.

.99

[poynt99]

More Comments

Harvey Post: http://www.energeticforum.com/63139-post1519.html

From where I sit, it seems we want a higher amplitude ring if possible. 1 The frequency of that ring will depend on the battery and inductor combined resonance. Perhaps Hoppy or Milehigh can offer suggestions on methods of determining the battery capacitance so we can actually engineer a resonant circuit. My time is divided greatly over the next few days, but I will help where I can. 2

The Avalanche circuit inside the HEXFET serves to snub the ringing, but under certain conditions can actually exacerbate it when it encourages an increased current draw at the start of the spike but on the next cycles of the ring the Avalanche diode turns off for some reason. That would result in an increased field energy that dissipates in the ringing. 3

Rosemary, were any voltage measurements taken across the HEXFET? Did they ever exceed 1KV?

EDIT:
How can I measure the capacitance (NOT capacity) of a battery? - Yahoo! Answers
Chapter 9: Internal Battery Resistance

1. Commented on in above post.

2. The resonant ringdown frequency of the coil is mainly determined by its own self-inductance and self-capacitance. It is not logical to state that the battery is involved here because during resonant ringdown, the MOSFET is OFF and there is no path through the battery, except for the stray, parasitic and junction capacitances I have mentioned several times.

3. There is no purposely-designed avalanche circuit in MOSFETs. This is analogous to reverse breakdown of a diode. Avalanche diodes are in fact designed this way.

If the MOSFET should ever avalanche in the OFF state due to excessive voltage, I'm not certain it would enhance the ringing effect at all. This would be analogous to the MOSFET turning back ON for a brief instant.

Also, keep in mind here that the Ainslie circuit includes a flyback diode, and with this diode in place there is no big voltage spike at any time on the MOSFET.

.99

[poynt99]

Some Comments

Harvey Post: http://www.energeticforum.com/63036-post1486.html

The image I have in my mind is the diagram found in the datasheet Fig 12a. labled Unclamped Inductive Test Circuit (which is nearly identical to Rosemary's diagram in Quantum)1

So, our circuit is ON, Voltage is present across our load and current begins to flow...what comes first? Well, if the voltage comes before the current, how long do we have to wait before current starts flowing through our IRFPG50? 2 Fig. 12b. gives us a clue. It all depends on the inductor and the Vdd value. But you can see it ramping up there, but what is happening at the peak? That's where we turn OFF the HEXFET, and notice all the activity that is still going on. 3 That's the test circuit, which is designed to fully deplete the stored energy of the inductor. But Rosemary's circuit doesn't do that. In her circuit, the current stops flowing through the IRFPG50 before the field is fully drained and the Vds < V(br)dss, now we have a voltage on the junction of the inductor and transistor that is higher than the voltage on the junction of the inductor and battery. So now we see a reverse current through the load. 4

What do you think, will the current going backwards through the load have any effect on the heating of it? 5 During the subsequent ringing that occurs, the over voltage causes repetitive passes of current through the inductor, and each pass adds heat as the excess voltage is exchanged for heat. 6

Since we suspect that the COP is a result of the ring energy, steps should be taken to enhance it. 7

1. Indeed the circuit is not unique, nor mysterious.

2. There is no delay--current begins to increase immediately.

3. As I mentioned in my diagram, the Ainslie circuit is being driven ON far longer than is required if only the reactive effects are desired--about 72% too long.

4. The reverse current pulse is about 1/3 the duration and 1/20 the amplitude compared to the forward current pulse in the Ainslie circuit with the 1N4007 in place.

5. There will be marginal heating added due to this reverse current pulse.

6. With such a low inductance to resistance ratio (i.e. low Q-factor) in this circuit, there is really very little ringing and it does not add much.

7. To enhance the ringing effect, you need to increase the Q of this circuit, namely by changing out the wirewound resistor for a real inductor with heavy wire.

.99

[poynt99]

So it is your belief then, that if two switches existed on each side of the inductor, and you opened both switches simultaneously, that the inductor would still ring?

I didn't say at all that it was my belief; I stated it as fact.

.99

[Harvey]

Here's something to get it started. This is the inductive circuit.

Now, who can come up with an explanation for OU of this circuit, and also a circuit employing a capacitor as per the original question in the first post?

From the attached diagram below:



.99

So it is your belief then, that if two switches existed on each side of the inductor, and you opened both switches simultaneously, that the inductor would still ring?

[morpher44]

belief

It is my strong belief that all the "inventions" that work today only work because the people that created them had a belief they would work.

I think if you actually go try to invent something that you believe will work -- you will rapidly discover that it does not work the way you believe it works.
Edison & team believed they could make a light-bulb. It wasn't as easy as they thought -- taking many many man hours, experiment after experiment ... with lots of materials, etc. Here is a dramatic case of belief NOT intersecting with reality.

I suspect success with invention has more to do with "intuition" ... or that ability to get messages from your more brilliant unconscious mind. But more than that you also have to TRY to make something work ... and probably fail hundreds of times before succeeding.

[morpher44]

coil-cap marriage

The Tesla bifilar coil has capacitive properties according to Tesla own words.
Michel

Yes. All coils will have a bit of capacitance .. not just the bifilar variety.
There is a point of self-resonance which is the resontant
frequency using L & C where C is the implicit capacitance in the coil.

In terms of a marriage, though between coils and capacitors,
I was thinking more that a new design would EXPLOIT this relationship
much more seriously. Either a coil can be wrapped around a
capacitor that has inner plates that are LOW resistance
OR
the coil can be built INSIDE the capacitor with the plates OUTSIDE
the coil.

The advantages that may occur would appear in the form of
negative resistance. The frequency response at various frequencies
might show that the capacitance side of the reactance is NOT
following the traditional inverse curve predicted by the capacitance,
but rather is BENT a bit by the extra charges conveyed --
van de graff -like- on to the capacitor.

It would be ideal if you could go to your neighborhood electronics store
and buy a coilpacitor with a known resonant frequency and Q
and other specifications CLEARLY identified so that you could
design with it.

[poynt99]

I would agree Gyula,

that conventional EM theory may not be able to explain "overunity effects".

However, if one does not possess an understanding of conventional EM, how dos one know the difference between conventional and unconventional effects?

Thanks, I had forgotten about Bearden's step-charged capacitor storage concept. He is correct in one regard, but also forgetting where the original energy came form. Yep, we can transfer energy from a battery or capacitor to another capacitor with almost no loss, but energy is still being used from the source battery or capacitor. Isn't that a huge oversight?

With Tito, I would not get too excited or worried about that one. It's the pattern/circumstance we've seen many times before.

What I'm asking about really, is an equivalent scenario to the inductive kickback circuit everyone thinks is so mysterious. The point is that there IS an equivalent capacitive circuit which complements it. I want to see if anyone can visualize and illustrate what it is, and explain how it works.

Capacitors and Inductors are truly inversions of each other, so logically there must be a circuit.

.99

[gyula]

Hi poynt99,

Though I wait for any meaningful report on this inductive circuit, I do not think conventional EM theory is able to explain any COP>1 possibility should it occur...

Regarding a capacitive version, Bearden occurs to me, referring to a so called stepwise charging of a capacitor, see here:
Additional Information On The Final Secret Of Free Energy

quote
...Rigorously one can use a normal capacitor as the collector, if one step-charges it in several hundred small incremental rectangular voltage steps (stair-step-charging). The proof that this can freely charge a capacitor with energy, without having to do appreciable work, is already known in the literature. You can charge the capacitor without entropy and essentially without drawing electron mass current. [Ref 2,Ref 3,Ref 4,Ref 5]
unquote

I also refer to a thread on overunity.com where a member wrote: ...discharging a capacitor in a coil is the real source for free energy..., see here:
Real OU-Effect to Share with everyone!!!

Unfortunately he does not seem to share his findings (if they are real, I hope they are, lol).

rgds, Gyula

[poynt99]

A Couple Additional Points

In the posted circuit diagram, we should not forget about the small dissipation in the flyback diode as well. Most of the energy will be dissipated as heat in Rcoil, but some in the diode too during the OFF phase when it is conducting.

Note also that even when SW2 is in the B position, i.e. the flyback energy is supplying power to an external load such a a charge battery, the return path is still at the +'ve side of the coil, which will again cause additional heat in Rcoil. Not as much this time, but additional nonetheless.

.99