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Inductive Resistor Open source development of highly efficient inductive resistor circuits.

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  #601  
Old 07-14-2009, 05:43 PM
RAMSET RAMSET is offline
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coils and frequency by Don Smith

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  #602  
Old 07-14-2009, 06:19 PM
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Originally Posted by gotoluc View Post
Hi Hoppy,

I don't have much time since I now have to work today.

I don't believe the input power to be zero since like I showed the filament is glowing. Read the Pdf document that .99 has so amazingly put together: Effects of Recirculating BEMF to Coil
and that will explain in an EE standard way what is going on.

Got to go.

Luc
Hi Luc

Yes, I had read Poynt99's document thoroughly before I posted the reference to it. I concur with his conclusions.

Hoppy
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  #603  
Old 07-14-2009, 06:20 PM
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Originally Posted by Aaron View Post
Thanks Poynt99 - yes, that is what I'm asking and it is not a test to see if you know the answer.

If the current/voltage relationship is locked - yet it is able to be changed at high frequency (or other mentioned conditions), what in the classic textbook teaching allows for this exception?
Aaron,

IF we can bend the "rules" regarding the accepted behaviour of electron flow and all associated (and possibly hidden) fields, then as far as I know you will not find the "method" or exposť in any classical text book.

.99
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Last edited by poynt99; 07-14-2009 at 06:23 PM.
  #604  
Old 07-14-2009, 06:21 PM
witsend witsend is offline
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Gotoluc - this is for you to think about.

The principle is this. On our experiment we generate counter electromotive force - not at a coil, as that would complicate the measurement protocol - but on an inductive load resistor. I believe they're referred to as 'curly wounds'. This still gives a kick - not as extreme as your coil - but enough for these purposes. This 'kick' is taken straight through the flyback to the postive terminal of the battery. We then see this. The battery is now able to recharge. And it is effectively being recharged from the very energy that it first discharged.

This, in itself is problematic. We are now saying, as Dr Stiffler pointed out, that we've withdrawn $2.00 from the bank - spent $2.00 and - from somewhere out of the ether - then redeposited something in excess of $1.00 back to the bank. That dollar - or that something more than a dollar? Where did it come from? This is the problem that classicists wrestle with. With good reason.

So. When we use a battery - in line with all good and classical measurements we can concede that the battery only delivers energy during the 'on' period of the switching cycle. We don't have to go into too much detail about the 'inductive component' as the inductive componenet here is in the resistor itself. Whatever it measures as 'heat' WILL also reflect the energy that was delivered.

The question then is how to compute the energy that was first delivered. Well there's no coil. So, presumably if we use a 'shunt' on the negative rail on the supply - then we can measure the voltage on that supply. We know that what the battery delivers can only relate to the ON period of the switch as this is the only time that the battery can, in fact discharge any energy. So that part is relatively easy. We take the vbatt divided by the Ohm's value of the shunt to compute current - and then multiply the current by the battery voltage to compute wattage. This is correct and strictly conforms to good classical measurement protocol. Whatever that number it precisely represents the wattage delivered by the battery.

Then comes the tricky part. The duty cycle changes. The battery is, in effect, no longer able to deliver any current. The fields on the resistor collapse to zero. And the strength of that collapse relates to the applied energy from the On cycle. Energy is energy. What comes in must go out. What gets kicked must be moved. And when magnetic fields collapse, all they are doing is changing in time. Magnetic fields changing in time induce an electric field. But the strength of those changing magnetic fields from some voltage to zero - also needs to discharge. There is only a very short period during which it can discharge being moments between the 'on' and the 'off' time of the switch. It takes full advantage of that 'moment' and developes a spike to carry the full force of the energy applied in the On time and kicks it back in nano seconds as a 'spike' that is always evident 'beween' the on and off period. That's the counter electromotive force.

But that energy is evidently also measurable on the shunt. And where does it go? It goes first through the battery, thereby recharging it - and then to the load thereby heating it up. So. We're back to that problem. How can we have found something more than $1.00 when we've already spent $2.00 to give anything back to the bank.

And yet. The paradox. Our measurement of that delivery of energy is determined by classical measurements protocol. Not by me.

I think you'll have picked up the significance as it relates to your utility supplier (edit) and to your test. But I'll get back to it.
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Last edited by witsend; 07-14-2009 at 07:46 PM.
  #605  
Old 07-14-2009, 06:26 PM
poynt99 poynt99 is offline
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Originally Posted by witsend View Post
Hoppy and .99 Regarding the significance of the measurement protocol - we make an assumption. When the battery delivers current flow it is at a loss to the battery. But when we return that energy it is a gain to the battery. The sum of the energy, therefore delivered by the battery is the difference between these two cycles.

This has been variously approved by every academic who has seen the paper and I think is substantially correct. But I would, nonetheless, be glad of some endorsement from both or either of you. It's a critical point to the claim.
Rosemary,
I am working at the moment so will try to respond later on today/tonight.

.99
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  #606  
Old 07-14-2009, 06:31 PM
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Michael John Nunnerley Michael John Nunnerley is offline
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Back again

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Originally Posted by RAMSET View Post
Hi Rose and all, I'm back again after not being able to access the forum here in Spain.

Ramset it is funny you have posted a link to this on Don Smith, this is one of my lines of investigation and part of my magnetic flux manipulation thread.

To keep things on track here, I think there is a link with Roses circuit in a round about way but at the moment I do not think it is apropriate in this thread, may be farther down the line, my opinion, but a fascinating topic

Mike
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  #607  
Old 07-14-2009, 06:36 PM
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An electronic circuit to free energy

Hi Rosemary,

I hope this will help you, forum members and guests ..... It was one of those web sites referenced in a post that ended up in French and I just scanned over for interesting photos, images, drawings and saw the schematic and scope shots it looked real similar to what is going on. I'm glad you and others liked it great find from "NerzhDishual" a OU member .... I'll PM him over there and see If he wants to partake.

Best Wishes
Glen



------------------------------------------------------------------------

An electronic circuit to free energy

An electronic circuit to free energy
Quanthomme thank Zoltan Szili
we have entrusted the distribution of his work.


For all correspondence here is the email address Zoltazn Jean Szili

p_baril@sympatico.ca

Introduction

"By publishing this very simple circuit, I would give fans the opportunity to prove that the extraction of electrical power vacuum exists.

The purpose of my letter is to inform my work and help develop quantum generators for use by people and industries.

I decided to make public a first circuit, and fans will be able to test it. "

Presentation of the author

"My name is John ZS All my life I worked in scientific research. For seven years I am retired.

Thirteen years I lived in France. Five years in Lyon and Paris to eight years. In Paris, I worked at the University of Orsay in Physics of Plasmas. I was born in Hungary. I emigrated to Canada (Quebec) in 1969 and I found work at the Institut de Recherche d'Hydro Quebec in the field of scientific research. I stayed twenty-eight years until I retire.

One day at the restaurant of the institute, I sat at a table in front of two engineers and their invited experts processors. I listened to the conversation, and engineers have told their guest, they could calculate the parameters of transformers with an accuracy of 0.1%, but they are 2% of power still in too.

This surplus of power remains inexplicable while taking account of ohmic losses and so on. I left the restaurant thinking that this 2% of excess power could come from an unknown source.

A few years later I became aware of the theory of Zero Point Energy. I assumed that there should be a way to extract a part of this phenomenal energy. Since my retirement, I work hard to find circuits that could extract energy from the zero point. Quickly, I left out the laboratory work for the computer simulation, in order to expedite my work. I have tried thousands of channels before finding one and then several of these circuits that say in English "overunity".

Method of work

"When we explore an unknown field, one must use trial and error empirical. Gradually, we learn about this new unknown world of science and use this knowledge to find circuits more efficient.

Currently, I know quite well how to extract the ZPE. The power of my circuits ranges from a few milliwatts to more than one hundred kilowatts. These circuits must be powered to operate, but the output circuits, gains power (COP = energy output / energy input) may be very high. The gain or COP may be 2, 10, 100, 1000 and beyond.

In these simulations, I always used electronic components existing in the trade. This may facilitate the realization of these circuits in the laboratory. This work remains to be done.

First, this circuit has been developed on a computer simulation. This allows for the simulation, the Jiles-Atherton model of electromagnetism. This model was designed to meet the reality test and not the law of conservation of energy. I can not give guarantees regarding the functioning of this circuit, but until proven otherwise, have confidence in the simulation. This program is used across the world in electronic and physical laboratories. It is very close to reality, although in experimental 99.999% of 100 cases energy is conserved. "

The mechanism of extraction

"There are special conditions for there to be extracting energy from quantum vacuum fluctuations. The mechanism of extraction, in my opinion, is based on ferromagnetic resonance. The free electrons exchange energy with electrons virtual vacuum. Normally the exchange is completely symmetrical. Outcome: No energy extraction. In addition, electrons are fermions, and quantum mechanics, two electrons can not be the same. The result is a dispersion very large in the frequency of ferromagnetic resonance, might be called semi-collective. This resonance is between a few hundred MHz and several GHz. Nevertheless, the circuits operate at a frequency less than one MHz. As against the circuit must be built as a circuit operating at least 25 MHz. Avoid loops, using more appropriate component. In these circuits, you must use special methods to make the asymmetric exchange of energy. In this case only, there will extract energy.

In publishing this first tour, my goal is to stimulate research ZPE. I think especially for amateurs.

A new about my research: I think I have found a physical law, connecting the electric current in a coil (as used in the circuit is sent) and the inductance of the coil. This relationship seems to inductor current and fundamental in the extraction of electrical power. This relationship is reciprocal. A variation of current varies the inductance and inductance variation varies the current.

When electric current increases, it decreases the inductor, and a decrease in the inductance increases the current ... and so on. It is the effect of avalanche. The decrease of inductance has a certain limit determined by the electronic circuit. After stopping the avalanche, the process is reversed. The inductance increases and this lowers the power ... and so on. This return takes less time because there was no extraction of electric power vacuum during the return. Normally, because of electrical losses, the Avalanche will not begin without stimulation (eg a pulse above a certain value). The physical form of this relationship is as follows:

I (L1) = (K1 / L (L1)) - or I K2 (L1) is the current flowing through the coil and L (L1) is the inductance of the coil. K1 and K2 are two constants (unchanging) depending components of the circuit.

By adjusting the constants K1 and K2, the curves of current and inductance are virtually identical. For other circuits, the formula could change, but what is important is that the current relationship and the inverse of the inductance should remain valid. "

The electronic circuit free energy

Mr Zoltan Szili

"This circuit is very simple in appearance, but to successfully make it work, then you need to take drastic precautions. True, it operates at a relatively low frequency of 20 kHz. As against, the signal generator pulse must be a square, positive, with a rise time of 10 nanoseconds from 0 volts to +5 volts.

The simulation shows very clearly that if the rise time of the square is longer than 10 nanoseconds, the extraction decreases very quickly and completely canceled between 50 and 100 nanoseconds.

The simulation also shows that a parasitic capacitance at the point of connection of the transistor (M1), the inductor (L1) and output resistance (R1) with a value of 100 picofarads to ground, completely destroyed the extraction (capacity of 100 picofarads may be the ability of an oscilloscope probe).

The parasitic inductances can also prevent the extraction, if it exceeds 10 micro Henry.

In assembling the circuit, we must minimize the loops, as if the circuit were fonctio.nner to 25 Mhz

In fact, the element is the extraction of the ferrite toroid inductor (L1). "

Schematic and measures

Not need an oscilloscope to demonstrate on the unit: it is enough to measure currents ex 2: between the earth and the resistance R2, and between the land and the resistance R1 "






-1 --



-2 --



-3 --

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  #608  
Old 07-14-2009, 06:38 PM
Joit Joit is offline
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@witsend.
I dont know, if Tinselkoala did modify his timer Circuit, that he dont get the same Result.
My Scopeshot shows the same, as his last 'Way it is"
My Offset line is also the the low long bright Line from upper Channel.
The On Time shows up as the Peak.

Just the different, he use his Functiongenerator and need a Dutycycle of 96%, to get the Mosfet for ~3% ON.
With the Timer Circuit from the Quantum Article it does the same at the Mosfet, it sets the Mosfet ~3% to on.
Anyhow it sounds like, there is a Missunderstandig where the Duty cycle should be.
Maybe someone did read some Posts to fast? But i think, we can agree, that the 3% Cycle should be at the Mosfet ON time.

And may we should look different at the Puls from the Timer to the Gate, not as a rising Voltage,
more like a Unit of Energy, what saturate the Gate at a certain Point.

At last, his last Shot from the Scope and my Shots are identical.
So, his Statement, that the Timer Circuit dont works, still cant been right.
I am still waiting for it, that he do rebuild it with the right Parts, and it works, as he showed in his last Shot.
OR he goes finally into the Kitchen and bakes a Cake for us.

Maybe he did connect it different. I did it like
Plus - 10W Bulb - 2. Probe - Drain - Source - Minus
|-same Plus -> Diode - Timer - Pin3 - 1.Probe - Potentiometer - Gate
And from Scope Ground to Ground.


@Poynt99
Anyhow i dont get your Point, can you speak free please, at what you are thinking about, what the output from the Timer do/should cause?
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  #609  
Old 07-14-2009, 06:42 PM
poynt99 poynt99 is offline
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Originally Posted by witsend View Post
YouTube - Electric OU 7: Up is OFF

This link has been provided by a friend from OU.Com. It's been posted by TK and apparently shows that a MOSFET is constitutionally unable to apply any effective duty cycle to a battery supply source.

Yet again, .99 - as you're on line and with us at the moment. Could you be so good as to take a look at this and tell us what's wrong? Could it be that our experiment is somehow negated because we cannot ever 'switch off' the battery supply source as is now being argued? For some reason, and at the moment, I cannot get into the site myself. But I would be glad of some comment. It seems that we have been deluded into thinking that a switching cycle using even a MOSFET can ever interrupt the supply from a battery. It's also the same objection to the evidence of the 'on vs off' displayed by the 555. But this new slant seems an extraordinary argument as a basis to disprove the experiment.

EDIT I see .99 is not with us at present. But Hoppy. Could you take a look?
TK is precisely correct with his demonstration, as he has been throughout.

The problem that seems to be occurring here is folks are associating "voltage" with "ON", and "no voltage" with "OFF".

We are dealing with a simple switch here, and this is pretty much how the MOSFET should be viewed...as a simple single-pole, single-throw switch.

Now the question I will ask is this: What is the voltage across the two terminals of an ideal switch when it is turned "ON" (i.e. when the switch contacts are closed)?

It is zero volts correct?

What is the current through this same switch when it is turned "ON"?

The MOSFET acts exactly the same as this switch. So, when the MOSFET is "ON", there is zero volts across the MOSFET D-S leads, and since the S is connected to ground, we should measure almost 0V at the D lead when the MOSFET is "ON". This is also the state which will allow current through the load.

So, monitoring the voltage is confusing folks here. If the current was being monitored (i.e. place a bulb as the load), then it would become clear what is going on.

.99
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  #610  
Old 07-14-2009, 06:55 PM
Joit Joit is offline
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Quote:
Originally Posted by poynt99 View Post
We are dealing with a simple switch here, and this is pretty much how the MOSFET should be viewed...as a simple single-pole, single-throw switch.

Now the question I will ask is this: What is the voltage across the two terminals of an ideal switch when it is turned "ON" (i.e. when the switch contacts are closed)?

It is zero volts correct?

What is the current through this same switch when it is turned "ON"?

The MOSFET acts exactly the same as this switch. So, when the MOSFET is "ON", there is zero volts across the MOSFET D-S leads, and since the S is connected to ground, we should measure almost 0V at the D lead when the MOSFET is "ON". This is also the state which will allow current through the load.

So, monitoring the voltage is confusing folks here. If the current was being monitored (i.e. place a bulb as the load), then it would become clear what is going on.

.99

Umm, can you repeat it at the case, you did write once a ON instead a OFF to often?
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  #611  
Old 07-14-2009, 07:05 PM
Hoppy Hoppy is offline
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Quote:
Originally Posted by witsend View Post
Hoppy and .99 Regarding the significance of the measurement protocol - we make an assumption. When the battery delivers current flow it is at a loss to the battery. But when we return that energy it is a gain to the battery. The sum of the energy, therefore delivered by the battery is the difference between these two cycles.

This has been variously approved by every academic who has seen the paper and I think is substantially correct. But I would, nonetheless, be glad of some endorsement from both or either of you. It's a critical point to the claim.
Rosemary,

Poynt99 has explained in detail EE terms how power is calculated and shared in the circuit. He has not attempted to calculate the effect of the flyback energy on the battery itself. I agree that the sum of energy delivered to the circuit by the battery is the difference between the two cycles. The problem is working out exactly what effect the flyback has had on the capacity of the battery. The stored capacity in ampere hours is very difficult to measure with a good degree of accuracy, even using battery capacity meters (BCM's). If this were possible, we would have a viable method of calculating the true COP of the system under test by taking a capacity measurement before and after the system test run. However, A BCM should be capable of providing a reliable indication of whether or not a given battery has gained or lost capacity over a given period of time and this will at least show if a system is operating at unity or better.

Hoppy
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  #612  
Old 07-14-2009, 07:16 PM
witsend witsend is offline
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Ramset - the video link was so interesting. It's amazing to think that here we are battling to get a really, really small effect accepted, and there's Don Smith selling ZPE to the arAabs, Chinese and just about everybody. We're already obsolete and not even our work is accepted by mainstream.

Brings to mind the Wright brothers flying over the the universities where the students are being taught that flight is impossible.

We're living in exciting times folks.
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  #613  
Old 07-14-2009, 07:17 PM
Joit Joit is offline
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Oh Men, i think his biggest Enemy is his own Ego.
Seems he want to have a Battle about, Who is right and Who is Wrong, no matter, from the Results of it.

Well, a Mosfet is not really total off, that why you can use it for a AC Circuit too. But it opens still only in one Direction with a Pulse.
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  #614  
Old 07-14-2009, 07:24 PM
witsend witsend is offline
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FuzzyTomCat - you've done it again - pulled the cat out the bag. I've just seen the post. MANY MANY THANKS.

Can you now really perform a miracle and see if you can reach the guy and see if he'll talk to us. I've been trying to email him and nothing. He's got to be an asset. He's looking for us amateurs to test the theory. And I'm dying to talk theory.

I'd be so grateful if you could, perhaps try.

Thanks Glen - very much
Rosemary

EDIT - and everybody - PLEASE READ THAT PAPER. IT'S HUGELY SIGNIFICANT.
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Last edited by witsend; 07-14-2009 at 07:39 PM.
  #615  
Old 07-14-2009, 07:28 PM
RAMSET RAMSET is offline
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Don Smith

Rosemary I had to share that with you
I just got that today from user bolt at OU
should be required viewing for EVERY ONE
Seems like he was in the room during the event

Don Smith Free Energy - Video

Chet
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  #616  
Old 07-14-2009, 07:36 PM
witsend witsend is offline
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Quote:
Originally Posted by poynt99 View Post
The problem that seems to be occurring here is folks are associating "voltage" with "ON", and "no voltage" with "OFF"....

So, monitoring the voltage is confusing folks here. If the current was being monitored (i.e. place a bulb as the load), then it would become clear what is going on.

.99
.99 with respect. Who is confused? Even I, with my limited knowledge, understand this. Is TK giving us a public lecture? Is he asking the public the obvious significance of the voltage? Is he using this new point to throw doubt as to whether we are delivering energy while the switch is off? What is his point? I cannot work it out. EDIT - nor, I may add can anyone I've spoken to. And I've referred many to that video of his.
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  #617  
Old 07-14-2009, 07:43 PM
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magnetic field strength

Quote:
Originally Posted by poynt99 View Post
Aaron,

IF we can bend the "rules" regarding the accepted behaviour of electron flow and all associated (and possibly hidden) fields, then as far as I know you will not find the "method" or exposť in any classical text book.

.99
Thanks and I appreciate the honest answer.

Like Luc and so many others here, I do not have any training in physics or electronics. When I need to calculate something, I use use online calculators or I ask my friends and most of us can be resourceful enough to learn what we need to learn for whatever project we happen to be working on in the moment.

Anyway, there appears to be quite a few examples of the current:voltage relationship being violated like operating certain transistors in their negative zone where current drops while the voltage is increasing.

There are various electromagnetic circuits that produce a magnetic field with no voltage - in a way, they imitate a permanent magnet that has current but no voltage.

Capacitors can be alternately discharged into opposite ends of a coil in a novel way that charges the coil with all benefits of the amperage but there is no voltage - violating ohms law, lenz law, etc...

Anyway, I'm glad to see you're open to the possibility.
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Old 07-14-2009, 08:23 PM
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Mosfet question

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Originally Posted by poynt99 View Post
I have tested in PSpice the 555 timer circuit as per the Quantum article, and my findings are that the 555 timer output is indeed at about a 96% duty cycle when adjusting for about 2.4kHz frequency. The full range of the 2 potentiometers were swept and the lowest duty cycle achieved was about 73%. There was no setting that could achieve 3.7% duty cycle. Obviously the flyback diode was missed in the article, but it has no bearing on the 555 duty cycle output.

It seems quite clear from your posts and from the EIT paper that the desired duty cycle for driving the MOSFET is 3.7%, and I can accept that the circuit in the Quantum article achieves quite the opposite, because as you say you had no involvement with the design or testing of that circuit. No problem there for me.

It appears the fellow that did the 555 timer design for the article was doing so to achieve 3.7% duty cycle at the MOSFET Drain rather than the MOSFET Gate . Errors and mis-communications do happen, so I think it's safe to say that the 555 timer circuit shown in the Quantum article can be dismissed as valid for your experiment, and that those wanting to replicate your experiment do so with either a function generator or some other oscillator capable of achieving the desired 2.4kHz/3.7% duty cycle wave form, and driving the MOSFET.
"It appears the fellow that did the 555 timer design for the article was doing so to achieve 3.7% duty cycle at the MOSFET Drain rather than the MOSFET Gate"

I'm just wondering about something and anyone feel free to jump in.

Q1) The bottom line is that for each pulse/cycle, the battery should only be delivering current for 3.7% of the time. Is this correct? That is the point of having the duty cycle to begin with is that power is delivered for x% of the time per cycle.

Q2) If the Quantum article circuit has a 3.7% duty cycle at the Mosfet DRAIN - doesn't that mean that the Mosfet is allowing current to pass through from the battery at 3.7% of the time per pulse?

Q3) In the below diagram, if the Mosfet drain is conducting 3.7% of the time per pulse will the LED be on 3.7% of the time?

------------------------------------------------------------------



Example application of an N-Channel MOSFET. When the switch is pushed the LED lights up.

The LED only lights when the switch is closed to send power to the gate, which allows voltage and current to move through the mosfet (source/drain). If the mosfet wasn't off when the switch is disconnected, the led would stay lit at a 100% duty cycle making the mosfet obsolete meaning that the semiconductor industry has tricked everyone into thinking it is a switch. lol
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  #619  
Old 07-14-2009, 08:33 PM
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Aaron, Loved the example. I'm now going to sleep. It's been a great evening on the thread. But I feel my AGE. I need to walk the dogs. OH HOW I LOVE THIS SUBJECT.
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  #620  
Old 07-14-2009, 09:29 PM
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An engineers opinion

Aaron with reference to your question and diagram: imagine a voltmeter or oscilloscope connected at the mosfet drain. Or at the top of the resistor where it connects to the positive rail. When the gate signal is ON and the mosfet is conducting, what is the voltage at these points? When the gate signal is OFF and the mosfet is NOT conducting, what is the voltage at these points

When the mosfet is conducting the voltage at those points is LOW, not high. When the voltage at those points is HIGH the mosfet is not conducting, it is off. Joit's trace shows the voltage going HIGH at the mosfet drain for short periods. The transistor is OFF at these times. just hook up a bulb like I have done. Try it!


.
Slow the freqs down, and think about what you are seeing. Carefully.

Aaron the drawing is fine, and yes, as shown the mosfet STATE (on or off) clearly exactly follows the gate drive state: ON is ON, for sure. But that's not the issue: the issue is HOW LONG it's on, and how that on time is measured. If you are looking at the Point A in Ainslie's circuit or the drain of the mosfet in your diagram, the VOLTAGE that the scope is measuring--what it uses to give the duty cycle figure...that voltage is HIGH when the MOSFET ( and the light) IS OFF. Use that exact circuit and put a meter in the exact place Rosemary does. Push that button and tell us what the voltage is on your meter.
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Last edited by RAMSET; 07-14-2009 at 09:36 PM. Reason: clean up
  #621  
Old 07-14-2009, 09:40 PM
Hoppy Hoppy is offline
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I would like to return to Luc's last video demo where he pointed out that the input lamp was very dimly lit. The output lamp on the other hand was brightly lit. It had been strongly suggested by some prior to this test that the input lamp was a good indication of the total power being consumed in the circuit. Assuming this is the case and keeping things simple without EE equations involved, where is the additional energy coming from to light the lamp connected as the load? Logically, I think we have two possibilities, either from the supply battery or direct from the environment.

If we accept that the energy is being tapped from the environment and in some way finding its way to the load without affecting the input lamp, then surely we are in an OU situation big time! If we accept that all energy is being derived from the battery, then logically this energy must be present in the input circuit prior to being transferred to the load. If the energy is being derived from the battery, why is the lamp so dimly lit if it is supposed to be indicating the total power input to the circuit?

I apologise if this sounds rather confusing but I feel it is very important to apply a bit of logic to what we are observing.

Hoppy
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Last edited by Hoppy; 07-14-2009 at 10:05 PM.
  #622  
Old 07-14-2009, 09:55 PM
poynt99 poynt99 is offline
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Quote:
Originally Posted by Aaron View Post
"It appears the fellow that did the 555 timer design for the article was doing so to achieve 3.7% duty cycle at the MOSFET Drain rather than the MOSFET Gate"

I'm just wondering about something and anyone feel free to jump in.

Q1) The bottom line is that for each pulse/cycle, the battery should only be delivering current for 3.7% of the time. Is this correct? That is the point of having the duty cycle to begin with is that power is delivered for x% of the time per cycle.

Q2) If the Quantum article circuit has a 3.7% duty cycle at the Mosfet DRAIN - doesn't that mean that the Mosfet is allowing current to pass through from the battery at 3.7% of the time per pulse?

Q3) In the below diagram, if the Mosfet drain is conducting 3.7% of the time per pulse will the LED be on 3.7% of the time?

------------------------------------------------------------------



Example application of an N-Channel MOSFET. When the switch is pushed the LED lights up.

The LED only lights when the switch is closed to send power to the gate, which allows voltage and current to move through the mosfet (source/drain). If the mosfet wasn't off when the switch is disconnected, the led would stay lit at a 100% duty cycle making the mosfet obsolete meaning that the semiconductor industry has tricked everyone into thinking it is a switch. lol
Aaron, some very good questions ...ones that could finally put an end to the confusion here.

1) YES! "Duty Cycle" in general refers to the percentage of a given time period that something is delivering power to a load. In this case Rosemary has specified that the MOSFET should be driven at the GATE with 3.7% duty cycle wave form such that the MOSFET conducts current from the battery for only 3.7% of the full 2.4kHz cycle.

2) NO. Remember my question above about the ideal switch and what the voltage is between its two terminals when the contacts are closed? If the contacts are OPEN, i.e. the switch is "OFF", then you WILL measure a voltage between its terminals. It is the same with the MOSFET. When the MOSFET is "OFF", you can measure a voltage at its DRAIN. So in this case if we are measuring a voltage for only 3.7% of the time (the rest of the time we measure almost zero volts), what percentage of the total cycle is the MOSFET "OFF"?

If that is still unclear, then imagine removing the MOSFET from the circuit. You now have an open space where the MOSFET was and there is no possible connection between the coil and ground. If you place your DC voltage meter between the coil - terminal (the coil + terminal is connected to the battery +) and the circuit ground, you will measure a DC voltage equal to the supply voltage, correct? Let's say your supply voltage is 12VDC. So you will measure 12VDC on your meter. Now, while your meter is still connected to the same points, take a length of wire and connect the coil - terminal to one end, and the circuit ground to the other end of the wire. If you use a heavy enough wire, your meter will now read a very low voltage, probably very close to zero (and your wire may become very hot, so be careful LOL).

3) YES! If the MOSFET Drain is conducting for 3.7% of the total cycle PERIOD, then yes the LED will be ON also for 3.7% of the time. This is of course assuming that you are using a very low frequency so you can actually see the LED turn discretely ON and OFF.

If you can sort out answer #2 in your mind, you will have mastered how to read a scope with these types of circuits and wave forms, and you'll know what the true duty cycle is.

Hope that was helpful.

.99
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  #623  
Old 07-14-2009, 10:04 PM
gyula gyula is offline
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Hi Hoppy,

You can consider the input lamp and the rest of the circuit as a voltage divider, fed from the input DC source. The upper member of the voltage divider is the input lamp, the lower member of the divider is all the rest of the circuit.

What constitutes the individual dissipation of these two members?

Obviously the voltage across their own two legs and the common current flowing in them, I say common current because these two members are connected in series.
So the input lamp is hardly lit because the voltage drop across its two legs is small due to the divided DC source voltage, the bigger voltage part gets across the two legs of the rest of the circuit (in which the the output lamp is too).

And because the current is the same in the two members, the voltage drop is very different, this explains the hardly lit input lamp's case.

Simply put it. Study voltage dividers a little to understand more, if needed.

rgds, Gyula
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Last edited by gyula; 07-14-2009 at 10:06 PM.
  #624  
Old 07-14-2009, 10:11 PM
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Aaron Aaron is offline
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simple question

Ramset,

That is right but that does not answer the specific questions.

A closed switch is like taking a voltmeter and putting both leads next together on the same wire, there will be no potential difference even though current is flowing and there is voltage moving. That much is common sense. Actually, there will be small milivolt reading because there is a small potential difference but for all practical purposes, there is no voltage.

If the DRAIN has a 3.7% duty cycle, IS IT or IS IT NOT conducting current from a power source for 3.7% of the time per pulse?

It really is a simple yes or no question.
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  #625  
Old 07-14-2009, 10:28 PM
Hoppy Hoppy is offline
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Quote:
Originally Posted by gyula View Post
Hi Hoppy,

You can consider the input lamp and the rest of the circuit as a voltage divider, fed from the input DC source. The upper member of the voltage divider is the input lamp, the lower member of the divider is all the rest of the circuit.

What constitutes the individual dissipation of these two members?

Obviously the voltage across their own two legs and the common current flowing in them, I say common current because these two members are connected in series.
So the input lamp is hardly lit because the voltage drop across its two legs is small due to the divided DC source voltage, the bigger voltage part gets across the two legs of the rest of the circuit (in which the the output lamp is too).

And because the current is the same in the two members, the voltage drop is very different, this explains the hardly lit input lamp's case.

Simply put it. Study voltage dividers a little to understand more, if needed.

rgds, Gyula
Hi Gyula

If you read my earlier posts you will see that I have explained exactly what you are saying. I'm trying to get others to understand what we are both saying and also what Poynt99 is trying to convey in his report. The alternative is to accept that this circuit is running at rampant OU and I don't think many of us really believe that this is the case with this circuit. This is why I asked Luc to run this lamp demo as it shows an extreme difference between the brightness of the lamps, which some have strongly suggested represents true power level in a circuit.

Hoppy
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  #626  
Old 07-14-2009, 10:51 PM
gyula gyula is offline
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Hi Hoppy,

Ok, and sorry for not being aware of your full contributions ( I was also blocked out from the energeticforum.com since last Saturday like some other innocent member here (Jetijs, Michael Nunnerley) from Europe..., for unkown reason for me.)

I would be happy to read Peter's experiences with this circuit, for his version of schematics appeared on the very first page of this thread, and since then nobody has addressed his circuit.

Thanks, Gyula
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Last edited by gyula; 07-14-2009 at 10:54 PM.
  #627  
Old 07-14-2009, 11:50 PM
poynt99 poynt99 is offline
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Quote:
Originally Posted by witsend View Post
.99 with respect. Who is confused?
Rosemary,

Where there is one, there are likely many more.

Quote:
Even I, with my limited knowledge, understand this. Is TK giving us a public lecture? Is he asking the public the obvious significance of the voltage?
No.

Quote:
Is he using this new point to throw doubt as to whether we are delivering energy while the switch is off? What is his point? I cannot work it out. EDIT - nor, I may add can anyone I've spoken to. And I've referred many to that video of his.
Actually the point he and I and others have been trying to make is unchanged from the beginning. I don't think there is a new point at all in his new video. I'm sure TK will correct me if I'm wrong.

I think what you may be getting from the videos and portions of some posts is that TK is solely attempting to debunk your claims of energy gains from your circuit.

That may be partly true, but as far as I can tell the main point lately has been to establish that the 555 circuit everyone is building in an attempt to replicate your circuit, will produce the wrong drive output from the get-go.

If there is any chance of replicating your results the oscillator driving the MOSFET switch needs to be set to exhibit a 3.7% ON duty cycle, and from what I, TK and others have seen, folks are building the 555 circuit from the Quantum article, which is not producing the drive you have specified.

So again, the point lately has been to let everyone know that they should be using a function generator or a corrected 555 circuit, and I am quite certain that if asked, TK would design one so that folks can get on with testing the circuit as specified by you.

Folks are also asking questions and trying to understand WHY the published circuit won't give the right result's and that is a big positive step forward for everyone here.

If TK is unwilling to design a corrected 555 circuit (he probably already has), then anyone else that wishes to, feel free to step forward. Luc? In fact it wouldn't be a bad idea for all interested to take a crack at it, perhaps in a new thread so as not to throw this one any further off track than it already is. There are several folks around (myself included) that could help out with the design as questions come up. Ramset, you want to head this 555 project up perhaps?

Regards,
.99
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Last edited by poynt99; 07-14-2009 at 11:57 PM.
  #628  
Old 07-15-2009, 12:00 AM
RAMSET RAMSET is offline
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Reply to Aaron

NO.
When the drain signal is at battery voltage (high) the mosfet is OFF and the load is non-conducting. Clearly. You can wire up that circuit, using a real bulb instead of an LED, and if your voltmeter is sensitive enough to pick up the small voltage drop, you will see it. What looks like a big signal on the scope traces is really a very small fluctuation sitting on top of a large DC offset---the battery voltage. This is why 'AC coupling' must be used to resolve it. If you use DC coupling, if the screen shows a line at battery voltage the fluctuation will be too small to resolve--the voltage doesn't drop very much when the mosfet is ON. But it drops, for sure. If an oscilloscope is relatively dumb, like the Fluke 199, it needs to be TOLD whether you are calling the "high" signal "OFF" or "ON" to give a duty cycle output. There are 2 separate controls in the FLuke that must be set properly: the trace invert function AND the duty cycle definition. Only ONE of the Four possible combinations of these controls is correct in this experiment.

PS

If you want to call the drain signal a 3.7 percent duty cycle, that's OK as long as you know the MOSFET is OFF at that time. Not ON.

PPS
Jolt how is that light bulb test going
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Last edited by RAMSET; 07-15-2009 at 12:20 AM.
  #629  
Old 07-15-2009, 12:20 AM
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mosfet switch

Quote:
Originally Posted by poynt99 View Post
Aaron, some very good questions ...ones that could finally put an end to the confusion here.

1) YES! "Duty Cycle" in general refers to the percentage of a given time period that something is delivering power to a load. In this case Rosemary has specified that the MOSFET should be driven at the GATE with 3.7% duty cycle wave form such that the MOSFET conducts current from the battery for only 3.7% of the full 2.4kHz cycle.

2) NO. Remember my question above about the ideal switch and what the voltage is between its two terminals when the contacts are closed? If the contacts are OPEN, i.e. the switch is "OFF", then you WILL measure a voltage between its terminals. It is the same with the MOSFET. When the MOSFET is "OFF", you can measure a voltage at its DRAIN. So in this case if we are measuring a voltage for only 3.7% of the time (the rest of the time we measure almost zero volts), what percentage of the total cycle is the MOSFET "OFF"?

If that is still unclear, then imagine removing the MOSFET from the circuit. You now have an open space where the MOSFET was and there is no possible connection between the coil and ground. If you place your DC voltage meter between the coil - terminal (the coil + terminal is connected to the battery +) and the circuit ground, you will measure a DC voltage equal to the supply voltage, correct? Let's say your supply voltage is 12VDC. So you will measure 12VDC on your meter. Now, while your meter is still connected to the same points, take a length of wire and connect the coil - terminal to one end, and the circuit ground to the other end of the wire. If you use a heavy enough wire, your meter will now read a very low voltage, probably very close to zero (and your wire may become very hot, so be careful LOL).

3) YES! If the MOSFET Drain is conducting for 3.7% of the total cycle PERIOD, then yes the LED will be ON also for 3.7% of the time. This is of course assuming that you are using a very low frequency so you can actually see the LED turn discretely ON and OFF.

If you can sort out answer #2 in your mind, you will have mastered how to read a scope with these types of circuits and wave forms, and you'll know what the true duty cycle is.

Hope that was helpful.

.99
Yes, very helpful and I understand it perfectly because any closed switch has no potential difference on it if the ground and positive lead of a voltmeter is at both ends of the switch. I give the example in response to Ramset before I saw this post.

But my question 2 is not in reference to a scope shot? Just asking about an authentic 3.7% conducting time.
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  #630  
Old 07-15-2009, 12:22 AM
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Ground - Grounded - Grounding

Hi everyone,

There has been something that I have noticed looking at many older documents and illustrations of the term "Ground" as in Earth or "Terra firma" Terra firma - Wikipedia, the free encyclopedia . As you all know common voltages from countries vary, the UK, Europe, Africa parts of Asia many more not to be named for example use 240 volt ..... but the United States and other North and South America countries use 120/240 volt ..... the big difference is how the "Ground" is connected and referenced.

Where 240 volt is and the only voltage available meaning "NO" 120 volt, a ground conductor is actually Grounded to "Ground", Earth or "Terra firma".

Where 120/240 volt is a ground conductor is bonded to the Neutral wire in a Electrical Service Panel (circuit breaker or fuse) whether in a residence, commercial or industrial application. This would also include "Bonding" of any ground rods, water pipes ( if metal ) and natural gas lines all bonded to the Service neutral conductor. SO A GROUND WIRE AND EVERYTHING CONNECTED TO IT ( neutral wire, pipes ) IS NOT A TRUE GROUND and anything connected to it can be subject to unwanted frequency's or harmonics induced into the grounding system through the neutral conductor.

How To Fix -

A separate "Ground" Earth or "Terra firma" connection must be used generally called a "Isolated Grounding System" using one 8'-0" ground rod a minimum of 6 (six) feet from any other ground rod system or underground water and gas lines. It must be totally isolated using a minimum of a #8 AWG insulated green conductor and must not be connected in any way to you existing grounding system ..... any questions you should contact a qualified person (disclaimer).

Testing equipment causing harmonics can be somewhat isolated during operation using a "UPS" power battery back up supply ( AC to DC to AC ) but that does not solve the ground reference problem.

An electronic circuit to free energy

Quote:
Not need an oscilloscope to demonstrate on the unit: it is enough to measure currents ex 2: between the earth and the resistance R2, and between the land and the resistance R1 "
Best Regards,
Glen
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