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Author Topic: Is joule thief circuit gets overunity?  (Read 600309 times)

plengo

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Re: Is joule thief circuit gets overunity?
« Reply #30 on: November 23, 2012, 06:03:42 PM »
Fausto:

Not extra energy.  When you first put voltage across an inductor no current flows.  Electrical work has to be done to get the current to flow through an inductor (voltage x current x time).  The current starts from zero amperes and slowly climbs.  Once the current is flowing and you put a high resistance load across the inductor then you get a back-EMF spike.  The energy in the back-EMF spike is identical to the work (voltage x current x time) that was expended by the battery to get the current flowing in the first place.

This is in contrast to a resistor where the current starts to flow the instant you put voltage across the resistor.  There is no "extra work" required to get the current to start flowing through the resistor.  Likewise, there is no back-EMF spike from a resistor.

MileHigh


so if I had a resistor and the flow is constant I get heat in exchange for the voltage x current x time I got that. But the inductor ALSO has the same resistance so it has to also create the same heat as the equivalent resistor and on top of that returns a BEMF for the same input time of power/energy.


What I see here is free energy as BEMF, it is not the same 100% input of energy but it is there, while the resistor does not have that same effect (at least in the same level).


How can this not make you think differently? An Inductor has the ability of providing what the resistor is not. This to me is free energy. The work done by an equivalent resister is the SAME as the work done by an inductor concerning the input power/energy.


Btw, on the BEMF we also have more resistive heat loss on the indcutor which goes in line with another experiment the shows more heat is generated over a resistor when pulsed many times.


The inductor is doing twice the work that a resistor does with the same input of power/energy.


Fausto.




MileHigh

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Re: Is joule thief circuit gets overunity?
« Reply #31 on: November 24, 2012, 04:14:43 AM »
Fausto:

You are not getting it.  What you are doing is showing your wishful thinking about inductors.  The truth is that every electronic circuit that you see discussed here is under unity.  I am not trying to stop anybody from experimenting with electronics.  You notice that when people get excited about an electronic circuit and think it might be over unity, they never attempt to explain where the over unity is produced.  What is implicit in all of this is that there is a belief system that says that coils "just might" get extra energy from the "environment."  Sterling Allen states that all the time and in a sense you are stating it now.

Let's review the concept of inductors using energy and power as the frame of reference.

Let's assume that the inductor has a resistance of 10 ohms.

Now, using a simple way of describing it and omitting some details for clarity:

1)  You apply voltage to the inductor and it takes some time for the current to reach the full current flow.   During this "start up" time the battery expends 10 Joules of energy to get the current flowing.  That means that 10 Joules of energy are stored in the magnetic field.

2)  While the inductor has the full current flow let's say it is dissipating 5 Watts of power continuously.  Of course we know that power is coming from the battery.

3)  When you stop supplying battery power and you put a high resistance across the terminals of the inductor you get a high-voltage back-EMF spike discharge of energy.  The amount of energy in the back-EMF spike is 10 Joules.

Do you get it?  The battery expends 10 Joules of energy to get the current flowing.  The 10 Joules of energy are stored in the magnetic field.  Then when the inductor discharges it discharges 10 Joules of energy.

In other words:  + 10 Joules - 10 Jules = ZERO.

i.e.; The energy in the back-EMF spike from a coil comes from the battery, and nowhere else.  Nor is it "radiant energy," nor is it "cold electricity."

The inductor is a temporary energy storage device.  It temporarily stores energy that was supplied by the battery and discharges it at a later time.  There is no such thing as "double usage" of energy for an inductor.

That is the reality Fausto.

MileHigh

ltseung888

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Re: Is joule thief circuit gets overunity?
« Reply #32 on: November 24, 2012, 07:36:12 AM »
Happy Thanksgiving to you all.  I think that we can settle the question of whether a JT can be overunity once for all.  Please refer to Nov 22a.zip in the following:
http://www.overunityresearch.com/index.php?topic=1516.msg26471#msg26471

If we believe in the formula:
Instantaneous Power = Instantaneous Voltage x Instantaneous Current
and that the Instantaneous Voltage value can be obtained from the DC Coupling of the Oscilloscope, then all we need to do is to examine the oscilloscope results.
 
The oscilloscope result for case 4 was obtained with the battery removed.  The capacitor was connected for some minutes.  The Output Voltage frequency increased from 1.4KHz when the battery was disconnected.  As the both the Input and Output Voltage dropped, the frequency increased.  At this state, the Output Power was negative and the numerical value of the COP was greater than 1.  This can be reproduced from the Lead-out Energy Research kit components from BSI.
 
God Bless.

plengo

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Re: Is joule thief circuit gets overunity?
« Reply #33 on: November 24, 2012, 05:51:26 PM »
Fausto:

You are not getting it.  What you are doing is showing your wishful thinking about inductors.  The truth is that every electronic circuit that you see discussed here is under unity.  I am not trying to stop anybody from experimenting with electronics.  You notice that when people get excited about an electronic circuit and think it might be over unity, they never attempt to explain where the over unity is produced.  What is implicit in all of this is that there is a belief system that says that coils "just might" get extra energy from the "environment."  Sterling Allen states that all the time and in a sense you are stating it now.

Let's review the concept of inductors using energy and power as the frame of reference.

Let's assume that the inductor has a resistance of 10 ohms.

Now, using a simple way of describing it and omitting some details for clarity:

1)  You apply voltage to the inductor and it takes some time for the current to reach the full current flow.   During this "start up" time the battery expends 10 Joules of energy to get the current flowing.  That means that 10 Joules of energy are stored in the magnetic field.

2)  While the inductor has the full current flow let's say it is dissipating 5 Watts of power continuously.  Of course we know that power is coming from the battery.

3)  When you stop supplying battery power and you put a high resistance across the terminals of the inductor you get a high-voltage back-EMF spike discharge of energy.  The amount of energy in the back-EMF spike is 10 Joules.

Do you get it?  The battery expends 10 Joules of energy to get the current flowing.  The 10 Joules of energy are stored in the magnetic field.  Then when the inductor discharges it discharges 10 Joules of energy.

In other words:  + 10 Joules - 10 Jules = ZERO.

i.e.; The energy in the back-EMF spike from a coil comes from the battery, and nowhere else.  Nor is it "radiant energy," nor is it "cold electricity."

The inductor is a temporary energy storage device.  It temporarily stores energy that was supplied by the battery and discharges it at a later time.  There is no such thing as "double usage" of energy for an inductor.

That is the reality Fausto.

MileHigh


MileHigh,


thank you again for the explanation and this time you said it yourself exactly what I said but you are still not seeing.


Let go slow: you said: energy goes from the battery to the inductor. Good. Inductor discharges that stored energy back to us (or another system, whatever). Good, all the same, what goes in, goes out.


Is that correct? Good, now, think again, battery give 10 watts of power in a second, so 10 joules. Great. Inductor discharges and gives back (minus losses) 10 joules or less back.


Now, this process heats up the resistive wire TWICE, not once like a regular resistor. Can you explain that? How come a resistor for the same 10 joules of input power ONLY heat once ? in others words a regular resistor will give you only an equivalent amount of 10 joules of heat, while the SAME resistance in a inductor will give you more than that in heat.


I would love to hear how you going to explain that out!


Wishful thinking is YOURs in thinking that you can deny that simple fact that power has nothing to do with the amount of magnetic energy stored or produced by an inductor.


Fausto.

Neo-X

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Re: Is joule thief circuit gets overunity?
« Reply #34 on: November 24, 2012, 07:02:05 PM »
Hmmm... I can imagine how a transformer gets overunity but in inductor im still double thinking. If the inductor really has more power in bemf or when the magnetic field is collapsing, then it will self oscillate when a capacitor is added because the extra energy from the bemf sustain the oscillation.

plengo

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Re: Is joule thief circuit gets overunity?
« Reply #35 on: November 24, 2012, 09:58:32 PM »
Hmmm... I can imagine how a transformer gets overunity but in inductor im still double thinking. If the inductor really has more power in bemf or when the magnetic field is collapsing, then it will self oscillate when a capacitor is added because the extra energy from the bemf sustain the oscillation.


I am not saying that BEMF is more power back than we put it in. BEMF will be much less.


BTW, it is funny how IGNORANT people are to overunity. Look at this video from Newman (he has been doing this for decades) http://www.youtube.com/watch?v=RMYo1QlvK5g&list=PL4E7C4E1A7713B3D8&index=1&feature=plpp_video, it is very clear that his machine 2000 pounds is running in voltage only.


I want to see how any one can explain this to not be overunity. 2000 pounds doing work, pumping water is NOT fake and it is not running on current, since ALL the batteries are in series.


If this is not the power of induction at work, I don't know what is power than.


Fausto.

plengo

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Re: Is joule thief circuit gets overunity?
« Reply #36 on: November 24, 2012, 11:07:12 PM »
and to show more work: http://youtu.be/N1Fj4r-we84


This video is a trial replication of Lasersaber Joule Ringer Cross Over. I did not replicate it 100% BUT I did replicate the input energy growing effect. I had to burn about 30 or so transistors.


Very difficult this circuit. Extremely sensitive.


Fausto.

ltseung888

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Re: Is joule thief circuit gets overunity?
« Reply #37 on: November 24, 2012, 11:36:51 PM »
Happy Thanksgiving to you all.  I think that we can settle the question of whether a JT can be overunity once for all.  Please refer to Nov 22a.zip in the following:
http://www.overunityresearch.com/index.php?topic=1516.msg26471#msg26471

If we believe in the formula:
Instantaneous Power = Instantaneous Voltage x Instantaneous Current
and that the Instantaneous Voltage value can be obtained from the DC Coupling of the Oscilloscope, then all we need to do is to examine the oscilloscope results.
 
The oscilloscope result for case 4 was obtained with the battery removed.  The capacitor was connected for some minutes.  The Output Voltage frequency increased from 1.4KHz when the battery was disconnected.  As the both the Input and Output Voltage dropped, the frequency increased.  At this state, the Output Power was negative and the numerical value of the COP was greater than 1.  This can be reproduced from the Lead-out Energy Research kit components from BSI.
 
God Bless.
A Picture is worth a thousand words.  Four pictures may be worth 4 thousand words.  Refer to the above link and the nov 22a.xlsx file for full details.
All Glory and Honor to the Almighty.  I am just the humble Server of the Divine Wine.

TinselKoala

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Re: Is joule thief circuit gets overunity?
« Reply #38 on: November 25, 2012, 02:05:50 AM »
@LTseung: I'm glad you are taking me seriously. Thanks for posting the scopeshots.
Yes, you are right that instantaneous power = instantaneous voltage x instantaneous current, at each _instant_.  This automatically takes into account phase angles, complex waveforms and the lot. However there are other considerations.

First, and perhaps most importantly, Power is NOT energy. Power is not necessarily "conserved" because you can take lots of small chunks of power and concentrate them into a bigger chunk. ENERGY is conserved. To sustain a claim of overunity performance you must show that energy out > energy in.
How do you find the energy? By integrating the instantaneous power curve over an appropriate time period.
So your COP claims based on power are still incomplete.

Second..... if you load samples from your DSO into your spreadsheet and have the spreadsheet integrate the values making up the instantaneous power curve, you need to make sure you aren't "looking at your data thru a picket fence".  That is, all DSOs are _sampling_ instruments, they do not read continuous values from your signal. They sample it at discrete intervals. Digital Sampling Oscilloscope.  Many DSOs... high end ones....have very large sample rates like 1 Gs/sec. Clearly..... if you are sampling a waveform at 1 Gs/sec..... you are going to have a +lot+ of samples per cycle. A lot of data for a spreadsheet CSV file. On the other hand if your scope is a basic one, it probably has a much slower actual sample rate, and may even be interpolating between sample intervals rather than giving you actual measurements.
This means that using a spreadsheet to integrate a complex signal with fast risetimes, based on data dumps from a low-end digital scope is a process that is.... er.... um...... "fraught" as someone once said. Fraught with difficulty and opportunities for error, that is.
It is better, in general, to use an integrating power meter like the Clarke-Hess power analyzers, or in especially problematic cases, even bolometric power analyzers, or to beg or borrow a high-end scope with huge sample rates and built in integration math, and have the scope perform the necessary integrations on-board.

Finally (for this post anyway) you must use the _input power used to charge the capacitor_ as your "input" for your COP calculations on capacitor-powered devices. This is actually the benefit of using caps: the input power can be much more precisely known than otherwise by measuring the charging voltage and current and, again, doing the inst. mult. and integration. The _energy_ that is on the cap at any given time can be known simply by computing E in Joules = (CxVxV)/2 where C is the capacitance in Farads and V is the measured voltage in volts.
So you charge a cap to a certain voltage. You know the energy it took to charge the cap.... and you know the energy ON the cap. Then you compare these to your output _energy_ value over a suitable period of time.

TinselKoala

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Re: Is joule thief circuit gets overunity?
« Reply #39 on: November 25, 2012, 03:01:06 AM »
and to show more work: http://youtu.be/N1Fj4r-we84


This video is a trial replication of Lasersaber Joule Ringer Cross Over. I did not replicate it 100% BUT I did replicate the input energy growing effect. I had to burn about 30 or so transistors.


Very difficult this circuit. Extremely sensitive.


Fausto.

Here's my first complete demo:
http://www.youtube.com/watch?v=Ro36zIieP3w

And another one coming in an hour or so.

But I've only burned out two transistors.....I must be doing something wrong.....   :(

 ;)
« Last Edit: November 25, 2012, 06:13:50 AM by TinselKoala »

ltseung888

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Re: Is joule thief circuit gets overunity?
« Reply #40 on: November 25, 2012, 05:59:54 AM »
@LTseung: I'm glad you are taking me seriously. Thanks for posting the scopeshots.
Yes, you are right that instantaneous power = instantaneous voltage x instantaneous current, at each _instant_.  This automatically takes into account phase angles, complex waveforms and the lot. However there are other considerations.

We shall send at least 30 oscilloscope-test-ready boards to the top universities of the World.  They have the very high end 4 channel scopes to do very thorough tests.  That was the objection 3 years ago when I first took the simple FLEET to the Hong Kong University.  The immediate reaction was - your China made Oscilloscope may not give the right result.

*** Another major objections was that with a 2 channel oscilloscope, one cannot measure the Input and Output waveforms simultaneously.  In the particular experiment, I took pictures of the Output and then the Input.  Thus there were no guarantee that the LHS corresponds to the RHS exactly. ***
 
I believe PhysicsProf also met with similar comments.  He had to drive many miles to BYU to use the high end scope.
 
Both G-LED and BSI have produced Forever Lighted Lamps based on the resonance tuning of the Atten Oscilloscope.  From the many repeated peaks, I believe the oscilloscope gives reasonable answer.  I do not need exact values to confirm the lead-out energy theory.  Let the Universities and their Professors shine.  Let them come out with the top scientific papers.
 
Let the Almighty bring out the top Wine Servers.

ltseung888

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Re: Is joule thief circuit gets overunity?
« Reply #41 on: November 25, 2012, 06:17:22 AM »

First, and perhaps most importantly, Power is NOT energy. Power is not necessarily "conserved" because you can take lots of small chunks of power and concentrate them into a bigger chunk. ENERGY is conserved. To sustain a claim of overunity performance you must show that energy out > energy in.
How do you find the energy? By integrating the instantaneous power curve over an appropriate time period.
So your COP claims based on power are still incomplete.

The COP claims are based on Average Output Power over Average Input Power with 11250 sample points. 

TinselKoala

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Re: Is joule thief circuit gets overunity?
« Reply #42 on: November 25, 2012, 06:27:33 AM »
Lawrence, you do not absolutely need high end digital oscilloscopes to do accurate power measurements! They are nice, can save a lot of time,  but not required. What do you think people did before there +were+ DSOs? The folks that designed the first DSOs "only" had analog scopes to work with, and they made accurate power measurements of complex waveforms just fine.

I want to see if I can check your present  computations, if I may. Can you please tell me the following little bits of data, for _one_ JT that you think is OU.... (and let's continue to speak only of that _one_ that you select, please.)

What is the oscillation frequency and the approximate duty cycle (the ratio of the "on" output time to the total period of a full cycle)?
What is the sample rate setting of your DSO when making your measurements?
How large (how many samples) is your CSV file for a set of _four_ complete cycles in the waveform?
May I examine your present spreadsheet computations (not just results, I'd like to see your formulae and procedures if I may.)

Now, I am mostly concerned with the output waveforms (voltage, and current represented by voltage drop thru a monitoring resistor)  here. We will consider how you are measuring the input power a bit later on. You may not even need a scope at all for that.



TinselKoala

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Re: Is joule thief circuit gets overunity?
« Reply #43 on: November 25, 2012, 06:34:55 AM »
The COP claims are based on Average Output Power over Average Input Power with 11250 sample points.
And that is where you likely are going wrong. Please see my post above. I think you may be undersampling your waveform on the output, and I think you might not be doing your input measurements correctly. But without closer examination I can't really tell.


Still, for a quick check, you could always daisy-chain your 30 boards together into a real fleet of FLEETS.... and have them run themselves in a big Ring of Power. If they are even the slightest bit OU.... by the time you turn on the 30th one you will be in Overunity Heaven.
With all those OU Wine Servers, no doubt.
 ;D

ltseung888

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Re: Is joule thief circuit gets overunity?
« Reply #44 on: November 25, 2012, 10:13:08 AM »

We shall send at least 30 oscilloscope-test-ready boards to the top universities of the World.  They have the very high end 4 channel scopes to do very thorough tests. 


The Top Universities include TsingHua University (the MIT of China), MIT, Columbia, Stanford Universities of USA, Imperial College of England etc.  They will receive the Lead-out Energy Research Kit and an oscilloscope-test-ready board.
 
All the Lead-out Energy Research Kits from BSI will have enough components to build an oscilloscope-test-ready board.  BSI and G-LED will be meeting early December to decide on the final details.
 
Initially, I intended to wait until Oct 2013 before providing the full tuning information.  One event at the hospital changed my mind.  Someone used the "survival of the fittest" statement.  That statement is a thorn in the heart for me.  That was what the Colonialists used to justify murdering other races.
 
The gun and the atomic bomb were paradigm shift technologies.  The Lead-out Energy technology will be a similar paradigm shift technology.  No one group or nation is allowed to own and control this technology exclusively.  The Divine Wine is for all humans.
 
I do not care whether someone believes or does not believe my posts.  The analogy was the thousands of objections before the Wright Brothers demonstrated their plane.  The moment the 30 top Universities publish their reports; when any one can purchase the Lead-out Energy Research Kit; when the Forever Lighted Lamp appears on TV etc., it will be reality.
 
May the Almighty guide us to benefit the World.  Amen.