Is joule thief circuit gets overunity?

[TinselKoala]

http://answers.yahoo.com/question/index?qid=20110630015105AAuZR0y
Can power factor of an electrical system be measured as a negative value?
A negative power factor would indicate that current leads or lags the voltage by more than 90 degrees. Physically, that can happen if the load is returning power to the source rather than using power. For example, if an induction motor is used to operate an electric railway locomotive, the power factor will be lagging by less than 90 degrees as the locomotive climbs a hill. If the motor is nearly loaded, the power factor might be about 0.85. If the track is level at the top of the hill, the motor's load will be reduced and the power factor might drop to 0.65. As the locomotive begin to go downhill, the power factor drops more and may be come negative as the motor begins to operate as a generator returning braking power to the source.
Source(s):
I have worked as an engineer in the manufacture of variable frequency drives (VFD). We tested drives under load by using the VFD under test to control an induction motor which drove a second induction motor that was connected to utility (mains) power. Using the VFD to slightly overspeed the second motor forced that motor to become a generator and return power to the utility.

Another Answer:

The power factor itself, will be positive, between 0 and 1, and is given from cos (theta).

If it is 1, your load is purely resistive.
If it is 0, your load is purely reactive.

The angle theta, however, could be negative or positive.

If it's negative, then the reactive portion is capacitive, and your voltage is lagging behind the current.

If it's positive, then the reactive portion is inductive, and voltage is leading current.

Lawrence this is garbage. It is true, but it does not apply to your instantaneous power computations. POWER FACTOR is not POWER. Your instantaneous power curves that go negative at the end have nothing to do with POWER FACTOR or returning power to the source.

When you do instantaneous multiplication of current and voltage, as you are doing, your POWER FACTOR is automatically compensated for. It is only when you are dealing with average values of regular sine waves of current and voltage, phase shifted because of reactance in the load, that you need to consider power factor and do all of that trig algebra.

http://www.tina.com/English/tina/course/29power/power

  There are several different definitions of power in AC circuits; all, however, have dimension of V*A or W (watts).
1.      Instantaneous power: [/i] p(t) is the time function of the power,  p(t) = u(t)*i(t). It is the product of the time functions of the voltage and current. This definition of instantaneous power is valid for signals of any waveform. The unit for instantaneous power [/i]is VA.
2.      Complex power: [/i] S [/font]

   (http://www.tina.com/English/tina/course/29power/power.1.gif)

Complex power is the product of the complex effective voltage and the complex effective conjugate current. In our notation here, the conjugate is indicated by an asterisk (*).Complex power can also be computed using the peak values of the complex voltage and current, but then the result must be divided by 2. Note that complex power is applicable only to circuits with sinusoidal excitation because complex effective or peak values exist and are defined only for sinusoidal signals. The unit for complex power [/i]is VA. [/font]3.      Real [/i]or average power[/i]:  P can be defined in two ways: as the real part of the complex power or as the simple average of the instantaneous power.[/i] The  second definition is more general because with it we can define the instantaneous power [/i]for any signal waveform, not just for sinusoids. It is given explicitly in the following expression

   (http://www.tina.com/English/tina/course/29power/power.2.gif)

The unit for real[/i] or average power[/i]  is watts (W), just as for power in DC circuits. Real power is dissipated as heat in resistances.
4.      Reactive power:  [/i]Q is the imaginary part of the complex power. It is given in units of volt-amperes reactive[/i] (VAR). Reactive power is positive[/i] in an inductive[/i] circuit[/i] and negative[/i] in a capacitive circuit[/i]. This power is defined only for sinusoidal excitation. The reactive power doesn't do any useful work or heat and it is the power returned to the source by the reactive components (inductors, capacitors) of the circuit [/font]
  5.      Apparent power: [/i]S is the product of the rms values of the voltage and the current, S = U*I. The unit of apparent power is VA. The apparent power[/i] is the absolute value of the complex power[/i], so it is defined only for sinusoidal excitation.[/font]

Well, the formatting and the formulae got screwed up in the quote. Please go to the link and read it there. I have highlighted the part you are misunderstanding. Your discussion of Power Factor is irrelevant because you are obtaining your instantaneous power curve by multiplying the instantaneous values of voltage and current.

[ltseung888]

Mark Dansie and team visited G-LED USA and were impressed.  They saw the FLEET without battery lighting during the entire meeting.  They also saw the 2KW unit lighting 2,500 LEDs.

Their plan is to fly to Hong Kong to meet me and the owners of BSI.  They wanted to purchase a number of Lead-out Energy Research kits for testing now and may go into business arrangement afterwards.

Apparently, they have much better equipped laboratories and can do the 4 channel, maths function tests.  My indication experiments will be improved to a much higher degree of accuracy.

The owners of G-LED and BSI will also meet shortly to decide on the marketing strategy.  Hopefully, that will happen before Christmas.

The Divine Wine will be tasted by many this Christmas.

[TinselKoala]

I said,
I am astounded !! Lawrence.... that is even worse! You CANNOT take output data from one twenty minute run and compare it "minute by minute"  with input data from another, different twenty minute run!!! Who told you that was legitimate?
I am flabbergasted that you would even consider doing such a thing. It's outrageous. It invalidates ALL of your comparison data.
LTseung replied:

*** As I mentioned, I do indication experiments.  The top universities will have the DSO and experts to do the real experiments.  That is why we need them.

You are getting them all excited over your claims, which are based on your invalid data from your "indication experiments". You are crying wolf when there is no wolf. The more times people do this, the less credibility we all have. You cannot do "indication experiments" INCORRECTLY and then draw conclusions and make claims of overunity! It give everybody a bad name.

I asked,
How are you "discharging thoroughly" your capacitor? Please give me the exact procedure you use for this, including times.

And LTseung answered,

*** Use a wire to directly connect the two ends of the capacitor for 5 minutes.  Put it back on the JT circuit and the LED should show no light.  Check the Output voltage waveform on the Atten Oscilloscope.  It should show NO more characteristic pulsing waveform.

That might barely be acceptable, but I should think that a much longer time "shorted" by the jumper would be necessary for a low-voltage supercap. I only have 3 Farads, but I'll do some experiments and see what I can learn.
However, the issue is moot because you simply cannot use output and input data from different runs for comparison purposes. You do not have control of all the conditions that could affect your measurements. To be able to make the claim that input power was less than output power in a circuit with constantly varying parameters,  you have to measure them at the same time! You might be able to get somewhere by measuring input and output voltage on one run, then measuring input and output current on another run, and then computing your power curves from data taken that way. At least that way you will have simultaneous measurements of _something_ on your input and output. But I still doubt the validity of comparisons made this way.
You don't need to even use the scope for your input measurements, though. Simply monitor the input current and voltage with DMMs and the time with a precise timer. Your graphs reveal that average values will accurately reflect your input power function.

But you must not conclude overunity performance based on your present data gathering methodology. You are comparing apples grown in one season, with apples grown in the next season, and you have no idea of the variables that changed over the winter.

[TinselKoala]

Mark Dansie and team visited G-LED USA and were impressed.  They saw the FLEET without battery lighting during the entire meeting.  They also saw the 2KW unit lighting 2,500 LEDs.

Their plan is to fly to Hong Kong to meet me and the owners of BSI.  They wanted to purchase a number of Lead-out Energy Research kits for testing now and may go into business arrangement afterwards.

Apparently, they have much better equipped laboratories and can do the 4 channel, maths function tests.  My indication experiments will be improved to a much higher degree of accuracy.

The owners of G-LED and BSI will also meet shortly to decide on the marketing strategy.  Hopefully, that will happen before Christmas.

The Divine Wine will be tasted by many this Christmas.

Lawrence, I have circuits sitting here right in front of me, running right now,  that FAR outperform the circuits you are describing here. The reason I am not being wined and dined by Divine Wine and being showered with money and accolades is because I AM NOT MAKING CLAIMS THAT CANNOT BE SUPPORTED BY ACTUAL DATA.

[ltseung888]

Dear All,

I am facing the following problem and need a reasonable solution:
How to use ONE oscilloscope to compare the Input Power and Output Power when both are decreasing with time?
The Atten Oscilloscope has 2 channels.  We can use Ch1 to measure Instantaneous Voltage and Ch2 to measure Instantaneous Current.  Such data can be captured as CSV files to be analysed by EXCEL.  The important equation is:
Instantaneous Power = Instantaneous Voltage x Instantaneous Current
We can use the Atten Oscilloscope to measure Input waveforms and output waveforms separately but not simultaneously.

Changing the connections from Output measurement to Input measurement will take approximately 1 minute.

At present, I try to produce the same initial condition – same procedure in turning the DC Power Supply ON with the capacitor totally discharged (as checked on the oscilloscope).  Then take reading at 1 minute intervals.  Use two separate runs - one for Input and one for Output.

Is there a better method assuming that you only have ONE 2 channel oscilloscope?