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Author Topic: Is joule thief circuit gets overunity?  (Read 529524 times)

Offline picowatt

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Re: Is joule thief circuit gets overunity?
« Reply #570 on: April 13, 2013, 02:31:40 AM »
@PW,
 
You probably missed it with the hugh number of posts.  See reply 552.  It contains the full DSO analysis for Board 80.  The first sheet is the Output.  The second sheet is the Input.  The various wavforms such as Voltage, Current and Power are also displayed.  The saved BMP files are included for comparison.  The COP was calculated to be 1.41 at the end of first sheet.
 
@All,
 
I also tried to get CH2 Vavg to 0V by adjusting the DC Power Supply.  The readings were fluctating and I could not get a perfect steady reading.  However, when CH1 Vrms was at 680mV, CH2 Vavg fluctuated between 4mV and -2mV.

Lawrence,

I have looked at the DSO Analysis.xls attached to post 552 several times.  It appears to be only the output data calculated for average output power, but no where do I see the input power data or calculation.

It should not be this difficult to communicate.  A while back you posted Board 80's screen captures as slide 12 and slide 13 in your post 503.  You said those captures indicated OU.  I have asked several times for the input and output power calculations from the data captured during that same test and represented by slides 12 and 13.  The calculations in post 552 appear to be from slide 13, but no where have I seen the slide 12 input power calculations. 

Actually, to date, I have not been able to find any posts wherein you provide input AND output power calculations from input and output data collected at the SAME time during any given test you have presented.  With some tests you post your output calculations, with others you post input calculations.

PW 




Offline Pirate88179

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Re: Is joule thief circuit gets overunity?
« Reply #571 on: April 13, 2013, 02:49:32 AM »
Lawrence,

Although your Board 80 input current trace does show some excursion above the zero ref line, overall, there is much more area under the zero ref line.  Everything under the zero ref line is current being drawn from the battery.  Looking at that board's input and output power captures demonstrates an efficiency of about 75 to 85%.

TK's reply 522 definitely looked like it was OU by a significant amount.  The apparent equal area above and below the zero ref line on the input current trace made me wonder, however, if he had inadvertantly switched to AC coupling.  Although the jury may still be out on that completely, it is the most likely explanation for that anamalous capture.

PW

I still stand with my original prediction made  March 31, 2013, 02:53:13 PM,  partially quoted below:

" I want to make a prediction here.  I believe that .99's tests will show somewhere between 60 and 69% efficiency.  Not bad for a simple JT circuit but no where near 100%, much less 200%+ as has been discussed."

Those toroids are not the best ones to use and he is losing a lot of energy to those resistors as heat.  No way is it COP  of 1,221.  We will see if I am close to being correct.

Bill

Offline picowatt

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Re: Is joule thief circuit gets overunity?
« Reply #572 on: April 13, 2013, 02:58:31 AM »
I still stand with my original prediction made  March 31, 2013, 02:53:13 PM,  partially quoted below:

" I want to make a prediction here.  I believe that .99's tests will show somewhere between 60 and 69% efficiency.  Not bad for a simple JT circuit but no where near 100%, much less 200%+ as has been discussed."

Those toroids are not the best ones to use and he is losing a lot of energy to those resistors as heat.  No way is it COP  of 1,221.  We will see if I am close to being correct.

Bill

Bill,

My "eyeball" guesstimate of efficiency for board 80 in post #503 slide 12 and 13 is tad bit better than your estimate, but not a COP of 1,221...

PW

 

Offline Pirate88179

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Re: Is joule thief circuit gets overunity?
« Reply #573 on: April 13, 2013, 04:38:51 AM »
Bill,

My "eyeball" guesstimate of efficiency for board 80 in post #503 slide 12 and 13 is tad bit better than your estimate, but not a COP of 1,221...

PW

Stefan should start a contest to see who gets the closest, ha ha.  We could all agree to go by .99's numbers when they come in.  I guess we will just have to wait and see.  I thought I was guessing on the high side but...oh well...we will see.

Bill

Offline Groundloop

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Re: Is joule thief circuit gets overunity?
« Reply #574 on: April 13, 2013, 06:00:02 AM »
The oscilloscope analysis for Board 107.

 
 Average Output Power
12.05105778 watt
 Average Input Power
0.009865636 watt
 
COP = 1221.52
 
This Board 107 will be carefully studied and preserved.
 

ltseung888,

Please post a photo of this board.

GL.

Offline ltseung888

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Re: Is joule thief circuit gets overunity?
« Reply #575 on: April 13, 2013, 06:00:59 AM »
Lawrence,

I have looked at the DSO Analysis.xls attached to post 552 several times.  It appears to be only the output data calculated for average output power, but no where do I see the input power data or calculation.

It should not be this difficult to communicate.  A while back you posted Board 80's screen captures as slide 12 and slide 13 in your post 503.  You said those captures indicated OU.  I have asked several times for the input and output power calculations from the data captured during that same test and represented by slides 12 and 13.  The calculations in post 552 appear to be from slide 13, but no where have I seen the slide 12 input power calculations. 

Actually, to date, I have not been able to find any posts wherein you provide input AND output power calculations from input and output data collected at the SAME time during any given test you have presented.  With some tests you post your output calculations, with others you post input calculations.

PW
@PW
 
I am not sure what you were looking at.  I attach the top and bottom of both Input and Output for you to check.  The same file is again attached.

Offline ltseung888

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Re: Is joule thief circuit gets overunity?
« Reply #576 on: April 13, 2013, 06:23:45 AM »
ltseung888,

Please post a photo of this board.

GL.
@GL
 
Here it is.

Offline Groundloop

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Re: Is joule thief circuit gets overunity?
« Reply #577 on: April 13, 2013, 06:57:26 AM »
@GL
 
Here it is.

ltseung888,

Thank you for posting the photo of the board number 107.

Ouote
  The oscilloscope analysis for Board 107.
   Average Output Power
  12.05105778 watt
   Average Input Power
  0.009865636 watt
  COP = 1221.52
End Quote

Based on common sense, do you really think your measurement of 12 Watt output is correct for this board?

You did made a output measurement over a 1 Ohm resistor in series with that small 0,1 Watt LED. Do you think
that a small LED like that will survive 12 Watt through it?

GL.

Offline picowatt

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Re: Is joule thief circuit gets overunity?
« Reply #578 on: April 13, 2013, 08:05:27 AM »
@PW
 
I am not sure what you were looking at.  I attach the top and bottom of both Input and Output for you to check.  The same file is again attached.


Lawrence,

I am concerned that the data presented contain different record lengths for the input and output data.  Both scopes should be set to sample identically when performing your tests.

As a calibration/verification test of your scope settings, consider connecting both scope one and two to measure only the input power of a board.  That is, CH1 of both scope 1 and 2 connected to the input V (battery +) and CH2 of both scope 1 and 2 connected to the input current (batt- and CSR junction).

Perform captures as you usully do and then calculate the average power.  See if both scopes give you the same average power result.

Alternately, try performing an input/output power measurement as you normally do using scope 1 for input power and then repeat the test with the scopes swapped so that scope 2 is now measuring input power.  The calculated results from the two tests should be similar (providing you can swap the scopes and perform the second test before the circuit's operating parameters change too much).

PW

Offline ltseung888

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Re: Is joule thief circuit gets overunity?
« Reply #579 on: April 13, 2013, 08:43:29 AM »
ltseung888,

Thank you for posting the photo of the board number 107.

Ouote
  The oscilloscope analysis for Board 107.
   Average Output Power
  12.05105778 watt
   Average Input Power
  0.009865636 watt
  COP = 1221.52
End Quote

Based on common sense, do you really think your measurement of 12 Watt output is correct for this board?

You did made a output measurement over a 1 Ohm resistor in series with that small 0,1 Watt LED. Do you think
that a small LED like that will survive 12 Watt through it?

GL.
@GL,
 
That is why I put it aside.  Much more research is needed.  The present work is focused on the normal boards such as Board 80.

Offline ltseung888

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Re: Is joule thief circuit gets overunity?
« Reply #580 on: April 13, 2013, 08:47:19 AM »

Lawrence,

I am concerned that the data presented contain different record lengths for the input and output data.  Both scopes should be set to sample identically when performing your tests.

As a calibration/verification test of your scope settings, consider connecting both scope one and two to measure only the input power of a board.  That is, CH1 of both scope 1 and 2 connected to the input V (battery +) and CH2 of both scope 1 and 2 connected to the input current (batt- and CSR junction).

Perform captures as you usully do and then calculate the average power.  See if both scopes give you the same average power result.

Alternately, try performing an input/output power measurement as you normally do using scope 1 for input power and then repeat the test with the scopes swapped so that scope 2 is now measuring input power.  The calculated results from the two tests should be similar (providing you can swap the scopes and perform the second test before the circuit's operating parameters change too much).

PW
@PW:
 
Thank you for your suggestions.  There is much work on Board as suggested by a Hong Kong Government Consultant.  His suggestions take higher priority.

Offline ltseung888

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Re: Is joule thief circuit gets overunity?
« Reply #581 on: April 14, 2013, 12:09:36 AM »
 
I have done the systematic test on Board 80.  The test started with the DC Power Supply set to as low as possible but enough to light the LED (dimly).  The Input Current CH2 Vavg was almost all positive (all values were on the positive side of the zero axis).  I then increased the Input DC Power Supply slowly.  Sure enough, the CH2 Vavg started to cross the zero axis to become negative.  At higher Input DC Power Supply(1.5V), most of the CH2 Vavg values became negative.
 
 
 
I did the full DSO analysis on five points first.  The full analysis files are in:
 
http://www.overunityresearch.com/index.php?topic=1516.msg30693#msg30693
 
 
 
I also managed to get CH2 Vavg = 0.  This occurred at the same displayed CH1 Vrms value of one of the above points (400mV).  That file is “DSO analysis6.xls” at the same above thread.
 
 
 
The summary is displayed here.  I managed to get COP values from -10.57 to 9.17 on the same Board 80 just by tuning with the DC Power Supply.  If Poynt99 has a DC Power Supply in addition to his 4-CH DSO, he can easily reproduce the above.  That will confirm that the Zhou Boards are Overunity with this particular oscilloscope analysis.  When such results are widely published, more resources will come to Lead-out Energy Research.
 
All the well known "free energy" researchers will get DSOs and DC Power Supplies.  They will be invited to special conferences.  (TK will get his platinum card!)  The floodgate is now open.
 
 
 
God Bless
 

Offline picowatt

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Re: Is joule thief circuit gets overunity?
« Reply #582 on: April 14, 2013, 01:35:02 AM »
Mainly .99, but everyone else too,

If you are still following along on this thread, please consider the following:

I believe I have found the measurement errors that cause Lawrence's JT's to measure OU.  I am not fully certain I have the formulae completely worked out, but at the very least this should be food for thought.  Possibly you can assisit in working them out a bit further.  I believe the error to be in the way the output power is calculated.

Although it is a bit spread out, there has now been sufficient data presented regarding Board 80 that an analysis can be made.

To better follow along, I would print, or at least open, the three files below:

Schematic from reply 469:    Two DSO config.jpg

Input screen capture from reply 503:    Slide12.jpg

Output screen capture from reply503:    Slide13.jpg


Also, the calculated input and output data is in reply 575, which shows an input power of 31mw and an output power of 44mw.  I have scanned the .xls files and the data and calculations appear correct as presented, and are representative of the data derived from slide 12 and 13.


Please look at the schematic, note the locations of A1-A4 and B1-B4 test points, and follow along.  When referring to the toroid secondary, it will be the winding connected to the Q1 collector.

Input voltage, Vbatt, is equal to (A1-A4) and input current is equal to (A4/1). 

Instantaneous input power calculation is Vbatt times the input current.  Using the test points as per the schematic, the instantaneous input power is Pin(inst)=(A1-A4)x(A4/1), and Pin(avg)=Pin(inst)1+Pin(inst)2+...Pin(inst)n/n, where the numbers 1 thru "n" denote contiguous numbers of data sample points.

These are the formulae used to calculate the input power for the Board 80 data as represented in slide 12 and its associated .xls file, and I believe these to be correct.


However, instantaneous output power is being calculated as simply Pout(inst)=(B1)x(B3/1), and this I believe is wrong.

Referring to Slide 13, the only time there is actual output power is when the output voltage, B1, is greater than Vbatt.  Any time B1 is less than Vbatt, current flow thru the LED and CSR string, and/or, current flow thru the toroid and Q1, represent current being drawn from the battery.  These currents are fully accounted for in the input power calculations, but are mistakingly being added back in during the Pout calculations when using Pout(inst)=(B1)x(B3/1) to calculate output power.

I believe the correct formula for calculating the output power is closer to the following:

When Q1 is off, Pout(inst)=(B1)x(B3/1)-(Vbatt)x(B3/1)

When Q1 is on, Pout(inst)=(B1)x((B3/1)-(LEDleakpwr), where the expressions (B1)x(B3/1) and (LEDleakpwr) are interchangeable and, therefore, when Q1 is on, Pout(inst)=0

Possibly this can be better stated by:  If B1>Vbatt, Pout(inst)=(B1)x(B3)-(Vbatt)x(B3) and if B1< or = Vbatt, Pout(inst)=0

The formulae need a bit more work, but for now, they are at least food for thought.

Please read on and come back to this after completing the read.

Looking at Slide 13, Q1 turns on at the onset of the sharp falling edge of the B1 waveform shown by CH1.  As Q1 turns on, the output voltage, B1, falls rapidly to the Vce of Q1, which, from the raw data, is 80mV.  During the time when Q1 is on, 12ma of the input current is flowing parallel to the Q1 current thru the LED and output CSR string.  During this time when Q1 is on, the 12ma of current flow thru the LED and output CSR string is current being drawn from the battey.  The simple Pout(inst)=(B1)x(B3/1) formula is incorrectly adding this current to the output power calculation as output power, when in reality it is input loading that is already accounted for by the input power calculations and represents no output current.

As well, looking again at slide 13, Q1 turns off at the sharp rising edge of the B1 waveform.  The voltage at this point in time rises rapidly.  As the output voltage rises, the LED turns on harder eventually clamping the output voltage at the LED forward voltage.  As well, there is an overshoot in the output voltage above the LED's V(fwd) due to the turn on delay of the LED.  This is the narrow spike that rises quickly and then falls quickly to the point where a more exponential fall begins.

Looking at the output current waveform B3, which is CH2 in slide 13, there is a rapid rise (spike) in output current, most likely due to the LED junction capacitance, and then, as the LED turns on, a quick drop to a point where a more exponential fall in current begins, closely matching the curve of the falling output voltage.

My premise is that at any time if Vout is at or below Vbatt, current is being drawn from the battery.  If at any time Vout exceeds Vbatt, the toroid is "discharging" and is acting as an additional battery in series with Vbatt.  LED forward current increases to beyond that which Vbatt can alone produce only when Vout exceeds Vbatt.  Any time Vout is equal to or less than Vbatt, Vbatt is actually supplying all power.   


As additional supporting arguments:

Looking at the schematic, consider Q1 as being turned on and in a DC state.  The B1 (output voltage) will be equal to the Vce of Q1, and from the data we see that to be 80mV.  Looking at the input raw data files, this 80mV causes a current to flow thru the LED and output CSR string equal to 12ma (this woud be the LED's forward leakage current at 80mv).  There is also current flowing thru the toroid secondary and thru Q1 via its collector and emitter.  As well, there is the base current flowing thru the toroid primary and Q1 base.

Now, if we wish to measure the current drawn from the battery under these DC conditions, we consult the input data and calculate Vbatt as (A1-A4)=Vbatt and the input current as (A4/1)=Iin.  The input power is then simply (Vbatt)x(Iin)=Pin.  These are the formulae used to make the instantaneous Pin calculations for the Board 80 data as depicted by slide 12 and appear to be correct.  Averaging the product of of the instantaneous Pin calculations using that formula produces an accurate calculation of Pin(avg).

However, let's consider that 12ma being consumed in the LED and output CSR leg during this same DC condition when Q1 is on.

As we can see from the schematic and collected data, when Q1 is on (Vce=80mV), there will be current flow thru Q1 and the torroid, and, as well, there will be a smaller portion of the input current being dissipated thru the LED/output CSR string (equal to 12ma).  None of this matters regarding the calcultion of input power, because the use of the data collected using A1 and A4 as above, will account for all power being drawn from the battery.

The output power calculations, however, are being calculated as simply (Vout)x(Iout)=Pout.  Under the above DC conditions wherein Q1 is on, Vout=B1=80mV and Iout=B3/1=12ma.  The output power would therefore be calculated as approximately 1mw.  It should be readily apparent that under these DC conditions, there is no "output power", at least in the sense of something beyond what is being supplied by the battery.  There is power being disspiated in the toroid and Q1 and as well, there is approximately 1mw flowing thru the LED and output CSR.  All of this power, however, under these conditions, is power being drawn from the battery.  Under these DC conditions, Pin=(Vbatt)x(A4/1) and the totality of that power is being dissipated in the windings of the toroid, Q1, and the LED/output CSR.  Using the eroneous formula Pout=(B1)x(B3/1) would indicate that there is 1mw of power being generated at the output, when in reality Pout=0, and this 1mw is merely a portion of the load on the battery.


Now, referring again to the schematic, consider the alternate DC condition wherein Q1 is off.  Under this static condition, the output voltage, B1, will, for the most part, be equal to the voltage at Vbatt+, which is point A1 on the schematic.  Because Q1 is off, only Q1 leakage and a bit of base current will be flowing thru Q1 and the 1K resistor.  As we know the LED allows 12ma of current to flow when B1 is 80mv, it is logical to assume that when B1 is equal to Vbatt, or approximately 1.5 volts, a bit more leakage current will flow thru the LED at this time.  However, 1.5 volts remains below the LED forward voltage, so it is doubtful the LED leakage is a large amount of current.  Under these DC conditions wherein Q1 remains off, if we measure Pout using the erroneous formula Pout=(B1)x(B3/1), it will appear that again we have some amount of output power, moreso than when the Vout was 80mV due to increased LED leakage, when again, in reality all power is being drawn from the battery under these conditions.

Under these same static conditions wherein Q1 is off, let's replace the toroid secondary with a virtual 4 volt battery whose ESR is roughly equivalent to the toroid's impedance.  This virtual battery is connected so that its negative terminal is connected to A1 and its positive terminal connected to the Q1 collector and LED junction (effectively, B1).  The output voltage at B1 will now measure as a positive 5.5volts.  This voltage exceeds the LED turn on voltage causing the LED to conduct and current to flow thru the CSR, limited, for the most part, by the virtual battery's ESR (or, in reality, under AC conditions, the toroid's impedance) and current flow increases.  If, during this static condition, Pout is again measured using the erroneous formula Pout(inst)=(B1)x(B3), the measured output power will be (5.5)x(B3).  Clearly, under this static condition, 1.5 volts of the output voltage is actualy being provided by the input battery, and 4 volts is provided by the virtual battery (which, in reality, is the collapsing toroid secondary).  Therefore, output power, under these static conditions, is actually (B1)x(B3/1)-(Vbatt)x(B3/1) and simultaneously, (Vbatt)x(B3/1) is being drawn from the battery.



Again, I believe that the actual output power during dynamic conditions is closer to the following:

When Q1 is off, Pout(inst)=(B1)x(B3/1)-(Vbatt)x(B3/1)

When Q1 is on, Pout(inst)=(B1)x((B3/1)-(LEDleakpwr), where the expressions (B1)x(B3/1) and (LEDleakpwr) are considered interchangeable and, therefore, when Q1 is on, Pout(inst)=0

Possibly .99 can verify this in simulation.  Possibly these modified formulae also need a bit more work.



I know this has been a long post, but as I have followed Lawrence's work over what must be years now, I have been puzzled as to why he measures OU with his circuits.  Although the above may not be fully complete, it is food for thought and I believe the road to an answer.  Hopefully it is written well enough to allow the points to be grasped.

All comments welcome!

PW

Offline poynt99

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Re: Is joule thief circuit gets overunity?
« Reply #583 on: April 14, 2013, 02:12:30 AM »
Pw,

Your analysys is brilliant and detailed as usual, but is your effort made in vain?

I have not looked at Lawrences xls files, but I will take your word for how he is computing Pout. That method is incomplete. The correct method to obtain Pout (avg) is to average the instantaneous Pout readings exactly the same as is done for Pin. Now this of course assumes that Lawrence feels that Pout is in fact the power in the LED only.

Offline picowatt

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Re: Is joule thief circuit gets overunity?
« Reply #584 on: April 14, 2013, 02:20:58 AM »
Pw,

Your analysys is brilliant and detailed as usual, but is your effort made in vain?

I have not looked at Lawrences xls files, but I will take your word for how he is computing Pout. That method is incomplete. The correct method to obtain Pout (avg) is to average the instantaneous Pout readings exactly the same as is done for Pin. Now this of course assumes that Lawrence feels that Pout is in fact the power in the LED only.

.99,

After looking at all the data, I believe the input power is being calculated correctly.  However, regarding the output power, I believe it is more complex than just correcting Vout by subtracting the voltage at the output CSR, as was done for the input.  Possibly Vbatt needs to be subtracted.

If you look at CH1 in slide 13, draw a line at 1.5volts.  The only output power, is I believe, the area above that line.  Whenever Vout is less than or equal to Vout, there is no Pout.  Only when Vout is greater than Vbatt is there a net Pout.

PW