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Author Topic: Is joule thief circuit gets overunity?  (Read 561414 times)

picowatt

• Hero Member
• Posts: 2038
Re: Is joule thief circuit gets overunity?
« Reply #585 on: April 14, 2013, 02:29:43 AM »
Pw,

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,

And no, I dont believe it was in vain.  It is just a thought puzzle that has been gnawing away at me for some time, and I feel confident that what I have presented sheds a bit of light on the puzzle.

Glad to see you are still "hangin' around" even if it is from Pheonix...

PW

ADDED:  Possibly you can help me work out the correct formula for Pout based on what I have presented.  You're much better at that!

picowatt

• Hero Member
• Posts: 2038
Re: Is joule thief circuit gets overunity?
« Reply #586 on: April 14, 2013, 02:41:22 AM »
.99,

I have looked through the raw input/output data files for board 80 as presented in slide 12 and 13 and am confident that the raw data is representative of slide 12 and 13.  There is a smaller error in that there are approximately 3 full input cycles of collected data and closer to 4.75 cycles of collected output data.  A difference in the integral number of cycles of data collected will not affect the Pin(avg) versus Pout(avg) calculation, but that extra .75 output cycle will produce a slight error in favor of Pout(avg).

However, I believe this to be an error of 5% or less (the extra .75 of an output cycle did not contain a Pout power peak).

PW

picowatt

• Hero Member
• Posts: 2038
Re: Is joule thief circuit gets overunity?
« Reply #587 on: April 14, 2013, 02:58:37 AM »
.99,

Lawrence's Pout(avg) calculations and raw data are in the file "DSO Analysis1.xls" in his post #552.

The raw input data is the "Input.xls" file at the bottom of his post 503.  That is just the raw input data with no math performed, but in his post 575, he did do the input calculations and a beginning an end portion of the xls file is provided there as well (along with the Pin(inst) math).

PW

ltseung888

• Hero Member
• Posts: 4363
Re: Is joule thief circuit gets overunity?
« Reply #588 on: April 14, 2013, 08:27:11 AM »
@PW,

Please study reply 581.  If we vary the Input Voltage, we can get different COP (-10 to +9).  Vout (B1) is the voltage including LED and the 1 ohm resistor.  Iout(B3) is the current (voltage across the 1 ohm resistor).  We can change the load easily by connecting a component at B1 and B3.  That is effectively connecting it in parallel with the LED.  We can also take out the LED and the Load will be whatever is across B1 and B3.  Board 71 has this feature.

In reply 581, we can get different COP just by tuning the Input DC Power Supply.  We can also avoid all this discussion on possible Output Power Measurement Error by using a secondary winding on the Toroid.  The Output can then be calculated by the Secondary Voltage x the Secondary Current.  I used to do that.  Recently, I found that I can get COP >1 without the use of the secondary Coil.  I shall dig out some old winding-toroids and do the Output Power Measurements.

*** I have to admit that I could not follow your logic in the long Output Measurement posts.  Normally, in instantaneous measurements, we do not need to worry about the Load and the Voltage/Current fluctuations.  We just take the Instantaneous Voltage value and multiply it by the Instantaneous Current value to get the Instantaneous Power.  I thought that is simple and solid Physics.....

Do you have a 2-CH or 4-CH DSO and a DC Power Supply?    If so, I can send you one of these Zhou boards and you do not have to "guess".....  I believ TK now has an equivalent Zhou Board.  He is seeing the crossing of the zero axis by Input Current (CH2 Vavg).  If he had a DSO and a DC Power Supply, he could have performed the same experiments as reply 581...

ltseung888

• Hero Member
• Posts: 4363
Re: Is joule thief circuit gets overunity?
« Reply #589 on: April 14, 2013, 08:53:32 AM »
Pw,

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.

@poynt99

Please hook up your 4-CH scope to get the Input and Output waveforms as the first experiment.  That will immediately solve the problem of "non-simultaneous" capturing of Input and Output values.  They should correspond to the "separate Input and Output" waveforms from the 2-CH Atten Scopes.

I blieve that you also have a DC Power Supply.  Vary the DC Power supply from zero to just lighting the LEDs and slowly increase it to 1.5V.  See if you can detect the Iin (Input CH2 average value) changing from positive to negative value.  That will confirm that there are no experiment or equipment errors.

We can try to understand PW's logic later.  I read it six times now and still could not grasp it....  With the experimental set up, we can always make Iin (Input Current Average) very low.  That will force a high COP.  COP can be greater than 1 no matter what his new formula says....

picowatt

• Hero Member
• Posts: 2038
Re: Is joule thief circuit gets overunity?
« Reply #590 on: April 14, 2013, 09:13:08 AM »
@PW,

Please study reply 581.  If we vary the Input Voltage, we can get different COP (-10 to +9).  Vout (B1) is the voltage including LED and the 1 ohm resistor.  Iout(B3) is the current (voltage across the 1 ohm resistor).  We can change the load easily by connecting a component at B1 and B3.  That is effectively connecting it in parallel with the LED.  We can also take out the LED and the Load will be whatever is across B1 and B3.  Board 71 has this feature.

In reply 581, we can get different COP just by tuning the Input DC Power Supply.  We can also avoid all this discussion on possible Output Power Measurement Error by using a secondary winding on the Toroid.  The Output can then be calculated by the Secondary Voltage x the Secondary Current.  I used to do that.  Recently, I found that I can get COP >1 without the use of the secondary Coil.  I shall dig out some old winding-toroids and do the Output Power Measurements.

*** I have to admit that I could not follow your logic in the long Output Measurement posts.  Normally, in instantaneous measurements, we do not need to worry about the Load and the Voltage/Current fluctuations.  We just take the Instantaneous Voltage value and multiply it by the Instantaneous Current value to get the Instantaneous Power.  I thought that is simple and solid Physics.....

Do you have a 2-CH or 4-CH DSO and a DC Power Supply?    If so, I can send you one of these Zhou boards and you do not have to "guess".....  I believ TK now has an equivalent Zhou Board.  He is seeing the crossing of the zero axis by Input Current (CH2 Vavg).  If he had a DSO and a DC Power Supply, he could have performed the same experiments as reply 581...

Lawrence,

The measurements you made for board 80 related to the slides 12 and 13 appear to be correct, the problem is in the math used to calculate Pout.  If you priint out the schematic and the two sliides 12 and 13, it is a bit easier to follow along with my "long" post while referring to those printouts.  However, I am confident that .99 was able to understand my points in that post and, after pondering it a bit, hopefuly he will arrive at a more elegant solution for a proper Pout equation than my present efforts.

Basically, the use of the formula Pout(inst)=(B1)x(B3/1) causes an error in the Pout calculation because some of the power in that expression is actually input power.  This error causes the Pout calculations to be higher than than they really are.

Whenever B1 is less than Vbatt, there is no output current flow beyond that which the battery is supplying, and all currents flowing at that time are from the battery (this is a bit simplified for the actual AC condition, but for now I am just trying to make a point for discussion).

I believe the input power calculations are correct, that is; Vbatt=(A1-A4) and Pin(inst)=(Vbatt)x(A4/1), which is the formula you are using in your excel spreadsheet for input power.

As I said, print out the schematic and the slides 12 and 13, and at least read the supporting arguments given in my long post wherein I discussed the circuit under several static DC conditions.

PW

Pirate88179

• Moderator
• Hero Member
• Posts: 8366
Re: Is joule thief circuit gets overunity?
« Reply #591 on: April 14, 2013, 09:21:23 AM »
Lawrence,

The measurements you made for board 80 related to the slides 12 and 13 appear to be correct, the problem is in the math used to calculate Pout.  If you priint out the schematic and the two sliides 12 and 13, it is a bit easier to follow along with my "long" post while referring to those printouts.  However, I am confident that .99 was able to understand my points in that post and, after pondering it a bit, hopefuly he will arrive at a more elegant solution for a proper Pout equation than my present efforts.

Basically, the use of the formula Pout(inst)=(B1)x(B3/1) causes an error in the Pout calculation because some of the power in that expression is actually input power.  This error causes the Pout calculations to be higher than than they really are.

Whenever B1 is less than Vbatt, there is no output current flow beyond that which the battery is supplying, and all currents flowing at that time are from the battery (this is a bit simplified for the actual AC condition, but for now I am just trying to make a point for discussion).

I believe the input power calculations are correct, that is; Vbatt=(A1-A4) and Pin(inst)=(Vbatt)x(A4/1), which is the formula you are using in you excel spreadsheet for input power.

As I said, print out the schematic and the slides 12 and 13, and at least read the supporting arguments given in my long post wherein I discussed the circuit under several static DC conditions.

PW

PW:

Is there any data on the duty cycle of this JT circuit?  I have a scope but I am not that good with it.  What is the duty cycle and can that be determined by the scope shots?  On 50% and off 50% or some other variation?  This always made a huge difference in our JT circuits that we experimented with in the JT topic here.

Bill

picowatt

• Hero Member
• Posts: 2038
Re: Is joule thief circuit gets overunity?
« Reply #592 on: April 14, 2013, 09:37:20 AM »

PW:

Is there any data on the duty cycle of this JT circuit?  I have a scope but I am not that good with it.  What is the duty cycle and can that be determined by the scope shots?  On 50% and off 50% or some other variation?  This always made a huge difference in our JT circuits that we experimented with in the JT topic here.

Bill

Bill,

The duty cycle for the JT operating as depicted by slide 12 and 13 for Board 80 is pretty close to 55/45.  That is, during a complete cycle, Q1 is on 55% of the tme and Q1 is off 45% of the time.  The total width of one complete cycle is roughly 333 microseconds (for a rep rate of 3Kc).  Therefore, during a complete cycle, Q1 is on for about 183us and Q1 is off for about 150us.

I highly recommend that anyone wishing to discuss this print out the schematic and slide 12 and 13 for board 80 from the locations given in my long post.  It will make things a bit easier...

PW

ltseung888

• Hero Member
• Posts: 4363
Re: Is joule thief circuit gets overunity?
« Reply #593 on: April 14, 2013, 11:10:10 AM »
Lawrence,

The measurements you made for board 80 related to the slides 12 and 13 appear to be correct, the problem is in the math used to calculate Pout.  If you priint out the schematic and the two sliides 12 and 13, it is a bit easier to follow along with my "long" post while referring to those printouts.  However, I am confident that .99 was able to understand my points in that post and, after pondering it a bit, hopefuly he will arrive at a more elegant solution for a proper Pout equation than my present efforts.

Basically, the use of the formula Pout(inst)=(B1)x(B3/1) causes an error in the Pout calculation because some of the power in that expression is actually input power.  This error causes the Pout calculations to be higher than than they really are.

Whenever B1 is less than Vbatt, there is no output current flow beyond that which the battery is supplying, and all currents flowing at that time are from the battery (this is a bit simplified for the actual AC condition, but for now I am just trying to make a point for discussion).

I believe the input power calculations are correct, that is; Vbatt=(A1-A4) and Pin(inst)=(Vbatt)x(A4/1), which is the formula you are using in your excel spreadsheet for input power.

As I said, print out the schematic and the slides 12 and 13, and at least read the supporting arguments given in my long post wherein I discussed the circuit under several static DC conditions.

PW

@PW
You accepted that the Input is correct.  That means the calculation of Average Input Power has no problem.  Assume that P_in_avg is X watts.

We can take any Branch of any circuit.  If there were no other energy source, the Average Power through that Branch must be less than X watts.  Assume that P_branch_avg = Y watts.  Do you agree that Y must be less than X with conventional physics?

Instead of thinking about Average Output Power, please look at the B1 to B4 connections carefully.  Can it be treated as a Branch in the circuit.  If the Average Power through this Branch is greater than X, does that mean we detected something strange?  Does that strange thing mean more energy is coming in from somewhere?

I can easily produce and reproduce "branches" with Y greate than X.  The existing data on Board 80 are examples.

Hope that it clears your confusion.  We can always wait for poynt99 to do his experiments and interprete his results.  PhysicsProf should be ready to post his results in the near future as well.

I shall not post any more until poynt99 or PhysicsProf post their results.  More posting is likely to cause more confusion.  You have confused me already.....

picowatt

• Hero Member
• Posts: 2038
Re: Is joule thief circuit gets overunity?
« Reply #594 on: April 14, 2013, 05:20:36 PM »
@PW
You accepted that the Input is correct.  That means the calculation of Average Input Power has no problem.  Assume that P_in_avg is X watts.

We can take any Branch of any circuit.  If there were no other energy source, the Average Power through that Branch must be less than X watts.  Assume that P_branch_avg = Y watts.  Do you agree that Y must be less than X with conventional physics?

Instead of thinking about Average Output Power, please look at the B1 to B4 connections carefully.  Can it be treated as a Branch in the circuit.  If the Average Power through this Branch is greater than X, does that mean we detected something strange?  Does that strange thing mean more energy is coming in from somewhere?

I can easily produce and reproduce "branches" with Y greate than X.  The existing data on Board 80 are examples.

Hope that it clears your confusion.  We can always wait for poynt99 to do his experiments and interprete his results.  PhysicsProf should be ready to post his results in the near future as well.

I shall not post any more until poynt99 or PhysicsProf post their results.  More posting is likely to cause more confusion.  You have confused me already.....

Lawrence,

Please don't stop posting, your OU results have bee both fascinating and puzzling for some time.  I would like to undersand why your measurements produce results that demonstrate OU.  If the measurements are all correct and OU remains, then that would be a good thing.  OU or not, however, it is just a search for answers.

This has been a puzzler for some time and possibly I lhave ooked at the data too long until I was cross eyed.

In hindsight, I had forgotten that in the end we subtract the measured inpt power from the measured output power.

That is, we use Pout(avg)-Pin(avg)=Pnet , where Pnet, is expected to be zero.

So, the expression Pout(avg) should, contain all power derived from the input so that when the Pnet calculation is performed and Pin(avg) is subtracted from Pout(avg),  Pnet should equal zero.  Any Pnet that is a poitive number is "OU".

When analyzing the circut's operation, I lost track of the minus sign that is used in the end to calculate Pnet so much of what I said regarding the Pout calculation is very likely wrong.

So, in the end, it apears that I am now arguing against myself!

I do, however, have a favor to ask of you.  If you would, I would like to see the results of an edit performed on your output excel file for board 80 (DSO analysis.xls of post 552).

First, delete all data points beyond line 5775.

Second, delete all data points between line 14 and line 735.

Then rerun the Pout(avg) calculation using just the remaining data.

If you could do this and post the results, it would be greatly appreciated.

Thanks,

PW

ADDED:  Lawrence, hold off on the data edits until I look at the data sets a bit more.  I see in the post 575 screen shot of the input power set that there is more data there than depicted in slide 12 that I was looking at.

picowatt

• Hero Member
• Posts: 2038
Re: Is joule thief circuit gets overunity?
« Reply #595 on: April 14, 2013, 05:50:59 PM »
Lawrence,

If you are willing to perform the excel file edits, here they are.

On the output data set used for post 575 (which I believe is the same as DSO analysis.xls from post 552) perform the following edits:

First, delete all data points beyond line 7020.

Second, delete all data points between line 14 and line 736

Considering the 14 line offset in the data list, what you should have left are sample points 722 to 7006.

Perform your Pout(avg) calculations using only those sample points.

Please post the result if you will.

Thanks again,
PW

poynt99

• TPU-Elite
• Hero Member
• Posts: 3582
Re: Is joule thief circuit gets overunity?
« Reply #596 on: April 14, 2013, 07:43:33 PM »
Pw,
Please let me know why in your opinion the Pout computation will not be correct if done as follows:

Pout (avg)=avg ( iout (t) x vout (t))

No consideration for Q on time etc be necessary, is it?

picowatt

• Hero Member
• Posts: 2038
Re: Is joule thief circuit gets overunity?
« Reply #597 on: April 14, 2013, 10:05:04 PM »
Pw,
Please let me know why in your opinion the Pout computation will not be correct if done as follows:

Pout (avg)=avg ( iout (t) x vout (t))

No consideration for Q on time etc be necessary, is it?

.99,

I now believe it is correct.

I stared at it too long yesterday and in the midst of it all forgot that in the end, Pin(avg) is removed from Pout(avg) when the net power calculation is performed.

PW

picowatt

• Hero Member
• Posts: 2038
Re: Is joule thief circuit gets overunity?
« Reply #598 on: April 14, 2013, 10:09:24 PM »
Pw,
Please let me know why in your opinion the Pout computation will not be correct if done as follows:

Pout (avg)=avg ( iout (t) x vout (t))

No consideration for Q on time etc be necessary, is it?

.99,

As for Q on time, no, I have never considered that an issue because of the averaging.  Bill had asked about the duty cycle and the post I think you are referring to is my response to his question.

PW

picowatt

• Hero Member
• Posts: 2038
Re: Is joule thief circuit gets overunity?
« Reply #599 on: April 14, 2013, 11:15:42 PM »
Pw,
Please let me know why in your opinion the Pout computation will not be correct if done as follows:

Pout (avg)=avg ( iout (t) x vout (t))

No consideration for Q on time etc be necessary, is it?

.99

I am beginning to wonder if Lawrence's scopes need to have their input channel offsets checked.  Looking at the raw output data listing, when Q1 is on, Vout is 80mV.  At that same time, there is 12ma being indicated as the output current.  I have checked the current flow through several LED's of various "colors" and cannot find any that indicate anywhere near 1ma at that applied voltage.  Possibly his LED is different than those I have tested, but if that channel is applying a 12ma offset to all Pout calculations, that would be significant.

PW