Is joule thief circuit gets overunity?

[picowatt]

@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]

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]

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]

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]

@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.