[lyner]

jg38, what you are saying is probably somewhere near what actually happened. By your reasonable reasoning, the measurements show that the AC component of average power is a small fraction of the total power delivered to the load. That was interpreted as over unity.

[jerrygg38]

I would like her to show the data. Then we can bring the positive voltage level upward by bringing the minus voltage upward to zero. In this way we can produce the correct readings.  JerryGG38  I'm here.  Why do you refer to 'her' and talk over my head? 

  Sorry. I guess I was speaking to the general audience. Do you have the data?

[jerrygg38]

jg38, what you are saying is probably somewhere near what actually happened. By your reasonable reasoning, the measurements show that the AC component of average power is a small fraction of the total power delivered to the load. That was interpreted as over unity.

Unfortunately all my electrical textbooks were thrown out long ago. I only have the physics books and Calculus by Thomas.
  The fourier series for a square wave where the bottom pusle is at zero and the top pulse is at Vb has a general DC component of
  DC = Vb/2
 Then it has a fundamental at the 555 frequency. I will have to look up the series on the internet but the fundamental AC component is almost as large as the DC. Then comes the third harmonic which is reasonable large.

  Yes. Your words are very good. The probe is seeing the AC and Witsend and company modified the AC component and turned it into a DC component. The probe blocked the DC. Without the data and assuming that the AC component was approximately 80 percent of the DC component, we get from Witsends data

67.6 kilojoules for AC component
81.12 kilojoules for DC component

148.7 kilojoules delivered by battery

 The load dissipated by Witsends numbers

122 kilojoules (note Witsend states 1.22 megajoules)

  The efficiency of the circuit is

122/148  = 82%

  The duty cycle was on 3.7 percent but that merely changes the problem from a symettrical square wave to a pulse. The results will be basically the same. The error as we pointed out that they used the AC probe which eliminated the DC term.

[witsend]

Sophiecentaur.  Could you please define your definition of charge.  Otherwise we will be missing each other forever.  When I have got this I will then I hope to be able to explain what I mean.  Clearly my use of the term is not conventional.  I apologise.

And JerryGG38 - please could you read my earlier post.  You are arguing the same thing that I tried to cover.  I have NO authority to say that it is wrong to average the two cycles.  But I do have the authority of those academics who have been associated with this exercise.  While I am very happy to acknowledge your understanding in the field, I will not do so at the expense of their's.

The post is dated 08.06.2009. 

[jerrygg38]

Look in any e/m theory textbook and all of that stuff has been dealt with back to front and sideways. Sophiecentaur

Had I done this I would never have tried that circuit.  It strikes me that people go to extraordinary lengths to refute those very Laws that are irrefutable.  I think the idea is to try and keep the efficiency at less than 1.  I can think of no other reason.  But that's science for you.

EDIT - The difference being that I did not need to 'lose energy' as a result of the induced current flow.  I only needed to lose charge.  And there again, I could not understand why the boffins could not see that benefit.  Fortunately, also at the start, I had only read Zukov and Dyson and they both stated that current flow does not comprise a flow of electrons.  But I had no idea that ACTUALLY electrical engineers, to a man, seem insist that it does.

EDIT yet again.  My model was pretty well developed by the time I proposed current flow was magnetic fields.  But I actually thought that no-one knew what it was.  That's why I felt brave enough to come forward with my explanation.  The first time I tried to wrap my mind around how you guys thought of it was this year.  Before that the question never came up.  If it did it certainly wasn't in discussion with me. 

No one is trying to deny the overunity circuit due to some prior understandings. This circuit has been built billions of times in telephone circuits and military circuits. It is just a common circuit and poor test techniques produced an error in thinking.Therefore this whole discussion of the circuit is meaningless. Interest but meaningless.

  Now your ideas of magnetic fields producing current flows is a worthy discussion. I suggest you drop the overcurrent circuit and specify your ideas concerning current flows.

   The flow of an electron through a vacuum tube is a current flow where the quanta of charge is 1.602E-19 coulombs. It is certain that the current flow is exactly the number of electrons per second. There is no reason to believe it is subdivided any finer.

   Now we take the current flow in a wire. Conventional theory specifies that an electron flows from one atom to another and pushes each atom along. That always seemed okay to me but it is possible that the quanta could be smaller than the electron.

  In my Dot-wave theory my smallest high energy dot wave has an energy and wavelength equal to the standard AM radio band.
  To me a moving dot-wave is a magnetic field. A stationary dot-wave is a stationary electric field. The minute the dot-wave moves it is part of the magnetic field.

  Therefore you may very well be correct that an electric current is the flow of a magnetic field. So I suggest you start a new topic and discuss your ideas which have a degree of merit.

  

[jerrygg38]

Sophiecentaur.  Could you please define your definition of charge.  Otherwise we will be missing each other forever.  When I have got this I will then I hope to be able to explain what I mean.  Clearly my use of the term is not conventional.  I apologise.

And JerryGG38 - please could you read my earlier post.  You are arguing the same thing that I tried to cover.  I have NO authority to say that it is wrong to average the two cycles.  But I do have the authority of those academics who have been associated with this exercise.  While I am very happy to acknowledge your understanding in the field, I will not do so at the expense of their's.

The post is dated 08.06.2009. 

I suggest you copy our comments and return them to your friends. Then they will see their errors.

[witsend]

jerryGG38 - Are you saying, in simple terms that we must average the voltage during both periods of the duty cycle?  Please just say yes or no.  If it's No, then explain what you mean, if you don't mind.  Thanks.

Edit:Jerry, I'm holding my breath here.  Have I got this right.  Do you mean us to average the voltage measured across the resistive load?  If this is not what you mean then may I ask you to explain what you mean? 

Edit again - sorry, I mean over the shunt resistor.

[jerrygg38]

jerryGG38 - Are you saying, in simple terms that we must average the voltage during both periods of the duty cycle?  Please just say yes or no.  If it's No, then explain what you mean, if you don't mind.  Thanks.

I just cannot say yes because it could be misinterpreted. There are two ways of solving the problem.

Solution 1 requires you to raise the lowest level of the waveshape to zero. This will raise up the highest level. For example if the waveshape has a low level of -1 volts and a high level of 4 volts, we must bring the -1 level to zero volts and the 4 volt level to 5 volts.

   Then you can calculate the data correctly. You will have nothing to subtract.

 The second method is by turning the waveshape into DC and AC components. The DC component is the average level. The AC components are what you see on the scope. The problem is you need to produce a fourier series solution for the waveshape.
  Alternatively there are devices called spectrum analyzers. These instruments (they are expensive) will tell you all the AC components. The most important will be the first, third, and fifth harmonic.
  It is easier to calculate the series if it is a reasonable pulse or square wave type.

 The solution is the square root of the sum of the squares of all the components. This is a tough way to go without a spectrum analyzer.

  Since you used the first method, it is only necessary for you to move the bottom voltage up and then recalculate the problem. Never subtract anything!!!!!!!!!!

[witsend]

So.  Is it reasonable to assume that the voltage measured above zero comes from the battery?  And, in the same way is it also for some reason unreasonable to assume that the Negative voltage comes from the inductance on the load resistor?

EDIT - what I'm actually asking is this.  Do you know of any lead acid battery that is able to deliver a negative current flow?

[jerrygg38]

So.  Is it reasonable to assume that the voltage measured above zero comes from the battery?  And, in the same way is it also for some reason unreasonable to assume that the Negative voltage comes from the inductance on the load resistor?

If you had used a DC probe on the oscilloscope, the shunt resistor voltage would go from zero in the off cycle to a positive value in the on cycle.

  It is very unreasonable to assume that the negative voltage (flat level) comes from the inductor.

  The only thing the inductor could do is produce a positive spike. However if the circuit is  underdamped due to extraneous capacitance, you could get a secondary negative spike.

  The big problem is the inductor produces spikes and not a flat negative level. That negative level comes from the oscilloscope probe.
Check to see if they used an AC probe. The probe itself produces the negative level.

[witsend]

It is very unreasonable to assume that the negative voltage (flat level) comes from the inductor. jerryGG38

Ok.  Let me try this again.  When the battery is disconnected - courtesy the switch and the MOSFET - then we find that there is a very large voltage.  This is evident on the oscilloscope.  I'm sure you will agree with me that it does not come from the battery.  It is generally known as Back EMF - in motor driven circuits.  We don't have a motor - so here it is, apparently and correctly referred to as counter electromotive force.  This counter electromotive force is known to be caused by the field collapsing in inductive components in the circuit.  The only outrageously inductive component in our circuit is the resistor.  So why then is it very unreasonable to assume, in line with well known circuit theory, that this counter electromotive force comes from the inductor?

EDIT: Just ignore, for now, whether it is above or below zero.  Don't even think of measuring it.  Let's establish where this 'spike' comes from.

[Vern]

So why then is it very unreasonable to assume, in line with well known circuit theory, that this counter electromotive force comes from the inductor?

It is not unreasonable to assume. In fact the counter electromotive force does come from the inductor. It was put there in the on cycle. It comes back in the off cycle. But you never get as much back as you put in. []

Keep in mind that it is power that is conserved. Voltage is only potential. Power is voltage times current.

[witsend]

Vern - I'm so glad you're there.  You're right.  But I need someone to acknowledge that this counter electromotive force is only evident during the off period of each switching cycle. 

Do you see I'm trying to master your own impeccable style of brevity? []

Vern, or someone, please acknowledge this.  Still holding my breath.

[witsend]

JerryGG38 - I've just heard from my co-author.  He asked me to point out the following.  He conducted his own tests on a TK TEKTRONIX TDS 3054B 4 channel 500Mhz with a sample rate of 5GS probes.  Measurements on attenuation by 10.  Also used a spectrum analysr LG rated at 5 x 4 GHtz.  I think I've got that right.  He's been following this thread but has no spare time to join in.

He duplicated the tests to determine the validity of the initial claim.  We used the quantum test publication because it had the written permission of the accreditors. 

[jerrygg38]

It is very unreasonable to assume that the negative voltage (flat level) comes from the inductor. jerryGG38

Ok.  Let me try this again.  When the battery is disconnected - courtesy the switch and the MOSFET - then we find that there is a very large voltage.  This is evident on the oscilloscope.  I'm sure you will agree with me that it does not come from the battery.  It is generally known as Back EMF - in motor driven circuits.  We don't have a motor - so here it is, apparently and correctly referred to as counter electromotive force.  This counter electromotive force is known to be caused by the field collapsing in inductive components in the circuit.  The only outrageously inductive component in our circuit is the resistor.  So why then is it very unreasonable to assume, in line with well known circuit theory, that this counter electromotive force comes from the inductor?

EDIT: Just ignore, for now, whether it is above or below zero.  Don't even think of measuring it.  Let's establish where this 'spike' comes from.

 I assume that you are measuring the spike at the shunt resistor. Channel B on the scope. When the Mosfet opens up a voltage of 2 times the battery voltage will instanteously appear at the junction of D1 and Q1. As high as 48 volts is possible. this voltage is the sum of the battery voltage and the inductive spike voltage which can be as high as 24 volts also.
  However the zener diode in the mosfet will pass this voltage to the shunt resistor. therefore the shunt resistor could see a spike of 48 volts minus the zener voltage. Let us say that the zener voltage is specified at 10 volts (each mosfet has different zeners)
  Then the spike passing unto the shunt resistor would be 36 volts. In general due to stray capacitances it will be less than that.
  Therefore the spike comes from the series circuit of the battery and the inductor. which flows through the mosfets zener diode.

  At the same time, diode D1 starts to conduct and clamp the spike. You will get different answers for the spike value depending upon all the circuit parameters.Each circuit will produce different results depending upon their specifications.

[jerrygg38]

So why then is it very unreasonable to assume, in line with well known circuit theory, that this counter electromotive force comes from the inductor?

It is not unreasonable to assume. In fact the counter electromotive force does come from the inductor. It was put there in the on cycle. It comes back in the off cycle. But you never get as much back as you put in. []

Keep in mind that it is power that is conserved. Voltage is only potential. Power is voltage times current.

Vern you answered this too quickly. Energy is conserved. Voltage times current times time.

Also nothing is coming back in that circuit. The inductor charges up and diode D1 conducts. This causes the inductor energy to flow around a circle. It discharges unto itself and not the battery. The mosfet is open after the spike and the battery has no part of the discharge through the diode D1 and inductor.

[jerrygg38]

JerryGG38 - I've just heard from my co-author.  He asked me to point out the following.  He conducted his own tests on a TK TEKTRONIX TDS 3054B 4 channel 500Mhz with a sample rate of 5GS probes.  Measurements on attenuation by 10.  Also used a spectrum analysr LG rated at 5 x 4 GHtz.  I think I've got that right.  He's been following this thread but has no spare time to join in.

He duplicated the tests to determine the validity of the initial claim.  We used the quantum test publication because it had the written permission of the accreditors. 

I am not familar with the latest scopes. It has been 16 years since I did any lab work. He mentions all high frequency stuff. Was it a sampling scope?

  In any event I see no mention of a DC probe. It all appears as AC stuff. Therefore the entire electrical spectrum has been invalidated by the improper scope. You need just a simple DC scope and DC probe, not the fancy scope he used.

[witsend]

Ok JerryGG38 - then we're on the same page.  But that voltage spike can and in fact does exceed 34 volts.  It depends on the duty cycle.  Now Vern is absolutely spot on.  The amount of energy that is returned by the counter electromotive force may very well have been stored on the resistor in the first instance.  And, also correct, is that it never seems to exceed the amount of energy that was first delivered during the ON period of the switching cycle.  But here's the thing.  It always returns some very small fraction less.  Not much difference.  If the duty cycle is on for 10% or 90% - however much energy is first stored is then returned - less that fraction in that spike.  I'm sure you're right.  It's probably because of the diode in the MOSFET or even the diode in parallel to the resistor - or, indeed, both.

So.  If it was stored - or - if the energy that delivered the counter electromotive force was courtesy extra energy from the battery, then it would be evident how?  We measure the voltage across the resistor - during the 'ON' period of the duty cycle to follow Ohm's Law.  In other words the amount of voltage divided by the Ohm's value of the load resistor, over the time of the duty cycle, conforms to whatever would have been determined according the same measurement applied to a simple load placed in series with a battery without the complication of a switch.

So.  If the energy was stored at some extra cost from the battery, where do we find this extra energy?  Is it something that's there, but hidden?

[witsend]

Sorry Jerry - I've just seen your post.  I should have added it was dc coupled.  The spectrum analysis was done on a separate machine.  Please give me an answer to the last post - you or Vern.

EDIT It's sampling range is 10 000 samples.  I know this because it took forever to scroll down the XL page to get to the end.  In other words you choose your range 1 waveform or any number required (0bviously within some limit, not sure what) and it will give you a sample range of 10 000 voltage measurements taken across that range.

[lyner]

These posts are coming too fast to cope with AND make the dinner! Yeah - I know - get your priorities right man!

witsend
My definition of charge is what you will find in all textbooks.
We have a serious problem if you refuse to do that because "they don't make sense" or some such reason. Every time you are pressed in the direction of 'knowing' some proper Science, you react badly. If you come on a Science Forum, then the least you could do is to get some fundamental knowledge of the common language we all try to speak. We can't be expected to adopt a special language for a 'one-off' discussion.

Tell me you don't really mean that "Energy is stored in the Resistance". Please tell me you mean "in the inductance of the resistOR".

Neither I nor jg38 would imply that you 'average' anything. The average of a sinewave is zero! That's why we use RMS. This means adding up all the I²R s (i.e. integrating), which gives a non zero value and corresponds to the sum of the energies during the infinitessimal instants over the time you are observing. The 'root two' factor which is commonly used refers to a sine wave and, for any other waveform is it either necessary to analyse in terms of the harmonic content or, in the case of your waveform, which doesn't have a fixed frequency (I gather?) then you have to RMS the samples.

The AC probe problem is well known, The mean value of what an AC probe shows you will be Zero because it will 'float' up or down until it is zero.

I have just noticed your comment that the spikes varied in appearance on the Scope. This sounds very much like the effects of under-sampling, which is something I suggested earlier.  A 20MHz Scope is very 'slow' for RF work - it's main application would be for so-called video work.  Sampling scopes are lovely as far as they go but their sharp cut off, due to the anti-aliasing filter means that you are totally blind above their specified operating frequency. A tired old analogue scope will give you glimpses of frequencies well above its 3dB point. If you cannot rely on the samples being correct then you can't do valid calculations with them.

jg38 (And Vern, who just wrote the same thing whilst I was cogitating)
I have just had second thoughts about the operation of the diode D1 when the Mosfet switches off. The only path for current to flow is, in fact, in a loop through the diode and the resistor RL. All the magnetic energy will be dissipated in the resistance. The Mosfet is off so the battery is no longer in circuit. That's correct, isn't it? How can charge return to the battery - apart from through some parasitic component? The Drain Source capacitance is a few thousand pF, according to the data sheet.