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Author Topic: a circuit that produces overunity results.  (Read 98961 times)

Offline witsend

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« Reply #250 on: 10/06/2009 16:50:04 »
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.
« Last Edit: 10/06/2009 16:54:29 by witsend »
 

Offline Vern

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« Reply #251 on: 10/06/2009 16:57:25 »
Quote from: witsend
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.

« Last Edit: 10/06/2009 16:59:03 by Vern »
 

Offline witsend

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« Reply #252 on: 10/06/2009 17:03:08 »
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.
« Last Edit: 10/06/2009 17:05:21 by witsend »
 

Offline witsend

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« Reply #253 on: 10/06/2009 17:32:42 »
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. 
 

Offline jerrygg38

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« Reply #254 on: 10/06/2009 17:38:27 »
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.
 

Offline jerrygg38

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« Reply #255 on: 10/06/2009 17:45:40 »
Quote from: witsend
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.


 

Offline jerrygg38

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« Reply #256 on: 10/06/2009 17:49:27 »
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.
 

Offline witsend

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« Reply #257 on: 10/06/2009 18:05:16 »
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?
 

Offline witsend

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« Reply #258 on: 10/06/2009 18:09:38 »
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.
« Last Edit: 10/06/2009 18:14:35 by witsend »
 

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« Reply #259 on: 10/06/2009 18:13:51 »
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.
 

Offline witsend

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« Reply #260 on: 10/06/2009 18:18:05 »
Sophicentaur, just quickly, I have not read through your post.  I do mean resistor.  It was an error.  And I have been so careful not to take offense.  I have NOT been OFFENDED in any way at all. If you're referring to the fact that JerryGG38 did not ask me directly - it is simply because I need to be involved in any discussion. 

Where do you say that I've reacted badly.  I am trying my very best to answer questions.
 

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« Reply #261 on: 10/06/2009 18:22:42 »
jg38
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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)
The zener acts presents zero resistance for voltages in excess of its Vz. I think that you won't find a huge voltage across the 'shunt' resistor, the voltage will be IR, where I is, at most,the original current flowing and R is 0.25ohms. How could it be more than that, when 'held down' by such a low resistance? But the action of D1 should catch the voltage spike long before this happens - it only needs to be 0.7V above the potential of the + battery terminal for it to conduct all the current round through the Rl again.
 

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« Reply #262 on: 10/06/2009 18:25:59 »
witsend
When I say you have reacted badly, I mean badly for the discussion - you just say you don't understand the Science or the Maths, when it suits you but you expect everyone else to trust your judgment because it is obvious to you. Or, you bat on about conventional Science not being willing to listen and just being reactionary for the sake of it.
It really isn't the way to conduct a Scientific dialogue.
 

Offline jerrygg38

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« Reply #263 on: 10/06/2009 18:26:14 »
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?

I think the inductance has complicated the problem. You could have used a non-inductirve resistor for your experiment. The inductance merely adds complications but has little meaning for this circuit.

  What was the value of discharge resistor for the battery. RL is ten ohms and the duty cycle was 3.7 percent as shown on page 5.

  The corresponding draw down resistor should have been

  R = (100/3.7) (10 ohms) = 270 ohms

A 270 ohm resistor across the battery should draw down the battery in the same time as your switching circuit. The inductance does not change the draw down very much.

  What was the value of your draw down resistor???????
 

Offline jerrygg38

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« Reply #264 on: 10/06/2009 18:37:34 »
quote author=sophiecentaur link=topic=23243.msg257259#msg257259 date=1244654562]
jg38
Quote
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)
The zener acts presents zero resistance for voltages in excess of its Vz. I think that you won't find a huge voltage across the 'shunt' resistor, the voltage will be IR, where I is, at most,the original current flowing and R is 0.25ohms. How could it be more than that, when 'held down' by such a low resistance? But the action of D1 should catch the voltage spike long before this happens - it only needs to be 0.7V above the potential of the + battery terminal for it to conduct all the current round through the Rl again.
[/quote]

You are correct in general. I am just doing a worst case possibilitiy in which the shunt resistance and the wiritn path back to the battery has enought inductance to allow for such a large spike. The actual circuit will respond based upon many parameters. Most of the spiking will occur at the junction of D1 and Q1 and not at the shunt resistore.

  However the whole discussion is meaningless. They will get the same results even if the resistor RL has no inductance whatsoever. I am afraid that the inductance spiking merely complicated a more basic error.
Right now I believe that the 3.7 duty cycle required a draw down resistor of 270 ohms to replace the switched 10 ohms. I just asked Witsend for the value of their draw down resistor. I think they used the wrong resistor.
 

Offline witsend

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« Reply #265 on: 10/06/2009 18:43:31 »
If you cannot rely on the samples being correct then you can't do valid calculations with them. Sophiecentaur

I think that Spescom dealt with this problem in the paper.  They got Fluke to send a guarantee that the instrument was capable of sampling within the frequency range that we were testing. 
« Last Edit: 10/06/2009 18:49:38 by witsend »
 

Offline Vern

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« Reply #266 on: 10/06/2009 18:47:43 »
Quote from: jerrygg38
Vern you answered this too quickly. Energy is conserved. Voltage times current times time.
I realized I misspoke while at lunch.
 

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« Reply #267 on: 10/06/2009 19:00:02 »
If you cannot rely on the samples being correct then you can't do valid calculations with them. Sophiecentaur

I think that Spescom dealt with this problem in the paper.  They got Fluke to send a guarantee that the instrument was capable of sampling within the frequency range that we were testing. 

Fluke answered the question you asked them. It is not their job to tell you that it was not the right question to ask.
My question concerns the highest frequency in your waveforms. That requires an answer of the numerical kind. The relevant frequency is not the fundamental oscillation frequency of your circuit. It will be much higher. What is it?
 

Offline jerrygg38

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« Reply #268 on: 10/06/2009 19:01:55 »
These posts are coming too fast to cope with AND make the dinner! Yeah - I know - get your priorities right man!


Tell


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.


Ah ha. A few thousand pf. Yes. When you have a high accuracy analog to digital converter a few thousand pf hurts.

As far as the scope is concerned, a 20 MHz scope is too powerful for this simple circuit. We are only dealing with 2.4 KHz. If you go 100 times all your need is a 0.25 MHz scope for good results. It won't show up the small spikes through the mosfet but who cares???????

  A 60 year old scope with 1 MHz would be far too good for this circuit.
It looks like an old fashioned simpson voltmeter would work just as well.

The whole discussion is reduced to absurdity I am afraid. You are correct that the only path for the resistive discharge is through the diode and itself.
  It is hard to understand how anyone who can operate the fancy equipment for the test could come up with such incorrect answers. I am beginning to laugh at the meaningless ness of this discussion.

  I return to Union Square Park in 1956 in NYC to the man with the talking coconut. The coconut said that he had a simple switching circuit
that could power the world. The people did not believe the man with the talking coconut but every night he returned and stated that he had a simple switching circuit that could power the world. It was funny then and it is still funny today.

  Sorry Witsend. I cannot stop laughing!!! Sorry to offend but I cannot stop laughing.
 

Offline witsend

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« Reply #269 on: 10/06/2009 19:05:03 »
We look to use inductive resistors to ensure that there is counter electromotive force.  However, to the best of my knowledge, the only non-inductive resistors used were for the shunt.

What was the value of discharge resistor for the battery. RL is ten ohms and the duty cycle was 3.7 percent as shown on page 5.
The corresponding draw down resistor should have been
R = (100/3.7) (10 ohms) = 270 ohms
A 270 ohm resistor across the battery should draw down the battery in the same time as your switching circuit. The inductance does not change the draw down very much.
jerryGG38

I do not know what a corresponding draw down resistor is.  But if you mean a control then we did not use a control for the experiment as we were advised that battery draw down rates were meaningless.  However, we did these type of tests for BP.  And also, the rate at which our battery discharged in the experiment in that paper was indeed consistent with that amperage draw down.

What was the value of your draw down resistor???????

What resistor?
 

Offline witsend

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« Reply #270 on: 10/06/2009 19:08:48 »
 Sorry Witsend. I cannot stop laughing!!! Sorry to offend but I cannot stop laughing.

Not at all.  Glad to know your at least amused.
 

Offline jerrygg38

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« Reply #271 on: 10/06/2009 19:11:41 »
Quote from: jerrygg38
Vern you answered this too quickly. Energy is conserved. Voltage times current times time.
I realized I misspoke while at lunch.

I am beginning to get punchdrunk from this discussion. Right now I cannot stop laughing
 

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« Reply #272 on: 10/06/2009 19:15:12 »
jg38
Quote
I think the inductance has complicated the problem. You could have used a non-inductirve resistor for your experiment. The inductance merely adds complications but has little meaning for this circuit.

I see where you are coming from but wasn't the whole point to show that an inductance has the magical property of regenerating energy? Of course it would have made sense to buy (for a couple of quid) a high quality non-inductive resistor as a control. But there a lot of other things that could have been done in order to isolate the flaws and to account for the anomaly (Occam's Razor). I don't think the exercise was aimed in that direction, though.

AND you didn't read the info about the free-running frequency, which was about 150kHz!!
Really, I've got time to do the dinner PLUS read all the facts. I multitask so well I could be a woman!

It strikes me that the people who helped you with this venture, witsend, may not have been as commited as you were. It is much easier to agree with someone who is fired with enthusiasm  than to dig deep into the theory and spot the flaw.
 

Offline witsend

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« Reply #273 on: 10/06/2009 19:17:03 »
Sophiecentaur and JerryGG38 - make up your minds.  Nothing is fast enough for Sophiecentaur and everything is too fast for JerryGG38.  The assurance we had was that the accuracy of the flukemeter was adequate for the frequency of the experiment.

May I please have an answer from someone regarding the question - if energy was first stored in the load resistor, then where do we find that extra energy.  The voltage measured across the load resistor conforms to Ohm's Law.
 

Offline jerrygg38

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« Reply #274 on: 10/06/2009 19:28:54 »
We look to use inductive resistors to ensure that there is counter electromotive force.  However, to the best of my knowledge, the only non-inductive resistors used were for the shunt.

What was the value of discharge resistor for the battery. RL is ten ohms and the duty cycle was 3.7 percent as shown on page 5.
The corresponding draw down resistor should have been
R = (100/3.7) (10 ohms) = 270 ohms
A 270 ohm resistor across the battery should draw down the battery in the same time as your switching circuit. The inductance does not change the draw down very much.
jerryGG38

I do not know what a corresponding draw down resistor is.  But if you mean a control then we did not use a control for the experiment as we were advised that battery draw down rates were meaningless.  However, we did these type of tests for BP.  And also, the rate at which our battery discharged in the experiment in that paper was indeed consistent with that amperage draw down.

What was the value of your draw down resistor???????

What resistor?

Something is wrong. You have a circuit which turns on and off at a 3.7 percent duty cycle. Since the load resistor was 10 ohms, the circuit basically looks like a 270 ohm resistor to the battery.
  If you connect the circuit to one battery and at the same time connect a 270 ohm resistor to another sister battery, both batteries should draw down about the same rate. Then you can compare how much better the switching circuit did compared to its equivalent 270 ohm resister. The switching circuit should discharge the battery faster than the 270 ohm resistor.,
  That is a very simple test. Ultimately the inductance means nothing to the switching circuit. The circuit is ultimately a 270 ohm resistor.
 

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