Hi all,
The tests and test protocol I've outlined are based on an academic investigation of a particular mode of energizing an Inductive-Resistor Heater.
I have performed an analysis of the Heater's performance. The analysis DOES NOT rely on the use of expensive test equipment and can be replicated by anyone with the patience and diligence to follow through. A comprehensive testing protocol is outlined at the end of a Video Slide Show I have prepared. The Video Slide Show can be viewed on my YouTube channel at:
http://www.youtube.com/watch?v=q473lX-Zw1w (http://www.youtube.com/watch?v=q473lX-Zw1w)
You must put the following in context AFTER viewing the Video Slide Show. The analysis is of the 2nd circuit configuration below:
The PWM and MOSFET Gate Driver are powered from their own support battery common'd at GND to the circuit being tested. This is to isolate the controlling circuitry from the MOSFET Switch and the Heater Element. Therefore, the PWM and Gate Driver are being treated as sort of a 'friction-less commutator' for the purposes of this investigation. However, it is vital to consider the effect the Driver has on the entire setup. For this reason I performed several tests to determine if the Driver was contributing any appreciable energy in heating the Inductive-Resistor. The tests were performed for two circuit configurations with similar outcomes. Below are the two circuit configurations followed by scope captures showing MINIMAL (if any) contribution from the Gate Driver.
The difference in their voltage drop between Loaded and Not Loaded is 0.9mV across a 0.1 Ohm x 1% non inductive CSR ...SHDriver. The support battery runs nominally at 13VDC. That translates to 0.117 Watts and for the duration of my tests (8-hours) translates into 0.93W-Hrs of energy.
So ... the contribution from the MOSFET gate driver is negligible, but has been considered against the observed performance of the circuit and noted here. Also, the PWM's input to the Gate Driver is logic level and wouldn't be considered in the scheme of things any more than you'd consider the power a function generator is drawing from its wall socket.
Hi again,
Just in case someone will want to know what batteries I'm using, I shot some photos of them. I also took a shot of the chargers I use. The larger battery is a SEC1075, 12V, 7Ah AGM (Absorbent Glass Mat) and the smaller one is the SLA0810, 2V, 6Ah, AGM. I use two 12V and one 2V in series. The battery charger is a really cool 6V / 12V Charger, P/N SEM-1562A. I have two of these chargers. One I use for charging the 12V batteries in parallel. The other charger is used to charge the 2V battery, so I have two more 2V to make up the total of 6V, so those charge in series. This is so I can charge both 12Vs & 2V at the same time ... here they are:
Enjoy,
Greg
Greg,
Is it not more likely that all of your testing has merely proven the well known phenomenon that a lead acid battery, and indeed most battery chemistries, produce different amp hour ratings for different load profiles? Would you expect a 100% duty cycle 1 amp load, a 50% duty cycle 2 amp load, and a 25% duty cycle 4 amp load, applied to a given battery, to all yield identical amp hour ratings from that battery?
Moreover, with regard to your circuit, and in particular her circuit, the concept of desulphation wherein a continuous sequence of reverse polarity pulses is applied to a lead acid battery to break up large sulphate crystals and enhance battery performance has been around for some time. Additionally, pulse and reverse pulse plating has been used by electroplaters for years to produce finer grain metallic platings. As the action of a lead acid battery is also a "plating" process, it is very likely that similar finer grain platng occurs when pulsing or reverse polarity pulsing a lead acid battery while under load, thereby increasing the active area of the plated surfaces and hence, increasing its amp hour rating.
When claims of COP=infinity were beng made, an easy proof would have been to just let the circuit run "forever" to prove the battery never discharged. Now that the claims are with respect to merely demonstrating that a battery's amp hour rating is increased with respect to a given amount of energy extracted usig different load profiles, one would think that a claim of overunity must also account for and prove that less energy is required to recharge the battery than was witdrawn by the load.
If you search for desulphator schematics and look at various desulphator waveforms, you will see a great deal of similarity between the operation of those circuits and related waveforms and the circuits and waveforms utilized to produce claimed overunity with lead acid batteries. But again, I believe these claims are moreso related to claims of increased battery amp hour ratings under varying load profiles in concert with the actions of desulphation and reverse pulse plating than to claims of "overunity".
PW
Quote from: picowatt on April 28, 2013, 03:06:05 PM
Greg,
Is it not more likely that all of your testing has merely proven the well known phenomenon that a lead acid battery, and indeed most battery chemistries, produce different amp hour ratings for different load profiles? Would you expect a 100% duty cycle 1 amp load, a 50% duty cycle 2 amp load, and a 25% duty cycle 4 amp load, applied to a given battery, to all yield identical amp hour ratings from that battery?
Moreover, with regard to your circuit, and in particular her circuit, the concept of desulphation wherein a continuous sequence of reverse polarity pulses is applied to a lead acid battery to break up large sulphate crystals and enhance battery performance has been around for some time. Additionally, pulse and reverse pulse plating has been used by electroplaters for years to produce finer grain metallic platings. As the action of a lead acid battery is also a "plating" process, it is very likely that similar finer grain platng occurs when pulsing or reverse polarity pulsing a lead acid battery while under load, thereby increasing the active area of the plated surfaces and hence, increasing its amp hour rating.
When claims of COP=infinity were beng made, an easy proof would have been to just let the circuit run "forever" to prove the battery never discharged. Now that the claims are with respect to merely demonstrating that a battery's amp hour rating is increased with respect to a given amount of energy extracted usig different load profiles, one would think that a claim of overunity must also account for and prove that less energy is required to recharge the battery than was witdrawn by the load.
If you search for desulphator schematics and look at various desulphator waveforms, you will see a great deal of similarity between the operation of those circuits and related waveforms and the circuits and waveforms utilized to produce claimed overunity with lead acid batteries. But again, I believe these claims are moreso related to claims of increased battery amp hour ratings under varying load profiles in concert with the actions of desulphation and reverse pulse plating than to claims of "overunity".
PW
Hi PW,
Thanks for your observations. Although I'm aware of Rosie's tests that reference Battery ratings, I don't consider the amp-hour ratings of the batteries at all other than to 'range' my load and test duration. In my presentation I do not make any reference to some important data contained in the data sheets ... and that is the 'unloaded' battery voltages. The reason for that data was to compare the start and end voltage differences, loaded and unloaded, and you'll see that those numbers are very nearly the same for any given test. This is a good account of how the batteries drew down ... that is to say ... I was NOT recording 'surface' charge or 'fluff'. For my batteries, upon unloading, they recover to 95% of their 'new' resting voltage in 2-minutes. That's why I use that figure of 2-minutes (read the testing outline). And also, the times it took to recharge the batteries after the Circuit Test and after the 2nd Draw Down Test were within 45 minutes of one another. After the 1st Draw Down Test, the batteries took a good 3-hours longer to fully recharge.
The test I show in the presentation is a repeat of over a dozen similar tests. And as far as:
" ... Would you expect a 100% duty cycle 1 amp load, a 50% duty cycle 2 amp load, and a 25% duty cycle 4 amp load, applied to a given battery, to all yield identical amp hour ratings from that battery? ... "
I have no idea what you're getting at. I don't use any duty cycle values for any calculation. I'm dealing with equivalent heat in terms of "Energy" ... not power ... using 'power' which is an 'instantaneous component value' is incorrect to use in these analyses.
I used whatever frequency and duty cycle gave me the most heat on the test fixture. It's as simple as that. All I'll say is that I have no idea where the excess heat is coming from but it's there.
My intent for this thread is to post my progress and findings. I will not debate the integrity of the data or my methods with anyone. These tests have been performed many times with consistent results. The batteries always show the same characteristics and show no signs of degradation.
I'm forging ahead on this. Don't attempt to dissuade me because you can't. Thank you,
Greg
Quote from: gmeast on April 28, 2013, 06:22:39 PM
Hi PW,
Thanks for your observations. Although I'm aware of Rosie's tests that reference Battery ratings, I don't consider the amp-hour ratings of the batteries at all other than to 'range' my load and test duration. In my presentation I do not make any reference to some important data contained in the data sheets ... and that is the 'unloaded' battery voltages. The reason for that data was to compare the start and end voltage differences, loaded and unloaded, and you'll see that those numbers are very nearly the same for any given test. This is a good account of how the batteries drew down ... that is to say ... I was NOT recording 'surface' charge or 'fluff'. For my batteries, upon unloading, they recover to 95% of their 'new' resting voltage in 2-minutes. That's why I use that figure of 2-minutes (read the testing outline). And also, the times it took to recharge the batteries after the Circuit Test and after the 2nd Draw Down Test were within 45 minutes of one another. After the 1st Draw Down Test, the batteries took a good 3-hours longer to fully recharge.
The test I show in the presentation is a repeat of over a dozen similar tests. And as far as:
" ... Would you expect a 100% duty cycle 1 amp load, a 50% duty cycle 2 amp load, and a 25% duty cycle 4 amp load, applied to a given battery, to all yield identical amp hour ratings from that battery? ... "
I have no idea what you're getting at. I don't use any duty cycle values for any calculation. I'm dealing with equivalent heat in terms of "Energy" ... not power ... using 'power' which is an 'instantaneous component value' is incorrect to use in these analyses.
I used whatever frequency and duty cycle gave me the most heat on the test fixture. It's as simple as that. All I'll say is that I have no idea where the excess heat is coming from but it's there.
My intent for this thread is to post my progress and findings. I will not debate the integrity of the data or my methods with anyone. These tests have been performed many times with consistent results. The batteries always show the same characteristics and show no signs of degradation.
I'm forging ahead on this. Don't attempt to dissuade me because you can't. Thank you,
Greg
Greg,
Not trying to disuade you at all. Just saying, a battery under different load profiles will net a different amp hour rating and hence, a different discharge curve or rate.
From my read of your test methods, you utilize the differences in battery discharge rates with different load profiles, i.e., a pulsed load of a given duty cycle versus a fixed load of a given resistance to determine efficiency. So yes, duty cycles, or more specifically, load profiles, are indeed involved in your tests in that one of your loads is your pulsed circuit at less than 100% duty cycle and your other reference load is a 100% duty cycle load using an equivalent heat output resistor. Hence, my question, would you expect the same amp hour rating, i.e., discharge rate, from a given battery when loads of 1 amp at 100%, 2 amp at 50% or 4 amp at 25% are applied? Each of those load profiles would generate the same heat output, but I would not expect the battery to respond equally regarding discharge rates.
But, please forgive me, I was not aware we were not to post comments here.
PW
Quote from: picowatt on April 28, 2013, 07:28:04 PM
Greg,
Not trying to disuade you at all. Just saying, a battery under different load profiles will net a different amp hour rating and hence, a different discharge curve or rate.
From my read of your test methods, you utilize the differences in battery discharge rates with different load profiles, i.e., a pulsed load of a given duty cycle versus a fixed load of a given resistance to determine efficiency. So yes, duty cycles, or more specifically, load profiles, are indeed involved in your tests in that one of your loads is your pulsed circuit at less than 100% duty cycle and your other reference load is a 100% duty cycle load using an equivalent heat output resistor. Hence, my question, would you expect the same amp hour rating, i.e., discharge rate, from a given battery when loads of 1 amp at 100%, 2 amp at 50% or 4 amp at 25% are applied? Each of those load profiles would generate the same heat output, but I would not expect the battery to respond equally regarding discharge rates.
But, please forgive me, I was not aware we were not to post comments here.
PW
Hi PW,
You are correct with regards to your assessment of power and heat for ... "1 amp at 100%, 2 amp at 50%, 4 amp at 25% ... " and so on, but only for those nice, clean pure square wave pulses applied to a non-inductive, purely resistive load that cares not about rapid-edge transitioning pulses applied to it. Things change when you apply the same pulsing regimen to an inductor. An inductor stores energy in a magnetic field AND heats up when powered. When power is remove via a switch of equivalent, the magnetic field collapses, viciously slicing through the inductor and any and all background or vacuum energy fields and further induces, collapses, induces, collapses, etc, and eventually damps out (most likely manifested as heat). These are complex events that are non-linear and asymmetrical, and DO NOT TEST OUT so neatly as the theoretical and 'ideal' examples above. Heat produced on the test fixture strays far from the theoretical and 'ideal' for various combinations of frequency and duty cycle.
For clarity: Amp-Hours is an expression of capacity ... and can be quite arbitrary. My batteries are 7Ah and only at .350A for 20Hrs, or it can be a 10Hr rating or a 5Hr rating or 50Hr rating. I'm talking Watt-Hours ... "Energy".
Yes ... I assume that after every battery recharge, the batteries are returned to the same storage level each time ... and they are. It is easy to verify ... it is the reason for the rather complex charging and start-up procedure leading into every test. This is what is tested:
From the instant the load is applied, time is carefully tracked to make sure the test's starting voltage is reached in the same time as in the other tests. For a purely resistive load applied during this time, this proves the battery's capacity is the same as for previous tests. This is inarguable. A voltage drop or voltage rise from the same starting voltage across a known resistive load over the same measured time is the same amount of Energy. As long as the temperatures are also the same. Again ... this is the reason I also sample the unloaded battery voltages at the start and end of the tests .. again proving I'm recording actual resources used up and not 'surface charge' or 'fluff' ... "fluff" - what a stupid term! Who came up with that?
Thank you PW,
Greg
Quote from: gmeast on April 28, 2013, 09:20:24 PM
Hi PW,
You are correct with regards to your assessment of power and heat for ... "1 amp at 100%, 2 amp at 50%, 4 amp at 25% ... " and so on, but only for those nice, clean pure square wave pulses applied to a non-inductive, purely resistive load that cares not about rapid-edge transitioning pulses applied to it. Things change when you apply the same pulsing regimen to an inductor. An inductor stores energy in a magnetic field AND heats up when powered. When power is remove via a switch of equivalent, the magnetic field collapses, viciously slicing through the inductor and any and all background or vacuum energy fields and further induces, collapses, induces, collapses, etc, and eventually damps out (most likely manifested as heat). These are complex events that are non-linear and asymmetrical, and DO NOT TEST OUT so neatly as the theoretical and 'ideal' examples above. Heat produced on the test fixture strays far from the theoretical and 'ideal' for various combinations of frequency and duty cycle.
For clarity: Amp-Hours is an expression of capacity ... and can be quite arbitrary. My batteries are 7Ah and only at .350A for 20Hrs, or it can be a 10Hr rating or a 5Hr rating or 50Hr rating. I'm talking Watt-Hours ... "Energy".
Yes ... I assume that after every battery recharge, the batteries are returned to the same storage level each time ... and they are. It is easy to verify ... it is the reason for the rather complex charging and start-up procedure leading into every test. This is what is tested:
From the instant the load is applied, time is carefully tracked to make sure the test's starting voltage is reached in the same time as in the other tests. For a purely resistive load applied during this time, this proves the battery's capacity is the same as for previous tests. This is inarguable. A voltage drop or voltage rise from the same starting voltage across a known resistive load over the same measured time is the same amount of Energy. As long as the temperatures are also the same. Again ... this is the reason I also sample the unloaded battery voltages at the start and end of the tests .. again proving I'm recording actual resources used up and not 'surface charge' or 'fluff' ... "fluff" - what a stupid term! Who came up with that?
Thank you PW,
Greg
Greg,
Yes Greg, I can see that your waveforms are anything but square. They do not however represent a 100% duty cycle anymore than a sine wave does. Disregarding the semantics, the point was that different load profiles will produce different battery discharge curves. Your pulsed circuit is one load profile of less than 100% duty cycle and the fixed resistive load is a different load profile with a 100% duty cycle.
It is my understanding that you use the measured time between your battery start/stop voltages using two different load profiles to calculate your efficiency. That is, I thought you used the fixed resistance load in concert with start stop voltages versus time to determine input power.
Possibly I have not read your test protocols correctly, but it sounds like you are using the response of the battery to one load profile as a method to measure the "energy" dissipated by a different load profile.
PW
Quote from: picowatt on April 28, 2013, 09:45:02 PM
Greg,
Yes Greg, I can see that your waveforms are anything but square. They do not however represent a 100% duty cycle anymore than a sine wave does. Disregarding the semantics, the point was that different load profiles will produce different battery discharge curves. Your pulsed circuit is one load profile of less than 100% duty cycle and the fixed resistive load is a different load profile with a 100% duty cycle.
It is my understanding that you use the measured time between your battery start/stop voltages using two different load profiles to calculate your efficiency. That is, I thought you used the fixed resistance load in concert with start stop voltages versus time to determine input power.
Possibly I have not read your test protocols correctly, but it sounds like you are using the response of the battery to one load profile as a method to measure the "energy" dissipated by a different load profile.
PW
Hi PW,
Read it carefully. You are incorrect in your assessment, but I appreciate going through this exercise with you. The circuit load was a recorded value that 'was what it was' across SH3. It was NOT an established load, it was the load on the battery the circuit (seemingly) presented, or 5.6mV on 0.05-Ohm CSR SH3. It was a RESULTANT value on SH3 in the Circuit test. It was simply a value that was recorded. Nothing was 'set to it' in the Circuit Test. However, the 1st Resistive Load Test Was ADJUSTED to 5.6mV on SH3. So we have the same (seeming) load on the batteries ... one being the RESULT of loading the batteries and the other being SET AS THE LOAD on the batteries. What meters, scopes and instruments CANNOT DETECT are the energies providing either excess heat to RL or any energies being returned to the batteries, or both.
As it turns out, it doesn't matter what the load was for the 1st resistive load test. A heavier load would draw the batteries down sooner and a lighter load would draw them down later. A heavy load is a higher wattage (power) and a lighter load is lower wattage (power). What's important is where this discharge curve crosses the ENDING voltage of the Circuit test. Higher Power X Shorter Time = Lower Power X Longer Time. It ends up being the same Watt-Hours of Energy.
The 2nd Resistive Load Test was the PROOF. I used the quotient of the 1st Resistive Load Test's Energy / The Circuit Test's Heating Energy (on the test fixture provided by the precision DC power supply) as a factor to adjust this test's resistive load such that its starting and ending voltages were the same as for the Circuit test. And that simple ratio 'pegged' the proper loading 'dead-nutts-on' or 'spot on'. It's nearly exactly the energy as for the 1st Resistive Load Test. Of course I was able to get actual power from this because I could measure the resistive load, and I know the battery voltages, and a child can do the math.
I then 'Proved' the Proof by applying this power 'DIRECTLY' to RL (using the precision DC power supply) which resulted in less heating on the test fixture.
Thanks PW,
Greg
Quote from: gmeast on April 28, 2013, 10:39:59 PM
Hi PW,
Read it carefully. You are incorrect in your assessment, but I appreciate going through this exercise with you. The circuit load was a recorded value that 'was what it was' across SH3. It was NOT an established load, it was the load on the battery the circuit (seemingly) presented, or 5.6mV on 0.05-Ohm CSR SH3. It was a RESULTANT value on SH3 in the Circuit test. It was simply a value that was recorded. Nothing was 'set to it' in the Circuit Test. However, the 1st Resistive Load Test Was ADJUSTED to 5.6mV on SH3. So we have the same (seeming) load on the batteries ... one being the RESULT of loading the batteries and the other being SET AS THE LOAD on the batteries. What meters, scopes and instruments CANNOT DETECT are the energies providing either excess heat to RL or any energies being returned to the batteries, or both.
As it turns out, it doesn't matter what the load was for the 1st resistive load test. A heavier load would draw the batteries down sooner and a lighter load would draw them down later. A heavy load is a higher wattage (power) and a lighter load is lower wattage (power). What's important is where this discharge curve crosses the ENDING voltage of the Circuit test. Higher Power X Shorter Time = Lower Power X Longer Time. It ends up being the same Watt-Hours of Energy.
The 2nd Resistive Load Test was the PROOF. I used the quotient of the 1st Resistive Load Test's Energy / The Circuit Test's Heating Energy (on the test fixture provided by the precision DC power supply) as a factor to adjust this test's resistive load such that its starting and ending voltages were the same as for the Circuit test. And that simple ratio 'pegged' the proper loading 'dead-nutts-on' or 'spot on'. It's nearly exactly the energy as for the 1st Resistive Load Test. Of course I was able to get actual power from this because I could measure the resistive load, and I know the battery voltages, and a child can do the math.
I then 'Proved' the Proof by applying this power 'DIRECTLY' to RL (using the precision DC power supply) which resulted in less heating on the test fixture.
Thanks PW,
Greg
Greg,
Possibly I misunderstood.
I thought you were running your pulsing circuit, noting the stabilized temp, and measuring time between battery start and stop voltages.
Then, usng the bench supply, you adjust for a similar stabilized temp and note the V and I from the supply.
Then, using the V and I figures from the bench supply test, you select a fixed resistor value that applies a similar load to the recharged and stabilized battery as the bench supply indicated and again note the time between battery start and stop voltages.
Efficiencyis then determined by comparing the first and last portions above.
If that is incorrect, I will have to take some more time when available to reread your protocol.
PW
Quote from: picowatt on April 28, 2013, 11:13:29 PM
Greg,
Possibly I misunderstood.
I thought you were running your pulsing circuit, noting the stabilized temp, and measuring time between battery start and stop voltages.
Then, usng the bench supply, you adjust for a similar stabilized temp and note the V and I from the supply.
Then, using the V and I figures from the bench supply test, you select a fixed resistor value that applies a similar load to the recharged and stabilized battery as the bench supply indicated and again note the time between battery start and stop voltages.
Efficiencyis then determined by comparing the first and last portions above.
If that is incorrect, I will have to take some more time when available to reread your protocol.
PW
Hi PW,
Thanks for the exchange. All of the information is in that Slide Show. When you:
... " take some more time when available to reread your (my) protocol." ...
it would be time well spent. Actually there are so many technical and non-technical individuals who have contacted me and expressed gratitude for having explained these experiments in the detail and with the simplicity and clarity that I have, I don't see the need to engage you any further on this. Most everyone else seem(s) to 'get it'. My efforts are now aimed at conducting several dozen (more) experiments testing the reliability of my protocol and publishing the results as I have with the Heater Slide Show.
I do this for all of those experimenters out there that NEED a simpler way to conduct their experiments, take measurements, generate meaningful data and contribute technology that will help break the backs of the control mongers of this world.
Regards,
Greg
x
Quote from: gmeast on April 29, 2013, 09:38:02 AM
Greg,
I have been watching your video and remain a bit confused.
I thought the premise of your measurement method was that it did not require the measurement of the voltage/current of any complex or fast risetime waveforms. Yet apparently the voltage at SH3 is being used.
So far, this is what I gather you are doing, please correct me if I am in error:
Step 1: Run the burst heater and note the deltaT, SH3 Vdrop, and the Vbatt start/stop voltages.
Step 2: To determine the burst heater's output power, you use a fixed DC supply to drive Rload to the same delta T as in step 1 and note the supply's V and I.
Step3: To determine input power for step 1, use a rheostat as Rload adjusted to produce the same SH3 Vdrop as noted in step 1 and again measure Vbatt start and stop voltages.
Step4: Compare the battery discharge curves from step1 and step 3
Before I attempt to grasp this a bit more, how is SH3 being measured in step 1?
PW
Greg,
I originally thought you were:
Step1: Run BH circuit, note deltaT and Vbatt start/stop voltage
Step2: Use supply to produce same deltaT in step 1 and note the required power using supply's V and I
Step3: Apply a selected resistive load to Vbatt which produces a similar power load to Vbatt as determined by step 2
Step4: Compare discharge curves of step1 and step2
The reason I thought you were doing this way was to eliminate the need to measure complex waveforms.
Explin it a bit more if you would...
PW
Greg,
I see there is a "step 5", as further verification, wherein you use the rheostat to generate a similar discharge curve as observed in step one and note the V and I of that test.
From what I originally gathered, and still do, it appears that you rely heavily on the discharge characteristics of your lead acid batteries to make your determination of OU.
I can only restate what I said in my original post here, which is that under different load profiles, a battery will net a different amp hour (or watt hour) rating and, ultimately, different discharge curves, even if those load profiles produce similar average loads.
As well, the effects of desulphation and pulse/reverse pulse plating when using pulsed loads, as compared to DC loads, will very likely affect the discharge curves as well.
Is there some part of your testing that I have missed that rules out the effects of different load profiles, desulphation, and pulse/reverse pulse plating as the reason for the different discharge curves? One would think that if theses effects are not in play with your tests, that the battery could be eliminated altogether and only a DC supply used to prove that more heat is generated in the burst heater circuit than is produced under DC conditions.
Have you ever run the burst heater off of a well filtered supply, measured the V and I, and then applied that same amount of power from the supply directly to Rload to see if excess heat is produced in the first instance?
PW
Discharge 'characteristics "no" ... energy drawn from the batteries "yes". I must say, you are the only one stuck on this "characteristics" nonsense.
To attempt to clarify things further:
I use a scope to make sure the waveform is not doing weird things. I monitor the FET's drain for waveform shape and SH3 for one of the mean CSR values and I also use a DVM on SH3 as a check for agreement between the instruments ... and they always agree. Using a DVM is that poynty-head's thing he's so proud of.
BUT ... the values from SH3 do not represent the energy consumed by the circuit. NO instruments can detect the types of energies that account for the excess heating on RL nor can they detect anything going on inside the battery as a result of D1.
You are simply assuming that the final judgement has been rendered as to how to measure anything and everything and that everything is known that is ever to be known and there is nothing left to be learned. How terrible that you have limited yourself in that way.
Bye
Quote from: gmeast on April 29, 2013, 05:44:54 PM
Discharge 'characteristics "no" ... energy drawn from the batteries "yes". I must say, you are the only one stuck on this "characteristics" nonsense.
To attempt to clarify things further:
I use a scope to make sure the waveform is not doing weird things. I monitor the FET's drain for waveform shape and SH3 for one of the mean CSR values and I also use a DVM on SH3 as a check for agreement between the instruments ... and they always agree. Using a DVM is that poynty-head's thing he's so proud of.
BUT ... the values from SH3 do not represent the energy consumed by the circuit. NO instruments can detect the types of energies that account for the excess heating on RL nor can they detect anything going on inside the battery as a result of D1.
You are simply assuming that the final judgement has been rendered as to how to measure anything and everything and that everything is known that is ever to be known and there is nothing left to be learned. How terrible that you have limited yourself in that way.
Bye
Greg,
I am trying not to asume anything. In the slide show, you theorize that the observed excess is due to the magnetic collapse of Rload. If this is the case, why is a battery needed? If the observed effect is due to the collapsing inductance of Rload, should this effect not also be readily observed when the circuit is being fed from a DC supply?
If you ran the BH circuit on a well filtered DC supply and noted the circuit's power draw and Rload's stabilized deltaT and then applied that equivalent DC power directly to Rload and measured less stabilized deltaT, that would do much to support your claim of overunity from the collapsing inductance theory you propose. As all measurements would be at DC, they would be fairly easy to perform.
Have you ever performed such a test?
PW
Quote from: picowatt on April 29, 2013, 07:12:05 PM
Greg,
I am trying not to asume anything. In the slide show, you theorize that the observed excess is due to the magnetic collapse of Rload. If this is the case, why is a battery needed? If the observed effect is due to the collapsing inductance of Rload, should this effect not also be readily observed when the circuit is being fed from a DC supply?
If you ran the BH circuit on a well filtered DC supply and noted the circuit's power draw and Rload's stabilized deltaT and then applied that equivalent DC power directly to Rload and measured less stabilized deltaT, that would do much to support your claim of overunity from the collapsing inductance theory you propose. As all measurements would be at DC, they would be fairly easy to perform.
Have you ever performed such a test?
PW
Hi Picowatt,
I am NOT claiming anything relating to the source of the excessive heat on RL, I only suggested some possibilities. I have NO idea where the stuff comes from, only that more equivalent energy in the form of heat evolves than the energy supplied to it.
".. then why is a battery needed?"
That is perhaps the best question of all. Why?
Hooking up a DC supply to the circuit in place of the batteries yields completely different oscillations. I have tried this many times. It's just not the same thing, nor should it be regarded AS being the same thing. That's just foolishness. One is a power supply and the other is batteries. Power Supply DOES NOT EQUAL Batteries.
Example of the traction drive:
Two discs ... one driving, one driven and under load. Their peripheries are in contact. The two discs each have gear teeth on their peripheries. The driven disc can't slip against rotation because both discs have gear teeth. Now I'm going to make the gear tooth pitch greater with many more finer and smaller teeth. Again, no slippage because they both have gear teeth. I now make the gear teeth so fine with so many of them that by all measurement each disc's periphery seems smooth and polished and now resembles a traction drive. But now under a lesser load, the driven disc now slips. Why? ... because a gear drive is not the same a traction drive. In theory, there should not be any difference between a set of coarse-toothed gears and a set of gears with infinitely many very fine gear teeth each ... but there IS.
So when you imply that a Power Supply as a DC source is the same as a Battery as a DC source, you couldn't be more mistaken.
You simply refuse to accept the possibility ... the possibility that your grandiose, expensive and sophisticated scopes and other 'accepted' devices just aren't measuring the types of energies present in these systems because they can't.
At this point, I'm now wasting my time here. You are anti-free energy, anti-OU, anti-OO or whatever you want to call it. I should have known that one of you characters from the TK assassination squad would show up.
I'm done sharing here. I have better things to do with my time than to play into your condescendence.
Adios
Quote from: gmeast on April 29, 2013, 10:16:23 PM
Hi Picowatt,
I am NOT claiming anything relating to the source of the excessive heat on RL, I only suggested some possibilities. I have NO idea where the stuff comes from, only that more equivalent energy in the form of heat evolves than the energy supplied to it.
".. then why is a battery needed?"
That is perhaps the best question of all. Why?
Hooking up a DC supply to the circuit in place of the batteries yields completely different oscillations. I have tried this many times. It's just not the same thing, nor should it be regarded AS being the same thing. That's just foolishness. One is a power supply and the other is batteries. Power Supply DOES NOT EQUAL Batteries.
Example of the traction drive:
Two discs ... one driving, one driven and under load. Their peripheries are in contact. The two discs each have gear teeth on their peripheries. The driven disc can't slip against rotation because both discs have gear teeth. Now I'm going to make the gear tooth pitch greater with many more finer and smaller teeth. Again, no slippage because they both have gear teeth. I now make the gear teeth so fine with so many of them that by all measurement each disc's periphery seems smooth and polished and now resembles a traction drive. But now under a lesser load, the driven disc now slips. Why? ... because a gear drive is not the same a traction drive. In theory, there should not be any difference between a set of coarse-toothed gears and a set of gears with infinitely many very fine gear teeth each ... but there IS.
So when you imply that a Power Supply as a DC source is the same as a Battery as a DC source, you couldn't be more mistaken.
You simply refuse to accept the possibility ... the possibility that your grandiose, expensive and sophisticated scopes and other 'accepted' devices just aren't measuring the types of energies present in these systems because they can't.
At this point, I'm now wasting my time here. You are anti-free energy, anti-OU, anti-OO or whatever you want to call it. I should have known that one of you characters from the TK assassination squad would show up.
I'm done sharing here. I have better things to do with my time than to play into your condescendence.
Adios
Greg,
I have only been trying to understand and discuss your experiments and methodologies. Why all the attitude? You seem to think you know what I think, great, where did you get that from? I am not Harvey, I don't speak French, I am not "anti-OU" nor am I refusing to accept anything. But, that is another matter. In any event, is anyone that questions your methods or attempts to understand them subject to such behavior? If so, good luck with that. It will surely not assist you in gaining further acceptance. I, however, am doing my best to ignore it...
I have looked over your slides for some time. I believe I follow and can agree with your methods up to a certain point.
From your own measurements, it looks like you are saying that to produce the same deltaT as the BH, it takes 3.16watts from the DC supply. Using the rheostat to produce the same Vdrop at SH3 requires 3.09watts. Although I think there may be an error with regard to how you are determining the contribution of of the gate drive to the circuit, using your figure of .117watts, then apparently, from your numbers, the BH is using 3.2 watts to produce an output of 3.15 watts. Even that just under OU efficiency is, it itself, amazing, but as I said, I believe you need to study the contribution of the gate driver a bit further.
When you disconnect the drain Vsupply, the capacitances in the FET increase to their extreme maximums, i.e., the driver sees a maximum capacitive loading. When you reapply drain voltage, the FET capacitances reduce substantially. Merely comparing and using only the difference in driver power between those two conditions may not provide an accurate assessment of the driver's power contribution to the circuit overall.
But, where I am having the most difficulty accepting your results is when you use the time it takes the battery to discharge to 27.44V when loaded with the rheostat to determine total watt hours consumed. This is where my concern regarding a battery's capacity varying with different load profiles comes into play. If desulphation and pulse plating effects cause the battery to have an increased capacity when the load is pulsed above that capacity observed under a DC load, it would be improper to use the disharge curve as you do to determine watt hours.
To expect similar discharge curves from a battery under a given load for 8 hours and that same battery under that same given load for the same duration but with a desulphator circuit attached would be questionable at best.
Regarding the oscillations being different with a DC supply, have you attempted to produce the battery's equivalent circuit at the output of your supply? That is, isolate the supply with an equivalent circuit that models the measured ESR, ESL and C of the battery? That may allow you to produce the same oscillations using the DC supply and perform further investigations using just the supply.
PW
@PW: Of course you are Harvey. Who else could you be? You are Harvey, I am Brian (or Bryan) .... the Red Queen has declared it to be so, therefore it is so.
Note that she also continues to lie and to libel us both, as well as .99, in letters she is sending to other people. The outrageous lie about .99's simulation, and the absurd continuing fascination with "pickles"..... reveal her mendacity and her madness for all to see.
TK:
It looks like the domain name servers don't like Rosemary's baby anymore. Could it be temporary or something else??? Knocked off the air by a Zipon-Neutron bomb?
Gmeast:
When you speak to PW you are graced with the presence of Zen Master electronics and measurement guru. Harvey is not in the same league as PW at all. Rosie Posie can't qualify anyone because she has no knowledge base to work with.
I have only skimmed at some recent postings on this thread and I can tell you that you should take every single word that PW says very very seriously.
MileHigh
Quote from: MileHigh on April 30, 2013, 12:45:47 AM
TK:
It looks like the domain name servers don't like Rosemary's baby anymore. Could it be temporary or something else??? Knocked off the air by a Zipon-Neutron bomb?
Gmeast:
When you speak to PW you are graced with the presence of Zen Master electronics and measurement guru. Harvey is not in the same league as PW at all. Rosie Posie can't qualify anyone because she has no knowledge base to work with.
I have only skimmed at some recent postings on this thread and I can tell you that you should take every single word that PW says very very seriously.
MileHigh
Interesting. I also see an error now when I try to go there via my proxy server. It has happened in the last couple of hours, whatever it is. Maybe she finally got the letter from Bryan Little's lawyers.
You do know that she took down from YouTube the famous "demo" video-of-lies, right?
Quote from: picowatt on April 29, 2013, 11:32:45 PM
Greg,
I have only been trying to understand and discuss your experiments and methodologies. Why all the attitude? You seem to think you know what I think, great, where did you get that from? I am not Harvey, I don't speak French, I am not "anti-OU" nor am I refusing to accept anything. But, that is another matter. In any event, is anyone that questions your methods or attempts to understand them subject to such behavior? If so, good luck with that. It will surely not assist you in gaining further acceptance. I, however, am doing my best to ignore it...
I have looked over your slides for some time. I believe I follow and can agree with your methods up to a certain point.
From your own measurements, it looks like you are saying that to produce the same deltaT as the BH, it takes 3.16watts from the DC supply. Using the rheostat to produce the same Vdrop at SH3 requires 3.09watts. Although I think there may be an error with regard to how you are determining the contribution of of the gate drive to the circuit, using your figure of .117watts, then apparently, from your numbers, the BH is using 3.2 watts to produce an output of 3.15 watts. Even that just under OU efficiency is, it itself, amazing, but as I said, I believe you need to study the contribution of the gate driver a bit further.
When you disconnect the drain Vsupply, the capacitances in the FET increase to their extreme maximums, i.e., the driver sees a maximum capacitive loading. When you reapply drain voltage, the FET capacitances reduce substantially. Merely comparing and using only the difference in driver power between those two conditions may not provide an accurate assessment of the driver's power contribution to the circuit overall.
But, where I am having the most difficulty accepting your results is when you use the time it takes the battery to discharge to 27.44V when loaded with the rheostat to determine total watt hours consumed. This is where my concern regarding a battery's capacity varying with different load profiles comes into play. If desulphation and pulse plating effects cause the battery to have an increased capacity when the load is pulsed above that capacity observed under a DC load, it would be improper to use the disharge curve as you do to determine watt hours.
To expect similar discharge curves from a battery under a given load for 8 hours and that same battery under that same given load for the same duration but with a desulphator circuit attached would be questionable at best.
Regarding the oscillations being different with a DC supply, have you attempted to produce the battery's equivalent circuit at the output of your supply? That is, isolate the supply with an equivalent circuit that models the measured ESR, ESL and C of the battery? That may allow you to produce the same oscillations using the DC supply and perform further investigations using just the supply.
PW
Hi PW,
Read it again! There is NO Delta-T in either of the battery draw-downs. That's not in the data ... RL is not even hooked up during those draw-downs. Look ate the circuit diagrams.
FYI: I received a PM from a 12-year old that completely understands this, but had to explain it to her science teacher ... who finally 'got it'.
You are in error on the gate driver issue. I spent many hours testing this because of your insistence that it be included. It turns out to have a constant overhead whether it's driving the gate or not and therefore, for the sake of pure research, that overhead can be excluded just as the PWM's can. That's not even a point for discussion.
As far as everything else goes, as I said, "I'm done sharing here".
Regards,
Greg
Quote from: TinselKoala on April 30, 2013, 01:05:24 AM
Interesting. I also see an error now when I try to go there via my proxy server. It has happened in the last couple of hours, whatever it is. Maybe she finally got the letter from Bryan Little's lawyers.
You do know that she took down from YouTube the famous "demo" video-of-lies, right?
Hi TK,
Thanks (I guess) for your input. Rosie's website thing is a mystery because I was just there, my access is unhindered, my permissions remain active ... so who knows.
I don't know any thing about any "Harvey", any measurement 'guru' or anyone else. In fact, I don't really care about any "demo" video-of-lies'. I'm not 'feeding' anyone's agenda but my own ... which is to share my research.
I can't take anything PW has said seriously because in our last exchange, it is clear he has yet to actually and carefully study the presentation.
Example: he somehow determined that there were Delta-T Fixture measurements as part of my draw-down data. The Delta-T Fixture is not even hooked up in those tests and the circuit diagram shows that. As well I have shown that the gate driver has a constant overhead whether or not it is driving the gate and THAT overhead can be excluded from the performance calculations just as the PWM's can.
PW is not the final word, you are no the final word. I'm not so sure what any of you who are anti-Ainslie are actually up to. All I know is that I started this thread to share my findings (WHICH IS WHAT THESE FORUMS ARE FOR!), NOT to have it be an announcement platform from where to launch something about Rosie's site having some technical problems.
DON'T HIJACK THIS THREAD. But it doesn't matter anyway, I'm done sharing here.
Regards,
Greg
WHERE IN THE HELL ARE YOU HARTMAN? TAKE CARE OF THIS HIJACKING CRAP NOW! YOU ARE STILL PERMITTING THE SAME NON-PRODUCTIVE BEHAVIOR ON THE PART OF:
"THE SQUAD"
Let me moderate my own thread so I can delete obvious B.S.
Greg
Quote from: gmeast on April 30, 2013, 09:28:20 AM
Hi PW,
Read it again! There is NO Delta-T in either of the battery draw-downs. That's not in the data ... RL is not even hooked up during those draw-downs. Look ate the circuit diagrams.
FYI: I received a PM from a 12-year old that completely understands this, but had to explain it to her science teacher ... who finally 'got it'.
You are in error on the gate driver issue. I spent many hours testing this because of your insistence that it be included. It turns out to have a constant overhead whether it's driving the gate or not and therefore, for the sake of pure research, that overhead can be excluded just as the PWM's can. That's not even a point for discussion.
As far as everything else goes, as I said, "I'm done sharing here".
Regards,
Greg
Greg,
I never said that the delta T's were used with regard to drawdown (except for the initial BH run).
Based on your numbers, up to the point where you do compare drawdowns, you demonstrate an efficiency of about 2.5% OU, or just under OU if your calculated driver contributions are added in.
This is determined by comparing the power required from the DC supply to produce same deltaT as the BH, which is used as "output power", to the power calculated from either the rheostat test or the measured SH3 voltage and the average Vbatt, which is used as "input power".
Am I correct so far?
The drawdown numbers from the rheostat test are then used to determine the time required for Vbatt to cross the BH run's end voltage, and the length of time at which that occurs is then used to calculate the watt hours consumed by the BH.
Is this correct?
PW
Quote from: picowatt on April 30, 2013, 11:39:12 AM
Greg,
I never said that the delta T's were used with regard to drawdown (except for the initial BH run).
Based on your numbers, up to the point where you do compare drawdowns, you demonstrate an efficiency of about 2.5% OU, or just under OU if your calculated driver contributions are added in.
This is determined by comparing the power required from the DC supply to produce same deltaT as the BH, which is used as "output power", to the power calculated from either the rheostat test or the measured SH3 voltage and the average Vbatt, which is used as "input power".
Am I correct so far?
The drawdown numbers from the rheostat test are then used to determine the time required for Vbatt to cross the BH run's end voltage, and the length of time at which that occurs is then used to calculate the watt hours consumed by the BH.
Is this correct?
PW
You are simply wrong, and I'm done here.
Quote from: gmeast on April 30, 2013, 12:02:15 PM
You are simply wrong, and I'm done here.
Greg,
What part is wrong?
I have looked at the slide show several times. If I am wrong, please tell me where am I wrong.
Surely you would not want others to have the same misunderstanding that I am when looking at your slide show ...
PW
Quote from: picowatt on April 30, 2013, 01:25:54 PM
Greg,
What part is wrong?
I have looked at the slide show several times. If I am wrong, please tell me where am I wrong.
Surely you would not want others to have the same misunderstanding that I am when looking at your slide show ...
PW
But others don't.
Greg,
Again, this is what I get from the slide show:
(1). You run the BH (burst heater) and note the deltaT, SH3 voltage, and also log the battery discharge curve.
(2). To determine output power, you use the DC supply connected to Rload to produce the same deltaT at Rload as in (1) above and note the supply I and V. Using the supply I and V you calculate output power, which I believe was determined to be 3.16 watts (from memory, I don't have the slide video open right now)
(3). To determine input power, you use a rheostat in series with SH3 set to produce the same SH3 voltage as in (1) above (which I believe was 5.6mv) and calculate input power based on the total resistance used and the Vbatt average voltage. You also log the battery disharge curve over time while performing this rheostat test.
As alternate input power verification, you also use the current calculated from the SH3 voltage measured in (1) above multiplied by the average Vbatt voltage and note that the two methods agree closely. I believe that fom the input power tests and calculations you arrived at figures that were in close agreement, being 3.09 versus 3.099 watts or something like that, depending on the method used to determine input power.
At this point, by your your own meaurements and calculations, you demonstrate 3.16 watts of output using only 3.09 watts of input, which is just over unity by 2.5% or so. If your calculated driver contribution is added to the input power, then the figures shift slightly to 3.16 watts out for 3.20 watts in, which is just slightly under OU.
Am I correctly following your slide show up to this point?
PW
Quote from: picowatt on April 30, 2013, 01:46:18 PM
Greg,
Again, tis is what I get from the slide show:
1. You run the BH and note the deltaT, SH3 voltage, and log the battery discharge curve.
2. To determine output power you use the DC supply connected to Rload to produce the same deltaT at Rload as in (1) above and note the supply I and V. Using the supply I and V you calculate output power, which I believe was determined to be 3.16 watts (from memory, I don't have the slide video open right now)
3. To determine input power, you use a rheostat in series with SH3 set to produce the same SH3 voltage as in (1) above (which I believe was 5.6mv) and calculate input power based on the total resistance used and the Vbatt average voltage. You also log the battery disharge curve over time while performing this rheostat test.
As alternate input power verification, you also use the SH3 voltage from (1) above multiplied by the average Vbatt voltage and note that the two methods agree closely. I believe that from the input power tests/calculations you arrived at 3.09 versus 3.099 watts or something like that, depending on the method used to determine input power.
At this point, by your your own meaurements and calculations, you demonstrate 3.16 watts of output using only 3.09 watts of input, which is just over unity by 2.5% or so. If your calculated driver contribution is added to the input power, then the figures shift slightly to 3.16 watts out for 3.20 watts in, which is just slightly under OU.
Am I correctly following your slide show up to this point?
PW
This is where you have gone wrong. You cannot use REAL-TIME measures of Power to determine EFFICIENCIES of these things. You MUST use 'Energy'. The energy consumed from the batteries is 25.28 Watt-Hours heating RL (3.16W X 8-Hours). That heating over 8-hours drew the battery down .4V and at 8-hours the battery voltage was 27.44V. For the 1st draw-down, the same Starting Voltage after battery re-charge and stabilization, the Rheostat load drew the batteries down to the SAME 27.44V in 6.38 hours. The Rheostat load was 3.1Watts for ONLY 6.38 Hours for an Energy of only 19.78 Watt-Hours. The 5.6mV SH3 was simply a reference for adjusting the load rheostats. I could have used anything. I could have used 10mV and the batteries would have drawn down quicker, but the energy would still have been around 19.78 Watt-Hours. I could have used 3mV and the draw-down would have lasted longer than 8-hours, but still would have been around 19.78 Watt-Hours ... at the point where the batteries hit 27.44V.
The 2nd Rheostat load test simply used the ratio of the Energies from the first two tests to adjust the rheostats such that the starting and ending voltages were the same as the 1st test (the circuit test) ... 27.84V to 27.44V. 19.78Wh / 25.28Wh = 0.78 So 5.6mV (SH3) X 0.78 = 4.4mV for the new SH3 voltage drop and I adjusted the rheostats to produce that load. The 2nd draw-down test at 4.56mV(avg) (SH3) resulted in identical starting and ending battery voltages as the circuit test ... 27.84V to 27.445V. Then I measured the load rheostat resistance and calculated the power which was 2.52Watts .... THIS IS THE INPUT POWER. I then applied 2.52Watts DIRECTLY to RL and it produced a significantly lower Delta-T, and this simply proved everything out.
You are like everyone else that has assumed you can simply use poynty-head's PIN POUT nonsense crap for determining efficiency. YOU CANNOT USE REAL-TIME MEASURES OF POWER FOR THIS STUFF. YOU MUST USE MEASURES OF ENERGY!
The reason I know you haven't done any more than skim my presentation is that ALL OF WHAT I SAID ABOVE IS IN THAT SLIDE SHOW.
Take off your BLINDERS you guys.
Regards,
Greg
Quote from: gmeast on April 30, 2013, 02:50:30 PM
This is where you have gone wrong. You cannot use REAL-TIME measures of Power to determine EFFICIENCIES of these things. You MUST use 'Energy'. The energy consumed from the batteries is 25.28 Watt-Hours heating RL (3.16W X 8-Hours). That heating over 8-hours drew the battery down .4V and at 8-hours the battery voltage was 27.44V. For the 1st draw-down, the same Starting Voltage after battery re-charge and stabilization, the Rheostat load drew the batteries down to the SAME 27.44V in 6.38 hours. The Rheostat load was 3.1Watts for ONLY 6.38 Hours for an Energy of only 19.78 Watt-Hours. The 5.6mV SH3 was simply a reference for adjusting the load rheostats. I could have used anything. I could have used 10mV and the batteries would have drawn down quicker, but the energy would still have been around 19.78 Watt-Hours. I could have used 3mV and the draw-down would have lasted longer than 8-hours, but still would have been around 19.78 Watt-Hours ... at the point where the batteries hit 27.44V.
The 2nd Rheostat load test simply used the ratio of the Energies from the first two tests to adjust the rheostats such that the starting and ending voltages were the same as the 1st test (the circuit test) ... 27.84V to 27.44V. 19.78Wh / 25.28Wh = 0.78 So 5.6mV (SH3) X 0.78 = 4.4mV for the new SH3 voltage drop and I adjusted the rheostats to produce that load. The 2nd draw-down test at 4.56mV(avg) (SH3) resulted in identical starting and ending battery voltages as the circuit test ... 27.84V to 27.445V. Then I measured the load rheostat resistance and calculated the power which was 2.52Watts .... THIS IS THE INPUT POWER. I then applied 2.52Watts DIRECTLY to RL and it produced a significantly lower Delta-T, and this simply proved everything out.
You are like everyone else that has assumed you can simply use poynty-head's PIN POUT nonsense crap for determining efficiency. YOU CANNOT USE REAL-TIME MEASURES OF POWER FOR THIS STUFF. YOU MUST USE MEASURES OF ENERGY!
The reason I know you haven't done any more than skim my presentation is that ALL OF WHAT I SAID ABOVE IS IN THAT SLIDE SHOW.
Take off your BLINDERS you guys.
Regards,
Greg
Greg, I understand what you are saying with regard to your watt hour calculations, which was the next step beyond those I covered in my previous post.
But, what I was asking, to make sure I understood your methods, was, up to the points discussed in my previous post, did I follow your methods correctly and the measurements you made? (understanding, of course, that you did not use them for any in/out comparisons at that point)
PW
Using CSRs to measure power in certain types of pulsing circuits returns inaccurate and unreliable results. It's fine for sine waves, nicely packaged square waves, saw-tooth waves, triangle waves, but not the types observed here.
Quote from: picowatt on April 30, 2013, 03:01:45 PM
Greg, I understand what you are saying with regard to your watt hour calculations, which was the next step beyond those I covered in my previous post.
But, what I was asking, to make sure I understood your methods, was, up to the points discussed in my previous post, did I follow your methods correctly and the measurements you made? (understanding, of course, that you did not use them for any in/out comparisons at that point)
PW
Good ... then we're done.
Quote from: gmeast on April 30, 2013, 03:20:23 PM
Using CSRs to measure power in certain types of pulsing circuits returns inaccurate and unreliable results. It's fine for sine waves, nicely packaged square waves, saw-tooth waves, triangle waves, but not the types observed here.
Greg,
Please just answer yes or no as to whether I was following along correctly up to the points in my previous post.
You say "I don't get it", so I am trying. Did I follow along correctly up to that point?
As to using CSR's with complex waveforms, it seems you relied on the voltage measured at SH3 to determine input power, if I have followed along correctly up to the point in my previous post.
PW
Quote from: picowatt on April 30, 2013, 03:25:16 PM
Greg,
Please just answer yes or no as to whether I was following along correctly up to the points in my previous post.
You say "I don't get it", so I am trying. Did I follow along correctly up to that point?
As to using CSR's with complex waveforms, it seems you relied on the voltage measured at SH3 to determine input power, if I have followed along correctly up to the point in my previous post.
PW
Generally "yes" up until your last statements (3). I said the draw-down test will give you 'some but not all' of the information you need to determine the input 'power' of the circuit. I did not say that this 1st draw-down test would give you THE input power. Also you sited the 3.09, 3.099 values. Those only relate to the DC draw-down in which case the SH3 values are extremely accurate and reliable. So you were only correct in your understanding up until (3).
It's clear that you are still hung up on using real-time power values to determine efficiencies ... and you are wrong to assume that's correct with these types of circuits.
So, 'no', I'm afraid you still don't 'get it' and I've answered the "why" since that posting and posted jpegs from the slide show.
gme
Quote from: gmeast on April 30, 2013, 05:49:37 PM
Generally "yes" up until your last statements (3). I said the draw-down test will give you 'some but not all' of the information you need to determine the input 'power' of the circuit. I did not say that this 1st draw-down test would give you THE input power. Also you sited the 3.09, 3.099 values. Those only relate to the DC draw-down in which case the SH3 values are extremely accurate and reliable. So you were only correct in your understanding up until (3).
It's clear that you are still hung up on using real-time power values to determine efficiencies ... and you are wrong to assume that's correct with these types of circuits.
So, 'no', I'm afraid you still don't 'get it' and I've answered the "why" since that posting and posted jpegs from the slide show.
gme
Greg,
So, all is well regarding my understanding of step 1 and 2. You seem to be saying I am incorrect about step 3, but are you actually saying that my understanding of the tests and measurements are correct, but only that you do not want that data to be used to calculate efficiency? At least let me know I am grasping your test methods correctly.
So,again, is my understanding correct regarding your methods and measurements in step 1 thru 3?
If I am correct, but you are actually just "hung up" on not using any of that data "as is" to measure efficiency, fine, just say so and I can move on to step 4.
So, as I understand it, in step 4 you use the step 3 drawdown data (using the rheostat set for 5.6mv across SH3) and note the time it takes the battery to discharge to the step 1 end voltage (27.44V I think it was). The step 3 drawdown reached 27.44V in less time than the step 1 test, so you use the shorter time interval of that drawdown to calculate the watt hours used in the step 1 BH circuit run.
Is this correct?
PW
Another important note here regarding the Gate Driver: The power the gate driver is drawing either in-circuit or on its own support battery is NOT supplementing the HEATING of RL. Aside from its overhead from 'just being there', as it does its 'driver thing' the increase in power draw is from doing what it has to do to source and sink the current required in charging and discharging the gate capacitance in order to maintain the required gate charge voltage whether turning it 'ON' or 'OFF'. This 'sourcing and sinking' is isolated from the rest of the MOSFET except for a leakage of 100nA ... 100 billionths of an Amp ... the published data for the UCC2732x drivers.
So, this is the reason why the gate driver's power is not considered in Exploring the Inductive Resistor heater. The ultimate goal is to have the components self-oscillate instead of relying on controlling circuitry such as a PWM and Driver arrangement ... something I've almost figured out how to do ... but not quite.
Everything is contained in that slide show, and in enough detail, to answer any and all questions. That's the reason I made it. No more 'on the witness stand' "yes" or "no" questions ... capiche?
Regards,
Greg
Greg,
Is my understanding correct up to and inclusive of step 4 in my previous post?
PW
Quote from: gmeast on April 30, 2013, 07:16:16 PM
Another important note here regarding the Gate Driver: The power the gate driver is drawing either in-circuit or on its own support battery is NOT supplementing the HEATING of RL. Aside from its overhead from 'just being there', as it does its 'driver thing' the increase in power draw is from doing what it has to do to source and sink the current required in charging and discharging the gate capacitance in order to maintain the required gate charge voltage whether turning it 'ON' or 'OFF'. This 'sourcing and sinking' is isolated from the rest of the MOSFET except for a leakage of 100nA ... 100 billionths of an Amp ... the published data for the UCC2732x drivers.
So, this is the reason why the gate driver's power is not considered in Exploring the Inductive Resistor heater. The ultimate goal is to have the components self-oscillate instead of relying on controlling circuitry such as a PWM and Driver arrangement ... something I've almost figured out how to do ... but not quite.
Everything is contained in that slide show, and in enough detail, to answer any and all questions. That's the reason I made it. No more 'on the witness stand' "yes" or "no" questions ... capiche?
Regards,
Greg
Greg,
Your comments regarding the operation of a FET fall short of the true picture. While it is true that under static DC conditions the gate leakage is typically very low, in the pico or nanoamp range, the dynamic conditions during turn on and turn off are quite different.
As you say, to turn the FET on or off the gate capacitance(s) must be charged or discharged. To do this quickly requires a significant amount of current. The driver you are using is capable of sourcing/sinking 4 amps for this purpose, but there are gate drivers capable of 20-40amps also available.
The most significant capacitances at the gate are the gate to source capacitance, Cgs, and the gate to drain capacitance Cgd. Cgs is almost always the largest, and it is equivalent to a capacitor connected between the gate and source of the FET. Cgd is similarly equivalent to a capacitor connected between the gate and drain of the FET. To charge Cgs from zero to 12 volts, the charging current must flow between the gate and source. Simultaneously, to charge Cgd, charging current must flow between the gate and drain. During hi speed switching, these charging currents can have very large peak currents.
If you are using the PG50, Cgs and Cgd are fairly large values. From the data sheet you will see that Cgs and Cgd vary with the drain to source voltage, Vds, and when the Vds is near zero volts (as it is when the FET is turned on in your circuit for example), these capacitances are at their maximum value. As the gate driver's waveforms typically contain a large amount of high frequency content (i.e., fast rise and fall times), the reactance of Cgs and Cgd can be quite low at these high frequencies causing large peak currents (many amps) to flow from the gate to both source and drain. This is why a gate driver is typically used, to provide the amps of current needed to charge these capacitances rapidly.
In your circuit, however, you are limiting the gate drive current by using the 100 ohm resistor between the driver and the gate. Assuming there is only small amount of stray capacitance between the driver output and the gate, whose reactance would effectively be in parallel with the 100 ohm gate resistor, your 100 ohm resistor limits the peak current available to charge the gate capacitances and hence greatly slows your on/off switching times as compared to what your driver is capable of.
It is a shame you are unwilling to discuss your circuit further.
PW
Quote from: picowatt on April 30, 2013, 08:00:56 PM
Greg,
Your comments regarding the operation of a FET fall short of the true picture. While it is true that under static DC conditions the gate leakage is typically very low, in the pico or nanoamp range, the dynamic conditions during turn on and turn off are quite different.
As you say, to turn the FET on or off the gate capacitance(s) must be charged or discharged. To do this quickly requires a significant amount of current. The driver you are using is capable of sourcing/sinking 4 amps for this purpose, but there are gate drivers capable of 20-40amps also available.
The most significant capacitances at the gate are the gate to source capacitance, Cgs, and the gate to drain capacitance Cgd. Cgs is almost always the largest, and it is equivalent to a capacitor connected between the gate and source of the FET. Cgd is similarly equivalent to a capacitor connected between the gate and drain of the FET. To charge Cgs from zero to 12 volts, the charging current must flow between the gate and source. Simultaneously, to charge Cgd, charging current must flow between the gate and drain. During hi speed switching, these charging currents can have very large peak currents.
If you are using the PG50, Cgs and Cgd are fairly large values. From the data sheet you will see that Cgs and Cgd vary with the drain to source voltage, Vds, and when the Vds is near zero volts (as it is when the FET is turned on in your circuit for example), these capacitances are at their maximum value. As the gate driver's waveforms typically contain a large amount of high frequency content (i.e., fast rise and fall times), the reactance of Cgs and Cgd can be quite low at these high frequencies causing large peak currents (many amps) to flow from the gate to both source and drain. This is why a gate driver is typically used, to provide the amps of current needed to charge these capacitances rapidly.
In your circuit, however, you are limiting the gate drive current by using the 100 ohm resistor between the driver and the gate. Assuming there is only small amount of stray capacitance between the driver output and the gate, whose reactance would effectively be in parallel with the 100 ohm gate resistor, your 100 ohm resistor limits the peak current available to charge the gate capacitances and hence greatly slows your on/off switching times as compared to what your driver is capable of.
It is a shame you are unwilling to discuss your circuit further.
PW
The driver I'm using supplies as much as 9Amps required to negotiate the challenges of the Muiller Plateau. The current is needed as I said it is and does NOT pass through to the drain. As I said the PG50 can only leak 100nA. My assessment does not fall short. I have had several good conversations with TI on the subject of the 321 firing the PG50 and you're all wet on this one. It's amazing how you clowns try and muddy the water when it comes to this Inductive Resistor Heater.
It's amazing how I start this thread and ass holes like you hikack it. FUCK YOU AND YOU TOO HARTMAN FOR LETTING THIS SHIT CONTINUE.
Quote from: gmeast on April 30, 2013, 02:50:30 PM
This is where you have gone wrong. You cannot use REAL-TIME measures of Power to determine EFFICIENCIES of these things. You MUST use 'Energy'. The energy consumed from the batteries is 25.28 Watt-Hours heating RL (3.16W X 8-Hours). That heating over 8-hours drew the battery down .4V and at 8-hours the battery voltage was 27.44V. For the 1st draw-down, the same Starting Voltage after battery re-charge and stabilization, the Rheostat load drew the batteries down to the SAME 27.44V in 6.38 hours. The Rheostat load was 3.1Watts for ONLY 6.38 Hours for an Energy of only 19.78 Watt-Hours. The 5.6mV SH3 was simply a reference for adjusting the load rheostats. I could have used anything. I could have used 10mV and the batteries would have drawn down quicker, but the energy would still have been around 19.78 Watt-Hours. I could have used 3mV and the draw-down would have lasted longer than 8-hours, but still would have been around 19.78 Watt-Hours ... at the point where the batteries hit 27.44V.
The 2nd Rheostat load test simply used the ratio of the Energies from the first two tests to adjust the rheostats such that the starting and ending voltages were the same as the 1st test (the circuit test) ... 27.84V to 27.44V. 19.78Wh / 25.28Wh = 0.78 So 5.6mV (SH3) X 0.78 = 4.4mV for the new SH3 voltage drop and I adjusted the rheostats to produce that load. The 2nd draw-down test at 4.56mV(avg) (SH3) resulted in identical starting and ending battery voltages as the circuit test ... 27.84V to 27.445V. Then I measured the load rheostat resistance and calculated the power which was 2.52Watts .... THIS IS THE INPUT POWER. I then applied 2.52Watts DIRECTLY to RL and it produced a significantly lower Delta-T, and this simply proved everything out.
You are like everyone else that has assumed you can simply use poynty-head's PIN POUT nonsense crap for determining efficiency. YOU CANNOT USE REAL-TIME MEASURES OF POWER FOR THIS STUFF. YOU MUST USE MEASURES OF ENERGY!
The reason I know you haven't done any more than skim my presentation is that ALL OF WHAT I SAID ABOVE IS IN THAT SLIDE SHOW.
Take off your BLINDERS you guys.
Regards,
Greg
Greg,
I have reread the above again, and I believe I now understand your method. It is actually quite brilliant.
However, does this not presuppose that the batteries have the same amp hour rating, and hence draw down characteristics, with a purely resistive load versus a pulsed load? (as per my original concern regarding load profiles versus battery capacity)
As an analogy, suppose you have a flooded lead acid battery on a motorized table that gently rocks the battery so that the electrolyte is being stirred constantly. A load resistor is applied and the voltage and current is monitored and its draw down from a start and stop voltage is noted over a measured time period. The same test with the same load is again performed but this time the "stirring table" is turned off. Would you expect the battery to necessarily measure the same capacity in both tests?
As well, suppose a pulsing desulpator is connected to a lead acid battery driving a resistive load so that sulphate crystals formed during discharge are maintained at a smaller size, and as well, the effects of pulse plating produce a finer grain structure, that is, a greater conductive area, during discharge, so that the battery appears to have a larger amp hour rating than it does when similarly loaded without the desulphator connected. Would this prove that the desulphator produces overunity or would it only prove that the capacity of a lead acid battery can be increased by pulsing the battery during discharge?
I bring this up because for many years there have been various claims of overunity with pulsed circuits, but for some reason the "overunity" always requires a battery. And a lead acid battery appears to be, for the most part, the most popular battery chemistry used.
This is why I asked if you had ever attempted to operate your circuit using only a well filtered supply to determine if the circuit itself is truly overunity or if the observed effect is moreso related to the battery having a different capacity under different load profiles.
You say that your circuit will not oscillate when operating from the DC supply. This could be further investigated by installing a network between the supply and battery that models the measured equivalent series resistance, inductance, and capacitance of your batteries. If necessary, the supply can be isolated at AC by installing inductors between the supply and network. Doing so might allow you to operate your circuit from the supply and make your input measurements at DC. This would assist in determining if the observed OU is due to the operation of the circuit, or moreso, to the increase in battery capacity under a pulsed load profile.
As an aside, you seem to have a certain disdain for modern test equipment regarding the ability of any equipment being able to measure the voltage at SH3 because of its "complex" waveform. The waveform there is not all that complex nor particularly fast, and direct measurement there, given a very low inductance CSR (due to your use of .05ohms), can be accurately performed. I believe you stated that you measured the SH3 voltage using both a scope and .99's multimeter approach and had close agreement with both methods. So why then, do you dismiss that measurement out of hand as inaccurate?
In any event, as it appears that your input power determined by direct measurement and by use of the drawdown method are in significant disagreement with each other, would you not at least agree that a third method is in order to determine which method of measurement is more accurate?
Thank-you for your patience... us old guy's are slow on the uptake (just wait till you get there!)
PW
Gmeast:
i belive PW is a seasoned retired analog electrical design engineer with perhaps 40 years of design and test experience under his belt. That's just my gut feel I am not asking him to divulge any details about himself. He has more knowledge about this stuff on the tip of his pinky then you will ever have in your entire life. He could figure skate circles around you backwards and blindfolded. You are an insignificant pea in a darkened pod and PW is a mountain under the stars.
PW is being very polite to you and you act like a rude, vulgar, ignorant and stupid wanker. It's pathetic.
Go ask Rosie, even she knows about AC current coupling through the gate-source capacitance of a MOSFET. You can have a very nice and very uncomfortable conversation with her. She will stroke you and tell you how brilliant you are.
Jackass.
MileHigh
P.S.: Everybody I apologize for my strong comments and I won't repeat them again. I just wanted to set the record straight.
Quote from: gmeast on April 30, 2013, 11:17:14 PM
The driver I'm using supplies as much as 9Amps required to negotiate the challenges of the Muiller Plateau. The current is needed as I said it is and does NOT pass through to the drain. As I said the PG50 can only leak 100nA. My assessment does not fall short. I have had several good conversations with TI on the subject of the 321 firing the PG50 and you're all wet on this one. It's amazing how you clowns try and muddy the water when it comes to this Inductive Resistor Heater.
It's amazing how I start this thread and ass holes like you hikack it. FUCK YOU AND YOU TOO HARTMAN FOR LETTING THIS SHIT CONTINUE.
Greg,
You only gave your driver part number as a UCC2732x. As you are limiting the gate current via a 100R gate resistor, I just picked the 4 amp version as I thought the 9 amp version would just be even further overkill.
In any event, even with the 9 amp version, that 100 ohm resistor limits gate drive to 120ma at 12volts. So either way, 4 amp or 9amp, with the 100R in series to the gate, the driver never approaches anywhere near its drive current rating nor does the FET's switching speed ever come close to the driver's switching speed. It would take 100 volts across that 100R at the gate just to produce 1 amp of gate drive current. So, just because the specs say your driver can switch in nanoseconds, your FET is switching way, way slower than that.
Irregardless of how fast the driver switches, it is the 100R and the gate capacitance time constant that is determining the FET's switchng speed. However, that 100R likely allows for, in concert with Cgs and Cgd, a bit of positive feedback making the circuit less stable and allowing it to ring as it does.
But feel free to ask TI what your FET's switching speed would be if you use their driver and whatever FET you are using with a 100R in series between the driver output and the FET gate. They can tell you how slow it would be, but will probably ask you why you even need a high current driver when such a low gate drive current results from using that 100R.
PW
Wouldn't you love to see Gmeast standing next to his demonstration at a poster session at a physics conference? With his attitude, he'd get arrested before the first coffee break. Or punched in the nose.
Using a 100R resistor in series with a gate driver chip is just laughable, especially with a mosfet with such high internal capacitances as the PG50. It's like driving a Ferrari with the throttle limited to 10 percent by sticking a brick under the accelerator pedal.
Well, maybe it can't go very fast... .but it's a FERRARI, stupid, so fuck you and where are the moderators, I want to whine and cry instead of facing the issues. And instead of demonstrating that different discharge schedules produce different "total energy content" values for used lead-acid batteries -- something that has been known for some time -- the results presented MUST indicate OU performance.... because that is what GMeast set out to +prove+.
But of course the Red Queen has defined Gmeast's "work" as a replication of her claims...... even though the components are different, the circuit is different, the waveforms are different, the operation is different, the behaviour is different, the testing is different, and even the claims made are different.
While at the same time, true replications that use the same components as Ainslie, in the same (several different) circuits claimed by Ainslie, with the same waveforms and behaviour as Ainslie's circuit, tested by the same methods that Ainslie mis-uses, and that give the _same objective results_ as Ainslie's...... are dismissed with false statements and bogus insults by Ainslie.... because they clearly show that her own claims are wrong and are full of lies and misrepresentations.
It's amazing, isn't it? Proper use of "Pin, Pout" measurements made with inline CVRs have been used for many years and have resulted in the ability to design things like robot spacecraft that hit their targets and send pictures home from a billion miles away, things like digital storage oscilloscopes, home computers, and even things like efficient electric automobiles. But now we learn that this method isn't valid. And what's the evidence for that claim of invalidity? Why.... simply that the method doesn't give the results that GMeast desires. The "Pin, Pout" measurement shows ordinary underunity performance... so therefore it's invalid, because we already know this system is OU.
Quote from Greg "over there":
QuoteI mean what I said about those jack-offs over on OU. I've invested too much to shelve all of this. I will continue to test what I have and post anything significant here as I will on EF and Heredical since I have moderator privileges there at H. I am so surprised that Stefan Hartman is such a dolt as to allow the TKs and the poynty-heads and the picowatts to hijack all of OU. I wonder who's paying these pricks to sit on that site? Picowatt is so self absorbed in thinking he knows everything about MOSFETS ... it's amazing. I've talked to the engineers at TI about the UCC27321 MOSFET Gate driver driving the gate of the IRFPG50 mosfet, and they assure me that the only thing that can get through and add heating to RL is the 100nA as published in the PG50's data sheet (gate to drain leakage). They (at TI) will not answer questions (officially) about stuff like this, but will get back to you 'unofficially' in time ... like a month. picowatt has got (loser) TK convinced of his authority on these topics and it's laughable. I have found out that they contacted many (so-called) circuit experts to attest that the gate driver is adding heating to the 'element'. They just kept fishing until one agreed to attest to it. One that wouldn't attest to it contacted me today (of all times). He wishes NOT to get embroiled in any of this and I respect that.
I hate "free-energy"!
Greg
Such a nice young man, and what a load of crap regarding the operation of his MOSFET (and some paranoid BS about contacting "circuit engineers")
For static DC conditions, it is true that only picoamps or nanoamps of current typically flow thru the gate. This is the gate's DC leakage current value. The amount of gate leakage current that flows is both voltage and temperature dependent (and given in data sheets).
However, the gate of a FET/MOSFET appears to the outside world as a capacitor that must be charged or discharged to turn the FET on or off. There are several "capacitances" related to the gate capacitance, but the primary gate capacitances appear as two variable capacitors connected between the gate and source (Cgs) and the gate and drain (Cgd). The value of these two variable capacitors varies as the voltage between the drain and source (Vds) varies. Typically the values of Cgs and Cgd increase as Vds decreases. As well, the gate to source capacitance, Cgs, is typically the largest of the capacitances appearing at (connected to) the gate.
To turn the FET on or off requires charging or discharging these intrinsic capacitors to a voltage value above or below the gate turn on voltage. Because the ends of these "capacitors" opposite their gate connections are connected to the drain and source, the current required to charge or discharge these capacitors must flow between both the gate and source and the gate and drain.
The amount of current that flows thru these gate capacitances while charging or discharging them (turning the FET on or off), depends upon the value of the gate capacitance(s) and the rate at which these gate capacitances are being charged or discharged (how fast the MOSFET is being switched on/off).
To turn the MOSFET on and/or off very quickly, that is, to charge or discharge the gate capacitance very quickly, requires a large amount of current, and this current flows between the gate and both the source and drain. Because of its typically larger capacitance value, the greatest current flow during dynamic conditions is between the gate and source. However, during hi speed transitions, i.e., when high frequencies are present at the gate, very significant currents can flow thru both Cgs and Cgd.
A gate driver IC is used to drive the gate when fast switchng is desired. A gate driver IC can typically sink/source several amps to tens of amps to charge and discharge the gate capacitances very quickly. A gate driver IC also typically has a very low output impedance to reduce losses during high peak currents at its output. The current the driver IC drives the gate with during dynamic conditions flows between the gate and source, and as well, between the gate and drain. Large currents only flow during dynamic conditions, wherein the voltage at the gate is changing and are rate dependent. These currents can be tens of amps being dependent upon the desired/designed switching speed, the MOSFET gate capacitance, any resistance or reactance in series with the driver output and gate, and the driver's output currentcapability. Once a stable DC voltage is reached at the gate, gate current, i.e., gate leakage current, reduces to a very small value, typically being picoamps or nanoamps depending upon the device.
Greg is using a 100 ohm resistor between his driver and gate. Assuming the 100R resistor is low inductance and that there is little stray capacitance between the driver output and the MOSFET gate (whose reactance would be in parallel with the 100R resistor), irregardless of the driver IC's 9 amp current capability and very fast switching times, at 12volts, the 100R resistor limits his peak gate current to 120ma. This limited current available to drive the gate (i.e., charge/discharge the gate) greatly reduces the rate (speed) at which his MOSFET can switch. His diriver may be transitioning very quickly, but the rate at which the gate voltage can charge or discharge the gate capacitnce is dependent upon the time constant created by the 100R resistor in concert with the value of the gate capacitance.
It is also very likely that, in concert with Cgs and Cgd, the 100R resistor allows for a bit of positive feedback to the gate which assists in causing the circuit to ring as it does.
Greg, please feel free to ask any EE, at TI or elsewhere, if anything I have written above in my somewhat "simplistic" description of MOSFET operation is untrue. I am quite confident you will be unable to find any EE that would disagree with anything I have written above.
Possibly, by adjusting your attitude a bit and realizing that it may actually be YOU who is acting like the "know it all", you might learn a bit more than what your apparently closed mind is allowing.
Good luck...
PW
Greg,
You just have to ask yourself, if only nanoampers of current ever flow thru the gate, why on Earth would a gate driver capable of 9 amps of drive current ever be required? And when all that current from a gate driver does flow into or out of the gate, where exactly does it go? There are only two additional terminals on the MOSFET thru which to complete the circuit...
Think about it...
PW
It's pretty clear that Gmeast doesn't have the knowledge of components and how they behave, that one might expect a "free energy" experimenter to have.
Here's a video that I published last July, illustrating in a very simple manner that the gate-drain and gate-source capacitances CAN and DO pass substantial currents, when the mosfet is actually operating in a circuit, or when it's not. This is part of a series of 10 or so videos that severally and individually refute several of the absurd claims that Ainslie makes about mosfets and function generators and circuit behaviour in general, and now I see that they also refute Gmeast's misconceptions about mosfets and how to drive them and what happens when you do.
I think it's completely laughable that he spews his childish insults and lies and misinformation about me, and about PW and others, when he apparently doesn't even understand the basics of circuit performance, circuit measurement, or how a mosfet even works.
http://www.youtube.com/watch?v=WzUcx3haZbA (http://www.youtube.com/watch?v=WzUcx3haZbA)
ETA: This series of ten or so "MOSFETs... How Do They Work?" videos was made in an effort to educate Rosemary Ainslie about her own circuit's performance and to illustrate that the absurd claims she made are just that: absurd. (Claims that a function generator can't act as a power source or pass current from an external source from its "probe" to its "terminal" to use her terms; claims like Gmeast's that a mosfet can't pass current when it is "off"; claims that a mosfet is strictly a "switch" and can't act in a linear conductance mode being partially on; misconceptions about the nature of the gate charge that turns a mosfet on and off; and etc.) Ainslie promised long ago to review these videos and refute my demonstrations point by point.... and we are still waiting for those refutations. No doubt I overestimated the level of the pitch.... instead of tenth grade level I should have pitched the demos at sixth grade comprehension level, using little cartoons, fuzzy animal toys and words of single syllables. Then perhaps she could have followed along.
Gmeast, of course, will respond, if at all, with more insults, and will still try to deny the obvious refutation of his absurd "nanoamps" claim.
Quote from: picowatt on April 30, 2013, 11:37:18 PM
Greg,
I have reread the above again, and I believe I now understand your method. It is actually quite brilliant.
However, does this not presuppose that the batteries have the same amp hour rating, and hence draw down characteristics, with a purely resistive load versus a pulsed load? (as per my original concern regarding load profiles versus battery capacity)
As an analogy, suppose you have a flooded lead acid battery on a motorized table that gently rocks the battery so that the electrolyte is being stirred constantly. A load resistor is applied and the voltage and current is monitored and its draw down from a start and stop voltage is noted over a measured time period. The same test with the same load is again performed but this time the "stirring table" is turned off. Would you expect the battery to necessarily measure the same capacity in both tests?
As well, suppose a pulsing desulpator is connected to a lead acid battery driving a resistive load so that sulphate crystals formed during discharge are maintained at a smaller size, and as well, the effects of pulse plating produce a finer grain structure, that is, a greater conductive area, during discharge, so that the battery appears to have a larger amp hour rating than it does when similarly loaded without the desulphator connected. Would this prove that the desulphator produces overunity or would it only prove that the capacity of a lead acid battery can be increased by pulsing the battery during discharge?
I bring this up because for many years there have been various claims of overunity with pulsed circuits, but for some reason the "overunity" always requires a battery. And a lead acid battery appears to be, for the most part, the most popular battery chemistry used.
This is why I asked if you had ever attempted to operate your circuit using only a well filtered supply to determine if the circuit itself is truly overunity or if the observed effect is moreso related to the battery having a different capacity under different load profiles.
You say that your circuit will not oscillate when operating from the DC supply. This could be further investigated by installing a network between the supply and battery that models the measured equivalent series resistance, inductance, and capacitance of your batteries. If necessary, the supply can be isolated at AC by installing inductors between the supply and network. Doing so might allow you to operate your circuit from the supply and make your input measurements at DC. This would assist in determining if the observed OU is due to the operation of the circuit, or moreso, to the increase in battery capacity under a pulsed load profile.
As an aside, you seem to have a certain disdain for modern test equipment regarding the ability of any equipment being able to measure the voltage at SH3 because of its "complex" waveform. The waveform there is not all that complex nor particularly fast, and direct measurement there, given a very low inductance CSR (due to your use of .05ohms), can be accurately performed. I believe you stated that you measured the SH3 voltage using both a scope and .99's multimeter approach and had close agreement with both methods. So why then, do you dismiss that measurement out of hand as inaccurate?
In any event, as it appears that your input power determined by direct measurement and by use of the drawdown method are in significant disagreement with each other, would you not at least agree that a third method is in order to determine which method of measurement is more accurate?
Thank-you for your patience... us old guy's are slow on the uptake (just wait till you get there!)
PW
Greg has done a great job in his calculations and very honest in his proof.
PW is doubtful and cautious, but he can not show any calculations, so there is no way to assess how significant is the deviation in energy calculation due to the factors listed by him. It could be that the conclusion from Greg's calculation still stands even if those factors are totally ignored.
It seems to me, the "increase in battery capacity under pulsed load" argument is conducing rather than deducing to Greg's calculations, because when a battery is increasing in its capacity while draining in its stored energy, then the voltage drop should be an exaggeration (because there is a bogus voltage drop due to increased capacity without change in stored energy). So under that observation, Greg's calculation of input to the original circuit (a pulsing load) is an exaggeration, and the OU is even more pronounced than what his calculations shows.
Quote from: TinselKoala on May 03, 2013, 02:42:43 AM
(snip)
Gmeast, of course, will respond, if at all, with more insults, and will still try to deny the obvious refutation of his absurd "nanoamps" claim.
So predictable. And so very vile. I think you struck a nerve, there, Picowatt.
@lanenal: You do realize that Gmeast is saying that the most current that can pass through a MOSFET gate to the drain or source is 100 nA, right? And that PW has explained from the circuit theory standpoint, and I have illustrated empirically in the video above, that the 100 nA claim is definitely not true for the kind of signal that is being applied to the gate by the gate driver which is capable of supplying 9 amps (if not restricted by the inline resistor.) Right?
Yet you see how he responds. He is refuted time after time but cannot deal with the refutations; instead he says "Fuck them all dead". That is his argument!
Where is the evidence for his paranoid claim that "circuit experts" were contacted by either myself or PW? There is none. We have referred him to the data sheet where the capacitances are clearly cited and we have shown him, or tried to show him, demonstrations and clear explanations. I personally have contacted no one on this issue.... because it is basic to mosfet design and usage. Watch my video and then see if Gmeast has an explanation that jives with his "100 nA" claim!
Hi Greg,
I have watched your video, and I have a question for you about the measured voltage drop over SH3 when the circuit is hooked up. Since the voltage should be oscilating, what you measured must be some sort of average voltage, right? Is it RMS? or is it something else?
great job, and thanks for sharing.
lanenal
Quote from: TinselKoala on May 03, 2013, 10:48:38 AM
@lanenal: You do realize that Gmeast is saying that the most current that can pass through a MOSFET gate to the drain or source is 100 nA, right? And that PW has explained from the circuit theory standpoint, and I have illustrated empirically in the video above, that the 100 nA claim is definitely not true for the kind of signal that is being applied to the gate by the gate driver which is capable of supplying 9 amps (if not restricted by the inline resistor.) Right?
Yet you see how he responds. He is refuted time after time but cannot deal with the refutations; instead he says "Fuck them all dead". That is his argument!
@TinselKoala: I am not a judge between you.
Greg in his first post already estimated the maximal energy contribution from the Gate Driver, which seems to be no big deal. As for the Gate current, Greg might be talking about gate leakage current, not the current to induce the voltage difference.
Transistors are gated by currents, while MOSFETS are gated by voltage difference, and to develop voltage difference there is a "hidden cap" (so to speak) at the Gate. The current depends on the frequency, max voltage difference, the current limiting resistor, and the capacity of the hidden cap. The gate leakage current is due to the imperfect insulation inside the "hidden" cap.
regards,
lanenal
Quote from: lanenal on May 03, 2013, 09:27:35 AM
Greg has done a great job in his calculations and very honest in his proof.
PW is doubtful and cautious, but he can not show any calculations, so there is no way to assess how significant is the deviation in energy calculation due to the factors listed by him. It could be that the conclusion from Greg's calculation still stands even if those factors are totally ignored.
It seems to me, the "increase in battery capacity under pulsed load" argument is conducing rather than deducing to Greg's calculations, because when a battery is increasing in its capacity while draining in its stored energy, then the voltage drop should be an exaggeration (because there is a bogus voltage drop due to increased capacity without change in stored energy). So under that observation, Greg's calculation of input to the original circuit (a pulsing load) is an exaggeration, and the OU is even more pronounced than what his calculations shows.
Lanenal,
I am not sure how you figure that an increase in capacity would somehow cause an exaggeration in Vdrop. If the capacity of a battery is increased, it will hold a given voltage for a longer period of time under a given load. There would be no "bogus voltage drop". The larger capacity would just allow the battery to sustain a given load for a longer period of time before reaching a given stop voltage.
As the bulk of the OU is only measureable when comparing the battery discharge characteristics of a lead acid battery under a dynamic versus a static load profile, it would seem wise to devise an alternate measurement method to reconcile the difference between this "battery rundown" method and his direct measurements.
As well, a further investigation of any possible FET driver contribution to the circuit should be considered. But, one must first understand that under dynamic conditions, the gate of a MOSFET can indeed pass more than nanoamperes into the circuit, before one can devise methods to quantify this.
In my original post, I merely stated that as it seems that many of these "OU" pulsed circuits require a battery, and that a lead acid battery is, for the most part, the chemistry of choice. The possibility exisits that the observed OU is moreso related to differences in battery capacity under static versus dynamic loading, than to some previously unobserved phenomenon related to inductors. Desulphators have been around for some time, and as well, the effects of pulse plating in the world of electroplating are also well known.
PW
Quote from: picowatt on May 03, 2013, 11:41:35 AM
Lanenal,
I am not sure how you figure that an increase in capacity would somehow cause an exaggeration in Vdrop. If the capacity of a battery is increased, it will hold a given voltage for a longer period of time under a given load. There would be no "bogus voltage drop". The larger capacity would just allow the battery to sustain a given load for a longer period of time before reaching a given stop voltage.
As the bulk of the OU is only measureable when comparing the battery discharge characteristics of a lead acid battery under a dynamic versus a static load profile, it would seem wise to devise an alternate measurement method to reconcile the difference between this "battery rundown" method and his direct measurements.
As well, a further investigation of any possible FET driver contribution to the circuit should be considered. But, one must first understand that under dynamic conditions, the gate of a MOSFET can indeed pass more than nanoamperes into the circuit, before one can devise methods to quantify this.
In my original post, I merely stated that as it seems that many of these "OU" pulsed circuits require a battery, and that a lead acid battery is, for the most part, the chemistry of choice, the possibility exists that the observed OU is moreso related to differences in battery capacity under static versus dynamic loading. Desulphators have been around for some time, and as well, the effects of pulse plating in the world of electroplating are also well known.
PW
PW, let me first explain the exaggeration thingy. If I understood you correctly, from what you have said, increased capacity would allow the battery to last longer for the same voltage drop under a given load. (BTW, this looks like an OU statement: if the battery increases its capacity under pulsing load of equivalent wattage, then it lasts longer, which means that the total energy output is larger as time X watt = output energy).
So if the stored energy remains constant (by the law of energy conservation), AND if the base voltage does not change, then the initial voltage must drop. In math:
Battery Capacity = Energy Output / Delta Voltage.
If the battery starts from full charge to full drain:
Battery Capacity = Total Stored Energy / (Initial Voltage - Base Voltage)
Now if Battery Capacity increases, and Total Stored Energy and Base Voltage remains constant, then Initial Voltage must drop to keep the identity.
regards,
lanenal
Edit: I am not making a strict argument, as batteries not not linear. If we assume that the point-wise Battery Capacity (as a function of current voltage) is bloated up by a constant factor, then the above argument can be easily transformed into integrations, with the conclusion unchanged.
Quote from: lanenal on May 03, 2013, 12:15:02 PM
PW, let me first explain the exaggeration thingy. If I understood you correctly, from what you have said, increased capacity would allow the battery to last longer for the same voltage drop under a given load. (BTW, this looks like an OU statement: if the battery increases its capacity under pulsing load of equivalent wattage, then it lasts longer, which means that the total energy output is larger as time X watt = output energy).
So if the stored energy remains constant (by the law of energy conservation), AND if the base voltage does not change, then the initial voltage must drop. In math:
Battery Capacity = Energy Output / Delta Voltage.
If the battery starts from full charge to full drain:
Battery Capacity = Total Stored Energy / (Initial Voltage - Base Voltage)
Now if Battery Capacity increases, and Total Stored Energy and Base Voltage remains constant, then Initial Voltage must drop to keep the identity.
regards,
lanenal
Lanenal,
That is exactly my point. You think that the example I gave appears to be OU.
If one connected a desulphator to a battery and applied a load and noted the discharge time to a given stop voltage, and then did the same using the same load but without the desulphator connected and noted a shorter discharge time, there would be those who would claim that the desulphator produces overunity.
But, who ever said that a battery is "unity" to begin with? A battery produces waste heat during both charge and discharge. Suppose the charge/discharge efficiency of a lead acid battery is 70%. Suppose the use of a desuplhator can increase this to 85%. Though better, still not OU.
Desulphators supposedly produce finer grained sulphate crystals effectively increasing plate area. As well, pulse and reverse pulse plating similarly are known to produce finer grained structures. Visualize large clumps of sulphate crystals forming at the plates as opposed to a much smoother, finer grained structure. As well as an increase in surface area, less resistive losses occur over the finer grains. This is, supposedly, how a desulphator functions, and the waveforms used to desulphate have much in common with the waveforms of these OU circuits.
Do a search for "battery desulphator". There are both construction articles as well as commercial units available.
I am not stating with any certainty that these effects are the reason for the observed OU, but, as these effects have been observed regarding lead acid batterys when being pulsed or reverse pulsed, should they not, at the very least, be considerd and ruled out as the reason for the observed OU?
As well, has anyone ever demonstrated the ability to discharge more energy from a battery than was required to charge it to begin with?
PW
Quote from: lanenal on May 03, 2013, 11:10:58 AM
@TinselKoala: I am not a judge between you.
Greg in his first post already estimated the maximal energy contribution from the Gate Driver, which seems to be no big deal. As for the Gate current, Greg might be talking about gate leakage current, not the current to induce the voltage difference.
Transistors are gated by currents, while MOSFETS are gated by voltage difference, and to develop voltage difference there is a "hidden cap" (so to speak) at the Gate. The current depends on the frequency, max voltage difference, the current limiting resistor, and the capacity of the hidden cap. The gate leakage current is due to the imperfect insulation inside the "hidden" cap.
regards,
lanenal
I see that you, too, do not bother to watch my videos. Nor do you seem to understand the issues. The current flowing from the Gate to the Drain or Source of a mosfet under the dynamic conditions illustrated in my video and experienced by circuits such as Gmeast's and Ainslie's is NOT the 100 nA steady-state leakage current, nor is it "the current to induce the voltage difference" necessary to switch the gate, nor is it due to "the imperfect insulation" inside the "hidden" cap. It is a normal consequence of presenting a pulsed or AC signal to a capacitor.
Please watch my video above, and then explain how and why the light bulb attached to the function generator "probe" lights up when I have the FG connected between Gate and Source, or Gate and Drain of the mosfet... whether the mosfet is working in a circuit or not.
Quote from: picowatt on May 03, 2013, 01:08:40 PM
Lanenal,
That is exactly my point. You think that the example I gave appears to be OU.
If one connected a desulphator to a battery and applied a load and noted the discharge time to a given stop voltage, and then did the same using the same load but without the desulphator connected and noted a shorter discharge time, there would be those who would claim that the desulphator produces overunity.
But, who ever said that a battery is "unity" to begin with? A battery produces waste heat during both charge and discharge. Suppose the charge/discharge efficiency of a lead acid battery is 70%. Suppose the use of a desuplhator can increase this to 85%. Though better, still not OU.
Desulphators supposedly produce finer grained sulphate crystals effectively increasing plate area. As well, pulse and reverse pulse plating similarly are known to produce finer grained structures. Visualize large clumps of sulphate crystals forming at the plates as opposed to a much smoother, finer grained structure. As well as an increase in surface area, less resistive losses occur over the finer grains. This is, supposedly, how a desulphator functions, and the waveforms used to desulphate have much in common with the waveforms of these OU circuits.
Do a search for "battery desulphator". There are both construction articles as well as commercial units available.
I am not stating with any certainty that these effects are the reason for the observed OU, but, as these effects have been observed regarding lead acid batterys when being pulsed or reverse pulsed, should they not, at the very least, be considerd and ruled out as the reason for the observed OU?
As well, has anyone ever demonstrated the ability to discharge more energy from a battery than was required to charge it to begin with?
PW[size=78%] [/size]
PW, you don't seem to understand what I am talking about. Your very argument about battery capacity is against the law of energy conservation. Your lengthy argument only seem to mislead and misinform people.
lanenal
Quote from: TinselKoala on May 03, 2013, 07:13:15 PM
I see that you, too, do not bother to watch my videos. Nor do you seem to understand the issues. The current flowing from the Gate to the Drain or Source of a mosfet under the dynamic conditions illustrated in my video and experienced by circuits such as Gmeast's and Ainslie's is NOT the 100 nA steady-state leakage current, nor is it "the current to induce the voltage difference" necessary to switch the gate, nor is it due to "the imperfect insulation" inside the "hidden" cap. It is a normal consequence of presenting a pulsed or AC signal to a capacitor.
Please watch my video above, and then explain how and why the light bulb attached to the function generator "probe" lights up when I have the FG connected between Gate and Source, or Gate and Drain of the mosfet... whether the mosfet is working in a circuit or not.
If you really understood what I have wrote to you, you won't say all that nonsense. You are amazing.
Quote from: lanenal on May 03, 2013, 10:48:44 AM
Hi Greg,
I have watched your video, and I have a question for you about the measured voltage drop over SH3 when the circuit is hooked up. Since the voltage should be oscilating, what you measured must be some sort of average voltage, right? Is it RMS? or is it something else?
great job, and thanks for sharing.
lanenal
It would be great to know more about your set up. For example, what is the frequency and duty cycle of your signal sent to the gate? What is the resistance/inductance of your RL? From that I can actually calculate the battery energy spent on the RL. Also, please share about what material is used to construct the RL and what other things to note in the construction. What kind of Diode and MOSFET is in use, etc. Thanks!
Quote from: lanenal on May 04, 2013, 12:49:41 AM
If you really understood what I have wrote to you, you won't say all that nonsense. You are amazing.
In the video I CLEARLY SHOW THE MOSFET PASSING FAR MORE THAN 100 nA between the Gate and the Source and also between the Gate and the Drain. I do this with the mosfet operating in a circuit from a different power source, and I also do this with the "bare" mosfet disconnected from any power source except the FG. Then I hook the mosfet back into the powered circuit to show that it is still operational and still works as a normal mosfet.
According to you, and to Gmeast, this is impossible. So I want your explanation as to how I did it.
You are amazing with your nonsense, that's for sure. You won't even deal with the issues, nor will you bother to educate yourself, nor do you deal with the points I raise, or that PW raises, directly.
I'll put my knowledge of MOSFETS and their operation up against yours any day, any time, lanenal. Where are your demonstrations, where are the devices you've built using mosfets, what are your qualifications to even be in this discussion? As far as I am aware, you have NONE.
Quote from: TinselKoala on May 04, 2013, 03:09:54 AM
According to you, and to Gmeast, this is impossible. So I want your explanation as to how I did it.
[size=78%]
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[size=78%]This clearly shows you didn't understand. You were not even able to discern gate current and leakage current.[/size]
Quote from: lanenal on May 04, 2013, 03:29:08 AM
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[size=78%]This clearly shows you didn't understand. You were not even able to discern gate current and leakage current.[/size]
DID YOU WATCH THE VIDEO? Apparently not.
You really have no idea what you are talking about. I am showing MUCH MORE CURRENT than "gate current" or "leakage current" flowing through the mosfet from GATE to SOURCE and from GATE to DRAIN. Enough to make a 12 volt light bulb light up brilliantly.
Explain how this is done, since you claim that the mosfet CANNOT pass more than the leakage current. You might also get off your high horse and watch the video where I switch the mosfet using only the charge from my fingers, illustrating that only a tiny current, providing a sufficient CHARGE, is needed to switch the mosfet. So what is your explanation for the hundreds of milliAmps of current that I show passing through the mosfet's GATE-SOURCE and GATE-DRAIN capacitances?
Come on, Mosfet master, I want to hear your explanation that explains how ONLY 100 nA can pass, yet the bulb will light nevertheless.
Quote from: lanenal on May 04, 2013, 03:29:08 AM
[/size]
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[size=78%]This clearly shows you didn't understand. You were not even able to discern gate current and leakage current.[/size]
I DID NOT MEASURE, NOR DID I ATTEMPT TO MEASURE, either of those currents. What I DID DO, is to show that the Gate-Drain and Gate-Source capacitances are essentially transparent to currents oscillating at the frequencies being used, and that they will easily pass FAR MORE than the 100 nA leakage current.
Your posts are showing more and more that you do not understand the operation of mosfets. "Gate current" applies to bipolar transistors and is different from the gate CHARGE that must be applied at the mosfet gate to turn it on. Gate leakage, for a mosfet, is just that, and in DC conditions the 100 nA figure is reasonable. The current supplied by a highspeed mosfet driver is there to charge the gate fully at high speed in spite of the gate capacitances, which is not properly "gate current" in the same sense as for a bipolar transistor.
It is definitely the case that power from the gate drive circuit can make its way to the load by way of the gate-drain and gate-source capacitances, at far higher currents than 100 nA, and this is what I showed in my video. It's really too bad that you continue to deny what is being demonstrated right before your eyes. Or would be, if you'd actually watch the video.
Does this sound familiar?
QuoteThis circuit should be the easiest to tune as well (Reason: when the pulse is at 6V, all transistors should be open switch, so logically, there will never be battery shorts). The pulse voltage source V5 can be replaced by a 555 astable circuit. V6 can be obtained by using an equal voltage divider (two 5K resisters, for example) sharing the same power as the 555 circuit. All the JBT transistors can be replaced with equiv. MOSFETs. I used voltage dependent current source and a resistor to mimick the behavior of a transformer. You can use 4 transformers instead in your implementation (MOSFETs are probably better than JBTs, I am affraid, as MOSFETs have no gate current). I used no diodes for switching purpose in this circuit. LTspice simulation is performed and the plot of the current over the load R3 is given below (I omitted the two capacitors at each end of the load R3 for the sake of simulation, in your implementation, you should include them).
Do they, or don't they? Why is a 9 amp driver needed then? Because of the 100 nA leakage?
Quote from: lanenal on May 03, 2013, 11:10:58 AM
Transistors are gated by currents, while MOSFETS are gated by voltage difference, and to develop voltage difference there is a "hidden cap" (so to speak) at the Gate. The current depends on the frequency, max voltage difference, the current limiting resistor, and the capacity of the hidden cap. The gate leakage current is due to the imperfect insulation inside the "hidden" cap.
regards,
lanenal
What happens when the "hidden cap" is getting charged? Is the "hidden cap" between the Gate and GND, or between the Gate and Source and the Gate and Drain?
Quote from: lanenal on May 04, 2013, 12:48:09 AM
PW, you don't seem to understand what I am talking about. Your very argument about battery capacity is against the law of energy conservation. Your lengthy argument only seem to mislead and misinform people.
lanenal
Lanenal,
If I understand you correctly, you wish to treat a lead acid battery as if it were an ideal capacitor.
There are a lot of EV companies out there that wish this were true! But, sadly, it is not.
There are many factors associated with most battery chemistries that affect the total energy they can deliver, or their efficiency during the charging or discharging process.
To treat them as ideal capacitors falls very short of depicting their true nature.
PW
Quote from: TinselKoala on May 04, 2013, 03:39:37 AM
Explain how this is done, since you claim that the mosfet CANNOT pass more than the leakage current.
Can you show me where did I claim that? You probably misunderstood me somehow. I was only saying that Greg's 100nA could be referring to leakage current. I also said that gate current depends on quite a few factors.
Quote from: picowatt on May 04, 2013, 10:22:00 AM
Lanenal,
If I understand you correctly, you wish to treat a lead acid battery as if it were an ideal capacitor.
There are a lot of EV companies out there that wish this were true! But, sadly, it is not.
There are many factors associated with most battery chemistries that affect the total energy they can deliver, or their efficiency during the charging or discharging process.
To treat them as ideal capacitors falls very short of depicting their true nature.
PW
PW,
My initial layout of math is linear, in my PS I only briefly touched on how to treat the nonlinear case. In fact, there I put forth a capacity expansion ratio. And that's still a simplification. In its most general form, the ratio should also change with voltage, and the conclusion still holds even in this most general form as long as the ratio is always greater than 1, which is the mathematical equivalent of saying: the capacity expands with pulsing load.
lanenal
Quote from: poynt99 on May 04, 2013, 09:03:59 AM
What happens when the "hidden cap" is getting charged? Is the "hidden cap" between the Gate and GND, or between the Gate and Source and the Gate and Drain?
Good question, and my answer was:
The current depends on the frequency, max voltage difference, the current limiting resistor, and the capacity of the hidden cap.
Quote from: lanenal on May 05, 2013, 01:29:41 AM
Can you show me where did I claim that? You probably misunderstood me somehow. I was only saying that Greg's 100nA could be referring to leakage current. I also said that gate current depends on quite a few factors.
Did I misunderstand, or did you? This is what Gmeast is claiming and what you apparently are defending:
QuoteAnother important note here regarding the Gate Driver: The power the gate driver is drawing either in-circuit or on its own support battery is NOT supplementing the HEATING of RL. Aside from its overhead from 'just being there', as it does its 'driver thing' the increase in power draw is from doing what it has to do to source and sink the current required in charging and discharging the gate capacitance in order to maintain the required gate charge voltage whether turning it 'ON' or 'OFF'. This 'sourcing and sinking' is isolated from the rest of the MOSFET except for a leakage of 100nA ... 100 billionths of an Amp ... the published data for the UCC2732x drivers.
So, this is the reason why the gate driver's power is not considered in Exploring the Inductive Resistor heater. The ultimate goal is to have the components self-oscillate instead of relying on controlling circuitry such as a PWM and Driver arrangement ... something I've almost figured out how to do ... but not quite.
Yet as PW has described and I have shown empirically, this is manifestly NOT TRUE. The Gate-Source and Gate-Drain capacitances of the mosfet act like any other capacitances: they pass AC or pulsed DC currents. A substantial part of the power at 1.5 MHz applied to the Gate by the gate driver CAN INDEED make it "through" the gate to be dissipated in the load, or in the mosfet itself as heat.
Gmeast seems to be basing his claim that the gate driver power can be neglected as input power, on the 100 nA DC leakage current in the datasheet. He does not appear actually to have measured, to see if his assumption is correct or not, and I don't believe he could have described the situation accurately to his "consultants" who told him that no more than 100 nA could possibly be conducted across the Gate-Source or Gate-Drain capacitances.
Hopefully by now you will agree, lanenal, that this is not proper. It would seem that you must either agree that a lot of current can indeed pass, as I show in the video, and therefore Gmeast is wrong, OR.... if Gmeast is right, you need to explain how I did what I did in the video.
Oh, and Gmeast: If you want your circuit to self-oscillate, that is easy. Just throw out all your nice construction techniques, and do what Rosemary Ainslie did: Use lots of excess wiring, especially on the Gate wiring, and use loose connections like clipleads. I know you don't care about my experience... even though it's a lot greater than yours... but that's exactly how I got Tar Baby to self-oscillate, after a proper tight layout would not. I looked at Ainslie's photographs and added the "rat's nest" of wires. Then the circuit self-oscillated without difficulty.
The magic oscillations you are trying for are nothing more than feedback.
Quote from: TinselKoala on May 05, 2013, 06:07:29 AM
Did I misunderstand, or did you? This is what Gmeast is claiming and what you apparently are defending:
TK, please read my post carefully. I didn't say that gate current can only get to 100nA. What I was saying is that that 100nA could be referring to the leakage current. Is that clear?
BTW, Greg already measured the a loose upper bound to the possible energy injection through the gate signalling. Even if he was wrong on the gate current, it does not invalidate his result.
Quote from: lanenal on May 05, 2013, 01:44:13 AM
Good question, and my answer was:
The current depends on the frequency, max voltage difference, the current limiting resistor, and the capacity of the hidden cap.
Well, read my two questions carefully; you didn't actually answer the two questions that I asked.
Quote from: lanenal on May 05, 2013, 09:31:03 AM
TK, please read my post carefully. I didn't say that gate current can only get to 100nA. What I was saying is that that 100nA could be referring to the leakage current. Is that clear?
BTW, Greg already measured the a loose upper bound to the possible energy injection through the gate signalling. Even if he was wrong on the gate current, it does not invalidate his result.
This is a direct quote from YOU. Please read your post carefully, including the part that I have highlighted in
RED:
QuoteThis circuit should be the easiest to tune as well (Reason: when the pulse is at 6V, all transistors should be open switch, so logically, there will never be battery shorts). The pulse voltage source V5 can be replaced by a 555 astable circuit. V6 can be obtained by using an equal voltage divider (two 5K resisters, for example) sharing the same power as the 555 circuit. All the JBT transistors can be replaced with equiv. MOSFETs. I used voltage dependent current source and a resistor to mimick the behavior of a transformer. You can use 4 transformers instead in your implementation (MOSFETs are probably better than JBTs, I am affraid, as MOSFETs have no gate current). I used no diodes for switching purpose in this circuit. LTspice simulation is performed and the plot of the current over the load R3 is given below (I omitted the two capacitors at each end of the load R3 for the sake of simulation, in your implementation, you should include them).
http://www.overunity.com/6793/simplest-solid-state-tesla-switch/msg156706/#msg156706 (http://www.overunity.com/6793/simplest-solid-state-tesla-switch/msg156706/#msg156706)
You here state that Mosfets HAVE NO GATE CURRENT. That is what the words "MOSFETs have no gate current" which you wrote in the post above mean in English. IS THAT CLEAR?
Now, once again, please explain how that can be true, at the same time that I can light up a light bulb on current passing through the Gate-Source and Gate-Drain capacitances of a mosfet.
The 100 nA cited in the mosfet data sheet as leakage current is leakage current, nobody has ever said that it wasn't. The issue is that both YOU (as shown above in the direct quote from YOU) and Gmeast both seem to think that is the only current that can pass between the gate and the other pins of the mosfet. Now, you are waffling about, since you have been presented with incontrovertible proof that your claims are false, now you are trying to pretend that you didn't make them. But as the DIRECT QUOTATION above proves.... you did indeed claim that "MOSFETs have no gate current." Is that clear?
What did you mean by that statement? Why does the lightbulb probe light up in my video? Is there current flowing through my mosfet's gate-drain and gate-source capacitances, or not? Is this the current required to charge the gate and switch the mosfet, or is it in excess of that?
QuoteEven if he was wrong on the gate current, it does not invalidate his result.
Of course it does. Crippling the gate driver by using the large inline resistor helps his cause by limiting the current here, but he is still wrong not to include the contribution of the driver to the circuit's overall power dissipation.
Quote from: TinselKoala on May 05, 2013, 06:20:24 PM
This is a direct quote from YOU. Please read your post carefully, including the part that I have highlighted in RED:http://www.overunity.com/6793/simplest-solid-state-tesla-switch/msg156706/#msg156706 (http://www.overunity.com/6793/simplest-solid-state-tesla-switch/msg156706/#msg156706)
You here state that Mosfets HAVE NO GATE CURRENT. That is what the words "MOSFETs have no gate current" which you wrote in the post above mean in English. IS THAT CLEAR?
Now, once again, please explain how that can be true, at the same time that I can light up a light bulb on current passing through the Gate-Source and Gate-Drain capacitances of a mosfet.
The 100 nA cited in the mosfet data sheet as leakage current is leakage current, nobody has ever said that it wasn't. The issue is that both YOU (as shown above in the direct quote from YOU) and Gmeast both seem to think that is the only current that can pass between the gate and the other pins of the mosfet. Now, you are waffling about, since you have been presented with incontrovertible proof that your claims are false, now you are trying to pretend that you didn't make them. But as the DIRECT QUOTATION above proves.... you did indeed claim that "MOSFETs have no gate current." Is that clear?
What did you mean by that statement? Why does the lightbulb probe light up in my video? Is there current flowing through my mosfet's gate-drain and gate-source capacitances, or not? Is this the current required to charge the gate and switch the mosfet, or is it in excess of that?
Of course it does. Crippling the gate driver by using the large inline resistor helps his cause by limiting the current here, but he is still wrong not to include the contribution of the driver to the circuit's overall power dissipation.
TK, what a great find -- I wonder what tool you have used, looks like you've got CIA team behind you :).
In that post, I was talking about the time average current (cause that's what matters in that case) while ignoring the leakage current, in that case, it is obvious that the time averaged gate current is zero.
lanenal
Quote from: poynt99 on May 05, 2013, 09:57:51 AM
Well, read my two questions carefully; you didn't actually answer the two questions that I asked.
The first question I answered indirectly, but you don't seem to understand. The second question can be answered the same way -- just try to understand my very post from which you raised your question, because it can be inferred from it. If you can't, I am sorry I can not help you either.
I have to say that we should focus on the validity of Greg's conclusion, not about nitpicking all those small things. That only distracts and disinforms.
Quote from: lanenal on May 06, 2013, 12:53:19 AM
The first question I answered indirectly, but you don't seem to understand. The second question can be answered the same way -- just try to understand my very post from which you raised your question, because it can be inferred from it. If you can't, I am sorry I can not help you either.
That's just the kind of non-answer we have come to expect from people who don't know what they are talking about, but attempt to "explain" things to people who DO.
Quote
I have to say that we should focus on the validity of Greg's conclusion, not about nitpicking all those small things. That only distracts and disinforms.
In other words.... stop picking on the ignorance of lanenal, and get back to discussing the bogus claims of someone else who is so ignorant that he uses a big resistor in line with a gate driver chip.
Meanwhile.... here's something that is in Gmeast's "direct lineage" of descent. This is the "work" that caused Gmeast to undertake his project, and this is the "work" that is at the basis of all his and Ainslie's claims.
Take a look at what Rosemary Ainslie thinks her circuit does.
Note carefully how she misrepresents the explanations that she has been given, while at the same time failing to understand HOW MOSFETS WORK and how the capacitances come into play. She still maintains, in spite of literature from Agilent, Tektronix, and others, and in spite of several very clear demonstrations to the contrary, that a Function Generator cannot pass current from an external source "from its signal terminal and probe"..... Further, she clearly STILL doesn't understand the linear operation regime of mosfets, something that plays a critical role in her circuit's behaviour.
It's easy to make fun of this ignorant and arrogant woman, because her own words, captured exactly in images like this, amply illustrate her paranoia, willful ignorance, arrogance, and insulting manner.
She libels anyone who disagrees with her, she lies about the events that happened concerning FuzzyTomCat and Harvey, and I am looking forward to seeing what she comes up with on June 1. Or is it July 1? Her original demo video, posted on one of her FOUR DIFFERENT YouTube accounts, contained so many lies and "smoking gun" reveals that she finally tried to remove it... but it is still available in spite of her coverups.
Quote from: lanenal on May 06, 2013, 12:43:06 AM
TK, what a great find -- I wonder what tool you have used, looks like you've got CIA team behind you :) .
Do you think so? Or is it more likely that I simply used the forum's "search" function?
Quote
In that post, I was talking about the time average current (cause that's what matters in that case) while ignoring the leakage current, in that case, it is obvious that the time averaged gate current is zero.
lanenal
Is it really?
Well, then..... you are in the position of having to explain my results, then.... Your flailing about is becoming amusing.
My light bulb attached to the FG output lead LIGHTS UP when the circuit is made through the mosfet's capacitances .... even though the "time averaged current" is zero, it being an AC signal that is applied by the FG. Yet, in spite of "time averaged current" being ZERO..... power is dissipated in the light bulb.
Why does my light bulb light up, if the "time averaged current" is zero and the mosfet cannot pass anything except the leakage current?
Why do we focus on these small things? Because they are at the root of the larger things, like claims of free energy/overunity performance, when there is actually none there.
Come on, lanenal, explain to me why my light bulb lights up. Time averaged current = zero, signal applied to a mosfet that is functioning in a circuit, and to one that is isolated, with the same results: signals that are oscillating at or around the frequencies used in the circuits we are discussing here, pass through the gate-drain and gate-source capacitances with hardly any attenuation, and are fully capable of dissipating power in a load, even though the "Time averaged current" may be zero.
Here's a video that presents a demonstration that Ainslie denies is even possible. She has claimed many times, even as recently as this morning, that a Function Generator cannot do this: Pass current from an external battery source to power a load, or act as a power supply itself. You wonder why we continue to examine and stress the basics? It is because of things like Ainslie's ridiculous and ignorant-- willfully ignorant--- claims about circuits and test equipment behaviour. And Gmeast is her protégé!
http://www.youtube.com/watch?v=GuBWVmRmUtc (http://www.youtube.com/watch?v=GuBWVmRmUtc)
And here's another video illustrating some more things that Ainslie thinks are impossible. MOSFETs switch by gate CHARGE, and they can in many applications be used NOT as a switch but as an amplifier, with an output resistance that depends on the magnitude of the gate CHARGE, not some poorly defined and even less well understood (by lanenal and gmeast) "gate current". The "linear" mode of operation is very important in the behaviour of Ainslie's circuit.... yet she refuses even to acknowledge that it is even possible.
http://www.youtube.com/watch?v=tKstLQYayNA (http://www.youtube.com/watch?v=tKstLQYayNA)
And while I'm on the subject..... note Ainslie's claim in the post image above about running without a separate supply for the signal generator (which supplies the negative bias necessary to produce her magic oscillations) in her circuit. Guess who was the first to show how to do this, well before Ainslie's claim (never demonstrated) to have managed it. And also note her lie about what it is that I "claim". All of us have shown that a negative bias current is producing the oscillations in her circuit; we have shown how to make the oscillations without any FG at all, in a steady state, by supplying the required bias current from an external power supply, an external battery, and finally-- MY WORK-- from the circuit's run batteries themselves by using a charge pump inverter. And I have shown, by reproducing her measurements, that her measurements do not mean what she thinks they mean at all and that there is no battery recharging, or lack of discharging, or any other anomalous effect happening in her circuit. Now, Gmeast's circuit either IS, or IS NOT, a replication of Ainslie's work, depending on the day of the week or the phase of the moon.... but everything we have learned from dealing with Ainslie and her circuit(s) over the past three or four years applies directly to his circuit work as well.
http://www.youtube.com/watch?v=OHxstOJkFtM (http://www.youtube.com/watch?v=OHxstOJkFtM)
A demonstration of the necessary bias current to obtain oscillations in the Ainslie "Q-array" circuit, no signal generator needed:
http://www.youtube.com/watch?v=hbHo3CCJtaw
Quote from: TinselKoala on May 06, 2013, 03:34:16 AM
Do you think so? Or is it more likely that I simply used the forum's "search" function?Is it really?
Well, then..... you are in the position of having to explain my results, then.... Your flailing about is becoming amusing.
My light bulb attached to the FG output lead LIGHTS UP when the circuit is made through the mosfet's capacitances .... even though the "time averaged current" is zero, it being an AC signal that is applied by the FG. Yet, in spite of "time averaged current" being ZERO..... power is dissipated in the light bulb.
Why does my light bulb light up, if the "time averaged current" is zero and the mosfet cannot pass anything except the leakage current?
Why do we focus on these small things? Because they are at the root of the larger things, like claims of free energy/overunity performance, when there is actually none there.
Come on, lanenal, explain to me why my light bulb lights up. Time averaged current = zero, signal applied to a mosfet that is functioning in a circuit, and to one that is isolated, with the same results: signals that are oscillating at or around the frequencies used in the circuits we are discussing here, pass through the gate-drain and gate-source capacitances with hardly any attenuation, and are fully capable of dissipating power in a load, even though the "Time averaged current" may be zero.
Why are you keep repeating some trivial common sense which you pretend that I don't understand? And as I have already pointed out, it is not pertinent here because that energy has already been accounted for by Greg in his first post.
It appears Lane is simply here to troll again. I've had similar dealings with him in the past.
He's clearly demonstrated here that he has no intention of having a reasonable discussion.
That's clear enough. But what do you expect from that crowd of mutual apologetics?
Remember this blast from the past?
http://www.youtube.com/watch?v=P8AIRkWF55k (http://www.youtube.com/watch?v=P8AIRkWF55k)
What is the correct circuit diagram? Where is the Black lead from the function generator located? Why was one battery removed in the second half, leaving only 48 volts, when the claim is made that the circuit AS SHOWN can withstand 72 volts input?
Sorry, I posted in the wrong thread...
Quote from: PicoWatt
Desulphators supposedly produce finer grained sulphate crystals effectively increasing plate area. As well, pulse and reverse pulse plating similarly are known to produce finer grained structures. Visualize large clumps of sulphate crystals forming at the plates as opposed to a much smoother, finer grained structure. As well as an increase in surface area, less resistive losses occur over the finer grains. This is, supposedly, how a desulphator functions, and the waveforms used to desulphate have much in common with the waveforms of these OU circuits.
Lead Sulfate is the normal electrochemical
product of battery discharge (lead-acid battery)
as the Sulfuric Acid electrolyte combines with
the active materials of both the positive and
negative plates. Initially, the lead sulfate crystals
are very tiny and nearly amorphous.
If the partially discharged battery with such lead
sulfate on the plates is re-charged immediately
and fully that lead sulfate will be almost completely
converted (by electrochemical decomposition) back
into active plate materials and regenerated sulfuric
acid.
If, however, the partially discharged battery is left
unattended for some time the lead sulfate crystals
will begin to undergo a phase change to become
larger "hardened" sulfation which will resist conversion
back into active plate materials and regenerated
sulfuric acid by normal charging voltages.
Typically, this hardened sulfation will gradually increase
in volume and mass as the battery ages and will deprive
the plates of active material capacity. The battery becomes
weak with reduced capacity and will not hold a charge.
The desulfator circuit applies to the battery very sharp
and very short pulses of considerable overvoltage at
a frequency of about 1000 Hz. The higher than normal
voltage is necessary to rejuvenate the hardened sulfate
crystals by decomposing them back into active plate
material and sulfuric acid; thus restoring the integrity of
the plates to near original condition. The plate area which
had been lost to the hardened sulfation is recovered and
made available for normal discharge/charge cycles.
It is a slow process which evolves considerable heat and
there are invariably some losses in the plate structure,
especially with automotive batteries which are by design
much less robust than a deep cycle battery. But if the
battery hasn't sustained significant damage to the plate
structures substantial capacity and usefulness will be restored,
Fortunately, desulfators are easy and inexpensive to make
and they can be very effective.
"Reverse Pulse Plating" is a unique process which is similar
to "burping" a battery during charging in order to reduce
the buildup of dendrites.
Regarding the matter of inductive discharge, it was long
ago discovered that a free-wheeling diode across a DC
Motor which is pulse driven both reduces the possibility
of transient generation and significantly increases the
efficiency of the motor.
It stands to reason that a free-wheeling diode placed
across an inductive resistor would produce the same
result when pulse driven.
Quote from: lanenal
Why are you keep repeating some trivial common sense which you pretend that I don't understand?
They do it because they are unable to resist the
temptation for mischief. They quite literally are
unable to help themselves. The adrenaline rush.
Boys will be boys! :o
Will Rosie ever come back to put them in their
place? 8)
Ainslie is promising another of her famous demonstrations of incompetence on June 1. But she is already snivelling about what is to be shown. She's not going to be showing what she promised to show LAST APRIL and failed to do so, for instance. Maybe you can figure out what she intends to try to demonstrate. I can hardly wait.
Of course, based on her past record, the chances of there actually being anything demonstrated on June 1st is very small. After all, she has told us how she is constantly being hacked, her computers stolen, and other sabotage perpetrated against her. What will she blame her next failure-to-deliver on? Is MrSean2K really vacationing in South Africa this month? Are her lawyers prepared to meet with Bryan Little to discuss a settlement for all the slanders and libels against him she's committed? The world is watching carefully.
And I am laughing so hard I can hardly see the screen.
Quote from: TinselKoala on May 11, 2013, 04:40:26 AM
Ainslie is promising another of her famous demonstrations of incompetence on June 1. But she is already snivelling about what is to be shown. She's not going to be showing what she promised to show LAST APRIL and failed to do so, for instance. Maybe you can figure out what she intends to try to demonstrate. I can hardly wait.
Of course, based on her past record, the chances of there actually being anything demonstrated on June 1st is very small. After all, she has told us how she is constantly being hacked, her computers stolen, and other sabotage perpetrated against her. What will she blame her next failure-to-deliver on? Is MrSean2K really vacationing in South Africa this month? Are her lawyers prepared to meet with Bryan Little to discuss a settlement for all the slanders and libels against him she's committed? The world is watching carefully.
And I am laughing so hard I can hardly see the screen.
TK,
You do persists. You should read some of the comments in threads on respectable energy sites. They are all about you. Free energy has been here all along in the form of your blubbering. If only we could harness the abundance of energy flowing from that trap of yours, the would would finally be at peace. I've already contacted Stefan. When he returns from vacation, he'll see that YOU, once again, turned a thread meant to share results of research into a forum to promote your own malicious bullshit. He locked one of your threads and now he'll lock this one. It's that p-nis problem I guess.
Regards
The main problem with Rosemary's (and gmeast's) test protocols, is they assume that a battery's A-h rating is constant and linear, independent of time or current during discharge.
Take a 100 A-h battery for instance. The A-h rating on a battery is usually given at a 20 hour discharge to 10V. At this rate of discharge the battery will output roughly 5A of current. 5 x 20 = 100; all seems good.
Now, decrease the discharge time to 2 hours; would we expect the battery to supply 50A for the duration? In practice, the battery would deliver about 40A, and therefore the actual A-h rating at the 2 hour discharge is 80 A-h.
Go to the extreme and decrease the discharge time to 2 minutes. In practice the battery would deliver about 680A for an actual A-h rating of 22 A-h.
So it's quite evident that we can't linearly extrapolate the A-h rating when changing from one discharge current to another. So what are the implications?
If we are to properly compare the A-h rating and capacity of a battery,
the battery needs to be discharged at the same average current for all cases. One battery rating called "Reserve Capacity" is tested by measuring the length of time the battery can deliver 25A (starting current) until it discharges to 10V.
References:
http://www.dcbattery.com/faq.html (http://www.dcbattery.com/faq.html)
http://www.allaboutcircuits.com/vol_1/chpt_11/3.html (http://www.allaboutcircuits.com/vol_1/chpt_11/3.html)
http://www.answers.com/topic/amp-hour-ratings (http://www.answers.com/topic/amp-hour-ratings)
http://overlandresource.com/what-is-an-amp-hour-and-how-to-calculate-battery-capacity (http://overlandresource.com/what-is-an-amp-hour-and-how-to-calculate-battery-capacity)
http://www.odysseybattery.com/documents/US-ODY-TM-001_0411_000.pdf (http://www.odysseybattery.com/documents/US-ODY-TM-001_0411_000.pdf)
Quote
How Are Amp Hours Determined? Battery manufacturers complete tests on their batteries to give them an Amp-Hour rating. A typical time period for a test is 20 hours, but it varies - batteries are tested over different periods, such as 24 hours, 75 hours, even 100 hours. But as you'll see below, a 100 Amp-hour battery tested over a 100 hour period (i.e. 1 amp drawn for 100 hours) will not have the same capacity as a 100 Amp-hour battery tested over a 20 hour period (i.e. 5 amps drawn for 20 hours). It turns out that if you drew 1 amp continuously from the 20 hour test battery, it would last for 110-120% longer than on the 100 hour test battery.
Quote from: gmeast on May 11, 2013, 08:43:13 AM
TK,
You do persists. You should read some of the comments in threads on respectable energy sites. They are all about you. Free energy has been here all along in the form of your blubbering. If only we could harness the abundance of energy flowing from that trap of yours, the would would finally be at peace. I've already contacted Stefan. When he returns from vacation, he'll see that YOU, once again, turned a thread meant to share results of research into a forum to promote your own malicious bullshit. He locked one of your threads and now he'll lock this one. It's that p-nis problem I guess.
Regards
Preserved for posterity.
You are the blubbering one, Gmeast. You can whine and insult all you like, but the record is clear: You cannot refute anything that .99, Picowatt, or I have said. You started two threads whose only purpose was to insult and mock me, nothing to do with any topic on this forum at all, in clear violation of this site's TOS and in violation of civil behaviour. You are a troll.
You have used and continue to use foul language like a ten year old child. You cannot even discuss things rationally with Picowatt, who has never said a harsh word to anyone.
Anyone who reads the thread you are talking about -- this thread -- can see for themselves that I DIDN'T EVEN POST IN IT until after you started with your insults and disrespect of your betters, specifically PicoWatt. So you are lying again when you blame your difficulties on me.
Your famous "FUCK THEM ALL DEAD" quotation will go down in history as typical argumentation from you and your ilk. Some people might even interpret that as a physical threat. I just find it very childish and insulting. Nobody here has spoken to you like that, except in direct response to your own foulness.
Your whining to Stefan is also typical. You've said many times that you don't care about this forum, that you aren't going to post here any more, and then you start up your threads just to insult and denigrate those who know their topic far better than you do. What was the last thread that was locked that I posted in? It was the Wayne Travis thread, and he asked for that thread to be locked because he could not answer the points I and others raised .... and he STILL cannot show what he claimed to have two years ago. And the thread before that was the Tar Baby thread.... my own thread, the only one that I've started here.... locked because Ainslie deliberately flamed and lied and even threatened Stefan with her lawyers. I keep the pressure on, and claimants who cannot support their claims with facts wind up whining, pleading with the site owner (even after insulting him over and over.... ) and flaming me whenever they can, just like you do.
Stefan might not have time or patience to review all the facts. And you distort them as you see fit, and avoid the truth like it would melt you. But I know this much: YOU, Gmeast, have contributed nothing to this site, you've never helped anyone, and you have been incredibly and increasingly insulting and foul-mouthed. You've insulted Stefan, complained about his administration, and you even use other forums to insult this site and its members. Why would anyone want you around? You don't even understand your own topic.
In case the message wasn't clear, allow me to spell it out a bit:
As the data shows, outlasting the rated 20 hour discharge figure of a battery is NOT proof that the circuit it is powering is OU or applying charge back to the battery.
The same goes for an apparent negative input power measurement. That one is easy to achieve, even in a simulation as I have done.
The only way Rosemary and gmeast will ever prove to anyone that their circuits are operating in some extraordinary mode, is to either perform proper power measurements, which they have not done, loop their devices and remove the power source, or demonstrate that the power source will outlast its amp-hour rating, based on the drawn operating current, also not yet done.
For the latter, the circuit would have to be run for at least 200 hours continuously (assuming a 100 A-h battery), based on a 1 ampere or so current draw typical of Rosemary's circuit. If the battery has significantly declined in open circuit voltage at the end of the run, then obviously there was no recharge occurring.
The so-called "control" is not even required if these tests are performed properly.
Well, you know that Ainslie is claiming that she will perform some kind of demonstration on June 1st, right? A demonstration of just what, I suppose we shall have to wait and see.
But here's a simple challenge for her and her "team": Simply repeat the demonstration shown in the second part of their earlier Demo Video, with the exact same circuit and construction layout they used then, but with all six batteries, providing 72 volts, instead of using only four at 48 volts as they did in the video (without explaining why at all, no matter how many times they have been asked.) Bring 700 mL of water "to boil" with that exact same circuit, built the same way, the same load, and 6 full 12 volt batteries. After all, that is what their "paper" claimed to do, NOT with only 48 volts.
But they won't be doing this, even though it's easy to set up, reasonable, and is what is claimed in their "paper" to have been done.
Quote from: TinselKoala on May 13, 2013, 12:42:06 PM
Well, you know that Ainslie is claiming that she will perform some kind of demonstration on June 1st, right? A demonstration of just what, I suppose we shall have to wait and see.
But here's a simple challenge for her and her "team": Simply repeat the demonstration shown in the second part of their earlier Demo Video, with the exact same circuit and construction layout they used then, but with all six batteries, providing 72 volts, instead of using only four at 48 volts as they did in the video (without explaining why at all, no matter how many times they have been asked.) Bring 700 mL of water "to boil" with that exact same circuit, built the same way, the same load, and 6 full 12 volt batteries. After all, that is what their "paper" claimed to do, NOT with only 48 volts.
But they won't be doing this, even though it's easy to set up, reasonable, and is what is claimed in their "paper" to have been done.
TK,
I could be wrong, but the way I read it, she is likely going to show that the battery voltage drops less in a given amount of time when two different load profiles are applied. This I gather from her desire to abbreviate the time required to perform the tests.
Is this proof of OU? No. It is only proof that a lead acid battery can have different capacities under different load profiles, even if the two load profiles produce similar average loads.
Would not proof of OU with this type of test, using batteries, require that more energy is able to be drawn from the batteries than is required to charge them? That would require several well measured charge/discharge cycles to prove. Not having access to the electrolyte to measure its specific gravity would make this somewhat difficult, although doing a CC charge cycle terminating with a CV charge to a predetermined minimum current might be an acceptable alternative, if repeated sufficiently. Accurate sampling of Vcharge and Icharge would be required and calculations made to accurately determine the energy required to recharge the battery. An RC battery A-Hr/wattmeter might be useable as charge current would be measured at DC.
However, if the "extra energy" is coming from the magnetic and material properties of the inductor as per her thesis, then why is a battery even required? If the "extra energy" can only be observed when using batteries, does it not seem more sensible to conclude that what is actually being proven is moreso related to the vagaries of the batteries themselves?
It amazes me how little desire either of them seem to have in making their circuits work with a well filtered DC supply. As I have suggested, a network can be inserted between the supply and circuit that would mimic the AC and DC characteristics of the battery so that the circuits will oscillate or ring as they do with a battery. Alternately, a DC supply could be connected to the batteries and isloated with an inductor in series therewith to prevent AC loading, and the steady state voltage and current required from the supply to maintain battery voltage/charge measured and used to provide Pin.
Proving that the circuits produce more heat when driven with a well filtered, easy to measure DC supply, than is produced when that same amount of power is applied directly to a resistor, would be much more interesting than proving that the batteries have different capacities when different load profiles are applied.
PW
Quote from: picowatt on May 13, 2013, 03:06:50 PM
Would not proof of OU with this type of test, using batteries, require that more energy is able to be drawn from the batteries than is required to charge them?
Yes, or equivalently, that the batteries do not discharge or that they even increase in charge level during the experimental run. Of course neither of these things occur in Ainslie's or Gmeast's experiments. The batteries discharge normally, and the total heat energy output by the load and the circuit does not exceed, or even approach, the energy required to charge the battery in the first place. Neither Ainslie nor Gmeast have ever shown any data that indicates otherwise. And of course when capacitors are used to power the circuits, it's easy to see that there is no magic, no overunity, no "advantage" from an oscillatory discharge, and of course the capacitors run down normally and make the "magic waveforms" all the while until their voltage drops enough for the circuit to die.
Quote from: TinselKoala on May 13, 2013, 04:03:59 PM
Yes, or equivalently, that the batteries do not discharge or that they even increase in charge level during the experimental run.
TK,
Even though she still does not understand how Q2 is biased on by having its source pulled negative wrt the gate, or how the DC bias current flows thru the FG, or how AC current flows thru the intrinsic MOSFET capacitances, she has, at least, apparently backed off on the cop=infinity claim.
PW
My impression is that Rosemary is aiming to show that for the same heat rise on the load resistor, the DUT battery will outlast the control experiment battery, using the 10V mark as the cutoff.
She also appears to be claiming that the amp-hour (although she incorrectly states watt-hour) rating of the battery will be exceeded when used in the DUT.
Quote from: poynt99 on May 13, 2013, 08:24:36 PM
My impression is that Rosemary is aiming to show that for the same heat rise on the load resistor, the DUT battery will outlast the control experiment battery, using the 10V mark as the cutoff.
She also appears to be claiming that the amp-hour (although she incorrectly states watt-hour) rating of the battery will be exceeded when used in the DUT.
.99,
I assume you mean to take each 12V nominal battery down to 10V. 10V is a bit low to take a 12V battery down to. As well, I thought I read she was going to do all this in 17 hours. She must be using a new circuit, I thought the batteries stayed up for months upon months with the original circuit.
So how does demonstrating that a battery can have different capacities with different load profiles provide proof of OU? Would you not agree that the energy required to charge the battery would have to be consistently exceeded in order to claim OU?
PW
Quote from: picowatt on May 13, 2013, 09:14:02 PM
.99,
I assume you mean to take each 12V nominal battery down to 10V.
Yes.
Quote10V is a bit low to take a 12V battery down to.
Agreed, but they do this very thing for generating extended discharge characteristic curves as per the 5th link I provided and the snapshot showing the A-h ratings vs. discharge currents.
QuoteThe following twenty graphs show detailed discharge characteristics of the entire ODYSSEY battery line. The end of discharge
voltage in each case is 10.02V per battery or 1.67 volts per cell (VPC). Each graph shows both constant current (CC)
and constant power (CP) discharge curves at 25ºC (77ºF). The table next to each graph shows the corresponding energy
and power densities. The battery run times extend from 2 minutes to 20 hours.
Quote
As well, I thought I read she was going to do all this in 17 hours. She must be using a new circuit, I thought the batteries stayed up for months upon months with the original circuit.
Probably not a new circuit per se, but most likely a set of much smaller capacity batteries (as we recommended years ago).
Quote
So how does demonstrating that a battery can have different capacities with different load profiles provide proof of OU?
Rhetorical question I know, but it doesn't. That was the purpose of my post.
Quote
Would you not agree that the energy required to charge the battery would have to be consistently exceeded in order to claim OU?
Of course, but that would require work...and know-how. At the present, even a proper Pin and Pout measurement is asking too much.
It's completely futile to expect anything reasonable or rational from her. Just take a look at her latest. It's clear that she learned absolutely nothing from the months of discussion with .99. She still cannot fathom the linear operation mode of a mosfet, she still thinks the battery is "disconnected" during the oscillations, she still has no clue about her circuit's operation or how a mosfet works.
"Take water to boil". What a fool. She cannot use her circuit, as described in her "paper" and shown in the second part of her last demo, with 72 volts input, to "take water to boil" as she has claimed. She cannot reproduce the problematic scopeshots we have noted with the circuit she published and intact mosfets. She can, however, reproduce those scopeshots easily enough.... simply by running her system as shown in the second half of the demo, trying to boil water, but with the full 72 volts input instead of the mere 48 she was forced to use then. Because this will blow the Q1 mosfet in short order... and then the scopeshots will look like what she published.
If she shows anything at all on June 1 -- which I sincerely doubt -- it will be even more of a farce than her last "demo".
Quote from: poynt99 on May 13, 2013, 08:24:36 PM
My impression is that Rosemary is aiming to show that for the same heat rise on the load resistor, the DUT battery will outlast the control experiment battery, using the 10V mark as the cutoff.
She also appears to be claiming that the amp-hour (although she incorrectly states watt-hour) rating of the battery will be exceeded when used in the DUT.
I want everyone to remember just what kind of highjacking assholes this gang of 3 really are. I started a thread about MY research of a variant of the Inductive Resistor Heater. I started that thread so I could share MY research. But what happened instead? This gang of 3 ... being TinselKoala, picowatt and poynt99, used MY thread to criticize Rosemary Ainslie's work. They opitomize a gang of 5th-grade bullies. What's amazing is Stefan Hartmann ... this forum's owner, seems to think that's OK to do.
This didn't happen just once ... it happended twice. I'm just going to keep reminding everyone of this gang's behavior as often as I can.
These three are self-proclaimed experts on just about everything ... and that's all they are ... remember that.
Please just view my video slide show and decide for yourself. Not one of this gang of 3 ever made a valid challenge of my work other than to try and demean me by saying "... it's not work at all ... ". That was a quote from the all-knowing TK aka TinselKoala.
My video Slide Show here:
http://www.youtube.com/watch?v=q473lX-Zw1w (http://www.youtube.com/watch?v=q473lX-Zw1w)
Thank you
QuotePlease just view my video slide show and decide for yourself. Not one of this gang of 3 ever made a valid challenge of my work other than to try and demean me by saying "... it's not work at all ... ". That was a quote from the all-knowing TK aka TinselKoala.
Resurrect a thread that has been dead since mid May, just to troll and stalk? And to complain about the site owner? Amazing, but true to form.
Prove it by giving a link to where I said "it's not work at all".
Meanwhile, I can prove that YOU say some pretty wild things, GMeast.
You want me to stay out of "your" threads, fine .... then don't mention me at all, especially not with one of your baseless distortions.
And, in case you haven't noticed.... I'm not melting down at all. In fact, it's rather the other way around, isn't it. You and your patron saint have fallen out, she produced the most hum-dinger of a "demonstration" imaginable where she soundly refuted herself much better than I ever could do.... and everything I ever said about her circuit is now known to be true. And the demo has given me even more material to illustrate just how wrong Ainslie and Donovan Martin really are, and how mendacious and devious they are. You can hear Ainslie melting down live on cellphone, just by clicking a few links, and you can see her own disproofs of her claims, in spite of her trying to withhold the screen images, on "my" thread having to do with her June 29 demo. Yet I am perfectly calm.... and perfectly correct.
That infuriates you, doesn't it. And it makes me LOL.
There is a recent Gmeast quote that I find kind of funny:
QuoteThe waveform put out by the Keppe motor would handily act as the controlling signal for an Ainslie Heater. So there's a potential for a SUPER efficient heater/fan system product.
For starters a Keppe motor if I recall is just a plain old boring dumb pulse motor hooked up to a fan. So how the "output waveform" could make for a "controlling signal" for an Ainslie heater is beyond me.
As a sidelight, the fan itself is quite a creation. They take a regular fan propeller and glue wood veneer to the fan blades so you can make yourself feel good about spending $250 for a "natural" electric fan.
Going back to Gmeast's comments, what pray tell is a "SUPER efficient heater/fan?" What is THAT? Last time I looked you could buy ANY space heater that uses ANY fan and it will be 100% efficient at producing heat.
Like wow man! You can go to the local Big Box store and get a HEATER THAT IS 100% EFFICIENT. Call Sterling up Gmeast and tell him the news! He will get all excited and write an article about it.
And it goes without saying that TK, Poynt, and PW are highly qualified individuals with tons of knowledge and experience. So when Gmeast calls them all nasty names and says they are stupid he is just talking crap like some angry man with no better place to dissipate his frustration. It's a spectacle where almost everybody that reads his rants against these good knowledgeable people knows that they are full of crap. It's just a bunch of mendacious ugly nasty negativity with no redeeming qualities except perhaps for Gmeast himself. Meds might be a better alternative or perhaps get a Nerf bat and flail away at the walls, floor and ceiling until you are completely exhausted.
MileHigh
Quote from: TinselKoala on July 07, 2013, 07:22:57 PM
Resurrect a thread that has been dead since mid May, just to troll and stalk? And to complain about the site owner? Amazing, but true to form.
Prove it by giving a link to where I said "it's not work at all".
Meanwhile, I can prove that YOU say some pretty wild things, GMeast.
You want me to stay out of "your" threads, fine .... then don't mention me at all, especially not with one of your baseless distortions.
And, in case you haven't noticed.... I'm not melting down at all. In fact, it's rather the other way around, isn't it. You and your patron saint have fallen out, she produced the most hum-dinger of a "demonstration" imaginable where she soundly refuted herself much better than I ever could do.... and everything I ever said about her circuit is now known to be true. And the demo has given me even more material to illustrate just how wrong Ainslie and Donovan Martin really are, and how mendacious and devious they are. You can hear Ainslie melting down live on cellphone, just by clicking a few links, and you can see her own disproofs of her claims, in spite of her trying to withhold the screen images, on "my" thread having to do with her June 29 demo. Yet I am perfectly calm.... and perfectly correct.
That infuriates you, doesn't it. And it makes me LOL.
Your response shows your colors. Your response also shows that you are indeed melting down. " ... I am perfectly calm... and perfectly correct." + "I am, I am, I am." It does not infuriate me, it makes me laugh.
Thank you.
Quote from: MileHigh on July 07, 2013, 07:43:59 PM
There is a recent Gmeast quote that I find kind of funny:
For starters a Keppe motor if I recall is just a plain old boring dumb pulse motor hooked up to a fan. So how the "output waveform" could make for a "controlling signal" for an Ainslie heater is beyond me.
As a sidelight, the fan itself is quite a creation. They take a regular fan propeller and glue wood veneer to the fan blades so you can make yourself feel good about spending $250 for a "natural" electric fan.
Going back to Gmeast's comments, what pray tell is a "SUPER efficient heater/fan?" What is THAT? Last time I looked you could buy ANY space heater that uses ANY fan and it will be 100% efficient at producing heat.
Like wow man! You can go to the local Big Box store and get a HEATER THAT IS 100% EFFICIENT. Call Sterling up Gmeast and tell him the news! He will get all excited and write an article about it.
And it goes without saying that TK, Poynt, and PW are highly qualified individuals with tons of knowledge and experience. So when Gmeast calls them all nasty names and says they are stupid he is just talking crap like some angry man with no better place to dissipate his frustration. It's a spectacle where almost everybody that reads his rants against these good knowledgeable people knows that they are full of crap. It's just a bunch of mendacious ugly nasty negativity with no redeeming qualities except perhaps for Gmeast himself. Meds might be a better alternative or perhaps get a Nerf bat and flail away at the walls, floor and ceiling until you are completely exhausted.
MileHigh
OOPS! I was wrong .... it's the Gang Of 4. What ... are you a shrink or something? I'm not frustrated at all, in fact I have gotten great joy watching the responses. BTW ... talk about calling nasty names, I suppose you condoned the filth that TK spewed in his rhetoric about Ainslie ... not that I care now. At the time, that was OK though ... right? Double standard. Why am I not surprised by that.
Gmeast:
I don't condone the nasty name calling on any side. But I made a conscious decision to not involve myself beyond a certain limited point in this affair because this is just history reviving itself.
I am just interested in the truth coming out. The truth is that a self-described old lady that saw some oscillations on a scope display due to measurement error and got all excited for nothing. The whole thing is completely meaningless when you look at the big picture. However, there are some important principals at stake that are worth defending if you choose to go there. Once this nonsense reaches it's proper conclusion Rosemary will be spent and the whole thing will quickly be forgotten.
Going back to the trash talk, you step it up a notch such that it is extra ugly and corrosive and at the same time just more silly untrue nonsense waiting to go into the memory shredding machine.
This will all be forgotten before too long. The academics will never "get engaged."
MileHigh
Quote from: MileHigh on July 07, 2013, 08:59:40 PM
Gmeast:
I don't condone the nasty name calling on any side. But I made a conscious decision to not involve myself beyond a certain limited point in this affair because this is just history reviving itself.
I am just interested in the truth coming out. The truth is that a self-described old lady that saw some oscillations on a scope display due to measurement error and got all excited for nothing. The whole thing is completely meaningless when you look at the big picture. However, there are some important principals at stake that are worth defending if you choose to go there. Once this nonsense reaches it's proper conclusion Rosemary will be spent and the whole thing will quickly be forgotten.
Going back to the trash talk, you step it up a notch such that it is extra ugly and corrosive and at the same time just more silly untrue nonsense waiting to go into the memory shredding machine.
This will all be forgotten before too long. The academics will never "get engaged."
MileHigh
I don't care about Ainslie's stuff, her oscillations, 'the academics' or anything else for that matter. I care about the variant I built, tested and my attempt to post my results here ...... PERIOD ... and the right to do that without being attacked by " " and " " and " " and you because I have a variant of Ainslie's heater. I didn't mention the Gang Of 3's names (or yours) in this reply because " " said he'd stay out of this thread if I don't mention his name. And that's fine ... he's irrelevant to my work and irrelevant to most everything else honest experimenters are doing.
Well Ainslie HASN'T been 'forgotten' for more than 10 years, so don't think she's going anywhere soon ... and as I said, "I don't care". I've conducted dozens of identical tests who's results justify the further exploration of a heater using an element possessing Inductive and Resistive characteristics. I have no working theory, but I have consistent test results which I have attempted to share.
Stay out of this thread. I don't need your feedback or input. You have already made up your minds about many OU projects that people have attempted to share in good faith and conscience in these forums. Someone needs to tell you that NO ONE HAS ASSIGNED YOU GUYS TO BE THE WORLD'S FILTER for technologies that have potential to benefit mankind ... though it seems you have self-appointed yourselves to that task.
Thank you
Do what you want with your circuit Gmeast but there is a salient point that I am not sure you are getting: No matter what kind of heater circuit you are building, when you look at it from the battery terminals onward, it will be a 100% efficient heater. You can change the design, change your operating frequency or do whatever you fancy, you will always get a design that is 100% efficient in producing heat. You may as well as buy a big ceramic resistor, some heat sink compound, and a heat sink. Connect your battery to that, no switching or MOSFET required, and you will have a 100% efficient heater.
Quote from: GMEast
I've conducted dozens of identical tests who's results justify the further exploration of a heater using an element possessing Inductive and Resistive characteristics. I have no working theory, but I have consistent test results which I have attempted to share.
I share your interest and enthusiasm.
The circuit lends itself well to potential
enhancement with external inductance
(in addition to the heating element)
and high efficiency switching techniques.
Quote from: MileHigh on July 07, 2013, 10:26:42 PM
Do what you want with your circuit Gmeast but there is a salient point that I am not sure you are getting: No matter what kind of heater circuit you are building, when you look at it from the battery terminals onward, it will be a 100% efficient heater. You can change the design, change your operating frequency or do whatever you fancy, you will always get a design that is 100% efficient in producing heat. You may as well as buy a big ceramic resistor, some heat sink compound, and a heat sink. Connect your battery to that, no switching or MOSFET required, and you will have a 100% efficient heater.
As I said earlier, you have a mindset and are convinced of your conclusion. I have proven that additional energy is entering the system to produce more heat for the energy consumed than should normally occur.
DO NOT POST HERE ANY MORE. Or go ahead and do so and waste your time. I'll not argue with you. You are wrong in you conclusion with respect to the anomalies I've observed, recorded and reproduced. I'll not be deterred in the least. Your guys' tactics are well recognized and it's becoming clear to more and more in this community that your are FOES of over unity performance technologies. These systems derive supplemental energy from alternative sources that are not easily measured. It's as simple as that.
I'll thank you to leave now.
P.S.
If you are basing your conclusions on current theory alone, then there is NO basis for discussion because you have not built, tested and observed the anomalies yourself. If this is so, then you are ONLY an armchair engineer (or whatever you call yourself) and your conclusions are empty, invalid and without merit.
Quote from: SeaMonkey on July 07, 2013, 11:17:42 PM
I share your interest and enthusiasm.
The circuit lends itself well to potential
enhancement with external inductance
(in addition to the heating element)
and high efficiency switching techniques.
Thank you SeaMonkey!
Quote from: MileHigh on July 07, 2013, 10:26:42 PM
Do what you want with your circuit Gmeast but there is a salient point that I am not sure you are getting: No matter what kind of heater circuit you are building, when you look at it from the battery terminals onward, it will be a 100% efficient heater. You can change the design, change your operating frequency or do whatever you fancy, you will always get a design that is 100% efficient in producing heat. You may as well as buy a big ceramic resistor, some heat sink compound, and a heat sink. Connect your battery to that, no switching or MOSFET required, and you will have a 100% efficient heater.
Probably you are right cause this is the common view and knowledge !
Probably not :
http://www.al-bernstein-industries.com (http://www.al-bernstein-industries.com/)
Sincerely
CdL
its funny how steorn is now fully concentrating on such heaters.@gmeast dont listen to milehigh,there is a gap in the thermodynamic cycles here for intake of ambient heat.just gotta polish it up..
Quote from: profitis on July 08, 2013, 12:45:16 PM
its funny how steorn is now fully concentrating on such heaters.@gmeast dont listen to milehigh,there is a gap in the thermodynamic cycles here for intake of ambient heat.just gotta polish it up..
Hi profitis,
WOW ... thanks, yes there is a breach of 'something' for sure. And yes ... it is interesting that Steorn is looking to a product of excessive and anomalous heat production. Something 'rings right' about this approach to COP>1 performance. Even I have a problem with Electrical Energy 'IN' to COP>1 Electrical Energy 'OUT' but not COP>1 Heat Energy 'OUT'.
Regards,
Greg
P.S.
I'm re-posting the link to my Video Slide Show which shows how I determined COP>1.25:
http://www.youtube.com/watch?v=q473lX-Zw1w (http://www.youtube.com/watch?v=q473lX-Zw1w)
nice @gmeast. I just want to point out that if anyone is lucky enough to get their inductor circuits tailored just right to tap into that narrow thermodynamic gap that any ambient heat inflow is going to happen in the inductor core rod itself and the energy be spat out on the current in the resistor load coil or lightbulb or whatever it is that one uses to perform work,ontop of the battery source reciprical current of course(inductive piggyback riding).
Quote from: profitis on July 08, 2013, 12:45:16 PM
its funny how steorn is now fully concentrating on such heaters.@gmeast dont listen to milehigh,there is a gap in the thermodynamic cycles here for intake of ambient heat.just gotta polish it up..
Are you referring to Steorn's "HephaHeat" technology?
PW
@pw..yes
Quote from: picowatt on July 08, 2013, 07:39:07 PM
Are you referring to Steorn's "HephaHeat" technology?
PW
Steorn is NOT the topic here. If you want to ask someone about that or something else then do it via a personal message to that person.
Thank you
@gmeast..he probably wants to shit on steorn and use that as a weapon here.@pw steorn isnt the issue here.i just said that they are doing inductive research thats all.and maybe signing multimillion dollar contracts thats all.we know nothing thereof.
Quote from: profitis on July 08, 2013, 07:20:47 PM
nice @gmeast. I just want to point out that if anyone is lucky enough to get their inductor circuits tailored just right to tap into that narrow thermodynamic gap that any ambient heat inflow is going to happen in the inductor core rod itself and the energy be spat out on the current in the resistor load coil or lightbulb or whatever it is that one uses to perform work,ontop of the battery source reciprical current of course(inductive piggyback riding).
How refreshing! You are so right. You can't use just 'any' inductor constructed with 'any old' wire. As is being realized in the LENR work and other work such as Manelas' and the former Floyd Sweet's, it is the combination of nano-particles, suitable alloys and RAPID EDGE TRANSITIONING wave forms that contribute to the observed performances. In my case, I'm using a wire alloy comprised of Nickel, Chromium and Iron. I have no core, but the iron in the wire serves to 'fill the bill'.
There is so much more going on in these systems than can be argued against using 200-year old theories and principles ... though THEY are trying, oooohhh so hard.
Quote from: profitis on July 08, 2013, 08:54:21 PM
@gmeast..he probably wants to shit on steorn and use that as a weapon here.@pw steorn isnt the issue here.i just said that they are doing inductive research thats all.and maybe signing multimillion dollar contracts thats all.we know nothing thereof.
As Steorn's HephaHeat technology has little or nothing to do with the "technology" discussed in this thread, or any claims of "OU", I just wanted to make sure there was not some other Steorn technology being discussed that I was unaware of.
Carry on...
@gmeast ..im glad you pointed out the nichrome composition to me as i had earlier thought that becoz ainslies circuit didnt have a ferrite core through the coils that there was no vehicle for creation of magnetic domains but now i see otherwise.nickel and iron in the wire itself indeed are magnetized.
@pw i wouldnt be surprised if steorn is fully aware of the thermodynamic discrepencies i described on my thread which you ran away from in terror.
Quote from: profitis on July 08, 2013, 11:07:00 PM
@pw i wouldnt be surprised if steorn is fully aware of the thermodynamic discrepencies i described on my thread which you ran away from in terror.
I did not run away at all. I walked away from someone who states their "beliefs" and theories as if they are proven facts. Your compressed air analogy was a good one. It clearly demonstrates the conservation of energy.
Do you understand Steorn's HephaHeat technology? The main point is the use of steel as opposed to water for thermal storage. It allows for higher temp storage, power usage off peak, etc. The inductive heater part of it is rather trivial. No claims of OU. Probably the first sensible thing to come from Steorn.
But to mention the HephaHeat technology as if it somehow bolsters or proves your beliefs regarding inductors? That's a bit of a stretch...
I better go now, wouldn't want the OP to start spouting foul language...
PW
Quote from: picowatt on July 08, 2013, 11:32:50 PM
I did not run away at all. I walked away from someone who states their "beliefs" and theories as if they are proven facts. Your compressed air analogy was a good one. It clearly demonstrates the conservation of energy.
Do you understand Steorn's HephaHeat technology? The main point is the use of steel as opposed to water for thermal storage. It allows for higher temp storage, power usage off peak, etc. The inductive heater part of it is rather trivial. No claims of OU. Probably the first sensible thing to come from Steorn.
But to mention the HephaHeat technology as if it somehow bolsters or proves your beliefs regarding inductors? That's a bit of a stretch...
I better go now, wouldn't want the OP to start spouting foul language...
PW
TOO LATE! You just can't avoid contaminating my thread can you? Your conduct opitomizes that of a bully or of a spoiled brat ... actually there's little difference. Stay the f--- out of my thread. You offer little of any consequence. You obviously have no (working) knowledge of how certain alloys, nano-particles, and rapid edge transitioning pulses interact to produce the anomalous excess energies that have been observed by so many competent experimenters and researchers. As I said, "Stay the f--- out of my thread". Your perpetual 'buzzing around' is annoying. You and your 'gang members' have referred to me as a 'troll'. What the f--- does your conduct make you? ... a F-----G TROLL ... that's what!
PLEASE DON'T ANSWER THE QUESTION.
Thank you
What's wrong with you Hartmann to let this crap take place in these forums ... HUH? What's wrong with you? ??? ??? ??
Quote from: gmeast on July 09, 2013, 12:52:12 AM
TOO LATE! You just can't avoid contaminating my thread can you? Your conduct opitomizes that of a bully or of a spoiled brat ... actually there's little difference. Stay the f--- out of my thread. You offer little of any consequence. You obviously have no (working) knowledge of how certain alloys, nano-particles, and rapid edge transitioning pulses interact to produce the anomalous excess energies that have been observed by so many competent experimenters and researchers. As I said, "Stay the f--- out of my thread". Your perpetual 'buzzing around' is annoying. You and your 'gang members' have referred to me as a 'troll'. What the f--- does your conduct make you? ... a F-----G TROLL ... that's what!
Thank you
What's wrong with you Hartmann to let this crap take place in these forums ... HUH? What's wrong with you? ??? ??? ??
I was respondng to Profitis..
Carry on with your tantrum...
Quote from: picowatt on July 09, 2013, 12:57:18 AM
I was respondng to Profitis..
Carry on with your tantrum...
I'm glad EVERYONE can see who the antagonist is here. What a child you are. I told you if you you want to carry on a conversation with someone else, do it via a private message. The respect you're going to garner from your bullshit here will be valuable to you I'm sure. I think I'll just see how far you'll go with this crap.
For anyone interested in the information I've assembled about my continuing research, you can find it on both Energetic Forum and Heretical Builders. I have moderator privileges on Heretical Builders so I can easily eliminate PicoWatts's bullshit. On Energetic Forum, I'm sure I can convince Aaron to do the same on my behalf.
Thank you
@pw..enough about steorn.my beliefs and theories on my thread are founded on a very well established fact,,magnetic order is temperature dependant,,as simple as that.
@gmeast..it seems there is a discrepency even in gaseous thermodynamics,ive just been reading about the proell-effect,realy interesting.ive heard rumours on this forum,even from antagonists, about heat pumps exceeding 100percent efficiency but having no way to convert the heat gains into useful work exceeding the effiency threshold.with inductor circuits acting as the heat pump we are theoreticaly able to do this all in one step,ie.direct conversion of ambient heat into electrical energy.
Quote from: profitis on July 09, 2013, 12:49:24 PM
@gmeast..it seems there is a discrepency even in gaseous thermodynamics,ive just been reading about the proell-effect,realy interesting.ive heard rumours on this forum,even from antagonists, about heat pumps exceeding 100percent efficiency but having no way to convert the heat gains into useful work exceeding the effiency threshold.with inductor circuits acting as the heat pump we are theoreticaly able to do this all in one step,ie.direct conversion of ambient heat into electrical energy.
Heat pumps are a unique case when it comes to efficiency. Unlike electric, gas and fuel oil heaters which all create heat in a single environment, a heat pump is merely 'moving' heat from one environment to another. The only work the heat pump's compressor must do is to create a sufficient temperature differential at the condenser so that heat can be transferred from it, and the same thing at the evaporator ... only sufficient temperature differentials for heat flow to it. Really huge evaporators and condensers make these things very efficient, but not cost effective.
I've tried drawing a parallelism between the Inductive Resistor Heater and the Heat Pump, but I have been unsuccessful in doing so in my attempts to develop a working theory for the heater.
Quote from: gmeast on July 09, 2013, 01:41:53 AM
I have moderator privileges on Heretical Builders so I can easily eliminate PicoWatts's bullshit. On Energetic Forum, I'm sure I can convince Aaron to do the same on my behalf.
Thank you
That's great, can you please post a direct links to your moderated threads?
@gmeast..google 'steven j.smith magnetothermodynamics 3'i cant put a link here but you will see a brief one-page analysis of the thermodynamic cycle of a standard inductor core and how it can theoreticaly breach the 2nd law.it may help you to formulate a reference for future research.this is where i borrowed my theory from and tried to put in language that most can understand.
Quote from: orbut 3000 on July 09, 2013, 07:43:17 PM
That's great, can you please post a direct links to your moderated threads?
Not much has been posted there lately and they contain practically identical content. I'm presently attempting to gather some resources so I can take the next steps: 1) ultracapacitor bank 2) alter the circuit to be self-oscillating ... this one is the biggest challenge of all.
The links:
http://www.energeticforum.com/inductive-resistor/13925-exploring-inductive-resistor-heater.html (http://www.energeticforum.com/inductive-resistor/13925-exploring-inductive-resistor-heater.html)
http://www.hereticalbuilders.com/forumdisplay.php?f=21 (http://www.hereticalbuilders.com/forumdisplay.php?f=21)
My Channel and the Inductive Resistor Heater Video Slide Show:
http://www.youtube.com/user/gmeast (http://www.youtube.com/user/gmeast)
Quote from: profitis on July 09, 2013, 07:55:25 PM
@gmeast..google 'steven j.smith magnetothermodynamics 3'i cant put a link here but you will see a brief one-page analysis of the thermodynamic cycle of a standard inductor core and how it can theoreticaly breach the 2nd law.it may help you to formulate a reference for future research.this is where i borrowed my theory from and tried to put in language that most can understand.
Why can't you put that link here? I just looked at it ... pretty heady stuff. Thanks for the reference.
Quote from: gmeast on July 09, 2013, 01:43:37 PM
Heat pumps are a unique case when it comes to efficiency. Unlike electric, gas and fuel oil heaters which all create heat in a single environment, a heat pump is merely 'moving' heat from one environment to another. The only work the heat pump's compressor must do is to create a sufficient temperature differential at the condenser so that heat can be transferred from it, and the same thing at the evaporator ... only sufficient temperature differentials for heat flow to it. Really huge evaporators and condensers make these things very efficient, but not cost effective.
I've tried drawing a parallelism between the Inductive Resistor Heater and the Heat Pump, but I have been unsuccessful in doing so in my attempts to develop a working theory for the heater.
Clarifying: that the size of the Evaporator is pretty closely tied to the compressor's ratings/capacities ie. pressure, flow, refrigerant whereas the condenser can be huge thus requiring a lower temp differential for the same rate of heat rejection/transfer, but again, not cost effective.
@gmeast.. cant post links coz im on opera mini compression on a phone.are you using lead acid batteries there gmeast.do they have openings that you can open up.theres a very nice way to gauge total power in lead-acid systems.
Quote from: profitis on July 10, 2013, 07:54:57 PM
@gmeast.. cant post links coz im on opera mini compression on a phone.are you using lead acid batteries there gmeast.do they have openings that you can open up.theres a very nice way to gauge total power in lead-acid systems.
Presently I'm using the batteries I show on the forums. They are SLA and/or AGM (Absorbent Glass Matt) and there is no access to the interior. What were you suggesting anyway ... specific gravity?
I'll post the Steven J. Smith link for you:
http://www.whale.to/b/magneto_thermodynamics3.html (http://www.whale.to/b/magneto_thermodynamics3.html)
yes i was going to suggest specific gravity or chemical test or conductivity measurement for varification of unchangeing electrolyte concentration post-run.chek out the 'self-charging electric car'thread where i give details for a spot chemical test for sulfuric acid gauging in the bats.if you want total absolute varification of overunity to yourself or to the public this is best way gmeast,to use lead-acid open type cells as the totality of power usage or gain is directly proportional to acid concentration.
Quote from: profitis on July 11, 2013, 09:23:45 AM
yes i was going to suggest specific gravity or chemical test or conductivity measurement for varification of unchangeing electrolyte concentration post-run.chek out the 'self-charging electric car'thread where i give details for a spot chemical test for sulfuric acid gauging in the bats.if you want total absolute varification of overunity to yourself or to the public this is best way gmeast,to use lead-acid open type cells as the totality of power usage or gain is directly proportional to acid concentration.
I'm not sure this would be definitive in that the batteries are expected to discharge. It's just that they discharge less than would be expected. I've been very careful to run the Inductive Resistor Heater tests head-to-head with straight-up ohmic loading. It's the comparison of these two tests that are vital.
@gmeast.i see what you mean yes.is your circuit recharging the bats at all? If its straight electric to heat comparison then these electrolyte tests wont do yes.if any circuits are selfchargn bats then its a genius method for total proof.even for sake of improved efficiency proofs.
Just an update:
I'm still going to replace the batteries with a capacitor bank at some point, but the bank will have to be huge just to get a small enough voltage drop on the caps to allow for a a decent test duration ... hopefully 8-hours. If you examine the discharge curve for any capacitor you see it's the inverse of it's charge curve. On discharge, the cap voltage drops very rapidly over time for a given load. So the cap bank must be very big (wide - that is "parallel") if the voltage is to drop only .5VDC to no more than 1VDC over the test duration. As the voltage supply drops much below 24VDC to 25VDC, the performance drops off considerably. Lower voltage also affects the gate driver circuitry.
I'll first build a single 'series' bank for the supply voltage requirement (for 30VDC) and load it as I did during the battery tests ... around 3.2Watts. From this loading I'll be able to determine the rate of discharge of the bank for the particular capacitor specie. It then should be only a matter of arithmetic to scale the 'real' capacitor bank based on the same specie of capacitor. It could take well over 150 capacitor$ to get a sustained 3.2Watt draw for 8-Hours with only a 1VDC(max) voltage drop. It's possible to figure that out now just from the above requirements, but actual data is always best to have.
That's all for now ... only takes money.
Again, my video slide show "Preliminary Study of The Inductive Resistor Heater" is on my YouTube Channel:
http://www.youtube.com/user/gmeast (http://www.youtube.com/user/gmeast)
Thank you
Quote from: gmeast on July 29, 2013, 02:19:44 PM
....
It then should be only a matter of arithmetic to scale the 'real' capacitor bank based on the same specie of capacitor. It could take well over 150 capacitor$ to get a sustained 3.2Watt draw for 8-Hours with only a 1VDC(max) voltage drop. It's possible to figure that out now just from the above requirements, but actual data is always best to have.
That's all for now ... only takes money.
....
Hi Greg,
Here is a useful link for estimating the Farads for the capacitor test: http://www.circuits.dk/calculator_capacitor_discharge.htm (http://www.circuits.dk/calculator_capacitor_discharge.htm)
My playing with some numbers brought this: you would need a capacitor bank in roughly 7750 Farad total capacity value and then starting the test from 25V and running it for 8 hours (i.e. for 28800 seconds) the cap bank voltage would drop to 24.5V and if you change this to allow the discharge down to 24V from the initial 25V, then the cap bank would need to have only half of the 7750 F i.e. roughly 3824 Farad. I used Imax=128000 uA discharge current (3.2 W at 25V) and for capacitor ESR I used 0.1 Ohm (you have to consider later the real capacitors ESR value how they calculate in the series and parallel assembly and recalculate the result with the real ESR value).
I suggest looking for prices at ebay... ::)
rgds, Gyula
Edit: I found this offer, perhaps a good one at present, still not enough in numbers of the caps though: http://www.ebay.com/itm/20x-Maxwell-2600F-2-5V-Supercap-2600-Farad-Super-Ultra-Capacitor-BCAP0010-USA-/150699241350?pt=LH_DefaultDomain_0&hash=item23165fef86
Quote from: gyulasun on July 29, 2013, 06:00:05 PM
Hi Greg,
Here is a useful link for estimating the Farads for the capacitor test: http://www.circuits.dk/calculator_capacitor_discharge.htm (http://www.circuits.dk/calculator_capacitor_discharge.htm)
My playing with some numbers brought this: you would need a capacitor bank in roughly 7750 Farad total capacity value and then starting the test from 25V and running it for 8 hours (i.e. for 28800 seconds) the cap bank voltage would drop to 24.5V and if you change this to allow the discharge down to 24V from the initial 25V, then the cap bank would need to have only half of the 7750 F i.e. roughly 3824 Farad. I used Imax=128000 uA discharge current (3.2 W at 25V) and for capacitor ESR I used 0.1 Ohm (you have to consider later the real capacitors ESR value how they calculate in the series and parallel assembly and recalculate the result with the real ESR value).
I suggest looking for prices at ebay... ::)
rgds, Gyula
Edit: I found this offer, perhaps a good one at present, still not enough in numbers of the caps though: http://www.ebay.com/itm/20x-Maxwell-2600F-2-5V-Supercap-2600-Farad-Super-Ultra-Capacitor-BCAP0010-USA-/150699241350?pt=LH_DefaultDomain_0&hash=item23165fef86 (http://www.ebay.com/itm/20x-Maxwell-2600F-2-5V-Supercap-2600-Farad-Super-Ultra-Capacitor-BCAP0010-USA-/150699241350?pt=LH_DefaultDomain_0&hash=item23165fef86)
Hi Gyula,
I very much appreciate your response. I also appreciate the time you spent on your calculations. The exponential discharge curve does mess things up a little. As you can see, the capacitor bank will cost $$thousands$$ of dollars. That's something I am not able to swing. But perhaps I will be able to provide convincing evidence using a shorter test duration. My goal is to run a system cyclically ... discharge and generate heat, then disconnect for a recharge and then repeat the cycle. I hope to eliminate the "Battery Effect" as mentioned in other threads by researchers such as M.D.
An important note: Some of Tesla's systems were isolated systems, or closed systems, and did not necessarily run in real-time concert with the grid. When attempts were made to do so, the systems became ineffective ... no benefit. I can't cite the patent, but I recall a street lighting system like this ... no true ground or neutral ... just 'floated'.
Thanks very much for your input Gyula,
Regards,
Greg
Hi Gyula,
Thanks again. It would take 12 of those to meet the voltage requirement alone ... about $7200.00. I hope those 'so-called' "experts" out there actually realize what it takes to "simply run it on a capacitor bank" to circumvent the "Battery Effect". It's NOT as easy as "simply running it on a capacitor bank" ... unless you're rich$$$$$$!
Now, I may be reading the EBay offer wrong ... if that's 20 Macwell 2600F 2.5V Supercaps then that's a good deal. I might be able to swing something like that, but I'd still need the bank to be pretty wide. I'll look into it.
Thanks again Gyula,
Greg
P.S. Did you ever watch the movie "The last Star Fighter"? Centauri (actor Preston Foster) mentioned someone by the name of "Gyula" ... cool name!