You [nul-points] previously posted:
> ...a dielectric which can support ion flow, not just electron polarisation
  eg. water in its liquid phase can exhibit both these properties...

If a dielectric supports ion flow but not electron flow - it is a simi-conductor and not a true dielectric.

as you can see from my post you just quoted, i didn't say 'supports ion flow but not electron flow'

Dielectrics do not permit electron or ion flow unless the undergo breakdown and become conductive then they are no longer dielectrics.

you appear to know quite a lot about this subject - more than Cornell Dubilier Electronics**, it seems 

(** CDE Inc., once the world's largest capacitor company)

You may want to read up on dielectrics and ions.

not really, looking at my library of stored webpages on the subject, i see i already did some reading around the subject this summer


here's just a couple quotes (because i like you)

eg., stored 28/07/08
  The Perreault Capacitor has more than twenty times the capacitance density of conventional capacitors, essentially unlimited cycle life, and stable operating performance. Unlike many batteries, the Perreault Capacitor can be safely and reliably operated throughout the -55° to +85°C temperature range. The high capacitance of the capture capacitor results from an electrostatic charge stored at the interface between activated carbon and an aqueous electrolyte in the so-called electric double layer. Since ions are physically stored rather than through chemical reactions, extremely long cycle life results.

and these, stored 29/07/08, from Cornell Dubilier Electronics, not quite as knowledgable as you admittedly, but still interesting 
  "Electrolytes have lower conductivity than metals, so are only used in capacitors when metallic plate is not practical, such as when the dielectric surface is fragile or rough in shape or when ionic current is required to maintain the dielectric integrity"

  "The disadvantage of electrolytic capacitors is the non-ideal, lossy characteristics which arise from the semiconductive oxide properties"

  "The primary purpose of the electrolyte is to serve as a "plate" on the outer anode oxide surface and also to connect to the cathode plate. The electrolyte is a high-resistivity, high-dielectric-constant, high dielectric-strength organic liquid solvent with one or more dissolved, ionically conductive solutes."

all the best
s.

as you can see from my post you just quoted, i didn't say 'supports ion flow but not electron flow'


you appear to know quite a lot about this subject - more than Cornell Dubilier Electronics**, it seems 

(** CDE Inc., once the world's largest capacitor company)


not really, looking at my library of stored webpages on the subject, i see i already did some reading around the subject this summer


here's just a couple quotes (because i like you)

eg., stored 28/07/08
  The Perreault Capacitor has more than twenty times the capacitance density of conventional capacitors, essentially unlimited cycle life, and stable operating performance. Unlike many batteries, the Perreault Capacitor can be safely and reliably operated throughout the -55° to +85°C temperature range. The high capacitance of the capture capacitor results from an electrostatic charge stored at the interface between activated carbon and an aqueous electrolyte in the so-called electric double layer. Since ions are physically stored rather than through chemical reactions, extremely long cycle life results.

and these, stored 29/07/08, from Cornell Dubilier Electronics, not quite as knowledgable as you admittedly, but still interesting 
  "Electrolytes have lower conductivity than metals, so are only used in capacitors when metallic plate is not practical, such as when the dielectric surface is fragile or rough in shape or when ionic current is required to maintain the dielectric integrity"

  "The disadvantage of electrolytic capacitors is the non-ideal, lossy characteristics which arise from the semiconductive oxide properties"

  "The primary purpose of the electrolyte is to serve as a "plate" on the outer anode oxide surface and also to connect to the cathode plate. The electrolyte is a high-resistivity, high-dielectric-constant, high dielectric-strength organic liquid solvent with one or more dissolved, ionically conductive solutes."

all the best
s.

Sandy,

You apparently have absolutely no understanding of the information you have quoted, so don't go around spouting off with stuff you do not even understand.

A dielectric that can support ion flow is either in a state of breakdown or has otherwise been rendered conductive - like they say in the info you just post where they add stuff to the dielectric fluid to make it an electrolyte.

The "Perreault Capacitor" sounds like a bunch of hype about a water capacitor.

Hi all,

This thread has gone a bit scew wiff?

Sandy has said that from the energy used from C1, there is more in total stored in C2 and used in R1 than was taken from C1.

So this is not about mis-understandings of 50% loss during charging with no inductive parts or differences with Switched Mode Power Supplies which can have theoretical efficiency up to %99.  Sandys test shows in the region of %147.

It is purely that according to his measurements there is more energy used in R1 and stored in C2 than was taken from C1 in the first place, but everyone is skirting around that and trying to pick out other things.

Measuring the starting energy in C1 and the energy taken from C1 cannot be at fault beacuse that is straight forward DC measurement and application of simple energy formula.  The energy stored in C2 after the test run is also a simple DC measurement and simple formula.  Just these two do not show any excess energy.

The excess shows up when you add the energy used in R1 to the energy left over in C2 and compare that to the energy taken from C1.  So the key lies in the scope measurement and calculation for energy used in R1.

It is an irregular waveform and Sandy has taken the average of the instantaneuous power in R1 over the complete test cycle.  In average measurements this may be perfectly acceptable, but here it could could be the difference between champagne and a red face.

.99 did a very nice PDF which showed how efficiency can increase (although it does not explain why the need to use more energy to charge a capacictor the more it is charged, can be over come (according to Sandy)).  But it does not explain how more energy can be used in total than was taken from the first capacitor.

The energy measurement in R1 must be confirmed.  I have performed a few tests and so far I cannot see any excess energy let alone confirm the measurement.  I have been very busy over the last few weeks but will try a few more tests before moving on.  We must not ignore things like the sampling rate of our digital scopes and the data we then exported to Excel which could give us different figures for the same test?

@.99, you mentioned that if you can charge a capacitor to a higher voltage it doesn't mean you have excess energy?  I am not sure of the circumstances, but if I have a circuit running off a cap that starts with say 10v and the circuit is able to use the energy stored in that cap and increase the stored voltage to 100v in the same cap and the capacitance value has not changed, then that would surely be excess energy would it not?  I am sure this is not what you meant when you said you had done that many times?



Regards,

Dave.

Sandy,

You apparently have absolutely no understanding of the information you have quoted, so don't go around spouting off with stuff you do not even understand.
...
The "Perreault Capacitor" sounds like a bunch of hype about a water capacitor.

i'm pleased to see that your sense of humour is as well-developed as your knowledge of hybrid capacitors 


well, gentlemen, much as i'd like to sit down with you guys, chewing the fat, the holiday is over and i've got experiments to run

all the best
s.

i'm pleased to see that your sense of humour is as well-developed as your knowledge of hybrid capacitors 


well, gentlemen, much as i'd like to sit down with you guys, chewing the fat, the holiday is over and i've got experiments to run

all the best
s.

You missed my point, but you are correct - I know nothing about hybrid capacitors.

Good luck with you experiments. 

Sandy has said that from the energy used from C1, there is more in total stored in C2 and used in R1 than was taken from C1.

No dispute here.

So this is not about mis-understandings of 50% loss during charging with no inductive parts or differences with Switched Mode Power Supplies which can have theoretical efficiency up to %99.  Sandys test shows in the region of %147.

I have no issue with this. If you check back a few pages (see link below) you will see that it was not I that brought up the discrepancy issue between normal capacitor charging efficiency vs. the efficiency of SMPS's. My response was to point out that there is no discrepancy. Both are as they should be. At any rate, I hardly think this is too off topic. It is all related and educational to reiterate.

http://www.overunity.com/index.php?topic=4419.msg145191#msg145191 (bottom of post)

It is purely that according to his measurements there is more energy used in R1 and stored in C2 than was taken from C1 in the first place, but everyone is skirting around that and trying to pick out other things.

Measuring the starting energy in C1 and the energy taken from C1 cannot be at fault because that is straight forward DC measurement and application of simple energy formula.  The energy stored in C2 after the test run is also a simple DC measurement and simple formula.  Just these two do not show any excess energy.

The excess shows up when you add the energy used in R1 to the energy left over in C2 and compare that to the energy taken from C1.  So the key lies in the scope measurement and calculation for energy used in R1.

If you check here: http://www.overunity.com/index.php?topic=4419.msg145134#msg145134 you'll see that the anomaly is actually in how quickly C1 gets discharged in the process. There is no dispute as to how much energy is dissipated in R1and left over in C2. Mine and Sandy's numbers agree quite closely on this. The question is why C1 discharges at only 1/2 the rate it should be in Sandy's case. In your case Dave, it may be discharging according to theory.

So Sandy either has a strange pair of capacitors (which could be possible), or his 2.5mH inductor is in fact tapping energy in the process and using it to partially power R1 and charge C2.

It is an irregular waveform and Sandy has taken the average of the instantaneous power in R1 over the complete test cycle.  In average measurements this may be perfectly acceptable, but here it could could be the difference between champagne and a red face.

As I stated above, the numbers all add up correctly, except for the C1 end voltage. There is no issue with the power in R1.

.99 did a very nice PDF which showed how efficiency can increase (although it does not explain why the need to use more energy to charge a capacitor the more it is charged, can be over come (according to Sandy)).

I have made subsequent posts that do in fact address this:
http://www.overunity.com/index.php?topic=4419.msg145897#msg145897
With ideal components (L and C), no work is required to charge a capacitor.

The only work performed in this entire process, is the work required to create the battery that charged the original capacitor. After this the energy can be transferred from capacitor to capacitor 1000 times and not a single bit of "work" will be done in the process.

When one understands this, it becomes quite clear why in the real world more energy is required to charge a capacitor.

But it does not explain how more energy can be used in total than was taken from the first capacitor.

As I explained above, there is some kind of anomaly occurring here, and it is related to either a strange capacitor effect, or an ou effect from the pulsing coil. I seem to recall there is a thread and video showing a Bedini type window motor running with only a large capacitor as its source of energy. It is this that allows me to believe that something odd may be happening in Sandy's setup, as shown by his C1/C2 discharge/charge scope shot. I have shown that his C1's discharge rate is 1/2 of theoretical, and this is where the mystery lies.

@.99, you mentioned that if you can charge a capacitor to a higher voltage it doesn't mean you have excess energy?  I am not sure of the circumstances, but if I have a circuit running off a cap that starts with say 10v and the circuit is able to use the energy stored in that cap and increase the stored voltage to 100v in the same cap and the capacitance value has not changed, then that would surely be excess energy would it not?  I am sure this is not what you meant when you said you had done that many times?

Mr. allcanadian is using a 12V battery, an inductor, and a charging capacitor. In your scenario, yes that would indicate excess energy entering the system, but what allcanadian is doing is using the inductive kickback to charge a capacitor to a voltage higher than 12V. The same can be done with a capacitor as the source rather than a battery, as long as the source capacitor is larger in value than the charging capacitor. A battery can be viewed as a capacitor with a huge value of capacitance and so almost any capacitor value used for the charging capacitor can be charged to a higher voltage than the battery. The final end voltage of the charging capacitor will be a function of the circuit LCR values.

Regards,
.99

A Simpler and More Reliable Way to Test for Overunity

Sandy,

The overunity you are currently seeing is either a result of some unknown capacitor effect or from extra energy being tapped by the inductor and added to the circuit, agreed?

If you agree, then the resistor R1 should have no effect on the net energy gain.

By eliminating R1, most of the net energy gain should be evident in C2's final voltage. Instead of losing energy in R1, we're now capturing it in C2.

So the proposed test involves removing R1 and just letting the circuit free-run so that C1 discharges, and C2 charges through L1. At some point the voltage on C1 and C2 will equalize, i.e. will be the same. There will still be some loss in the diode, but with 50% apparent excess energy, the relatively small loss in D1 will not kill the effect.

Use a scope on both C1 and C2 as you did before, and note the final voltage where they equalize. I figure it will take about 100ms to do so, but I would recommend you run it for about 200ms.

A final voltage of about 5.6 Volts is close to unity. If the final voltage of C1 and C2 is above about 5.6V, then that would indicate overunity. If you've got overunity, then I've got an idea how to extract the excess energy and use it to power a load in a continuous mode.

An excess of 50% energy would correspond with a C1,C2 final voltage of about 6.9V.

.99