OU/COP>1 switched cap PS cct like Tesla's 'charge siphoning'

[HEYDUDE]

Sandy, you have done a lot of hard work on the capacitor charging issue, but I would like to respectfully agree and disagree with a few statements you have made on your Doc Ringwood website in the "latest" section.

"text-book explanations of 'energy used' and 'energy stored' when charging a capacitor give equations which indicate that the external work done in charging a capacitor has the same value as the energy which gets stored in the capacitor as a result - NB. it is  not  the 'same' energy - it is just the same 'value' of energy."

"if a capacitor has just been charged, from empty, with 2 Joules of energy (for example) then the text-books state that an additional 2 Joules of energy must have been used as work to transfer the other 2 Joules into the capacitor - this means that a circuit has to convert 4 Joules of input energy to end up with 2 Joules of energy in the output capacitor"

(this is only true for a resistor feeding the capacitor from a voltage source. In this case one half the energy is spent in the resistor)

"it follows from this statement that a simple capacitor-charging arrangement cannot be more than 50% efficient."

(yes, a simple resistor from a voltage source cannot be more than 50% efficient in charging a capacitor, we agree this far)

"BTW this limit is not dependent on any series resistance in the charge path - larger resistances just reduce current and increase charge time, and smaller resistances do the opposite"

"it seems to be common knowledge, however - in the power-supply design industry, at least - that it's possible to achieve efficiencies > 50% if a series inductor is included in the capacitor charge-path, theoretically up to 100% efficiency"

(yes, I have worked in this industry and designed switchmode power supplies. The above phrase is correct, so far so good!)

"so are we now saying that a physical process which previously required us to do work to overcome the increasing internal polarisation producing the capacitor field has suddenly changed to NOT needing to overcome the increasing internal polarisation, just because an inductor has been added in the external current path?"

(here is the conceptual error and non-sequeter. No, the textbooks are not saying this, that is an assumption you have made. When using an inductor, it is a reactive component and in it's ideal form does not dissipate any power, just stores it and returns it to the source in the form of a pure current source. This is very different than a resistor and voltage source, hence the resistor looks really bad in comparison wasting 50% of the energy. This has nothing to do with your assumed "extra" work to overcome increasing internal polarization).

I would like to add that I believe there is a math error in your calculations of power dissipated in the resistor and inductor. I will point this out at a later time and give someone else a chance to find it.


Respectfully, HD

[Grumpy]

Where did the notion that Tesla "shuttled" energy back and forth originate?  Does this come from conceptions of resonant circuits?

If you wanted to use energy over and over, why not rotate it in a circle?

Eric Dollard seemed pretty convinced that the Tesla Transformer is a magnifier of transient impulses - nothing to do with shuttling.

Bedini seems to get good results charging and discharging caps - not sure why he does not see the losses that everyone else sees - perhaps he is polarizing the dielectric in an additional way - yeah that must be what he is doing. 


@HD
How many ways are there to polarize a dielectric?

[nul-points]

hi guys, thanks for the interest

@Grumpy, ...a quick answer which may help:
look for T. W Barrett's paper (PDF  format): Tesla's nonlinear oscillator-shuttle-circuit (OSC) theory

@HD
thanks for the review of the results - they're a bit of a moving target for readers, as i'm continually investigating new aspects of the setup and my conclusions on the website aren't always up to date with my latest findings

i've tried to present what i understand to be the 'official' line on the operation of such a circuit, but it's emerging that the equations appear to match the steady-state and the continuous current-flow cases better than they do the actual measured behaviour of the switched charge circuit

for example: i've just run some tests with a 30uF capacitor as C2 (C1 is 200uF; L is 2.5mH; R is 10ohm) - the results from this most recent test show a charging-to-stored energy ratio of approx 4:1 compared with the 'official' value of 1:1


> here is the conceptual error and non-sequeter. No, the textbooks are not saying this, that is an assumption you have made. When using an inductor, it is a reactive component and in it's ideal form does not dissipate any power, just stores it and returns it to the source in the form of a pure current source. This is very different than a resistor and voltage source, hence the resistor looks really bad in comparison wasting 50% of the energy. This has nothing to do with your assumed "extra" work to overcome increasing internal polarization

i think you misunderstand my statement about adding the inductor - i'm not talking about REPLACING the resistor with the inductor - i'm talking about ADDING an inductor to the existing C2 & Rload

the statement about extra work required to overcome increasing internal polarization is not mine - it is the official explanation about work being necessary to store charge in a capacitor, as given in a University text on the subject


my question is aimed at testing the statement, often repeated, that the inherent 50% efficiency achieved when charging a capacitor is changed by ADDING an inductor

i don't believe this is true

i believe my original question still holds - if work is needed to store charge in a capacitor (using an external series resistor and voltage source), then ADDING an inductor to these three components does not obviate the need to overcome the increasing internal polarisation within the capacitor as it charges

i agree totally that the energy stored in the inductor gets returned to the circuit when the switched charge pulse ends - it shows up clearly on my scope shots & in my measurements - you can see the current continues after the pulse ends (decreasing from the maximum value it reached at the final moment of the pulse, discharging down to zero)

i believe the I^2*R losses are still very much present - because external current is flowing into the capacitor - therefore external energy WILL be dissipated, either in an intentional resistive load, as in this test circuit, or else in the ESR of the capacitor, and the wiring, and the internal equivalent resistance of the voltage source

No - my results are showing that adding an inductor is not reducing dissipated loss (it actually increases it with the coil DC resistance) - it is also causing an excess of energy in the system - witness the charge gain anomaly in the circuit

the assumed boundary of influence for (at least) this circuit is incorrect - approximately 50% MORE energy is being converted (at the resistive load) than is being supplied by the input energy source

this energy is not being created out of nothing - i have to assume that Maxwell was correct in thinking that there is an energetic medium which is the ultimate source of energy in the Universe (call it: ZPE, aether, vacuum medium, etc) and we need to redraw the system boundaries to show that it should be included in our 'closed-systems' to allow the energy account to balance for circuits such as these

i look forward to hearing your thoughts on the error in the calculations

all the best
sandy
______________________________________________________________________________
Doc Ringwood's Free Energy site  http://ringcomps.co.uk/doc   ...bringing you measured Overunity results since May '08

[HEYDUDE]

Hi Sandy

Agreed, we have a slight communication problem. Indeed it takes additional energy for each incremental volt stored in a capacitor. That is why the equation is 0.5CV^2. The internal polarization does indeed resist further increases and more work must be done for each increment of voltage compared to the prior increment. However this work is all released upon discharge.

If this were not the case dual slope integrating type A/D convertors would not work or be extremely innacurate. I have designed these and have never had to take into account an "extra" work function because of polarization effects. The only real problem with these seems to be dielectric absorption phenomena causing slight  errors around the "zero phase" on the order of fractions of a percent.

When charging the capacitor from an ideal lossless inductor (no DCR, no eddy losses) that has been priorly charged from a voltage source, it now becomes a pure (lossless) current source and the magnetic energy in joules stored in the inductor, when released into the capacitor will charge it to the same quantity of joules. There is no "extra" amount of joules required to do this.

I have a whole stack of university level textbooks cluttering my desk, but common sense will tell you that the 50% loss is attributed to the non-ideal current source comprised of the voltage source plus charging resistor. This is the culprit and what the textbooks base the "simple charge equation" on. The 50% energy is lost as heat in the resistor, because in this case all of the current must flow through the resistor and it is a lossy device by nature.

In your latest test you might want to examine how you arrive at your power dissipated in the resistor and inductor DCR.  The duty cycle is 0.12mS on time, 0.28mS off time and 0.4mS cycle time. It appears you are taking the entire time (9.16 mS) to arrive at your power dissipated in the resistor and inductor DCR. It should be closer to 2.64mS.

You state there are 22 cycles of 0.12mS on time. This equals 2.64 mS. total on time, not 9.16
The total off time is then 0.4mS-0.12mS  x 22 cycles = 6.16mS off time.

Also it is difficult to compute the actual RMS current per cycle due to the irregular waveform. You might also want to examine the actual power per cycle based on a changing amplitude over 22 cycles.

Kind regards HD

[Grumpy]

@HD
How many ways are there to polarize a dielectric?