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

[poynt99]

when supplying current direct to the capacitor (no coils or rectification), the text books claim 50% energy efficiency

The text books are correct in this regard. I have proven this to myself with a test of my own here:
http://www.overunity.com/index.php?topic=6090.msg143650#msg143650

as you've seen from the results of the switched charge tests here, it appears that its possible to refine the coil/rectification approach to get up to 150% energy efficiency on a resistive load

At the moment I must reserve judgment on whether in fact it is possible or not. So far there are only two independent experiments using your circuit configuration; yours which seems to indicate overunity, and Dave's which seems to indicate underunity. Until more independent tests are done (including my own), for me the jury is still out on this one.

Regards,
.99

[nul-points]

> Quote from: nul-points on Today at 01:30:51 AM
> when supplying current direct to the capacitor (no coils or rectification), the text books claim 50% energy efficiency

The text books are correct in this regard. I have proven this to myself with a test of my own here:
http://www.overunity.com/index.php?topic=6090.msg143650#msg143650

ok, i think we're getting to the heart of this anomaly now...

you see, either a capacitor inherently always needs 50% external work performed in charging it or not

the text books say 'yes it does - a work function is required to over come the increasing polarisation of the dielectric as the cap contains increasing charge' - the energy losses, L, are around 50% of the input energy, E
  eg.  E - L >= 50% of E

in which case, improving the overall efficiency with an LC arrangement, like your sim and my tests show, must be adding excess energy, X, to the system - in addition to the inherent losses, L,  of charging the cap
  eg. X + E - L > 90% of E

if this is not the case, then the text books are wrong and it is possible, with a suitable arrangement, to charge a cap with minimal external work
  eg. L < 10% of E

this is the point of my earlier statement comparing the claimed efficiencies of SMPS and inherent cap charging


interesting, then, that your testing confirms that L is around 50%, don't you think?

all the best
s.

[nul-points]

...correcting typo above:-

    E - L = 50% of E

[poynt99]

ok, i think we're getting to the heart of this anomaly now...

you see, either a capacitor inherently always needs 50% external work performed in charging it or not

the text books say 'yes it does - a work function is required to over come the increasing polarisation of the dielectric as the cap contains increasing charge' - the energy losses, L, are around 50% of the input energy, E
  eg.  E - L >= 50% of E

in which case, improving the overall efficiency with an LC arrangement, like your sim and my tests show, must be adding excess energy, X, to the system - in addition to the inherent losses, L,  of charging the cap
  eg. X + E - L > 90% of E

if this is not the case, then the text books are wrong and it is possible, with a suitable arrangement, to charge a cap with minimal external work
  eg. L < 10% of E

this is the point of my earlier statement comparing the claimed efficiencies of SMPS and inherent cap charging


interesting, then, that your testing confirms that L is around 50%, don't you think?

all the best
s.

I sense a fundamental misunderstanding of how and where the losses occur, and how and why with the addition of a proper inductor one can increase the efficiency substantially, theoretically to 100%

If we assume the simple case without any inductors, the text books are correct and 50% of the energy used to charge a capacitor is lost mainly due to dissipation in the resistive elements in the circuit. This limits the energy transfer efficiency to 50%. We also know that it makes no difference at all how small or large the value is for the resistance in the circuit, the total energy lost will always be 50%.

With the addition of a properly selected inductor to the circuit, the energy transfer efficiency can be increased to a theoretical maximum of 100%, although in practice due to various resistive elements, some loss is exhibited and practical limits are 95% to 97%.

So the addition of the inductor promotes the reduction of losses. It does not add excess energy to the circuit.

How does the inductor accomplish this feat?

It's really quite simple and it comes down to resistance, resistance, resistance.

A superconductor does not dissipate any energy whence power flows through it. For all intents and purposes, it has zero resistance R and hence P=I²R = 0 Watts. If no power is dissipated in this superconductor, then it is also true that no energy is lost in it.

Unfortunately, superconductors are still hard to come by in a practical sense, so for now we will have some finite resistances and their associated losses to deal with in our circuits. There is hope however.

By introducing an inductance into the circuit, we now have a storage element as well as a dissipative element (our finite resistances) in series. What we want to do is make the ratio between the inductance and resistance as high as possible. This offsets the balance between the dissipative element and storage element to the point where the storage element largely dominates. Remember a resistive element can only dissipate energy, it can not store it. The higher we can make this ratio, the higher will be the energy transfer efficiency (maxing out at 100%).

So now we have 3 storage elements in series; C1-L1-C2. Capacitors store energy in the form of charge, and inductors in the form of magnetic flux. Assuming ideal components, C1 dumps 100% of its energy to L1 where it is stored as magnetic flux, and after the switch opens L1 transfers 100% of this energy to C2 where the energy is once again stored in the form of charge. In this lossless process, 100% the energy has been transferred from C1 to C2 and it has changed form twice. L1 acted as the lossless transfer medium. L1 is ideal so it did not lose any of the energy in the process. It was charged, then discharged of its magnetic flux.

To be fair, we must re-introduce a tiny bit of resistance (r), because in reality that is what we have: C1-r-L1-r-C2. Because of this tiny resistance, the balance sways back a little to the lossy side, and the process becomes imperfect. The "r" is where the loss occurs (as small as it may be) and it's one we can't escape at the moment.

If we go back to the beginning scenario without any significant inductance we have C1-R-C2. The "lossless" storage element is gone and we are left only with a dissipative one that consumes half of our applied energy, the ubiquitous circuit resistance!

.99

[nul-points]

hi Poynt

I sense a fundamental misunderstanding of how and where the losses occur

that's the nicest way i've seen of saying that someone doesn't know what they're talking about   

nice lecture, BTW


let's get back to the original point:  i commented that there's a discrepancy between the energy efficiency claimed for SMPS operation (>90%) and that claimed by text-books for the fundamental process of getting charge stored in a capacitor (50%)

we're told by text-books that external work has to be done to charge a capacitor because, as it charges, the polarisation of the dielectric increases and the applied source has to do work to overcome this increasing opposition

the same text-books do not appear to offer exceptions to this inherent behaviour - so IF this WERE a fundamental truth about the work function involved in charging a capacitor then it OUGHT TO apply in whatever circumstances a capacitor gets charged

i am not saying that the text-books are correct or incorrect on this - i am pointing out an inconsistency between their explanation of why work needs to be done in charging a capacitor (side effect: 50% efficiency) and our experience of real-world circuits, like some SMPS systems (90+% energy efficiencies), which may involve the frequent repetitive charging of a capacitor

IF the text-book justification for work expended in charging a capacitor is INCOMPLETE or INCORRECT then we can expect to be able, with a suitable circuit arrangement, to charge a capacitor with little penalty of energy loss

IF HOWEVER the text-books are CORRECT and COMPLETE in their statement about cap charging requiring work to be expended, then this is not consistent with the system efficiencies, >50%, we see when we include inductance and rectification or switching into the charge path

our experience does not tie-in with the text-book

my point is that either there is excess energy available or it IS possible to charge a cap without increasing work required - either way, the text-books need to be updated


all the best - happy holidays, all
s.