TK:

I believe a RAT conclusion is that here is a net reverse current flow back into the battery.  Certainly they allege that there is a net power flow back into the battery.  That is also synonymous with "COP infinity."  So that's why it's interesting.

Keep those delicious lip-smacking oscillations coming!

MileHigh

Ok, here is a very simple example.

The switch at the top right is held on til the cap fully charges through the load and then open the switch.

Now we close the switch in the middle of the circuit to discharge the cap into the load, without the battery. When the cap is fully discharged, the cycle repeats.

The scope shots show the source on the left and the load on the right.

We can recycle energy. Use it more than once.   

Did we gain anything? 


Mags

@Mags: Can you have your sim integrate those two sets of traces? Integrate over the time the negative waveform on the left, and compare that to the  total integration of the two positive waveforms on the right.

Ah... the wonders of modern computing. If I want to integrate a waveform on the HP180, I have to trace it on paper, cut out the tracing, and weigh it on an analytical balance.

TK:

Are we permitted a little diversion while the tar is brewing?

Scenario 1:  So, you have a 1 kg mass sliding happily along on a magic frictionless plane at 1 m/s.

Scenario 2:  The 1 kg mass sliding happily along at 1 m/s on a frictionless plane hits a stationary mass of 1 kg.  When they hit they stick together and keep sliding happily along.

Scenario 3:  The 1 kg mass sliding happily along at 1 m/s on a frictionless plane hits a stationary mass of 0.5 kg.  When they hit they stick together and keep sliding happily along.

So the question is, what are the energies in scenario 1, scenario 2, and scenario 3?

MileHigh

Yes i can. I see why now also.  The source trace is longer.

Also with the cap in series with the load, we have a voltage division of the source.

Sooo, with a 5v source, as the cap charges, the source -cap=vr

So the load only gets half of what the source lost, and the cap gets the other. Then the caps "half" delivers the other half to the load.

Yup. Looked good in the brain. ;] 

Soo is that why we seem to lose 50% of energy when we discharge a cap into another cap till they even out. Voltage division. Not a loss in heat.  No?

Mags

No, I don't think so. The losses will always wind up as heat, or maybe RF radiation, which is the same thing just lower frequency. The cap has an "ESR" or equivalent series resistance which is dissipative, and also it does take work to stretch lattices and move electrolytes around and jiggle ions and stuff like that there. But that's not what's causing the voltage drop equalization.
The Energy in Joules on a capacitor goes as the square of the voltage and linearly with the capacitance: E = (C)(V^2)/2. If the caps were perfectly lossless you wouldn't lose energy by the voltage division, just voltage. Energy is the conserved quantity. So you have a cap with known capacitance in Farads and you charge it to a certain Voltage. This gives you a certain amount of Energy in the cap. Then you discharge into another uncharged cap. The voltage will equalize. Now you have apportioned the original energy between the two, in ratio determined by the ratio of the capacitances, at the new equilibrium voltage. Minus some losses from heating and RF and such. If you know the second capacitance you can calculate the equilibrium voltage, and vice versa.

The hot trick, for me, is to charge caps at high voltage in series, then discharge them at lower voltage in parallel. This is how to extract energy from the Earth's electric field and "down-convert" it to useful power.

Well...

In scenario 1 the energy is 0.5 Joules but in scenario 2 the energy is 0.25 Joules.  What gives?

Well...

In scenario 1 the energy is 0.5 Joules but in scenario 2 the energy is 0.25 Joules.  What gives?

How do you figure that?

Mv initial = mv final. (CofM).
Scenario 1: 1 kg moving at 1 m/s = 1 kg-meter/second. Energy  = 1/2 (mv^2) or 0.5 Joules.

Scenario 2: 1 kg mass moving at 1 m/s = 2 kg mass moving at x meters per second. (CofM) Solving for x, we have x = 0.5 m/s.
Solving for energy, we have E = 1/2 (mv^2) or 1/2 (2 x 0.5 x 0.5) = 0.25 Joules... therefore aliens.

But.... you've made a trick. The simple Energy Conservation law applies to _elastic_ collisions. The part about the two weights sticking together and moving off together in the same direction means that your collision is inelastic. Conservation of momentum still applies simply. But the CofE part now needs to take into account the energy lost to sticking together, deformation, heat and so on. The fact that energy does NOT appear to be conserved in the easy naive calculation is the indication that the collision you are looking at is inelastic, and energy is lost to the moving system. If you could account for all the losses (by enclosing the whole shebang in a perfect calorimeter, for example) you would see that energy is still conserved.

This is sort of like all the little resistances and radiations that will suck the energy out of the batt-cap-switch system. A perfect inductor is sometimes easier to understand because we have "touchy feely" experience with storage of energy in a magnetic field and its conservative return, using permanent magnets. We don't have this same fingers-on experience with the electric field energy storage that happens in capacitors, so they seem especially mysterious. It's just another kind of spring, that's all, with its own set of losses that suck energy out of the spring's motion until eventually it's all gone, lost as heat.

But how do the systems know to lose exactly half their energies to heat, and keep half in KE or capacitance?