Can someone explain this?

[nul-points]

hey Cap'n

when you connect the two capacitors together (without a diode) the energy initially flows between them, from high to low energy state, to equalise between the two (imagine joining two similar large cans, one full with water, the other empty, with a fairly large-bore tube between the base of each)

the circuit will have some resonant frequency (depending on cap size, wiring resistance etc) - think of the energy flow as having some momentum

so the energy 'oscillates' slightly between the two caps (ie. before it settles to a steady equalisation between the two caps), with a little overshoot into the second cap which then decreases whilst overshooting back into the first cap, followed by an overshoot back into the second again, etc - each overshoot decreasing in amplitude until the voltages have equalised

[your results show different final voltages on Cap1 & Cap2, so either you disconnected the two caps to take a 'final' reading, or you have some appreciable wiring resistance, and the oscillation was still happening as you made the reading]

each time the energy flows back & forth through the connecting wires it dissipates some energy in the wire resistance, so the total energy in the circuit will reduce with time (& therefore the two cap voltages 'should' total less than the initial Cap1 voltage)

by adding a diode in series with the connection you allow the flow to occur in the first direction but now the energy can't flow in the reverse direction (imagine your tube between those two cans of water now has a one-way valve inline) - so you only dissipate an energy loss in the wiring once

different diodes (with different Forward volts drops & different reverse current leakage) and different initial voltage conditions will affect this scenario by altering the threshold at which energy is prevented from oscillating back

hope this helps

all the best
sandy
Doc Ringwood's Free Energy site  http://ringcomps.co.uk/doc

[captainpecan]

Thank you for your input, most of that I do already understand though. And how I am doing it is keeping it an open circuit...  discharging Cap 2 down to absolute 0v... Charge Cap 1 up using a 9v battery... then I briefly complete the circuit. I then disconnect the caps and measure them individually.

I expected to see somewhere in the line of a .7v drop by adding the diode to the circuit. Even if it wasn't that much, I expected to see SOME KIND of drop in voltage, larger than not having it in the circuit at all. This is where I am confused a bit, it appears that there is next to no loss at all by entering a diode into the circuit. I guess I am misunderstanding what the voltage drop of a silicon diode being .7v really means? I'm not seeing it when used in this manor.

[nul-points]

hi Captain

a diode is not a perfect uni-directional conductor

- firstly, it conducts both directions (it just happens to conduct a whole lot better one way than the other)

- secondly, it's voltage/current curve is not really discontinuous at a micro-level: it varies continuously from 0V upwards, and with a different curve from 0 downwards (the curves are non-linear, of course)

so in reality a regular silicon diode will only have a forward voltage drop of 0.7V at one particular current - vary the current & the voltage drop will vary

in this respect a diode is much like a resistor - it's just that in the case of a silicon diode the resistive Volt/Amp slope for the forward voltage polarisation is stacked around a 0.7V level

connecting one capacitor to another through a resistor won't affect the final voltage - if left long enough - it just affects the current flow & therefore the time before equalisation


so using a diode (which looks more or less like a unidirectional resistor as far as the caps are concerned) should affect the current flow (and therefore the charge rate)


but, as i mentioned previously, the diode also affects the oscillation of the energy back & forward thro' the wiring, so with only one forward flow of energy you only get one lot of energy dissipation - so, adding a component can give you less losses (because of it's rectifying behaviour)


instead of only briefly making the circuit, you could try your setup with the circuit left connected and compare the results with your original set

you should still see a difference in losses with & without a diode, but the final voltages across each cap should be approximately the same with & without the diode


you could also magnify the difference in losses by adding an additional resistor (say 1Kohm?) in series with the connecting wire (when testing with & without the diode)

monitoring the voltages on a storage scope (PC?) gives a better view of the dynamics of the energy oscillation and voltage charging behaviour than using a meter

all the best
sandy
Doc Ringwood's Free Energy site  http://ringcomps.co.uk/doc

[captainpecan]

That helps a lot. It makes more sense now to me. Basically, having the diode makes less loss in the circuit as there is only one shot through it. Then it adds some loss because of the diode being there. Adding them up in this scenario happens to show less loss, but in other scenarios it possibly could show more. This makes sense to me now, especially knowing that the .7v loss is not a constant.

Has anyone figured out a way to force more than half the charge over to the second cap? By theory of how they work, It doesn't appear to be a workable concept, but hey, I guess that's why I'm here!

[nul-points]

hi CP

> Has anyone figured out a way to force more than half the charge over to the second cap?


with suitable initial voltage on Cap1,  a DC to DC convertor could do the job

you could also pulse it thro' a transformer

...but hey, that's why I'M here! 


all the best
sandy
Doc Ringwood's Free Energy site  http://ringcomps.co.uk/doc