Joule Ringer!

[hartiberlin]

Okay Conrad and Bill,
great new scopeshots, now one can really see it all much better.

Now you can see, that the recharge spike, (when the voltage rises over the 12 Volts)
has  almost  exactly the same area as the first spike discharging the cap.
Only the next 3rd smaller discharge spike has to be avoided somehow.

To see it even more better we need to have 2 channel scopeshots.
So simulataneously show the input current waveform (scope channel B)
with the cap voltage ( scope channel A)

Please do it like this and post again the scope shots.

This way we will see, what the input current is doing while
the cap voltage rises or falls.

Then after this we need to find out via other scopeshots at the
transistor(s) and coils, where we can suppress the 3rd spike and
make the recharge spike faster...

You see, we really need the right scopeshots at the right locations to analyze this better
and to exactly produce the right waveforms, which will recharge the cap even better
and to know, from which pulses the cap is recharged the best way.

Regards, Stefan.

[hartiberlin]

Okay,
so here is another analysation from Conrad´s scopeshots,
where the cap voltage over time is shown:

When we open the transistor some current pulse will flow,
so that the cap voltage drops.
This is the red area in the attached scopeshot.

Now we have a recovery phase, where some coil components
fire BackEMF current back to the cap and this will recharge the
cap. This is the green area, where the cap voltage rises again.

Then there will be these blue marked areas, where the cap voltage
is decreasing via these 2 spikes.
These areas could be suppressed via some diodes or simular, so
that in these time intervalls no current will be drawn from the cap.

Regards, Stefan.

[hartiberlin]

As we are looking here at the cap voltage,
we could even switch the cap away from the circuit,
when the positive spike has charged the cap up over the
starting voltage !

This way we could start the next cycle a few milliseconds later
and the cap will charge up and up again and the voltage will rise
on it...

This only needs a sophisticated peak voltage comparator and a fast
on-off switch.

So you see what kind of joy already these waveform show us ,
if you analyse them the correct way !

[Pirate88179]

Stefan:

Great idea!  The only thing I can think of that can switch that fast might be a small transistor...say the 2N3904 or something that takes so little to run maybe?  This is getting more and more interesting.

Bill

[xee2]

So you see what kind of joy already these waveform show us ,
if you analyse them the correct way !

I do not think you are interpreting the waveforms correctly. What they are showing is the voltage developed across the internal resistance of the capacitor due to current flow. You can consider this to be a small resistor in series with the capacitor. When current is flowing out of the capacitor the voltage developed across the resistor subtracts from the static voltage of the capacitor. When current is flowing into the capacitor the voltage adds to the static voltage of the capacitor. The area under the curves is the energy leaving the capacitor and the energy returning to the capacitor. The interesting thing that the waveforms show is that the energy going back into the capacitor is almost the same as the energy coming out of the capacitor during each cycle. Thus the net loss of capacitor energy is very small in spite of the currents flowing out of it (since most of the energy is returned).

If you try to put a switch on the capacitor, the voltage will disappear as soon as the switch is opened since there would no longer be any current to generate the voltage.