Capacitor Discharge Circuit with Energy Recuperation

[Rerouter]

you could possibly get away with a small dumbed down fly-back converter, short out the cap to ground through a small transformer via a transistor or fet, and on the fly-back kicks out a large enough voltage to feed most of it back into the supply rail. while at the same time pulling a negative on your cap which provided you limit it with a diode, should be able to discharge it in a sufficient time frame,

[NiHaoMike]

What about make the charging converter bidirectional? Just adding a MOSFET and drive circuit to the secondary side might be all that's needed.

[Simon]

How about using the power to run a 30-50V motor ?I mean 5mS charge and 5mS dicharge will be something like 30-50% duty

[KedasProbe]

Some other numbers to help ideas.
The average current from the capacitor will be about 1A.
100V*47µF/5ms = 940mA
100²*0.000047 / 2= 0.235 Joules (1 charge)
0.235J/0.005s = 47W (average inside the 5ms, apply 'duty cycle' to this)

[mzzj]

How about using the power to run a 30-50V motor ?I mean 5mS charge and 5mS dicharge will be something like 30-50% duty

Energy does not disappear in the motor. Motor converts electrical energy to mechanical energy and small portion goes to losses mainly as thermal energy.
Then you have to get rid of the mechanical energy 
Install brake to motor and then you got unwanted thermal energy 

More crazy ideas (instead of dumping energy back to charging converter or supply rail):
Convert it to RF-energy and let it radiate to free space 

[G7PSK]

Use the motor to drive a dynamo and feed that power back int the system. But to my mind unless you are testing thousands of these capacitors simultaneously it would be more economical to just dump the waste power.

[Hugoneus]

I skimmed through this, so I apologize if I am missing something.

It seems that you are charging and discharging a cap at a rate of 100Hz. If you are charging a cap, by any traditional means, you will dissipate as much power as you put into the cap. E = (1/2)*C*V^2.

Note how this equation does not have a resistance term in it! This means, even if you have two ideal capacitors (with ideal 0-ohm resistors) and one is fully charged and the other fully discharged, as soon as you connect them in parallel, some charge sharing will happen. After that, the total energy stored in both caps is half of what used to be stored in the first fully charged cap. The second half will disappear! In the case of ideal caps and wires, the energy will dissipate in the form of light, sound, and EM waves (from a spark). If there is a resistor in the middle, it will be dissipated mostly as heat.

Considering that, I don't see how you can charge and discharge that cap at 100Hz without dissipating (at least) as much power as you put into the cap.

[Simon]

that is what we have all been trying to explain to him

[luky315]

The old story that you loose 1/2 of the energy if you connect 2 capacitors is only true if you connect them directly. With 2 Diodes and an inductor you can discarge one capacitor into another with >90% efficiency.

[Kremmen]

WTF???
What story is this? Do you seriously propose that energy _disappears_ somewhere? Certainly not - you must have heard about conservation of energy? (or mass-energy if you insist on picking nits).
Assuming ideal caps and conductors and forgetting sparking, the energy goes nowhere else than into the caps.
Further assuming you have equal caps, one discharged and one charged to voltage V, connecting them to parallel will result in both having half the original energy resulting in voltage V/sqrt(2) for both.

[jeroen74]

No, half the voltage. There are quite a few discussions and papers on this on the net. It's an age old problem.

[Kremmen]

Do you have link(s)? That would be real interesting, because the energy has to go somewhere. Stating that half disappears just is not an answer.

P.S. Ah, i see it now. You are right as far as the voltage goes. In my consternation over the energy loss i neglected to account for all physical effects actually occurring in the cap.
Indeed, you only see half the voltage, but that definitely does not result from energy being lost somewhere. The "missing" energy is needed to maintain the charge concentration in the electrode plates of the cap. So no mystery after all.
But please, do not say that some of the energy is lost because that is not the case. The total system of 2 ideal caps maintains its energy perfectly, some of the energy is just in a form that is not visible as voltage across the cap.

P.P.S. Regarding charging and discharging the OP's cap is a different question and there of course you need to transfer the full cap energy back and forth to accomplish the goal. That energy must go to / come from somewhere at each cycle.

[Hugoneus]

Do you have link(s)? That would be real interesting, because the energy has to go somewhere. Stating that half disappears just is not an answer.

P.S. Ah, i see it now. You are right as far as the voltage goes. In my consternation over the energy loss i neglected to account for all physical effects actually occurring in the cap.
Indeed, you only see half the voltage, but that definitely does not result from energy being lost somewhere. The "missing" energy is needed to maintain the charge concentration in the electrode plates of the cap. So no mystery after all.
But please, do not say that some of the energy is lost because that is not the case. The total system of 2 ideal caps maintains its energy perfectly, some of the energy is just in a form that is not visible as voltage across the cap.

P.P.S. Regarding charging and discharging the OP's cap is a different question and there of course you need to transfer the full cap energy back and forth to accomplish the goal. That energy must go to / come from somewhere at each cycle.

If you connect a fully charges cap across a fully discharged cap the total energy that remains in both caps after everything is settled is half as much as you started. If the wires are ideal and there are no resistances, the energy dissipates into the atmosphere in the form of EM energy, light, heat and sound when a spark happens.

[Hugoneus]

The old story that you loose 1/2 of the energy if you connect 2 capacitors is only true if you connect them directly. With 2 Diodes and an inductor you can discarge one capacitor into another with >90% efficiency.

Can you provide a circuit for this?

[Simon]

Do you have link(s)? That would be real interesting, because the energy has to go somewhere. Stating that half disappears just is not an answer.

P.S. Ah, i see it now. You are right as far as the voltage goes. In my consternation over the energy loss i neglected to account for all physical effects actually occurring in the cap.
Indeed, you only see half the voltage, but that definitely does not result from energy being lost somewhere. The "missing" energy is needed to maintain the charge concentration in the electrode plates of the cap. So no mystery after all.
But please, do not say that some of the energy is lost because that is not the case. The total system of 2 ideal caps maintains its energy perfectly, some of the energy is just in a form that is not visible as voltage across the cap.

P.P.S. Regarding charging and discharging the OP's cap is a different question and there of course you need to transfer the full cap energy back and forth to accomplish the goal. That energy must go to / come from somewhere at each cycle.

If you connect a fully charges cap across a fully discharged cap the total energy that remains in both caps after everything is settled is half as much as you started. If the wires are ideal and there are no resistances, the energy dissipates into the atmosphere in the form of EM energy, light, heat and sound when a spark happens.

you sure ? light ?

[Kremmen]

If you connect a fully charges cap across a fully discharged cap the total energy that remains in both caps after everything is settled is half as much as you started. If the wires are ideal and there are no resistances, the energy dissipates into the atmosphere in the form of EM energy, light, heat and sound when a spark happens.

That cannot be right. Assuming the system is ideal there by definition will not be any losses. So no heat, no light no nothing related to any imperfection in the connection itself.
If we are extremely strict, then yes all accelerated electrons do radiate EM quanta and you can't move the electrons from one cap to another without accelerating and decelerating them.

Hmm, let's think about that for a moment... OK, heat and light are EM radiation so while there are no losses in an ideal system, the acceleration radiation technically qualifies because it is a property of quantum mechanics and EM in nature. Also now that i am thinking about it, there are 2 further possibilities; either we assume the connection to be totally lossless or then there are wires that have a distributed inductance and capacitance. The wire capacitance can be assumed to be not significant but the inductance will create a lossless resonant circuit with the original caps.
So in the case of no inductance and no resistance we will have one hell of a current pulse when the charge carriers balance (no losses remember). I am too lazy to think how to calculate the current magnitude, but probably you need to do it the hard way from first principles. Electric field force accelerating the mass of an electron or whatever you want to use as the charge carrier. Nope, too lazy to start that.  The peak current will anyway be _high_.
And if we do have inductance, there will be a lossless resonant circuit. Only not totally lossless, because accelerating the charges back and forth will again cause EM radiation only on a vastly lower scale than in the previous case. So eventually this damped oscillation will stabilize (provided the universe lasts long enough) and the charges in the caps will be equal. Of course i just say so; a lossless oscillation never stops but at least you can calculate the limit value where the stable point would be.

Without actually calculating it i am prepared to accept that the remaining charge in the caps is only half the original. The main thing was that the energy does not "disappear" into nevernever land.
It is a bit surprising but on the other hand it shouldn't. I vaguely recall some theory related to power transmission, where you can only get maximum of half to the other end. Been too long, can't really recall any more but must be the same thing...

[jeroen74]

Infinitely short current pulse of infinite magitude (Dirac pulse).

[Kremmen]

In pure theory yes, but if you include the mechanism of current flow (i.e. mobile charge carriers) then no, the pulse will not be infinite even in the ideal case. Dirac delta function is an abstraction that cannot be realized outside pure math. Not even in a lossless ideal electric circuit, provided you account for the necessary physical phenomena, i.e. provide a mechanism for the events in the circuit but neglect losses. And you must provide the mechanism, otherwise the fundamental explanation of energy dissipation does not work.

[luky315]

A circuit from another Forum:
http://www.mikrocontroller.net/attachment/91031/umlad.png