Silly question about capacitors

[poynt99]

How about one more time. 

See the attachment for more info on capacitor energy transfer and how to maximize it.

[Farmhand]

Thanks Poynt99, I have that filed appropriately now so i can link it in future.

....

To anyone, does anyone have the time to simulate my circuit example drawing in Falstad simulator or something ? I am trying to learn LT spice, but I don't have a lot of time lately to learn new stuff with my medical issues and chores ect. If I had time to learn it I would simulate it myself. All we need is a DC one shot pulse to the mosfet or switch, the pulse width can vary, but should not allow full current through the series inductor, the 10 Ohm load would limit the current to 1 Ampere max anyway with the circuit pre-charged to 10 volts.

Maybe a longer pulse than 1.5 mS might work better or more or less load resistance to vary peak current

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Mags, pretty good explanation, the inductor acts as a low loss load which when the switch is opened discharges it's stored energy into the smaller capacitor and raises it's voltage above that of the supply capacitor, that can happen due to the inductors properties.

Cheers

[Magluvin]

To expand on it a bit further, how can we get an actual 50% of the 300psi(total energy) into each tank? Meaning, the 2 tanks together equal round abouts the amount of energy of the original 300psi tank.

We would need the 2 tanks to round about 212psi each, and used together can do as much work as the single 300psi tank. Even though the pressures are different, 300psi(1 tank) and 212psi(2 tanks), the same amount of work can be done with each. The 2 tanks of 212psi may seem like less than 300psi, but we have more total air in the 2 tanks than the 1 tank at 300psi. Like the difference between a 12v batt at 12ah and a 24v batt at 6ah.

But how can this be done?   

We have our 2 tanks, 1 full(300psi) and the other empty, and we have our air motor with flywheel setup. Now we open the valves. The pressure from the full tanks drives the motor/flywheel and begins to pressurize the empty tank. But, we have an added redirect valve between the source tank and the air motor. When we see the source tank get down to 212psi, we turn the added valve. This valve closes off the source tank and opens the input to the air motor to the open air instead. Now, at this point, we stopped the source tank from depressurizing and it now contains 1/2 the energy of the original 300psi. But our flywheel is still going, and pumping the originally empty tank with outside air. And if there wasnt losses in the motor and such, it would be able to pump the second tank to right about 212psi. Now we have 2 tanks that have the same energy potential of the original single tank that was 300psi.      But one thing has changed. There is more total air in the 2 tanks than was in the single 300psi tank, because we pulled in air from the outside once we turned our newly added redirect valve to do so.

With caps, in this situation, we are basically doing the same thing, except once we turn our redirecting valve, the flywheel(inductor) is just transferring from one plate of the cap to the other.
So my analogy would be better to have 2 air tanks for the source and 2 tanks for the empty cap. The problem would be that the source tanks would need to be 1 full at 150psi and the other source tank at a negative 150 psi to have a clearer comparison between the 2 ideas. But the single tank to single tank gives a basic and simple understanding.

Mags

[Magluvin]

Mags, pretty good explanation, the inductor acts as a low loss load which when the switch is opened discharges it's stored energy into the smaller capacitor and raises it's voltage above that of the supply capacitor, that can happen due to the inductors properties.

Hey Farmhand

The inductor/flywheel is a way of capturing energy that is lost because of a stupid way of losing it.
Think. Do you believe that the energy lost in the direct transfer between 2 caps can be captured by storing the heat created during that transfer? Or by capturing the 'radiation' ? Maybe a little will be recovered, but definitely not all or near 50%, no way no how. Lets say the air tank analogy produces heat in the hose.  Does that in any way explain how we took 1 tank full, directly to 1 tank empty till they are equal, has anything to do with the 50% loss in initial source energy??

There is something a miss there with the ideals of the so called losses of direct cap to cap transfer. I have some new ideas on this lately that may blow your mind, but I will reveal them when I have it all down pat myself.    It took me a while the get this far to be able to explain what I have here.

Mags

[Marsing]

hi all..  nice explanation

what is formula the time needed to transfer from c1( full cap ) to c2 ( empty cap)?
according to VAsic041 pict, i took about 200ms