How to fast charge capacitor 2200uF/400v in fewer seconds?
I lot has been discussed elsewhere on the web regarding a Baxandall style design. You might look that up for an efficient approach. Personally, I find them harder because of the requirements for probably having to do a custom transformer, which I'm not fully adept at making. So I wouldn't go that route and would pick something easier to do for a hobbyist.

An approach I've used before, but not for this peak voltage, is a discontinuous mode approach. You store energy in the inductor and then release that energy. And there are plenty of inductors available that are designed to work this way, too. Plus, they are not hard to wind or make. At a low frequency like this, I might just use an iron core and file a gap into it for these purposes. But ferrite can work fine here, too.

There are lots of design details that go into the inductor. You have to be sure that the wire can handle the peak current and you want to use an even larger diameter wire than that to reduce the resistance still more to reduce simple resistive losses. But thicker wire also means that the windings require a larger core window. To go with that, there are limitations on the core volume given the energy of each pulse. And using ferrite, with its own limitations, means even bigger still. (Which is probably why I'd go with iron and gap it, instead.)



Let's start with power considerations. Those are simple. You need the energy on the capacitor to be \${1\over 2}CV^2\approx 100\:\text{J}\$. You need that done in \$10\:\text{s}\$. So this works out to \$10\:\text{W}\$. That an easy calculation. At a guess, I'd say you'd be lucky if you can do this with an average of \$25\:\text{W}\$ from the power supply. That doesn't say anything about the peak current compliances that are also required, though. So keep that in mind. it's just that I think you might be lucky getting 40% efficiency and you should plan on it being that bad (or worse, perhaps.) This isn't an easy problem if you also include high efficiency into the equation and it is perhaps beyond me to help, then. (I am just a hobbyist and I have no special experiences in this area, besides.)

So that's one detail. And already you know that the average current must be more than \$2\:\text{A}\$ during this period (counting inefficiencies noted above.) You can arrange things so that the difference, between this average and whatever peak requirements are worked out, is handled by some largish capacitor on your power supply. But this is a later consideration. Hold the thought for now.

Now, we know that you need to deliver at least \$10\:\text{W}\$ for a period of \$10\:\text{s}\$ via the inductor itself. (There is no escaping this minimal fact, regardless of the efficiency or inefficiency of the rest of the circuit.) Assuming a fixed frequency (you can do better by varying the OFF period because this changes as the capacitor charges up; but this makes the circuit more complex and I want to avoid that problem), where \$f={1\over t_\text{ON}+t_\text{OFF}}\$, this is a relatively easy equation to write:

$${{{1\over 2} \:I_\text{PK}^2\: L}\over t_\text{ON}+t_\text{OFF}}\ge 10\:\text{W}$$

We still have to solve for the ON and OFF times (and therefore also the frequency), the size of the inductor, and the peak current. And again, keep in mind that the above equation relates only to what the inductor itself must be doing to push energy onto the capacitor. It does NOT address itself to losses elsewhere in the circuit, which is a separate subject.



Here is the circuit I'm imagining for the purposes of the remaining discussion:







simulate this circuit &ndash; Schematic created using CircuitLab

There is something (an MCU here) driving what amounts to a Darlington switch. There is an inductor attached to this and a free-wheeling diode that is supposedly capable of handling the peak currents as well as the reverse voltage requirements (which are not small.) A Schottky diode might be appropriate for its lower barrier voltage, but it's not easy to find ones capable of standing off that high of a voltage difference. So I'll leave the selection of the diode up to you.

Approximately \$11\:\text{V}\$ will appear across the inductor when the circuit is ON. This will cause a rise in current and, held long enough, will reach our desired \$I_\text{PK}\$ along a nearly linear ramp upwards. This will take time, which I call \$t_\text{ON}\$ above. Once this peak is reached (and this time is known a priori by whatever the design is here), the MCU then turns the circuit OFF and this allows the inductor to self-discharge into the charging capacitor.

Initially, the \$+12\:\text{V}\$ source voltage will just go through the inductor and diode and charge up the capacitor to \$\approx 11.5\:\text{V}\$. Or thereabouts. That will be the starting point for the rest of the charging cycle. This detail helps determine the allowable frequency, as it allows us to make some improved predictions about the worst-case OFF time, \$t_\text{OFF}\$. Another determination of that might also be the so-called time constant that exists between the inductor and the capacitor. However, it turns out that while this matters a little bit at the very early moments, it won't really matter so much over the long haul of charging. It may impact calculations about the total worst-case Webers. But I'm not even going to worry about that for reasons I don't want to waste time on, here.



At this point, some design questions start to enter into this. For example, what is the peak current that your batteries can reliably support? You've already mentioned that you are using a \$12\:\text{V}\$ lithium battery that is rated at \$5500\:\text{mAH}\$. Frankly, I don't know what that means. Is it a "3S?" Or a "4S?" Also, these are probably going to be limited to just an amp or two. And frankly? If that's all it is capable of then you aren't going to make it in time. You need something capable of more than a couple of amps to get from A to B in this project. (Because I think we need to plan on more than an average of \$25\:\text{W}\$ from your battery, so that it can support substantial peaks above that average from time to time while operating.)

So now I'm in need of a datasheet for your battery, because the design cannot continue without important details about your battery system and what it is reasonably and rationally capable of delivering.