Inrush current of high voltage power supply
The problem with a transformer/rectifier is that it's a low impedance voltage output power supply. Connected to a discharged capacitor, a very high current could flow. You can take the sting out of this by using a series resistor. However, this is not a 'nice' solution as it (a) wastes power and (b) provides a very low charge rate as the capacitor gets to a similar voltage to the supply.

Charging a capacitor is really a job for a current source. At low power, a conventional linear regulator configured for output current control would do. At high power, you need some sort of inductive energy storage somewhere.

One solution is a boost converter, configured for current output control. Related to this is a flyback. In both cases, you store energy in an inductor, and it feeds this out to a high voltage at a controlled current. They differ in isolation, and the degree of step-up that they can achieve efficiently.

An alternative is the semi-resonant voltage doubler. It's strongly related to the boost converter, but is automatically operating.







simulate this circuit &ndash; Schematic created using CircuitLab

Assume C1 starts off uncharged. Close SW1. The supply voltage appears across the inductor, and current starts to build. If you want to describe the waveforms for the current and C1 voltage, then L1 and C1 are most easily understood as a resonant circuit. After one quarter cycle of the L1C1 resonance, C1 voltage is equal to the supply, and the current has stopped increasing. The next quarter cycle of the resonance sees L1 continuing to push current into C1, whose voltage rises to twice that of the supply, while L1 current falls back to zero. 

As L1 current tries to go negative, it's blocked by D1. The charging cycle is now finished. No energy has been dissipated. The peak current is controlled and predictable. The charging time is predictable and takes exactly one half cycle (hence the name semi-resonant) of the L1C1 natural frequency.

If C1 starts off with some initial voltage, then it doesn't rise to double the input voltage, only increases by twice the difference of the supply and its initial voltage.