What can you say about the current limiting circuit charging an electrolytic capacitor?

Question

Do it yourself

The main question concerns the current limiting circuit used on the bipolar transient (or soft start, as it would be more correct). When the transistor is fully open (after the capacitor is charged), the voltage drop across it is less than 1V. I plan to power the lighting network from LED lamps in the house (about 150W). I would like to hear an outside opinion. First of all, the correctness of the scheme is of interest. It is also worth noting that it works at the temporary hut (now exactly as it is drawn).
Scheme:

Answer 1

Cool fireworks you have laid. As soon as a couple of LEDs break through, the voltage on the LM317 jumps to the limit and there will be a cool broads.
1000 microfarads at 450v = 80 joules. In case of problems, the capacitor will dry out so that it doesn’t seem a little. But there will be problems, since you put the capacitor completely without a margin into the environment, where even 1kV can be caught in the input pulse.
Tip - make a normal impulse driver. And not this circle of "skillful hands" without galvanic isolation and filters.
Even if we conditionally accept this circuit as correct, you need to put ceramic capacitors around the LM317 so that it does not ring.
And yes, current limiting by a transistor is done differently - in your circuit it will simply explode because initially a network will be applied to the E-K junction.
And your divider will apply 236 volts to the EB transition, which will also lead to an explosion of the transistor.

Answer 2

After a few clarifications, it finally became clear what you want to achieve: a common power supply for several circuits of series-connected LEDs. You considered the node for the smooth charge of the filter capacitor to be the main problem. In my opinion, in such a scheme there are several much more critical places. But first, on the topic.
1000 microfarads - this value is suitable for a load current of 0.5 ... 3 amperes, and not tens of milliamps (22 ... 50 microfarads are enough there). The transistor can be installed if you need to make a smooth, for 4 ... 20 seconds, increase in brightness - but you have several garlands! Do they really have to start in the whole apartment at the same time? And what about the switches - do you want to switch the circuit = 310 volts instead of the regular ones, switching the circuit ~ 220 volts, placing the switch between the capacitor and the garland? Such a solution looks at least somehow justified for a "smart home" (and even then not everything is clear in it), but in an ordinary apartment it makes no sense to do so. It is more correct to install its own separate PSU for each garland - and then it is much more profitable to use ordinary super-cheap (and much more reliable!) Tapes with parallelLEDs at 12 volts, and not with self-made series, in which the burnout of one diode completely deprives you of light.
Another purpose of the soft charge unit is to protect rectifier diodes from multiple overload at the moment of switching on, when the capacitor is completely discharged. But this problem is completely solved by a much simpler method - instead of T1 and R1, R3, you need to insert a thermistor with a resistance of several tens of ohms, which decreases when warmed up to 0.5 ... about the same load current as yours. You can get such a thermistor from any dead computer PSU.
And finally, about something that is not in your question, but it catches your eye - about the current stabilizer on the LM317, which absorbs excess mains voltage. The fact is that such a stub is only operable in the range from 3 to 40 volts. Tolerance for mains voltage in the city serviceable network is 10%, i.e. from 198 to 242 volts. So, if you calculated the stub for the lower limit (and this is usually done), then at the upper limit the voltage on the stub will go beyond the allowable 40 volts. If you set it to the top of the range (i.e., 242), then at the lower limit the voltage on the stub will drop below 3 volts, and it will no longer stabilize the current. And I’ll keep silent about what will happen to this scheme in rural areas, where mains voltage fluctuations are much wider. So such a circuit will only work normally with a stable mains voltage - but with a stable network, a stabilizer is not needed,

Answer 3

in the current version, it was not possible to achieve stability when changing the load of 15-150W (I tried to add a zener diode to the base circuit). after some experiments, I came to this option by replacing the transient with a field
one, which made it possible to achieve stability of the control voltage, and also reduced the voltage loss on the transistor itself (in this case, the drain-source resistance is less than 0.5 Ohm in the open state)
the time until the transistor is fully opened is about 15- 20 seconds, during this time the capacitor manages to charge up to 250 V. I also plan to add filtering to the network circuit (at the moment of switching on, there are parasitic current surges that create interference) and it may be necessary to reduce the resistance R3 to accelerate the discharge of capacitor C2 (you need to check the operating mode for a short-term power off)
important!this circuit is applicable only for powering LEDs, because they have a high resistance in the inactive state, unlike conventional incandescent lamps, the low ohmic resistance of which does not allow the capacitor C1 to be charged.