What you've found is a simple transformerless power supply. It is also known as a capacitive dropper.
The way it works is that it uses the impedance of the series capacitor (the big red capacitor marked 225 for 2.2uF) to limit the current.
A 2.2 uF capacitor has an impedance of about 1445 ohms at 50Hz. At 220VAC, that's a current of about 150 milliamperes. If you short circuit the DC output, it will only draw 150mA from the AC side. The DC current would be around 200mA.
The circuit is an approximation of a constant current source. It has a fairly high input voltage in series with a (relatively) high impedance.
It isn't really a good current source, however. The output current varies with the input voltage and with the load.
If you use such a circuit, you must make sure that it is completely insulated. The supposed low voltage side is connected directly to the mains. Touching the output can kill you just as easily as touching the mains on the input side.
Contrary to what other folks have found, I have found such cheap circuits in brand name 220V LED bulbs.
Here's the circuit from a bulb I disassembled a few years ago:
The circuit doesn't work well. The bulb quit working a few months after I bought it.
One of the LEDs in the bulb is damaged. Whether from too much current all the time or because the circuit let through a high voltage (and higher current) surge I couldn't say.
The inductor is a half-hearted attempt at preventing high voltage spikes from getting to the LEDs. I doubt it really does much.
R1 is to discharge C1 when the bulb is disconnected from AC. Likewise, R2 is to discharge C2.
