The 220v lighting power supply wants to supply power to the LED lamp of the portable lamp.

First of all, your charger is a resistance-capacitance step-down method, and the output voltage is not necessarily accurate, because it is stepped down by the capacitance reactance of the capacitor, and the capacitance value is biased. You have not measured the output voltage.

You didn't break this line. You look so dizzy. ...

1. Use 120K to connect two parallel resistors and four parallel LED lamps. But the lights are dim.

2. Connect four LED lamps in series with a resistor of 10k, and then connect them in parallel. Add a 50K resistor before the parallel resistor. The light is still dim.

3. An LED lamp is connected in series with a 6K resistor, and the resistor smokes 1 s after being electrified. Fortunately, the LED lights and resistors were not broken, so I cut off the power supply. The light is "more than twice as bright as before" (is it a wrong number), but it is still not half as bright as the single light of the portable lamp.

Note: the diode may not be completely damaged when it exceeds the current, but it will damage the light-emitting PN junction and will never become very bright again.

You may want to light a few LEDs with DC power supply.

First, determine the output voltage of your charger. General chargers are 12V, 5.5V, etc. And the mobile phone charger is 5V.

The white LED of your straw hat is about 0.5 or 0.8, and the maximum current is about 20 mA.

Let me estimate the voltage of your charger:

1. Use 120K to connect two parallel resistors and four parallel LED lamps. But the lights are dim.

If the lamp is dim, assuming that the current is only 1 mA, the resistance is 60k (parallel minus half), and the voltage should be 60V.

2. Connect four LED lamps in series with a resistor of 10k, and then connect them in parallel. Add a 50K resistor before the parallel resistor. The light is still dim.

This one is still black. Forget it. You didn't say it was better than the first one.

3. An LED lamp is connected in series with a 6K resistor, and the resistor smokes 1 s after being electrified.

The rebels are smoking. If a resistance of 0.25W (quarter watt) is used, smoking means that the power exceeds 1W at least, and the resistance of 6000 ohms can exceed the power of 1W, and the voltage can reach 1W at least at 80V.

I estimate that the voltage of your charger is about 70 V.

Therefore, if you want to light a diode, you need a resistor:

(70V-0.6V)/20mA = 3470Ohm = about 3.5K Power 1.4W You need a resistor of 3.5K2W (brightest).

If there is only a resistance of 0.25W, it is necessary to reduce the current to 3.5 mA and use a resistance of 20K0.25W (dark color).