But that's not the point. In the comments, it was suggested that this thread would not survive more than one ignition. Let's check it out. The article dealt with the following power sources:
-Battery 18650, 3.7 V – 2 pcs;
– and an unnamed VCC + 5V, which we will forget about.
We consider our source two 18650s connected in series. When setting up with a 220 Ohm resistor used, the current through 10 (!) Parallel LEDs will be (2 * 3.7V – 3V) /0.22 = 20 mA. It's all? Not a lot … Not everything is clear with the filament: “… pulls out one copper core from the wire. Installs connecting wires instead of LED . © »
What kind of a core that heats up and does not burn out when a voltage of 3 V is applied to it? Or is it a current of 20 mA?
Okay, take the power supply unit, put 7.5 V and start poking different veins without any resistors.
Naturally, they burn out with a flash. We limit the current supplied by the PSU to the stage when the copper wire with a diameter of 0.2 mm does not burn out.
In this case, a current of about 7 amperes flows through the wire with a voltage drop of about 1.2 volts. We can calculate the required resistor in the original circuit: (7.5 V – 1.2 V)/7 A is equal to less than one Ohm, not 220. By the way, you can also power the power circuit from parallel connected batteries.
We wind 2 turns of wire (to hold) on the head of the match, turn it on. Nothing happens, nothing, suddenly there is a flash, I did not have time to photograph and did not notice when the wire broke.
We repeat, now it's more accurate.
The wire is still intact. We repeat with the same wire.
Wire again intact, repeat? No, the matches are out and the burned finger hurts.
Yes, the author is right, the filament is reusable, or rather, it can be. Everyone, comment!