Lighting and soldering iron dual-purpose circuit

In order to use high-brightness light-emitting diodes to illuminate the soldering iron , the effect
is very good. Use a 35W internally heated soldering iron, measure the resistance with a digital meter to be
1. 43kfl, and then calculate the current; I= U/R-220V/1430fl=0. 154A.
The working voltage of high-brightness light-emitting diodes is about 3V. Therefore, the terminal voltage of Ri is
3.5V, which is the terminal voltage provided by the light-emitting diode. The calculated value is Rl-3.5W
0.154A=22.7fl. The island is a current limiting resistor, and its resistance can be selected according to the required brightness
. If you use one light-emitting diode and it doesn't light up, you can add another one and connect it in reverse parallel,
like this. Under AC conditions, both diodes can light up. Since
the operating current of the light-emitting diode is more than 20 mA - most of the current flows through R. , calculate the power of Rl as P-0. 13×3.5V=O. 455W. Therefore, R. It can be
about 2W, and R3 and C are pulse absorption circuits .
    During production, weld the light-emitting diode and current-limiting resistor firmly outside the soldering iron and wrap them with insulating tape. R． And R3, C must be wrapped with insulating paper and
put into the handle.