There are switches on sale with LED backlight, it shines at night and we find it unmistakably in the dark.You can assemble such a backlight yourself and I will give you three simple, correct and working schemes.
The very first diagram on the photo from above.It consists of a current-limiting resistor and an LED, and this circuit is not very correct, although it is used in technology and by amateurs.Let’s figure out what this scheme is bad about.The resistance of the resistor has chosen 220 ohms and I supply power to about a little above 7 volts AC.The signal at the output of the transformer is a sine wave with a positive and negative half-wave with an amplitude value of about 10 volts each.
We look at the signal on the resistor.The positive half-wave is cut off, and the negative one passes.What is the disadvantage of this scheme? And the disadvantage is that the positive half-wave enters the LED in the form of reverse voltage, and the reverse voltage of the LED is on average about 10-20 Volts, there are less.If this circuit is connected to a 220 Volt network, in which the amplitude value is 310 Volts, then this reverse voltage of 310 Volts can disable the LED.It is necessary to make sure that this reverse voltage does not flow to the LED.
And the first correct scheme will help to do this.We connect a diode in series to the LED, and now the positive half-wave will pass through the diode and enter the LED, and the negative one will not.In this way there will be no reverse voltage problem.
Now we put another LED in parallel with the LED.At each half-wave, its own LED will shine and the reverse voltage will not flow to the LEDs.
With such a counter-parallel switching on of the LEDs on the resistor, there will not be a half-wave, but quite a whole sine wave.
The third circuit I connected to the 220 Volt network, this circuit with a 1n4007 diode connected in parallel to the LED.The resistor is selected with a resistance of 180 kOhm and a power of 0.5 watts.The current per LED depends on the resistance of the resistor and the formula will help to calculate it: R=U/I. We take 310 Volts and divide by the current, let’s say 2 mA: 310/0.002 = 155 kOhm and get a resistor with a resistance of 155 kOhm.But it must be borne in mind that the more current per LED, the greater the voltage drop on the resistor will be and it will heat up, so you will have to put a more powerful resistor.
That’s how the LED shines in the dark from a 220 Volt network, not bright but visible.
