LED on mains voltage?

[peranders]

diyer, you are pretty much right. Insulation is very important. They main advantage of this circuit is that it can be made rather small and this is used in 22 mm DIN time relays.

An another alternative is using a 0.3 VA Hahn transformer, very cute

[sivan_and]

What abt having an blinking LED ? You don't need high voltage capacitor,bleeder resister and negligible power wastege..
See the circuit attached:
Brief discription:
When 230volt a/c is applied, it charges C1 (100Mfd/16Volts) until the transister breaksdown and charge stored in the capacitor is transferred to Led which lights,until the capacitor discharges and the transistor goes to off state and the LED is off,until the next cycle is started again.
For 110 volts reduce resistor to 47Kohms 1/4 watts.
for 230 volts use 1/2 watt resistor.
Any small signal transistor will work for T1 Eg:2N2222 etc



IT IS DANGEROUS TO TOUCH THE CIRCUIT WHEN POWERED.IT IS DANGEROUS TO TOUCH THE CIRCUIT WHEN POWERED.

THIS Led will blink only when the voltage is above 160volts AC.

[sivan_and]

Herez the circuit:

[fever16]

how can i make this circuit to work for 1watt led at 230v.
imean 1watt LED which draw 350ma for full brightness.

i have some requirements where i need to run the 1 watt alone and a series 1 watt leds(around 4-5) in series.

need some desperate help here
Regards

[AndrewT]

Hi,
you have a problem.
1.) you need a >400V diode in series with the LED.
2.) the voltage drop across the LED+Diode ~4V.
3.) the voltage drop across the series resistor ~ 226Vac half wave rectified.
4.) the power to be dissipated ~ 226V * 50% * 0.35A ~40W
5.) the power resistor is at mains voltage.

[martin clark]

Fever - I would definitely use a transformer / wallwart, and turn that into a conventional low-voltage DC project.

One little mistake and you are dead, and then some.

NB. 350mA at 240VAC is over 80W ...even with current limiting, that's a significant amount of power, and therefore a fire risk too...

[burbeck]

hi
the simple solution is 240V panel neon indicators

[Elvee]

Hi,

You could use a capacitive supply with a full wave rectifier: see picture.
For 350mA at 230V/50Hz, C2 would be 5.6µF, D1 to D4 a 1A rectifier bridge, R2 a 10ohm/3W WW resistor and R3 a 220K/0.5W.
5.6µF is a bit large for "electronic" 250Vac X capacitors, but can easily be found in electrotechnical applications for induction motors, fluorescent tubes ballasts, etc.
Rl would be your LED(s). The beauty of this circuit is that it is an almost perfect current source, and you can use any number of LEDs at the output.
Don't omit C1, it absorbs the current surge at the power-on.
Almost no power is wasted with this PSU.

[shushz]

Hi,

I'm very new to the field of electronics and trying to learn from online resources so forgive me if the questions are stupid, but could someone please explain what is the purpose of R2 in the above diagram proposed by Elvee and why does it have to be WW?
Also, to my understanding using capacitor to drop voltage with such low resistance would result in rather poor power factor of the circuit. Am I right? Is it something to be concerned about, or it can be disregarded for such low currents?
If not, how can the power factor be corrected?

[Andy5112405]

It doesn't have to be WW but large resistors are generally WW.