In my home I am getting 220 AC Volts of electricity. I want to connect white color Light Emitting Diod (LED) that requires 3 volts output with the resistor which will be connected to 220 Volts power. So can any body please tell me what is the formula to find that how much voltage is output from a resistor when 220 volts of AC voltage is applied. I want to use this so that I'll be able to have only this much light in the room where I sleep. I'll be very thankful for anyone who'll answer.

Also if anyone knows some relevant useful links then do please let me know. I searched and found a very interesting windows application in which if you enter color codes' information of the resistor, it'll tell you what the resistance value of this resistor is. This is the link of the website you can download this application from here http://www.doctronics.co.uk/download.htm.

But I want an application such that if I tell it to tell me resistor's value of a resistor which gives output of 3 volts i.e I'll input 3 volts in the edit box of the application and then on pressing the calculate button it'll tell me the resistor's value and all the color codes as well. I don't know whether anyone has developed this application, if not then it should be as it is something very useful.

Once again anyone's help will be very much appreciated.

You might find this site useful: LED Calculator - Current limiting resistor calculator for LED arrays

It may not be all that simple. Remember that your 220V line is AC, whereas LEDs work off DC current. So I can't tell you that this approach will definitely work.

Also, simply using an arrangement of dropping resistor that feeds an LED with 50 milliamps of current (a pretty standard figure for LEDs) from a 220 volt source you will find that the power that has to be handled by the dropping resistor is huge: 220 volts * .05 amp = 11 watts, which is a lot of power for one resistor to handle.

You'd be better off using an AC power supply - something like one of those 9 volt power dongles that are used by small electronics. You probably have a few lying around the house. Hook your 9 volt supply to two resistors in series - one being 120 ohms and the second 60 ohms - and then connect the LED to the two leads of the 60 ohm resistor. This system will consume about 0.45 watts of power (9 volts into 180 ohms) and will drive the LED at 3 volts.

ebaines:

In your diagram, the LED may BARELY come on if it comes on at all.

LED's are current driven and now with high power LEDS in the picture who knows what he wants. The HIGH power LEDS really need to be driven by a special driver.

Run of the mill LED's are typically 20-50 mA and the burst of 1/2 cycle will be pretty high. About root 2 * 220. The LED won't like that. It also won't like the reverse voltage ether.

A diode (1n400?) with a PRV of 400V+ would prevent the LED from self destructing on the reverse cycle.

Although it should work with R = (220-3-0.6)/20e-3 and the series diode, I don't recommend it. The 3 is the drop across the LED and the 0.6 is the drop across the diode. They can be ignored in this case.

The LED tyically drops 3 volts. So if you had a 12 VDC source, R ~= (12-3)/2e-3

Need to know current?

Instead of using a single LED, use many in a series, At least more than 60. Be sure to add them with correct polarity or else even a single attached in wrong order will make the series defunt.

Use 4diode bridge rectifier to use only DC. Though LEDs themselves are diodes but are rarely used that way because of the delicate filament which burns so easily if the passed voltage is a bit a high.