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<blockquote data-quote="thylantyr" data-source="post: 1333370" data-attributes="member: 560358"><p>You would probably need to do this as a bare bones solution to making it work.</p><p></p><p>You insert a generic diode like 1N4003 and place it in series with the LED because</p><p></p><p>the LED reverse voltage is only rated for 5V. These 1N4003 are rated for 200V.</p><p></p><p>AC input -&gt; 1N4003 -&gt; LED -&gt; resistor</p><p></p><p>Diagram 1</p><p></p><p><a href="http://library.thinkquest.org/16497/projects/beginner/p3/led-diode.gif" target="_blank">http://library.thinkquest.org/16497/projects/beginner/p3/led-diode.gif</a></p><p></p><p>The diode and LED have a polarity and notice how they have it wired.</p><p></p><p>Diode cathode to LED anode.</p><p></p><p>Replace the 3V battery in the circuit with 120v AC.</p><p></p><p>Last, recalculate the new resistor value for the increased voltage.</p><p></p><p>AC volts peak = 120v rms x 1.414 = ~ 170.</p><p></p><p>Figure out what the nominal LED current is to safely operate it. Lets use</p><p></p><p>0.02 amps.</p><p></p><p>The shortcut math that is 99% accurate is;</p><p></p><p>170v / 0.02 A = 8500 Ohm resistor.</p><p></p><p>Replace that 470 Ohm resistor in that drawing with one that is 8500 Ohm,</p><p></p><p>but this resistance value is not common so you can round up to 10,000 Ohm.</p><p></p><p>Since the AC mains are alternating, the LED will be turning off/on many times</p><p></p><p>per second. It may be doing this so fast that you may not notice it.</p><p></p><p>Diode specs;</p><p></p><p><a href="http://www.fairchildsemi.com/ds/1N/1N4001.pdf" target="_blank">http://www.fairchildsemi.com/ds/1N/1N4001.pdf</a></p><p></p><p>If you want to make a better circuit, there is more ways. This is the bare bones</p><p></p><p>method.</p><p></p><p>If you have more than one LED then you insert them in series and because there is</p><p></p><p>a small voltage drop across each LED and diode, eventually you will have to redo the math</p><p></p><p>to get an appropriate resistor so you don't get severe dimming.</p></blockquote><p></p>
[QUOTE="thylantyr, post: 1333370, member: 560358"] You would probably need to do this as a bare bones solution to making it work. You insert a generic diode like 1N4003 and place it in series with the LED because the LED reverse voltage is only rated for 5V. These 1N4003 are rated for 200V. AC input -> 1N4003 -> LED -> resistor Diagram 1 [URL="http://library.thinkquest.org/16497/projects/beginner/p3/led-diode.gif"]http://library.thinkquest.org/16497/projects/beginner/p3/led-diode.gif[/URL] The diode and LED have a polarity and notice how they have it wired. Diode cathode to LED anode. Replace the 3V battery in the circuit with 120v AC. Last, recalculate the new resistor value for the increased voltage. AC volts peak = 120v rms x 1.414 = ~ 170. Figure out what the nominal LED current is to safely operate it. Lets use 0.02 amps. The shortcut math that is 99% accurate is; 170v / 0.02 A = 8500 Ohm resistor. Replace that 470 Ohm resistor in that drawing with one that is 8500 Ohm, but this resistance value is not common so you can round up to 10,000 Ohm. Since the AC mains are alternating, the LED will be turning off/on many times per second. It may be doing this so fast that you may not notice it. Diode specs; [URL="http://www.fairchildsemi.com/ds/1N/1N4001.pdf"]http://www.fairchildsemi.com/ds/1N/1N4001.pdf[/URL] If you want to make a better circuit, there is more ways. This is the bare bones method. If you have more than one LED then you insert them in series and because there is a small voltage drop across each LED and diode, eventually you will have to redo the math to get an appropriate resistor so you don't get severe dimming. [/QUOTE]
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