If you use a current limiting resistor, it will get extremely hot, and will waste energy. By using a capacitor, no heat is dissipated in the capacitor. My question is, how much electricity will I use? To get 20mA through the LED, surely I am going to use 20mA at 110V. So where does the energy go?
And can someone offer a calculation for the capacitance to run a 2v 20mA LED. Thanks.
2007-07-23 03:37:32 · 2 answers · asked by Experimentor 2 in Science & Mathematics ➔ Physics
You can use a capacitor; it's quite easy. Just put a reverse-conducting diode in parallel with the LED. There is no net DC through the capacitor, and the LED will blink at 60 pps. Start with a very small cap and work higher until you get usable output. Remember peak LED currrent will be 2*sqrt(2) times higher than average, so don't drive it to normal brightness.
Where the power goes: You won't be using power equal to voltage * current, since voltage and current will be nearly 90 degrees apart in phase. You will be changing the power factor of your house's load on the mains, but the amount will be infinitesimal.
Warning: Remember this is a danger device. The "grounded" side will actually be directly connected to the mains. Keep kiddies and cats away!
2007-07-23 04:19:22 · answer #1 · answered by kirchwey 7 · 1⤊ 0⤋
You cannot use either a capacitor or a resistor. For one thing, an LED is a DC device.
But basically you need to reduce the voltage and rectify it. That means a transformer or a thyristor circuit.
2007-07-23 10:48:25 · answer #2 · answered by Anonymous · 0⤊ 0⤋