DPRG List

 [DPRG] Building an LED nightlight Message index sorted by: [ date ] [ thread ] [ subject ] [ author ] Previous message: [DPRG] Building an LED nightlight Next message: [DPRG] Building an LED nightlight Subject: [DPRG] Building an LED nightlight From: Rick J. Bickle rbickle at swbell.net Date: Tue Jan 13 09:09:01 CST 2004 ```The capacitor LED circuit described earlier has some disadvantages. The capacitor will limit the current to the LED by stopping current flow once the capacitor is fully charged. This of course depends on the value of the capacitor, but the downside is that the LED will only see current flow for a small portion of the AC waveform 120 times per second. This means that the LED is only flashing 120 times per second and only using a small percentage of its light output power. How about a bridge rectifier and 50 or so LEDs in series with a current limiting resistor? (and a fuse) The LEDs (depending on the color) will drop around 2.4V, so 50 in series will drop 120 volts. The resistor then only has to dissipate 50mA at the voltage difference of 170-120 = 50V. Power is then 50V * 0.05A = 2.5Watts. Increase the number of LEDs further to decrease resistor power dissipation. Rick -----Original Message----- >From: dprglist-admin at dprg.org [mailto:dprglist-admin at dprg.org] On Behalf Of Ed Koffeman Sent: Tuesday, January 13, 2004 8:45 AM To: birt_j at earthlink.net; 'Chuck McManis'; 'DPRG' Subject: RE: [DPRG] Building an LED nightlight The capacitor won't actually dissipate any watts at all (ok, a little >from its internal Ohmic resistance, which is negligible). It is a reactive device, and its current is out of phase with the voltage applied, such that it has impedance without dissipation. This is why you would use a capacitor instead of a resistor with the same impedance. A resistor would indeed get very hot. As a matter of fact, it is has a minor benefit of counteracting the opposite effect that inductive loads have on the A.C. supply, and improves the power factor seen by the electric utility supply. Ed Koffeman -----Original Message----- >From: dprglist-admin at dprg.org [mailto:dprglist-admin at dprg.org] On Behalf Of birt_j at earthlink.net Sent: Tuesday, January 13, 2004 8:12 AM To: Chuck McManis; DPRG Subject: Re: [DPRG] Building an LED nightlight I too was impressed with the simplicity of this LED circuit. This arrangment however means that the cap will dissipate 1.2 watts whilw the LED's only dissipate .036 watts. Compared to the el-cheapo incandesent night light currently plugged into my wall which uses 4 watts, assuming close to the same number of lumens, the LED light is still using only 1/3 the power. We had several 4 watt flourescent night lights, which were about \$10 each from Lowes for a month 'till they all started to go bad. 4 watts of flourescent is a lot of night light too. The only idea I would have for making something more energy effiecient would be to wind a very small 33 to 1 transformer with a 1K current limiting resistor on the primary to drive the LEDs. Jeff Birt -----Original Message----- >From: Chuck McManis Sent: Jan 11, 2004 5:57 PM To: DPRG Subject: [DPRG] Building an LED nightlight > > What kind of capacitor should I use? I think > > electrolytic capacitors don't!! like AC, right? > > So ceramic capacitors, or what? > > Got a suggested capacitor size to start with? You'll need to refer back to my original schematic. But here is a brief discussion of the circuit. AC can pass through a capacitor just like DC can pass through a resistor, the resistance is known as "reactance" and is denoted X rather than R. So if you want to power your LED's with 10mA of current, and the house current is 120V AC @ 60hz, you use ohms law which states E=IR (or E = IX in this case) and divide E by I to get X. So you want an X that is 120/.010 = 12,000. The formula for reactance is X = 1/(2*pi*f*C) where f is in Hz and C is in farads. Re-arranging and solving for the unknown C you get: 1/(12,000*2*pi*60) = C or 1/4,523,893.421 = C or 221 x 10^-9F = C C needs to be about 221 nF, or more conventionally .22 uF. The capacitor should be rated for a voltage > 150V for safe operation. A .22 uF like the one from Kemet (rated at 200V, Digikey part #399-2105-ND is \$4.63 in quantities of 1. As an exercise you should compute the size capacitor you need to put 25mA through the LEDs. Now to the other side of this capacitor are two BACK TO BACK LEDs. There are two reasons for this : #1 - The LEDs only light when they are forward biased, this way you get light on both halves of the light cycle. #2 - The reverse voltage "seen" by the diode is limited to the forward voltage of the conducting diode. Anyway, like I said in my earlier e-mail, you really can't beat if for simplicity. --Chuck _______________________________________________ DPRGlist mailing list DPRGlist at dprg.org http://nimon.ncc.com/mailman/listinfo/dprglist _______________________________________________ DPRGlist mailing list DPRGlist at dprg.org http://nimon.ncc.com/mailman/listinfo/dprglist _______________________________________________ DPRGlist mailing list DPRGlist at dprg.org http://nimon.ncc.com/mailman/listinfo/dprglist ``` Previous message: [DPRG] Building an LED nightlight Next message: [DPRG] Building an LED nightlight Message index sorted by: [ date ] [ thread ] [ subject ] [ author ] More information about the DPRG mailing list