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[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 <cmcmanis at mcmanis.com>
Sent: Jan 11, 2004 5:57 PM
To: DPRG <dprglist at dprg.org>
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



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