How to Convert AC110V to 3.2V 200mA transformerless

How to Convert AC110V to 3.2V 200mA without transformer?

Trying to replace lamp with 10 White LED (Extreme Light LED).

 

Why don't you put the LEDs in series ?

Graham

 

Get the details on designing and building a transformerless power supply at
http://uk.geocities.com/ronj_1217/tless.html

Be certain that you observe every safety precaution !!!! The transformerless
supply is NOT isolated from the AC mains, and is LETHAL.
--
Dave M
MasonDG44 at comcast dot net (Just substitute the appropriate characters in the
address)

Some days you're the dog, some days the hydrant.

 

Put the LEDs in series so they drop 32 volts - that way they each get
the same current (LEDs) aren't going to drop exactly 3.2 but a range
from 3.2 to 3.8 (for white ones).. In parallel some will hog more
current than others.

Use a point six (0.6) microfarad non polarized (polyester is good) cap
with at least a 300 VDC rating or 120 VAC rating in series with one
leg of the full wave bridge rectifier that serves the LED string, The
rectifier only has to carry 20 ma but use something like a 1N4003 or
so. Add a 100 ohm 1/4 watt resistor in series (serves as fuse and to
limit inrush currents - in series with the cap)

You use ohms law to decide what capacitive reactance you need. In the
above example I used 4.4 K, at 120 volts AC 60 HZ, to drop 88 volts,
at 20 milliamps.

The formula to solve for Capacitance in Microfarads is

159200
C= ---------------------------
F Xc

F is frequency (your power mains)
Xc is capacitive reactance in ohms (to drop the required voltage)

Can't find a point six cap? a .56 will work it isn't too critical

There are several schemes for using caps to run LEDs from the power
mains. I like the FWB, but you could just put a shunting diode around
the string to keep the wrong polarity from frying the LEDs in their
reverse breakdown mode - the downside of that approach is it takes a
larger cap to get 40 ma through the string - at half wave power - and
the LEDs flicker a little bit, and the cap will cost more.

You could, of course, put a separate circuit for each LED and use the
appropriate cap for each (~.40 uf) but that costs more and takes up
more room.

I have an LED night light that's been on for around ten years now
(three red LEDs)

 

Actual circuit he's referring to is on
http://www.zen22142.zen.co.uk/Circuits/Power/tps.htm

 

110VAC motor, driving 3.2V generator.

As others have pointed out, it's much more practical to put the LEDs in
series and drive them through a bridge rectifier, driven through a
current-limiting impedance, while observing that there's no isolation
from the 110V (120V?) mains circuit. Or, get an inexpensive
transformer designed to drive halogen lamps, and use current-limiting
resistors to drive series strings of two or three LEDs; but be sure you
convert the AC out of the transformer to DC.

Cheers,
Tom

 

It is better to put the leds in series (and safer to use a transformer).

If your insist on connecting directly to the mains your can use this
program to find a circuit and calculate the components:
http://www.miscel.dk/MiscEl/miscelLeds.html

 

In other words, convert to DC after using a capacitor (low heat
production
but drops voltage) as a ballast component. Fusible resistor as fuse
and surge resistor was a brilliant addition. One capacitor, one
resistor,
four diodes, and ten LEDs.

It's going to flicker at 120 Hz, and you'll have average current about
70 percent
of what the LEDs are rated for (or the peaks will exceed rating). You
can
use a different rectifier (voltage doubler type) and a second (filter)
capacitor to
make it a 100 percent drive without flicker. Same light output
would take one ballast capacitor, one filter capacitor, two diodes,
two resistors,
and seven LEDs.

Voltage doubler rectifier uses a blocking capacitor (your ballast
capacitor performs
that function) into a diode which clamps (-) excursion to neutral and
a second
diode which passes (+) excursion to the filter capacitor. A second
resistor from the
filter capacitor to the series string of diodes (about 1k ohms) is
recommended,
because it will smooth the current fluctuations due to residual ripple
and allow
the filter capacitor to be smaller (something like 1000 uF/50V).

If you don't mind filter inductors, you could use that instead of the
second resistor (and no filter capacitor is required). Mainly, filter
inductors are
hard to get, heavy, and have cooties.

 

You can buy off-the-shelf, AC/DC adapters that do that. They might even
have some at Walmart. I know they used to years ago. Here's an example:

http://www.voltage-converter-transformers.com/switching-ac-dc-supply.html
VM 1898 - Digital Camera AC/DC Adapter - 100V/240V

» AC to DC Adapter
» Input: 100 - 240V. AC
» Output: 3/3.3/5/6/6.5/7 or 8.4V, DC
» Maximum Load: 2500mA.
» Polarity: Outer - Negative (-)
Inner - Positive (+)
» Ideal Power Supply for almost all Digital Cameras.
» Eliminating the use of expensive batteries.

Weight: 9 oz.
Dimensions: 3" x 2.3" x 1.2"
Price: $15.99

 

....because he is not a serial illuminator?

 

Put the LEDs in series, is a good idea.
And the transformerless power supply is alos a good solution.
Let me start with put the LEDs in series.
Thanks for all these information.

gi0001tw ´£¨ì:

 

Thanks for all of your information, I finally found the schematic of
"Line Powered White LEDs".

http://ourworld.compuserve.com/homepages/Bill_Bowden/page10.htm#ledlamp.gif

How much heat will produce from this circuit?

gi0001tw ´£¨ì:

 

Very little, your can use this program to analyze the circuits:
http://www.miscel.dk/MiscEl/miscelLeds.html

Allowing bigger inrush current will reduce the power loss.

I have "designed" a good circuit with 10 leds here:
http://www.miscel.dk/Led_on_110volt.png