How to Convert AC110V to 3.2V 200mA transformerless

Get the details on designing and building a transformerless power supply at

Be certain that you observe every safety precaution !!!! The transformerless supply is NOT isolated from the AC mains, and is LETHAL.

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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)

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Actual circuit he's referring to is on

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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:

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:

VM 1898 - Digital Camera AC/DC Adapter - 100V/240V

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

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

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".

How much heat will produce from this circuit?

gi0001tw ´£¨ì:

Very little, your can use this program to analyze the circuits:

Allowing bigger inrush current will reduce the power loss.

I have "designed" a good circuit with 10 leds here: