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How to Convert AC110V to 3.2V 200mA transformerless

Discussion in 'Electronic Design' started by gi0001tw, Oct 12, 2006.

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  1. gi0001tw

    gi0001tw Guest

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

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

    Eeyore Guest

    Why don't you put the LEDs in series ?

  3. DaveM

    DaveM Guest

    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.
    Dave M
    MasonDG44 at comcast dot net (Just substitute the appropriate characters in the

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

    default Guest

    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

    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)
  5. default

    default Guest

  6. Tom Bruhns

    Tom Bruhns Guest

    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.

  7. HKJ

    HKJ Guest

    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:
  8. Guest

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

    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

    » 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
  10. Robert Baer

    Robert Baer Guest

    ....because he is not a serial illuminator?
  11. gi0001tw

    gi0001tw Guest

    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:
  12. gi0001tw

    gi0001tw Guest

  13. HKJ

    HKJ Guest

    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:
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