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LED 240VAC

Discussion in 'Electronic Design' started by Phil, Dec 18, 2003.

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  1. Phil

    Phil Guest

    I want to drive an LED (3mm) from 240VAC
    any idea's? or is there a 3mm neon available.

    Phil
     
  2. Just a diode (reversed) across the LED and a resistor in series with these
    2. Resistor about 22kOhm.
    Jo
     
  3. WARNING : This could work but the resistor will need to be 5W at least
    (P=U2/R...) and should be certified for high voltage use (>500V). Two 10K/2W
    resistors in series are far safer. Add a fuse too... And keep your fingers
    out...
    An alternative is to use a small transformer, or one of these 220V-power
    supply ICs (maxim I think).
     
  4. John Fields

    John Fields Guest

    ---
    View with a fixed pitch font like Courier:

    +--[<1N4001]--+
    | |
    240VAC>---[12K]--+----[LED>]---+
    10W |
    |
    240VAC>------------------------+
    <---20mA--->
     
  5. For safety reasons the LED is going to require extra insulation if
    it's connected with a transformerless (either dropping resistor or
    reactance type) power supply.

    Adding a cheap series 1N4007 will cut the resistor dissipation in
    half. It still needs to be a high voltage rated type.

    Best regards,
    Spehro Pefhany
     
  6. Tilmann Reh

    Tilmann Reh Guest

    Better than using a resistor as current limiting part, use
    a capacitor (has to be rated for line voltage, though).

    240VAC -----||-----+----|>|-----.
    50 Hz 100n X2 | LED |
    | |
    `----|<|-----+
    ____ |
    240VAC ---|____|----------------´
    220R

    The additional resistor is to limit currents at higher frequencies
    (spikes, harmonics) for which the capacitor has rather low
    impedance. This circuit draws almost pure capacitive current,
    so power loss is very small. The resistor can be 0.5 or even 0.25
    watts. At more powerful surges, it acts like a fuse :)

    To get more light and less flicker, you could also place the LED
    inside a diode bridge (instead of using an antiparallel diode).
    You can also use a smaller capacitor then (while maintaining
    intensity).

    However, take care: the complete circuit is connected to mains,
    as others already have mentioned.

    --
    Dipl.-Ing. Tilmann Reh
    Autometer GmbH Siegen - Elektronik nach Maß.
    http://www.autometer.de

    ==================================================================
    In a world without walls and fences, who needs Windows and Gates ?
    (Sun Microsystems)
     
  7. Phil

    Phil Guest

    Thanks all for your thoughts, I do not have space
    for an X2 cap so its the diode & resistor solution
    maybe I should look at a low current LED to reduce
    the wattage of the resistor.

    Phil
     
  8. John Fields

    John Fields Guest

    ---
    Good plan. I'd also use a diode in anti-parallel across the LED instead
    of a diode in series with the LED in order to keep the mains voltage
    alternations from possibly destroying the LED. The reason is that the
    series diode will have some reverse capacitance which will appear in
    series with the reverse capacitance of the LED, forming a reactive
    voltage divider and possibly allowing enough reverse voltage to build up
    across the LED to puncture its junction. This capacitance and the LED's
    capacitance will be swamped by each other when they're in anti-parallel
    as the mains voltage alternates, totally eliminating the possibility of
    puncture.

    Depending on the current you can run through the LED to get minimum
    acceptable illumination, you can increase the size of the resistor to
    limit the current to that value.

    View with a fixed pitch font like Courier:


    +--[<1N4001]--+
    | |
    240VAC>---[12K]--+----[LED>]---+
    10W |
    |
    240VAC>------------------------+
    <---20mA--->

    From the example given previously (above), the value of the forward drop
    of the diode and the LED is so small it can be ignored and the value of
    the resistor can be determined by:

    R = Vs/Iled

    where R is the resistance of the resistor in ohms
    Vs is the RMS supply voltage in volts
    Iled is the LED current in DC amperes,

    yielding: R = 240V/0.020A = 12000 ohms.


    The power the resistor will dissipate can be determined from:

    P = Iled*Vs

    Yielding: P = 0.02A*120V = 4.8 watts.


    By using a high-efficiency 2mA LED like an HLMP-4700 the resistance of
    the resistor will increase to:

    R = 240V/0.002A = 120000 ohms

    And the dissipation will decrease to

    P = 0.002*240 = 0.48W

    So you could safely use a resistor rated for >= 1 watt, a much better
    choice than a 10 watter!


    If you were to use a high-brightness LED you could probably get away
    with using much less current for the brightness you required, and
    consequently enjoy being able to use use a much smaller wattage
    resistor.

    One caveat is that, no matter what size resistor you use, you have to
    make sure that it can stand the _peak_ mains voltage which, for 240VRMS
    mains, turns out to be 240*1.414 ~ 340V. If you can't do it with one
    resistor put two in series, each of which has a resistance of half of
    the total resistance you need and a wattage rating equal to the total
    wattage you need.
     
  9. <snip>

    It *is* possible to do *both*, John. ;-)

    (which was my suggestion..)



    Best regards,
    Spehro Pefhany
     
  10. John Fields

    John Fields Guest

    Sure is, sure was!

    I read 'Adding" and thought 'substituting'; mea culpa :-(
     
  11. WAW

    WAW Guest

    Hey Phil, I may just be an electronics student but I like Jo's idea
    best. It's simple and to the point and best of all, it works.
    There is one hitch though: that 22 KOhm Resistor is going to
    dissipate...

    P = I * E

    I = E / R

    so P = (E ^ 2) / R = (240v ^ 2) / 22 KOhm = about 2.5 watts.

    If you can't find a 2.5 watt or higher resistor you can make one. It's
    gonna be big but ...

    place ten 220 KOhm resistors in parallel. The result will be a 22KOhm
    resistor capable of handling 2.5 watts. Adding more resistors requires
    you to raise the ohmage of each but also divides the current amongst
    more resistors yeilding greater wattage capability.

    There's pretty much no way to get around it. 240V is a lot for
    risistors to drop.

    P.S. This things gonna get HOT!!! {:eek: Hot Hot Hot. I'm talkin about
    cookin' eggs on the sidewalk hot. I'm talkin' may-need-a-fan hot. I'm
    talkin' ... I'm talkin' too much.

    Anyway, God bless all and America too. <><
     
  12. Roger Gt

    Roger Gt Guest


    Perhaps I am late with this, but I usually run LED's from a 230/240 VAC line
    by connecting a 270 ohm 1/2 resistor in series with a 400 volt film cap with
    a pair of LED's back to back connected to the Neutral line. For 50 hertz
    the cap = .47mF For 60 Hertz the cap = .33 mF and there is only the
    dissipation in the LED's and the surge resistor (about .33 Watt) The cap is
    fairly large, but doesn't get hot! I currently use some Panasonic caps, and
    they are only about 16 X 12 X 6mm.

    Just a suggestion. But I know it works, I've used thousands of them!

    Roger Gt

    --
    Set wards, light torches, unfurl banners, play a joyous tune.
    Yes! Eat drink and be merry, for tomorrow --- You know the rest!
    The lot of all living things. To bide a bit, and pass!
    Leaving only foot steps in the sands of time!

    Happy Holidays to all! Celebrate as you will!
    May the Gods be kind to you and yours!
     
  13. Lee Leduc

    Lee Leduc Guest

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