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running led off 220V mains

Discussion in 'Beginner Electronics' started by Johan Wagener, Sep 27, 2003.

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  1. Is it possible to run a led off 220V mains supply without using a
    transformer? It draws very little current so I suppose a 1/2 W resistor with
    a very high resistance would do the trick? also do I need rectification
    considering the fact that a led is a diode
     
  2. UncleWobbly

    UncleWobbly Guest

    sure- you'll need to put it back to back or block it with a normal power
    diode (350v forward or reverse - whatever you decide) coz the reverse
    voltage will fry the LED otherwise, but work out the resistor as peak
    voltage-2 / 10mA so with 220V RMS, we get (to get peak from RMS divide by
    0.7071)

    220/0.7071 = 311V peak

    less 2V for the LED (practically you can ignore the LED voltage when driving
    from HV sources) driving at 10mA... 309V/0.01A=30900ohms so any resistor
    above 30K will do the trick - of course it will flicker (impersceptably?) at
    the mains frequency

    remember also that the other diode (you could always use two LEDs) will be
    shunting the current if mounted back to back (by which I mean the two diodes
    are mounted in parallel but opposite polarities) and it must be rated at
    350V forward voltage. If you decide to use a diode to block (by which I mean
    the two diodes are mounted in serial with the same polarities) you must
    ensure it has a reverse breakdown of 350V or more otherwise it'll all go up
    in smoke :eek:) remember also that 300V @ 10mA is 3Watts of power - resistor
    and power diode please take note. If you can get away with lower current
    you'll be better off - consider using a hyperbright LED and drop the current
    (by raising the resistor value) as much as possible. These LEDs are still
    very bright at just a few mA. You could half the current (and so the power)
    by doubling the resitor and still get a perfectly acceptable brightness from
    the LED - perhaps even more. Play around and get it right....be very careful
    rectifying mains - you'll get 320+volts of DC and across fingers it stings
    *big time* - Any voltage at that level (but particularly DC) from one hand
    to the other ***will kill*** even at very low currents - Don't take chances
    with it!!!
     
  3. It's a matter of power. Most LEDS operate best at a 20mA. Neglecting the
    voltage across the LED you need a 11KOhm 5W resistor to run the led. It will
    become pretty hot. You also have to put another diode antiparallel to the
    LED as the LED will not withstand the 220V (about 300V peek) reverse
    voltage.

    As the LED conducts for only one half cycle the real current is only about
    10mA. To improve the circuit you can place the LED in a diode bridge made by
    four 1N4148 for instance to get the full current.

    To reduce energy loss, you can lower the current. Most LEDs light at small
    currents of 1-2mA already. You may look at high efficiency ones to get more
    light for the current.

    The circuit I advise for the LED uses thee 4k7 0.25W resistors in series, a
    four 1N4148 diode rectifier bridge and a high efficienct red LED.

    For more light you may replace the 1N4148 by LEDs.

    But... for simplicity you can better use a 120kOhm 0.5W resistor and a small
    neon light.

    pieter
     


  4. You'll need a diode to half-wave rectify the mains, and a series resistor.
    Like this:

    .. ___
    .. o------>|------|___|------>----.
    .. 1N4004 R I |
    .. | /
    .. 220VAC LED V /
    .. RMS -
    .. |
    .. |
    .. o------------------------------'


    The half-wave rectifier protects the LED from excessive reverse voltage, and
    R limits the current. For 220V RMS, there will be about 310V peak across R.
    So, for I = 20mA peak, you'll need a 15 kOhm resistor. Because of the
    half-wave rectifier, there will be about 155V RMS across R, so it will
    dissipate about 1.6 Watts. Note that the average LED current will be much
    less than 20mA, but most LEDs will light up just fine. You can also use
    low-current LEDs to minimise heat dissipation across R. Finally, BE EXTRA
    CAREFUL, you can get killed by directly handling live mains like this!

    cheers,
    Costas
     
  5. Brian

    Brian Guest

    The problem with using a series capacitor is that any transients will pass
    through the cap at full voltage. They tend to fry electronic components. So
    if using this method it is essential to include another capacitor across the
    electronic item. Usually around one tenth of the value of the series cap
    will be a good starting point.
     
  6. cpemma

    cpemma Guest

    Actually it doesn't, at least not quickly. I've run a led happily on 60v AC
    for several days (led didn't pack in, I did). On reverse voltage, the led
    will hit its avalanche (?zener?) region, and *provided the current is
    limited* will survive, just as a zener diode does. Current does need to be
    limited to a few mA, or the watts generated exceeds the led rating.

    That said, it flickers perceptibly on 50Hz half-wave, so a
    resistor/capacitor current limiter with a FW bridge rectifier system to the
    led is much better.
     
  7. I am afraid it can happen slowly. Some LEDs fade (apparently to me from
    diffusion of active ingredients) as a result of some combination of
    temperature and bias voltage (what is specifically mentioned in Agilent's
    relevant application brief is a combination of duty cycle and
    temperature). This makes me think that reverse bias, especially with
    heat, can be bad for some of them.

    Some other LEDs, as in GaN and InGaN types in general (in general UV,
    violet, blue, blue-green, non-yellowish-green, white, and Nichia's yellow)
    develop partial shorts nearly enough instantly from reverse breakdown.
    - Don Klipstein (, http://www.misty.com/~don/ledx.html)
     
  8. What current? Low current leds can do with 2 mA, normal ones 20mA,
    and big ones even much more. I will presume 20 mA here.

    Do not use a resistor, but a capacitor in series. For 20 mA you need
    an impedance of 220V/20mA = 11000 ohm. This would mean a
    capacitor of (at 50 Hz) 0,2894 u, lets say 270 nF / 400V
    To limit the inrush current when you switch on at the peak of the
    50Hz sine I recommend a series resistor of 2k2, 2 watt or so.

    Only a series resistor for 20mA would be 11kohm, 4.4 Watt

    And then, use a rectfier bridge that can handle these
    voltages and connect your led to the + and -.

    Success
    Pieter
     
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