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How to drive LED from AC240V?

Discussion in 'General Electronics Chat' started by VictorPS, Jul 26, 2005.

  1. olly_k

    olly_k Member

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    How does power factor relate to this issue i.e. the efficiency of the circuit in question (imagine the led reaaaaaally big!)
    Also to Ron H how did you arrive at 680nf I tried using formula 1\2 pi f Xc = C but only got 265nF????
     
  2. Roff

    Roff Well-Known Member

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    Remember it's 50Hz, 220V, and the LED current is half wave. Also, I'm lazy. I simulated it. The average current was 21+ mA with 680nF and 220 ohms.
     
  3. ym2k

    ym2k New Member

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    Most commercial products you seen that drive at 240vac and compact is using the capacitors method. I had tried it many times and it always works for me.

    If you need transients protection, get a varistors. So far my leds is fine without one.
     
  4. dave

    Dave New Member

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  5. VictorPS

    VictorPS New Member

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    For 100mA , what cap value I shall use? Will it be bulky?
    Thank you. :)
     
  6. ym2k

    ym2k New Member

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    May i know what kind of led you are using? Why the need of 100mA? A typical 3,5 & 8 mm led can be driven at 20mA for best performance. High powered led like luxeon I drive at 350mA and III series at 700mA. For high power led, using capacitor methods is not advisable.

    Typical commercial 240VAC led lightnings without transformers are all driven at 20ma. Previously, i post this diagram before. You can refer to it to light up your leds without the fickering cause by direct AC. For the 470nF, get those supperssor-type. Voltage rating should be above 400VAC for safety. The bigger the voltage rating, the bigger the cap size.
     

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  7. VictorPS

    VictorPS New Member

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  8. ym2k

    ym2k New Member

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    Hm, driving these kind of led without a transformer is not advisable. You can use a small audio stepdown transformer for your application. Step it down to 12v and rectify it to dc voltage and use a constant current driver for your application. LM317 works pretty well if efficent is not in concern.

    If you need more efficent, a flyback step down circuit may be use to control both voltage and current.
     
  9. VictorPS

    VictorPS New Member

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  10. Oznog

    Oznog Active Member

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    You would need a friggin HUGE resistor to do this. 100mA of whatever the voltage gap is between 240v and a low voltage is most of 24 watts.

    The best way to do this is a LED buck converter. SuperTex makes them:
    http://www.supertex.com/pdf/datasheets/HV9910.pdf

    That results in a smooth, regulated constant current output. There is no ballast resistor so it is both very efficient and insensitive to voltage changes.
    The trick may be in finding a MOSFET with enough voltage capacity. Actually, a bipolar can do the job here too.

    A problem I ran into with SuperTex is the initial current spike through the source from charging the gate triggered the latch off prematurely. SuperTex recommended a 300nS RC circuit on the current shunt signal. That works but the circuit cannot respond quickly enough and it means there is now a minimum on-time of around 300nS. Since you have a need for very short duty driving an LED off 240v, this may be an issue that affects your max freq.

    A bipolar wouldn't have the surge issue, the base current's contribution to what the shunt reads should be quite predictable.
     
  11. Oznog

    Oznog Active Member

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    Or you would just use 2 strings of LEDs, one in the forward direction and one in the reverse direction. Thus one string is always on and there is no problem with reverse voltage since the reverse biased side won't see more than its forward voltage.
     
  12. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

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    As this is presumably for a mains ON indicator?, how about using a simple neon bulb and series resistor?.

    Often the older techniques are far better than a more modern solution!.
     
  13. panic mode

    panic mode Well-Known Member

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    i agree with nigel. oznog's suggestion would work for leds but i would rather use tiny bridge rectifier and just one led string.
    the downside of two anti-parallel led strings you get only 50% duty cycle for each string and is if one led ever dies (open circuit), the leds in other string are doomed too.
     
  14. Oznog

    Oznog Active Member

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    The bridge rectifier is way far from practical though. What, you're going to use a many-watt resistor?
     
  15. panic mode

    panic mode Well-Known Member

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    i'm not sure what do you have in mind...
    same voltage drop / current limiting circuit would be used regardless
    if you use bridge and one string or two strings in anti-parallel.
    in fact it was already posted by ym2k
     
  16. spuffock

    spuffock Member

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    On the capacitor circuit: OK, the capacitor dissipates no power. But the current is still there, so the losses in distribution of the electricity are still there. I think Nigel is nearer right on this one. The bulk of the losses are still there, but the electricity company pays for them.
     

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