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How to drive LED from AC240V?

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

  1. VictorPS

    VictorPS New Member

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    I like to drive 5 LED (white) directly from 240V AC , the application is to replace normal lighting bulb..
    The LEDs are connect in series, the Vf for each LED is 3.2V, current can be 100mA, so total voltage require is about 16Vdc 100mA.
    Beside using transformer to step down the voltage, what is other possible way to drive the LEDs (100mA) directly from AC240V?
    Is there any dedicated IC for this application? :roll:
     
  2. bloody-orc

    bloody-orc New Member

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    connect a series 2k2 resistor and a diode in the other way the leds are. if the leds are from "+" to gnd then diode must be from gnd to "+".

    (gnd is one hole in AC and "+" is the otherone)

    the diode must take the voltage so use 1n400X series. It is there so that the reverse voltage wouldn't kill your diodes.
     

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  3. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

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    A transformer is the only sensible way, either conventional, or a switch-mode design. Any crude 'dropper' technique is going to waste most of the electricity used - in your case 93% wasted! - which rather makes a mockery of trying to use low energy LED's.
     
  4. dave

    Dave New Member

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  5. eblc1388

    eblc1388 Active Member

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    Looks at your circuit diagram again. If you connect AC1 and AC2 to 240V, prepare to put out a fire. :twisted: :twisted: :twisted:
     
  6. Someone Electro

    Someone Electro New Member

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    Yep it aint looking good for the "protection" diode.Conecting an diode directly to the mains is going to totaly fry it
     
  7. checkmate

    checkmate New Member

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    Not to mention that when diodes get fried, they usually die as a "short circuit" :twisted:
     
  8. Someone Electro

    Someone Electro New Member

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    yep.bye bye diode and probobly bye bye your house.

    I once tryed to conect an 100 ohm 1/8W resistor to an 40 000uF cap bank.The resistor cought fire in les then a second.

    EDIT:
    Oh an the cap bank was charged to 60V
     
  9. bloody-orc

    bloody-orc New Member

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    well, this schematic has always worked for me. nothing had birned down (jet). Only thing different is that in setead of this "protection diode" i used anothed LED with a resistor.

    If you are so STUPID to try this circuit out with a 1\8W resistor you shouldnt be operating with AC at all. it HAS to be 2W or more. (mine is 5W).
     
  10. Roff

    Roff Well-Known Member

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    Here's what we're trying to tell you (ignoring the issue of wasted power):
     

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  11. Roff

    Roff Well-Known Member

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    Why has no one suggested the usual method of using a capacitor to limit the current? I have usually seen it with a low-valued resistor also in series, for protection from transients. Of course, you still need the anti-parallel diode (or another LED) across the LED.
     
  12. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

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    It's not really any different?, it still wastes the same amount of power!.
     
  13. eblc1388

    eblc1388 Active Member

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    Why put the diode in parallel? Just put it in series and save 50% of energy and have a smaller resistor power rating.
     
  14. Styx

    Styx Active Member

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    Because although standard diodes can have very high blocking voltages (little BAT16 can block 50V) the proporties of an LED comes at the price of some of a diode characteristics. Namely blocking voltages

    It is not uncommon to have an LED that can only block 5V.

    You connect an LED to the mains (in the correct configuration) with no anti-parallel diode, the LED will have to block all the volts and it wont be albe to
     
  15. Roff

    Roff Well-Known Member

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    The capacitor won't dissipate any power. The resistor will, but it's much lower in value than it would be if it were doing all the current limiting. I'm thinking of maybe 220 ohms or so.
    A 680nF cap will give an average LED current of about 20mA.
     
  16. Roff

    Roff Well-Known Member

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    Without the anti-parallel diode, even if the LED were able to stand off the reverse voltage, no average DC could flow through the LED, so it wouldn't light up.
     
  17. VictorPS

    VictorPS New Member

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    Thanks for reply.
    1. I have seen the LED bulb that direct plug to 240VAC, they are compact , light weight, very little heat dissipate,
    Anybody know what is the circuit inside?

    2. If I want to use switch mode power supply that can deliver 16Vdc 100mA, or constant current mode 100mA,
    what IC you suggest?

    3. Is it possible to use SCR/Triac technique to perform above task?
     
  18. eblc1388

    eblc1388 Active Member

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    Hi Ron and Styx,

    I re-read the posts and now know I am referring to the case of resistor dropper and you guys referring to the capacitor dropper. I agree that in case of a capacitor dropper, an anti-parallel diode is needed across the LED for it to work.
     
  19. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

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    'Magic' doesn't work :lol:

    If you're feeding the LED at the same current, then EXACTLY the same amount of energy HAS to be dissipated somewhere! - as most of the voltage is dropped across the capacitor, the capacitor will dissipate it. The resistor is mainly intended as a 'fuse', if the capacitor goes S/C.

    Do the maths! - "dissipation = voltage across capacitor x current through capacitor", capacitors aren't some 'magic' component where standard rules don't apply :lol:

    Capacitors used to pass high currents DO get hot, and their failure is a common problem.
     
  20. JimB

    JimB Super Moderator Most Helpful Member

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    Sorry Nigel but you are way off on this one.
    Dont forget that in a capacitor the voltage and current are at 90 degrees to each other, a phase angle of 90deg.

    In an AC circuit Power = V x I x Cos(phi)

    Here phi is 90 deg so Cos(phi) = 0

    So no power dissipated in the capacitor.


    Ok, above I was referring to perfect capacitors, in practice there is a VERY SMALL Equivalent Series Resistance (ESR) to account for losses in the capacitor but for most capacitors it is very small.

    However, with capacitors like big electrolytics the ESR is a bit higher, if they are passing large currents they warm up and start to age, the ESR will increase so they get hotter and eventually just give up or explode.

    JimB
     
  21. Roff

    Roff Well-Known Member

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    I have only seen the capacitor current limiting scheme discussed in forums - I have never tried it. I don't think it is a good idea unless the line (mains) is really clean (i.e., no big transients). Otherwise, peak currents could be really high, and might destroy the LED.
    I stand by my power dissipation comments, which were explained well by JimB.
     

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