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LED pilot light powered from 230V

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by adrianbodor, Mar 28, 2011.

  1. adrianbodor

    adrianbodor New Member

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    I read a lot about this circuit, for many of you probably is a simple circuit but i want to use for a simple LED indicator to a socket outlet, which means it will stay continuously lit for a long period of time. I want to use a capacitor class X2 to eliminate the heat in case of use a resistor and with 10mA is enough for this purpose.

    I saw many variants of this circuit on a polish forum
    http://obrazki.elektroda.net/49_1202304608.png
    http://danyk.wz.cz/ledzar.html
    http://obrazki.elektroda.net/85_1187423053.gif

    Now, I understand that capacitor limits the current to the LED with a value of 150nF which gives about 10mA. the bleeding resistor across the cap discharge it when power is out. The resistor R2 1K serves to limit the current stroke when you turn. Bridge rectifier is obviously what it does. The elko limits the strobe effect to the LED. Till now I understood quite well.

    What I don't know is the use of the zener i the other circuits. I saw a description of this, it protect the circuit in case of LED failure

    I want to design an almost perfect circuit taking into consideration a LED failure because it happens after continuous use, CAP failure, etc. I don't want the circuit to became a firework when some components fail to operate due to long use.

    I accept any kind of suggestions to make this circuit better.
     
    Last edited: Mar 28, 2011
  2. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    In the first two examples the LED current is limited by the high voltage capacitor. It is effected by frequency (50/60) and line voltage.
    In the last example, It looks like the desire is to get 12 volts (regulated) then use the 180 ohm resistor to set the LED current. The light will be more constant with line voltage changes. (more parts)

    The high voltage capacitor sets the current. LEDs convert current to light.
     
  3. RODALCO

    RODALCO Well-Known Member

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    I have successfully used led's for more than 20 years with a 1N4007, and two resistors of 27 k.Ohm 1Watt in series, hardly any heat.
    A reverse 1N4448 diode to protect the led.
    The high efficiency led's need only 2 or 3 mA to give good light out put.
    Search under 230 v led light in this forum
     
  4. dave

    Dave New Member

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  5. adrianbodor

    adrianbodor New Member

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    I know that 2-3 mA is enough but i want to use 10mA for a decent light, this is equivalent with 150nF cap. I saw many circuit of this type with the resistor for the inrush current placed before cap in other circuit is placed after cap and bleed resistor. Does it matter where you put the resistor?

    I think i will do schematic based on this circuit and complete with a zener diode to limit the voltage and one elko to eliminate the strobo effect.

    Electronic Circuit Collection: AC Powered LED Circuit LED Circuit AC Powered compact design

    Another dilema is that i want to use 2 LEDs in series but with different colors, one orange and the other blue and the Vf is different for both, one has 2.1 and the blue 3.4V but if current will be set to 10mA it will do harm to one of the LEDs?

    Tomorrow I will post the schematic with simulation and power consumption.

    Thank you
     
    Last edited: Apr 19, 2011
  6. carbonzit

    carbonzit Active Member

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    Multiple LEDs in series isn't a problem, if they all use (approxmiately) the same current. It's when one or more draw significantly higher current than the others that you have trouble. Simply treat the entire series string as a single voltage drop.
     
  7. adrianbodor

    adrianbodor New Member

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    As I said this is the circuit modified. I added a zener diode and a filter cap. The circuit has two different colors LEDs. As carbonzit suggested i treated the entire LEDs series as a single voltage drop of 5.5V and used 5.6V zener to limit voltage across cap.

    The circuit consume aprox 250mW and current is aprox 7.5mA through the LEDs.
    If you have commentary please feel free to post anything will improve this. On the safety side, I think is OK but I'm not sure what will happen if some component fail; This circuit will operate full time.

    Thank you.

    [​IMG]
     
  8. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    hi,
    Interesting circuit.
    I would have used a miniature mains type 'bead' neon lamp with a resistor as an ON indicator.
     
  9. carbonzit

    carbonzit Active Member

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    The only part that could fail catastrophically would be the capacitor, and that seems unlikely, especially since you're using an X2 type. Anything else up there would probably just go open-circuit if anything, so it seems it should be pretty safe.

    The one part I'd be a little careful with is the 470Ω resistor, which may get a bit warm. Might want to run it a while and check the temperature.
     
  10. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    I would agree with CZ, the 470R is too high a value, IIRC the UL requirement is about 15R thru 47R.
     
  11. carbonzit

    carbonzit Active Member

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    So why do you think you need the bleeder resistor? You said up there that this is to discharge the capacitor if the power goes out. Well, who cares if the capacitor is charged? so long as you don't put your fingers across it ...

    I've had my little power-line (120V) powered LED up now for a week. It's just a 0.56µF cap, a 1KΩ resistor and a red LED in series. No rectifier, no bleeder, no nothing. Works fine. (I actually built it partly just to tweak some people here who said I shouldn't do it this way.) Cost me $0, made of parts from my junk box, cased in a Tic-Tac package. It's been plugged in next to my fridge, plenty of inductive-motor-starting transients, etc. We'll see how long it lasts.
     
    Last edited: Apr 24, 2011
  12. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    hi CZ,
    If you are asking me.??
    The discharge resistor is a UL requirement, the same as the 15R.
     
  13. adrianbodor

    adrianbodor New Member

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    Didn't know that but also acording to multisim simulation the power disipated on 470R resistor is very small, 0.100W and the resistor is 1W rated so I don't think heat will be a problem.

    Maybe a silly question but what IIRC the UL requirement represent?
     
  14. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    hi,
    IIRC... If I Recall Correctly......
    The UL, are the American Underwriters, they test and specify safety and other standards to industry.

    http://www.ul.com/global/eng/pages/corporate/standards/
     
    Last edited: Apr 24, 2011
  15. colin55

    colin55 Well-Known Member

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    carbonzit is a very bad design. Firstly there is nothing discharging the capacitor except the reverse voltage through the LED. This will damage the LED.
    The simplest circuit for two LEDs is the second diagram:
    [​IMG]
     
    Last edited: Apr 24, 2011
  16. adrianbodor

    adrianbodor New Member

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    My opinnion a full wave bridge is best choice allong with one elko cap to reduce flickering, in this current will flow in both direction through the bridge.
    but is my option to choose this.
     
  17. KMoffett

    KMoffett Well-Known Member

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  18. colin55

    colin55 Well-Known Member

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    To reduce flicker an electro can be added:
    [​IMG]
     
  19. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    Agree 100% Ken , my post #7.
    It seems an over complicated way for power indication.
     
  20. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    Colin,
    I would use a mains rated 10uF, just in case an LED fails o/c....loud bang and smoke.
     
  21. adrianbodor

    adrianbodor New Member

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    I'm thinking wich one is more reliable, this circuit or other one witch include a small encapsulated transformer rated at 6VAC and 80mA plus a voltage regulator, bridge and a filter cap.

    I'm talking reliable in terms of continuous use. and power consuption.

    What to you think?

    Thank you,
     

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