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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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    As ericgibbs suggested i modified the circuit with an inrush current limiter resistor of 47R Metal Oxide type flame proof and removed the elko cap and zener diode, because in case of zener fail the elko would explode. Also the safety cap is 0.1 witch leads to aprox 7mA to the LED, enough for this purpose especially if u use an high efficiency LED.

    So in terms of safety if the LED fails there is basically no load.
    If the X2 safety cap fails (this is self extinguishing) also the metal oxide inrush resistor will melt safely being flame proof, leaving an open circuit.

    [​IMG]

    The circuit consume aprox 100mW, but for all who wants to build it (in the test phase) be aware that you are working with lethal voltage, all leads must be isolated with heat-shrink tubing including the LED. After circuit is switched off (if u switch off) must wait several minutes until the bleeder resistor discharge the safety cap.

    Thank you all for posting. For any comment feel free to correct me if i'm wrong with something.
     
  2. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    adrianbodor,

    I have used your circuit. It works well. The diodes do not need to be 1n4007 they could be 1n4001 because there is only 4 volts on them. (1n914 or 1n4148)
     
  3. peterlonz

    peterlonz New Member

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    Thanks for showing a 230VAC/LED circuit.
    The parts count is increasing so it's no longer the dead simple circuit of "you will hate it", Carbonzit.
    If that circuit was better understood, & assuming it's still working, could we not design on that basis?
    Which boils down to answering the question why does it work with some LED's & not others.
    I happen to have "investigated" a commercial 230VAC/LED circuit. Can't be sure I can accurately draw the circuit but I should be able to provide a front & back pic if anyone is interested.
     
  4. dave

    Dave New Member

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  5. colin55

    colin55 Well-Known Member

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    That is not true. There is 325v across two of them in series.

    47R is far too low. The minumim is 2k2.
     
  6. ronv

    ronv Well-Known Member Most Helpful Member

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    From wiki:


    Electrical polarityMain article: Electrical polarity of LEDs
    As with all diodes, current flows easily from p-type to n-type material.[81] However, no current flows and no light is emitted if a small voltage is applied in the reverse direction. If the reverse voltage grows large enough to exceed the breakdown voltage, a large current flows and the LED may be damaged. If the reverse current is sufficiently limited to avoid damage, the reverse-conducting LED is a useful noise diode.

    Did the yellow diodes blow?
     
  7. adrianbodor

    adrianbodor New Member

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    In this post he said that a good value for the inrush current limiter is max 47R now u said

    Now what is the correct value
     
  8. carbonzit

    carbonzit Active Member

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    I'd just like to point out that my little line-powered LED (circuit given in this post) which is just a cap, a resistor and a red LED is still burning brightly and has been doing so for about six weeks now, showing no signs of failing.

    (I live in 120-volt land.)
     
    Last edited: May 18, 2011
  9. colin55

    colin55 Well-Known Member

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    I would just like to point out that the LED sees a reverse voltage of about 160v and LEDs don't like reverse voltages above 5v. It breaks down at about 5v.
    We are trying to be an educational electronics site.
    If you want to start up a site that teaches all sorts of incorrect material, go to India and join Electronics For You magazine, where they teach all sorts of rubbish to the readers.
    Lot's of things work, but that does not make them an engineering-correct alternative.
    You are discharging the capacitor via the LED and allowing a reverse voltage and current to be applied to the LED.
     
    Last edited: May 18, 2011
  10. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    Carbonzit; I normally put a small diode across the LED to short out the reverse current, that other wise would pass through the LED.

    Colin; How do you get 325 volts across the full wave bridge (4 diodes) when there is a 3.4 volt "Zener" (LED) across the bridge. I think anytime there is more than 4.6 volts across the diodes the LED turns on and limits the current.
     
  11. colin55

    colin55 Well-Known Member

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    You, you are correct, the voltage across the diodes is not 325v.
     
  12. adrianbodor

    adrianbodor New Member

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    I'm still confuse about current limiting resistor and its value

    How can I calculate this value?
     
  13. colin55

    colin55 Well-Known Member

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    The current limiting resistor is a protection-resistor for the time when the circuit is turned on and the wavefrom from the mains is at a peak.
    The capacitor is effectively uncharged and the current through the LED will depend on the value of the resistor. If the resistor is 2k2, the initial current will be about 150mA. This is only for a very short period of time but a LED cannot withstand a high current for more than a few milliseconds.
    Anything less than 470R is going to instantly damage the LED under these conditions.
     
  14. 4pyros

    4pyros Well-Known Member Most Helpful Member

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    You need to cycle power to see if it blows.
     
  15. peterlonz

    peterlonz New Member

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    Hi Carbonzit,

    I was interested in your unconventional & VERY simple LED "AC mains driver circuit".
    No I am not smart enough to contribute; caps in AC circuits have me bluffed.
    You said in your earlier posts (April 27th 2011)you intended to do an extended test to establish if early failure was likely.
    May I ask if you got around to doing that & what if any was the result.
    Thanks,
    Peter O
     
  16. peterlonz

    peterlonz New Member

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    Colin55,
    Only now have I just read your comment.
    I'm a bit disappointed, surely what we would all like to understand here is exactly why/how carbonzits super simple circuit works.
    I know its not been subject to rigorous tests, & because some LED's work & others do not, it should probably be operated within well defined limits.
    But it works against expectations, & if we knew why, surely circuits much simpler than some of the "safe & tested" options could perhaps be developed & used reliably?
    I don't think carbonzit deserves to be sent to India, .......... do you?
    Peter O
     
  17. buteman

    buteman New Member

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    How did you work out power consumption? 230volts x 7.5mA = 1725 mW
     
  18. RODALCO

    RODALCO Well-Known Member

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    You design the circuit for double the mains voltage with a capacitor.

    For 230 Volts the peak value is 325 Volts.

    If the cap is switched on at + 325 and the mains sine wave is at -325 Volts, twice the voltage is present which can destroy the lot.

    A 400 V rms rated cap or higher is needed.

    Then ensure that the current surge is limited to < 20mA at worst scenario to avoid popping the LED's / reverse protection diode or series resistor. Usually 1 k.ohm 1 Watt is adequate.

    I will do a duration test with a capacitive LED circuit and put it on YouTube in the next week or so.
     
  19. peterlonz

    peterlonz New Member

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    LED indication on AC mains supply

    Rodalco,

    Thanks for the contribution.
    Understand your capacitor selection criterion.
    However if I purchase a 250VAC cap am I not covered?
    How to could measure the voltage output & so "limit to 20MA" I'm not sure?
    I do not feel confident my DMM would survive if I attempted to use to measure ac output in the circuit.

    I think Carbonitz has already shown in his ultra simple circuit from 120VAC mains:
    " It's just a 0.56µF cap, a 1KΩ resistor and a red LED in series."
    Apparently (see his post) this works.
    At least under certain circumstances.
    Admittedly he has not since reported on his endurance test.

    What amazes me is how everyone has their own take on how to do this; there are after all, quite a number of "safe conservative" circuits published & available on the net.

    Why also is the "efficiency" of LED's referred to, this is a parameter that I never see in manufacturers Data-sheets. We normally know the power in but only mcd or lumens output.
    What I have found is that if you want a bright warning LED; then you need a minimum of 2000mcd, more likely 5000mcd particularly if you need to see the warning light in sunlight conditions. Normally the input for a 5000 mcd red LED is about 3.5 FwdV & about 25mA.
    I'm pretty sure Carbonitz red LED on test would be maybe only 12mA.

    I guess I am going too have to try this out somewhat blind, start with high value resistor & X@250VAC rated cap with 5000mcd red LED.
     
  20. RODALCO

    RODALCO Well-Known Member

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    Hi peterlonz

    I have set up my LED test circuit for durability.

    On for 57 seconds, off for 3 seconds.
    Event counter attached

    LED 1 with caps 4 x 47nF 1kV parallel 9 about 188 nF. I about 13 mA ( 1 k ohm R inrush limit )

    LED 2 with 0.1µF 300V rms cap I about 8 mA ( 680 R inrush limit )

    LED 3 with resistors 2 x 27k.ohm series 1 Watt

    A video will be posted on YouTube with the results after 1 week.
     
    Last edited: Oct 18, 2011
  21. RODALCO

    RODALCO Well-Known Member

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    If the cap is AC or RMS rated it should be OK.

    Often for extra insulation the 400 V Caps are preferred. although in appliances the X 2 caps are often 250 or 275 Volts

    the I limit R will act as a fuse and burn out when the Cap fails
     

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