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LED pilot light powered from 230V

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

  1. RODALCO

    RODALCO Well-Known Member

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    LED and Capacitor

    Just checked my workshop, desk lights off, tripped MCB

    The LED with the 0.1 µF 300 V ac rated cap out.
    Cap blown apart, led kaput ( conducts in both directions with No light )
    680 R series R burnt out.

    Other 2 led's going again.

    event counter was on 92.

    Not sure if the 300 V ac cap was X 2 as it was not on the label.
     
  2. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    hi Rod,
    With a 230Vrms supply, at peak of approx 325V, with a 680R limiting resistor, thats a 'pulse' of around 470mA.

    My guess would be, if the switch ON occurs on the negative half cycle peak, that 470mA reverse current will blow the LED.

    It would be interesting if when you have finished this test, rerun with a reverse parallel LED or Diode, just to 'confirm' that its the reverse LED current thats causing the failure.

    E.
     
  3. RODALCO

    RODALCO Well-Known Member

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    Hi Eric,

    The LED in this case had a 1N4448 reverse protection diode across it.

    I have more of those 300 V rms caps and will repeat it again , just in case that one off cap was dodgy.

    Then i found a 250 V rms X 2 cap which i will add to the test as well.
     
  4. dave

    Dave New Member

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  5. peterlonz

    peterlonz New Member

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    I also have just completed a few tests with unexpected results
    With 230V mains AC, I placed two 275VAC X2 rated caps in parallel to achieve a combined capacitance of 0.32uF.
    This should limit the circuit to 23mA (RMS) which I thought would suit most of my higher brightness LEDs.
    Typically they are spec'd for about 25 mA to 30 mA continuous, & fwd Voltage typically 3V. Brightness varies from 5,000 to 15,000 mcd, depending on which colour/type is selected.
    Now as I calculate the current in Carbonitz's circuit No 2 ("you're going to hate this") in which he removes the resistor altogether & relies only on a 0.3uF cap to control the current at about 21 mA - Very close to my set-up!
    I then proceeded to test individually, three different LED's in series with the parallel caps.
    All destroyed, not even a flicker of life.
    So clearly I have done something fundamentally different to others experimenting here, but what?
    BTW for the sake of safety I ran the mains via an RCD with a 230VAC CF lightbulb wired in parallel to my test circuit & illuminated when the circuit was live - just a precaution against any absent mindedness on my part, I can't see this affecting anything however.
    I'm not the sharpest in this area of electrical technology so I am baffled as to why I can't reproduce the results of others as a starting point at least.
    I suppose Carbonitz is LOL, he probably has 12 weeks up now on his rig?
     
  6. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    hi p,
    I would say you are getting the same results as the other experimenters, you are blowing LED's when connecting them to the mains using this badly designed circuit.

    The reason that you are getting 'slightly' different results are for the reasons given, ie: variation in manufacture, switching the mains ON near its cycle peak,, etc....
     
  7. Diver300

    Diver300 Well-Known Member

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    I think that it all depends on the reaction of the LED to reverse current. Some LEDs can take reverse current and survive, some don't. The data sheets are no help, because most LEDs are only tested to 5 V reverse, because it is generally very simple to design circuits that never exceed that.

    Some LEDs that I tested didn't breakdown at 40 V. However, it the LED did manage block all current at the peak mains voltage, the LED wouldn't light because the capacitor would never discharge, so it would be fully charged next cycle and there would be no forward current.

    As Eric said, it's a good idea to have a resistor to limit the turn-on current.
    If you assume that the LEDs can't take reverse current, and put a diode to take the reverse current, or use a bridge rectifier, just about any LED will work.
     
  8. RODALCO

    RODALCO Well-Known Member

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    Capacitive circuit

    Interesting replies here and great detail in the findings all with destructive results.

    I have done my survey but I keep the led currents well below the 20 mA.

    Everyone seems to want to run the leds at 20 mA.

    Most high brightness leds are already very bright at 10 - 14 mA.

    With the Cap circuit the inrush current is the problem here.

    I have done a week survey with nearly 15000 on off cycles and had two serious faillures with unknown but mains rated caps.

    But I use a 1 Watt series R 680 - 1000 ohms, ( these acted as a fuse when the fault occured and tripped a 5 Amp MCB )

    The other leds all survived

    I haven't edited the video yet but it will be uploaded today on YouTube.

    I will post the link when loaded.
     
  9. RODALCO

    RODALCO Well-Known Member

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    This link refers to the older type leds ( look at 2:14 )

    In this video I destroy some leds with a cable fault surge generator e.g. thumper

    http://www.youtube.com/watch?v=dnZ3TUQBimM

    See at 2:14 of the video.

    Some of those older red led's take currents up to 400 mA before going pop.

    The newer led's are a lot more surge sensitive.
     
  10. RODALCO

    RODALCO Well-Known Member

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  11. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    hi Rod,
    Many thanks for carrying out that test and posting your results.

    If I understand correctly, your advice would be to have a series capacitor, rated at X2, at ideally double the mains supply voltage.

    Series resistor and a reverse diode parallel diode across the LED.

    Eric
     
  12. colin55

    colin55 Well-Known Member

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    Putting a diode across the LED is 50% wasteful. Use 2 LEDs back to back.
    A current limiting resistor of about 1k or higher is also needed.
     
  13. RODALCO

    RODALCO Well-Known Member

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    Hi Eric, thanks for your kind comments.

    Exactly what you mentioned.

    I recommend a X2 Capacitor or a Cap rated at least twice the mains voltage to allow for a safety margin.

    and a reverse blocking diode to protect the LED. That can be a low power one 1N4448, 1N914 or eq.

    Series R which reduces the inrush current and it acts as a fuse in case of a serious fault.
     
  14. RODALCO

    RODALCO Well-Known Member

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    That is possible or use a small bridge rectifier.

    Also currents don't have to be always 20 mA. most high efficiency led's are very bright already at 7 - 15 mA.

    And we are talking a pilot indication light here.
     
  15. Mr RB

    Mr RB Well-Known Member

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    All the decent pro designed capacitive reactance supplies I've seen use a series resistor, it protects from turn on surge and also really reduces mains spike voltages. The resistor usually drops 5% to 10% of the total mains voltage.

    An X2 cap rated at the mains voltage is standard and fine, I've never seen a mains rated cap fail in a device that had the series resistor. X2 cap failing is common in nasty units that don't have the series resistor.
     
  16. Diver300

    Diver300 Well-Known Member

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    At the risk of opening a whole new can of worms, I'll post what I've just found.

    I came across an LED mains indicators that consisted of simply an LED and a 100 kΩ resistor. I now have a storage 'scope, and it can be battery powered so I can connect the 'scope earth to the mains neutral, so I recorded the trace. The yellow trace is the mains voltage, the red is the voltage across the LED. The forward voltage is so small that it can't be seen with the gain so low on the 'scope:-

    [​IMG]

    The LED is breaking down in reverse at around 140 V. This means that the peak current is around 2 mA, and the peak power is around 300 mW. I would estimate that the average power is around 75 mW. Running an LED like that with a steady current of 20 mA would result in around 40 mW of power being dissipated in the LED.

    Interestingly, it doesn't. As we have seen, when an LED is rated at a reverse voltage of 5 V, that means that it is guaranteed not to let current pass at 5 V. However, the reverse voltage rating of 5 V certainly doesn't imply that the LED will conduct at 6 V reverse. It also appears that if the current and power are not too big, forcing the LEDs to conduct in reverse does not cause immediate failure.
     
  17. KMoffett

    KMoffett Well-Known Member

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    What colour LED? Did you try it wit other colours?

    Ken
     
  18. Diver300

    Diver300 Well-Known Member

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    It was green. I haven't tried with any other colours. I guess that the reverse breakdown characteristics are not related to colour.
     
  19. KMoffett

    KMoffett Well-Known Member

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    Look at my post #56.

    Ken
     

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