220 volt led on 12 volt

Discussion in 'General Electronics Chat' started by jonnydolt, Jun 11, 2016.

  1. jonnydolt

    Thread Starter New Member

    Oct 23, 2015
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  2. BobTPH

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    Jun 5, 2013
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    You cannot just give it 12V and expect it to work.

    That said, the LED(s) operate a much lower voltage (less than 12V) and if you were to take it apart and replace the circuitry inside with something that would work at 12V you could make it work.

    Bob
     
  3. Tonyr1084

    Active Member

    Sep 24, 2015
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    Internally I suspect there's just a red LED (Red assumed) and a high value resistor.

    To run a red LED on 12 volts (at 20 mA) you'd need a 500 Ω resistor. Again, assuming it's a "Red" LED internally running on 220 VAC, I'd suspect the resistor in there is anywhere from 7K Ω to 11K Ω. The 7K would provide approximately 30 mA and the 11K should give you close to 20 mA. It's the current that is of most concern. Voltage is of little consequence, especially when at high voltages.

    Can you run it on 12 volts? Probably. But you'll have to change the internal resistance.

    Now, if they're using a super bright white LED then your current needs will be different.
     
  4. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    I suspect there's a capacitor internally, used as a dropper. (A simple resistor would have to dissipate over 4W at 20mA.)
     
  5. Tonyr1084

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    Sep 24, 2015
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    @Alec_t: Good point. I didn't stop to consider wattages. I have to agree with you.
     
  6. #12

    Expert

    Nov 30, 2010
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    I was all puckered up to make a schematic for you, then I remembered Wendy rapping my knuckles with a ruler for doing that. (Inside joke)
    It's against the rules of this site to tell you how to use a capacitor to run an LED straight off a 220 volt power line.
    Therefore, this Thread is in great danger of being shut down by a Moderator.
     
  7. Alec_t

    AAC Fanatic!

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    No prob. The TS wants to run the LED from 12V, not 220V.
     
  8. #12

    Expert

    Nov 30, 2010
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    Oops. You caught me making a mistake.:oops:
    Sorry. My head is a bit foggy right now from working in a heat index over 100 F for the last 3 hours.:(
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    And 20 mA through 500 Ω would be 1 W -- and that's at 12 V DC. If you have 30 mA through 7 kΩ you would be talking 6.3 W, though it would be more like half of that due to the rectification.

    There's no way of telling what current limiting circuitry is in there, but it is likely to be a capacitive voltage divider, in which case it won't work at DC at all.
     
  10. #12

    Expert

    Nov 30, 2010
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    I think you slipped a digit.
    P = I^2 R
    0.02A^2 x 500 ohms = 0.2W
     
  11. WBahn

    Moderator

    Mar 31, 2012
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    Oh, I did more than that. I just multiplied 20 m by 500 and, because I was doing everything in my head on the fly, ignored the units (which I usually don't do even then, but I certainly do do it from time to time). Further proof of how valuable units checks are. Thanks for catching me.

    Interestingly, I did apply the "ask if it makes sense" and thought that 1 W was an awfully high number, but I talked myself out of looking further by noting that I was looking at a circuit that was going to be about 15% efficient, so perhaps 1 W wasn't too unreasonable. Given that these kinds of sanity-checks are aimed at picking off order-of-magnitude errors, I'm not too surprised that I let myself rationalize this one away too easily.
     
  12. Tonyr1084

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    Sep 24, 2015
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    I'm glad my numbers worked out. However, I DID neglect to account for wattage. In this case a 1/4 watt resistor WOULD have been sufficient. Had the wattage been higher than 0.24 (240 mW) watts then I might have deserved being gnaw'd upon.

    Proof that even the wrong can get lucky and actually be right from time to time. Good discipline would be to work ALL the numbers, not just the current or just the resistance or just the voltage or just the wattage. ALL the numbers. Otherwise it's "Smokin' time."
     
  13. WBahn

    Moderator

    Mar 31, 2012
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    If you expect the resistor to dissipate 0.24 W, then the resistor should be sized for 0.5 W. Safety factors of at least two are recommended for most power computations unless some other consideration is a driving factor.

    Plus, it very likely is not possible (at least with reasonable effort) to bypass the internal current limiting without destroying the lamp assembly altogether. And what's the motivation for even attempting to do so? Instead of wasting time and effort to modify an indicator designed for 220 VAC, why not just purchase an indicator intended for 12 VDC?
     
  14. Tonyr1084

    Active Member

    Sep 24, 2015
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    I must have been staring at Cindy L during that part of the lecture in high school electronics.

    I would agree that keeping a safety margin is important. As for why do this? I guess it's like why we climb mountains. It's there. They say "Necessity is the mother of invention." Probably highly true. However, many inventions have come about by accident or by seeing IF something could be done. I once discovered a key while trying to create a loop antenna. And no - I won't elaborate. Only that I discovered a way to park in the doctors parking garage for free.
     
  15. jonnydolt

    Thread Starter New Member

    Oct 23, 2015
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    Thanks for the input guy's very much appreciated .These are being used for a child's toy ,switch's, lights ,fans etc.I had purchaced some 12 volt lights but they were crap and the soldered leads broke as soon as I picked them up .I had a heap of the 240 volt units lying around and hoped I could use them .I guess I will have to sit down with a magnifying glass and resolder the leads back on the 12 volt units.Once again many thanks.
     
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