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A few questions about LEDs

Discussion in 'General Electronics Chat' started by jmaf, May 20, 2009.

  1. jmaf

    jmaf New Member

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    Hi, my first post on the forum will be to ask about LEDs. I thought I knew all there was to know about LEDs until today I was fooling around in after hours with a LED circuit fed directly from my 220 VAC mains line and observed a few things I'd never known.

    1) I fed a LED from the 223 VAC RMS mains line in series with 2 10k resistors. I plugged it in and checked the voltage across the LED terminals and it was 76 VAC, the LED was cool, light was moderate, resistors were warm. Left on for over a minute. How can this voltage be possible and not blow the LED?

    2) I fed the same circuit on 122 VAC RMS and voltage drop across the LED was now 60V. If the proportion 122/223 was maintained, the voltage should be 41.5 V across the LED. How is a nonlinear circuit possible with a LED? Again, 60V and the LED was cool, resistor warm to the touch.

    3) So I exaggerated he resistor adding a 2M2 resistor, connected to a 223 VAC line. The LED didn't light up, but surprisingly the voltage drop across the LED was still 60VAC

    Other resistor values tested:
    110K VLED 66V
    10K VLED 84V

    Any ideas why the LED stays intact? Why is the LED voltage drop nonlinear?

    Thanks in advance for any pointers.
  2. Boncuk

    Boncuk New Member

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    Are you sure those were LEDs?

    Looks to me as if they were neon lamps.
    Last edited: May 20, 2009
  3. jmaf

    jmaf New Member

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    Hi Boncuk, yeah, LEDs for sure. Here are some photos of the frankentest...I added another 10k resistor to keep them warm so I had time to take the pictures. The closeup pics only show 1 10k resistor.

    How can this LED be cool, not even warm, dropping over 70 V AC directly from the mains line? The camera flash illuminated the led so it's not immediately obvious that it's glowing, you can see though its core glowing.

    [​IMG]

    [​IMG]

    [​IMG]

    [​IMG]
  4. Hayato

    Hayato New Member

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    Have you tried another meter?
  5. jimlovell777

    jimlovell777 New Member

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    Forgive my ignorance here but the LED is in essence a diode in this circuit. You have your meter set to measure AC. The LED should result in half wave rectified DC if I'm not mistaken. Therefore the your DMM reading is twice what it should be, I think. I'm only a hobbyist so someone with more experience will need to chime in.
  6. jmaf

    jmaf New Member

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    Hi Jim, thanks for the idea. If we had some other load after the LED, we'd have half rectified wave, yes. In this case the LED is the load and is consuming 1/2 wave. Tried it though, it reads 64.5 V on the DC scale.

    Hayato: did not try another meter, I only have this one, but other measurements seem right...
  7. birdman0_o

    birdman0_o New Member

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    (230V - 3V) / 20kOhm = 10mA This is why the led is only dim.
    My theory for the Voltage across the LED is that it is in fact the multimeter that is taking an average of the DC/AC components, because as previously stated your circuit is a half bridge. The led is only on half the time which makes the 10mA seem like 5mA. When a diode is reverse biased it sees the whole 223 volts or 122V. Perhaps the multimeter is showing some weird average. The 50-60Hz of half the wave must be messing with the calculations.

    Try this, another led in parallel with that led but hooked up in reverse with respect to this one, they should both look dim, and if you put the multimeter across the terminal now in DC mode it will maybe show 3V....just a guess from an amateur

    Mike
  8. jmaf

    jmaf New Member

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    Mike, you were right(as well as Jim above). At DC now I get 0.001 volts and on the AC scale I get 2.01 volts. What I was seeing was some weird average computed by the multimeter.

    Actually, both lit up pretty brightly, the reason it seems dim on the photo was the camera flash which overwhelmed the LED brightness.

    Edit: as we're now consuming the full wave, the resistors are burning hot, as expected.

    Many thanks to everyone who replied.
    Last edited: May 20, 2009
  9. birdman0_o

    birdman0_o New Member

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    The resistors are dropping about 227V at 0.01A / 2 thats a heck load of heat!!!! ~ 1.15W per resistor!!!!! The ones you are using are rated at 1/4 of a watt, be carefull.

    If you want the led's at theyr maximum brightness perhaps try 4 2k2 resistors in series :)

    Glad I could be of aid

    Mike
  10. jmaf

    jmaf New Member

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    Yes, will be careful if this ever becomes part of a real circuit. Thanks again.
  11. Hayato

    Hayato New Member

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    Jmaf,
    The thing is, you are not causing a 60 ~ 70 V drop.
    As the others have said, you have only rectfied the sinewave, and your multimeter took the half-wave RMS value.

    Your voltage drop is ~2V as any typical LED:

    For 120V
    v(t) = 169.71*cos(377*t)
    The rms value is given by the sqrt of the integral of v²(t) over one period.
    But as you have half-wave, the integration limits is going to be half period:
    sqrt((1/377)*int((169.71*cos(377*t))^2)) =~ 84V
  12. MikeMl

    MikeMl Well-Known Member

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    So far, no one has the right answer. Look up the reverse break down voltage of an LED. On the half cycle when the AC line polarity is pushing current through the LED in the forward direction, the forward drop in the LED will be the normal ~2V, with the rest of the voltage being dropped by the series resistance. During the half cycle when the AC line polarity is reverse biasing the LED, the LED breaks down, and begins conducting when its reverse voltage limit is exceed. THAT is what you are reading with your AC voltmeter...

    The correct way to light a LED using high voltage AC is to put a 1N4007 in series with the resistor, pointed the same way as the LED. This does two things: it protects the LED against the reverse voltage (because the 1N4007 blocks in that direction), and it reduces the power dissipation in the current limiting resistor (because current flows only in half cycles).
  13. RODALCO

    RODALCO Active Member

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    And to add to Mike's comment of the 1N4007 use 1 Watt resistors instead of ¼ Watts.
  14. jmaf

    jmaf New Member

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    Thanks Hayato, Mike and Rodalco for the additional information. Appreciate it.
  15. colin55

    colin55 New Member

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    You can't add a single diode.
    That's why it's best to use a 220n capacitor to limit the current.
    See circuit - next posting.
    Last edited: May 22, 2009
  16. MikeMl

    MikeMl Well-Known Member

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    If you use a capacitor to limit the current through the LED, you will have to put a silicon diode (1N4001 or 1N4148) in inverse-parallel with the LED. This is because in the capacitor, current flow must be AC, otherwise it will charge up and there will be no current. Note the two examples of driving the LED from 50Hz and 60Hz.

    Of course, if your goal is energy saving, then you might as well use a second LED in place of the diode. For that matter, you could put up to 30 LEDs in series, and then 30 more in the opposite direction :D That would take a slightly larger capacitor...

    Attached Files:

  17. colin55

    colin55 New Member

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    Here's a suitable circuit:
    [​IMG]
  18. Mr RB

    Mr RB New Member

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    Hmmm, just put the leds as 2 strings of 5 leds in inverse parallel. Then you can get rid of the bridge rectifier and just need 1 cap and 2 resistors.
  19. mneary

    mneary New Member

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    Circuit suggested in post #18 has fewer parts, but if you're sensitive to 120 Hz strobing, the circuit in post #17 has a much smoother light output.