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Discussion in 'LEDs and Optoelectronics' started by Martaine2005, Jan 27, 2021.

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  1. Martaine2005

    Martaine2005

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    May 12, 2015
    Hi guys, I am a little discombobulated.
    I don’t have the data sheet for these.
    I have 5mm RGB common anode and common cathode LEDs. I am only using RED and GREEN, BLUE leg is snipped off.
    I want GREEN on constantly dim and RED to over power the GREEN by being brighter when switched on.
    I’m supplying 14V and current limiting RED with 1K (11mA). This is bright enough and I’m assuming Forward voltage is 2 to 3 volts.
    Now, the for the GREEN. I have gone through my resistor values to get a nice dim glow. Calculations seem pointless now, 1mA was still too bright. I am now satisfied with 3MΩ. But wow, 0.00003μA?.
    Could this be a mistake?
    It’s the same with common anode and common cathode.

    Martin
     
  2. Audioguru

    Audioguru

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    Sep 24, 2016
    You made an arithmatic error.
    If the green LED is a very bright modern 3.4V one then it has the same chemistry as a white and blue LED. If the green LED is a dimmer old one then it is about 2.2V.
    Why 14V? In a car?
    (14V - 3.4V)/3M= 3.4uA, not 3.4pA. Maybe your 3M resistor is actually only 30k ohms?
     
  3. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    I second Audioguru's assumption about the resistor value. 3.4 µA is way too little for an LED to glow visibly. Very good LEDs require around 1 mA minimum. Why not measure the actual current and/or resistance instead of assuming?
    Why assume? Measure it.
     
  4. Martaine2005

    Martaine2005

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    May 12, 2015
    Oh dear me.
    That should have been 0.000003, I omitted a 0.

    I have done some measuring this morning and double checked resistors. Forward voltages are 1.8V red, 2.1V green. (Blue 2.9V).
    The result for a 2MΩ are:
    (14-2.1)/2M=0.000005μA.
    The GREEN LED is visible in daylight and a perfect glow in the dark.

    What am I doing wrong?. I never would have believed a LED would be visible at such a low current.

    And yes, it’s for a door open and closed indicator in my sons car.

    Martin
     
  5. Harald Kapp

    Harald Kapp Moderator Moderator

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    Measured current is ?
     
  6. Martaine2005

    Martaine2005

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    May 12, 2015
    Measured current is 05.3μA
    and calculation is 0.00000595.

    Martin
     
  7. kpatz

    kpatz

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    Feb 24, 2014
    Assuming a 2.1Vf on the green LED and a 14V supply, that leaves you with 11.9V across the resistor. 11.9/2M = 0.00000595 Amps, or 5.95 μA (not 0.0000595μA). Remember in Ohm's Law, the units are Amps, Volts, and Ohms, and you have to convert units accordingly.

    Modern LEDs are efficient, they can glow (dimly) at these currents.

    Those insanely bright blue LEDs you see everywhere are running on just a few mA of current. Gone are the days you need to give a small indicator LED 20mA.
     
    Last edited: Jan 29, 2021
  8. Harald Kapp

    Harald Kapp Moderator Moderator

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    I agree to a few mA, but a few µA? Well, Martin measured so it seems to work.
    How dim is dim? Still visible in daylight? Or from an angle other than 180 °?
    Personally I'd use a higher current, especially in a car. What with changing lighting, temperature etc.? But of course if your satisfied you're entirely entitled to use as small a current as you like.
     
  9. Martaine2005

    Martaine2005

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    May 12, 2015
    Yes, visible in daylight on my breadboard and more so inside the car on a black dashboard. At night, it’s a perfect glow to be seen but not dazzle.
    They are diffused so visibly good all round.
    An example of dim/glow is a TV standby LED or smoke detector LED. Although these obviously vary from manufacture etc, they are quite visible indicators.
    And yes, 0.00000595A not μA. My mistake.

    Martin
     
  10. Tha fios agaibh

    Tha fios agaibh

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    Aug 11, 2014
    Interesting.
    Doing the math the other way would suggest a 3.4Vf although at current that low, i would expect Vf below 2v given the low current.
    I believe the 2.1Vf is based at its current rating of 10ma or so, not a miniscule 5.3uA.

    Try measuring the voltage drop across the led.
     
  11. Martaine2005

    Martaine2005

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    May 12, 2015
    Hi John,
    I’ll do more measurements tonight.
    It’s my math I doubt. I get light headed when moving the decimal point!.

    I will try less resistance too, but 820K was too bright in the car at night. Very strange.

    Martin
     
  12. Harald Kapp

    Harald Kapp Moderator Moderator

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    I'd accept 0.5 mA as a reasonable value.
     
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    The LED could be visible in daylight due to UV exciting the phosphor (assuming it's a LED with phosphor)
     
  14. Martaine2005

    Martaine2005

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    It’s clearly off without a power source and once connected, it’s clearly on.
    Later I will try my bench supplies as I’ve been using my RS powered breadboard.
    Although I was keeping a close eye on voltage, a second test won’t hurt.

    Martin
     
  15. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Well, your eye is most sensitive to green, and depending on the wavelength of the red LED, you're eye is quite insensitive. Also, your perception of brightness is not linear, so this may have some bearing on it.

    I've connected a LED and resistor across a supercap and it was visible for several days, although only in darkness after a day or so. The current through that LED was extremely low, although I can't remember if I measured it (or tried to) when it was very dim.
     
  16. Martaine2005

    Martaine2005

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    May 12, 2015
    Well, I have measured several scenarios and came to a happy compromise.
    I’ve added an extra power source to the GREEN LEDs via diodes and via the car lights auto dimming circuit. We now have brighter LEDs in daylight which dim when the car lights are turned on. And all are dimmer with the engine off (12.7V).
    Daylight (14 - 2.1)/820=14.51μA. Actual measurement 12.68μA.
    Dark 2MΩ and 5.95μA. Actual measurement 5.71μA.
    I am still very surprised at the brightness at such low current.

    Martin
     
  17. Tha fios agaibh

    Tha fios agaibh

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    Aug 11, 2014
    I think you mean 14.5mA.

    6uA surprises me too. It seems hardly enough current to get it to turn on.

    Also, If you look at a Vf vs current chart, the voltage drop is far below 2.1v @ 6uA. How is it that the math works out?
     
  18. Audioguru

    Audioguru

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    Sep 24, 2016
    The math is wrong again. (14V - 2.1V)/820 ohms= 14.5mA, or 14.5 thousand micro-amps.
    Maybe he used 820k ohms for 14.5uA?
     
  19. Martaine2005

    Martaine2005

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    May 12, 2015
    I feel like a silly school boy!.
    It is 820K.

    Martin
     
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