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Can 2.5V to light up a typical 3mm LED?

Discussion in 'Electronic Basics' started by Mr. Man-wai Chang, Jun 24, 2013.

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  1. I supposed it should work fine...

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  2. Was considering to connect two 3mm LEDs to the 5V of a molex connector
    inside a PC chassis. Nothing fancy!

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  3. Daniel Pitts

    Daniel Pitts Guest

    Always use a current limiting resistor. If you don't, and you're lucky,
    you'll just burn out your LED. If you're unlucky, it'll burn out your
    power supply and/or other circuitry in the path.

    Diodes work differently than other types of loads. They always drop a
    pretty consistent voltage, regardless of current. This means that if
    nothing else resists the current, then it basically looks like a short.

    That's a simplified model of diodes, but for the most part a very useful
    one. As others have mentioned, the voltage that LEDs drop depends on the
    chemistry. As does the color.

    Usually, when you buy LEDs, the spec sheet tells you the important
    characteristics:

    1. max forward current, If(max).
    2. Typical forward current, If(typ)
    3. Typical forward voltage drop. Vf

    To figure out how much resistance to use, you want to make sure that the
    total current is less than the max forward current, and preferably
    closer to the typical forward current.

    Let's use a concrete example. From the first google result for "Blue LED
    spec sheet" <http://www.lc-led.com/products/500tsb4d.html>

    Peak current is necessary to consider if you have capacitor in the
    circuit, or some other kind of timing dependent circuitry. Let's ignore
    that for now.

    Max continuous forward current is 30ma. So we want to keep it below
    that. Let's aim for 20ma.

    Typical forward voltage drop is 3.6 (already 2.5v is too low).

    Take your total supply voltage (5v in your case), and subtract the rated
    typical forward voltage drop. 5v - 3.6v = 1.4v

    Now we know that the LED drops 3.6v, and the "rest" of the series
    circuit needs to drop 1.4v. Using ohm's law V=I*R, we know that we want
    R such that 1.4v=20ma*R. 1.4v/20ma = 70ohms. That is the minimum
    resistor you need. The next standard size up from 70Ω is 75Ω. You could
    also go up to 100Ω and end up with 14ma through the LED, which may be
    bright enough.

    In professional applications, you'll also need to consider the power
    dissipation, but that's another few pages of discussion ;-)

    Hopefully this helps,
    Daniel.

    P.S. I'm not an expert in the field, so I may have forgotten something
    or gotten something wrong. Anyone else should feel free to correct me.
     
  4. Hey, that raises an interesting question. I always figured LED's are like other diodes (I-V curve is an exponential.)
    So I've got an amber (590nm) LED... here's the voltage drop vs bias current..

    Current Forward Voltage
    10nA 1.273V
    100nA 1.387V
    1uA 1.498V
    10uA 1.605V
    100uA 1.706V
    1mA 1.806V
    10mA 1.923V

    With a dark room I can just see the light at 1uA. 10uA is no problem.

    I've also heard tell you can get a feel for Plancks constant by plotting
    forward voltage vs led color (for a bunch of different leds).
    e*V = h*freq = h*c/wavlength
    But this is not right.... first off, for the above 590 nm LED I get 2.09V
    And secondly what current do I pick?

    Hey it's gotta be that the LED is just using those electrons at the high energy end of the Boltzmann distribution... Can you see any LED cooling of the junction? At 1mA and ~0.2 V (2.0 - 1.8) that's ~0.2mW

    George H.
     
  5. Jamie

    Jamie Guest

    depends on make, model and color!

    Jamie
     
  6. It is working! A 3mm green LED in series with a 5mm red LED connected to 5V.
    Thanks!

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  7. Thanks

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  8. Do I need that resister if I am to connect *TWO* LEDs in series over the 5V?

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  9. Thanks

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  10. What if one LED in the parallel circuit died? Would the remaining LED
    receiving more than its share of current, even with a current-limiting
    resistor?

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  11. Daniel Pitts

    Daniel Pitts Guest

    No, that's not how parallel circuitry works. In a parallel circuit, the
    voltage is equal across the branches. In your case, that's the 5v will
    always be across each of the LED+resistor circuits. If you had equal
    current going through both branches, and then one of the branches
    "died", resulting in an open circuit, your total current would just be
    cut in half.

    Now, if you had a series circuit, and one LED died, then both LEDs would
    appear off, since no current could flow at all.

    OTOH, in a series circuit, if either LED failed by shorting, you'd end
    up with too much current through the other remaining LED, and likely fry
    them both. In a parallel circuit, the current wouldn't likely go
    through the other LED.
     
  12. Daniel Pitts

    Daniel Pitts Guest

    Yes. Unless the voltage drop equals exactly 5v, *and* the voltage source
    will never go above 5v, you will have some extra voltage left over. If
    you have no resistance (or the very low resistance of the wire), you
    will have a run-away current.

    It may even appear to work for a while, but you could be damaging your
    power-supply.
     
  13. I remember now: 1/R(parallel) = 1/R1 + 1/R2

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  14. Would a LED suddenly short itself? Not very likely, I supposed...

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  15. Really? It didn't short itself, did it?

    Another question:

    LED is a diode.

    Can large current make a diode to behave like conductor and hence short
    the +5V and GND?

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  16. But would the short piece of string melt like a FUSE **BEFORE** the
    large current could get through?

    If it did melt *FAST* enough, then there was nothing to worry about... . :)

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  17. BTW, are we talking about circuitry or material science? ;)

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  18. Of course not!!!!

    **IF** an LED could become a conductor by slight fluctuation in the +5V
    supply, a limiting resistor would not help, would it? It would still
    short itself even whether you connected it serial or in parallel, with
    or without a resistor!

    So the question would demand an answer relating 3 objects:

    1. diode/LED (properties)
    2. conductor
    3. fuse

    How would you explain the relationship between these words?

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  19. Anyway, I agree that a limiting resistor is a safer bet.

    That still does NOT explain whether the LED would short itself.... :)

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  20. I acknowledge the safety measure, but is the threat real?

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