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hmmm... quandry....

Discussion in 'Electronic Basics' started by feebo, Mar 11, 2007.

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  1. feebo

    feebo Guest

    OK, I have been mucking about with electronics and stuf for years and
    there is not much that stumbles me... but...

    LEDs in series, OK? say I want to run 6 LEDs in series from a 12v
    power source. if each LED has a 2V drop, that amounts to the 12v
    supply... how do I limit the current to whatever I like?

    bog standard question, driving me mad for a week or more.
     
  2. Well, if you have no voltage margin, then a series resistor is not going to
    work.
     
  3. as far as i can tell, the only way to boost current would be increase
    voltage, or reduce the number of leds.
    R constatnt so for I to change V must change
     
  4. The voltage drop of the LEDs will not be exactly 2 volts, as it depends on
    current, temperature, age, etc. If you have a power supply that is at least
    about 1 volt more than the maximum of all series LEDs under all conditions,
    then a current regulator can be made. An LM317 works very well, if you can
    spare 1.2 volts for its reference and maybe 1 volt more for it to drop. You
    just put a resistor from its output to its reference at GND for R =
    1.2/Iled. So for 20 mA LEDs, R= 60 ohms.

    V=2VDC
    14V---LED1---LED2---LED3---LED4---LED5---LED6-*-| LM317 |-----+ =1.2 VDC
    | |
    | |
    | |
    GND-------------------------------------------------+-60 ohms-+

    You can also make a current regulator using an NPN transistor, a 30 ohm
    sense resistor, a bias resistor (about 1k), and two diodes. I leave it as
    an exercise for the reader to complete the circuit. It can also be done
    with a PNP transistor if you want the LED string to go to ground.

    Paul
     
  5. me

    me Guest

    feebo wrote in
    a small resistor would probably work. Find the IV curve for LEDs, get
    voltage drop at desired current, add up voltage drops, subtract from
    supply, divide answer by current desired to get resistor value...
     
  6. Chris

    Chris Guest

    LEDs, like all diodes, don't exactly have one standard voltage drop.
    The forward voltage drop is dependent on the forward current, and your
    2V nominal forward voltage drop at a nominal 20mA will be more like
    1.7V at 1mA and 2.4V at 60mA. This is similar to a 1N4001, which will
    have a forward drop of a little less than 0.6V at 1mA, 0.7V or so at
    100mA, and more than 0.8V at 1 amp of forward current.

    Let's assume you put your 6 LEDs in series across a power supply of
    exactly 12V. Rather than being limited by a resistor, the forward
    current will actually be limited by the increase in forward voltage of
    the six LEDs in series. This might look like you can save a componet
    here, except that the forward voltage of LEDs is somewhat chancy. It
    changes with changes in temp and also changes over time. There is
    also quite a bit of variation between forward voltage drops of LEDs,
    even within the same manufacturing lot.

    This means you're risking either to little light from the LEDs, or
    excessive current (which can lead to early failure). Either way, not
    the best solution -- even if you hand pick teh LEDs for an individual
    circuit.

    As Mr. Popelish says, it's obviously best to use two strings of 3
    LEDs, and then use a series resistor which will limit the current
    through each string to the desired amount. Let's say you want 20mA
    through each string, and you know the nominal forward voltage of each
    LED is around 2V. Take your 12V supply, and subtract the 6V across
    the three LEDs, Then use Ohms Law to select a resistor which will
    take up the remaining 6V when 20mA is passing through it. That would
    be:

    6V / .02A = 300 ohms, a standard 5% value.

    If you absolutely have to have the six in series, you might want to
    use a 555 voltage doubler to increase your nominal source voltage
    across the LEDs to 23V or so, and then select a series resistor which
    will do the job. Here's a good link to help with the voltage doubler:

    http://www.reconnsworld.com/power_voltdoubler.html

    Good question!

    Cheers
    Chris
     
  7. If each led is EXACTLY 2V drop, and your power supply is EXACTLY 12V
    then you can't, and you risk going insane or something bad like that :-
    That's a theoretical question and LEDs are practical devices. LEDs
    have V/I curves that can vary from device to device, so the answer
    ain't easy when you try to "push the limits" like this.

    Good engineering design in this case would mean you would drive only
    say 5 LEDs to give you some margin to play with and be able to
    calculate a suitable resistor at the current you want.

    If you *really* want to run six 2V drop LEDs from 12V you would use
    some form of DC-DC converter that would allow you to do this.

    Dave :)
     
  8. jasen

    jasen Guest

    where did you find the low drop-out LM317? :^)

    you'll not get V at the * down to 2V, 3V maybe.


    Bye.
    Jasen
     
  9. Eeyore

    Eeyore Guest

    Assuming 6 identical lEDS, theur Vf will only be exactly 2V at one specific
    current. If does in fact vary with Vf.

    Graham
     
  10. And a good one. you can't regulate current if there is no
    extra voltage to discard in the regulating mechanism.

    If you change the arrangement to two strings of 3 LEDs, it
    is pretty simple to put a resistor in series with each
    string that wastes the extra 6 volts while passing the
    desired current to the LEDs.

    If they have to all be in series, you need a voltage
    boosting circuit of some sort, like a flyback current
    regulator drive them efficiently. But that extra bit of
    power would have to be pretty expensive to justify that
    additional cost and complexity.
     
  11. Rich Grise

    Rich Grise Guest

    You can't do it and guarantee that the LEDs will light and won't fail.
    You could replace one of the LEDs with a constant current source or sink,
    which would drop the "excess" voltage, and make the whole thing less
    voltage-dependent. An LED is a current-operated device - the current is
    related to the exponent (e^x) of the voltage drop - the standard diode
    graph; if you try to drive an LED with voltage, the current can vary
    wildly from very minor variations in the drive voltage.

    If you really need six LEDs, then use two strings of 3 each, each with
    its own current source/sink.

    Hope This Helps!
    Rich
     
  12. feebo

    feebo Guest


    phew!

    Thanks to everyone for their answers - I thought I was going mad but
    seeing it has taxed others is a re-assurance - if you get what I mean
    :eek:)
     
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