Orange And Yellow LED's

Discussion in 'The Projects Forum' started by jj_alukkas, Mar 21, 2009.

  1. jj_alukkas

    Thread Starter Well-Known Member

    Jan 8, 2009
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    I need to run abt 10 Yellow or Orange normal LED's from a 8.06v mobile charger.. What is the best possible configuration if I arrange it in strings??
    Will 4 each in series be ok? Can I assume these LED to drop 2v?? Resistor vlaue Required?
     
  2. GioD

    Active Member

    Mar 20, 2009
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  3. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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  4. jj_alukkas

    Thread Starter Well-Known Member

    Jan 8, 2009
    751
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    That That array wizard was cool..
     
  5. SgtWookie

    Expert

    Jul 17, 2007
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    You'll need to look at the specifications for the particular LEDs you're considering first.
    Typical Vf @ current is the specification to go by.

    You're also going to need to find out what voltage your charger puts out under load. One way to do that is to make a constant current sink using an LM317 regulator with a suitable resistor connected between the OUT and ADJ terminals; then connect the IN terminal to V+ and the ADJ terminal to V-.
    Iout = 1.25 / R1. Conversely, R1 = 1.25 / Iout.

    Start off assuming you'll need 3 strings. If your LEDs are rated for 20mA, then your total circuit current will be 60mA.
    R1 = 1.25 / 60mA = 20.83 Ohms
    You can use this series/parallel resistance calculator to figure out what resistors to use to get close:
    http://www.qsl.net/in3otd/parallr.html

    Then once you've determined the charger output under load, subtract 10%, then divide the result by the Vf of the LEDs to determine how many you can have in a string.
    Let's say the charger puts out 7.8v under load, and your LED's Vf is 2.2v@20mA:
    NumberInString = 7.8 x .9 / 2.2 = 7.02 / 2.2 = 3.191
    The integer of 3.191 is 3.

    Then you can calculate the current limiting resistors:
    Rlimit >= (Vsupply - (VfLED x NumberInString)) / Desired Current
    Rlimit >= (7.8v - (2.2v x 3)) / 20mA = (7.8 - 6.6) / 0.02 = 1.2 / .02 = 60 Ohms.
    A table of standard resistor values is here: http://www.logwell.com/tech/components/resistor_values.html
    62 Ohms is the closest E24 value to 60 Ohms.

    To calculate the actual current you'd get:
    1.2/62 = 19.35mA
    Then, to calculate the wattage:
    Power(Watts) = 1.2 x 19.35mA = 23.22mW. We double that for the resistor's rating; 46.44mW or higher.
     
    Last edited: Mar 21, 2009
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