Ba9s LED

Discussion in 'The Projects Forum' started by roscirc, Sep 13, 2012.

  1. roscirc

    Thread Starter New Member

    Apr 30, 2012
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    0
    Hi All,

    The LED in the datasheet attached is a Ba9S type (130V) which has internally an array of 7 LEDS.

    I am trying to add a resistor to reduce the current into the LEDs.

    These are the things that I can not figure out:

    1. the array of 7 LEDs in asingle bulb is connected in series or parallel?
    2. The 5.9mA refers to the current of the single LED of the whole array (if in parallel)?
    3. Can I assume the internal resistor is 22kOhm (130/5.9)?
    Thanks,

    Ros
     
  2. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Where does the voltage come from.
    If it is directly from the mains, you are violating the ToS.

    Bertus
     
  3. roscirc

    Thread Starter New Member

    Apr 30, 2012
    9
    0
    They are connected directly to a 110V battery.
     
  4. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    In the calculation of your resistor, you did not take the voltage drop of the leds in account.
    White leds will have a voltage drop between 3.5 and 4 Volts each.
    So when the voltage drop is 3.5 Volts each, the resistor will be (130 - 7 * 3.5) / 5.9 = 17.88 KOhm (so probably 18 KOhm).

    Bertus
     
  5. roscirc

    Thread Starter New Member

    Apr 30, 2012
    9
    0
    Hi,

    So you are assuming that they are connected in series?
     
  6. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Yes, that is a practical way to connect them, but if one fails, all will go out.

    Bertus
     
  7. MikeML

    AAC Fanatic!

    Oct 2, 2009
    5,450
    1,066
    Whoa! That lamp is designed to work off the AC line voltage. It has circuitry in the base to take care of current limiting.
     
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