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DC LED and resistor project help

Discussion in 'Lights - Store Bought and Home Made' started by nato595, Aug 10, 2013.

  1. nato595

    nato595 Let's do this!

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    Oh wow, I'm confused!
    I'm still in the beginning stages of learning about resistors, LEDs, milliamp-hours, power draw, etc. so I apologize if I ask a dumb question or say something completely wrong! :D
    I was hoping to get a little advice before I start building something wrong!
    I am looking to power 4 5mm or 10mm blue LEDs, using a battery pack.
    Specs for LEDs:
    DC forward current: 30 mA
    Forward voltage: 2.7 V
    Again I have four LEDs, and want a battery pack that is not huge (please no C or D batteries). I am hoping for a few AA, A23, or 9V. I am having a hard time figuring out the following:

    1. Which battery will last the longest with this setup? I am needing at least 5 hours of constant use.
    2. Series or Parallel, and based on this how/where to add resistors?
    3. Based on the above, what will be my resistor(s) of choice?
    4. Based on the above, what will my battery life span be while fully lighting the LEDs?
    I have used the resistor calculators on several websites but they are lacking the information of mAh (specifically how long will my setup last on one battery pack), and/or how much of the battery am I wasting in my resistors.

    Thank you in advance for any help and advice!!
     
  2. SmartAlecLights

    SmartAlecLights Im a SmartAlec what can i say! Community Project Designer

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  3. kenoman742

    kenoman742 New Elf

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    That is a really cool calculator. Thank you for posting the link.
    Ken
     
  4. David_AVD

    David_AVD Bite my shiny metal ass!

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    I'd say that AA cells will give you a fairly good energy density. In series, 4 of them they would be a nominal 6 Volts.

    Your LEDs are 2.7V, so two in series is 5.4V which is less than (but close to) your 6V supply. All good so far.

    So, your supply (6V) minus the 2 LEDs in series (5.4V) leaves you with 0.6 to drop across a series resistor.

    If you want to drive the LEDs at 30mA, the equation is 0.6V / 0.03A = resistance in Ohms. That comes to 20 Ohms and the nearest (but higher) E12 (common) value is 22 Ohms.

    Your 6V supply may be slightly higher when the batteries are fresh so some tweaking of the resistor value may be required.

    Now you have the method for hooking up two LEDs and one resistor in series, you can put several of those "series strings" in parallel across the 6V supply.

    ACL wiki reference here.
     
  5. damo1271

    damo1271 Full Time Elf Generous Elf

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    The data sheet for a typical energiser battery can be found here:
    http://data.energizer.com/PDFs/E91.pdf
    The capacity of the battery discharging to 0.8V at 100mA is 2500mAh. (in theory 25 hours use)

    But. ..
    With a forward voltage drop across the LEDS of 5.4V (2@2.7V) you can only afford the batt voltage to drop to 1.35V in each battery. After that the LEDs may not light. So the capacity will be considerably less than the 2500mAh!

    To get a 5 hour duration means the maximum current draw that would allow this would be 200mA at 1.2V and again the LEDs probably wont light at 4.8V.

    Therefore if you want 5 hours duration then the maximum current draw based on the charts (and some guessing of the line position for about 1.3V) will be around 100mA - give or take. That would be 3 series strings in parallel if each string draws 30mA. (6 LEDs and 3 resistors)
     

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