LEDs Wiring

Discussion in 'The Projects Forum' started by jeremyhaberman, Aug 7, 2013.

  1. jeremyhaberman

    Thread Starter New Member

    Aug 7, 2013
    3
    0
    Hello everyone,
    I am build an overhead panel for my home cockpit and I need some help. I will have some DPST switches that will have one set of poles going to a usb interface board my my flight simulator to recognize the command (this part I understand). I want to wire a LED to the second set of poles. My confusion is that I want to have multiple LEDS all controlled separately (by their own on/off switch) but powered by one power supply, is it possible and how can I do it. I really need the help.

    :confused:

    Much Thanks,
    Jeremy Haberman
     
    Last edited: Aug 7, 2013
  2. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
    4,769
    969
    yes its simple.
    wire each led with its own current limiting resistor and done.
     
    jeremyhaberman likes this.
  3. tubeguy

    Well-Known Member

    Nov 3, 2012
    1,157
    197
    If you need help on calculating current limiting resistors, please post your LED specs and LED power supply voltage.
     
    Last edited: Aug 7, 2013
  4. jeremyhaberman

    Thread Starter New Member

    Aug 7, 2013
    3
    0
    I want to use 5 red LEDs at 2.2 v and 20mA, 5 Blue LEDs at 3.2v at 20mA, and 3 green LEDs at 3.2v at 20mA. I want to use a 5vdc 300mA power supply. Each LED will have is own toggle switch. What type of resistors do i need.
     
  5. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
    4,769
    969
    Resistor values are calculated as follows.

    R= (Vin-Vf)/A
    Where
    Vin is power supply voltage
    Vf = LED forward voltage drop
    A=led current in amps

    Then you need to calculate the wattage for the resistor where
    P=2 x (I^2 x R)
    Where P watts
    I = led current in amps
    R=resistance

    Example (for red LED)
    Resistance
    5-2.2 = 2.8
    2.8/.020 = 140 ohm resistor or slightly larger

    Wattage
    .02^2 = .0004
    .0004 x 140 = .056
    .056 x 2 = .112Watts (1/8W resistor)
    The times 2 is a "safety factor" to keep the resistor from running too hot.

    In general LED's do fine anywhere from 10ma to 20ma (more current the brighter they are)
    BUT once you start going over 20mA the lifespan is usually decreased.. So always shoot for no more than 20ma
     
    jeremyhaberman likes this.
  6. jeremyhaberman

    Thread Starter New Member

    Aug 7, 2013
    3
    0
    Thanks, I like how instead of just telling me straight up which one to use yoU tell me how to calculate using the formula because I need to learn this stuff.
     
  7. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
    4,769
    969
    No problem..
    Have fun with your cockpit :D
     
Loading...