Circuit Load Calculation

Discussion in 'Math' started by FFtravism, Sep 30, 2009.

  1. FFtravism

    Thread Starter Active Member

    Aug 21, 2008
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    I am trying to calculate the load of a circuit. I have running LED's and resistors. I am running 48 LEDs in groups of 3 with 252ohm resistors (2x 510ohm parallel). The math was figured at 25mw per LED. I am trying to figure out the size of wire to use. Do I just take the 25mw * 48 = load?
    Thanks
    Travis
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    Calculate the current through each group and then multiply by the number of groups to get the total current. Use the total current to find the appropriate wire.
     
  3. FFtravism

    Thread Starter Active Member

    Aug 21, 2008
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    So do I have to account for any resistor calculations?
     
  4. mik3

    Senior Member

    Feb 4, 2008
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    Calculate the current through each parallel combination of resistors (252 Ohm) and multiply by the number of branches to get the total current of the circuit. Then according to the total current value you can find the proper wire.
     
  5. FFtravism

    Thread Starter Active Member

    Aug 21, 2008
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    So (.25mA + .25mA + .25mA) * 8 = 6mA or .006A correct?


    [​IMG]
     
  6. FFtravism

    Thread Starter Active Member

    Aug 21, 2008
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    hrmm that does not seem right but atleast the circuit is posted.

    LED's are 2.2v 25mA resistors are 510ohm each.
    Voltage is figured at 12.9v

    Travis
     
  7. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    Each Led branch will draw about 25mA. You said you have 48 Leds, thus there are 16 Led branches.

    Total current=current through each branch*number of branches

    Thus

    Itotal=25mA*16=400mA

    A 33 AWG wire is enough but it is very thin and it is not practical to use. You can use a lower AWG wire which is thicker and more practical to use like 25 AWG.
     
  8. FFtravism

    Thread Starter Active Member

    Aug 21, 2008
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    Ok that makes more sense.
    Thanks.
     
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