126 led lamp

Discussion in 'General Electronics Chat' started by alank878, Feb 21, 2016.

  1. alank878

    Thread Starter New Member

    Feb 21, 2016
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    Hi Guys,

    I have an under bonnet lamp for working on my car, it seems to have a burned resistor (larger black one) am i correct in saying there should be 2 for this type of circuit? can any one tell me which size it should be so i can get a replacement?

    Specs are

    Battery Type 3.7 V 4400mAh Lithium-Ion
    Brightness 700 lumens
    Charge Time, hours 5
    Charger 110 V AC 50/60 Hz
    Charging Voltage/Current 4.2 V/1 A
    Description 128 LED Under Hood Light
    Discharge Time, hours 3

    Any help would be great :)
     
  2. dl324

    Distinguished Member

    Mar 30, 2015
    3,231
    619
    Welcome to AAC!

    Did you intend to post a picture and a schematic?
     
  3. alank878

    Thread Starter New Member

    Feb 21, 2016
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  4. ScottWang

    Moderator

    Aug 23, 2012
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    767
    If the Leds are in parallel directly then it is not so good for the Leds, assuming that the spec of led is 3V/20mA then the total current of 128 Leds is :
    I = 20 mA*128 = 2.56 mA,
    R = (4.2V-3V)/20 mA = 1.2V/20 mA = 60 Ω,
    W = V* I = 1.2V*2.56A = 3.072 W
    So the values of W should be at least 3 times of calculation and that is 9W, but if using more than 5 times is better and that is 15W.

    But I prefer to use 80% of rating current as 16mA, the product didn't shows the current, now we only can guessing.
     
  5. alank878

    Thread Starter New Member

    Feb 21, 2016
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    0
    Hi scott,

    maybe this is why the resistor is burned? so you say this resistor should be changed for a 60 ohm part?
     
  6. ScottWang

    Moderator

    Aug 23, 2012
    4,853
    767
    The 60 Ω just guessing can calculation, do you remember how many hours do the leds can be light up continuing?
     
  7. alank878

    Thread Starter New Member

    Feb 21, 2016
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    hi its around 3-4 hours max
     
  8. ScottWang

    Moderator

    Aug 23, 2012
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    767
    I = 4400 mA/3 hrs = 1467 mA = 1.467A
    I = 4400 mA/4 hrs = 1100 mA = 1.1A

    If the battery can be used for 3 hrs:
    R = V/I = 1.2V/1.467A = 0.82 Ω

    If the battery can be used for 4 hrs:
    R = V/I = 1.2V/1.1A = 1.09 Ω

    Can you recognize the value of resistor?
     
  9. alank878

    Thread Starter New Member

    Feb 21, 2016
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    0
    hi scott, the resistor is un recognisable as its very burned, i will take a photo today and show you
     
  10. ScottWang

    Moderator

    Aug 23, 2012
    4,853
    767
    I checked the spec and it showed that Discharge Time, hours : 3.
    I = 4400 mA/3 hrs = 1467 mA = 1.467 A/hrs
    If the battery can be used for 3 hrs:
    R = V/I = 1.2V/1.467A = 0.82 Ω
    W = V*I = 1.2 V*1.467 A = 1.76W
    As above calculation, I assuming that the Vf of LEDs is 3V.

    I_1led = 1.467 mA/128 Leds = 11.46mA, each led draw current 11.46 mA.
    If the assuming is right then maybe could choose 1Ω/10W when the 0.82 Ω/10W is not easy to buy.

    For the safety reason, if we assuming that the Vf of LEDs is 2V, and then
    If the battery can be used for 3 hrs:
    R = V/I = (4.2V-2V)/1.467A = 1.5 Ω
    W = V*I = 2.2 V*1.467 A = 3.23W, so it may choose 1.5 Ω/15W

    HS15-1R5J -- Resistor: wire-wound with heatsink; screwed; 1.5Ω; 15W; ±5%
     
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