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Supply for 30 LEDs

Discussion in 'LEDs and Optoelectronics' started by Arul Raju, Jul 25, 2016.

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  1. Arul Raju

    Arul Raju

    28
    2
    Mar 5, 2014
    I'm trying to use damaged emergency power supply (Battery and few resistors are gone). I do not have data for 30 LEDs (unknown). I do have 6V battery (4.5Ah Sealed Lead Acid Battery). Could some body advice me how to operate these LEDs with this battery. How to calculate resistor values for unknown LED parameters? I tried using 1K resistor, but LEDs are glowing and after few seconds one of the LED got damage due to over current. And PCB is getting to much heat after supplying 6V.

    Thanks in advance.
    Regards,
    Arul
     
  2. Sadlercomfort

    Sadlercomfort Ash

    424
    55
    Feb 9, 2013
    Hi Arul,

    Do you still have the damaged resistors and battery? If so what are the resistor values and battery specs?

    Include some pictures of the whole unit, any damage and the LEDS used if possible.

    What colour are the LED's?

    Thanks,
    Ash
     
  3. Arul Raju

    Arul Raju

    28
    2
    Mar 5, 2014
    Hi Ash,
    Thanks for the response
    I attached photographs for the complete unit. The resistor is burnt so color code is not visible to check resistor value.
    30 white color LEDs are used.
    If you look at the photography, on the right side is the existing batteries and the left side is the battery that I'm going to replace. Please help me on this.

    Thanks,
    Arul
     

    Attached Files:

    Sadlercomfort likes this.
  4. BobK

    BobK

    7,682
    1,687
    Jan 5, 2010
    A 6V supply and a 1K resistor should not damage any LED. Are you sure you connected it right.

    Bob
     
  5. Arul Raju

    Arul Raju

    28
    2
    Mar 5, 2014
    Yes, i made the connection right..i connected battery positive terminal to one end of resistor and resistors' another end is connected to positive (anode) end of LED..battery negative end is connected to cathode end of the LEDs. How could you tell 1K is sufficient for 30 LEDs? How long the 5Ah battery can supply 30 LEDs? Does it affects LEDs life snap if i use continuously?

    Thanks,
    Arul
     
  6. BobK

    BobK

    7,682
    1,687
    Jan 5, 2010
    A 1K resistor will allow no more than 6 mA of current through it from a 6V source. And white LEDs drop around 3.3V themselves, so that means 2.8 mA max through a white LED. Those LEDs are designed to handle at least 20 mA and probably more like 50mA to supply enough light for emergency lighting. So I am convinced you did something wrong. Are you sure it is a 1K resistor?

    Also, how are the LEDs wired? If they are in parallel they must already have resistors. If they are more than 2 in series, then they would need a boost converter to get enough voltage to run them at all from a 6V battery.

    Bob
     
    Sadlercomfort likes this.
  7. Arul Raju

    Arul Raju

    28
    2
    Mar 5, 2014
    Hi Bob,
    There was a PCB (see photography) on the emergency kit itself but i was not using them. Also, there were two batteries (unknown specifications), i haven't used them as well. I used separate 6V battery to power LEDs. Can i use 1K resistor for longer time? How long this battery can supply all the LEDs?

    Arul
     

    Attached Files:

  8. Arul Raju

    Arul Raju

    28
    2
    Mar 5, 2014
    I did the connection once again.Now i see all the LEDs lights very dim.
     
  9. Sunnysky

    Sunnysky

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    123
    Jul 15, 2016
    how are the LEDs connected inside?
     
  10. Sunnysky

    Sunnysky

    483
    123
    Jul 15, 2016
    if battery is 6.5V X 5 Ah. it can supply 32.5 Wh
    in theory each 5mm LED can consume 75mW max. thus 30x. = 2.25W thus 15 h.
    But without a DC-DC converter, it won't work if all the LEDs are in two parallel banks of 15 as each LED draws3-3.2V @20mA typ. so two In series is 6-6.4 V so there is insufficient voltage to run more than a few hrs bright with a 2W 2 Ohm resistor

    If you connected the battery direct to the LEDs they are likely in an array of 15P2S and would have gone into overcurrent and thermal runaway burning out the brightest LED.

    So look for a 2W resistor around a few Ohm's or 2 minimum.

    A better solution is a 7.2 LiPo pack with 1V drop at 300mA max or 3.3 Ohm's.
     
  11. BobK

    BobK

    7,682
    1,687
    Jan 5, 2010
    If you cannot tell us how the LEDs are connected on the board, we cannot help you (or at best, we can only guess, as Sunnysky has done.)

    If you cannot determine this yourself, a least post a picture of the back of the LED board that we can see clearly.

    Bob
     
  12. Arul Raju

    Arul Raju

    28
    2
    Mar 5, 2014
    Hi Bob and Sunnysky,
    Here are the pictures of PCB (LED side and solder side).

    Arul
     

    Attached Files:

  13. BobK

    BobK

    7,682
    1,687
    Jan 5, 2010
    It looks like they are all simply in parallel, which is not good, but it is what you have to work with. Unless you want to try to add a resistor to each one.

    Assuming 3.2V for the forward voltage, and 20mA current, I calculate a 140Ω resistor. Try a 150Ω. The power rating would be 0.6 * 2.8 = 1.68W, so a 2 Watt resistor. And expect the resistor to get hot.

    Bob
     
  14. BobK

    BobK

    7,682
    1,687
    Jan 5, 2010
    Whoops, too early in the morning for me to do math. Previous post is totally wrong, I forgot to multiply the current by 30 when calculating the resistor. That would be the right resistor for 1 LED not 30.

    New calculation:

    (6 - 3.2) = 30 * 0.02 * R
    2.8 = 0.6 R
    4.66 = R

    Use a 4.7Ω.

    Power calculation was correct, so 2 Watt.

    Bob
     
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  15. Sunnysky

    Sunnysky

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    Jul 15, 2016
    Agreed but, Since they are all parallel 3.1V*20mA =620mA is ideal if constant , but battery is not constant and also 2X Vf so efficiency loss is 50%.
    Better to use 16850 3.6V LiPo and have spares then use Ohm's Law on R.
    e.g (3.6-3.1)V/0.62A= 1Ω. losing 620mW of power so 2W or 5W for cooler operation.

    ESR of 30*5mm LEDs 15Ω each is 0.5Ω at 20mA FYI. This rises ~20% at half current so Vf drops a bit towards 2.85= Vth, the threshold of very "dim". You can verify if you like.

    Thus at 3.3V , current will be ~0.3V/1.1Ω =0.27A or half brightness then dim much faster as battery drops to 3V then 2.85V if left on, which ages battery faster, so replace battery when brightness drops >50% in an hour or few.
     
    Arul Raju likes this.
  16. Arul Raju

    Arul Raju

    28
    2
    Mar 5, 2014
    Thanks a lot Sunnysky.
    I would like to know what will happen if i apply per Bob's response? I would need 1 hr (Max.) after power fails. If yes, i will use 4.7ohm resistor. If not, i will procure new battery and 15 Ohm resistor per your suggestion.

    Arul
     
  17. Sunnysky

    Sunnysky

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    Jul 15, 2016
    .15 OHm? for which battery? is that 29 LED's now?
     
  18. Arul Raju

    Arul Raju

    28
    2
    Mar 5, 2014
    for 3.6V battery and 15 ohm resistor..sorry, didn't mention about latest no. of LEDs. LEDs now are 27 (three LEDs are not working).
     
  19. Sunnysky

    Sunnysky

    483
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    Jul 15, 2016
    Ok know measure Vled and Vbat.
    If I Assume Vled=3.1, Vbat=3.6 then

    ( 3.6-3.1)/15R= 333mA /27 then average current per LED is 12.3 mA is safe and will last longer but not full brightness.
    I think you can calculate this for desired Ah and current so it does not exceed 20mA per LED. Since power at 333 mA x3.1V= 1W on large array, it will get warm but not too hot unless sealed in a small box.

    15 Ω R will be dissipating 0.5V X 1/3 A = 167 mW using these assumptions.

    So 1W/(1+ 0.167W) *100= 85% efficient
     
  20. Arul Raju

    Arul Raju

    28
    2
    Mar 5, 2014
    I have not yet purchased 3.6V as Sunnysky recommend..meanwhile, i made connection per you recommendation (6V battery and 4.7 ohm resistor is used), the resistor is getting hot and hot (smoke came after few seconds)..but LEDs were glow brightly.i measured voltage, it was 3.32V.
    What could be the reason for resistor heat?

    Arul
     
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