my situation

Discussion in 'General Electronics Chat' started by highwayhonkie, Apr 24, 2011.

  1. highwayhonkie

    Thread Starter New Member

    Apr 23, 2011
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    i have 9 10mm Triple Chip 200mW Infrared IR LED Night Vision and i have 3 10 ohm 1/4 watt +/- 5% resistance resistors mounted on with one 9 volt standard battery. there is three strings of three LED's one resistor on each string. switch is on negative side and resistors are on positive sides. the 9 volt battery only lasts for about 20 minutes. what can i do to prevent this from happening.


    Each light has this info on them: Voltage is 1.5~1.6, Current is: 140mA at peak, (pulse): 700mA, Power is: 200 mW, Size is: 10mm, Wavelength is: 850nm red glow

    the lights i bought are from lightobject.com
     
  2. designnut

    Member

    Apr 21, 2011
    33
    1
    !0 ohns will draw too much current and the battery is probably 140 MA hours capacity. Use the largest resistor you can as a compromise between light and battery life. Sounds like a rechargeable cell is in your duture.
     
  3. marshallf3

    Well-Known Member

    Jul 26, 2010
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    No way will a 9V battery power those, they put out very low current.
     
  4. Adjuster

    Well-Known Member

    Dec 26, 2010
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    If the 9V battery is the common PP3 type, it is far too small. Think about it: the LEDs are rated at 140mA DC, so with three strings you would need a total of 420mA. This is far too much for a PP3. What you need is something like a stack of six AAA cells (or, better, AA cells) in a holder.

    You must also check the resistor values. If your battery voltage was really 9V, and the three LEDs at (say) 1.5V Vf dropped 4.5V, there would be 9V - 4.5V = 4.5V across the resistor. With a 10Ω resistor you would expect 450mA which would fry your LEDs. A resistance more like 4.5V/140mA = 32Ω (use 33Ω) would be safer. You might want to go a bit higher resistance to allow for a decent size battery starting above its nominal voltage, or to economise on current.

    Probably up to now your LEDs may have survived, because the poor little 9V battery could not hold its voltage up under the heavy current drain. If you get a better battery with things as they are you can expect them to fail, so don't forget to get the resistors right.
     
  5. highwayhonkie

    Thread Starter New Member

    Apr 23, 2011
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    i have 8 AA batteries now with 3 10 ohm resistors on each string of 3 led's and putting a 390 ohm resistor by the battery on the positive side will this work.
     
  6. Adjuster

    Well-Known Member

    Dec 26, 2010
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    Now you are starting with 12V, (a bit less if your cells are NiMH). The 10Ω in line with each string would drop 1.4V at the rated 140mA, so each string of 10Ω plus three LEDs would add up to a minimum of 1.4V + 3 * 1.5V = 5.9V. This would leave 12V = 5.9V = 6.1V across the resistor. That's rather wasteful - more so than with 9V.

    The total current would be 3*140mA = 420mA, so the extra resistor for full current comes to 5.9V/0.42A = 14.5 Ω. You might want to go up a preferred value or two, at least 15Ω or even 18Ω.

    Your 390Ω would give a total current estimated at (12V-4.5V)/(390Ω+3.3Ω) = 19mA, which is too low. How did you get that value?
     
  7. marshallf3

    Well-Known Member

    Jul 26, 2010
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    6V lantern battery or a 4 cell holder (preferably AA size or better) would be more in line.

    Even the AAs wouldn't last very long driving that kind of load.
     
  8. highwayhonkie

    Thread Starter New Member

    Apr 23, 2011
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    hey sorry the last resistor is not 390 its 270 and holding 3.99 volts on the lights and holding how can weget to 4.5 volts
     
  9. Adjuster

    Well-Known Member

    Dec 26, 2010
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    Alkaline AAs would last a few hours at 420mA: I'm assuming the OP does not want a huge battery as this sounds like a portable application.

    Eight cells certainly sounds wasteful, but whether a 4 cell battery would be sufficient is debatable - it depends on what voltage per cell you are willing to accept as an end point. 1.2V/cell gives only 4.8V which would not give much current, so with only 4 cells you would not be able to use such a big percentage of the total capacity.

    What this really points up is the fact that running LEDs from batteries with resistive ballast is pretty inefficient. A current regulator - preferably switched mode - would be much better but I would think this might be difficult for the OP.
     
  10. Adjuster

    Well-Known Member

    Dec 26, 2010
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    I was assuming that the 10Ω resistors were still there, so the minimum string voltage at full current would be 140mA*10Ω + 3*1.5V = 5.9V, of which 4.5V would fall across the LEDs

    The quoted Vf is for the rated current of 140mA, but are you sure you have the correct value? I've visited the web site concerned, and the LEDs of this type seem to be quoted at 1.4V to 1.5V. There are a few similar types - is yours this one? http://www.lightobject.com/10x-10mm-3-Chips-200mW-Infrared-IR-850nm-LED-15-degree-P449.aspx
     
  11. Adjuster

    Well-Known Member

    Dec 26, 2010
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    By the way, if you do try to use a 12V battery, and run your LEDs at 140mA each string, watch out for the resistor power rating.

    A single resistor dropping about 6V at 0.42A in line with the battery would dissipate over 2.5Watts. If you try this with an ordinary little 1/4 Watt resistor, it will burn out very fast, and maybe even set something alight. You could use a wire-wound resistor rated at 3W or more, I suppose.

    The 10Ω resistors would burn about another 0.2W each, bringing the grand total to just over 3W. This shows how horribly wasteful it is to use a supply voltage more than twice what the LEDs use.

    It would be better to have a separate resistor for each chain, in which case a 1W resistor for each would just about do, but prefer 2W. Preferably, don't use 12V in the first place. Even 9V is probably on the high side, but that would reduce the dissipation for a three-resistor solution to under 1W each.

    I wonder if this makes any sense to you?
     
  12. highwayhonkie

    Thread Starter New Member

    Apr 23, 2011
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    no the angle on my lights are 30 degrees
    so what resistors should i use how many what wattage and what battery please answer for me need to get this illuminator working fast
     
  13. Adjuster

    Well-Known Member

    Dec 26, 2010
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    The 30 degree lamps also appear to be 1.4V to 1.5V Vf. http://www.lightobject.com/10x-10mm-3-Chips-200mW-Infrared-IR-850nm-LED-30-degree-P91.aspx

    I would suggest you try 6V first and see how the battery life comes out. You can always add a bit more resistance and switch to 9V later.

    Assuming 6V input, 3 LEDs at 1.4V, 140mA gives R = (6V-4.2V)/0.14A =12.86Ω Nearest preferred value is 13Ω, 15Ω might be safer.

    Power rating is about 1.8V*1.8V/13Ω = 0.249W, a bit too close for 1/4W so use 1/2W.

    So for 6V, three 15Ω 1/2W resistors, one in series with each string of three LEDs.

    If you only have 1/4W to hand, you could make each 15Ω from two 30Ω 1/4W in parallel (or one 27Ω and one 33Ω 1/4W).

    To use 9V the resistance per chain rises to (9-4.2V)/0.14A = 34.3Ω, so you could use 36Ω, or about 20Ω added to the 15Ω used for 6V.

    For the full 36Ω the rating is 0.65W so a 1W resistor would be needed. The 20Ω addition would burn 0.45W.
     
  14. highwayhonkie

    Thread Starter New Member

    Apr 23, 2011
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    so if i doublee the resistors can i make it double lights and 12 volts
     
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