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Multiple LED's on circuit

Discussion in 'LEDs and Optoelectronics' started by Peelster, May 9, 2016.

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  1. Peelster

    Peelster

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    May 9, 2016
    Hi,

    I need some help on a project that requires 6 x 8mm LEDs to be lit up on a circuit by a 9v battery. I've tried a range of ideas but at the moment we have 3 LED's running on 2 parallel lines. The picture probably shows it better than I can explain. I think I might need 2 x 260 omh resistors instead of 2 x 160 ohm but I'm not sure. Any advice would be appreciated
     

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  2. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Welcome to EP.

    Your drawing is not how a typical schematic looks like, therefore not easy to read ;).
    The resistance you need depends on the current you want to flow through the LEDs. The voltage across the resistor is V=(Vbat-3*Vled), Vbat being the battery voltage, Vled being the operating voltage of the LEDs.
    From R=V/I you get the required resistance.
    We have a ressource on this topic explaining more details.

    The operating voltage of a typical white LED typically is around 3V...3.3V. Therefore putting 3 LEDs in series leaves no headroom for the voltage across the resistor. Or, in other words, the unavoidable voltage drop across the resistor will reduce the operating voltage for the LEDs such that they will not light very bright.
    This is made even worse when the battery voltage drops due to discharge. Although a typical 9V battery has a comparatively small capacitance (mAh), so your circuit will soon start to dim.

    You're probably better off using a switch mode LED driver which can not only control the LED current but also use more of the battery's capacity by boosting the voltage as required even when battery voltage is low.
     
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  3. Peelster

    Peelster

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    May 9, 2016
    i can't seem to find that much reading material on the switch mode led driver. What exactly do it do and do I just need to fit it in to my current circuit or does it need redesigning?

    Sorry about my drawing! Lol just thought it might explain it better visually than me waffling on .

    All these 6 LEDs are doing is proving light which in inset into a wooden letter to act as a wall piece. I thought it would be a lot simpler

    Thank you so much for getting back to me
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

    10,025
    2,138
    Nov 17, 2011
    You could roll your own, but I guess that's a wee bit above your level - no offense meant ;)
    Get an off-the-shelf model, e.g. here. The main parameter is the max. power you need. Plus how many LEDs can be driven by one driver module.

    Or scavenge an inexpensive LED flashlight like this one. It operates on 3*AAA batteries and contains all the necessary 'electronics' plus 6 LEDs (I think there are 6). All you have to do is to remove the housing and re-wire the LEDs with some longer wire to fit in the positions you want them to occupy. A battery holder and switch are also included in the flashlight.
     
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  5. Peelster

    Peelster

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    May 9, 2016
    Do you reckon the switch mode led driver is the simplest method of achieving what I want?

    All I want is to power the 6 LED's with a affordable battery instead of mains powered and a switch.

    Thanks again for your advice.
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    if the LEDs are not high power, then using resistors is fine.
     
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  7. Peelster

    Peelster

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    May 9, 2016
    As shown in my rubbish diagram I'm currently using 2 resistors, 1 on each of my 2 parallel lines but my 9v battery doesn't seem to be enough power to light more than 3 and the resistor I have, which if 1 led is 3v then it makes sense. I was just wandering if changing my resistor might help
     
  8. BobK

    BobK

    7,682
    1,686
    Jan 5, 2010
    If you have 3V LEDs and a 9V battery, the best you can do is 3 strings of 2 LEDs in series. You would need a resistor in each of the 3 strings.

    What currernt are the LEDs rated for at 3V? You would use this to calculate the value of the resistor from Ohms law. You need the resistor to drop 3V at the rated current, the other 6V going to the two LEDs. Can you see the problem with using 3 in series? There is no voltage left to drop across the resistor, and therefore the resistor would be 0 and you have no control over the current.

    Bob
     
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  9. chopnhack

    chopnhack

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    Apr 28, 2014
    Welcome Peelster!

    Harald has given you some great advice here. The three AAA batteries should give you a pretty decent lifespan. The electronics bit in that flashlight is contained in the top of the flashlight, so you may have to cut and peel open the flashlight housing which is made of aluminum. I have one of those very same models, very bright light and seems to last forever. I am not sure if you have a 'rubbish tool store' as we call them here, but if you do, take a gander. You should be able to easily mortise the back of the letter to hide the electronics.
     
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  10. Peelster

    Peelster

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    May 9, 2016
    Thanks for everyone's help, I'm going to try out a few of your suggestions and see if I can get these things working! Thanks again
     
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