Quick LED array question

Discussion in 'The Projects Forum' started by Rob1132, Apr 5, 2011.

  1. Rob1132

    Thread Starter Member

    Nov 4, 2010
    36
    0
    Alright, so I'm trying to build an ultraviolet LED light. I have the parts and everything, but I'm still new to a lot of this stuff. I used an LED calculation wizard to figure out what kind/how many resistors I would need for the array I wanted to build and it came up with this...

    +----|>|----|>|---/\/\/----+ R = 150 ohms
    +----|>|----|>|---/\/\/----+ R = 150 ohms
    +----|>|----|>|---/\/\/----+ R = 150 ohms



    I'm running 6 LEDs at 3.3v forward voltage and 20mA with a 9 volt battery. Now, the part I'm unsure about with the solution is where everything gets soldered. I think that it looks like I'd have to solder the 3 resistors all together to the negative end of the 9v battery and then 3 positive sides of the LEDs to the positive side of the 9v. Basically, I'm just having trouble understanding the diagram so any help would be appreciated.
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    You don't solder to the battery terminals.
    You use a battery terminal clip for a 9v PP3 battery, or a battery holder for AAA, AA, C and D-type batteries. If you're in the USA, your local Radio Shack stocks a variety of them.

    I don't recommend 9v batteries for most projects, as they are very expensive and don't last long. There are six or seven "aaaa" or button cells inside the package, all wired together. An alkaline PP3 battery might be rated for 500mAh, which is (500mAh/20)mA for 20 hours until the battery is down to around 7v (dead).

    500mAh/20 = 25mA. If you draw more current than that, the battery will last a much shorter time.

    LEDs are sensitive to polarity. They are diodes. However, you will damage them if you subject them to more than about 5v in reverse.

    The cathode (shorter wire, often has a flat on 3mm and 5mm LEDs) goes towards negative, and the anode (longer wire) goes towards the the positive battery terminal.
     
  3. Rob1132

    Thread Starter Member

    Nov 4, 2010
    36
    0
    Yeah I have 9v battery clips; that's what I meant. I'm just having trouble reading and understanding the diagram/schematic. Here's a pic of it...

    [​IMG][​IMG]http://img9.imageshack.us/i/ledj.jpg/

    The other LED light project I've seen was quite different than that so I'm a little confused. From what I understand, that's telling me to just connect the positive and negative of 3 of the LEDs and then string the remaining 3 anodes to the positive side of the 9v snap. The other 3 cathodes would have the resistors on them and then go to the negative of the 9v battery snap. Am I understanding that correctly?
     
  4. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    OK, I copied your image and moved stuff around a little bit. Note that I added "-9V" by the junction on the right, and moved the "150 Ohms" labels over the resistors.

    [​IMG]

    Does this make more sense to you?
     
  5. saikat36

    Member

    Jan 28, 2009
    16
    0
    'Rob1132' ur schematic is 100% OK, don't change anything.
    If u put -9V then LED will be burnt out.
    1/4W Resistor is OK.
    Every branch will draw 16mA current.
    Whatever power requirement calculated practically requirement will be much less than that.

    So without any confusion go ahead.
     
  6. Rob1132

    Thread Starter Member

    Nov 4, 2010
    36
    0
    Ok, thanks for the help guys. I'm fairly sure I was right about how to do it now. Just wanted to check here first. Better safe than sorry after burning out some LEDs.
     
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