Best way to power ~150 LEDs from 9V or 1.5V batteries?

Discussion in 'The Projects Forum' started by white_nitro0, Apr 4, 2012.

  1. white_nitro0

    Thread Starter New Member

    Apr 4, 2012
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    Hi all,

    First of all let me apologise in advance if I've broken any rules, I'm new to this forum and not sure if I'm posting in the right forum or if my questions have already been answered on page 8...

    Basically what I am looking to do is to connect in the region of 150 LEDs together to flash alternatively 75 at a time. I have a basic understanding of electronics but it's been a long time since I've done any seriously/practically.

    I am using a mixture of (apparently standard) 3mm and 5mm LEDs, and would like the batter(y/ies) to last for a few hours (although as long as it lasts long enough to make an impression, perhaps an hour or two would be fine).

    By looking at some of the other threads, it appears connecting 3 LEDs in series with a 75 Ω resistor then connecting many of those in parallel is the general way forwards. What I am really wondering is, how can I connect many of these circuits together in a safe manner, and how can I get half of them to flash alternatively?

    Thanks in advance, any constructive criticism is welcome!
     
  2. Audioguru

    New Member

    Dec 20, 2007
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    You will have 75 LEDs turned on at a time. Maybe 3 are in series in 25 strings on each side.
    If each string is bright with a current of 20mA then the battery sees a current of 20mA x 25= 500mA.
    A little 9V alkaline battery supplies 9V at 500mA for about 30 seconds. Then its voltage drops to 7V in 7 minutes then drops slower to 6V in 30 minutes when it is dead and the LEDs are so dim that you can barely see them.

    Use six AA alkaline cells to make 9V which will last 5 or 6 times longer.
     
  3. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    Resistor in series with LEDs wastes power and the LED brightness varies as the battery voltage declines. I would probably use a curent source.
     
  4. KJ6EAD

    Senior Member

    Apr 30, 2011
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    We'd need the LED specifications to develop specific recommendations but a worst case estimate would be that you need 6Wh/hr of battery capacity.
     
  5. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    I have some superbright red LEDs that are REALLY bright with only about 2mA through them. In a project like yours, that can really save size and money on the battery. When we releases the LM2731, I built a sign out of these LEDs using about 60 of them spelling out "LM2731" and driven by one LM2731 IC and a couple of AA batteries.
     
  6. white_nitro0

    Thread Starter New Member

    Apr 4, 2012
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    Awesome thanks guys! So sticking six AA 1.5V batteries together and find some low power LEDs? How would that affect the resistors I had to use?

    Another point I could use some advice on, if I were to build a flip-flop circuit to make 75 flash at a time, could I connect 25 strings in parallel of 3 LEDs and a resistor (in series) and put them all going into the same transistor? What size/type of transistor/capacitors/resistors would I need? I'll try uploading a (badly drawn) image soon!

    [Edit]

    Here is the promised image, obviously haven't included all the LEDs for clarity. Also I have no idea if this is even the best way to do what I want!


    [​IMG]
     
    Last edited: Apr 4, 2012
  7. Audioguru

    New Member

    Dec 20, 2007
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    Your load current might be 50mA for each side or 500mA for each side.
    The transistors must be able to handle it (the BC547 max allowed current is only 100mA).
    The transistors saturate pretty well with a base current that is 1/20th the collector current so your circuit is completely wrong.
     
  8. white_nitro0

    Thread Starter New Member

    Apr 4, 2012
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    how do I go about making it not wrong?
     
  9. Audioguru

    New Member

    Dec 20, 2007
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    1) How much current do you want in each group of 75 LEDs?
    2) Divide it by 20 for European transistors to calculate the base current.
    3) Use Ohm's Law to calculate the base resistor value.
    4) Calculate the capacitor values.
    Simple arithmatic for 10 years old kids.
     
    white_nitro0 likes this.
  10. white_nitro0

    Thread Starter New Member

    Apr 4, 2012
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    So the basic set up of the circuit is correct, I just need to adjust the values of capacitors/resistors?

    I think it works out as having 500 mA of current in each group of 75 resistors, so 25 mA base current (I'll look for a transistor that can handle that much current).

    Base resistor = 9V / 0.025 A = 360 R

    How do I work out the capacitor values from this?
     
    Last edited: Apr 4, 2012
  11. #12

    Expert

    Nov 30, 2010
    16,257
    6,758
    You've confused the meaning of "superbright" with "low power". Be careful about that when selecting your LEDs. You're looking for "superbright". The result will be low power.
     
  12. Audioguru

    New Member

    Dec 20, 2007
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    The original circuit uses 10,000 ohm timing resistors and 100uF timing capacitors. If the resistors are changed to 360 ohms for power transistors then the capacitor values must be increased the same amount to 2700uF each.
     
  13. Adjuster

    Well-Known Member

    Dec 26, 2010
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    This circuit may be a bit too simplistic really, as with the LEDs in the oscillator collector loads the frequency is likely to vary a lot as the battery voltage runs down.

    Perhaps a 555 oscillator driving MOSFETs would be a better bet.
     
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