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Connecting 20 3v LEDs in series?

Discussion in 'LEDs and Optoelectronics' started by gfurther2, Jun 21, 2012.

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  1. gfurther2

    gfurther2

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    Jun 21, 2012
    I need to connect 20 3 volt 20mA white LEDs. I am very new to electronics, but I want to connect them in a series circuit because they're spaced about 5 feet about and the whole circuit makes a circle, so I could save half the wire. I'm using a 3 volt power source (2AA batteries), but this is only temporary to see what the LEDs look like until I get a power adapter. Here's my question: I connected 5 LEDs in series to just make a mini model, and the LEDs didn't light! But when I connected them in parallel they lit up. So I did some research online and found out that to connect 20 of these 3v LEDs in a series circuit I'd need a 60 volt power source? (3 volts for each LED). I want to use my 2AA batteries because I know I can't get hurt using the 2AAs, but I don't know about 60 volts. So do I have to connect these LEDs in parallel and waste another 100 feet of wire, or is there a way to wire them in series. Or any other way to do this without using 2x the amount of wire!
     
  2. GreenGiant

    GreenGiant

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    Feb 9, 2012
    you need to connect them in parallel yes, but you dont have to waste that much wire, it was making a loop yes, but if you make a loop to the farthest one, and back, then tie off of that loop for the rest you should achieve you goal

    you should note though that if the LED's require a full 3 volts to turn on then you may not see anyhting even in parallel, batteries are rated under no load, under load they will read lower
     
  3. gfurther2

    gfurther2

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    Jun 21, 2012
    So I've decided to use the parallel method, and I want to use a 9 volt power source (wall adapter). I know all the formulas, V=IR and all that from high school. I just have no idea how to apply it? And these current limiting resistors... Where do I put them? What size do I need. Been trying to figure this stuff out for years.
     
  4. gfurther2

    gfurther2

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    Jun 21, 2012
    9 volt supply voltage, 20 3 volt LEDs at .02 amps. So (9-3)/.02 = 300 ohm resistor

    Or

    (9-3)/(.02 x 20) = 15 ohm resistor? I multiplied the .02 amps by 20 because I'm using 20 LEDs.


    AND


    Since I'm doing 20 LEDs in parallel would it look like this:

    Vs(neg-) ----> Resistor ----> | ----> Resistor ----> | ----> (and so on)
    Space:::::::::::::::::::::::::LED::::::::::::::::: :::LED
    Vs(pos+) --------------------> | --------------------> | ----> (and so on)

    Or would the resistor be on the positive rail? Or does it matter?
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    No, the V+ and V- go straight across. Between them you connect a resistor in series with a LED.

    The way you've wired this up, the first LED would light normally, the second very dimly, and the next few would get progressively (and very rapidly dimmer) beyond a hand-full of LEDs the chance of you seeing any light is almost zero.
     
  6. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Connecting LEDs directly in parallel without individual resistors is not recommended because there are variations in forward voltage between LEDs, even LEDs from the same batch, that will cause some LEDs to draw more current than others and therefore glow brighter. Also the temperature of the LEDs will affect their forward voltage and cause the same problem.

    The best way to drive 20 3V LEDs is in series, from a constant current source that can deliver at least 60V. Although you can generate 60V from two AA cells using a boost converter, you can't realistically power 20 LEDs from AA cells because of the power required (150 mW per LED, assuming 50 mA forward current).

    You can connect them in parallel if you use a separate resistor for each LED, as Steve suggested, but a simpler and probably better option would be to break the LEDs into several series strings, with a resistor for each string, and connect the strings in parallel using a power source of say 15V or so. For example, five strings of four LEDs each.

    Do some googling; there's plenty of material out there.
     
    Last edited: Jun 22, 2012
  7. gfurther2

    gfurther2

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    Jun 21, 2012
    I want to put these LEDs around the pool (I attached a fast drawing). I also attached the schematic for what I'm going to do now (a parallel circuit) unless there's a better way to do this. If I were to use a boost converter how much would one cost? I started this project because I thought it would be fast and cheap (just one big series loop of LEDs), I still think that I'll finish the project using the parallel design, but when making suggestions just please consider cost.
     

    Attached Files:

  8. KJ6EAD

    KJ6EAD

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    Last edited: Jun 22, 2012
  9. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    If it's going to be near a swimming pool, definitely keep the voltage as low as possible, for safety. There's NO WAY I would put 60V anywhere near a swimming pool; even 15V could be dangerous. I think Steve's suggestion would be best - a two-wire bus carrying around 3-4V, with an individual current limiting resistor for each LED. It's not ideal from the point of view of efficiency, but safety is the most important concern.

    So you would make up individual modules containing an LED and a limiting resistor, and run a two-wire bus around the pool, with the modules connected across the bus at various points along the length of the bus.
    This also means you can add and remove LEDs fairly easily.

    I'd suggest around 6.8 ohms for the limiting resistors. That means that at 50 mA LED current, they will drop 0.34 volts, which is many times the forward voltage variation from one LED to another, and this would make the brightnesses very even.

    To get the LEDs to light to a good brightness, you'll need at least 20 mA per LED, or 400 mA total, which is quite a lot for AA batteries. I would use three rechargeable D cells in series, which gives 3.6V. That might be too bright; if so, you can add a switchable low-resistance voltage dropper in series with the battery, using rotary switch with several low-value resistors. This will let you select the brightness you want, and by dropping the current it will slightly lengthen the amount of time before the batteries need to be recharged.

    I would remove the batteries from the circuit and recharge them separately. You don't want anything that's connected to mains power to be connected to your LED loop at any time.

    Also your loop is quite long, so you should consider the resistance of the wire in the loop. If the wires are thin, you may get significant voltage drop across the length of the bus, which will affect the evenness of the LED brightness. Remember there will be voltage lost across EACH wire. If you connect your voltage to BOTH ends of the loop (i.e. make it an actual loop), you will greatly reduce the voltage loss.

    For a bad case example, for a twin wire 100 feet long (not a loop) with 100 mA loading at the far end, if your wire has a resistance of 0.01 ohms per foot the total resistance to the end will be 0.01 ohms x 100 feet x 2 conductors = 2.0 ohms, so the voltage loss will be V = 2.0 ohms (resistance) x 0.1 (current) = 0.2V which would be noticeable. So don't use very thin wires for your loop, and connect it as a loop, i.e. connect both ends of the long wire together.

    You might also want to add a fuse (e.g. 2A) in your battery supply in case the bus gets shorted out by being squashed or trodden on. And consider the possibility of people tripping on the wire, or hurting their feet by standing on the LEDs. This is especially important because there are many danger factors: pools are often used at night where there's poor visibility, there are hard edges and an immediate drowning danger, there may be alcohol involved, people are relaxed and may not be fully aware of their envornment, and so on.

    Good luck and always think of safety!
     
    Last edited: Jun 22, 2012
  10. gfurther2

    gfurther2

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    Jun 21, 2012
    Thank you everybody for the help! Since the last time I was on here I installed only two LEDs and after only 2 days you can see the corrosion. This has made me realize the real work of this project and I will be now holding off until I find a more reasonable outdoor solution.

    Thank you all for the help though, I really learned about electronics from you guys and will be posting a lot more questions because you guys really know how to explain things unlike Wikipedia!
     
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    I find that the leads of LEDs must be made of the most easily corroded metal known to mankind.

    I'm not sure why, but looking at LEDs I've used in a breadboard have leads that can be almost black.
     
  12. KJ6EAD

    KJ6EAD

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    Aug 13, 2011
    It's tarnish. LED leads are silver plated.
     
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