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LEDs in series to 240V?

Discussion in 'LEDs and Optoelectronics' started by nicksticks, Apr 14, 2012.

  1. nicksticks

    nicksticks

    4
    0
    Apr 14, 2012
    Hi I am trying to build an aquarium light using 30 star LEDs (its going to be very bright!). I've had some trouble with 12V power supplies that I bought on ebay (big surprise) and I have had the idea of simply cutting the 240V supply in half to 120V with a diode, smoothing it with a capacitor and running the LEDs connected in series from that. I have attached a diagram of the circuit.

    The LEDs are 3W each and are rated to use 3-4 V but they can tolerate 6V for at least 30 seconds. If they use 3W at 3V thats 1 amp. Their resistance is therefore 3 ohms minimum. 30 in a row is 90 ohms. Connected to 120V that is 1.3 A, a bit much but close enough at least for a test run. If it works I will buy a dimmer to control the voltage (I can get one on ebay :D)

    One of the problems that I can see is the fact that the capacitor will initially draw a very large current as it charges. It is an electrolytic capacitor and so should have a reasonably high internal resistance which might make it ok. I am also not sure that the capacitor will sufficiently flatten the voltage peaks and maybe the voltage peaks will kill the LEDs.

    Can I please have some opinions and advice for my circuit?
     

    Attached Files:

  2. gorgon

    gorgon

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    Jun 6, 2011
    The main problem here is that you will not get 240V DC out of this, but more like 340V peak with ripples down to whatever the capacitor manage to supply between the peaks.
    The second problem is that LEDs does not behave like resistors, they don't increase the forward voltage very much when the current increase, so you'll need more LEDs or an external drop resistor to bleed off the extra voltage.

    Even if you add more LEDs to fill the gap, you should add a small resistor to capture unwanted voltages spikes.

    TOK ;)
     
  3. duke37

    duke37

    5,257
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    Jan 9, 2011
    As gorgon says , the capacitor will hold the voltage up, near 340V. If you eliminate the capacitor, then the average power in the leds will be less than halved. The leds will flash but this will not be noticed by humans. Don't know about fish.
    You should always use a series resistance to control the current.
     
  4. nicksticks

    nicksticks

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    Apr 14, 2012
    Thanks Gorgon and duke37,

    I looked up the 240V AC waveform and I see what you mean, I thought maybe the capacitor would clip off those peaks a bit but I don't know how to calculate it so I will not rely on it.
    I don't want to use resistors because they will be passing a lot of current and will emit a lot of heat.
    I think I will buy more LEDs and a dimmer. I will put in a bridge rectifier instead of the diode so as not to assymmetrically load the mains supply. I will measure the AC voltage from the dimmer with my multimeter before I connect it to the LEDs and then divide it by 0.707 to calculate the peak voltage. I will modify the dimmer so that it cannot be set to exceed the voltage that I want by wiring a resistor in series with its own variable resistor (it is switchmode so no large current here).

    What do you think?
     
  5. gorgon

    gorgon

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    24
    Jun 6, 2011
    The dimmer could be ok, but if it is just clipping the sinus curve, so measuring it with a AC voltmeter can fool your calculation, since it is made to show a RMS value.

    If you want to see the true peak value you should use an oscilloscope an verify the value.
    A capacitor will never 'clip' the peak, but it will try to hold the value constant over the period.
    If you use a bridge rectifier, you'll get a double of the frequency and a higher effective voltage for the same capacitor value. The peak value will be the same.

    TOK ;)
     
  6. CocaCola

    CocaCola

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    Apr 7, 2012
  7. nicksticks

    nicksticks

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    Apr 14, 2012
    OK I understand about the capacitor now, I will leave it out, it is pretty bulky anyway. I'm trying to find out about fish vision too :)
    This calculator http://led.linear1.org/led.wiz has recommended that I use a 1 ohm resistor in series that will dissipate 1W of heat with a supply of 90V.

    I think the dimmer will be the type that just cuts out the lower voltage part of the wave :mad: so still ~340V peak! It seems that peak current is more important than RMS for LEDs :( What can I do to make it more like 90V DC?

    I like the neatness of the series capacitor solution. The problem is the large ripple current that the capacitor will be dealing with, I don't think I can get a capacitor with the capacitance, voltage and ripple current rating that I need. Does that make sense or have I misunderstood?

    Thanks again for the help :D
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,220
    2,695
    Jan 21, 2010
    Running LEDs from 240V is not a great idea.

    It's far better to get a safer source of power (lower voltage), ESPECIALLY when working around water.

    Some Chinese LED strings operate from 240V like this, but they tend not to last long and also tend to end up dissipating way more power in series resistors than they do by the LEDs. They're not wonderful, but they are cheap and generally reasonably insulated.
     
  9. CocaCola

    CocaCola

    3,635
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    Apr 7, 2012
    Several people (found via Google when I needed a solution) have reported great success with that design... I'm in the US on 110v so I followed this guys lead http://www.candlepowerforums.com/vb/showthread.php?260956-Fluorescent-to-LED-night-light-conversion his schematic has worked just fine for me, built, two of them as a night/work light replacement for my microwave that liked to blow two $3 'utility' bulbs ever other month... If you research the concept a little more with Google hits you will find that others have hesitation on the capacitor, but it appears that caps that work are not as hard to find as you might think, just make sure to over-spec it quite a lot...
     
    Last edited: Apr 15, 2012
  10. nicksticks

    nicksticks

    4
    0
    Apr 14, 2012
    Thanks for all the advice guys. My LED array just needs too much power for the capacitor option to be practical. It is also not really possible to drive it from the mains using resistance to limit the current without either getting a lot more LEDs or having a resistor dissipating a silly amount of heat.
    I have just gone through my junk box and found an old laptop transformer, 18.5V DC and 3.5A. I used the calculator here http://led.linear1.org/led.wiz again to plan the circuit for me, the result is attached. I bought the resistors and it works like a charm and is brilliantly bright! It gets very hot though so I need to add a heatsink and a fan, but thats a job for next weekend!
    Thanks again! :D
     

    Attached Files:

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