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Class 2 battery charger as a LED power supply?

  1. Jul 6, 2015 #1
    Hello,

    I'm planning to light up a tile chess board with 64 LEDs (blue and white) set between the tile squares, and I was wondering if this Class 2 Battery charger output: 12VDC 900mA, would work to power 16 parallel sets of 4 LEDs + 1 ohm 1/8W resistor and a SPST toggle switch? The current drawn by the circuit should be around 320 mA according to this: (http://ledcalculator.net/default.aspx?values=12,3,20,64,0).

    Thank you.
     
  2. jcsd
  3. Jul 6, 2015 #2

    berkeman

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    Staff: Mentor

    With the LED forward voltage drop at 3V, and 4 LEDs in series with a 12V supply, how does using a 1 Ohm resistor give you your 20mA? The math doesn't seem to work...
     
  4. Jul 6, 2015 #3

    berkeman

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    Also, you will want to be using a lot more than 1 Ohm in series to set the current. You can't count on dropping a few mV across a resistor to give you a steady & reliable current setting.
     
  5. Jul 6, 2015 #4
    Sorry, I'm pretty new to this. I was mainly going off of the diagram on the link. I thought the 1 ohm resistor was just to keep the flow to the LED smooth? How should I change the circuit set up?
     
  6. Jul 6, 2015 #5

    berkeman

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    No need to be sorry :smile:

    In LED circuits like that, you drop a volt or two out of the 12 volts across a current-setting resistor. Say you are using red LEDs with Vf around 2V. Then you could put 5 of them in series which leaves 2V to drop across a current-setting resistor. If you want 20mA for the LED current, you would set the resistor value to 2V/20mA = 100 Ohms.

    The forward voltage of the LEDs will vary some with temperature and forward current and manufacturing tolerances -- look in the datasheet to see what their range of Vf is to help you do your calculations for the resistor value.

    I can believe that the white LEDs will be around 3V, so you will need to stack only 3 of them in each series stack. What is Vf for the blue LEDs from the datasheet?
     
  7. Jul 6, 2015 #6
    The blue is listed at 3.2~3.6 and the white is listed at 3.2~3.4 both at 20mA.
     
  8. Jul 6, 2015 #7

    berkeman

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    Yikes! Would you consider stacking two of those voltage sources in series to give you 24V to start with? That might be more efficient...
     
  9. Jul 6, 2015 #8
    I can see if I have another battery charger lying around somewhere. I also have a 30VDC 5A adjustable power supply that I can use to test things.

    Would it be possible to just use lower voltage? I tried using a single LED with a two C battery pack and it lit up fine.
     
  10. Jul 6, 2015 #9

    berkeman

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    So the issue is efficiency. You would like to burn as little extra power across the current-setting resistors as possible. So if you have lots of parallel strings with only one LED and resistor per string, you are burning close to 50% of your available power across the resistors. But if you have a higher available voltage, you can series-connect more resistors, and have fewer current-setting resistors for the overall setup.

    You need to drop a volt or two across the current-setting resistor to accommodate the changes in series Vf across the LEDs, so it's best to series connect as many LEDs as possible. Makes sense?
     
  11. Jul 6, 2015 #10
    Yup, it makes sense.

    I didn't mean doing a bunch of single LEDs though, I meant along the lines of treating the LEDs as 3 volts instead of the 3.2-3.4-3.6 since the lone one seemed to work fine with the 3 volt battery pack. And then go with the x3 series stack you mentioned earlier with the 12 volt. They'll only be a bit dimmer if the voltage is a bit low, right?
     
  12. Jul 6, 2015 #11

    meBigGuy

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    Your best bet is an LED string driver or string driver module.

    http://www.linear.com/product/LT3598, for example.

    There are lots of modules and chips on the net. Search for "led string driver" or " led string module".

    If you still want to roll your own with resistors, then three LEDs in series at 3.6v gives 10.8V worst case, so the remaining 1.2V across 60 ohms gives 20ma.
    (note that the remaining voltage and the resistor is what sets the operating current)

    64/3 =~ 22 strings @ 20ma = 440ma

    But, if the 3 LEDs were 3.2V , that gives 9.6v, so then you would need 2.4v/.02ma = 120 ohms to limit the current to 20ma.
    So, depending on the LED's voltage (which varies from led to led) you would get different currents.

    I would try 12V driving 3 LED's with a 120 ohm resistor (this will give between 10 and 20 ma depending on the LED drop). You can then play around with your variable supply and decide on a final resistor value.

    You can do the same exercise for 2 LEDs in each series circuit. At 3.2V you need 5.6/.02 = 280 ohms.
     
  13. Jul 26, 2015 #12
    Late reply, sorry.

    I just got the 120 ohm resistors in the mail so I'll try them out. Thank you for the help :D.
     
  14. Jul 26, 2015 #13
    Yup looks like it'll work, thanks a bunch guys.
    11750704_906151656125073_997725313132321591_n.jpg
    11011570_906151586125080_4450347123983668437_n.jpg
     
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