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What to optimize in LED array?

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  1. May 13, 2015 #1
    I'm making an array of 11 3.3V LEDs, and I've got multiple various setups possible depending on source voltage (batteries supplying either 6V, 9V, or 12V) and whether I decide to put a bunch of the LEDs in a combination of series/parallel or all in parallel.

    I need to maximize battery life, so obviously current drawn from the source is a concern, but I've got a dilemma. Is it better to run with a lower voltage supply and a higher current drawn, but large battery capacity/mAh? Or better to stick to a lower current with fewer parallel bits, but higher voltage and lower capacity/mAh?

    My setup right now is with a 6V source with every light in a parallel circuit (and appropriate resistors), which draws 220mA if I did my calculations right, and (assuming I use four alkaline AA batteries, which have 2700mAh according to Wikipedia) will last the longest compared to the 9V or 12V. That's the highest current drawn out of all my possible setups, though...?

    That does not look very efficient. I'm new to the world of hobby electronics, though, so it might be a bit of newbie paranoia, but I thought I'd throw it out there anyway before I start putting anything together permanently.
     
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  3. May 13, 2015 #2

    phinds

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    Figure out what the power dissipated in the resistors are for each of your various possible setups and choose the one that has the least power wasted (power dissipated in the resistors is wasted as far as your circuit functionality is concerned).
     
  4. May 13, 2015 #3

    berkeman

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    Why are the LEDs 3.3V? Do they already have internal current limiting resistors for use with a 3.3V supply? Or are they Blue LEDs or some other version that has a very high Vf?

    The most efficient way to drive them is in series, if they are matched well. You could do a boost DC-DC converter with low-side current sensing, for example...
     
  5. May 13, 2015 #4
    They're white LEDs, which I believe are functionally the same as blue ones anyway, right?

    I know series is more efficient, but I'm trying to maximize the battery life while minimizing power loss. With a low-voltage battery (which is really going to be multiple AA batteries together), I have more mAh but I can't put them in series because each LED needs a voltage of at least 3.3V to light up, right? If I put them in series, I need a higher voltage supply (which, due to battery availability/costs, will be either 9V or 12V). However, neither of those have very high mAh, as far as I know.

    As far as simple circuit design goes, I'm unsure of which is more important to overall longevity of the battery. If I've got a big power loss, but the battery is technically able to supply enough power for long enough, is that a problem?

    I've never used a boost converter before. I'll have to take a look. I'm assuming it draws a larger current, though...?

    As I said, it might be a bit of a newbie question, but alas, I'm only learning.
     
  6. May 13, 2015 #5

    phinds

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    Minimizing power loss and maximizing battery life are the same thing (under anything like normal conditions), so you are not trying to do both, really, you are trying to do either one since if you do one, the other follows. Focus on minimizing power loss, as I said in my first response.
     
  7. May 13, 2015 #6
    Hm. Okay.

    The setup with the least power lost is with a 12V battery and as many LEDs in series as I can fit. I did a quick calculation and got that the battery wouldn't even last an hour, though. (12V battery that I'm using has 55mAh or something teensy, and the circuit setup has a current of 80mA.)

    So--it's efficient, but weirdly enough, maybe isn't my best option--unless I did my math wrong? The problem is that I've got multiple options for batteries, I guess. That, and putting a bunch of LEDs in parallel with a bunch of resistors and calling it my best option seems really counterintuitive...
     
  8. May 13, 2015 #7

    phinds

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    Don't rely on your intuition until you have done the math on this and similar problems enough times that your "intuition" is really just an educated guess that will generally be right.
     
  9. May 13, 2015 #8

    tech99

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    You need to decide how low the battery voltage is allowed to fall before the light goes out. This means having an initial battery voltage maybe twice the LED voltage, so that it still works reasonably when the battery falls to, say, 66%.
     
  10. May 14, 2015 #9

    meBigGuy

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    You need only to consider the ma-hr of the batteries at the discharge current you design for that battery and determine which configuration lasts longer.

    The comparative wasted power means little if the batteries don't have equivalent energy capacity. It's just ma-hr divided by ma gives hours.

    Remember though that the rated ma-hr is given for a (usually) 0.1C discharge rate, and goes down as the discharge is increased. Different battery technologies respond to higher discharges differently.

    For example, a 55ma-hr rated battery at 5.5ma (0.1C) discharge will last 10 hours. At 55ma it may only last 45min instead of 1 hour.
     
  11. May 16, 2015 #10
    LEDs are not voltage driven, they are current driven. Because of this you need to decide how bright you want them to be. You then need to figure how much current this requires. You can do this with a bench top power supply that has a current limited mode. Set the current limit to 1/2 the nominal driving current (and the voltage to at least 3.3V). Then hook up the forward biased diode. Adjust the current upward being careful not to exceed the absolute maximum rating on the diode. When the diode is bright enough, read the current flow. This is the current you want.

    Now you need a supply that drives that current. There are lots of ways to get this, some easy and inefficient (like a resistor) some more complex. The most complex I've heard of is a pulsed system with a frequency matching the relaxation time of clorophyl for grow lights. Slightly simpler are switched current supplies sometimes called LED drivers. You need to be careful with these because I've never found a switching supply that worked like it was supposed to. They are all finicky and some of them don't seem to work at all. (I've seen data sheets that require large value capacitors the size of a grain of sand -- clearly the manufacturer never used one.) Buy several and test until one works.

    Which driver you settle on will determine what your input voltage should be.

    If you decide that is too much work for your project, use one current limiting resistor with each of several (5 and 6?) LEDs. Hand match them for the proper current if you only need a few units. If you need about a hundred units, I feel for you. It's too many for hand matching and too few to spend a week finding an LED driver. You will need to limit your LEDs to smaller strings (3, 4, 4?) since overdriving them can be a problem with longer strings. (They have a positive feedback effect as they heat up with one trying to hog all the power.) Any current through the resistor is pure loss, so keep them to a minimum.

    P.S. If anyone has an easy to use switcher they like, please let us know. I've only tried a few, but I never found a favorite.
     
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