LED array circuit

Discussion in 'The Projects Forum' started by Kefka666, May 16, 2008.

  1. Kefka666

    Thread Starter Active Member

    Mar 4, 2008
    I'd like to make a very large array of 100-200 high power LEDs (as an energy-efficient grow light for my plants). I've been trying to work out the best way to do this. Let me know if I'm correct about this:

    The trouble with a low voltage (<6V) parallel array, which I've made, is that one can only have so many LEDs due to the current. Even at 2A, it's only possible to have up to 100 20mA LEDs [(2A/0.1A per LED) = only 20 LEDs at full power].

    So how do I get around this and have up to 100 - 200 of them? I was thinking that I could use a higher voltage and a different setup. I found this site, http://led.linear1.org/led.wiz, but was confused for several reasons.

    1) The wizard doesn't mention the current, and isn't that important? or does the current not matter in such a series circuit. (I used 3.4V forward voltage of LED at 100mA, 36V DC source, and 200 LEDs in the array). According to http://led.linear1.org/led.wiz, that should work, but I'm confused about how. Try this: the wizard, with those settings, lets you have a potentially infinite amount of LEDs in the array, and apparently it will still work? Try 1000. It works. Is there some basic fact I'm missing about electronics (I'm a beginner)?

    2) How come the resistors are on the downhill (-) side? Shouldn't they be in front of the LED? I thought that 36V coming at an LED would fry it.

    Anyway, if anyone can answer those questions it would help me learn some important information about circuits.
    Last edited: May 16, 2008
  2. SgtWookie


    Jul 17, 2007
    That's true. When you're running them in parallel, you need 100mA times however many LED's you want to hook up. You also need a current limiting resistor for each LED. That's a lot of wiring.

    When you are powering LED's in series, the current passing through one LED, also passes through all other LED's in the "string" - including the current limiting resistor.

    Everything is relative. :) I prefer to have the current limiting resistor at the end of the string that is furthest from ground, ie; closest to the source of current. However, you could actually put the resistor ANYWHERE in the series string, and it would function exactly the same.

    There are a some additional things you should be aware of:
    1) LED's are usually supplied with a MAXIMUM Voltage @ current and a TYPICAL Voltage @ current. Odds are that most of your LEDs will have a Vf (forward voltage) close to the "typical" voltage, but there will be some that may be quite a bit higher or lower. Since you are wiring them in series with a resistor in each string, the law of averages will tend to distribute the "oddball" LED's evenly in your mix, and you won't see difference in intensity.

    Where one gets in trouble is if they try to parallel strings of LEDs and only use a single current limiting resistor. Since the Vf of LEDs vary (even in the same batch!) one string of LEDs would get much more current than the others, rapidly burn out, and then there would be a domino effect of blazingly bright LED chains burning out, sort of like a nuclear meltdown. :eek:

    In that same train of thought the LED array wizard has a bunch of LED's in series - it's taken you at your word; that each LED has a Vf of precisely 3.4v, and that your supply voltage is precisely 36v, and that each LED requires precisely 100mA of current. So:
    36/3.4 = 10.588235294117647058823529411765
    It rounded it down to 10 LED's in a string.

    So, how much resistance do we need to limit the current to 100mA?
    Rlimit = (VoltageSupply - VfTotalLED) / LEDCurrent
    Rlimit = (36 - (3.4 x 10) / 100mA
    Rlimit = (36 - 34) / 0.1A
    Rlimit = 2/.1
    Rlimit = 20
    The closest standard value is 22 Ohms
    The actual current you'll get through your LED's is then somewhat less than 100mA:
    I = E/R (Current in Amperes = Voltage / Resistance in Ohms
    I = 2/22
    I = 90.909...mA
    Wattage required:
    P = EI (Power in Watts = Voltage x Current in Amperes)
    P = 2*.090909...
    P = 182mW
    You can use 1/4 Watt resistors, but 1/2 Watt would run much cooler.

    Now, if the Vf of the LEDs varys by much, you might get a surprise on the current being consumed! You might measure a good sample of them to see what your real numbers turn out to be.

    You can make a constant current source using an LM317 voltage regulator IC and a single resistor connected from the output to the adjust terminal, supplying power to the input terminal, and taking the output current from the adjust terminal. This is documented in National Semiconductor's datasheet for the LM117/LM317, bottom of page 18.

    The LM317 is available from your local Radio Shack. They also carry a 25 Ohm 3 Watt rheostat, which will work just fine for the resistor you need.
    In order to get 100mA out of the LM317, the rheostat needs to be set to 12 Ohms.

    You will need a power supply capable of putting out 6 to 12 volts in order to test your LED's with the LM317 current regulator circuit.
    Last edited: May 16, 2008
  3. Kefka666

    Thread Starter Active Member

    Mar 4, 2008
    Ok, I think I'm getting it. So even if I mixed and matched different LEDs together (say, 20mA mixed in with 100mA on the same string) they would still light up fine. Makes sense since current is conserved in a circuit and has to pass through each element. I hadn't considered that the resistor can be at any place in the series circuit.

    I like your comment about variations in LED Vf. I'll try testing some and if I have the time I can sort the lines by their apparent Vf groups.

    I'm still unsure what current the DC adapter should use when I power this. 36V would work for the setup of 100 LEDs that I want, but how high should the current be? I would assume 2A would be enough, but I'm not sure how to tell in this type of circuit. Thanks for the help btw
  4. SgtWookie


    Jul 17, 2007
    No, you shouldn't do that! :eek:
    This is the "Any chain is no stronger than it's weakest link" scenario. If you attempted to combine 20mA and 100mA LEDs together in a series string, and then put 100mA through it, the 20mA LEDs would rapidly be cooked/burned out, and the string would be broken.

    It's like a kink in a garden hose. Doesn't really matter where it happens; it limits the flow of water out the end of the nozzle wherever it occurs.

    Actually, you only really need to match them if you are planning on running them in parallel. Since you are running them in long strings, all you would need to do is take a statistical sample to get your median Vf.

    OK, you have strings of LEDs, each string has 10 LEDs in it.
    You have 100 LEDs, so that means you'll have 10 strings of 10 LEDs.
    Each string requires 100mA current.
    So, 10 strings of LEDs that require 100mA at 36 volts will use 10 x 100mA = 1 Ampere of current.

    Now, you need to use a REGULATED 36v power supply. Wall-wart type adapters are not typically regulated. If you simply try to plug in your strings to an unregulated supply, you will likely fry something, or lots of somethings. :eek:
  5. Kefka666

    Thread Starter Active Member

    Mar 4, 2008
    Thanks again! It makes a lot of sense now.
  6. Kefka666

    Thread Starter Active Member

    Mar 4, 2008
    I have one more question. I've changed some features of the array. Now I'm using 30V source, ninety six 3.6V LEDs at 100mA each. When I run the wizard (cheap, I know, but I'm learning), it says "the array draws current of 1200 mA from the source." Why is this the case? I haven't been able to figure out how to compute that value. My guess is that the array requires 1200mA DC source because there are 12 parallel rows each using 100mA? I suppose if there were 10 rows it would require 1000mA, then.

    For the limiting resistors, I used Ohm's Law to find:
    Rlimit = V x I.
    With 8 LEDs per row: Rlimit = (30 - (3.6 x 8)) / 100mA
    Rlimit = 12 Ohms.
    Last edited: May 21, 2008
  7. SgtWookie


    Jul 17, 2007
    Yes - since each row uses 100mA current, and there are 12 rows, that's 1200mA - because current in parallel is additive.

    That's cutting it a tad close. I like to leave a bit more of a "fudge factor" than that. 10% is comfortable. 5% is kind of pushing it. You're at 4%. (1.2v is 4% of 30v) Try using 7 LEDs per string instead - but see below first.

    Have you measured a sample of your LEDs to find out what their Vf at 100mA actually is? Because that's the number you really need to be using.

    If you don't feel like measuring them, then you should be using the "typical Vf" number.
    If there wasn't a typical Vf specified, you had best measure a statistical sample to find out what it is.