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Wiring LEDs together

Discussion in 'Electronic Design' started by alokw, Mar 30, 2007.

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  1. alokw

    alokw Guest

    Hi all,

    I have a quick question that should be simple to answer.
    I have 54 LEDs that require 3.4 V DC with a maximum of 30 mA that need
    to be wired together.
    I have 9V batteries to power them.

    I did the calculations and ended up wiring them in series of 2 LEDS
    with an 82 ohm resistor for each pair and hooked up the end to a
    single 9V battery. With the 27 pairs, it was suppsed to draw 810 mA.

    When I switched it on, everything was dim. I didn't think about how
    many mA the battery could provide.

    My alternative idea is to wire 2 9V batteries together and wire 55
    LEDs in series of 5 LEDs with a 39 ohm resistor with each series. This
    *should* draw 330 mA. Right?

    I think this will work, but want to verify it with you people that
    know what you're doing before I buy more 9V batteries and new
    resistors. I just want all the LEDs to be as bright as possible.

    Thanks for your help!
     
  2. forget about PP3 9V batteries, they are unsuitable.

    From the Duracell MN1604 datasheet:

    Rated capacity 580mAH.
    At just 25mA load the volts drop to about 7.5V in ten hours, and you
    want over 10 times this current


    martin
     
  3. Guest

    A great opportunity to do my 9V battery rant. Never design anything
    that needs a 9V battery. Use a DC/DC if you have to. OK, there are
    exceptions, notably if the current required is near the self-discharge
    of the 9V battery.

    Long live the AA cell.

    Back to the LED question, the experiment would be to power one series
    pair, then two, three, etc and watch the source impedance of the 9v
    battery become part of the equation.

    Back to the rant, two AA cells are about the same size as one 9V
    battery. AA cells are about 2.5AH. Now granted, alkalines drop their
    voltage linearly, but as a figure of merit, do this comparison:
    9x.58=5.22
    3*2.5=7.5
    so you get 40% more operating time. AA cells can easily be purchased
    for under 25 cents, while it is hard to get a 9V battery for less than
    a buck. So you get 40% more operating time at 1/4 the price.
     
  4. On 30 Mar 2007 14:04:29 -0700, in sci.electronics.design
    ARGHH wimp republican stuff

    checkout the F cell from GP batts
    GP1300FH

    monster stuff, but can't be charged in one hour, the cables will melt


    martin
     
  5. alokw

    alokw Guest

    Okay, so it sounds like everyone pretty much agrees that I shouldn't
    use 9V batteries. So... what would you all recommend? Should I just
    pick up some of those 4x AA battery holders and use those?

    Anybody have any idea what the mAh on a standard AA would be like? I
    mean - will they be able to drive 50+ LEDs at the required voltage and
    mA?

    Or... would it be better to go with something like the F cell from GP
    that martin recommended? I do need to keep costs as low as possible,
    but I'll do what's best.

    Thanks a lot for your replies, guys!
     
  6. Just found the F cell today, havent got a price, but it may be worth
    looking at 12V lead acid motor bike batteries.
    But you havent said how long you want the leds to stay on for, 1 hour,
    10 hours etc.

    zz need sleep, bye


    martin
     
  7. alokw

    alokw Guest

    Hey all!

    Thanks for the reply. These LEDs are for a theatre production. They
    have to stay on each night for about 1 minute and theres about 20
    performances. If I can power them for 45 minutes on one battery, I'll
    be happy.

    The LEDs are also installed inside hats, so there's not a lot of room
    to place a big lead acid battery. I'd like to stay smaller than around
    8 cubic inches or so (they're top hats).

    Thanks! Let me know if there's any other info that might be helpful. I
    really appreciate your help!
     
  8. Chuck Olson

    Chuck Olson Guest

    The Flashlight Manufacturers in China are doing this by paralleling 52 LEDs
    and driving them with three AA cells in series and not a single resistor
    anywhere (except, of course the internal resistance of the cells and the
    LEDs). But of course, they have control over the LED manufacturing process
    and can assure that all the LEDs have about the same forward voltage drop so
    that current hogging is minimized. I'm also amazed at the color temperature
    match they are able to achieve within each flashlight. Maybe this all comes
    about through automated testing and binning of raw LED manufacturing output,
    but for instance, a complete 32 LED flashlight with three AAA cells in
    beautiful anodized aluminum winds up costing the end user just $9.50
    (probably less if you buy it at the monthly Electronic Flea Market in
    Cupertino CA - - See this vendor's website
    http://www.ledwholesalers.com/store/index.php?act=viewCat&catId=20). It
    hardly pays to mess around with LEDs in a do-it-yourself effort any more.

    If you have an LED set that's not well controlled for forward voltage, then
    you're probably going to have to use resistors and higher voltage, probably
    losing some power in resistor dissipation. I'd go for the inexpensive
    finished flashlight instead. But if what you're planning involves a
    non-standard configuration, than I would still order a finished flashlight
    and just grab the matched LEDs out of it in order to avoid using resistors.

    Good luck

    Chuck
     
  9. alokw

    alokw Guest

    Thanks for the reply.

    Unfortunately, this is kind of a custom rig, and we do require the use
    of LEDs that we purchased from a special retailer. They are super
    bright 12000 mcd white LEDs that run at a typical voltage of 3.4 V and
    no more than 30 mAh. We have 9 hats and each hat requires somewhere
    between 35 and 65 of these LEDs.

    I have another question. I did some research and found these lithium
    9V batteries at Radios Shack, which advertise 2000 mAh. So I picked
    one up, and hooked it up to the configuration that I described in my
    original post (54 LEDs, 27 pairs w/ 82 ohm resistors on each pair).
    Only about 10 of the LEDs were at their full intensity!

    I was baffled by this, because I figured, if the LEDs were pulling 800
    mA, the 9V battery should be able to supply it in full capacity,
    right? It made sense that my normal alkaline batteries dimmed because
    they pushed under 600 mA, but these lithium ones were pushing the full
    load in theory. Am I totally misunderstood on the concept of
    electricity? Can anyone explain this to me? It seems like I'm
    providing the LEDs with more than enough power, but they're still
    dimmed - I'm sure it just comes down to the fact that I don't know
    what I'm doing - but I'm going to keep trying anyways - with your help
    of course!

    Thanks again for your help!
     
  10. Joel Kolstad

    Joel Kolstad Guest

    I think you mean "mA", not "mAh" -- the first (milliamp) is a measure of
    current, the second (milliamp-hour) is current integrated over (multipled by)
    time.
    Do you have a multimeter? Have you measured the current going through each
    LED string? Depending on the battery itself, it may or may not have been able
    to deliver 800mA. A cheapy carbon-zinc 9V generally can't (without
    significant voltage drop), whereas a NiCad can -- I'm not sure about lithium
    cells.

    Note that the battery being 2000mAh has nothing whatsoever to do with whether
    or not you can (readily) draw 800mA from it. What it *does* tell you is that,
    assuming you *do* draw 800mA from it, it'll last approximately
    2000mAh/800mA=2.5 hours.
     
  11. DJ Delorie

    DJ Delorie Guest

    mAh != mA

    A battery's milli-amp-hour rating is a measure of total energy, not
    the maximum current. For example, a 2000 mAh 9v battery might only be
    able to push 20mA, but for 100 hours (20 mA * 100 h = 2000 mAh) (for
    example). A "D" cell with the same capacity (2000mAh) might be able
    to push 2000 mA for 1 h (2000 mA * 1 h = 2000mAh).

    What you want is *peak* current. Bigger batteries offer more peak
    current, with 9v being the least (no matter what the mAh is) and D
    cells being the most (of the corner-store type batteries). Mostly
    this is due to internal resistance; the resistance of the battery
    itself is acting like an additional resistor in series with your LEDs,
    reducing the current.
     
  12. alokw

    alokw Guest

    Thanks again for all of your replies!

    That makes a lot of sense. I did indeed mean that each of the LEDs
    require 3.4 mA, not mAh. I think you're also right about the 9V not
    being able to provide enough "peak current" (look at how much I'm
    learning!).

    After some more research I came up with this: What if I use a total of
    (16) AAA batteries in each hat? This would provide around 24V and I
    could wire 56 LEDs in series of 7 LEDs with a 6.8 ohm resistor with
    each series. This *should* use 240 mA. Assuming the AAAs give 1250 mAh
    (from Energizer's site), they should last 5 hours. This is all under
    the assumption that they will be able to provide at least 240 mA of
    peak current.

    So, does that sound like a smart way of going about this project? I
    can get all of the batteries and holders for them for under $80, so
    the price isn't terrible. Also, how do I know what peak current a
    battery can supply? Can i be sure that the AAAs will provide at least
    240 mA?

    If I'm still being an ignorant idiot, just say so. I just have to get
    this done quick, right, and at a reasonable price - and all those LEDs
    have to light up real bright.

    Thanks so much for your help!
     
  13. Guest

    I know nicads are very low impedance. NiMH is a bit higher, but still
    better than alkaline cells. You would have to check, but I recall a
    nicad AA cell is around 30mOhms. The advantage to nicads is that while
    the initial voltage is lower (1.25 versus 1.5V), a nicad hangs in
    around 1.2V until it is exhausted, then it dies quickly. So not only
    is the impedance lower, but the voltage will be better regulated. I
    don't think I ever pulled more than 200ma from AA cells, so I don't
    know how well your circuit will react, but give it a shot.

    BTW, 30ma sounds high for continuous operation (i.e. non-pulsed).
     
  14. ehsjr

    ehsjr Guest

    5 or 10 alkiline cells. The better solution is 10
    cells. Both are described below. First, the 5 cell:

    5 alkaline AA 7.5 volts - enough for 28 strings of 2
    LEDs at 6.8 plus a 24 ohm resistor per string. With 28
    strings at ~29 mA per string, that's a total of ~816.6 mA.
    If you can't add an LED to get 56 LEDs, your last string
    would use a ~137 ohm resistor (made with a 1500 ohm in
    parallel with a 150 ohm). Energizer AA is rated at 2850
    mAh under light load, and 1500 mAh at 500 mA discharge.
    Since you are almost double that, figure ~ 750 mAh and
    you need ~ 570 mAh for 45 minutes.

    Give yourself a 50% margin - replace the cells after
    10 performances, and you should be good for 20 performances
    with room to spare. If you consider the 5 cells together as
    1 battery, it meets your "happy criterion", but why chance it?
    Buy 10 cells and be safe.

    The 10 cell solution: If you have room for 10 cells, you
    can reduce the number of strings to 14, using 4 LEDs and
    a 40 ohm resistor per string. If your last string has to
    have 3 leds instead of 4, it will need a 160 ohm resistor.
    Current draw in that configuration will total 420 mA, and
    your cells are rated at about 1500 mAh with a 500 mA draw,
    so you're golden for 45 minutes of run time across the 20
    performances.

    The performance numbers for the Energizer are found here:
    http://data.energizer.com/PDFs/E91.pdf

    Ed
     
  15. ehsjr

    ehsjr Guest

    ^^ 47

    Should be 47

    Ed
     
  16. ehsjr

    ehsjr Guest

    They'll last *way* less than an hour. As the
    discharge rate goes up, the mAh rate goes down.
    At 25 mA discharge, they'll provide 1250 mAh.
    But at 240 mA discharge, the mAh rating looks like
    it's about 500 mAh. And the voltage decreases
    to .8 in those ratings. With 7 LEDs in series,
    you can't really afford any drop in the cells. You
    have only .2 volts above the 23.8 the LEDs need -
    that's only .0125 volts per cell that you can
    afford to drop, and cell voltage drops rapidly.
    You are neither ignorant nor an idiot. To be ignorant,
    you need to ignore something you know - and this is not
    your area of expertise, so you are simply unfamiliar
    with it. And far from being an idiot, you are checking
    first to look for the pitfalls, rather than charging
    forward blindly, as an idiot might do.

    Love to see a picture of one of the top hats when
    you're done!

    Ed
     
  17. Werty

    Werty Guest

    Harbor Fright 18vdc , $10 on special
    but get the 1.8ah , not the smaller one .

    White LEDs are .02 amps , for highest
    eff' . DONT RUN AT .03 amps .

    LM431 and some parts to make a flyback
    to drive any voltage you need .
    Flybacks need low inductance , or the
    frequencey is too low .
    Start with less than .001 Henry , more than
    10 microHen'

    I just saw a nice MOSFET to drive it ..

    only $.10 each .
    PMN45EN , Id=1.8 amps @ Vgs =2.5 v .
    about .05 Rds .

    BTW , anyone into ARM s/w ?
     
  18. alokw

    alokw Guest

    Hi everyone, Thanks again for all of your very helpful replies!

    I looked over everything you all said and did some thinking. I like
    Ed's (10) AA battery solution and I'm pretty sure we have the space
    allowance in the hats for them. However, before I commit to purchasing
    the battery holders, resistors, batteries, and such, I want to be sure
    that the AA batteries will be able to push 500 mA of consistent
    current - I can't have these guys going dim on me. I wasn't able to
    find the maximum peak current on any battery manufacturer's web sites.
    Do you all know for a fact that 10 AAs can push at least 500 mA?

    In response to Werty's post, I checked out Harbor Freights web site,
    but the only 18vdc batteries I found were drill batteries. If these
    are what you were referring to, they definately won't work, just due
    to their size and weight. I've listed the exact specs of the LEDs we
    have below.

    Emitted Colour : WHITE
    Size (mm) : 5mm T1 3/4
    Lens Colour : Water Clear
    Peak Wave Length (nm) : N/A
    Forward Voltage (V) : 3.2 ~ 3.8
    Typical Voltage (V) : 3.4
    Reverse Current (uA) : <=30
    Luminous Intensity Typ Iv (mcd) : Average in 13000
    Life Rating : 100,000 Hours
    Viewing Angle : ±10°
    Absolute Maximum Ratings (Ta=25°C)
    Max Power Dissipation : 80mw
    Max Continuous Forward Current : 30mA
    Max Peak Forward Current : 75mA
    Reverse Voltage : 5~6V
    Lead Soldering Temperature : 240°C (<5Sec)
    Operating Temperature Range : -25°C ~ +85°C
    Preservative Temperature Range : -30°C ~ +100°C

    Should I really try to run these at 20 mA as opposed to the 30 mA that
    I've been calculating for?
    I also didn't quite understand the other solutions you suggested. I
    googled LM431 and found Shunt Regulators, but I'm not sure how I would
    go about using one of these. I think the whole shunt and MOSFET idea
    is a little beyond me. If there's a really good tutorial of how to use
    these online somewhere, it might be more feasible, but given my
    current knowledge level, I'd be more comfortable sticking with using
    LEDs, resistors, and batteries.

    Thanks again for all of your helpful replies and your knowledge! I
    couldn't do this without your help!
     
  19. Guest

    No no no no no! Death to the AAA battery. It is nearly as bad as the
    9V battery in terms of economics. AA cells are your biggest bang for
    the buck. The maH capability of a battery is a function of volume. The
    AA and AAA cells are the same length, but the AA cell has a larger
    diameter. The volume increases as the square of the radius, so a
    little more radius means a lot more capacity.

    The resistor represents wasted power, so I can see why you are going
    for more cells in series, but stick to AA cells. I still think you
    should try AA nicad cells. Since you have the LEDs on hand, it will be
    a simple experiment.

    http://www.batterystore.com/Sanyo/SanyoPDF/KR1100AAU.pdf
    These Sanyo AA cells are only 19mOhm.

    There are discharge curves on the datasheet. I'm not sure what the
    "lt" or "It" unit means, but I think the 1It discharge curve is at
    1100ma. Often in batteries, you refer to the current as a function of
    C, the cell capacity. That is, If you charge 1100ma for 1 hour, you
    have a capacity of 1100maH. [Technically this is not correct as the
    efficiency of charging isn't 100%. You can see the charge conditions
    are 110ma for 16 hours, not 10hours, as would be the case for 100%
    efficiency.) Thus I'm guessing the 1It rate is 1100ma. The battery
    holds up for about 50minutes at this rate.
     
  20. alokw

    alokw Guest

    Hi everyone,

    Today, I took some of your suggestions, and decided on trying out a
    (8) AA version for a hat requiring only 24 LEDs.
    I wired 8 series of 3 LEDs with the same 82 ohm resistors I already
    had, which theoretically allowed 25 mA through the LEDs (I compromised
    between the 30 mA I was using, and the 20 mA that was suggested). I
    used 8 AA batteries because I was able to buy a prebuilt 8 AA holder
    from Radio Shack. At 25 mA, in total it used only 200 mA.

    Overall, I was pleased with the results. I posted a picture here:
    http://www.alokw.com/hats/hats.jpg
    The only thing that might become a problem is the weight of 8 AA
    batteries. It's a bit heavy and may hinder the actors movement with
    the hats. I'm going to run this by the costume folk and see what they
    think.

    Now, assuming the costume department will be okay with the weight,
    would I be safe to use 8 AA batteries to run as many as 66 LEDs? That
    *should* draw 550 mA at 25 mA per LED. Is this okay? Also, should I
    keep using resistors to keep the current at 25 mA? or should I stick
    with 20? or 30?

    Thanks for all of your help again guys. I wouldn't be able to do it
    without you all.

    I have another unrelated question. I hooked up my micrometer to the
    leads on the 8 AA battery holder to check how much current it was
    pushing. I set it to 10A and the display read somewhere around 1.200.
    Does that mean it pushes 1200 mA? The snaps on the battery holder also
    started smoking. Does that mean I was drawing too much current for the
    battery holder to handle? or the snaps? or something else?
    Also, I hooked up each of the micrometer leads to an end on one of the
    LEDs while it was turned on to try to check how much current it was
    actually getting. I assume I did something wrong here, because the
    display said 0.05. How would I go about checking this correctly? Or
    did I do something else wrong?

    Thanks so much everyone!
     
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