Need help with a battery powered IR LED Array

Discussion in 'The Projects Forum' started by Jumper, Nov 26, 2009.

  1. Jumper

    Thread Starter New Member

    Nov 25, 2009
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    Hi AAC -

    I stumbled across this site while doing some research for a project and I am very impressed with the friendly help I have seen. I've searched a lot of threads and I've found enough info to be dangerous as I am just a hobbyist with a good grasp of the fundamentals of DC but only the VERY basics of electronics. I can barely read a schematic... However I have assembled a few velleman kits and I'm handy with a soldering iron.

    I have a simple project really, but what I've read has made me aware that there may be a much better way to do what I want than what I have in mind.

    I need to build 2, 15 IRLED array's and power them with a 12V battery.

    I have ordered 30 of these http://www.goldmine-elec-products.com/prodinfo.asp?number=G2318B (not the 1000 pack, just 30)

    and 30 of these http://www.goldmine-elec-products.com/prodinfo.asp?number=G13661


    I wasn't sure which one would be best and they were cheap so I figured better to pay shipping once rather than twice.

    Based on what I've read here, it looks like my array will consist of 4 series strings of 4 LED's with resistors for current limiting. I'll add a standard green LED to be the 16'th LED and it will light to give a visual indication of power. Seems pretty simple. Too simple in fact. What am I missing?

    Also, I would like to add a red LED that will illuminate when the battery voltage drops below ~11.5. I have no idea how to make this happen.

    Thanks for any help.

    Jumper
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    Hi Jumper,
    Unfortunately, the IR LEDs you ordered have no real specifications. Other than one was for Zenith remote controls and is very dark. The other is 5mm and light gray, and "operates from 2.8v" is mentioned, but the all important "typical Vf @ current" has been omitted.

    Since there is no specification for Vf @ current, you won't be able to start planning on how many to put in an array until you know what the Vf @ current actually IS.

    So, when you get them, you'll have to test a number of them.

    Since you do not know what their current specification is, I suggest that you use 20mA.

    Then you will have to measure several of them to find out what their typical Vf is when 20mA current is run through them.

    One easy way to make a constant current regulator is by using an LM317 adjustable voltage regulator and a resistor. You can order LM317's online from places like Digikey, Mouser, and many other suppliers - or your local Radio Shack probably carries them.

    To use an LM317 as a current regulator to test an LED:
    1) Connect a resistor from the OUT terminal to the ADJ terminal.
    2) Connect a positive voltage source (positive battery terminal) to the IN terminal. You will need in the range of 7 to 15v.
    3) Connect the anode of the LED (longer lead) to the ADJ terminal of the LM317.
    4) Connect the cathode of the LED (shorter lead) to the voltage source return (negative battery terminal).

    Calculating the resistor value:
    R = 1.25/Desired Current, where 10mA <= Desired Current <= 1.5A
    Since you want 20mA current, then:
    R = 1.25/20mA = 1.25/0.02 = 62.5 Ohms.
    62 Ohms is a standard E24 value of resistance.
    A table of standard resistor values is here: http://www.logwell.com/tech/components/resistor_values.html
    Bookmark that page, you will refer to it often.

    If you don't happen to have a 62 Ohm resistor available, but have several other resistors, you could use two or more in series and/or parallel to add up to 62.5 Ohms, but 62 Ohms will be close enough.
    A series/parallel resistor calculator is here:
    http://www.qsl.net/in3otd/parallr.html#
     
    Last edited: Nov 27, 2009
  3. Jumper

    Thread Starter New Member

    Nov 25, 2009
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    0
    Ha! I knew there was more to it than I thought! Thanks SgtWookie for the instructions. I'll test the LED's when they arrive and see what I've got.
     
  4. Jumper

    Thread Starter New Member

    Nov 25, 2009
    16
    0
    Well, my local Radio Shack didn't have the LM317. There is another way to do it isn't there? I used my bench power supply, multi meter, LED and resistors. The known values are 2.76vDC from supply under load, 65.1 ohm resistor (had to use two resistors in series to get close to 64 ohms). The voltage measured across the IR LED anode and cathode under load dropped to 2.25vdc. At that voltage the measured current was 7milliamps.

    Does this make my Vf ~.5v?
     
  5. Jumper

    Thread Starter New Member

    Nov 25, 2009
    16
    0
    I did the same test with the other IRLED type I ordered. I'll call this LED type "B". Same supply voltage and resistance. This one measured 1.22vdc across the anode and cathode. And it drew 19.5milliamps. Big difference....
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Yes. Known resistance, known voltage.
    I suggest using a 12v supply and a 560 Ohm resistor in series with the LED.
    12v across a 560 Ohm resistor will allow 21.4mA current.
    10v across a 560 Ohm resistor will allow 17.8mA current.

    An ATX form-factor computer power supply has +5v and +12v available on the 4-pin MOLEX connectors. You will get +5V between the red and black wires, +12V between the yellow and black wires, and if you use the yellow as +, red as -, you will see 7v difference between the two.

    No, that would make your Vf=2.25v at 7mA.

    It would be easier if you used a higher voltage and higher resistance.
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    Yep, that's a big difference. Much more realistic value for Vf @ current.
     
  8. Jumper

    Thread Starter New Member

    Nov 25, 2009
    16
    0
    OK as SGTWookie suggested I measured them again at a higher voltage. I used 10V, a 550Ω 1/4w resistor in series with the IR LED.

    I measured 5 of the first type LED and I got 2.31V across the anode/cathode and a current of 13.6mA. They all measured very close to each other. It looks like I have a forward voltage of 2.31v????

    I also measured 5 of the type "B" IR leds with 10V and 550Ω. I got 1.18v across the anode/cathode and 15.64mA. Again, they all measured very close to each other. And this LED has a forward voltage of 1.18v???

    So if I use the type "B" LED's and a 12v supply I could run 10 in series and limit the current with a ~550Ω 1/4w resistor? I think there is a safer way that won't risk my 10 LED string if one burns out. Isnt there?
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    They have Vf=2.31v @ 13.6mA.

    Your supply was actually measuring 9.79v under load.
    You want to know what your LED's Vf will be with 20mA flowing through it.

    For the moment, assume a Vf of 2.4v, and your Vsupply to be 9.79v.
    Rlimit = (9.8-2.4)/20mA
    Rlimit = 7.4/0.02
    Rlimit = 370 Ohms.
    Try measuring Vf again with as close as you can get to a 370 Ohm resistor.
    Hint:
    270+100 = 370
    220+150 = 370
    330+39 = 369

    They have a Vf of 1.18v @ 15.64mA.
    Your 10v supply sagged to 9.782v with the 15.64vmA load.
    Try testing these LEDs again with a new resistor:
    Rlimit = (9.78 - 1.2) / 20mA
    Rlimit = 8.58/0.02
    Rlimit = 429 Ohms
    430 Ohms is a standard value; close enough.

    No, that's not how it works.

    MaxLEDs = INT((Vsupply - 1)/Vf_LED)
    MaxLEDs = INT((12v - 1)/1.2)
    MaxLEDs = INT(11/1.2)
    MaxLEDs = INT(9.1666...)
    MaxLEDs = 9
    Rlimit >= (Vsupply - Total(VfLED)) /DesiredCurrent
    Rlimit >= (12 - (9 x 1.2)) / 20mA
    Rlimit >= (12 - (10.8)) / 0.02
    Rlimit >= 1.2 / 0.02
    Rlimit >= 60
    60 is not a standard value of resistance.
    Table: http://www.logwell.com/tech/components/resistor_values.html
    62 is the closest standard E24 value.
    1.2/62 Ohms = 19.35mA


    If one burns out, there's no problem - the entire string stops emitting light.

    If one shorts out, that could be a problem. The rest will get more current.
     
  10. Jumper

    Thread Starter New Member

    Nov 25, 2009
    16
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    Thanks SgtWookie! I'll draw up a diagram, post it and go from there?

    One other question: What about a parrallel circuit? Would each LED have to have its own resistor? Does one type of circuit have an advantage over the other in this type of application?
     
  11. tkng211

    Well-Known Member

    Jan 4, 2008
    65
    2
    You can try to use the attached circuit as the low voltage detector. If you don't want to use 2 LEDs, you may replace either LED by 2 small signal diodes such as 1N4148 connected in series.
     
    Last edited: Dec 6, 2009
  12. SgtWookie

    Expert

    Jul 17, 2007
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    You need to do the 2nd Vf test on the "A" and "B" LEDs, using the values of resistance I suggested for each type.
    Record
    1) The exact value of resistance used
    2) The exact voltage of the supply when under a 20mA load
    3) The Vf of at least 5 of each type LED.

    Each series string of LEDs, whether 1 or 9, needs it's own resistor.
    You can wire as many such strings of LEDs in parallel as your power supply will support.

    A single LED across a voltage supply requires a current limiting resistor.
    Several LEDs in series across a voltage supply requires a current limiting resistor.
    Multiple strings of LEDs in parallel require at least 1 current limiting resistor in series with each string.

    Having many LEDs in series is efficient, as most of the power dissipation occurs in the LEDs rather than in the current limiting resistors.

    Having a single LED with a single current limiting resistor across a medium voltage source (like 12v) would be inefficient, as the current limiting resistor would need to drop most of the voltage, thereby dissipating most of the power.

    Note that using many LEDs in series requires the power supply to have good voltage regulation. If your power supply is not well-regulated, you will need to use fewer LEDs in series.
     
  13. Audioguru

    New Member

    Dec 20, 2007
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    IR LEDs have a forward voltage drop of 1.0V to 1.5V.
    So the ones you have with a 2.3V forward voltage must have two IR LEDs in series inside.
     
  14. Jumper

    Thread Starter New Member

    Nov 25, 2009
    16
    0
    That would make a lot of sense because they have a short third electrode between the anode and cathode. And they were advertised as a "super" LED.

    SGT Wookie - I sampled 6 of each of the LED's to get the values that I posted. Essentially they were all the same with very little variation between them. I'm confident the "B" LED's have a Vf of 1.18 tested at 550Ω and 10v with a current draw of 15.64mA.

    tkng211 - Thanks for the battery monitoring circuit. I will probably incorporate it into the project last.

    I'll post a pic of the diagram here shortly.
     
  15. SgtWookie

    Expert

    Jul 17, 2007
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    You only have a Vf at a low current.

    If you put more current through them, the Vf will increase.

    That is why I asked you to re-test them with different value current limiting resistors.
     
  16. Jumper

    Thread Starter New Member

    Nov 25, 2009
    16
    0
    Hmm. Sorry must have missed that.

    I re-measured LED "B" at 10V (under load) 370Ω resistor (variable type, set with Multimeter). The voltage measured 1.19 across the anode-cathode and the current draw was a measured 23.7mA.
     
  17. SgtWookie

    Expert

    Jul 17, 2007
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    OK, good. So, use 1.19v @ 20mA for your calculations.

    Apparently, you want to power this array from a 12v lead-acid battery, right?

    What is the AH rating of the battery?

    What does it's voltage measure after it's been charged up and sitting for awhile?

    Are you planning on connecting it to a solar array for charging?

    You should plan on not discharging the battery below about 30% from full charge, or it will have a short life.
     
  18. Jumper

    Thread Starter New Member

    Nov 25, 2009
    16
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    It is a 7Amp/hr sealed lead acid battery. It has a rated standby voltage of 13.5 - 13.8. I do not plan on using it with a solar array, I'll simply charge it off a wall charger.
     
  19. Jumper

    Thread Starter New Member

    Nov 25, 2009
    16
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    OK using a Vsupply-1 = 12v, and 7 IR LED's with a Vf of 1.19 ea, for a total Vf of 8.3v for each series circuit I calculate I need a resistor >= 183.5Ω for each 7 LED circuit. This is for 20mA for each LED.

    I measured the battery voltage after it had been sitting for awhile off the charger and it tested 13.37v. Assuming this voltage is my battery's nominal value, is this resistor calculation safe?
     
    Last edited: Dec 7, 2009
  20. SgtWookie

    Expert

    Jul 17, 2007
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    I've had a pretty busy day today.

    OK, if your battery is fully charged, and just came off the charger, it'll probably have a "float" charge of around 13.8v. That won't last for long, though. Still, you need to plan for it.

    At a normal full charge, it'll read about 12.8v-12.9v.
    When it gets down to around 30% discharged, it will measure around 12v to 12.2v.

    So, you have a total range of about 12v to 13.8v.

    If you want to run a bunch of LEDs in series and still be pretty efficient, you could build a switching "boost-type" DC-DC converter. These are not terribly hard to build, but it will require a number of components. You will also need a "battery low" detector circuit, that will turn off the boost DC-DC converter when the battery gets low.

    Or, you could just go the really simple route, use a single resistor per string, and not worry about using the switching boost DC-DC converter or turning off the battery. If you do it the latter way, your light output will vary a fair amount between a freshly charged battery and a 30% discharged battery.

    But, I don't know how skilled you are at building circuits. There won't be a lot of components - maybe about a dozen - but if you're a complete n00b at electronics, it'll be a real challenge.

    You tell me if you think you might be up to a challenge or not.
     
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