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Help with Jumbotron

Discussion in 'General Electronics Discussion' started by thefstopjedi, Jul 25, 2013.

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  1. thefstopjedi

    thefstopjedi

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    Jul 25, 2013
    In short, I am building a very detailed rod hockey table from scratch and Im trying to build a replica jumbotron, and I need some help lighting it with SMD LEDs.

    Im inquiring about this product:
    http://www.superbrightleds.com/more...d-120-degree-viewing-angle-6000-mcd/316/1248/

    I am wondering if anyone can help me understand how I would input the correct information for this product into a resistor calculator such as the one below in order to get the correct resistors for my project?

    Id like to know what to input for BOTH in order to wire in Series as well as in Parallel (as i understand they may be different), so i might get a better idea as to what other products I might need to purchase.

    Risistor Calculator:
    http://www.hobby-hour.com/electronics/ledcalc.php

    Also, will i need a specific kind of resistor for SMD led's?


    In case it helps:
    Id like to test a series of 3 of these to see how and where to place them first maybe using a 9v bat as a source.

    and the finished product will include 25 SMD's lighting the screens,signs, and ice surface.

    and I can vary the power source for what ever is needed/recommended.
     
  2. BobK

    BobK

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    1,688
    Jan 5, 2010
    These appear to be 3 separate LEDs in a single package. If connected in parallel you would use 60mA and 3.2V, if connected in series you would use 20mA and 9.6V in the calculator.

    Bob
     
  3. thefstopjedi

    thefstopjedi

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    Jul 25, 2013
    Bob,
    Thank you, that is a big help. However, one question, when imputing this into the calculator, (because they are 3 to each unit) would I (in the case of 3 units) select 3 or 9 leds?
     
  4. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    How are you planning on connecting them? If the calculator is designed to do series and parallel connections, just treat each individual LED as 3.2V 20mA. Each LED, not each chip.

    Bob
     
  5. thefstopjedi

    thefstopjedi

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    Jul 25, 2013
    What ever way works. the way the structure is designed I can go either way. it just really depends on power as one way does not save space over the other.
     
  6. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    May 8, 2012
    I didn't see anything in the data sheet that even hinted that all three LEDs are matched. So each LED requires it's own dedicated limiting resistor. I mention this because parallel wiring will take up more real estate because of this.

    Chris
     
  7. thefstopjedi

    thefstopjedi

    59
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    Jul 25, 2013
    So if I was to wire it in series. how do you suggest I do this? and with what resistor(s)?
     
  8. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    4,960
    652
    May 8, 2012
    Use the calculator as stated by Bob. Wire the LEDs from the Cathode of one LED to Anode of the next LED. The resistor will be 1 per Jumbotron and can be placed anywhere in the series string.

    Chris

    EDIT: Your supply voltage must be higher than 3.2V * 3 = (9.6V).
     
    Last edited: Jul 26, 2013
  9. thefstopjedi

    thefstopjedi

    59
    2
    Jul 25, 2013
    Ok, thank you
    I checked a few dozen online calculators useing:

    12v power
    3.2
    20ma
    60 led's (20 chips)

    however, most all cap me at 32 led's. IF Im to count the leds individually with 20 of those chips, ill need to power 60.

    only one allowed me to input for 60, but said i did not need a resistor.

    any thoughts on what to do/use/change?
     
  10. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    4,960
    652
    May 8, 2012
    Are you sure that you understood what Bob and I posted?

    Chris
     
  11. thefstopjedi

    thefstopjedi

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    Jul 25, 2013
    I thought I did. what am i missing?
     
  12. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    4,960
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    May 8, 2012
    Then your limiting resistor for each chip (3 LEDs in series) will be 120 Ohms. You will need 20 of them.

    Chris
     
  13. thefstopjedi

    thefstopjedi

    59
    2
    Jul 25, 2013
    I thought I did, What am i missing?
     
  14. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    4,960
    652
    May 8, 2012
    If this is true you can string 2 chips (6 LEDs) in series and use a 24V source. The limiting resistor would be 240 Ohms per string. This would cut your current demand and resistor count by 50%

    Chris
     
    Last edited: Jul 27, 2013
  15. thefstopjedi

    thefstopjedi

    59
    2
    Jul 25, 2013
    Oh ok, so, stringing all 20 chips together in a series, one 120 resistor before each chip. and thats assuming a 12v source?
     
  16. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    4,960
    652
    May 8, 2012
    That's not what I said!!

    Chris
     
  17. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,271
    Nov 28, 2011
    You mentioned a 9V power source. If you were hoping to use a 9V "PP3" battery (https://en.wikipedia.org/wiki/Nine-volt_battery) to power 25 LEDs, be aware that you'll probably only get a few minutes of operation before the battery goes flat.

    Each LED component contains three actual LED elements. Each element has a forward voltage of 3.2V +/- 0.2V and a recommended maximum continuous current of 20 mA.

    If you connect the three elements in series, the forward voltages add together, but the currents don't. So you end up with a device with a forward voltage of 9.6V +/- 0.6V that needs 20 mA current flowing in it.

    A 9V supply doesn't have enough voltage. A 12V supply does; the series resistor (that limits the current through the three LED elements) will have 2.4V +/- 0.6V across it. Its value is calculated from a rearrangement of Ohm's Law: R = V / I where V is the voltage across the resistor, in volts, and I is the current, in amps. In this case, V=2.4 and I=0.02 so R=120 ohms.

    So in this setup, the three LED elements in each LED component are connected in series, and each LED component needs one 120 ohm current limiting resistor. The circuits are all connected (in parallel) across a 12V supply. This seems like a tidy arrangement. Each circuit draws 20 mA, so for 25 LED components, the total current is 500 mA. You can easily find a 12V, 1/2A power supply. (Use one rated for 1A, for safety.)

    That's a reasonable design, and it should work well for you.

    One factor that MAY be significant is the variation in LED forward voltage. If you're concerned about this, keep reading.

    I calculated the 120 ohm current limiting resistor assuming a forward voltage of 9.6V per LED component, but the actual total forward voltage (according to the specs) could be between 9.0V and 10.2V. This means the remaining voltage across the resistor (assuming a 12V supply) could range from 1.8V to 3.0V. This will affect the current; you can calculate this using Ohm's Law again: I = V / R where V is the voltage across the resistor, in volts, and R is its resistance in ohms.

    With those values, the current could range from 15 mA to 25 mA. It's very unlikely to vary that much, and that variation isn't even very extreme. But if you want, you can improve things by increasing the power supply voltage and increasing the current limiting resistors. The more voltage you drop across the resistor, the less the current will be affected by variations in the forward voltages of the LEDs.

    For example, let's increase the power supply to 15V and recalculate the resistor. R = V / I where V = (15 - 9.6) and I = 0.02 gives R = 270 ohms. Now a total LED forward voltage between 9.0 and 10.2V gives a current variation from 17.777 mA to 22.222 mA - less variation.

    Edit: I'm aware that those forward voltage limits are specified at 20 mA and the actual current will not be 20 mA if the forward voltage is different. I'm using 3.0V and 3.4V per element, and pretending that the current is always 20 mA. This isn't totally accurate but it is close enough.

    I think you would be fine with a 12V supply and 120 ohm resistors. You might want to try a few LEDs with those values and see how much voltage variation you actually get.
     
    Last edited: Jul 27, 2013
  18. thefstopjedi

    thefstopjedi

    59
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    Jul 25, 2013
    Thank you Krisbluenz,

    I think what I was confused by is perhaps the internal wiring of the component that i did not take into consideration.
    Please correct me if I mis-understood what you are suggesting.
    But I should be connecting each if the 25 components together in series with one 120ohm resistor before each?

    Im attaching an image of the Frame work, where ill be attaching the leds. there is a ledge where they will shine down illuminating screens, using 4 for each each side, with the remaining leds on both top and bottom. for the test I wired 3 to a 9v. Is this basic setup more or less going to work for all of them? or should I be looking to break them into groups or wire each of them separately?
     

    Attached Files:

  19. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,271
    Nov 28, 2011
    Each component has six leads and contains three LED elements. You need to connect these three elements in series, with the cathode of one to the anode of the next, using two slightly diagonal tracks. This will form a single functional LED with a forward voltage of 9.6V nominal (voltages in series add together) and an operating current of 20 mA (current in all parts of a series circuit is the same).

    Each component needs a series resistor to limit the current, and to make up the remaining voltage between the total LED forward voltage and the power supply voltage. It doesn't matter where in the series circuit this resistor is placed. The anode end of the series string of LED elements goes to the positive side of the power supply.

    Each of these series circuits is self-contained and will light when connected across the supply voltage. If you have 25 LED components (with 25 resistors), these circuits can all be connected across the same power supply. Each circuit will draw 20 mA. Currents in parallel add together, so your total current drain on the power supply will be 500 mA.
     
  20. thefstopjedi

    thefstopjedi

    59
    2
    Jul 25, 2013
    KrisBlueNZ,
    Ok, that will make this a bit more difficult to wire. Is what I ruffly have sketched out about what your talking about? (sorry Im a very visual person who cant draw) :)

    And when you say that the resistor can be anywhere in the series, are you speaking with in the leads on each chip, or is it possible to connect them in between each chip as pictured?
    I ask because it will change the depth of its placement.

    And I guess secondary, is there a good way to hook these up where I can just hook up all the anode and all the cathode leads to make 2 simple leads while using the same 12v power supply? As these things are really small.
     

    Attached Files:

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