Making a 32x24 LED matrix and need some assistance with the logistics and schematic

Discussion in 'The Projects Forum' started by Inglorious89, Aug 3, 2016.

  1. Inglorious89

    Thread Starter New Member

    Feb 12, 2016
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    Hello everyone.

    I'm in the planning phase for a project. Currently the plan is for a 32x24 LED matrix that's only going to show one image: an American flag. I'm thinking the size is going to be about the same size as a piece of paper (8.5x11) and will be 32 LEDs long and 24 LEDs tall. For the schematic, my first idea is just to have a resister with 32 LEDs in parallel going to the power supply. This design will be repeated 24 times to give me the full schematic. My first question is this: will such simple circuit work? Will I have issues with the LEDs on the far end not lighting up completely/at all, and if so then how can I correct for this? My second questions concerns the current draw; I'll be using 768 LEDs for this if it's going to be 32x24, and with each LED taking about 20ma, that'll be 15.26A. I know there's a lot of LEDs, but 16A seems like too much current for a bunch of LEDs to draw. Is there something fundamental about current draw that I'm not understanding that will cause the draw to be less severe, or will I really need a 16A power supply?

    And help or advice will be appreciated, thanks!
     
  2. cmartinez

    AAC Fanatic!

    Jan 17, 2007
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    Exactly what type of LEDs are you planning to use, can you post a datasheet?
    I'm not sure I'm following, are you planning to use a single resistor for those 32 LEDs in parallel (bad idea)? Or are you going to use one resistor per LED (good idea)?
    A schematic of what you're planning, however simple, would be of much help here. Even if it's drawn by hand and posted as a pic.
     
  3. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
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    If your power supply has enough voltage to drive several
    LEDs in series, then you can reduce the total current draw.

    Let's consider one row of 32 LEDs, each with a resistor. If the LEDs draw 20mA, then this one row requires 640 mA.

    Now consider a power supply with four times the voltage. It could light 4 LEDs in series with a resistor and still only draw 20 mA. You'd need 8 such strings to make up a row of 32 LEDs, and hence this time your row would only draw 160 mA.

    The resistor value needed would depend on the final configuration, the specifications of the LED, and what you had as a power supply.

    Note that in this simple example, your 16 A draw would drop to 4 A.
     
  4. bertz

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    Nov 11, 2013
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    Why design for a 20 mA draw when 10 mA will light the LEDs quite nicely. I assume you will be using standard T-1 3/4 LEDs.
    You don't even discuss what voltage you are considering for your power supply.
    Have you experimented with individual LEDs to determine what current level will give the desired brightness?
    Here's a clue:
    Blue LEDs have a Vf of 3.5 volts
    White LEDs have a Vf of 3.3 volts
    Red LEDs have a Vf of 2.0 volts
    Remember your basic laws of electricity. The current at all points in a series circuit is the same. So, if you have a power supply of 15 volts,
    you can string 4 blue or white LEDs in a series without a current limiting resistor and the current in the string will be around 12 mA.
    By combining series and parallel circuits you can keep the current requirements to a minimum.
    You get the idea mate?
     
  5. cmartinez

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    Jan 17, 2007
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    That was my line of thought too. But I was waiting for him to reach that conclusion on his own.
     
  6. Inglorious89

    Thread Starter New Member

    Feb 12, 2016
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    Here's the datasheet of the LEDs for I'm considering: http://www.semicon.panasonic.co.jp/ds4/LN21RPX_E.pdf
    And yes, my original plan was to use one resistor for the 32 LEDs. In theory, shouldn't I only need one resistor?
    And attached is the rough idea of my schematic.

    Appreciate the response and any feedback.
     
  7. Inglorious89

    Thread Starter New Member

    Feb 12, 2016
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    Apologies for not giving the voltage. My current thinking is 5V, but the point you're making with upping the voltage to light more LEDs while using less current is, of course, a great idea. And I have not experimented with various LEDs as I'm not sure which ones I'm going to use, and finding some that I'm determined to use would involve buying them (and Digikey shipping isn't cheap), so I was hoping to get as much information as I could before I make and purchasing decisions. If it's vital to the schematic that I know the type of resistor needed, then I will purchase a few LEDs and experiment.
     
  8. #12

    Expert

    Nov 30, 2010
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    I believe that is a mistake.
    4 x 3.3V = 13.2V
    15v-13.2v = 1.8v
    1.8v/0.02A = 90 ohms

    By the way, @Inglorious89 , that's how you do the math. Add up your LED volts, subtract that from the power supply volts, and divide by the current you want.
     
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  9. cmartinez

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    Jan 17, 2007
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    Sorry, but that is a very bad idea.

    The precise explanation is a bit lengthy, but here's a link with graphs, calculations and all that pretty much sums it up.

    My advice: use one resistor per LED. If you use SMT components it will save space and maybe be it will be much easier to build, depending on your PCB etching skills.
     
  10. blocco a spirale

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    Jun 18, 2008
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    My piece of paper has different dimensions.
     
  11. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Connecting LEDs directly in parallel does not work well. The voltage across an LED when it is on is called its forward voltage, or Vf. For equal brightness, LEDs do not have precisely matched operating voltages, so when connected as in your schematic you would see obvious brightness variations along the 32 LED string. OTOH, LED brightness is proportional to the current through it, If. Placing LEDs in series guarantees that all devices will have exactly the same current. Even though the individual Vf's will vary, the brightnesses will be *much* more evenly matched. The tradeoff is that you need a power supply with a higher output voltage. The more LEDs in a series string, the higher the required voltage.

    The data sheet you posted has a red LED with Vf = 2.2 V at If = 20 mA. 32 LEDs in parallel would need 2.2 V at 0.6 A. 32 LEDs in series would need 70.4 V (!!!) but at only 20 mA (0.02 A). So you see the tradeoff, higher system voltage means lower system current. Remember that since you are not multiplexing the array, the total system power in the LEDs will be the same no matter how you power them. What changes is the power dissipated in the current limiting resistors. 768 LEDs with 768 resistors is a lot of unnecessary heat. 768 LEDs with 24 resistors is the same LED power, but a lot less current limiter power. But that 70 V power supply just isn't practical, so a better granulation is needed.

    One approach is to start with the power supply and work backwards. An old computer ATX power supply makes lots of current at 12 V. 5 LEDs in series would be a total Vf of 11.0 V. Using Ohm's Law, a resistor to limit the current to 20 mA with the remaining 1.0 V across it would be 50 ohms, and dissipate 20 mW. An OK approach, but 4 LEDs and a 160 ohm resistor will make the brightness of the strings less dependent on the absolute accuracy of individual LED Vf values. This is a 4:1 improvement over the 768 resistor starting point, but not great.

    Next, let's hop to a 24 V power supply and 9 LEDs in a string. Total Vf is 19.8 V, resistor is 210 ohms, resistor power is still under 0.1 W, very comfortable for a 1/4 Watt part. 9 doesn't go into either of your dimensions evenly, so there will be one string with different parameters. But this gives you a pattern to follow as you combine LEDs of various Vf values, consider other power supply options, etc.

    BTW, a 24 x 32 square grid of LEDs, where the horizontal and vertical spacing of the lights are equal, will not be the correct aspect ratio. The US flag ratio of height to width is 1:1.9. For a square grid with a height of 24 LEDs, the width would have to be 45-46 LEDs, not 32. 17 x 32 would be smaller, but closer to ideal. Exact ratios would be 20 x 38 and 30 x 57.

    ak
     
    Last edited: Aug 5, 2016
  12. bertz

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    Nov 11, 2013
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    Yep, you're right. Too much of a hurry, however I would still drive the LEDs at 10 mA, thus 180 ohms.
    1.8 Volts x .01 amps = .018 watt 1/4 watt resistors are good to go
     
    Last edited: Aug 4, 2016
  13. hp1729

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    Nov 23, 2015
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    I second the 10 mA suggestion. You don't need to light up the room with these LEDs.
    What power sources do you have to work with? You can wire a few in series and get to any voltage you want and get your total current down. 768 LEDs at 10 mA, figure the total wattage at the suggested voltages and buy the power supply according to wattage.
    I missed a count of how many red, blue and white LEDs you will need. I couldn't get a design of 32 x 24 LEDs that works. Spacing of the LEDs on the field of stars did not match up with spacing of the stripes. There are seven stripes in the same space as 9 (or 11 if you count blue rows) rows of stars. Do you have a drawing of your planned layout?
     
    Last edited: Aug 4, 2016
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  14. Inglorious89

    Thread Starter New Member

    Feb 12, 2016
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    I'm liking this approach. From what I can tell, I'll only need 1 resistor per group of LEDs in connected in series, correct? I'd just like to make it clear as that's a lot of resistors to purchase (less then one per each LED of course), so I just want to be sure. Besides that I love you idea and I'll probably go with your suggestion.
     
  15. cmartinez

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    Jan 17, 2007
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    Yes, that is a feasible approach.
    Question, have you made a layout of what you're trying to build? With each led properly dimensioned and in its place in the circuit?
    You may find that the restrictions you've mentioned are not that easy to comply with.
     
  16. bertz

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    Nov 11, 2013
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    Here you go. This is a layout using 5 mm LEDs on .5" spacing. The strips are somewhat offset from the blue field and of course the black dots represent white LEDs. The overall size is 15.5" x 9.65". You could reduce the overall size by using 3 mm LEDs on .4 mm spacing. You might get a better effect by putting the LEDs behind a plastic diffuser. American Flag LED.jpg
    You will need a total of 758 LEDs; 327 Red, 338 White, and 93 Blue.
    758 x 10 mA = 7.58 amps
    You will need a power supply capable of delivering at least that much current. Unfortunately, you're going to find that they ain't cheap!
     
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  17. cmartinez

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    Nah, he could use a computer's power supply and won't need to spend more than $20.00 dlls on it
     
  18. bertz

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    Nov 11, 2013
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    You can probably pick up an ATX power supply from a computer repair shop for free. But as AnalogKid correctly pointed out, he will need around 165 parallel circuits to light this thing up. If you buy a power supply the shipping might be as much as the price of the supply.
     
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  19. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    In very round numbers, let's say the average Vf of red, white, and blue LEDs is around 3 V. For all LEDs in parallel and 7.58 A of current, you're only two options with a computer power supply are the 5.0 V and 12 V outputs. For the 5 V output, the total power dissipated in the current limiting resistors is 15.16 W, or 40% of the total system power of 38 W. For the 12 V output, the total power dissipated in the current limiting resistors is 68.22 W, or 75% of the total system power of 91 W.

    Or, with a single output 24 V supply he could reduce the total current limiter power to 3.27 W, or 12.5% of system power. System power is under 30 W, so the supply is under $20 on ebay.

    OR (new thought), scale everything for a generic 19 V laptop supply. A 20 x 38 array with an average of 5 LEDs in each series string, 152 strings, 6 W in the current limiters, 21% of 29 W total power.

    ak
     
    Last edited: Aug 5, 2016
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  20. cmartinez

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    And then, you can always scavenge one from an old broken desktop :)
     
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