Toggle LED's between blink and solid

Discussion in 'General Electronics Chat' started by Frank88, Feb 6, 2012.

  1. Frank88

    Thread Starter New Member

    Feb 6, 2012
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    Hello,

    I am looking to do something similar as mousepad's 10,000 leds in parallel but a bit less ambitious.

    I'd like to flash 36 LED's, actually want to use a switch to flash them (like a blinking traffic light at 2am) with the other state being solid.

    I'm actually looking for two circuits driving 36 leds, one for 2v red led, 20mA (although they are superbrights and rated as 30mA), and one for 3.8v blue led, 20mA (also superbrights rates as 30mA).

    Configuring for 30mA is ok, but I'm reading 20mA is standard and what should be accounted for - but I'm not sure.

    I have a 12v, 16amp power supply. The LED's will be about 3 feet apart on average.

    If I understand correctly a 555 8-pin IC in astable mode will handle the blinking side of the project. Although 36 leds at 20mA is about 3/4 of an amp and the 555 needs help to power this?

    Also, I'm afraid the power supply will damage the 555.

    I'm looking for the most efficient circuit design to handle both the blinking and solid sides of the circuits. Any design assistance is greatly appreciated.

    Frank
     
  2. iONic

    AAC Fanatic!

    Nov 16, 2007
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    It appears you have plenty of power for your project.
    The 12V supply is fine for powering the 555 as the 555 will not draw any more current than it needs, you wont get 16A thru the 555 unless you wire it wrong. You would power the LED's by having the 555 trigger on-off a power transistor. You would be better off stringing 7 parallel legs of 5 LED in series for a voltage drop of near 10V, then use a resistor to regulate the current.
    For the Blue LED's you might be able to do 12 parallel legs of 3 Leds in series and use a limiting resistor to drop the remaining voltage.
     
  3. Frank88

    Thread Starter New Member

    Feb 6, 2012
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    For the red I have an array of 6 x 6 with 1Ω resistors and the blue 3 x 12 with 33Ω. I'm more confused by the 555 astable mode to blink and then the switch to solid.

    I'm thinking: a) power supply can first go to two SPDT switches (one for red string of 36 leds and one for blue string of 36 leds), where one side of the switch simply sends the power to the array's as described and the lights will stay on solid. Correct assumption?

    b) The other side of the switch would direct power through the 555 to get the blinking desired.

    Does this initial assumption to direct power sound correct? This would put two positive lines to the array, with only one active at a time - would this present a problem to the 555 when the switch is set for solid?

    My next issue is the design for the circuit for the 555 to 1) handle additional needed power to run 36 led's and 2) set the 1/2 sec blink rate.

    This is new material for me - dumbing it down will be helpful. :cool:
     
  4. Wendy

    Moderator

    Mar 24, 2008
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    I get the feeling you are not too informed to LEDs and thier characteristics (it was the 1Ω that suggested this to me).

    Let go through what I think you wanted.

    You want a 6 X 6 grid of LEDs, each individually controllable. The mechanism to flash or turn on completely is pretty important, if you are after individual control. This strongly suggests some kind of computer control, the exact nature of which is to be determined.

    Making a 555 blink or stay solid is simplicity, but we have to know the exact criteria for the blinking. The 555 will need a simple transistor circuit just to handle the current, but the power supply you have is major overkill. A good thing for reliability.

    I have written a tutorial for LEDs you might find helpful...

    LEDs, 555s, Flashers, and Light Chasers

    Once we define the specifics a bit better then we can help with the schematics.
     
  5. Frank88

    Thread Starter New Member

    Feb 6, 2012
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    Yes, you are correct about my novice skillset (which is why I am here and it's been inspiring to say the least).

    I got the 1ohm from an ohm calc wizard. Seems to make sense for the 2v led's with a 6 parallel having 6 lights in series as this takes up 12v of power and the 1ohm was thrown in for good measure? The blues are 3.3v fwd voltage instead of the 3.8v i mentioned earlier.

    All 36 LED's in either line will all be either solid, or blinking, desired rate of 1/2 sec. The power supply is an extra one I have from a personal computer, so it's surplus and if the excess power doesn't cause a problem I'm ok with using it.

    Thanks for your interest.
    Frank
     
  6. Frank88

    Thread Starter New Member

    Feb 6, 2012
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    Bill, thank you for the link!

    Also the design is a line of 36 led's each light an average of 3' apart (not a grid), sorry to misinform here. Like Christmas lights but spaced further apart. When I mention array I'm using the term from the ohm calculator at LED Center. And all the lights in the line will either be on or manually switched to blinking. They're state does not need to change individually.

    Frank
     
  7. Wendy

    Moderator

    Mar 24, 2008
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    The 1Ω number is totally bogus. It is one of the reasons we tend not to recommend the various LED wizards out there, the programmers show no common sense.

    If you want to do these as a cluster there are really easy ways to do it. I can draw what you need pretty easily. I tend to be a draftsman here as well as a moderator. The moderator is still pretty new actually.

    So lets lay out a hypothetical design. You want 2 strings of 36 LEDs (distance not important for the electronics). Each string can blink or stay on. One string is red, one string is blue. Correct?

    You will find trying to describe something like this verbally is murder. I've gone through several pages with other people only to find I still have it wrong.
     
  8. Frank88

    Thread Starter New Member

    Feb 6, 2012
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    Yes, that is correct, and i sympathize with the verbal gymnastics. I'll try to be clear and simple.
     
  9. Frank88

    Thread Starter New Member

    Feb 6, 2012
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    The normal state will be on. From time to time I will manually throw the switch to cause the string of LED's to blink until I throw the switch back. That is what is desired.
     
  10. Frank88

    Thread Starter New Member

    Feb 6, 2012
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    Bill, my research into this project has unfortunately brought on more confusion as I'm told to use 20mA for the diode fwd current and 3.8V forward voltage, so just computing the resistors needed has become a challenge at this point. General concensus as you put it - the wizards are bogus.

    I'd sure like to get this resolved. I've attached links to exact specs for the Red and Blue LEDs. The 36 LED's is a set quantity for each color (there will actually be four colors, 36 each. I didn't include the Yellow and Green as their spec's are identical to the Red and Blue respectively).

    Then there is the power supply which is a Sparkle Power Supply, Model FSP235-60GI that a few years ago a friend of mine put red and black wire posts onto the exterior for each the 5V and 12V outputs On the panel the 5V + indicates 22.0A (Red) and - indicates 0.3A (Blue); the 12V + indicates 8.0A (yellow) and - 0.8A (Brown) on the panel. There is also 3.3V output but he didn't put any posts on the outside to connect to for supply. He's since moved to Chicago (too bad). I'm a Coloradan now, but loved my time in Grapevine/Dallas many years ago.

    Hopefully this is helpful.
    Regards, Frank

    Red
    http://www.superbrightleds.com/more...egree-viewing-angle-flat-tipped-1200-mcd/279/

    Blue
    http://www.superbrightleds.com/more...egree-viewing-angle-flat-tipped-1200-mcd/265/
     
  11. Wendy

    Moderator

    Mar 24, 2008
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    If you look at the specs you will note a number referred to as Forward Dropping voltage (AKA Vf). This is the real number the LED drops (it tends to be a constant for a particular LED), but each LED is different by a little (sometimes a lot). So when designing you pick a range of numbers. It is possible due to variations of Vf that the currents will be off quite a bit.

    For the red LED I used 2.2 Vf, for the blue I used 3.3 Vf. I aimed for 20ma. First I'll show the LED matrix. Each circuit using one resistor is referred to as a leg. The resistors can be tweaked to increase or decrease the LED currents. Current is what is important to the LED, the voltage is incidental and is only important when designing the circuit.

    Because of the general uncertainty of LED Vf I would recommend building one leg, then measuring the voltage across the LED resistor. At that point you can fine tweak the design if you so choose. Most cases people just go with it if it works.

    You answered one question. We generally like to know your location so we can refer you to parts sources. I use Radio Shack in my tutorials because they are the closest thing to a universal parts source, but for quantity and (more importantly) pricing they are my second choice. Dallas has a large number of electronics parts houses, but this is an unusual situation for most cities. So now for the second question. Who is your preferred parts source?

    [​IMG]

    At this point I am treating the LED matrix as a separate circuit. It will simplify the following discussion about flashers. The resistors shown are ¼W 5%.
     
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  12. elec_mech

    Senior Member

    Nov 12, 2008
    1,513
    193
    Here is my take on it. I was going to put the red and yellow LEDs in strings of five, but Bill has far more experience than I, so I defer to his wisdom.

    You can change the flash rate by adjusting VR1. Closing SW1 puts the 555 in astable mode which will flash the LEDs. Opening SW1 causes the 555 to go into reset mode, effectively off. R3, R16, R29, and R39 are pull-up resistors used to turn the MOSFETs on and thus the LEDs when the 555 is off.

    I believe this will work, but welcome critiques.
     
  13. Wendy

    Moderator

    Mar 24, 2008
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    Try adding the Vf voltages, 5 is an illegal number. If the red Vf is 2.5 it can't go to 4 either. Those little changes in voltage matter a lot.

    As for blue, if it is 3.3Vf 3 in a chain is it. If it is 3.6Vf 3 is pushing it with a 12V power supply.

    Depending on the OPs needs, I have a much simpler solution for the flash circuit. I'm going slow to try to teach.
     
  14. Mark_T

    Member

    Feb 7, 2012
    47
    8
    With the red LED's you must use 5 LED's in series and with the extra 2V and you require 20mA then use a (R=V/I R=2/0.02=100Ohms). Do this 7 times to use 35 LED's. If running in a cable the you will require 8 wires as ground is common to all. The 555 timer controls a MOSFET that can handle 7x20mA. I'm sure the 555 can be easily swithed to flash or on with a switch across the timing capacitor. If the switch switches the LED's off rather than on then invert the 55's O/P before the MOSFET.
    Please use a fuse, 12V 16A is overkill and can generate a lot of heat in the event of component failure.
     
  15. Wendy

    Moderator

    Mar 24, 2008
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    Heck with it, no sense in pulling teeth.

    [​IMG]

    You can replace R1 and R2 with one resistor to simplify things.
     
  16. Wendy

    Moderator

    Mar 24, 2008
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    Uhhh, no.

    If the LEDs drop 2.2Vf, then the total Vf for 5 in series is 11V. This is too close to the power supply. The tolerance on these LEDs can go up to 2.4Vf, which would put the total voltage at exactly 12V. This is a non starter. It is always a mistake to challenge tolerances, Murphy waits for such chances.

    It is also why online LED calculators are looked on with disfavor, anytime a resistance of less than 47Ω is used the odds are the design is flawed. LEDs are not precision components. I usually recommend measuring real world numbers with a voltmeter to verify the design.

    Look at the current requirements on my matrix designs. Both are very close or exceeding 200ma. The 555 is rated for 200ma max, so a transistor is required.

    LEDs may short out, but baring wiring error there is little chance of a major short. Therefore the fuse is optional. If the CMOS 555 shorts it will not smoke at such currents, but make a dandy fuse itself. The power supply may have something already, it may be practical to drop its fuse rating. A 2A power 12VDC power supply would work well with this circuit, given the max draw is under 1A, but given a power supply is available it is OK to go with what is available. Overkill is another word for very reliable in this case.
     
    Last edited: Feb 7, 2012
  17. Frank88

    Thread Starter New Member

    Feb 6, 2012
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    0
    Bill,
    After understanding the wizards for this are for interest only I worked up a spreadsheet and came up with 9 x 4 array for the Red as you did but used 12.6 for power supply adding a 5% tolerance. But my resistors computed at 190 ohms, using 2.2V fwd voltage and 20mA diode fwd current. 190 is probably not a resistor size.
    (12.6V – (2.2Vf x 4)) = 3.8V residual for dissipation 3.8V / (20mA/1000) = 190 ohms
    But I see you have 150 ohm resistors for this array. I’m interested in knowing what I did wrong. Using 12V power supply dropped my resistors to 160 ohm.
    And I also came up with the 12 x 3 using 3.3V fwd voltage, but came up with resistor size of 135 (also not a standard size); and your matrix shows 100 ohm. Also interested in how this computed out. Dropping power supply to 12V computes 105 ohm. Are you just taking the standard size just below?
    Testing a leg first is a welcome idea.
    I’ll use your specs. I have no preferred parts source here in greater Denver, but a recommendation is very welcome. We have Radio Shacks but the too closest ones are staffed with newbies (like me).
    This is all amazing and the comments and attachments greatly appreciated!
    Frank
     
  18. Wendy

    Moderator

    Mar 24, 2008
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    I used 12.0V in my calculations. I also used a standard resistor, one you can buy off the shelf. The nearest for 190Ω is 180Ω. Given the basic tolerance problems with LEDs a small change will not matter much.

    12.0V - 8.8Vf = 3.2V (Red)
    3.2V / 0.02A = 160Ω ≈ 150Ω

    12.0V - 9.9V = 2.1V (Blue)
    2.1 / 0.02A = 105Ω ≈ 100Ω

    But like I said, after it is all said and done I like to measure the real world values, then adjust if I feel it is needed. While 160Ω is on the list, it is not commonly available.
     
  19. Frank88

    Thread Starter New Member

    Feb 6, 2012
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    Also does the resistor wattage matter? 1/4w, 1/2w? Is this based on total dissipation?
    Thank you
     
  20. Wendy

    Moderator

    Mar 24, 2008
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    You can calculate it, but for 20ma ¼W is usually adequate.
     
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