Wig Wag LED with 555

Discussion in 'The Projects Forum' started by the3dman, Nov 6, 2009.

  1. the3dman

    Thread Starter New Member

    Nov 6, 2009
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    I have created a circuit to alternate between 2 leds flashing in a wig wag pattern the problem is that my 1st led flashes fine "on,off,on" but my second led does not go all the way off between flashes. It just goes "dim, bright, dim" Any way to fix this. I have included a diagram of my circuit.
     
  2. hgmjr

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    Jan 28, 2005
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    Take a look at the 555 circuits covered in Bill Marsden's Blog. He has one that is 50% duty cycle with the LEDs connected like your circuit.

    hgmjr
     
    Last edited: Nov 6, 2009
  3. SgtWookie

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    Jul 17, 2007
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    I believe you mean that LED1 does not go completely off.

    This is because BJT 555's output (pin 3) doesn't go all of the way to Vcc; it's maximum is about Vcc-1.3v. Some LEDs will still glow dimly with 1.3v across them.

    Try adding a 1k resistor from the junction of R3 and LED1's cathode to +V.
     
  4. the3dman

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    Nov 6, 2009
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    No led 1 is fine its led 2 that does not go off between flashes
     
  5. hgmjr

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    Exactly what is the value of your power supply?

    hgmjr
     
  6. the3dman

    Thread Starter New Member

    Nov 6, 2009
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    9 volts is what I am using
     
  7. SgtWookie

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    It does not make sense that LED1 would turn completely off, but LED2 would not.
    It should be the exact opposite.

    bjt (bipolar transistor) 555 timers' outputs (pin 3) can get within 0.2v or so of ground, but can't get within about 1.3v of Vcc.

    CMOS 555 timers can sink about 10x as much current as they can source; but not as much as bjt 555's can. The CMOS versions can also output very close to the power rails, if under a light load.

    I can't imagine how LED2 is staying dimly lit when LED1 is on.
     
  8. the3dman

    Thread Starter New Member

    Nov 6, 2009
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    I couldn't tell you I just know thats what it is doing. If I put a second led between the cathode of led 2 and the negative of the circuit then everything works fine (both 1 and 2 go completely out before flashing again) except the new 3rd led which flashes very dim.
     
  9. Wendy

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    Mar 24, 2008
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    Add a regular diode in series with the LED that is not going out all the way. It sounds like you have some older LEDs, newer LEDs drop a lot more voltage, but are a LOT brighter.
     
  10. the3dman

    Thread Starter New Member

    Nov 6, 2009
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    I will try that thankyou
     
  11. the3dman

    Thread Starter New Member

    Nov 6, 2009
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    I did try the diode and and it did not seem to help much, just slowed down the flashing, it didn't change the problem with LED 2. I am using brand new packages of leds so I don't think that is the problem. I even tried another 555 circuit from the blog recommended to me and I get the same issue. Not sure what to do from here.
     
    Last edited: Nov 7, 2009
  12. SgtWookie

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    Are you certain that both R3 and R4 are 470 Ohms? Yellow, Violet, Brown (gold/silver)
    It's easy to mistake a 47 Ohm resistor (Yellow Violet Black) for a 470 Ohm.

    If the resistors are correct, then try going to 680 Ohms: blue, grey, brown (gold/silver)
     
  13. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    In the circuit shown, with 9V, the current would flow to light both LED's from the battery, with the 555 disconnected, or am I reading this wrong?
     
  14. Wendy

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    If the 555 physically were not there you would be correct. The 555 has no tristate condition, the output is either one side or the other (with the provision the the + side is 1.4V away from the positive power supply voltage), so only one LED will be lite at a time.
     
  15. SgtWookie

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    As Bill said, both LEDs would light with the 555 removed.
    However, their brightness would be reduced.

    Let's say that each LED had a Vf of 2v @20mA. 9v-2(2v)=5v.
    There are two 470 Ohm resistors in series, for a total of 940 Ohms.
    5v/940 Ohms = 5.3mA current flow through both LEDs and both resistors.
    This is not exact, because if the LEDs have a Vf of 2v @ 20mA, their Vf at 5.3mA will be substantially lower.
     
  16. the3dman

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    Nov 6, 2009
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    I will try with the resistors. Could it be a bad 555?
     
  17. SgtWookie

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    It is certainly possible that the 555 is defective, however it is more likely that the resistors being used are not correct.

    Changing the resistors to 680 Ohms will only require 10mA source/sink capability from the 555, which is well within it's specifications. So is the original 15mA current draw, but I'm thinking that your upper resistor is 47 Ohms instead of 470 Ohms. That would be too much of a load for the 555 - and also would rapidly burn out LED1.
     
  18. Wendy

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    What is the exact part number of the 555 you're using?
     
  19. the3dman

    Thread Starter New Member

    Nov 6, 2009
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    33crjfm
    ne555p
     
  20. Wendy

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    You can take pin 2/6 on your circuit and either ground them or connect them to the battery power supply, the output will invert this signal. You should see your battery voltage - 1.2V or less than 0.1V respectivly, and the LEDs should switch accordingly.
     
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