LED chaser

Discussion in 'The Projects Forum' started by zacknie, Feb 3, 2011.

  1. zacknie

    Thread Starter New Member

    Jan 26, 2011
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    hi.. im new in electronics... can i see or give me a circuit diagram of LED chaser using 4017?..im designing a LED chaser that are more that 10 LEDs.. im getting problem how to connect 2 or more 4017 IC..plsss,..tnx... any ways.. this is great site for electronics fanatics.!! GOD BLESS!
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    Welcome to AAC!

    You are in luck, as it happens it is one of my specialities.

    LEDs, 555s, Flashers, and Light Chasers Chapter 11 -Making Patterns

    If you want 10 LEDs back and forth then you will need at least 3 4017s, and some extra circuitry.

    If you are brand new then you probably need to cover Chapter 1 and 2 in detail, and really try to understand it. It will hold you in good stead later.
     
  3. zacknie

    Thread Starter New Member

    Jan 26, 2011
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    thank you very much sir!.can i connect more than 5 4017ic?? to make more running LEDs?..
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    You could, if you have that much time to wire everything up!

    See the 3rd schematic down on this page:
    http://forum.allaboutcircuits.com/blog.php?bt=684
    Three 4017's that are daisy-chained together, giving 9+8+8 = 25 outputs.
    If you want to add more, you would basically copy the wiring for U3 and insert it between U3 and U4.
     
  5. Wendy

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    Mar 24, 2008
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    I'm thinking seriously of drawing up a PCB for the three 4017 sequencer. It would be handy for a lot of folks.
     
  6. nerdegutta

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    Dec 15, 2009
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    Question about fig 11.4 in link from above:

    Is pin 3, out 0, in U4 intentionally left unconnected?

    (Link from above.... could mean a lot of things....:D)
     
  7. Wendy

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    Mar 24, 2008
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    Yes, if we are talking Figure 11.4. It is a parking bit, never used. Remember, only one bit can be one of the bunch, and each chip will have one bit on. So one bit of each chip has to be thrown away. With the second 4017 it is used for part of the cascade logic, for the 3rd 4017 it is simply not needed.
     
  8. nerdegutta

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    Dec 15, 2009
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    OK, so if we add another 4017, U5. PIN 3 from U4 would connect to PIN 15 on U5, and PIN 3 on U5 would be unconnected?
     
  9. Wendy

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    Mar 24, 2008
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    Sounds right, but I don't have time to verify the schematic at the moment. If you look at the data sheet it will have the same schematic, only showing the AND gates (whereas I used diode AND gates).

    The first 4017 has 9 bits, the second adds another 8, as do all the rest that follow.
     
  10. nerdegutta

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    Dec 15, 2009
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    Think I'll try to draw it in Eagle during the weekend. Could be a fun weekend project. :)
     
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  11. Wendy

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    Mar 24, 2008
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    Cool. May I redraw the schematic to make it more general purpose? It really could use a diode per output to allow the user to make special patterns. If the user doesn't want diodes they don't have to use them, but in many cases they would be handy.
     
  12. magnet18

    Senior Member

    Dec 22, 2010
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    Do I get a prize if I can do it with only two 4017's? :)

    (can it be done with two?)
     
    Last edited: Feb 4, 2011
  13. SgtWookie

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    Jul 17, 2007
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    Rather than using all of those discrete transistors & resistors on the output, you'd save parts count and soldering by using three ULN2804's, like this:

    [​IMG]

    To keep the Vce's more or less matched, Q1 should be something like an MPSA13 Darlington, and R3 should be 10k Ohms.
     
  14. nerdegutta

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    Dec 15, 2009
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    I think I go with the "transistor" solution. Eventually I might build this on a PCB, and I have the transistors. I have some ULN2003, but then I guess I have to either double the ULN's or loose some LEDs.

    Ooops. Did I just hijack a thread...? If so, I'm sorry. :)
     
  15. SgtWookie

    Expert

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    ULN2003's have a 2.3k input base resistor, which is too low for 4000-series CMOS. You could use a 7.5k resistor between the 4017's outputs and the ULN2003's inputs, but that makes your parts count climb - and you'd need an additional three discrete drivers like shown as Q1/R3 above.

    You can certainly use discrete resistors/transistors for everything; it's just a lot more parts and connections.
     
  16. nerdegutta

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    In fig 11.4, in the link, C1 is polarized. Is it still polarized?
     
  17. SgtWookie

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    Yes, the cap is still polarized. You could use a non-polarized if you wished; but if you use a polarized cap, the - side goes to ground.
     
  18. nerdegutta

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    Dec 15, 2009
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    That's what I thought.:)

    What would happen if we grounded PIN 5, U1, using a capacitor? Would that prevent false-triggering when the power is applied?
     
  19. SgtWookie

    Expert

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    What you describe is not grounding pin 5 (which you normally wouldn't want to do) but providing a more stable threshold voltage.

    Whatever "glitches" there are during start-up should sort themselves out after a couple of clock pulses.

    The 4017's will start up in unpredictable settings, but by the time 10 or so clock pulses have been issued, they will be sorted out. If the RST pin (15) on U1 is held high momentarily on power-up, then the other two cascaded 4017's will also be reset. This could be implemented by putting a 0.1uF cap from U1 pin 15 to Vcc, and inserting a 10k Ohm resistor in series with the wire going to U4 pin 11.
     
  20. magnet18

    Senior Member

    Dec 22, 2010
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    [EDIT] - The proposed circuit in this post is ineffective

    Hi there,
    I know you guys know more about this than me, but i think there might be a simpler solution for our OP
    If the OP doesn't mind more than one LED lit up at one time, the 4017's could simply be daisy chained using the pin 12 carryouts, with the LED's going right from the outputs to ground with, say, a 330Ω resistor for maybe each group of 10?

    This would save a LOT of parts and as long as the OP doesn't mind one LED out of 10 on, should work fine and will be much simpler.
     
    Last edited: Feb 5, 2011
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