LED chaser

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zacknie

Joined Jan 26, 2011
9
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!
 

Wendy

Joined Mar 24, 2008
23,415
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.
 

Wendy

Joined Mar 24, 2008
23,415
I'm thinking seriously of drawing up a PCB for the three 4017 sequencer. It would be handy for a lot of folks.
 

nerdegutta

Joined Dec 15, 2009
2,684
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.
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)
 

Wendy

Joined Mar 24, 2008
23,415
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.
 

Wendy

Joined Mar 24, 2008
23,415
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.
 

Wendy

Joined Mar 24, 2008
23,415
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.
 

SgtWookie

Joined Jul 17, 2007
22,230
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:



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

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nerdegutta

Joined Dec 15, 2009
2,684
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. :)
 

SgtWookie

Joined Jul 17, 2007
22,230
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.
 

nerdegutta

Joined Dec 15, 2009
2,684
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:



To keep the Vce's more or less matched, Q1 should be something like an MPSA13 Darlington, and R3 should be 10k Ohms.
In fig 11.4, in the link, C1 is polarized. Is it still polarized?
 

SgtWookie

Joined Jul 17, 2007
22,230
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.
 

nerdegutta

Joined Dec 15, 2009
2,684
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?
 

SgtWookie

Joined Jul 17, 2007
22,230
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.
 

magnet18

Joined Dec 22, 2010
1,227
[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.
 
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