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How can I cascade 4017s?


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Cascade ?: one way is to connect c_out to c_in of the next one
or else if u want then to count in sequency, try

May be use 1 master 4017 which gets 1/10th of system clock and its output used as display en-able for all other '17s which are all clocked from same (I say system) clock. Unfortunately disp-en function isn't onchip so u have to plan ur external circuit. Transistors could be used to power individual '17s.

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The "carry out" pin of the CD4017 cannot be used for cascading them.

The CD4017 datasheet from Texas Instruments shows them cascaded using AND gates. But you need to figure out how to reset everything when the power is applied.

Bill Bowden has already done the reset circuit for you and also uses a few diodes instead of AND gates.

post-1706-14279143218326_thumb.png

post-1706-14279143218523_thumb.png

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The carry out will allow you to use multiple 4017s as ones and 10s type of counter. You do not want to use this for a continuous counting sequence.

Bill Bowden's circuit does not make a very good tutorial and I would recommend using the AND gates. I would also use gates for the clock instead of 555 if accuracy is required. Of course, if you are only blinking lights, neither of these two points matter, since the inaccuracies cannot be seen by the eye.

Below is the link to a site that shows several 4017s cascaded correctly. If you follow the way that each triggers the next and the last step resets the circuit, you will learn how to cascade as many 4017s as needed. I think it will act as a pretty good tutorial for those who have questions about cascading this chip for multiple LED strings.

http://etronics.free.fr/montages/eflum/EF01/EF01.htm
On this site, click on the link "Schema ef01.pdf" to get a larger schematic in pdf.

MP

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The French circuit uses dozens of current-limiting resistors for the LEDs. Only a single current-limiting resistor is needed for all the LEDs if the supply voltage is high, because only a single LED lights at a time. Bill Bowden's circuit shows how.

If the supply voltage is 9V or less then a current-limiting resistor is not needed because the output current of a CD4017 is low when its supply voltage is low.

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Good subject to expand on. Again, if you are only blinking LEDs, it might not matter. The individual resistors in the design from the French website give you a more universal device that is not limited to using the same common for all outputs. Thus, it has the capability of other applications. In an LED circuit, the individual resistors in this design give you the capability to use different types of LEDs where they will need different currents. If you will always use the same LEDs and the same common, you can save a few pennies by using a common resistor such as in the Bowden design.

MP

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Whats up with C2 - C6 in the french circuit?

They are supply bypass capacitors. One 0.1uF ceramic disc capacitor directly across the supply pins of each of the five logic ICs.

What is the J1 and C7 area on the french circuit too?

J1 is the 12V supply input jack.
C7 is the 100uF low frequencies supply bypass capacitor for the 555, but it should also have an additional 0.1uF ceramic disc capacitor for high frequencies.
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I look at the 1k resistor going to MR pin 15 of the upper 4017. It isn't needed since a piece of wire will do the same thing. But it allows you to connect the MR pin 15 to the positive supply for a moment to reset both 4017s.

So maybe the circuit isn't finished and an OR gate should be connected to MR pin 15 to make it high when power is applied, or when the 1k resistor goes high.

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