4017 Cascade, One more try

Discussion in 'General Electronics Chat' started by mxabeles, May 20, 2010.

  1. mxabeles

    Thread Starter Active Member

    Apr 25, 2009
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    Hey Everyone,
    so I posted a question about cascading 4017s. If there is really no answer to this problem, I'm cool with that. I just figured it may have gotten overlooked tho.
    Last post in this regard :

    I am still struggling with cascading the 4017s.
    I used this model : http://images.elektroda.net/8_1178380869.gif

    I have a one step delay at the end of the chain (all leds off). Is that considered the propogation delay, and if so is there any known way around this? Again, it is somewhat irrelevant for light display, but for audio work it is unacceptable.
    Thanks again,
    Max
    __________________
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    Hi Max,
    Sorry that your question was overlooked.

    For clarification, here is the basic schematic for cascading 4017's, excerpted from Motorola/ONsemi's datasheet:

    [​IMG]

    I suggest that there is a break between the rightmost 4017's Q9 and the RESET input for the first 4017, or much more likely that you have a bad connection or dead output from Q8 on the rightmost 4017.

    Note that Q0 is available for use only on the 1st 4017.
    Q9 is not available for use on any of the 4017's.

    When Q9 goes high on the last (rightmost) 4017, it should reset the leftmost 4017, which in turn should reset the 2nd 4017. Humans should not be able to detect the change, as it will be extremely quick.
     
    Last edited: May 20, 2010
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  3. mxabeles

    Thread Starter Active Member

    Apr 25, 2009
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    Thanks for the reply SgtW.
    I'll trying hooking it up again (breadboarded 4 the third time :))). The one strange thing is that it can't be a dead Q8 because what happens is that it cycles through every led correctly and then pauses for 1 pulse with no lights on and returns to Q0 of left most 4017. It must have something to do then with Q9 - Reset.

    Just to make sure the last 4017s Q0 is unattached, correct?
    I just got a new shipment of ICs, maybe one the 4017s or AND gates was fritzed somehow....
     
  4. Wendy

    Moderator

    Mar 24, 2008
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    Another variation of the circuit Wookie showed looks something like this...

    [​IMG]

    It uses diode AND gates instead. A similar circuit was also shown in Bill Bowden's website.
     
  5. SgtWookie

    Expert

    Jul 17, 2007
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    It's quite possible that you have a breadboard with a damaged row of sockets, or perhaps a broken jumper wire. Try using a different area of the breadboard.

    When Q8 is high and a clock is received, Q8 immediately goes low and Q9 goes high. The moment Q9 goes high, the leftmost 4017's RESET input goes high, which turns off it's Q9 output and turns on it's Q0 output. The Q0 output going high causes the RESET input of the 2nd 4017 to go high, so that it's Q0 output goes high, resetting the last 4017.

    Another possibility is that you did not connect a light to the leftmost Q0 output. It is the only Q0 that you can use for a sequence output; the Q0's on the other 4017's are just used for the RESET input of following stages. If there is no following stage, then Q0 is not used.

    Yes.
    You mentioned LEDs lighting up, but you didn't mention HOW you were connecting LEDs to the outputs. 4000 series CMOS has pretty puny source/sink capabilities. You should really use transistors or some type of driver IC (like a ULN2804 or ULN2004) to sink current from your LEDs. If you have too much of a load on the leftmost 4017 output Q0, that could delay or prevent the reset from occurring.

    If you limit the maximum current source/sink requirement to <=2mA, you should be OK.

    Note that Bill's using discrete resistors and transistors for driving the LEDs on his 4017 outputs. This is perfectly fine, and it will keep the current within the 2mA limit I suggested above.

    Using the ULN2804A saves on wiring and space, like this:

    [​IMG]
     
    Last edited: May 20, 2010
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  6. mxabeles

    Thread Starter Active Member

    Apr 25, 2009
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    thanks everyone.
    Yes, I actually just ordered the 2803 i believe. definite space saver...
    Okay,I'll try the circuit again with the transistor array.
    Cheers,
    M
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    Correction; it's the ULN2804A if you are driving it from CMOS.
    If you are driving it from TTL, then you need the ULN2803A. The base resistors are different; the CMOS versions don't supply enough current to drive the 2803. You CAN use the ULN2803 with 4000 series CMOS, but you must add resistors between the IC's and the ULN2803 inputs; 7.5k resistors. It is so much easier if you you just order the correct driver IC's to begin with. This keeps your parts count low.
     
  8. mxabeles

    Thread Starter Active Member

    Apr 25, 2009
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    ok cool thanks. I just spent the entire night making my first 4017 sequencer....4017, 2803 and 555 timers. various pots hooked into 2803 collectors to change pitch for each step. simple but its working. Anyway, why would you need a resistor between the CMOS output and 2803 base if CMOS has LOWER current? That just seems counter intuitive.
    Let my tired brain know. You guys are the best,
    Max
     
  9. SgtWookie

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    It's because the 4000 series CMOS IC's output voltage will drop quite a bit if they are heavily loaded.
     
  10. Wendy

    Moderator

    Mar 24, 2008
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    And then die.
     
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  11. mxabeles

    Thread Starter Active Member

    Apr 25, 2009
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    ok, tnx.
    Sucks tho.
    Just ordered a ton of 2803. well, i can still use them with...a nevermind. damn haha
    M
     
  12. Kwarl

    New Member

    Jun 4, 2010
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    What's the little arrow symbol that's shaped like an s on wookie/marsden's schematics? it's between every diode.

    EDIT : Also, can I skip the 330 ohm resistor if I'm not cascading? if not, how is its value calculated for 1 or 2 4017-2804 pairs?
     
    Last edited: Jun 4, 2010
  13. beenthere

    Retired Moderator

    Apr 20, 2004
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  14. mxabeles

    Thread Starter Active Member

    Apr 25, 2009
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  15. SgtWookie

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    Jul 17, 2007
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    The basic difference between the ULN2803A and ULN2804A is the value of the resistor between the input and the base of the first transistor in the Darlington pair.

    In the ULN2803A, it's 2.3k to 2.5k. In the ULN2804A, it's about 10.5k.

    If you are operating your 4017 on 5v, you can use the 2803 as-is. However, if you are using >5v for Vdd, you should use an additional resistor (2k to 7k) between the 4017 output and 2803 input for any 4017 output that is also used as a logic input to another CMOS IC, otherwise the signal may not rise high enough to be interpreted as a logic 1 by the other logic device.
     
  16. mxabeles

    Thread Starter Active Member

    Apr 25, 2009
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    ok cool. just to clarify whether or not power to 4017 is 5v or higher if the 4017 is being used in a system (not the end output) an extra resistor is needed in series with 2803 inputs? Thanks sgt,
    M
     
  17. SgtWookie

    Expert

    Jul 17, 2007
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    If you are using 6v or higher for the 4017, you should add a 7.5k resistor between the 4017 output and the ULN2803A input.
     
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