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Driving 555 IC with a pulse stream

Discussion in 'Electronic Basics' started by Jag Man, Nov 14, 2004.

  1. Jag Man

    Jag Man Guest

    I have a decade counter (NTE4017B) that generates a pulse stream on each
    decoded
    output, a couple of which I will use. However, the width of these pulses are
    too wide at lower frequencies. So, I'd like to use a 555 IC operating in
    monostable mode to set the pulse width to what
    I want, regardless of the frequency and pulse width coming from the decade
    counter.

    My guess is I have to use the decade counter output to both reset the 555
    (pin 4)
    and trigger it (pin 2). Is that right? I know the triggering is by dropping
    pin 2
    to ground, perhaps by switching with a transistor, but what does the reset
    want,
    high or ground?

    TIA

    Ed
     
  2. You can just use it in monostable mode. Tie reset to high. Look at the
    datasheet (google 555 datasheet) for more information.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  3. John Fields

    John Fields Guest

     
  4. Bill Bowden

    Bill Bowden Guest

    The reset pin is active high but it doesn't need to
    be used. Just connect it to ground.

    The 4017 pulse is probably positive going, and the
    555 needs a negative going edge, so you will need
    a NPN transistor to invert the pulse and a cap
    to trigger the 555 on the leading edge. Something
    like the drawing, but you may also need a diode from
    the transistor base to GND if the input is more than
    6 volts.

    +V
    |
    \ 555
    / +------+
    5.1K \ | |
    | | 3|--> Out
    +--------|2 |
    C | |
    0.01uF |/ +------+
    In --/\/\/---| |--+--B| NPN
    5.1K C | |\
    \ E
    5.1K / |
    \ |
    | |
    GND GND

    -Bill
     
  5. Brian

    Brian Guest

    Why not use a one-shot, like a 74123. Then you can use either a positive
    or negative going edge triggering.
    Brian
     
  6. Other respondents seem to have no trouble with your spec, so the odds
    are that I've missed something - but I'm darned if I can make sense of
    your description. Perhaps if you could describe your end objective?

    Meanwhile, as I understand it, you have a 555 astable clocking a 4017,
    yes? You want the 4017's outputs to have a fixed duration? Then what's
    the problem with just fixing the 555 astable period accordingly, which
    then fixes each 4017 high output to the same duration?

    In which case, where does the need for a 555 (or any other) monostable
    arise?

    You don't mention any specific values, but FWIW here's a 5V circuit
    with a 555 clocking a 4017 so that each of its outputs flashes an LED
    for about 4 ms.

    http://www.terrypin.dial.pipex.com/Images/555+4017.PDF
     
  7. Hey, Terry.

    I think what he wants is to shorten the pulse coming out of the 4017.
    Thus, I believe he is asking how to set up a 555 as a one-shot, given a
    long pulse.

    Now that I think about it, he probably wants an edge detector to drive
    the 555 trigger input.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  8. John Fields

    John Fields Guest

    --- _____
    It's active low; RESET
     
  9. Jag Man

    Jag Man Guest

    John,

    I now have the input to the 555 going to BOTH TRIGGER
    and RESET, without differentiation (i.e., no capacitor in series),
    and it seems to work. I can definitely get pulses out of the 555
    that are shorter than the inputs. As I see it, I am resetting and triggering
    at the same time. Is that antithetical?

    Ed
     
  10. Jag Man

    Jag Man Guest

    Thanks for coming into the discussion, Terry.

    The project was first introduced about a month ago, 10/10/04,
    as "Special dual pulse generator circuit." I'm implementing a
    circuit basically as proposed by Chris Foley. Briefly,
    I'm trying to achieve a tester for the EFI system on a Jaguar
    V12. the car has a trigger board in the distributor that generates
    two switch closures per revolution, 180 degrees apart, that drive
    the electronic control unit (ECU). I intend to plug the tester
    into the harness in place of the trigger board, thus allowing me
    to drive the ECU, amplifier, and fuel injector solenoids
    without the engine running. When the engine is running there are variations
    in speed etc.
    that make interpretation of scope traces difficult. My hope is this will
    allow me to
    test the various components under more controlled conditions.

    So, with Chris' great design the a-stable 555 runs at 10 times the
    rotor speed, so to speak, and the decade counter outputs are
    ad one 10th that frequency, thus emulating the spinning distributor.
    By picking up two the the decade counter outputs, e.g. 1 and 6,
    I get a very good representation of what actually happens. Importantly,
    the speed can easily be varied while always having the two outputs
    exactly 180 degrees apart.

    Chris' design meets my original description of the problem, but I then got
    ambitious. I noticed that by putting another 555 IC (actually 2, one for
    each
    output) between the decade counter and the output stage (optoisolators)
    I could have control over both speed and pulse width, allowing the
    same tester to be used to emulate the trigger board and drive the ECU
    OR to emulate the trigger board and ECU.

    Meanwhile, the big cat lumbers in the garage!

    Ed
     
  11. John Fields

    John Fields Guest

    ---
    No, it's actually pretty slick!

    What's happening is that when the 4017 output goes low it
    unconditionally resets the 555, discharging the timing cap, then when
    it goes high and it gets to about 0.7V it releases the reset and drags
    the trigger pin high along with it. Now, since the trigger pin will be
    at less than about 1/3 Vcc when the 555 comes out of reset, the output
    will go high and the 555 will start timing out. What's slick is that
    it's only going to take ever how long it takes for the counter's
    output to get to higher than 1/3 Vcc to not be triggering the 555 any
    more, and then the 555 will time out during the time the counter's
    output is high.

    __________ __________ ______
    4017 OUT __________| |__________| |__________|
    _____ __________ __________ ______
    RESET __________| |__________| |__________|
    ____ __________ __________ ______
    TRIG __________| |__________| |__________|
    _ _ _
    555 OUT ___________| |___________________| |___________________| |___

    -->||<--- effective trigger pulse width
     
  12. Jag Man

    Jag Man Guest

    Thanks, John. I may have misled you though, as I actually have an NPN (NTE
    123AP) transistor
    between the 4017 and the 555. E.g., pin 2 (a decoded output) of the 4107
    goes a 10K resistor, thence
    to the base of the transistor. Another 10k resistor connects the base to
    ground. The collector
    goes to pin 2 of the 555, and the emitter is grounded. Pin 2 is also
    connected through a 10k
    resistor to +12. This is to present pin 2 with a negative pulse, which is
    the way
    I've triggered 555 in other projects.

    So, does the analysis still hold?

    Ed
     
  13. John Fields

    John Fields Guest

    ---

    Yeah, but now your timing will look like this:

    __________ __________
    4017 OUT _______________| |____||___| |______
    ||
    _____ _______________ ____||___ ______
    RESET |__________| || |__________|
    ____ _______________ ____||___ _____
    TRIG |__________| || |__________|
    _ || _
    555 OUT __________________________| |__||______________| |___
    ||

    That is, the 555 will trigger on the trailing edge of the 4017 output
    instead of on the leading edge.
     
  14. Because that would make sense. You're confusing logic with your
    thinking. (a line by Gene Burrows from ABC TV Hollywood.)
    gg
     
  15. OK, thanks, understood. A more complex spec than I'd thought!

    From your separate thread 'Special "dual" pulse generator circuit
    (follow up)', I see you're sorted, apart from one residual issue.

    FWIW, simulating an opto-isolator fed with a 0-12V fast rise time
    pulse via a 10k base resistor gave me much faster output rise times
    than those you observed. With both models in its library, a 'generic'
    and the OP4N25, CircuitMaker gave the following results (which square
    with Chris's point about lowering the resistor value).

    Rc Rise time
    ---- ---------
    10k 300 us
    1k 50 us

    So I'd suspect some other cause. Is your circuit, or at least that
    section, posted anywhere?
     
  16. Jag Man

    Jag Man Guest

    Thanks for your interest, Terry.

    I just uploaded a scan of my circuit (thanks to Chris Foley) to:

    http://sowell.ecs.fullerton.edu/jag/fuelingsys/tester111704.jpg

    The diagram shows a 400/500 ohm resistor and an LED in the 2N3094 collector
    circuit, in the path from pin 2 of the 4N32. At the moment the resistor is
    a 5% 400 ohm, measuring about 380 ohms, and the LED has been removed.

    BTW, what version of CircuitMaker are you using, student of full? The full
    version is kind of pricy for someone not making their living in this stuff.


    Ed
     
  17. Jag Man

    Jag Man Guest

    Robert,

    Would a small capacitor in the input to the 4017 serfe as an "edge
    detecter?"


    TIA

    Ed
     
  18. Jag:

    If you are worried about the length of the pulse coming out of the 4017,
    you can use a capacitor and resistor to detect either a rising or
    falling edge of the output.

    To detect the rising edge, connect an output to the input of the next
    stage through the cap, and use a resistor to ground on the far side of
    the cap, as so:

    .---------.
    | |
    | | C
    | |
    | | ||
    | 4017 Qx|-----||--o---- Out
    | | || |
    | | |
    | | |
    | | \
    | | / R
    | | \
    '---------' /
    |
    GND
    created by Andy´s ASCII-Circuit v1.25.250804 www.tech-chat.de

    Then, when without a transistion for a while, the output will be
    grounded. When the Qx output goes from GND to 5V, however, the output
    will momentarily go up to 5V, and then will decay back to ground
    depending on the values of R, C, and the impedance of the output. The
    voltage at Out will fall to about 37% of the input pulse in t = R*C.

    When the 4017 output goes from 5V to 0V, the output will drop below GND.
    However, you can add a diode from GND to Out to prevent this.

    For a falling edge detector, you can use this (with the diode drawn in)


    VCC o--o
    | |
    .---------. \ |
    | | R / - Diode
    | | \ ^
    | | / |
    | | || | |
    | 4017 Qx|-----||--o--o- Out
    | | ||
    | | C
    | |
    | |
    | | Pulse length is ~RC
    | |
    '---------'
    created by Andy´s ASCII-Circuit v1.25.250804 www.tech-chat.de

    For your problem, I'd say you can use the second circuit between your
    4017, and the trigger input of a 555. It will generate a nice negative
    pulse. Then, you can set the output pulse width using the forumlas given
    in the 555 datasheet for the monostable configuration.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  19. Bill Bowden

    Bill Bowden Guest

    No, you can't get it any shorter because the
    transistor is holding the trigger (pin 2)
    at a low level during the entire input time,
    and the 555 output will not return low until the
    trigger line goes back up.

    If you use the capacitor, the transistor will
    only turn on during the time it takes to charge
    the capacitor, so you get a very short pulse at
    pin 2 and the 555 output will end earlier determined
    by the R and C parts at pin 6.

    Try the capacitor to see what happens. You can
    try different values of .1uF, .047uF, 0.01uF, etc.

    There is also a mistake in the other post as John
    Fields pointed out. The reset line (pin 4) will
    reset the 555 output to a low state when it is
    connected to ground. So it normally connects to
    +V for the 555 to operate.

    -Bill
     
  20. Jag Man

    Jag Man Guest

    Thanks again, Bill. Just out of curiosity, though, why does it work
    when I simply connect pin 4 to the same voltage seen by pin 2? That
    is, connected this way the output of the 555 can indeed be set shorter
    than the pulse width at 2/4. Curious.

    Ed
     
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