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Driving 555 IC with a pulse stream

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

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  1. John Fields

    John Fields Guest

    When the driving signal pulls RESET and TRIGGER both low at the same
    time, RESET will dominate and the output pf the 555 will go low
    unconditionally. Then when the driving signal starts going high, it
    will release the reset when it climbs through about 0.7V (1.2V max).
    When the reset is released the driving signal will still be below
    2/3Vcc, so it will trigger the chip, forcing the output high and
    starting the charging of the timing cap. Then, when the driving
    signal goes higher than 2/3Vcc, the trigger will be released and the
    555 will time out according to the values of the timing resistor and
    capacitor. What's interesting about triggering the 555 this way (for
    your application) is that there's no need to cap couple to the trigger
    input since you want the thing to start and finish timing out when the
    RESET/TRIGGER signal goes high. What you _do_ have to be careful
    about, though, is that your input signal doesn't go low while the 555
    is timing out. But, for your application, I don't think that can
  2. John Fields explained this a few messages back, but it's been expunged
    from my server.

    It has to do with the fact that the reset pin is active low. So, when
    the input (connected to pins 2 and 4) is low, the chip is in reset, to
    the output is also low. When it starts to rise, the point at which the
    chip comes out of reset is apparently still below the point where the
    trigger activates. Thus, the output comes up for a single monostable cycle.

    Lowering the input back to 0 may cause the reset to be engaged before
    the trigger, which would mean the output would stay at 0. Even if it
    doesn't, the reset will quickly stop any output (ie, it might glitch for
    a very short while as the input goes down from 5V to 0V). You can filter
    out the glitch if it occurs using an RC lowpass filter.

    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. Bill Bowden

    Bill Bowden Guest

    As John Fields explained, it works because as the input at
    pin 2 and 4 moves up, the reset ends while the trigger is
    still active, so you get an output pulse starting at the
    same time as the input. Problem is, when the input moves
    from +V to ground, pin 2 may trigger another pulse before
    pin 4 resets the output. So you may get a very short pulse
    out at the end of the input pulse. But apparently, that's not
    happening so it looks like you solved the problem.

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