Connect with us

pulse generation problem

Discussion in 'Electronic Design' started by Fred, Feb 12, 2007.

Scroll to continue with content
  1. Fred

    Fred Guest

    I stumble a bit on the following design problem. Should be easy for you

    I have a signal coming from one half of a 4538A monostable that generates
    15uS pulses at 30uS interval. From this, I want to generate a second pulse
    which starts at the _end_ of a 15uS pulse and ends only at the _beginning_
    of the next 15uS pulse.

    However, sometimes the next 15uS pulse can be delayed by as much as 5mS, in
    that case the second pulse should also last that long, that is, until the
    beginning of the next 15uS pulse.

    Note that the other half of the 4538 is free to use.

    Any suggestions or ideas to put me on the right track are welcome.

  2. Fred

    Fred Guest

    I knew I forgot a important detail :(

    The second pulse MUST be triggered by the transition at the end of a 15uS

  3. D from BC

    D from BC Guest

    Translating... :)
    Source: Tp = 30uS Ton = 15uS

    Function required: Neg edge trigger&reset on pos edge

    Solution: Use inverter??

    That's so easy it's gotta be wrong.....must be misreading this
    somewhere... :)
    D from BC
  4. John Fields

    John Fields Guest

  5. legg

    legg Guest

    AC-couple the pulse signals into a schmitt inverter, with a 30uSec
    input-coupling time constant, discharging to the positive rail.

    At start-up, you might get a false 'second' pulse, unless the
    rise-time of the supply is limited to below the 30uSec charging
    constant. Local RC supply decoupling can prevent this.

  6. Gibbo

    Gibbo Guest

    Unless your description is wrong (or I misread it) I'm with D from BC on
    this one. Just invert what you already have.
  7. Gibbo

    Gibbo Guest

    I just looked at the datasheet. It's already got a Q bar output which
    gives you what you want. Or at least what you asked for.
  8. Jamie

    Jamie Guest

    Set and Reset Flip Flop ?
    sounds simply to me.
  9. ehsjr

    ehsjr Guest

    Me too. Inverter fits his needs - or we're both
    missing the point.

  10. Fred

    Fred Guest

    That's the answer!

    Even though an inverter may seem to be the obvious answer as other people
    said, it isn't. It's a bit tricky to explain but in the absence of first
    pulse, there must be no second pulse. That is, _both_ must be low.
    That's why the second pulse must start by the transition at the end of the
    first pulse (negative or positive, doens't matter as 4538's got Q and Qbar

    Thanks guys!
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day