Counter starting on digit "9"

Discussion in 'The Projects Forum' started by haxxx, Feb 14, 2009.

  1. haxxx

    Thread Starter Active Member

    Feb 14, 2009
    35
    0
    I'm new to electronics, i know only what i've read online.
    I built the attached circuit, with a few changes.
    http://www.hobbytron.com/electronic-lab-2.html
    (scroll to the bottom of linked page)
    I Changed the 555 to monostable, & added a push switch.
    When power is applied the counter
    boots up on the digit "9" instead of "0", which is undesirable.
    The state is set by the 4 outputs of the 4029.
    The 4029 can be reset by briefly setting the preset pin "1" high.
    how could i achieve this on power up.
    Also i've been studying the truth table for the 4511.
    Is there any danger in me using its inputs (eg.
    0 1 0 0 which drives the counter to the digit "4")
    to enter into a logic switching circuit. (NOR & AND Gates)
    And do any components need to be connected between them.

    Thanks
    Haxxx
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    Before we address your question: The circuit you referenced is set up as a decade down counter. is that what you want?

    EDIT: For power on preset, you can probably get away with a 1uF cap from vcc to pin 1 (Preset Enable), and 100k from pin 1 to GND. Add a 1N4148 diode across the resistor, cathode to pin 1.
     
    Last edited: Feb 15, 2009
  3. haxxx

    Thread Starter Active Member

    Feb 14, 2009
    35
    0
    Thanks Ron, I'll give it a try but could u please explain what actually happens in the state u outlined. yes i am using it as a decade counter.
    Thanx.
    Haxxx
     
  4. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    It sounds like you are getting a single false clock in the count-down mode at power up. You might be able to cheat and set up the power up jam value to 0001 instead of 0000. Then the false single down count would result in a count of zero at power up.

    hgmjr
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    When power is off, pin one will of course be at zero volts, and the cap will have zero volts across it. When power is switched on, the voltage on the cap cannot change instantaneously, so pin 1 rises abruptly to vcc, enabling the parallel load. The cap charges through the 100k resistor, slowly returning pin 1 to zero volts. The parallel load is disabled whenever the voltage on pin 1 is below ≈vcc/2 (this takes about 69 milliseconds).
    The diode is there to discharge the cap rapidly when the supply is switched off. This makes the circuit ready for another ON cycle, and also protects pin 1 from reverse bias. Note that I edited my previous post. The diode needs to be in parallel with the resistor, not the cap.
     
  6. haxxx

    Thread Starter Active Member

    Feb 14, 2009
    35
    0
    Thanks a lot, i appreciate it.
     
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