What does this do to the signal ?

Discussion in 'Homework Help' started by simmsdk, Aug 21, 2011.

  1. simmsdk

    Thread Starter New Member

    Jul 16, 2011
    Please look at the attached image for the schematic...

    My question is:

    What exactly does the schottky diode in parallel with the resister, and the capacitor do to the signal before it enters the last NAND.

    What is it good for ?
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    The capacitor is quickly charge through the Schottky diode.
    And slowly discharges through 2.2K resistor.
    simmsdk likes this.
  3. praondevou

    AAC Fanatic!

    Jul 9, 2011
    When pin 3 of IC1A goes HIGH the capacitor C1 is being charged rapidly through diode D1.
    When pin 3 of IC1A goes LOW the capacitor C1 is being slowly (more slowly than it was charged)
    discharged through R1.

    You create a delay on only one edge of the signal.

    If you join the inputs of the left gate and feed a digital signal to it, then the rising edge of the
    output signal IC1B pin 4 will be delayed, the falling edge remains almost unchanged.
    simmsdk likes this.
  4. simmsdk

    Thread Starter New Member

    Jul 16, 2011
    Aha.. a delay.. Thanks :)

    Good explanations, it will help me think electronically..
  5. hgmjr

    Retired Moderator

    Jan 28, 2005
    A better description of the effect of this circuit would be that the original pulse would be stretched so that the output is wider that the input.