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
    2
    0
    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
    3,962
    1,097
    The capacitor is quickly charge through the Schottky diode.
    And slowly discharges through 2.2K resistor.
     
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  3. praondevou

    AAC Fanatic!

    Jul 9, 2011
    2,936
    488
    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.
     
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  4. simmsdk

    Thread Starter New Member

    Jul 16, 2011
    2
    0
    Aha.. a delay.. Thanks :)

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

    Moderator

    Jan 28, 2005
    9,030
    214
    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.

    hgmjr
     
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