Debouncing mystery

Discussion in 'General Electronics Chat' started by mikeysela, Jul 30, 2010.

  1. mikeysela

    Thread Starter Member

    Jul 24, 2010
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    this debouncing RC circuit with the 2 resistor and a cap with the schmitt trigger here: http://www.ganssle.com/debouncing-pt2.htm produces a high output from the schmitt trigger doesnt it? If i wanted a low output to a 4017 clock, would i have to change that circuit to a pull down resistor instead of pull up?
     
  2. BMorse

    Senior Member

    Sep 26, 2009
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  3. mikeysela

    Thread Starter Member

    Jul 24, 2010
    87
    1
    I see that..now i just have to figure out how to incorporate a switch , that only when pressed provides a low input to the clock.
     
  4. mikeysela

    Thread Starter Member

    Jul 24, 2010
    87
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    is it as simple as connecting my switch in series between the + and - terminals of a 9v battery?
     
  5. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    No; if you do that, you will place a dead short across the battery.

    On that same page you referenced, look at figure 3. It works very well. The inverter could be a 4106 Schmitt-trigger inverter, or a single 4093 Schmitt-trigger NAND gate made into an inverter by connecting two inputs together.

    Note that you must never leave any INPUTS to a CMOS logic IC unconnected. They must have a current path to either GND/Vss or Vdd/+V; otherwise they can oscillate and cause difficult-to-solve problems. Unused OUTPUTS should be left unconnected.
     
  6. sage.radachowsky

    Member

    May 11, 2010
    241
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    You're talking about Figure 2, right?

    That is an inverting buffer, so the output is normally low, and when the switch is pressed, it will go high. The tiny circle means that the output inverts.

    If you want to reverse that, then you can use a non-inverting buffer, or else you can add a second inverting buffer after the first. You probably have a dual or quad gate chip anyway, so you'll have an extra inverting buffer for free.
     
  7. mikeysela

    Thread Starter Member

    Jul 24, 2010
    87
    1
    Yea you are right, I have a HEX inverter chip at the moment. Will the debounce work as effectively if i use the same chip to invert the output twice and send that off to the clock?
     
  8. mikeysela

    Thread Starter Member

    Jul 24, 2010
    87
    1
    Thanks for the tip. But how do i introduce the switch into that circuit to be able to only operate the clock on each push of the switch? SPST swtich
     
  9. mikeysela

    Thread Starter Member

    Jul 24, 2010
    87
    1
    Also, how do u know if to tie the unconnected inputs to Vcc or Ground?
     
  10. hgmjr

    Moderator

    Jan 28, 2005
    9,030
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    It really does not matter whether you tie the unused inputs to Vcc or ground as long as you do one or the other.

    hgmjr
     
  11. sage.radachowsky

    Member

    May 11, 2010
    241
    38
    Yes, sure it will work just as effectively. The extra inverter adds almost no time delay, noise, or power consumption. There's no downside in this case.
     
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