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Considering a +5V battery source, I have a set of input switches. These switches are push buttons which stay closed upon clicking and open upon clicking again.

One side of a switch is connected to the positive side of the battery while the other is connected to a pull-down resistor, 10 kΩ specifically, and my combinational circuit. This circuit contains 74LS-series DIP ICs.

circuit

With that set-up should I still consider debouncing my circuit or not?

I'm thinking whether this set-up can affect one of my sequential circuit which send pulses when one input is clicked with the use of XOR gates, capacitors and resistors.

Marcus Müller
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Orlando Lewis
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    obviously, that depends on your circuit! So, without you sharing that, it'll be hard to answer. Anyway, it's likely that debouncing is a good idea – but you've probably know that already, else you wouldn't ask here, would you? :) – Marcus Müller Dec 13 '16 at 10:41
  • Here's a sample of my circuit... https://www.dropbox.com/s/48nxpl10crki66p/Drawing1.png?dl=0 – Orlando Lewis Dec 13 '16 at 11:28
  • The output releases a pulse whenever one input is clicked. Without debouncing, will this have erroneous results? – Orlando Lewis Dec 13 '16 at 11:31
  • You copied that circuit from somewhere and didn't quite understand it, right? I'd like to push you in the right direction: what purpose does the third gate, and the resistor and capacitor serve? – Marcus Müller Dec 13 '16 at 11:39
  • It detects the positive and negative edges from the pulses of the preceding XOR gates. – Orlando Lewis Dec 13 '16 at 11:42
  • exactly. And what does "bouncing" mean in context of switches? – Marcus Müller Dec 13 '16 at 11:43
  • It's when the push button oscillates from high to low in short instances as you turn the switch close or open. – Orlando Lewis Dec 13 '16 at 11:46
  • If the bounce causes unacceptable behavior, yes; if not, no. – Scott Seidman Dec 13 '16 at 11:46
  • @OrlandoLewis - FYI that circuit as currently shown **will not work reliably** (or perhaps not at all). The 10kΩ pull-down resistor value is too high to guarantee that the input voltage to each LS TTL input will be below 0.8V (V(IL)), when the switches are open, so they may not be seen as "low". Note I(IL) for LS TTL is up to 0.4 mA. See detailed explanation on this previous question "[Pull down resistor on logic gate input](http://electronics.stackexchange.com/q/238224)". And similar (for "standard" TTL so I(IL) = up to 1.6mA) [on this answer](http://electronics.stackexchange.com/a/136572). – SamGibson Dec 13 '16 at 15:10

3 Answers3

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Yes, you should consider debouncing your push button if the push button does not do this for you.

MathiasE
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Generally, if you have a circuit that detects edges and triggers some action (counter, interrupt, etc) then you need to debounce, otherwise every edge of a bouncing switch will be counted.

Some applications simply poll logic level periodically looking for a high or low state. In those cases you might be able to get away without debouncing, but you need to study the specifics of the application.

Supa Nova
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You should debounce the switches if bouncing on opening or closing would cause trouble. This really should have been obvious.

We don't know what's in the "output" block in your diagram, so we can't say. It looks like you have a edge to glitch converter in front of it. Bouncing switches would therefore cause multiple glitches. Again, only you can say whether that's OK or not.

Olin Lathrop
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  • Whoever downvoted this, it would be useful to explain what exactly you think is incorrect, misleading, or badly written. I still believe this answer is correct. – Olin Lathrop Dec 15 '16 at 11:37