0
\$\begingroup\$

I have a push button switch (single-pole double throw) connected to a latch as follows...

push button switch connected to latch

I was thinking about the timing diagram of the circuit above (above image is the default state of the circuit when the switch hasn't been pressed down).

Would this timing diagram sufficiently describe the output given the signals below?

I would just like clarification of my diagram before I implement this circuit on a breadboard as an example of switch debouncing. P is basically Q.

\$\endgroup\$
  • \$\begingroup\$ I believe one of your +Vcc should be some -Vcc or gnd or whatever. \$\endgroup\$ – Vladimir Cravero May 14 '15 at 13:03
  • \$\begingroup\$ I don't believe it does. One gate will always receive a 0 (when the switch has settled on one of the two ends) during a press / when not pressed? I don't want two inputs of 0 at any time. \$\endgroup\$ – The Audience May 14 '15 at 13:20
  • \$\begingroup\$ Sorry, I just did not look your diagram with enough attention. You are perfectly right. \$\endgroup\$ – Vladimir Cravero May 14 '15 at 13:21
1
\$\begingroup\$

Your diagram is mostly right. The only minor thing I'd look into is whether the switch is break-before-make or not. It probably is, and that's what you want. In that case, the previously low input will go high a little before the new one goes low. If this is not the case, then you can get some of the bouncing coming thru on one of the transitions.

\$\endgroup\$
  • \$\begingroup\$ I haven't seen the switch, yet, but it's described as being normally open (NO) when pressed down. The switch contact is normally closed (NC). Also, thanks; I'd probably say it is sufficient for my level as I wasn't aware of the concept "break-before-make" until just now. \$\endgroup\$ – The Audience May 14 '15 at 11:48
  • \$\begingroup\$ @TheAudience - The good thing you will have going for you here is that almost all switches are the break-before-make type. On the other hand the "make-before-break" type switch is usually the unique type with a special construction to allow for this behavior. \$\endgroup\$ – Michael Karas May 14 '15 at 12:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.