I want to design a debouncing circuit to count the pulses of a water meter with pulse emitter (it behaves like a push-button). The microcontroller is the msp430g2553. I want to eliminate the rebounds completely by hardware, no software. I have considered the following circuit but unfortunately I can not test it with a real water meter. As the rate of sending pulses is several in a minute (very low) I have designed an RC with a time constant of about 1s. Do you think it is an appropriate design? Thanks in advance.


simulate this circuit – Schematic created using CircuitLab

Best regards, Fran Martin

  • \$\begingroup\$ Cripes. Ever use google? See: ganssle.com/debouncing.htm \$\endgroup\$ – jonk Aug 13 '17 at 18:21
  • \$\begingroup\$ @jonk Thanks for your answer. Yes I used google, maybe I have not explained well. I know the circuit works for a pushbutton but I do not know the type of bounds of an impulse emitter (for example its duration), so I asked if a Tau of a 1 second would work. Best regards, Fran Martin \$\endgroup\$ – FranMartin Aug 14 '17 at 14:29
  • \$\begingroup\$ You have two \$\tau\$s. Not one. Its appropriateness is really up to the person who knows the situation better -- you. I gather your pulse emitter is just a reed relay? If so, these may have some bounce (less so, with mercury wetted.) But not as much as the circuit suggests. It's easy to buy a reed relay and get a magnet and just wire it in, by the way. So you can do some modest testing, though of course it won't be exactly the same situation as using the actual device. It's very easy to adapt a circuit, regardless. So don't over-think this. Just buy the water meter you feel is best. \$\endgroup\$ – jonk Aug 14 '17 at 15:19
  • \$\begingroup\$ @jonk Thanks again. I'll get a pulse-emitting counter and test it. Thank you for your correction. Could you explain why they are 2s? I understood that the Tau is the RC product:(33K+6k8)*27uF. \$\endgroup\$ – FranMartin Aug 15 '17 at 17:04
  • \$\begingroup\$ You have one \$\tau\$ for when the switch is closed and a different \$\tau\$ for when the switch opens. I think you can probably see why. (But if not, just ask and perhaps this may mean I'll have to write some kind of answer.) \$\endgroup\$ – jonk Aug 15 '17 at 17:32

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