I have come up with the following circuit that can be used to momentarily generate either a LOW or HIGH pulse based on what push button is pressed.

Can you please tell me if there is any issues with it? Can I apply improvements or simplifications to my circuit? I feel like these two can be somehow combined into a single circuit, but I'm not sure how.

circuit that allows generation of a short pulse using push button, capacitor, resistor and diode

The intention is to make sure that if I press SW-ON, the voltage at OUT temporary goes all the way down to 0V. If I press SW-OFF, the the voltage temporary goes all the way up to VCC, which is about 12V DC.

(Note that the voltage at OUT is kept at about ½ of the VCC using a voltage divider, see below)

P.S. This circuit is used as part of a NE555 bi-stable latch that allows a single button to toggle between the ON and OFF states.

NE555 Bi-stable toggle push button to control a relay as latch circuit

As you can see above, the Trigger and Threshold pins are tied together and go to the voltage divider (two 10KΩ), and then goes to the net name SW. The wire named OUT in the first picture is directly connected to the SW net.

A small note regarding the RESET is that it's being used to receive an exclusive reset signal from another chip, so I don't want to use it here as an OFF signal. The 100nF cap is used to keep the initial state during powering up to be OFF as well.

Lastly, I'm aware of some other alternatives of doing what I want which are better. However, I want to learn about the potential flaws in my circuit design so I appreciate it if you could please help me in that regard.

  • 1
    \$\begingroup\$ Why don't you use a circuit simulator to try the ideas out? \$\endgroup\$
    – Andy aka
    Commented Oct 7, 2023 at 18:13
  • \$\begingroup\$ I attempted to use some that I found online but I couldn't quite figure out how to use them. I have no offline simulations tools, so currently I'm testing these on a breadboard. \$\endgroup\$ Commented Oct 7, 2023 at 18:17
  • \$\begingroup\$ @Andyaka Additionally I don't think I can learn from just using a simulator, rather I'm looking for others' experience on designing simple circuits such as these. I appreciate the thought, though. \$\endgroup\$ Commented Oct 7, 2023 at 18:18
  • 2
    \$\begingroup\$ Everyone's using them and, they are a godsend to anyone interested in circuit design despite the learning curve. Not understanding how to use one does not logically permit you to dismiss them as inapplicable to learning. \$\endgroup\$
    – Andy aka
    Commented Oct 7, 2023 at 18:18
  • \$\begingroup\$ @Andyaka Even though it might be against the rules, but I appreciate if you could suggest one or some that might be suitable in my case so I can try those as well. Thanks. \$\endgroup\$ Commented Oct 7, 2023 at 18:19

2 Answers 2


If you ask about your design - it is too complicated for this simple task. Also it does not solve a basic problem which is de-bouncing. You need just a 2x Schmitt-Trigger inverters and resistor + capacitor. The circuit is just an example. A pulse source is your switch.

enter image description here

  • \$\begingroup\$ Thanks, will try this out. Does this work based on the propagation delay, or is there some other reason to chain two Schmitt-Triggers? \$\endgroup\$ Commented Oct 7, 2023 at 18:35
  • \$\begingroup\$ Also, can you please explain why the caps in my circuit doesn't solve de-bouncing? I ask this because in my tests it seemed to work correctly. \$\endgroup\$ Commented Oct 7, 2023 at 18:36
  • \$\begingroup\$ Yes! you will take one, a positive pulse after the first gate, than a second one after the second gate. Your circuit generates both pulses at the same time from same output - that's probably not what are looking for? Listen to @Andy aka - use an LTSpice or other simulation software. Also, I don't understand why don't you make this thing on a breadboard and probe it with a scope? \$\endgroup\$ Commented Oct 7, 2023 at 18:39
  • \$\begingroup\$ What test? Did you probe your circuit with a scope? \$\endgroup\$ Commented Oct 7, 2023 at 18:44
  • \$\begingroup\$ While the RC filter smooths out rapid voltage changes, it doesn't have the distinct hysteresis thresholds of a Schmitt trigger. Thus, the RC circuit might be more susceptible to noise, especially if the noise amplitude is close to the voltage level being read as a digital transition. Because you're relying on the RC time constant to filter out bounces, there's often a trade-off between filtering effectiveness and response time. If the RC time constant is too long, the circuit might respond slowly to legitimate switch transitions. \$\endgroup\$ Commented Oct 7, 2023 at 18:49

This circuit gives you a trigger pulse of all supply range.
With 1uF it has a time constant 100ms.
If you want another pulse width just change capacitor.
A diode just quick discharges the cap when button is released. A diode also clamps the output voltage to -0.7V during cap discharge otherwise the output voltage goes negative to about -10V.


simulate this circuit – Schematic created using CircuitLab

Combined version:


simulate this circuit


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