While assembling the 555 timer based circuit below I noticed that 1) the intended delay on pin 3 only works if the switch on pin 2 is closed before attaching the circuit to the 9V power supply.

enter image description here Source: circuitdigest

2) If the power supply is already attached and pin 2 is opened, closing it will produce the desired turn on delay on pin 3, which is fine.

My first question is whether scenario #1 is indeed the expected behavior in monostable mode (or perhaps a flawed construction on my part). The second question is whether it is possible to provide a delay also in the case where pin 2 remained closed but the power supply would be removed and reattached later on.

PS: I have looked into a similar question but was not convinced it can help in this case.

  • \$\begingroup\$ That schematics is incorrect. For a start R1 is connected to the wrong side of the switch. There is also no current limiting resistor with the LED. \$\endgroup\$
    – HandyHowie
    Sep 21, 2018 at 11:43

1 Answer 1


Without the switch pressed, the TR input is open, hence undefined. With an undefined input, you can't expect any specific behaviour.

Maybe you intended R1 to be a pull-up? As drawn, it has no function.

What voltage do you expect on pin 6 when you apply power? As drawn, it is undefined. A diode could be added to discharge C1 to the power rail.

And as HandyHowie remarks, the LED needs a current-limiting resistor.

  • \$\begingroup\$ Thank you for your input Wouter! For TR, I was thinking as a next step to attach a 1n4148 diode to GND to prevent the undefined state. However, I am not 100% sure how I would discharge C1 via a diode across the power rail. Can you perhaps elaborate? \$\endgroup\$
    – gnzg
    Sep 21, 2018 at 12:55
  • \$\begingroup\$ The idea is: when power is removed, the circuitry will still try to draw some power, so it will draw the power rail doen to ~ ground (0V) level. Hence a diode to power will discharge your C when power is removed. \$\endgroup\$ Sep 21, 2018 at 13:05
  • \$\begingroup\$ Fantastic, I will try it out when I am back home this evening! Thank you! \$\endgroup\$
    – gnzg
    Sep 21, 2018 at 13:06
  • \$\begingroup\$ Well, sthat was only for a start... did you consider what will happen when you trim RV1 to 0 Ohm and the DC pin tries to discharge the C? \$\endgroup\$ Sep 21, 2018 at 13:32
  • \$\begingroup\$ IIRC there is a monostable circuit in the 555 datasheet. Why not use that?? \$\endgroup\$ Sep 21, 2018 at 13:32

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