I have a single-shot 555 Timer. After triggering, output goes high for a period of time.

What I would like to do now: After Vout goes high and then low again, I want to disable the 555 timer for a period of 10 seconds. During this period, if trigeer signal is activated, it would not set Vouth high.

My current circuit:

Waveforms: Repeatedly activated trigger (green) and corresponding Vout (blue)

My theoretical idea is to delay a trig signal for a length of anticipated Vout (50ms-100ms). This delayed trig signal would go to Trig pin of another 555 Timer. This timer would generate Vout2 10s long. Then, I would add a MOSFET - Drain to Reset pin of first 555 Timer, Source to GND. Vout2 would be connected to gate, and on activation it would pull Reset of first 555 Timer to GND and disable the circuit for 10s.

Problem with circuit: How to delay trigger signal?

Other suggestions also welcomed.



  • \$\begingroup\$ easiest way is to use another 555 \$\endgroup\$ Commented Oct 27, 2018 at 23:02
  • \$\begingroup\$ ... to hold the first one in reset. \$\endgroup\$
    – Dave Tweed
    Commented Oct 27, 2018 at 23:12
  • \$\begingroup\$ I also thought so, and I am trying to implement it with another 555 (See paragraph under images in my question) \$\endgroup\$
    – DenR
    Commented Oct 28, 2018 at 1:48

1 Answer 1


Use two 555 chips. The falling edge of the first one's output triggers the second one to hold the first one in reset for 10 seconds.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ This is exactly what I had in mind! I even made similar circuit, but I was getting R3/C3 network wrong. Thanks! \$\endgroup\$
    – DenR
    Commented Oct 28, 2018 at 2:12
  • \$\begingroup\$ Yes, the R3-C3 time constant needs to be significantly less than the period of the first timer. \$\endgroup\$
    – Dave Tweed
    Commented Oct 28, 2018 at 2:15

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