I would like to know if it is possible to design a circuit which requires a single input oscillating between voltage high and voltage low for its output to be voltage high? If oscillation is not detected output of circuit would be voltage low.

In my application voltage high would be 3.3V and voltage low would be 0V. The frequency of oscillation is not so important for me but I can reliably produce PWM of up to a few thousand Hz on the MCU gpio.

Application would be to use this as a watchdog for an MCU. I have one now, but it relies on seeing a static GND signal from the MCU gpio, if this voltage achieved the MCU is assumed to be in good health, but this is not always true. During some total failures the MCU, I believe, can maintain this output voltage and not trigger a watchdog failure.

Is there a way to make such a circuit using simple ICs?

  • \$\begingroup\$ The type of circuit you have described in the first paragraph is a retriggerable one-shot, not a WDT (watchdog timer). It looks like you are missing an important requirement: when the strobe input stops, the output shall go low for some time, then go high to re-enable the MCU. (2) Why do you require an external WDT? Most MCUs have a built-in WDT which are strobed in firmware. \$\endgroup\$ Oct 13, 2022 at 3:21
  • \$\begingroup\$ @NickAlexeev I am using an esp32 and I have witnessed several failures when the module must be manually rebooted and is unresponsive. To manually reboot I just pull EN gpio low. To automate this I have developed some external pulse circuitry but it doesn't cover everything as I mentioned \$\endgroup\$
    – Feynman137
    Oct 13, 2022 at 3:33
  • 1
    \$\begingroup\$ Sounds like you have hardware and/or software problems. Using a watchdog to work around the issue is just masking the root cause. Nevertheless there are specific watchdog chips like the ds1232 and max691. \$\endgroup\$
    – Kartman
    Oct 13, 2022 at 4:48

1 Answer 1


One idea is a charge pump. You have to keep switching between high and low to keep pumping the charge from the input to the output. When pumping, the output voltage is the input voltage plus the pumping voltage minus two diode drops. Otherwise, it's just the input voltage minus two diode drops. If you set the input to zero volts, you have to pump it to get any voltage at all.

Image uploading is currently broken, so you don't get a schematic diagram.

Or you could use a watchdog IC.


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