simulate this circuit – Schematic created using CircuitLab

I am trying to generate a single high pulse (+5V) in the node EN when power is switched on, and then I want EN to give a low signal (0V) for the rest of the time that power is on.

I found this answer which suggested this circuit. How does the capacitor C1 discharge when the power is switched off? This circuit doesn't seem to allow to current to flow in the opposite direction.

  • \$\begingroup\$ Note that the time constant (tau) of 470 Farads and 10kΩ = 470*10k = 4.7M seconds. If instead the C was 470µF, 0.00047*10k = 4.7s. \$\endgroup\$
    – rdtsc
    Sep 24, 2021 at 12:28

1 Answer 1


The normal way is to have a diode and discharge resistor like this: -

enter image description here

If there are other circuits attached to the +5 volt line then you can usually rely on those being the "discharge" resistor.

Or you can reposition the switch (and add a normally closed contact) such that when power is disconnected, it shorts out the +5 volt line to ground but, you still need the diode.

  • 1
    \$\begingroup\$ If I kept the new discharge resistor, why is the diode necessary? Can't the cap discharge through both R1 and the new resistor? \$\endgroup\$
    – NadAlaba
    Sep 24, 2021 at 11:17
  • \$\begingroup\$ @NadAlaba it sure can but the discharge time is then determined by both those resistors in series. So, if there is sufficient time period between charges, the capacitor will discharge to an adequate level for what you need. If on the other hand you have a short period between charges then you will need to use a diode to "short out" the 10 kohm resistor. \$\endgroup\$
    – Andy aka
    Sep 24, 2021 at 11:20
  • \$\begingroup\$ I get it now, thank you! \$\endgroup\$
    – NadAlaba
    Sep 24, 2021 at 11:29

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