Circuit to discharge capacitor after opening the circuit

Question

I apologize upfront for another dumb newbie question (I'm in the beginning of learning circuitry).

I have a schematic like this in mind (disregard the specific nominal values of the components for now, please). It works okay in simulators.

But I suspect the capacitor is going to stay charged even after the circuit is open (the switch is 'off').

So that next time the circuit it closed again the LED's blink will be way shorter (if any).

I had an idea I could discharge the capacitor if I 'short circuit' it to the batter after the switch is off (using some kind of transistor trick).

Then I read the capacitor can only be discharged when it's 'shorted' to itself, so that the electric potential on it becomes 0.

Do you think you might suggest a solution for the capacitor discharge after the circuit is open?

P.S. I apologize again if the terminology used is not absolutely correct or something is unclear. I will try to clarify if you ask.

Answer 1

You could try something like the below:

^{simulate this circuit – Schematic created using CircuitLab}

Q1 is optional but will speed up the discharge. Otherwise, discharge takes place through D2 + R1 + R2 (if you leave out Q1). When present, Q1 operates in reverse-active mode (necessary if the supply voltage is >5V, otherwise you could use forward mode which would discharge the cap faster). This circuit does draw current through R2 if the switch is held, even though the LED is off. D2 is to prevent exceeding -5V on the LED.

An even easier and better way (if you can use that kind of switch) would be to use a contact on the switch to discharge the capacitor.

^{simulate this circuit}

Answer 2

The capacitor will eventually self discharge but this may be too long of a period. You can connect a resistor from the non battery side of the switch to the cathode of the LED to obtain a predictable discharge period.

The value of resistance determines the discharge rate of the capacitor. You can adjust this value in your simulation. Keep in mind that this resistor will draw current from your battery when the button is pushed so you should check this in your final design.