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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.

  • \$\begingroup\$ Sorry, forgot to add a bit of context: this was a naive/unrealistic attempt to make a simple goal/score indicator. A somewhat power-saving indicator, where the current stops after a short (~1 sec) flash of an LED. \$\endgroup\$ Commented Jun 14, 2017 at 12:23

2 Answers 2


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

  • \$\begingroup\$ Thanks a lot for the suggestions! Given the requirements (see comment to the original question) I'm not sure I will be able to do it. I was looking for an extremely simple (self-made, probably) pair of contacts to short (on puck in the goal) which would disconnect after the puck is out. \$\endgroup\$ Commented Jun 14, 2017 at 12:26
  • \$\begingroup\$ Did you mean D2 + R1 + R2? Or may be I'm totally missing something (sorry, I'm in the very beginning of learning circuitry) \$\endgroup\$ Commented Jun 14, 2017 at 12:32
  • \$\begingroup\$ As an alternative, is there a 'kind of' transistor which is closed when there is current on the base and opens if there is 'no current' (ground?) on the base? Like a photo-transistor but with '+'/ground on the base instead of 'light/no light'? \$\endgroup\$ Commented Jun 14, 2017 at 12:33
  • \$\begingroup\$ @SomehowBuddy You could use a normally-closed SSR with a MOSFET output. Internally they use a depletion-mode MOSFET with a photovoltaic coupler. There might be a way to use an n-channel depletion MOSFET alone. \$\endgroup\$ Commented Jun 14, 2017 at 12:51
  • \$\begingroup\$ @SpehroPefhany Looking at your first circuit, I am confused. When the switch is first closed, C1 is discharged, the collector and emitter of Q1 are both at ground potential, and the base is at 9 V. I would expect reverse breakdown, but more importantly, I would expect it would be uncontrolled, because there is no current limiting resistance between Vcc and Q1's base. Am I missing something? \$\endgroup\$ Commented Jan 19 at 3:03

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.

  • \$\begingroup\$ Thanks. Those were my thoughts as well. But given the application of the circuit, it may happen the next circuit 'closing' may happen before the capacitor is discharged. This way it won't work. \$\endgroup\$ Commented Jun 14, 2017 at 12:34

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