How is this Schottky diode helping discharge the filter capacitor?

Question

I'm trying to filter the possible bounce of the mechanical switch S1. For a simulation of a sequence of pulses the voltage on C1 keeps adding, it discharges very slow, so after some 10 ms bounces the output signal is activated:

If I place the Schottky diode D1 then it works much better (saw this in some book) but I don't understand why. This is the result:

But anyway the voltage in InputBySW is still increasing, is there a way to improve this?

EDIT:

I guess the effect of the diode is just its leakage current helping discharge the capacitor:

So my question remains, how to decrease the discharge time of the capacitor?

Answer 1

It discharges very slowly because of the 20K resistor between it and the ground. What you need is a ON/ON switch which connects either to V+ or to GND. When it connects to GND, the capacitor discharges faster, whereas when connected to V+ it charges gradually thanks to R1. To discharge C1 immediately, you can add a schottky diode from from C1 to SW1 bypassing R1 (not visible on the schematic). R2 prevents bounces and spikes to hit the input before being swallowed by the capacitor.

Values of R1, R2 and C1 are aproximative and up to your discretion.

The Zener diode D1 is better than a voltage divider with two resistors because bounce spikes can be much higher than the supplied voltage. The zenner will always give a predictable voltage while a voltage divider will only divide the voltage it's given and may be higher than expected. Of course you can replace the zenner diode with a resistor if you want or doubling the zenner diode with a resistor in case the switch would be floating or if you can't put a ON/ON switch. If you can't use such a switch, then we must think about reducing the impedance of the resistor to ground. Or something else. But then, it will consume a little bit more power because the supply is 24V.

The schotky diode D2 prevents reverse current which happens inevitably with switch bounces.

^{simulate this circuit – Schematic created using CircuitLab}

^{simulate this circuit}

The second schematic shows a solution where ON/ON switch is not an option because we want to use a momentary switch. The idea is to fill C1 with already divided voltage. The zenner limits the voltage to the base voltage. C1 is only 1uF and R2 10K. Discharge time should 40x shorter than with 20uF + 20K.

Answer 2

I kept the zener on the circuit in order to use this leak current to help discharge the capacitor. I ended up using a different circuit because I needed a pulse on the output even if the switch is kept active.