how does this work? ( Schematic)

Here is a debouncing circuit using the 4093 Schmidt. Could someone please explain to me in depth as to how this debounced a signal from a switch and how it generally works.

Thank you very much or your help.


( Please do not include picture in quote as it makes scrolling and reading very tedious. Thanks)

When the switch closes, its contacts bounce for several milliseconds, so the switch is repeatedly on/off/on/.... When on, the gate input voltage is Vcc and the cap is shorted. When off, the cap begins to charge via the resistor, so the gate input falls slightly. Before it can fall to the level where the gate would change state the switch is on again and the cap is re-shorted, so the gate input rises almost instantly to Vcc again. This sequence repeats until the bouncing stops.
Here's a simulation to show the effect:-

Your debounce circuit does not look right to me. I am use to seeing the circuit more like in this image below:


So just for my sanity, I will explain using my circuit.
Problem: When pressing a button, the mechanical contacts springiness cause actual mechanical bounce which gives multiple false triggers _|_||||__|||||___ instead of __|_____.
Each bounce would appear as a switch press although you only pressed once.
Solution: Switch Debouncer, comprised of two functional blocks.
1. Bounce Filter (RC filter)
2. Schmidt trigger inverter

1. RC filter - Basically all the switch bounce is removed by the RC filter, this is due to the RC time constant which is large compared to the bounce time.
2. The Schmidt trigger Inverter - This is used due to the RC filter. The RC filter causes real slow rise and fall times which violates the rise and fall time specs of digital inputs. The schimdt trigger uses a different VIH input level than the VIL input level to prevent multiple false triggering on a slow noisy rising or falling edges.

I should have just used this link to begin with, this paper explains the switch bounce circuit very nicely, much better than me.
https://www.google.com/url?sa=t&rct...WieieFjFw&sig2=MaFDfb92rQrBgCZfU1NZ5w&cad=rja

The 470k is making the 10n one hundred times more effective by making the voltage on the input fall very slowly.
This means the voltage can rise very quickly via the switch but if the switch doesn't make good contact as the voltage is rising, the voltage will not fall to zero very quickly because the 10n has to gradually charge via the 470k and this is why the input does not see any "noise."