Is this a functional debouncing circuit?

Question

Is there anything wrong with this debouncing circuit, aside from the fact that it can be made with a single Schmitt trigger and a capacitor that is charged when the button is not pressed?

Answer 1

Yes, many things wrong. Or at least not entirely properly designed. It will work and be functional and may even be used in practice, but it still has problems that are easily solvable and make it better and more reliable.

Another thing is, if the hardware debounce is even needed at all. If this is for a generic logic input, and the bouncyness or slow input change does not affect the circuit, if for example it is filtered by software or it can trigger a single action, the debouncing or Schmitt trigger is not needed. If however it goes to an input which must have single well-defined sharp edge, such as to generate one single event or interrupt or counter clock pulse or something, then the hardware debouncing by filtering and Schmitt trigger is definitely needed, or in some cases, an even better debounce might be necessary which requires a SPDT button.

The 10k resistor on left is useless. The switch has no relation to it and it will constantly have 5V supply over it and it just wastes power. There is no reason why it is or why it should be present in the circuit.

As the resistor on right keeps the capacitor discharged, pushing the button will short circuit the capacitor with power supply, immediately charging it with high pulse of current.

While the current surge is short, it is only limited by internal resistance of capacitor, copper trace wiring and switch contacts, and can be multiple amps, it will degrade the switch contacts over time, and will likely exceed the safe current limits of the capacitor and the button.

So as already mentioned, pushing the button causes a fast voltage and fast current step. Even worse if the button connection is scratchy and there are bursts of steps instead of one. If the button is user accessible at suitable location and the capacitor is further away near the chip, the PCB wiring between them has stray inductance. The large current trasients will cause large voltage transients and this may cause overshoots, undershoots and ringing at the chip input, and it may unintentionaly stress the chip input structures.

Answer 2

You should remove the 10K resistor across the 5 V supply. It has no function and wastes current.

After that, the remainder of the circuit you have is OK, as long as you use a Schmitt trigger buffer/inverter. This answer covers the operation of that remainder of the circuit.

It includes the note that the capacitor will draw a spike current from the supply as the switch contacts connect. In theory, this is an infinite current at t0. But not in practice, where the components and track impedances stop this happening and local circuit decoupling will stop the supply rail dipping. You can subsequently observe that on a 'scope with a real circuit.

The notion that the capacitor current will damage the switch contacts or dip a typically/well-decoupled supply is false. This is a very common and standard circuit, used, without problems for years, in very large numbers of boards made every year since I first saw it 40 years ago. Those systems are all around us. I myself always avoid capacitor debouncing and use MCU/CPLD/FPGA sampling to debounce switches. But this method is harmless to do and proven as such.