Electronic – How does this piezo input circuit work

1. At first sight, I would say the \$1k\Omega\$ resistor is used in combination with the \$560k\Omega\$ resistor as a voltage divider. This might sound like it decreases the sensitivity but the voltage output level of a piezo depends highly on the frequency spectrum of a push and the load impedance. The internal impedance of a piezo element was also relatively high in my experience, so for perfect matching the load resistance would have to be high as well. For that reason it is indeed possible that the voltage divider would render a larger voltage at the positive input of the operational amplifier.

2. The piezo does generate a negative pulse, but it is being short circuited by the diode. Depending at which point you measure your signal, it is possible that you measure two positive pulses.

3. If you remove the \$1k\Omega\$ resistor you could say that this is the case. But as I mentioned in answer 1, I would say that these two resistors were intended as a voltage divider. And again: the output voltage of the piezo element is highly dependent on the load resistance. If you choose unsuitable resistors the output voltage can become arbitrarily small. I would suggest you to conduct several measurements where you can estimate which resistor combination is best.

4. With this circuit the sensitivity can mainly be improved by two factors:

   1. Adjusting the directly connected load resistors (now \$560k\Omega\$ and \$1k\Omega\$) for optimal source/load matching.
   2. Adjusting the noninverting amplifier by adjusting the \$68k\Omega\$ and the \$1.2k\Omega\$ resistor using: $$ U_{out} = \Big(1+ \frac{R_2}{R_1}\Big)U_{in}$$ where: $$U_{in} = \frac{560k\Omega}{1k\Omega + 560k\Omega} \cdot U_{piezo}$$, $$R_1 = 1.2k\Omega$$ and $$R_2 = 68k\Omega$$

Moreover I'd say that the schmitt trigger is not really necessary unless the LED has to specifically be switched on at a certain level. Additionally if the task is only to switch on the LED by using a piezo element, you would not need the amplifier, the schmitt trigger etc, you could only implement a simple passive half wave or full wave rectifier and connect the LED to its output. With this setup there is also no need for an external power supply since the piezo element supplies the LED directly.

I hope this helps.

-Switched off: It draws current through the bulb. The current it draws is not enough to visibly light up the bulb. Certain type of LED bulbs may cause trouble with this switch though, as they prevent the switch from drawing enough current.

-Switched on: In creates a voltage drop of a few volts to keep the electronics active. The drop in brightness is not noticeable to the average human.

Regular phase-cut dimmers work in the same way.