A switch COM is pressed and connected to high level, and then bounced to connect to the nmos tube below. The B+ and B- of the mos tube are connected to the + and-of the op amp. Can't understand whether the Nmos tube is connected to B+ or B- after it is turned on? What is the role of the op amp behind?
Andrew postato su July 31, 2020
Personal opinion, let's compare it. From the perspective of this circuit, if the switch is pressed, the B+ level is greater than the B- level, and the comparator outputs a high level 1. If it is not pressed, the B+ level is equal to the B- level, and it should output 0.
Samuel postato su July 31, 2020
B- is low from beginning to end, B+ is high when the switch is on, and low when the switch is not on, and the following circuit does not know.
David postato su July 31, 2020
The MOS gate voltage is 5V, the source is connected to GND through a resistor, and the MOS is turned on (depending on the turn-on voltage of the MOS). When A (COM) is connected to high (5V), 5V will charge or discharge the circuit of terminal A (depending on the circuit connected to terminal A), the drain stage of MOS (the negative terminal of the diode) is suspended, the source stage is connected to GND through a resistor, and the latter stage The op amp has no input and the output is 0. A is connected to the MOS (switch bounce) drain stage, and the A terminal signal is connected to the rear stage operational amplifier through MOS for amplification. But the strange thing is that the role of this MOS does not seem to be great, because it is always in the on state. Unless the voltage at the A terminal is too high, causing the MOS to not fully turn on or even turn off, it also means that the input of the subsequent op amp has voltage limitations.
