usually in circuit design an interrupt signal is through an open drain nmos. In applications a pullup resistor is connected on this open drain to indicate an interrupt.
I would like to know how does this open drain NMOS works. I tried looking online, however, could not comprehend.
When the gate of the open drain goes high, the drain is sinking current through the pull-up resistor and therefore it goes low. When the gate is low, the transistor is switched off (no current) and the pull-up resistor will set the voltage to high.