I am fairly new to electronics and just had a quick question. I saw the following suggestion for a one shot trigger to wake up an ESP from for example a PIR sensor with one shot trigger and to read the data on another pin:
I do not understand how current can reach the base of the transistor and make the collector emitter path conductive. I always was under the impression that capacitors do not let current flow in a DC circuit and just charge up. I would have expected the capacitor in parallel to the base emitter path, so it would charge up and delay the current flow to the base and the resistor in series from interrupt signal to base to limit the base current. Any answer is greatly appreciated!
I am sorry if my English is not the best, it is not my first language.
This isn't a DC circuit.
The input is a pulse, not a constant voltage.
When the rising edge of the pulse arrives, the capacitor voltage can't change instantaneously, so the transistor base voltage will follow the pulse. Then a momentary current will flow and the capacitor will charge, dropping the base voltage back toward zero over a few milliseconds. When the falling edge of the pulse arrives it will discharge the capacitor, requiring a momentary current in the opposite direction.
Assuming the duty cycle is low, the pulse rising edge will need be strong enough to bias the transistor briefly into forward operation. The falling edge will not affect the transistor, just pull the base slightly negative as it pulls current through the 5k resistor.