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The circuit you have used relies on a LOW PULSE
The switch must be open ALL THE TIME and when it is pushed, the uncharged capacitor puts a pulse on the pin of the 555. During this time the capacitor charges and the switch must be opened for the capacitor to discharge. When the cap is discharged, the switch can be pushed again and the circuit will operate.
This circuit has been presented as it will create a fixed HIGH on the output no matter how long the switch is pressed.
You will have to tell us exactly what you want to do, for us to supply the correct circuit.
Your 555 does not have a power supply voltage so it won't work.
The 3.3V input pulse has an amplitude that is too small for a regular 555 (because its minimum supply is 4.5V) but will be fine if it is a Cmos 555 with a 3.3V supply voltage.
IF i lower the supply voltage of the 555 to 5V does that means the trigger will occur when the input goes below 1.67V (1/3 VCC)?
Also would you recommend a buffer on the input of the 555 trigger?
The output of an ordinary 555 with a supply that is only 5V is a high of +3.6V and a low of +0.01V.
The first transistor does not need a resistor to ground because the output low from the micro-controller is 0.0V.
The two 100k resistors and the 0.1uF capacitor allow a DC input to ground but the pin 2 trigger input gets pulsed properly.
Because if pin 2 is held low too long then the output stays high and the timer will not time out.
The collector of each NPN only goes low, the resistors pull it high.
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