With the discharge pin no longer discharging the capacitor C1 begins to charge through R3 and the potentiometer. As the flip-flop output is 0V the output stage of the 555 inverts this and therefore the output switches to 9V. Even when the switch is depressed the transistor stays off because the flip flop remembers what the last input was (i.e. The flip flop output is connected to the base of the discharge transistor so when the switch is pressed the transistor stops discharging the capacitor. This short pules from COMP1 sets the flip flop and therefore the output of the flip flop will be 0V. Therefore the output of COMP1 is on for a short amount of time. This means that normally the output of COMP1 is off but when the switch is pressed, the inverting pin (for a short time), is at a lower voltage than the non-inverting pin. COMP1 inverting input is connected to the trigger pin and the non-inverting input is connected to 1/3VCC. This is done by pressing the switch (when the switch contacts are shorted the trigger input is connected directly to ground and therefore will have a voltage of 0V). To start the monostable the trigger voltage needs to quickly be dropped to 0V. The trigger input is at 9V (due to the pull-up resistor R1).The flip flop output voltage is therefore 9V.Both comparators inside the 555 chip have 0V on their outputs to the flip flop.
The voltage across the capacitor C1 is 0V.In the starting state of the circuit the following is true
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