Exam paper question:
What happens to the voltage of the driver’s input when the switch is pressed and released. With the help of the information in Fig. 8.1 draw and explain how the driver’s output voltage changes with time.


Exam paper answer given:
At time t = 0 VIN = 0 and therefore VOUT = 4.5 V.
t = 3s, VIN goes to +5 V and VOUT drops to +1.0 V.
when switch is released capacitor discharges and VIN falls. VOUT remains at +1.0 V until VIN drops below 1.5 V at which point VOUT returns to 4.5V. This occurs at 15 s.


My answer:
When switch is pressed the capacitor is then discharged through the switch, then when its released the capacitor re-recharges - not discharging?


Where have I gone wrong?


Thanks

 

Your answer is right.
Capacitor is start charging when you release the button.
And discharge when you press the button.

 

Thanks Jony.

That question and answer were taken from a specimen question paper and mark scheme for a past official examination.

A little worrying.