When the car door is open, the push to off switch of the door is ON and hence it charges the 22uF capacitor fully. The op-amp is acting as a voltage follower and its output is same as the voltage across the capacitor, which is 12V when the capacitor is fully charged. Due to a high voltage at the output of the IC, the transistor saturates, turning ON the bulb to full brightness.
Now when the door is closed, the door switch is pushed in and hence the switch goes OFF. When the switch is OFF, the capacitor starts discharging slowly through VR1 and the 10K resistor and the voltage across it decreases slowly. Hence at the output of IC 741 also the voltage decreases gradually, hence decreasing the base current to the transistor. This produces a slowly decreasing current through the bulb and the bulb fades out and finally when the capacitor is fully discharged, the bulb goes OFF
Note: 2N3055 power transistor needs proper heat sink.
saved this sometime back wanted to make the same but not getting enuf time...neways pull the wires off the door switches. If the light goes off when you remove the wire from either one,means the door switch is not working properly
or simply try out the 4th one.. and btw the 2nd circuit i have the same one too but some comps are different number wise..
