This matchless circuit makes your dome light look cold. Usually when the automobile door is closed, the dome light goes OFF. With this circuit, you can have our dome light fade slowly in brightness & finally go OFF. This slow dimming of the light gives an excellent feeling at night. It looks romantic!
The circuit can be explained as follows: When the automobile door is open, the push to off switch of the door is ON & hence it charges the 22uF capacitor fully. The opamp is acting as a voltage follower & 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 & hence the switch goes OFF. When the switch is OFF, the capacitor starts discharging slowly through VR1 & the 10K resistor & 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 & the bulb fades out & finally when the capacitor is fully discharged, the bulb goes OFF.
After building the circuit, with the push-to-off switch in ON position (not pushed in) i.e. the automobile door open, fine-tune the preset VR2 to the necessary preliminary brightness of the bulb. Then push the switch in to turn it OFF(or close the door) & fine-tune VR1 for the time to bring the bulb from full brightness to OFF.
I would recommend you set VR1 & VR2 to their maximum values.
The circuit can be explained as follows: When the automobile door is open, the push to off switch of the door is ON & hence it charges the 22uF capacitor fully. The opamp is acting as a voltage follower & 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 & hence the switch goes OFF. When the switch is OFF, the capacitor starts discharging slowly through VR1 & the 10K resistor & 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 & the bulb fades out & finally when the capacitor is fully discharged, the bulb goes OFF.
After building the circuit, with the push-to-off switch in ON position (not pushed in) i.e. the automobile door open, fine-tune the preset VR2 to the necessary preliminary brightness of the bulb. Then push the switch in to turn it OFF(or close the door) & fine-tune VR1 for the time to bring the bulb from full brightness to OFF.
I would recommend you set VR1 & VR2 to their maximum values.
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