R1______________10M 1/4W Resistor R2______________10K 1/4W Resistor C1______________47µF 25V Electrolytic Capacitor IC1____________7555 or TS555CN CMos Timer IC D1___________1N4148 75V 150mA Diode Q1____________BD681 100V 4A NPN Darlington Transistor LP1___________Existing Lamp Bulb, usually 12V 5W SW1____________SPST Existing Door-Switch SW2____________SPST Existing Bypass Switch
I recently forgot to close the door of my car after parking in the garage and I found the battery completely exhausted after the week-end, when I tried to start the engine on Monday morning.
This inconvenience prompted me to design a simple circuit, capable of switching-off automatically after a few minutes the inside courtesy lamp, the real culprit for the damage.
When the door is opened, SW1 closes, the circuit is powered and the lamp is on. C1 starts charging slowly through R1 and when a voltage of 2/3 the supply is reached at pins #2 and #6 of IC1, the internal comparator changes the state of the flip-flop, the voltage at pin #3 falls to zero and the lamp will switch-off.
The lamp will remain in the off state as the door is closed and will illuminate only when the door will be opened again.
The final result is a three-terminal device in which two terminals are used to connect the circuit in series to the lamp and the existing door-switch. The third terminal is connected to the 12V positive supply.
- With the values specified for R1 and C1, the lamp will stay on for about 9 minutes and 30 seconds.
- The time delay can be changed by varying R1 and/or C1 values.
- The circuit can be bypassed by the usually existing switch that allows the interior lamp to illuminate continuously, even when the door is closed: this connection is shown in dotted lines.
- Current drawing when the circuit is off: 150µA.