The power off circuit ensures that any load is automatically removed from a battery. It is particularly useful in cars so that if the driver forgets to switch off the headlights, he does not return to a car with a flat battery a couple of hours later.
The only operating control is a spring-loaded change-over switch with a test position. When the button of the switch is pressed, the +ve supply line is connected to the base of T1 via R1 and R4, so that the transistor switches on. The relay then energized and its contact links the -ve terminal of the load to ground, so that the load is powered.
At the same time, CI is charged via RI. This capacitor and R3 determine the time constant of the circuit. When S1 is set to its center position, C1 discharges slowly via R3. When the base potential of T1 drops below 1.2 V, the transistor switches off, whereupon the relay contact connects C1 to ground via R2. The capacitor is then rapidly discharged completely. The current drawn by the circuit in this situation is nil. When S1 is in its lowest position, that is, connected to ground, the discharging process is halted immediately.
Diode D2 shows the status of the circuit. Resistor R5 should not have too low a value, otherwise, the LED may not show up properly.
The value of C1 may be increased to 4700 pF: the power-off time is then increased from 10 min to about 30 min. This capacitor should be a type with very low leakage current.
The relay may be 6-V or 9-V type to prevent it clattering when the car engine is started or a large load is switched on. Do not omit protection diode DI.
Since the circuit has no polarity protection, make certain that correct polarity is observed.
