Switch-off delay for battery supply
A frequent annoyance with battery-operated equipment is that just after you switch it on you notice that the battery is flat. Quite probably, the last user (you?) has forgotten to switch it off.
How Switch-off delay works: This timer relies on a permanently connected supply in order for it to function. Triggering of the timer takes place by applying a signal to the trigger input (i.e. contact closing). Usually, this must come from the positive/live side of the supply powering the timer. From here, its the same in that the relay will energize when triggered. When the trigger is removed, timing commences but the relay remains energized. After the timing period has elapsed, the relay will de-energize.
The circuit described here makes sure that this will never happen again. A touch of the button, S1, is sufficient to let the equipment work for a predetermined period only.
An interesting feature of the circuit is that its quiescent current is 0.00 mA because T1 switches the timer off completely at the end of the cycle. Switching on is affected by the energy contained in the power-on pulse. When S1 is pressed, the supply voltage is available immediately across C2. Because of the differentiating action of R2-C2, the supply voltage is briefly connected to the V+ input of IC1 via D1.
This energy is sufficient to enable the IC and start the timer, whereupon T1 is switched on. This transistor provides energy to the IC for the remainder of the cycle. At the end of the cycle, T1 is switched off and provides no more energy. The one-time, t, is determined by:
t = (P1+R4) C3 seconds.
The maximum current switched by to must not exceed 350 mA. The supply voltage may lie between 5 and 15 V. The minimum trigger amplitude is 5 V. The switch-off time with the component value shown in the diagram is 1-100 seconds. The current drawn by the circuit during the switching interval is about 4 mA when the supply voltage is 6 V.