A previous Design Idea describes a simple way to automatically turn off a battery after a preset on period to save battery life (Reference 1). This Design Idea presents a simpler way to perform the same function (Figure 1). Two gates of IC1, a quad two-input NAND Schmitt trigger, form a modified flip-flop. When you apply a 9 V battery to the circuitry, the output of IC1A goes high because the initial voltage on C1 is zero. The output of IC1B is low, which feeds back to IC1A through R2. C3 charges up through R3. The output of IC1C goes high because R6 is connects to ground. A P-channel MOSFET switch, Q1, is off, and the output IC1D goes high, which in turn charges C4 through R2.
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| Figure 1. | An improved power-off circuit automatically disconnects the battery after a preset on period. |
When you push momentary switch S1, IC1A’s output goes low because both of its inputs are high, and this output forces IC1B’s output high. The value of R2 is much smaller than R3, so that C3 holds a logic-level high when S1 stays on. When S1 goes off, C3 discharges through R3.
You can turn off the MOSFET switch in one of two ways. When tantalum capacitor C2 is charged up such that the voltage on IC1C’s input becomes lower than its threshold V–, IC1C’s output changes from low to high; this action turns off the MOSFET switch. C2 and R6 determine the duration of this automatic turn-off. With the values shown, the turn-off takes approximately six minutes. Meanwhile, the high-to-low transition on IC1D’s output forces IC1A and IC1B back to standby status through C4.
Alternatively, you can manually turn off the MOSFET switch by pushing S1. Because the voltage on C3 is low, closing S1 forces IC1A’s outputs high and IC1B’s outputs low. The high-to-low transition on IC1B’s output forces IC1C’s output to be high, which turns off the MOSFET. Because the value of C2 is fairly large, D1 provides a quick discharge route, and R4 limits the discharge current.
This circuitry consumes less than 0.2 µA of power during standby operation. Because the MOSFET switch has a low on-resistance, it has only a 2-mV loss when the load current is 100 mA. Add an LED with a current-limiting resistor in series to the load side if you need a power-on indicator.
Reference
- Gimenez, Miguel, “Scheme provides automatic power-off for batteries,” EDN
