Over the years, various power-off circuits have been published. However, the circuit proposed here is very simple and has excellent characteristics. It features a fast power-off transition, with only 8 mV voltage drop at power on and sub-µA leakage current at power off.
Importantly, many circuits of this type produce a rather sagging power-off transition, which is undesired. The circuit proposed here is capable of conducting with very low voltage drop for almost half an hour and then suddenly switch completely off within a fraction of a second.
The circuit in Figure 1 employs positive feedback to make the power-off transition much steeper. It also employs another trick to make the leakage current in the off-state extremely small. The circuit includes an LED as a demonstrator for the switch-off behavior and can be easily tested on a breadboard.
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| Figure 1. | The circuit diagram of the auto power-off timer circuit. | |
The working principle is as follows: We start with the capacitors discharged and hence both MOSFETs switched off. With no DC path at all, the battery current draw is basically zero.
We now put the timer in the on state via pushbutton S1. The charging current is limited by the 9 V block’s internal resistance. Capacitor C1 is almost instantly charged to the battery voltage. MOSFET Q1 switches on with very low voltage drop. Importantly, C2 is also discharged via Q2; its role is explained further.
Releasing the switch starts the power-off timer. Gradually, C1 discharges via R2 and after a while MOSFET Q1 slowly has a little bit more voltage drop between drain and source, causing transistor T1 to start conducting, getting its base current via R1 and the previously discharged C2. This drastically speeds up the discharge process of C1, quickly resulting in even more of a voltage drop across MOSFET Q1. Hence, a positive feedback loop comes into action, causing a much faster transition to the off state.
The power-off time can easily be approximated as 1.5×R2×C1. With the proposed component values this equals 1500 seconds, or 25 minutes.
Figure 2 shows the switch-off behavior as measured with a scope over the load. Note the voltage loss in the on-state is insignificant except for a little drop in the last few seconds of the 25 minutes on-time. Once the feedback loop comes into play, instant switch-off action occurs.
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| Figure 2. | The switch-off behavior of the power-off circuit as measured with a scope over the load. |
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The extremely low current at power-off is only possible thanks to the use of C2 and Q2. The system would work without them, but then would have a power-off leakage current of about 8 µA, caused by the current through 1 MΩ resistor R1. However, with C2 in series, the current will stop once the capacitor is charged.
Hence, the feedback loop is only active during the shutdown process and is not allowed to further conduct current after shutdown. Q2 is essential to discharge C2 when the power-on button is pressed, allowing a new cycle. If desired, a power-off pushbutton switch can be added between the gate of Q1 and ground, preferably with a 1 kΩ series resistor to protect that switch.
