AC power failure alarm

This simple circuit will sound an alarm when the AC power is lost (or the voltage drops below 50V).

The AC power is half-wave rectified by diode D1 and connected in series with resistors R1, R2, R3 and R4 to form a voltage divider. A smaller voltage is obtained on R3 to control the working state of transistor T1 and MOS field effect transistor T2. Once the AC power is cut off or the voltage is too low, the buzzer Bz1 will sound an alarm.

Since diode D1 acts as a half-wave rectifier, a pulsed DC signal is sent to transistor T1. Under normal AC power supply voltage, the voltage on R3 can keep T1 turned on, and the field effect tube is in the cut-off state. Once the AC grid voltage is lower than 50V, the voltage on R3 drops below the threshold value required for T1 to turn on, T1 is cut off, and the gate voltage of T2 increases. It is enough to turn on T2, and the buzzer emits a strong alarm sound.

In order to ensure that the alarm consumes almost no power under normal AC power grid conditions, the resistors in the voltage divider are all high-resistance. The current flowing through these resistors is less than 10μA. T2 is a MOS field-effect tube. R5 can be selected with a resistance of 10MΩ (because the gate current of the MOS tube is very small). In this way, when T1 is turned on and T2 is turned off, the current passing through the circuit is only about 1μA. Ordinary batteries can be used for several years. The buzzer uses CEP-2260A. The 9V power supply consumes 5mA.

The alarm is very easy to test. After installation, plug in the AC power supply, and the buzzer should not sound. Then unplug it from the AC power outlet, and the buzzer should sound strongly, indicating that the circuit is working properly. But be careful: if the circuit is always plugged into the AC mains, never touch the battery!