AC power outage alarm circuit diagram

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

The AC mains power is half-wave rectified by diode D1, and is connected in series with resistors R1, R2, R3 and R4 to form a voltage divider. A smaller voltage is divided on R3 to control the working status 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, what is sent to transistor T1 is a pulsed DC signal. Under normal conditions of the AC power supply voltage, the voltage on R3 can keep T1 turned on, and the field effect transistor is in a cut-off state. Once the AC grid voltage drops below 50v, the voltage on R3 drops below the threshold required for T1 to turn on, T1 cuts off, and the gate voltage of T2 rises. It is enough to make T2 conductive. The buzzer will sound a strong alarm.

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

The test of this alarm is very simple. After installation, plug it into the AC power supply, and the buzzer should not sound. Then unplug it from the AC power socket, and the buzzer should make a strong sound, indicating that the circuit is working properly. But be careful: if the circuit is always plugged into the AC mains, never touch the battery!

Figure 1 Circuit schematic diagram