This simple circuit can sound an alarm when the AC power is off (or the voltage is below 50V).
The AC mains 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 across R3 to control the operating state of transistor T1 and MOS FET T2. Once the AC is powered off or the voltage is too low. The buzzer Bz1 will sound an alarm.
Since diode D1 acts as a half-wave rectification, it is a pulsed DC signal that is fed into transistor T1. Under normal conditions of AC power supply voltage. The voltage on R3 keeps T1 on, and the FET is off. Once the AC grid voltage is below 50V, the voltage on R3 drops below the threshold required for T1 to turn on, T1 is off, and the gate voltage of T2 is rising. Enough to turn T2 on. The buzzer sounds a strong alarm.
In order to prevent the alarm from basically consuming power under normal conditions of the AC grid, the resistors in the voltage divider are all high resistance. The current flowing through these resistors is less than 10μA. T2 selects MOS field effect tube. R5 can be selected to have a resistance of 10 MΩ (because the gate current of the MOS transistor is small), so that when T1 is turned on and T2 is turned off. The current through the circuit is only about 1μA, the ordinary battery can be used for several years, and the buzzer uses CEP-2260A. The 9V power supply consumes 5mA.
The test of the alarm is very simple. After the AC power is plugged in after installation, the buzzer should not sound. Then pull out from the AC power outlet, the buzzer should make a strong tone. Indicates that the circuit is working properly. But be careful: if the circuit is always plugged into the AC mains. Never touch the battery!
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