# Power Failure Alarm. Home Utility Circuit 29

This circuit generates an alarm when the Mains power fails. It can be used as a Lift Alert so that the technician can immediately attend the stopped Lift. It is also helpful to know the power failure in labs where many instruments are switched on continuously.

The circuit is too simple and can be enclosed in a plug in type Adapter case. To make the circuit compact, a Transformerless power supply is used. So when the power is available, the rechargeable battery charges and when the power fails, the buzzer beeps using the power from the battery. No complex circuitry is included; instead a single relay is used to change over from the Mains supply to battery. This is only a prototype version and to get a loud alarm, only change needed is a 12 volt battery and a high sound beeper. Here the battery used is the 4 V 1 Ah mini Lead acid battery and the relay is 6V 100 Ohms type. Buzzer is 3-12 V DC Continuous type.

Warning! This circuit uses 230 VAC and exists at Lethal potential that can give a Fatal shock if touched. Build this circuit only if you are competent to handle high volt circuits. If you are a beginner and not familiar with AC circuit, leave this for safety.
Take all adequate safety measures during its testing. Test only after closing the case. Troubleshoot only after removing from the Mains.

The circuit has two sections. A Transformerless power supply and a Relay driven buzzer. Tried many Capacitors that can drive the relay and to provide sufficient current and voltage to charge the battery. After experimentation, finally selected the 225 J 400V capacitor. Its 2.2 uF and as per the calculation, it can deliver 101 mA current to hold the relay in the energized state as wells to provide charging current for the battery.

As a general rule, 1.3 uF capacitance give 0.06 Amps or 60 mA current. So the 2.2uF gives

2.2 x 0.06 / 1.3 = 0.101 Amps or 101 Milli Amps.

Tested with the power meter and it is showing that, the power supply is giving 104 mA current similar to the calculation. So it fixed. But its voltage is between 32-33 Volts at 104 mA current as shown by the Volt meter. But this is the open circuit without any load. (The volt meter has its own battery, so it is not giving any load).

But when the relay is connected, the voltage dropped to 10 volts due to the load. In other capacitors tested, the voltage drop is severe with relay as load and the relay not energized.

The output from the 225 J capacitor passes to the full wave rectifier which converts low value AC to DC. Capacitor C2 removes ripples. It should be a high value one like 100 volts to handle the voltage.

So what happens is, when the mains power is available, relay energize and its NO (Normally Open) contact makes connection with the Common contact. The Battery is connected to Common and the output voltage from the power supply passes through D5 to the NO contact. So in the energized condition of the relay, battery charges. At this condition, buzzer is disconnected from the battery, since it is connected to the NC (Normally Connected) contact of the relay. So when the power fails, relay de-energizes and the Common contact makes contact with NC. Buzzer then gets battery power and it beeps. Enclose the circuit in a shock proof case preferably in a Plug in type adapter case. No contact point should be present outside the case to prevent shock. Troubleshoot only after removing from mains.