In this project, we are making a simple wire break or burglar alarm using 555 IC. This circuit is easy to make, inexpensive, and beginner-friendly as it only uses three components 555 timer IC, a resistor of 100K, and a buzzer or a relay. The 555 timer IC is the most versatile and inexpensive IC, you don’t need many components if you are using it, it gets pretty much all the work done. You can use this circuit for various purposes whenever an alarm or wire break switch is required.
Hardware Components
The following components are required to make Burglar Alarm Circuit
| S.no | Components | Value | Qty |
|---|---|---|---|
| 1. | Input Supply DC | 9-12V | 1 |
| 2. | Enameled Wire | – | 1 |
| 3. | IC | NE555 Timer | 1 |
| 4. | Buzzer | – | 1 |
| 5. | Relay | 9-12V | 1 |
| 6. | Resistor | 100KΩ | 1 |
NE555 IC Pinout
For a detailed description of pinout, dimension features, and specifications download the datasheet of 555 Timer
Burglar Alarm Circuit
Working Explanation
The operating voltage of this circuit is 9-12V DC. A thin enameled wire is used as a sensor between two points A and B as shown in the figure, the wire can be wrapped around any place such as doors, cupboards, etc. and its lengths can be meters long depending on the place. So when the wire gets cut or breaks from any place it will send an input signal to the 555 timer IC.
The IC will provide a high signal from its output pin that will activate the buzzer or relay connected at its output. You can either connect a buzzer directly to the output of the IC or you can connect a relay instead. With relay, you will have the choice to connect any load or appliance and operate it with this circuit. Since the output current of 555 IC is 200mA so we can not connect any high current alarm or buzzer with the IC, in such cases a relay can be helpful.
