There are different fire alarm projects. However, I’ll concentrate on simple fire alarm circuit that is based on (LDR) Light Dependent Resistor. It is also fitted with a lamp pair used to sense the fire. The alarm operates by sensing smoke produced during fire. When fire breaks out with smoke, the circuit produces a clear sound.
In absence of smoke, the light from the bulb falls directly on the LDR. The LDR resistance will go low and the voltage across it too (below 0.6V). The transistor will be OFF hence nothing will happen.
In the presence of sufficient smoke, the LDR resistance increases since the smoke resists the light from falling o it. This also increases the voltage across the LDR. This therefore causes the transistor to switch on. In return, it gives power to ICI and its output 5V. The tone generator IC UM66 (IC2) is powered to play music.
Resistor R6 is used to protect the transistor when R4 is turned to low resistance values. Resistor R1 and R2 will form a feedback network for TDA2002. Feedback signal from R1 and R2 junction to the inverting of the same IC input is coupled by C1.
The diode D1 and D2 combined drops 1.4 Voltage to give the rated voltage to UM66. UM66 is not able to withstand more than 4V.
Points to note
- A 32Q tweeter speaker can be used.
- You can use pot R4 to adjust the alarm’s sensitivity.
- General purpose NPN transistor like BC148, BC548 or 2N222 can be used for Q1.
- A bright LED with 1K resistor series can be used instead of a bulb.
- It’s possible to power the circuit from a 9V DC power supply or 9V battery.
A well regulated power supply is important for this circuit. This is because; a slight disparity in the power supply voltage can seriously affect its performance.