This doorbell system works such that when someone presses your calling bell switch during the night, not only the bell rings but the bulb connected to it also glows. In order to turn the bulb off, just press the reset pushbutton switch provided in the circuit. Place the bulb near the calling bell switch so that you can see the person pressing the calling bell before opening the door. So you can choose not to open the door to doubtful persons. During day time, the bulb doesn’t glow and only the calling bell sounds.
When calling bell switch S1 is closed, the bell rings and simultaneously transformer X gets AC supply. The output of this 6V-0V-6V/500mA step-down transformer is rectified by diodes D1 and D2. The rectified output is filtered by 1000µF, 25V capacitor C1 and fed to the collector of transistor BC547 (T1) via 2.2k resistor R1. A light-dependent resistor (LDR) is connected to the base of this transistor.
During the day, the LDR has a very low resistance as it receives continuous light. So when the calling bell switch is pressed, the transistor conducts and its collector is pulled to ground. Thus the next section of the circuit remains inactive and we hear the calling bell only.
The next section consisting of IC 7408 (IC1) and IC 7473 (IC2) gets a separate supply voltage of 5V from regulator IC 7805 (IC3) as shown in the figure.
During the night, as no light falls on the LDR, it has a very high resistance. So when calling bell switch S1 is pressed, transistor T1 doesn’t conduct. As a result, diode D3 is forward biased to make input pins 12 and 13 of IC1 high. Since IC1 is an AND gate, its output at pin 11 will be high. This output is fed to pins 1 and 14 of JK flip-flop IC 7473 (IC2).
For a given high input to the latch made up of IC2, its output pin 12 will be high. Thus transistor SL100 (T2) receives base current and conducts. This energises a 9V, 200-ohm relay (RL) and the 100W bulb connected to the relay glows. The bulb will glow until you press reset pushbutton switch S2.
This circuit costs around Rs 150.
When calling bell switch S1 is closed, the bell rings and simultaneously transformer X gets AC supply. The output of this 6V-0V-6V/500mA step-down transformer is rectified by diodes D1 and D2. The rectified output is filtered by 1000µF, 25V capacitor C1 and fed to the collector of transistor BC547 (T1) via 2.2k resistor R1. A light-dependent resistor (LDR) is connected to the base of this transistor.
During the day, the LDR has a very low resistance as it receives continuous light. So when the calling bell switch is pressed, the transistor conducts and its collector is pulled to ground. Thus the next section of the circuit remains inactive and we hear the calling bell only.
The next section consisting of IC 7408 (IC1) and IC 7473 (IC2) gets a separate supply voltage of 5V from regulator IC 7805 (IC3) as shown in the figure.
During the night, as no light falls on the LDR, it has a very high resistance. So when calling bell switch S1 is pressed, transistor T1 doesn’t conduct. As a result, diode D3 is forward biased to make input pins 12 and 13 of IC1 high. Since IC1 is an AND gate, its output at pin 11 will be high. This output is fed to pins 1 and 14 of JK flip-flop IC 7473 (IC2).
For a given high input to the latch made up of IC2, its output pin 12 will be high. Thus transistor SL100 (T2) receives base current and conducts. This energises a 9V, 200-ohm relay (RL) and the 100W bulb connected to the relay glows. The bulb will glow until you press reset pushbutton switch S2.
This circuit costs around Rs 150.
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