Introduction: Automatic Night Light
In our busy life, many of us forget to switch off the light morning. By using this simple circuit, the light will "automatically switch on" in night and "automatically switch off" in morning.
Step 1: Components Needed
1. Bread Board
2. LDR (Light Dependent Resistor)
3. Relay 6 volt DC (SPDT)
4. Diode 1N4001
5. Resistor 1K and 330R
6. Transistor BC547 (2)
Step 2: Setting the Circuit
Connect one end of the LDR with Vcc and another end with one end of Resistor 330 ohms. Connect the another end of resistor with Ground.
Connect LDR with Base pin (i.e., 2nd pin) of Transistor BC547.
Connect 1K resistor with Vcc and another end with Collecter pin (i.e., 1st pin) of Transistor BC547.
Connect the Emitter pin (i.e., 3rd pin) of Transistor BC547 with Ground Voltage
Connect the Collector pin of Transistor BC547 with Base pin of another Transistor BC547
Connect the Emitter pin with Ground Voltage
Step 5: Introducing Relay
Connect cathode side of Diode 1N4001 with Vcc.
Connect anode side of same Diode with Collecter pin of Transistor BC547.
Connect 'A' terminal of 6volt (SPDT) relay with Vcc and 'B' terminal with Diode's anode side.
Step 6: Setting the Bulb With Relay
Cut any one of the two terminals of the Bulb's wire in order to connect it with the relay.
Connect one wire with 'POLE' and another with 'NO' of the relay.
Step 7: Adding 6Volt DC
Connect the circuit with 6volt DC.
Don't switch on the circuit now.
Step 8: Adding 230 Volt AC
(Now you are going to play with 230 Volt AC current, this is the risky step in this Project, please do this with caution. Or otherwise please don't continue.)
Connect the Bulb with the 230v AC.
Step 9: Its Ready Now...
Now switch on the power supplies of the circuit, both 6v DC and 230v AC.
If your circuit is in dark, the Bulb automatically lights
If your circuit is in light, the Bulb automatically switch to off.
We have a be nice policy.
Please be positive and constructive.
what changes should i do if i have to use 9v battery as supply?