Today I am going to make a circuit of auto cut off using 2N2222A transistor.This circuit is very simple.
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Step 1: Take All Components As Shown Below
Components required -
(1.) Battery - 9V x1
(2.) Transistor - 2N2222A x1
(3.) Resistor - 2.2K x2
(4.) LED - 9V
(5.) Battery clipper
(6.) Diode - 1N4007 x1
Step 2: Pins of This Transistor
Step 3: Connect All Components According to Circuit Diagram
Step 4: Connect Resistor to the Transistor
Solder 2.2K resistor to Base and Emmiter pin of the transistor.
Step 5: Connect +ve of Diode and Resistor to Base of Transistor
Next we have to solder +ve of the diode to base pin of transistor and
also solder a 2.2K resistor to the base pin of transistor as solder in the picture.
Step 6: Take a 12V LED
Here this LED is not of 9V.Therefore I connected a 220 ohm resistor to its +ve pin as you can see in the picture.
Step 7: Connect LED to the Circuit
Next solder +ve wire of LED to 2.2K resistor which is connected to the base pin of transistor and
also connect -ve wire of LED to collector pin of the transistor as you can see in the picture.
Step 8: Connect Battery Clipper Wire
Now we have to solder battery clipper wire to the circuit.
Solder +ve wire of battery clipper to +ve of LED and
-ve of battery clipper to emmiter pin of the transistor.
As you can see in the picture when I give power supply then LED is start glowing.
Step 9: Connect Charger to the Circuit
Now connect charger wire to the circuit.
As you can see in the picture when I give power supply for charging then LED gone turn off automatically.
~ When light will be available then LED will not be glow and as light will not be available then LED will start glowing automatically.
This type we can make auto cut off circuit using 2N2222A transistor.
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