LED blinking: A simple project

The LED blinking project 

In this project we are going to create a simple led blinking circuit. No hard method is used in this circuit, simple few transistors and capacitors are used in this project. Anyone can make this project at home. It can be used to decorate your home or for some educational purpose. This project is made by using the principle of an astable multivibrator.

What is an astable multivibrator?

An astable multivibrator is a circuit that changes the DC input to a square wave of its own. Two transistors and two capacitors are used in this circuit. Which are the main components of this circuit. The capacitor gives a 180^○ phase shift. Tow capacitors are used in this circuit that’s why it gives a 360^○ phase shift. The transistors work as a switch in this circuit.

How astable multivibrator works?

Circuit diagram

In this circuit two transistors, Q1 and Q2 are used. The transistor has a property due to which the voltage below 0.7V can’t pass through base to emitter. That means it must be greater than 0.7V to pass any current from the base of a transistor to the emitter. If it is below that, the transistor will work as an open switch. The astable multivibrator is based on this principle. Among the two transistors, one of the transistor has high conducting capacity than the other. In this case, let us suppose the Q1 transistor has a higher conducting capacity than Q2. When the current is started to flow in the circuit the Q1 transistor works as a close switch and the current started to flow through RL1 and R2. In this stage, the Q2 transistor work as an open switch so no current passes through RL2 and R1. As long as the current flows through Q1 the capacitor C2 will be charging. When the value of the voltage of C2 becomes greater than 0.7V the Q2 will work as a close switch the current flows through the base and emitter junction of Q2. As current flows through Q2 there will be no sufficient voltage in the base-emitter junction of Q1 and it will be reverse biased.  The Q1 transistor works as an open switch (turn off). So, no current flows through the base and emitter of Q1. There will be no current flow through RL1 and R2. The current will flow through RL2 and R1. When this will be happening the capacitor C1 will start to charge and when it reaches 0.7V the Q1 will start to work as an open switch and Q2 becomes an open switch. The same loop will continue in this circuit.

What are the components required?

The components required are as

1)     Breadboard -1

2)     Capacitors - 2 (10 µF)

3)     Transistors - 2

4)     Jumper wires - a lot

5)     Resistors – 2(10 kΩ)

6)     LED – 2

7)     Power source – 9V

With help of those components we can design the circuit which is shown in below diagram.

Circuit designing on breadboard

In this circuit, two NPN transistors are used. The bases of both transistors are connected with opposite-sided capacitors and resistors. While the collectors are connected with the negative terminal of LEDs and capacitors of the same side. The emitters are connected to the ground. The positive terminals of LEDs are connected with the positive terminal of the battery and the negative terminals are connected with a terminal of capacitors.  Similarly, one terminal of both resistors is connected with the positive terminal of the battery and another terminal of both resistors is connected with a terminal of different capacitors.

How much will be the time period of blinking of two LEDs?

To calculate the time period, the following formula can be used.

T= 1.38 RC

Where R is the resistance and C is capacitance.

This formula is applicable when both resistors and capacitors have the same values.
The final circuit design will look like

Final circuit simulation

The value of resistors and capacitors we used in this circuit were 10 kΩ and 10µF respectively. By putting those values in the equation we get the time period for this circuit is about 0.138s. But in this simulation the time work as different way. This simulation’s time ratio is 40ms/s that means 1 sec in this simulation will be 40ms in real. So you will see around 3.46 s take to complete whole cycle in this circuit but in real its time period is 0.138s.