*Important : If there is any difficulty in the working of circuit, reduce the value of R2 and try the circuit again. Reduce the value in steps ( 1 MOhm, 10 KOhm, 1 Kohm, 0 Ohm ).*We had already seen the basics of working of 555 in astable mode and the basics of working of 4017 Johnson counter IC. Before starting these tutorial, you must be aware about the basics of 555 in astable mode and 4017 johnson counter. Here we will design a johnson counter, in which the clock pulse is given from a 555 astable multivibrator. Circuit is done as shown in the following diagram. 5V is given from a 5V regulator.

*Important : If there is any difficulty in the working of circuit, reduce the value of R2 and try the circuit again. Reduce the value in steps ( 1 MOhm, 10 KOhm, 1 Kohm, 0 Ohm ).***Pinout diagram of 555**

**Pinout diagram of 74HC4017 / 74HCT4017**

From the 555 circuit, we get a continuos stream of square wave of predetermined frequency. Frequency of the square wave depends on the values of R1, R2 and C1. By varying the values of R1, R2 and C1, square wave of desired frequency can be generated and is clearly explained in the working of 555 in astable mode.

Generated square wave is obtained from the OUT pin ( pin 3 ) of 555 and is given as clock pulse to the CP0 ( pin 14 ) of 4017. From the basics of 4017, we had seen that, if clock pulse is given to CP0, counter will count during the positive transition of clock pulse.That is, when clock pulse goes from LOW to HIGH, counter will increment count.

**Calculation of Frequency of Generated Pulses**

Frequency of the rectangular pulses generated depends on the values of R1, R2 and C1.

*Frequency, F = 1 / ( ln( 2 ) . C1 . ( R1+2R2 ) )*

**T**_{on}= ln( 2 ) . C1 . ( R1 + R2 )

**T**_{off}= ln( 2 ) . C1 . R2C1- Capacitance in Farad, R1 and R2 - Resistance in Ohms, F - Frequency in Hertz, T

_{on}- Time for which output will be HIGH ( LED will be ON ), T

_{off}- Time for which output will be LOW ( LED will be OFF )

**When S1 and S2 are OFF**

Only one capacitor will be included in the circuit. Then

C1 = 10 nano Farad = 0.01 micro Farad = 0.01 * 10

^{-6 }Farad

R1 = R2 = 10 Mega Ohm = 10 * 10

^{6 }Ohm

ln ( 2 ) = 0.693

Substituting these values in the equations, we get

F = 4.81 Hz, Ton = 0.1386 S, Toff = 0.0693 S

**When S1 is ON and S2 is OFF**

Two capacitors will be in parallel in the circuit. Then

C1 = 10 + 10 nano Farad = 20 nano Farad = 0.02 micro Farad = 0.02 * 10

^{-6 }Farad

R1 = R2 = 10 Mega Ohm = 10 * 10

^{6 }Ohm

ln ( 2 ) = 0.693

Substituting these values in the equations, we get

F = 2.405 Hz, Ton = 0.2772 S, Toff = 0.1386 S

**When S1 and S2 are ON**

Three capacitors will be in parallel in the circuit. Then

C1 = 10 + 10 + 10 nano Farad = 30 nano Farad = 0.03 micro Farad = 0.03 * 10

^{-6 }Farad

R1 = R2 = 10 Mega Ohm = 10 * 10

^{6 }Ohm

ln ( 2 ) = 0.693

Substituting these values in the equations, we get

F = 1.603 Hz, Ton = 0.4158 S, Toff = 0.2079 S

From the calculations, we can conclude that, when the value of capacitance increases, output frequency and hence speed of running LEDs decreases as shown in the animated image.

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