# Section Content

# Practical Circuit 1: Astable Multivibrator

The astable multivibrator, also known as a free-running oscillator, can be easily built using a 555 timer. The output of an astable multivibrator is a square wave, series of pulses or an On-and-off signal with adjustable frequency, turn-on and turn-off time. This type of signal finds use in many digital and analog circuits. Whenever there is a need for a square wave signal or to turn something (a light emitting diode – LED) on and off at a regular interval and frequency, consider using an astable multivibrator. The circuit below is that of an astable multivibrator using a 555 timer.

# Bill Of Materials

Bill of materials required to build the astable multivibrator circuit shown below.

Component | Label | Quantity | Buy |
---|---|---|---|

Breadboard | – | 1 | ₦1600 View Add to Cart |

1KΩ Resistor | R1 | 1 | ₦10 View Add to Cart |

6.8KΩ Resistor | R2 | 1 | ₦10 View Add to Cart |

470Ω Resistor | R3 | 1 | ₦10 View Add to Cart |

100uF Capacitor | C1 | 1 | ₦10 View Add to Cart |

0.1uF Capacitor | C2 | 1 | ₦10 View Add to Cart |

Red LED | D1 | 1 | ₦10 View Add to Cart |

9V Battery with connector | BAT 1 | 1 | |

555 timer IC | U1 | 1 | |

Connecting Wires | – | – |

The circuit can easily be tested on a breadboard. Follow the illustration above to assemble and test out the Astable Multivibrator circuit on your breadboard.

The circuit is a simple circuit and works as follows: BAT1 supplies power to the circuit. A 9volts battery is used here. Passive components R1, R2 and C1 set the frequency of the output square wave at pin 3 of the 555 timer. Light emitting diode D1 is connected through resistor R3 to provide us with a visual clue of what is happening in the circuit. Whenever pin 3 of the 555 timer goes high, current flows through D1 and it lights up. Resistor R3 limits the current flowing through D1 to avoid burn out due to over current. Typically, most small (3mm – 5mm) LEDs can stand currents of up to 30mA. if higher brightness is required on D1, R3 can be reduced to 220 Ohms. Capacitor C2 shields pin 5 from noise pickup since its not being used.

The component values of R1, R2 and C2 as shown in the above schematic have been chosen to output a square wave with a frequency of 1Hz on pin 3 output of the 555 timer. The formula for choosing the component values of R1, R2 and C1 for a particular frequency is given by:

Let’s say we want an astable multivibrator with a frequency of 50Hz. Using the above formula, we can derive the appropriate values of R1, R2 and C1. There are three variables that need to be computed; to make things easy, we will choose a value of 1K Ohm for R1 and 1uF for C1. Why! you asked? Well, from my experience using 1K Ohm ensures that the circuit will always work. Values lower than 1K Ohm for R1 might still work but far too low values can cause stress on the internal discharge transistor of the 555 timer which could result in failure of the integrated circuit. For C1, you can start with a value of 1uF. You can always adjust this value and recompute if the resulting resistor values after calculation are too low for the required frequency. Some manufacturers advertise their 555 timers to be capable of 1MHz frequency of operation; I will advice you do not go beyond 500KHz if you don’t have the datasheet of the particular 555 timer chip you are using.

So, back to our 50Hz circuit. Rearranging the initial formula gives us:

Substituting values of f = 50, R1 = 1K Ohm and C1 = 1uF into the equation gives:

Solving the equation gives the value of R2 as 14,400 Ohms or 14.4K Ohms. This resistor value is not a standard resistor value. You might not be able to buy a resistor of this value from an electronics shop. To solve this problem, connect a 4.7K Ohm resistor and a 10K Ohm resistor in series for R2. This will result to an R2 value of 14.7K Ohm which will cause the frequency to fall a little below the required 50Hz. Experiment with different resistor values in series to see which gives you the required 14,400 Ohms. Alternatively, you can use a variable resistor that allows for more room of adjustment. For more precise frequencies, ensure that the tolerances of R1, R2 and C1 are low; say 1%. You will not be able to test this 50Hz astable multivibrator through the light emitting diode D1. This is because a 50Hz square wave will blink the LED on and off at a fast rate that your eyes won’t notice. As a result, the LED will apear to be constantly turned on. You can test the circuit by connecting a frequency meter or a loud speaker across the light emitting diode.

When talking about a square wave, we don’t just talk about the frequency but also the duty cycle. The duty cycle of a square wave is the ratio of the turn-on time to the period of the square wave.

Duty cycle is often expressed in percentage. A duty cycle of 50% means that the turn-on time is half the period of the square wave or that the turn-on time and the turn-off time of the square wave are equal.

It is possible to calculate the time-on and time-off periods of the square wave generated by the 555 timer astable multivibrator. From the preceding astable multivibrator circuit diagrams, the formular for the time-on and time-off periods are:

From these formulas, you will realize that time-on and time-off periods of the preceding astable multivibrator circuits can never be equal. The time-on period will always be greater than the time-off period; hence, our duty cycle will always be greater than 50%. This is because the timing capacitor charges through R1 and R2 but is discharged through R2 alone. Although a near 50% duty cycle square wave can be achieved by making R1 much smaller than R2, For lower duty cycles (below 50%) a small signal diode such as the 1N4148 or 1N4001 is usually connected across R2 and another in series with R2. Below is a circuit diagram of a 50Hz 555 timer astable multivibrator with duty cycle adjustable below 50%.

In this circuit, the frequency is 50Hz and the duty cycle has been set to 50%. Making R1 larger than R2 increases the duty cycle while increasing R2 and reducing R1 reduces the duty cycle. When both R1 and R2 are equal, the duty cycle is at 50%. Always ensure that the sum of R1 and R2 is constant else the frequency will change. The formula for deriving the frequency of operation for this circuit is:

The formula for the time-on and time-off periods is given by:

Using these formulas, let’s build a 555 timer astable multivibrator operating at 100Hz with a duty cycle of 30%.

first, we must derive the value of R1 and R2. we will use a 1uF capacitor for C1. Rearranging the frequency formula and substituting values gives:

The sum of R1 and R2 is 14400. By dividing this value by 2, we get the the values of R1 and R2 for 50% duty cycle. For our design, we need a 30% duty cycle. The value of R1 for this duty cycle is:

The value of R1 is 4320 Ohms. Since we already know the sum of R1 and R2, getting R2 becomes easy.

These resistor values will be really difficult finding in an electronics component shop. For ease of design, lets use 4.3K Ohms for R1 and 10K Ohms for R2. The final circuit design and its output waveform is shown below.

From the image showing the output waveform, you can see that the square wave time-on is shorter than the time-off. The duty cycle is clearly at 30%. The waveform shown here is what you will get when you connect an oscilloscope to pin 3 output of the circuit. A loudspeaker has been shown connected across the output in place of the LED. The loudspeaker should produce a 100Hz tone when the circuit is powered up. You can easily generate any tone frequency by using the frequency formula to derive the component values of R1, R2 and C1. Try building a 1KHz 50% duty cycle 555 timer astable multivibrator.