This document discusses the astable operation of the 555 IC timer. It begins by stating the lesson and chapter objectives, which are to calculate the time delay and frequency of a 555 timer circuit. It then describes the specific objectives of being able to draw the astable multivibrator circuit using IC 555, calculate the mark and space timing, and calculate the oscillator frequency using a given formula. An example calculation is shown to determine the mark, space, and frequency values given resistor and capacitor component values in an astable multivibrator circuit.

1. IT2001PA
Engineering Essentials (2/2)
Chapter 16 - Astable Operation of the 555
IC Timer
Lecturer Name
lecturer_email@ite.edu.sg
Nov 20, 2012
Contact Number

2. Chapter 16 - Astable Operation of the 555 IC Timer
Lesson Objectives
Upon completion of this topic, you should be able to:
 Students should be able to calculate the time
delay/frequency of a 555 timer circuit with given
formula.
IT2001PA Engineering Essentials (2/2) 2

3. Chapter 16 - Astable Operation of the 555 IC Timer
Specific Objectives
Students should be able to :
 Draw the circuit of an astable multivibrator using IC 555.
 Calculate the mark and space timing of the astable
multivibrator.
 Calculate the oscillator frequency using the formula :
Fosc = 1.44 /(R1 + 2R2) C
IT2001PA Engineering Essentials (2/2)

4. Chapter 16 - Astable Operation of the 555 IC Timer
The Astable Multivibrator
 It is a free-running oscillator.
 It outputs a rectangular wave and does not require any trigger input.
• Application as oscillators, square-wave enerators etc.
Input Circuit Output
Astable
None
Multivibrator
IT2001PA Engineering Essentials (2/2) 4

5. Chapter 16 - Astable Operation of the 555 IC Timer
555 as Astable Multivibrator
+V Output
R1 Vcc
8
7 3 Output
t1 t2
R2
555
t1 = 0.693 (R1 + R2 ) C
6
5 t2 = 0.693 R2C
2
1 0.01µF
C
1.44
fOSC =
0.693 (R1 + R2 ) C
IT2001PA Engineering Essentials (2/2) 5

6. Chapter 16 - Astable Operation of the 555 IC Timer
Example : To calculate the Mark , Space and the frequency of oscillations of the
output waveform. Given that C = 0.001µf, R1 = 4 KΩ and R2 = 100KΩ.
Solution : Mark, t1 = 0.693 (R1 + R2)C
Output
= 0.693 (4 KΩ + 100KΩ) 0.001µf Vcc
= 0.693 (4 000 + 100 000) 0.001 x 10-6
= 0.000 0721s
= 72.1µs 0
Space, t2 = 0.693 R2C t1 t2
= 0.693 (100KΩ) 0.001µf tosc
= 0.693 (100 000) 0.001 x 10-6
= 0.000 0693s
= 69.3µs
1.44
Frequency, fosc =
(R1 + 2)C
2R
1.44
= (4k + 200 k ) 0.001 x 10-6
= 7059Hz
IT2001PA Engineering Essentials (2/2) 6

7. Chapter 16 - Astable Operation of the 555 IC Timer
Next Lesson
IT2001PA Engineering Essentials (2/2) 7