Tutorial 8 - 555 Timer Circuit

Learning Objectives



The 555 timer is a popular circuit used for timing purposes.  It can be used in:

The 555 timer is a small 8-pin integrated circuit, and we can add resistors and capacitors to an external circuit to make it act as a monostable or an astable. 





Maximum current

100 mA

200 mA

Voltage range

2 15 V

4.5 4.5 V

Operating current

120 mA

10 mA


The pin arrangement (pin-out) is shown below:



555 timer in Monostable Mode

The 555 timer in monostable mode is constructed using the circuit shown in this diagram.


The circuit will give a single output pulse like this:

The components R and C determine the time period T of the output pulse. 

T 1.1 RC


Worked Example

What is the time period of the pulse given by a monostable if the value of R is 100 kW and the capacitor has a value of 20 mF?

Use T 1.1 RC


T = 1.1 100 000 W 20 10-6 F = 2.2 s


In using a 555-timer we need to be aware that the circuit has one or two little quirks:


Question 1

(a) If the capacitor has a value of 2.2 mF, what value of resistor would you use to achieve an output pulse of about 3 seconds?    


(b) You use a 2.2 F electrolytic capacitor while using the 555-timer in monostable mode. You include a resistor of the same value as the one you worked out in part (a). Comment on whether the period of the pulse will be 3 seconds.




555 timer in Astable Mode

The 555 timer can be wired up to produce a train of pulses by ensuring that the circuit is astable, which means that it is not in a stable state.  We can make astable circuits from other components, but the 555 timer gives a train of digital pulses. The diagram shows the circuit.



The output of the circuit is a square wave, as shown.


We need to consider some definitions:

 tH= 0.7(RA + RB)C

tL = 0.7 RBC

Ratio = tH tL

T = mark + space = tL + tH

f  =  ____1.4_____

  (R1 + 2R2)C


The time tH will be longer than tL, unless R1 is very small compared to R2.  If this is the case, then tH will be approximately equal to tL, but not quite equal.  We can say to a first approximation that the mark to space ratio is 1.  This will result in a square wave output.




What is the frequency of the square wave output from the circuit?

Use the formula

            f = ____1.4_____

                   (R1 + 2R2)C


From the diagram we can see that:

  • R1 = 4700 W

  • R2 = 2200 W

  •  C = 2.0 10-6 F

  We need to substitute these values into the formula:


                        f = ____1.4_____  = ____           1.4____________

                               (R1 + 2R2)C      [4700 + (2 2200)] 2.0 10-6


                         =  ____ 1.4_______ = 77 Hz

                             9100 2.0 10-6



Question 2

An astable has a 2.2 mF capacitor, R1 value of 10 k, and R2 value 20 k.

(a) Calculate the astable frequency.

(b) Work out the mark time, the space time, and the mark to space ratio.

(c) Is output a square wave?






  • T = 1.1.RC



f =  __1.44__  

     (RA + 2RB)C


Mark time tH = 0.7(RA + RB)C


Space time tL = 0.7 RBC


Mark to space ratio = mark time space time


T = mark + space = tL + tH.



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