Tutorial 8 - 555 Timer Circuit

Learning Objectives

• recall that a monostable circuit has one stable output state and one unstable output state;

• recognise, draw and use the circuit diagram of a monostable based on a 555 timer circuit;

• describe the operation of a monostable based on a 555 timer;

• calculate the time period of such a monostable using T = 1.1RC

• recall that an astable circuit has no stable output states but continually changes;

• recognise, draw and use the circuit diagram of an astable based on a 555 timer circuit;

• describe the operation of an astable based on a 555 timer;

• calculate the time, tL, that the output is low using tL = 0.7RBC

• calculate the time, tH, that the output is high using tH = 0.7(RA+RB)C

• calculate the output frequency using f = 1.44 ÷ (RA+2RB)C

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

• Monostable mode in which the output is a high for a time determined by the external circuit.

• Astable mode in which the output changes continually.

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.

Parameter

CMOS

Bipolar

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.

• When the push switch S is closed and released, the voltage at pin 2 goes from high to low to high again.

• This triggers the output to go to high. Pin 7 is also disconnected from zero.

• When the voltage across C gets to about 2/3 of the supply voltage, the output goes low.

• The period of the pulse is given by a simple relationship:

T » 1.1 RC

• Once triggered, the circuit cannot be re-triggered to extend the period T.

 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 Answer 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:

• The trigger period must be less than the output pulse.

• The 5 nF capacitor connected to pin 5 is needed to prevent false triggering.

• The circuit can produce brief dips in the voltage of the supply.  This can be countered by placing a large value capacitor across the supply rails.  This eliminates the voltage change (called decoupling).

• If electrolytic capacitors are used in the RC circuit, leakage currents and poor tolerances can result in the output pulse being greatly different to the value predicted by the formula.

 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:

• The mark time [t(H)] is the time at which the output is a 1.

tH= 0.7(RA + RB)C

• The space time [t(L)] is the time at which the output is a 0.

tL = 0.7 RBC

• The mark to space ratio = mark time ÷ space time.

Ratio = tH ÷ tL

• The astable period T is the time taken for one complete cycle, the mark and the space times added together.

T = mark + space = tL + tH

• The frequency = 1 ÷ period.

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 Answer: 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?

 Summary Monostable T = 1.1.RC   Astable 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.
 Links Ideas for using 555 timer Pages from Doctronics Animation from Falstad Data sheet Archive Video tutorial