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UNIVERSITY OF MASSACHUSETTS DARTMOUTH
COLLEGE OF ENGINEERING
EGR 101 INTRODUCTION TO ENGINEERING THROUGH APPLIED SCIENCE
I
555 INTEGRATED CIRCUIT TIMER
THE 555 TIMER USED AS AN ASTABLE MULTIVIBRATOR
The outline and pin configuration for a 555 Integrated Circuit Timer is shown in
Figure 1. At this point, it is not necessary to discuss the internal devices in any
detail.
Figure 1. A typical 555 IC Timer.
In your project, the 555 timer is used in a configuration known as an Astable
Multivibrator, where an output voltage is delivered as a rectangular wave. The
pulse and space width can be controlled by properly selecting two time
constants.
In this configuration, the 555 timer is connected as shown in Figure 2. The time
constant that determines the pulse width depends on resistors RA, RB, and
capacitor C. The pulse width is the time it takes for the capacitor C to charge via
resistors RA and RB from 1/3VCC to 2/3VCC, as shown in Figure 3. While this
charging is taking place, the output voltage will be equal to VCC.
Figure 2. The capacitor C1 charges via RA and RB.
Capacitor C1 “charging”
VCC
Figure 3. The waveforms of the capacitor and output voltages during charging.
2
Inside the 555 timer there is a component known as a transistor, which is labeled
Q1. Think of the transistor as an ON/OFF switch, which is done electronically.
When the capacitor voltage reaches 2/3VCC, the transistor Q1 will be switched
ON, creating a short-circuit path to ground from pin #7 causing C1 to discharge
towards 0 Volts via resistor RB, as shown in Figure 4, and the output voltage to
fall to 0 Volt. When the capacitor voltage reaches 1/3VCC, as shown in Figure 5,
the internal transistor Q1 is switched OFF, causing the capacitor to charge
towards 2/3VCC again. This “cycle” repeats over and over again.
Figure 4. The capacitor C1 discharges via resistor RB.
3
Capacitor C1 “discharging”
0V
Figure 5. The waveforms of the capacitor and output voltages during discharging.
MATHEMATICAL DESCRIPTIONS OF THE OUTPUT WAVEFORM
Pulse Width
The mathematical expression for the capacitor voltage on the “charging” (pulse
width) portion of the output waveform is given by
t
 


vC (t )  VCC 1  e 


Solving for the pulse width (PW) yields
PW   ln 2 RA  RB  C1 .
Space Width
On the “discharging” (space width) portion of the output waveform, the capacitor
voltage can be expressed as
vC (t )  VCC e

t

The space width (SW) turns out to be
SW   ln 2 RBC1
4
Frequency
The frequency of the resulting rectangular output voltage (in Hz) is determined by
f0 
1
 RA  2RB  C1
Period
The period (in seconds) of the output voltage is the sum of the PW and SW, and
can also be determined as the reciprocal of the frequency.
T  PW  SW   ln 2 RA  2RB  C1
Duty Cycle
We can define a quantity that relates the output voltage’s “On-time” (the PW) to
the total period (T). This “Duty Cycle” is expressed as a percentage and is given
by
R  RB
%DutyCycle  A
100%
RA  2 RB
5
THE 555 TIMER CONFIGURATIONS USED IN THE PROJECTS
The 555 timer can be simulated using MultiSim, as shown in Figure 6. The
frequency of oscillation and/or duty cycle of the oscillator can be changed by a
combination of resistance and capacitance.
XSC1
G
T
VCC
9V
A
B
RA
10k
RB
68k
4
7
6
2
5
C1
0.01uF
8
U1
VCC
RST OUT 3
DIS
THR
TRI
CON
GND
LM555CN
1
LOAD
8
Figure 6. The basic 555 Timer Astable Multivibrator using MultiSim.
There are two modifications that can be made to this basic configuration which
allow independent control of the timing periods or frequency control by means of
an externally applied voltage.
Those circuits are discussed next.
6
THE 555 TIMER WITH INDEPENDENT CONTROL OF THE TIMING PERIODS
The frequency and duty cycle are mathematically related since each is a function
of both RA and RB. If we could “remove” resistor RB from the charging portion of
the output voltage, making its value equal to 0, the pulse width would be given by
PW   ln 2  RAC1 .
The space width would still be given by
SW   ln 2 RBC1 ,
In this case, the timing period of the output becomes independently adjustable,
allowing us to adjust the frequency and duty cycle individually.
This can be accomplished by connecting a diode across resistor RB, as shown in
Figure 7, in such a direction that it acts as a short circuit on the charging cycle
(eliminating resistor RB from the circuit) and an open circuit on the discharging
cycle (leaving resistor RB in the circuit).
XSC1
G
T
VCC
9V
65%
D1
A
B
RA
Key = A
100k
RB
68k
1N4007GP
C1
0.47uF
4
7
6
2
5
8
U1
VCC
RST OUT 3
DIS
THR
TRI
CON
GND
LM555CN
1
LOAD
8
Figure 7. A 555 Timer with independent control of the timing periods.
7
THE 555 TIMER AS A VOLTAGE-CONTROLLED OSCILLATOR (VCO)
Up until now, pin 5 of the 555 Timer IC has been left open. This pin is labeled
CONTROL (CON for short) and can be used to change the frequency of a 555
timer configured as an oscillator by connecting it to a variable voltage supply. In
the circuit shown in Figure 8, a potentiometer is connected end-to-end across the
9-V battery, and the potentiometer arm is connected to pin 5 of the 555 Timer.
XSC1
G
T
VCC
9V
A
B
RA
68k
RB
10k
Frequency
C1
0.01uF
4
7
6
2
5
8
U1
VCC
RST OUT 3
DIS
THR
TRI
CON
GND
LM555CN
1
LOAD
8
Key = A
100k
50%
Figure 8. A 555 Timer connected as a Voltage-Controlled-Oscillator (VCO).
The next example illustrates how to simulate and calculate in MultiSim the
frequency and duty cycle for a basic astable multivibrator.
8
MULTISIM EXAMPLE
A typical 555 Timer configured as an Astable Multivibrator is shown in Figure 9.
Let’s check the expected values of the frequency and duty cycle.
f 
1
 RA  2RB  C1
%DutyCycle 

1.44
10  (2)(68)103 (0.01106 F )
R A  RB
R A  2 RB
 100% 
10 K   68 K 
10 K   (2)(68 K )
 986 Hz
 100%  53.42%
The Oscilloscope data shows that the period of the output is 1.020ms, so the
frequency is
f 
1
T

1
1.02  103 s
 980 Hz
Figure 9a. Measuring the period of a 555 Timer configured as a 1-kHz oscillator.
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Checking the duty cycle,
Figure 9b. Checking the duty cycle of the 1-kHz 555 Astable Multivibrator.
%DutyCycle 
535.71 106 s
1.02  103 s
 100%  52.51%
10