Clock Signals: 555 Timer
Digital Electronics
© 2014 Project Lead The Way, Inc.
555 Timer
This presentation will
• Introduce the 555 Timer.
• Derive the characteristic equations for the charging and
discharging of a capacitor.
• Present the equations for period, frequency, and duty cycle
for a 555 Timer Oscillator.
Going Further….
• Detail the operation of a 555 Timer Oscillator.
• Derive the equations for period, frequency, and duty cycle for
a 555 Timer Oscillator.
2
What is a 555 Timer?
• The 555 timer is an 8-pin IC that is
capable of producing accurate time delays
and/or oscillators.
• In the time delay mode, the delay is
controlled by one external resistor and
capacitor.
• In the oscillator mode, the frequency of
oscillation and duty cycle are both
controlled with two external resistors and
one capacitor.
• This presentation will discuss how to use
a 555 timer in the oscillator mode.
3
Capacitor
• A capacitor is an electrical component that can
temporarily store a charge (voltage).
• The rate that the capacitor charges/discharges is
a function of the capacitor’s value and its
resistance.
• To understand how the capacitor is used in the
555 Timer oscillator circuit, you must understand
the basic charge and discharge cycles of the
capacitor.
4
Capacitor Charge Cycle
•
•
•
•
Capacitor is initially discharged.
Switch is moved to position A.
Capacitor will charge to +12 v.
Capacitor will charge through
the 2 k resistor.
Equation for Charging Capacitor
VC  VFinal  VInitial   1 - e
-t/RC
 V
Initial
Where :
VC  The voltage across the capacitor
VFinal  The voltage across the capacitor that is fully charged
VInitial  Any initial voltage across the capacitor as it begins to charge
5
Capacitor Discharge Cycle
•
•
•
•
Capacitor is initially charged.
Switch is moved to position B.
Capacitor will discharge to +0 v.
Capacitor will discharge through
the 3 K resistor.
Equation for Discharging Capacitor
VC  VInitial  VFinal   e -t/RC 
Where :
VC  The voltage across the capacitor
VFinal  The voltage across the capacitor that is fully discharged
VInitial  Any initial voltage across the capacitor as it begins to discharge
6
Capacitor Charge & Discharge
12 v
5V
VC
20 mSec
Time
0v
Switch has been at position B
for a long period of time. The
capacitor is completely
discharged.
Switch is moved to position
A. The capacitor charges
through the 2k resistor.
Switch is moved back to
position B. The capacitor
discharges through the 3k
resistors.
7
Block Diagram for a 555 Timer
Discharge (7)
Vcc (8)
Control Voltage (5)
-
Threshold Voltage (6)
+
+
T1
Flip-Flop
RESET
Q
SET
Q
COMP2
Output (3)
-
Trigger Voltage (2)
COMP1
8
Ground (1)
Reset (4)
Schematic of a 555 Timer in Oscillator Mode
5 Volts
RA
N/C
3.333 V
Discharge
RB
Output
1.666 V
Threshold /
Trigger
C
Ground
N/C
9
555 Timer Design Equations
tHIGH : Calculations for the Oscillator’s HIGH Time
THE OUTPUT IS HIGH WHILE THE CAPACITOR
IS CHARGING THROUGH RA + RB.
5v
3.333 v
Vc
1.666 v
0v

tHIGH

HIGH
Output
LOW
t HIGH  0.693R A  RB C
10
555 Timer Design Equations
tLOW : Calculations for the Oscillator’s LOW Time
THE OUTPUT IS LOW WHILE THE CAPACITOR
IS DISCHARGING THROUGH RB.
5v
3.333 v
Vc
1.666 v
0v

tLOW

HIGH
Output
LOW
t LOW  0.693RBC
11
555 Timer – Period / Frequency / DC
Period:
Duty Cycle:
t HIGH  0.693 R A  RB  C
DC 
t HIGH
 100%
T
DC 
0.693 R A  R B  C
 100%
0.693 R A  2RB  C
DC 
R
R
t LOW  0.693 RBC
T  t HIGH  t LOW
T  0.693 R A  RB  C  0.693 RBC
T  0.693 R A  2RB  C
A
A
 RB 
 100%
 2RB 
Frequency:
ƒ
ƒ
1
T
1
0.693 R A  2RB  C
12
Example: 555 Oscillator
Example:
For the 555 Timer oscillator shown below, calculate the circuit’s,
period (T), frequency (ƒ), and duty cycle (DC).
13
Example: 555 Oscillator
Solution:
R A  390 
Period:
RB  180 
T  0.693 R  2R  C
B
 A
T  0.693 390Ω  2 180Ω  6.8μ.
T  3.5343 mSec
Frequency:
ƒ
1
T
1
3.534 mSec
ƒ  282.941Hz
ƒ
C  6.8 F
Duty Cycle:
R  R   100%
R  2R 
390   180    100%
DC 
390   2  180  
DC 
A
A
DC  76%
B
B
14
Example: 555 Oscillator
Example:
For the 555 Timer oscillator shown below, calculate the value for
RA & RB so that the oscillator has a frequency of 2.5 kHz @ 60%
duty cycle.
15
Example: 555 Oscillator
Solution:
Frequency:
Duty Cycle:
1
1

 400Sec
f 2.5 kHz
T  0.693 R A  2RB  C  400Sec
T
T  0.693 R A  2RB  0.47F  400Sec
R A  2 RB 
400Sec
 1228.09 
0.693  0.47F
R A  2 RB  1228.09
DC 
R
R
R
R
A
A
A
A
 RB 
 100%  60%
 2RB 
 RB 
 0 .6
 2RB 
R A  RB  0.6R A  2RB 
R A  R B  0 . 6  R A  1 .2  R B
0 .4  R A  0 .2  R B
R A  0 .5  R B
Two Equations & Two Unknowns!
16
Example: 555 Oscillator
Solution:
Frequency:
Duty Cycle:
R A  2 RB  1228.09
R A  0.5  RB
Substitute and Solve for RB
R A  2 RB  1228.09 
0.5  RB  2 RB  1228.09 
2.5 RB  1228.09 
RB  491.23 
Substitute and Solve for RA
R A  2 RB  1228.09 
R A  2 491.23    1228.09 
R A  982.472   1228.09 
R A  245.618 
17
Going Further…
555 Oscillator
Detail Analysis
18
Detail Analysis of a 555 Oscillator
5v
3.333 v
Vc
1.666 v
0v
RESET
SET
HIGH
LOW
HIGH
LOW
HIGH
Q
T1
Q
LOW
ON
OFF
HIGH
LOW
19
Detail Analysis of a 555 Oscillator
5v
3.333 v
Vc
1.666 v
0v
RESET
SET
HIGH
LOW
HIGH
LOW
HIGH
Q
T1
Q
LOW
ON
OFF
HIGH
LOW
20
Detail Analysis of a 555 Oscillator
5v
3.333 v
Vc
1.666 v
0v
RESET
SET
HIGH
LOW
HIGH
LOW
HIGH
Q
T1
Q
LOW
ON
OFF
HIGH
LOW
21
Detail Analysis of a 555 Oscillator
5v
3.333 v
Vc
OUTPUT IS HIGH WHILE THE CAPACITOR IS
CHARGING THROUGH RA + RB.
1.666 v
0v
RESET
SET
HIGH
LOW
HIGH
LOW
HIGH
Q
T1
Q
LOW
ON
OFF
HIGH
LOW
OUTPUT IS LOW WHILE THE CAPACITOR
IS DISCHARGING THROUGH RB.
22
555 Timer Design Equations
tHIGH : Calculations for the Oscillator’s HIGH Time


 VInitial   1 - e RC   VInitial


VC  VFinal


RC 
1
2


e
1

V

V

V
  31 VCC

3
3
CC
CC
CC


t


VCC   VCC   1 - e RC   31 VCC


t
1
2

3 VCC  3 VCC

RC 
e
1



2
V


3
CC
t
2
3
2
3


 1 - e RC 


t
1
2


 1 - e RC 


t
t
1
2
 21  e

t
RC


1
ln 2   ln e RC 


t
 0.693  
RC
t HIGH  0.693 R C
t
t HIGH  0.693R A  RB C
23
555 Timer Design Equations
tLOW: Calculations for the Oscillator’s LOW Time


VC  VInitial  VFinal    e RC 




RC 
1
2


V

V

0

e


3
3
CC
CC


t


RC 
1
2


V

V

e


3
3
CC
CC


t
1

3 VCC

RC 

e


2
V


3
CC
t


  e RC 


t
t
1
2


1
ln 2   ln e RC 


t
 0.693  
RC
t LOW  0.693 R
t
t LOW  0.693 RBC


  e RC 


t
1
2
24
555 Timer – Period / Frequency / DC
Period:
Duty Cycle:
t HIGH  0.693 R A  RB  C
DC 
t HIGH
 100%
T
DC 
0.693 R A  R B  C
 100%
0.693 R A  2RB  C
DC 
R
R
t LOW  0.693 RBC
T  t HIGH  t LOW
T  0.693 R A  RB  C  0.693 RBC
T  0.693 R A  2RB  C
A
A
 RB 
 100%
 2RB 
Frequency:
ƒ
ƒ
1
T
1
0.693 R A  2RB  C
25