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# Digital Electronics

555 Timer
555 Timer
2
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.
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
5
¸ ) ¸ )
Initial
RC -t/
Initial Final C
V e - 1 V V V + - =
Equation for Charging Capacitor
charge to begins it as capacitor the across voltage initial Any V
charged fully is that capacitor the across voltage The V
capacitor the across voltage The V
: Where
Initial
Final
C
=
=
=
 Capacitor is initially discharged.
 Switch is moved to position A.
 Capacitor will charge to +12 v.
 Capacitor will charge through
the 2 KO resistor.
Capacitor Discharge Cycle
6
¸ ) ¸ )
-t/RC
Final Initial C
e V V V - =
Equation for Discharging Capacitor
discharge to begins it as capacitor the across voltage initial Any V
discharged fully is that capacitor the across voltage The V
capacitor the across voltage The V
: Where
Initial
Final
C
=
=
=
 Capacitor is initially charged.
 Switch is moved to position B.
 Capacitor will discharge to +0 v.
 Capacitor will discharge through
the 3 KO resistor.
Capacitor Charge & Discharge
7
0 v
12 v
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 2KO resistor.
Switch is moved back to
position B. The capacitor
discharges through the
3KO resistors.
20 mSec
5 V
V
C
Time
Block Diagram for a 555 Timer
Control Voltage (5)
Threshold Voltage (6)
Trigger Voltage (2)
Ground (1)
Vcc (8) Discharge (7)
Reset (4)
Output (3)
-
+
-
+
RESET
SET
Q
Q
COMP1
COMP2
Flip-Flop
T1
8
Schematic of a 555 Timer in Oscillator Mode
5 Volts
N/C
N/C
Discharge
Threshold /
Trigger
Ground
Output
1.666 V
3.333 V
R
A
R
B
C
9
555 Timer Design Equations
10
t
HIGH
: Calculations for the Oscillator¶s HIGH Time
¸ ) C R R 0.693
B A
+ =
HIGH
t
5v
3.333 v
V
c
1.666 v
0 v
p t
HIGH
n
Output
HIGH
LOW
THE OUTPUT IS HIGH WHILE THE CAPACITOR
IS CHARGING THROUGH R
A
+ R
B.
555 Timer Design Equations
11
C 0.693R
B
=
LOW
t
t
LOW
: Calculations for the Oscillator¶s LOW Time
5v
3.333 v
V
c
1.666 v
0 v
p t
LOW n
Output
HIGH
LOW
THE OUTPUT IS LOW WHILE THE CAPACITOR
IS DISCHARGING THROUGH R
B.
555 Timer ± Period / Frequency / DC
12
¸ )
¸ ) . J . J
¸ )C R 2 R 693 . 0 T
C R 693 . 0 C R R 693 . 0 T
t t T
C R 693 . 0 t
C R R 693 . 0 t
B A
B B A
LOW HIGH
B LOW
B A HIGH
+ =
+ + =
+ =
=
+ =
Period:
¸ )C R 2 R 693 . 0
1
F
T
1
F
B A
+
=
=
Frequency:
¸ )
¸ )
¸ )
¸ )
% 100
R 2 R
R R
DC
% 100
C R 2 R 693 . 0
C R R 693 . 0
DC
% 100
T
t
DC
B A
B A
B A
B A
HIGH
-
+
+
=
-
+
+
=
- =
Duty Cycle:
Example: 555 Oscillator
13
Example:
For the 555 Timer oscillator shown below, calculate the circuit¶s,
period (T), frequency (F), and duty cycle (DC).
Example: 555 Oscillator
14
Solution:
F 6.8 C 180 R 390 R
B A
Q = O = O =
¸ )
¸ )
mSec 534 . 3 T
F 8 . 6 180 2 390 693 . 0 T
C R 2 R 693 . 0 T
B A
=
- O - + O =
+ =
Q
Hz 282.941 F
mSec 534 . 3
1
F
T
1
F
=
=
=
¸ )
¸ )
¸ )
¸ )
% 76 DC
% 100
180 2 390
180 390
DC
% 100
R 2 R
R R
DC
B A
B A
=
-
O - + O
O + O
=
-
+
+
=
Period:
Frequency: Duty Cycle:
Example: 555 Oscillator
15
Example:
For the 555 Timer oscillator shown below, calculate the value for
R
A
& R
B
so that the oscillator has a frequency of 2.5 KHz @ 60%
duty cycle.
Example: 555 Oscillator
16
Solution:
¸ )
¸ )
¸ )
¸ )
¸ )
B A
B A
B A B A
B A B A
B A
B A
B A
B A
R 5 . 0 R
R 2 . 0 R 4 . 0
R 2 . 1 R 6 . 0 R R
R 2 R 6 . 0 R R
6 . 0
R 2 R
R R
% 60 % 100
R 2 R
R R
DC
- =
- = -
- + - = +
+ = +
=
+
+
= -
+
+
=
Frequency: Duty Cycle:
¸ )
¸ )
09 . 1228 R 2 R
09 . 1228
f 47 . 0 693 . 0
Sec 400
R 2 R
Sec 400 f 47 . 0 R 2 R 693 . 0 T
Sec 400 C R 2 R 693 . 0 T
Sec 400
kHz 5 . 2
1

f
1
T
B A
B A
B A
B A
= +
O =
-
= +
= + =
= + =
= = =
Q
Q
Q Q
Q
Q
Two Equations & Two Unknowns!
Example: 555 Oscillator
17
Solution:
B A
R 5 . 0 R - =
Frequency: Duty Cycle:
09 . 1228 R 2 R
B A
= +
Substitute and Solve for R
B
O =
O =
O = + -
O = +
23 . 491 R
09 . 1228 R 5 . 2
09 . 1228 R 2 R 5 . 0
09 . 1228 R 2 R
B
B
B B
B A
¸ )
O =
O = O +
O = O +
O = +
618 . 245 R
09 . 1228 472 . 982 R
09 . 1228 491.23 2 R
09 . 1228 R 2 R
A
A
A
B A
Substitute and Solve for R
A
18
Going Further«
555 Oscillator
Detail Analysis
Detail Analysis of a 555 Oscillator
19
5v
3.333 v
V
c
1.666 v
0 v
RESET
HIGH
LOW
SET
HIGH
LOW
HIGH
LOW
T1
ON
OFF
Q
HIGH
LOW
Q
20
5v
3.333 v
V
c
1.666 v
0 v
RESET
HIGH
LOW
SET
HIGH
LOW
HIGH
LOW
T1
ON
OFF
Q
HIGH
LOW
Q
Detail Analysis of a 555 Oscillator
21
5v
3.333 v
V
c
1.666 v
0 v
RESET
HIGH
LOW
SET
HIGH
LOW
HIGH
LOW
T1
ON
OFF
Q
HIGH
LOW
Q
Detail Analysis of a 555 Oscillator
22
5v
3.333 v
V
c
1.666 v
0 v
RESET
HIGH
LOW
SET
HIGH
LOW
HIGH
LOW
T1
ON
OFF
Q
HIGH
LOW
Q
OUTPUT IS LOW WHILE THE CAPACITOR
IS DISCHARGING THROUGH R
B.
OUTPUT IS HIGH WHILE THE CAPACITOR IS
CHARGING THROUGH R
A
+ R
B.
Detail Analysis of a 555 Oscillator
555 Timer Design Equations
23
¸ )
¸ )
¸ )
¦
'
+

'

=
¦
'
+

'

=

+
¦
'
+

'

- =
+
¦
'
+

'

- =
+
¦
'
+

'

- =

RC
t
2
1
RC
t
CC
3
2
CC
3
1
CC
3
2
CC
3
1 RC
t
CC
3
2
CC
3
2
CC
3
1 RC
t
CC
3
1
CC CC
3
2
Initial
RC
t
Initial Final C
e - 1
e - 1
V
V V
V e - 1 V V
V e - 1 V V V
V e - 1 V V V
¸ )
¸ )C R R 693 . 0 t
C R 693 . 0 t
RC
t
693 . 0
e ln ln
e
e - 1
B A HIGH
HIGH
RC
t
2
1
RC
t
2
1
RC
t
2
1
+ =
=
=
¦
'
+

'

=
=
¦
'
+

'

=

t
HIGH
: Calculations for the Oscillator¶s HIGH Time
555 Timer Design Equations
24
¸ )
¸ )
¸ )
¦
'
+

'

=
¦
'
+

'

=
¦
'
+

'

- =
¦
'
+

'

- =
¦
'
+

'

- =

RC
t
2
1
RC
t
CC
3
2
CC
3
1
RC
t
CC
3
2
CC
3
1
RC
t
CC
3
2
CC
3
1
RC
t
Final Initial C
e
e
V
V
e V V
e 0 V V
e V V V
¸ )
C R 693 . 0 t
R 693 . 0 t
RC
t
693 . 0
e ln ln
e
B LOW
LOW
RC
t
2
1
RC
t
2
1
=
=
=
¦
'
+

'

=
¦
'
+

'

=

t
LOW
: Calculations for the Oscillator¶s LOW Time
555 Timer ± Period / Frequency / DC
25
¸ )
¸ ) . J . J
¸ )C R 2 R 693 . 0 T
C R 693 . 0 C R R 693 . 0 T
t t T
C R 693 . 0 t
C R R 693 . 0 t
B A
B B A
LOW HIGH
B LOW
B A HIGH
+ =
+ + =
+ =
=
+ =
Period:
¸ )C R 2 R 693 . 0
1
F
T
1
F
B A
+
=
=
Frequency:
¸ )
¸ )
¸ )
¸ )
% 100
R 2 R
R R
DC
% 100
C R 2 R 693 . 0
C R R 693 . 0
DC
% 100
T
t
DC
B A
B A
B A
B A
HIGH
-
+
+
=
-
+
+
=
- =
Duty Cycle:

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

 In the time delay mode.  This presentation will discuss how to use a 555 timer in the oscillator mode. the delay is controlled by one external resistor and capacitor. the frequency of oscillation and duty cycle are both controlled with two external resistors and one capacitor. 3 .  In the oscillator mode.What is a 555 Timer?  The 555 timer is an 8-pin IC that is capable of producing accurate time delays and/or oscillators.

you must understand the basic charge and discharge cycles of the capacitor.  To understand how the capacitor is used in the 555 Timer oscillator circuit. 4 .  The rate that the capacitor charges/discharges is a function of the capacitor¶s value and its resistance.Capacitor  A capacitor is an electrical component that can temporarily store a charge (voltage).

resistor.Capacitor Charge Cycle     Capacitor is initially discharged. Switch is moved to position A. Equation for Charging Capacitor VC ! . Capacitor will charge to +12 v. Capacitor will charge through the 2 K.

VFinal  VInitial v .

.e 1 Where : VC ! The voltage across the capacitor -t/RC  V Initial VFinal ! The voltage across the capacitor that is fully charged VInitial ! Any initial voltage across the capacitor as it begins to charge 5 .

resistor. Capacitor will discharge to +0 v. Equation for Discharging Capacitor VC ! .Capacitor Discharge Cycle     Capacitor is initially charged. Switch is moved to position B. Capacitor will discharge through the 3 K.

VInitial  VFinal v .

e Where : -t/RC 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 .

resistors.Capacitor Charge & Discharge 12 v VC Time 0v 5V 20 mSec Switch has been at position B for a long period of time. Switch is moved to position A. Switch is moved back to position B. The capacitor is completely discharged. The capacitor discharges through the 3K. The capacitor charges through the 2K. resistor. 7 .

Block Diagram for a 555 Timer Vcc (8) Discharge (7) Control Voltage (5) Threshold Voltage (6) + COMP1 Flip-Flop T1 RESET Q COMP2 Trigger Voltage (2) Ground (1) + - SET Q Output (3) 8 Reset (4) .

Schematic of a 555 Timer in Oscillator Mode 5 Volts RA N/C Discharge 3.333 V RB Threshold / Trigger 1.666 V Output C Ground N/C 9 .

693 . 5v 3.333 v Vc 1.555 Timer Design Equations tHIGH : Calculations for the Oscillator¶s HIGH Time THE OUTPUT IS HIGH W HILE THE CAPACITOR IS CHARGING THROUGH RA + RB.666 v 0v p tHIGH n HIGH Output LOW t HIGH ! 0.

R A  RB C 10 .

666 v 0v p tLOW n HIGH Output LOW t LOW ! 0.693R BC 11 .333 v Vc 1.555 Timer Design Equations tLOW : Calculations for the Oscillator¶s LOW Time THE OUTPUT IS LOW W HILE THE CAPACITOR IS DISCHARGING THROUGH RB. 5v 3.

555 Timer ± Period / Frequency / DC Period: t HIGH ! 0.693 .

693 .R A  R B C t LOW ! 0.693 RBC T ! t HIGH  t LOW T ! ? .

R A  RB CA ?0.693 .693 RBCA 0 T ! 0.

R A  2RB C DC ! DC ! Duty Cycle: DC ! t HIGH v 100 % T 0.693 .

693 .R A  RB C v 100 % 0.

R A  2RB C .

R .

R A A  RB v 100%  2RB Frequency: F! F! 1 T 0.693 .

R A  2RB C 1 12 .

period (T). frequency (F). and duty cycle (DC).Example: 555 Oscillator Example: For the 555 Timer oscillator shown below. calculate the circuit¶s. 13 .

Example: 555 Oscillator Solution: R A ! 390 .8 QF T ! 0. C ! 6.693 . Period: RB ! 180 .

693 .R A  2RB C T ! 0.

8 QF T ! 3.941 Hz .534 mSec Frequency: Duty Cycle: DC ! 1 F! T 1 F! 3.390. v 6.534 mSec F ! 282.  2 v 180.

R  R v 100% .

R  2R .

 180 .390 . v 100% DC ! .

 2 v 180 .390 . A B A B DC ! 76% 14 .

Example: 555 Oscillator Example: For the 555 Timer oscillator shown below. 15 .5 KHz @ 60% duty cycle. calculate the value for RA & RB so that the oscillator has a frequency of 2.

5 kHz T ! 0.693 .Example: 555 Oscillator Solution: Frequency: T! 1 1 ! ! 400 QSec f 2.

R A  2R B C ! 400 QSec T ! 0.693 .

09 .R A  2R B 0.47Qf Duty Cycle: DC ! .693 v 0. 0.47Qf ! 400 QSec R A  2 RB ! 400 QSec ! 1228.

R .

R A A  RB v 100% ! 60%  2RB .

R .

6  2RB R A  R B ! 0.6.R A A  RB ! 0 .

09 Two Equations & Two Unknowns! 16 .4 v R A ! 0 .R A  2R B R A  R B ! 0 .2 v R B R A ! 0 .2 v R B 0 .6 v R A  1 .5 v R B R A  2 RB ! 1228.

5 v R B R A  2 RB ! 1228.09 . 2.5 v RB  2 RB ! 1228.23 . Substitute and Solve for RA R A  2 R B ! 1228.09 . R A  2 .5 RB ! 1228. RB ! 491.Example: 555 Oscillator Solution: Frequency: R A  2 R B ! 1228. 0.09 .09 Substitute and Solve for RB Duty Cycle: R A ! 0 .09 .

491. R A  982. R A ! 245.09 .618 .472 .23 .09 . ! 1228. ! 1228. 17 .

Going Further« 555 Oscillator Detail Analysis 18 .

666 v 0v RESET HIGH LOW SET HIGH LOW Q T1 HIGH LOW ON OFF Q HIGH LOW 19 .Detail Analysis of a 555 Oscillator 5v 3.333 v Vc 1.

666 v 0v RESET HIGH LOW SET HIGH LOW Q T1 HIGH LOW ON OFF Q HIGH LOW 20 .Detail Analysis of a 555 Oscillator 5v 3.333 v Vc 1.

666 v 0v RESET HIGH LOW SET HIGH LOW Q T1 HIGH LOW ON OFF Q HIGH LOW 21 .333 v Vc 1.Detail Analysis of a 555 Oscillator 5v 3.

666 v 0v RESET HIGH LOW SET HIGH LOW Q T1 HIGH LOW ON OFF Q HIGH LOW OUTPUT IS LOW W HILE THE CAPACITOR IS DISCHARGING THROUGH RB.Detail Analysis of a 555 Oscillator 5v 3.333 v OUTPUT IS HIGH W HILE THE CAPACITOR IS CHARGING THROUGH RA + RB. 22 . Vc 1.

555 Timer Design Equations tHIGH : Calculations for the Oscillator¶s HIGH Time VC ! .

e RC ¸  1 V 2 1 V ¹ 3 CC 3 VCC ! .e © ª 1 2  t RC ¸ ¹ º ¨ 1 .VFinal ¨ 1 .e RC ¸  V  VInitial v © ¹ Initial º ª t t 1 2 ! ¨1 .

e RC ¸  1 V 2 2 ¹ 3 CC 3 VCC ! . CC  3 VCC v © ª º ¨1 .

3 VCC v © ª º t 2 1 3 VCC  3 VCC ¨1 .e RC ¸ !© ¹ ª º t ¨ e RC ¸ 1 ln.e RC ¸ !© ¹ 2 ª º 3 VCC t 1 2  ! e  t RC ¨1 .

693 !  RC t HIGH ! 0.693 R C t t HIGH ! 0.2 ! ln© ¹ ª º t  0.693.

R A  RB C 23 .

555 Timer Design Equations tLOW: Calculations for the Oscillator¶s LOW Time VC ! .

VInitial 1 3 ¨ e RC ¸  VFinal v © ¹ ª º t t 1 2 ¨ e RC ¸ !© ¹ º ª t t VCC ! .

2 3 ¨ e RC ¸ VCC  0 v © ¹ ª º t ¨ e RC ¸ 1 2 ¹ 3 VCC ! .

3 VCC v © ª º t 1  3 VCC ! ¨ e RC ¸ © ¹ 2 ª º 3 VCC 1 2 ¨ e RC ¸ 1 ln.

2 ! ln© ¹ ª º t  0.693 RBC ¨ e RC ¸ !© ¹ º ª t 24 .693 !  RC t LOW ! 0.693 R t LOW ! 0.

555 Timer ± Period / Frequency / DC Period: t HIGH ! 0.693 .

R A  R B C t LOW ! 0.693 .693 RBC T ! t HIGH  t LOW T ! ? .

693 .R A  RB CA ?0.693 RBCA 0 T ! 0.

R A  2RB C DC ! DC ! Duty Cycle: DC ! t HIGH v 100 % T 0.693 .

693 .R A  RB C v 100 % 0.

R A  2RB C .

R .

693 .R A A  RB v 100%  2RB Frequency: F! F! 1 T 0.

R A  2RB C 1 25 .
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