1.2.5 ClockSignals_The555 Timer.pptx

Clock Signals: 555 Timer

© 2014 Project Lead The Way, Inc. Digital Electronics

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.

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.

Capacitor Charge Cycle

5

Equation for Charging Capacitor

• _{Capacitor is initially discharged.} • _{Switch is moved to position A.} • _{Capacitor will charge to +12 v.} • _{Capacitor will charge through}

the 2 k resistor.







-t/RC



_Initial

Initial Final

C

V

V

1

-

e

V

V









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 Discharge Cycle

6

Equation for Discharging Capacitor

• _{Capacitor is initially charged.} • _{Switch is moved to position B.} • _{Capacitor will discharge to +0 v.} • _{Capacitor will discharge through}

the 3 K resistor.







-t/RC



Final Initial

C

V

V

e

V







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 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 2k resistor.

Switch is moved back to position B. The capacitor discharges through the 3k resistors.

20 mSec 5 V

V_C

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

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

555 Timer Design Equations

10

t_HIGH : Calculations for the Oscillator’s HIGH Time

5v 3.333 v

V_{c 1.666 v}

0 v

 t_HIGH 

Output

HIGH

LOW

THE OUTPUT IS HIGH WHILE THE

CAPACITOR IS CHARGING THROUGH R_A + R_B.



R

R



C

0.693

_A



_B



HIGH

555 Timer Design Equations

11

t_LOW : Calculations for the Oscillator’s LOW Time

5v 3.333 v

V_{c 1.666 v}

0 v

 t_{LOW }

Output

HIGH

LOW

THE OUTPUT IS LOW WHILE THE

CAPACITOR IS DISCHARGING THROUGH R_B.

C

0.693R

_B



LOW

555 Timer – Period / Frequency / DC

12

Period:

Frequency:

Duty Cycle:

   



 



R 2R  C 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          

R R  C . ƒ T ƒ B A 2 693 0 1 1         

R 2R  100%

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          

Example: 555 Oscillator

13

Example:

For the 555 Timer oscillator shown below, calculate the circuit’s, period (T), frequency (ƒ), and duty cycle (DC).

Example: 555 Oscillator

14

Solution:

Period:

Frequency: Duty Cycle:

F 6.8 C 180 R 390

R_A   _B    





mSec 3.5343 T 6.8μ. 180Ω 2 390Ω 0.693 T C B 2R A R 0.693 T        _      Hz 282.941 ƒ mSec . ƒ T ƒ    534 3 1

1 _{ }

      % 76 DC % 100 180 2 390 180 390 DC % 100 R 2 R R R DC B A B A              

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:

Frequency: Duty Cycle:

Two Equations & Two Unknowns!

          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                          09 1228 2 09 1228 47 0 693 0 400 2 400 47 0 2 693 0 400 2 693 0 400 5 2 1 1 . R R . F . . Sec R R Sec F . R R . T Sec C R R . T Sec kHz . f T B A B A B A B A                      

Example: 555 Oscillator

17

Solution:

Frequency: Duty Cycle:

Substitute and Solve for R_B

Substitute and Solve for R_A

B A 0.5 R

R  

09 . 1228 R

2

R_A  _B 

           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                618 . 245 R 09 . 1228 472 . 982 R 09 . 1228 491.23 2 R 09 . 1228 R 2 R A A A B 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

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

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

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

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

t_HIGH : Calculations for the Oscillator’s HIGH Time













                                                 RC t 2 1 RC t CC 3 2 CC 3 1 CC 32 CC 31 RC t CC 3 2 CC 32 CC 31 RC t CC 31 CC CC 32 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

 



R R



C 693 . 0 t C R 693 . 0 t RC t 693 . 0 e ln ln e e -1 B A HIGH HIGH RC t 21 RC t 2 1 RC t 2 1                          

555 Timer Design Equations

24

t_LOW: Calculations for the Oscillator’s LOW Time













                                             RC t 2 1 RC t CC 3 2 CC 31 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 21 RC t 2 1                     

555 Timer – Period / Frequency / DC

25

Period:

Frequency:

Duty Cycle:

   



 



R 2R  C 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          

R R  C . ƒ T ƒ B A 2 693 0 1 1         

R 2R  100%

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          