**
555 Timer and Related Theory**

**ASTABLE
MULTIVIBRATOR USING 555 TIMER ::**

A multivibrator is a relaxation oscillator generating nonsinusoidal waveforms.
Basically, the multivibrator is a two-stage amplifier or oscillator operating in
two modes or states. Each amplifier stage feeds back the other such that the
active element of one stage is driven to saturation and the other to cut off. A
new set of actions, producing the opposite effects, then follows. Thus the
saturated stage becomes cut off and the cut off stage saturates. The operation
of the multivibrator is based on the fact that no two active elements have
exactly identical characteristics. The NE 555 is a widely used IC timer, a
circuit that car run in either of two modes:- Monostable and Astable.

In the astable mode, it can produce regular waves with a variable duty cycle.
The 555-timer chip has the following components:

`1. A voltage divider.`

2. Two comparators

3. An R-s flip-flop

4. An n-p-n transistor.

**Block diagram representation of the 555 timer
circuit.**

**OPERATION ::**

In astable operation the 555 timer has no stable states, which means that it
cannot remain indefinitely in either state i.e. it oscillates when operated in
the astable mode and it produces a rectangular output signal. Since no input
trigger is needed to get an output, the operating in the astable mode is
sometimes called *Free-Running Multivibrator*. Here we need
two external resistors and one capacitor to set the frequency of operation.

Figure above shows an astable multivibrator implemented using the 555 IC together with an

external resistor R

_{A}, R

_{B}and an external capacitor C.

**1-Initial State S=1 R=0 -> Q=1 Q.=0 (C begins to charge)**

Initially capacitor is discharged or empty. At this time V

_{TH}> V

_{C}causes output of the

comparator 1 to be 0 so R=0 and V

_{TL}>

_{VC}causes output of the comparator 2 to be 1 so S=1.

For S=1 and R=0 , Q=1(high,Vcc) and Q.=0(low,0V).Thus Vo is high and transistor is OFF.

Capacitor C will charge up through the series combination of R

_{A}and R

_{B}, and the voltage

across it , Vc , will rise exponentially toward Vcc.

**2- Vc = VTL , comparator 2 -> Low S=0 R=0 -> Q=1 Q.=0 (no change , C is still charging)**

As Vc crosses the level equal to V

_{TL }, the output of the comparator 2 goes low.(Vc = V

_{TL}, comparator 2 -> Low ). This however has no effect on the circuit operation because this will make the inputs of the flip-flop as S=0 and R=0 (no change state) which means outputs of flip-flop will remain same. This state continues until Vc reaches and begins to exceed the threshold of comparator 1, V

_{TH}.

**3- Vc = V**

_{TH}, comparator 1 -> High S=0 R=1 -> Q=0 Q.=1 (C begins to discharge )When Vc reaches and begins to exceed V

_{TH}, the output of the comparator 1 goes high and

resets the flip flop( S=0 R=1 -> Q=0 Q.=1).Thus Vo goes low , Q. goes high and so transistor is

turned ON.

The saturated transistor causes a voltage of approximately 0V to appear at the common node

of R

_{A}and R

_{B}. Thus C begins to discharge thru R

_{B}and the collector of the transistor.

Note that R = 1(flip-flop input) for a very short time.

**4- Vc = V**

_{TH}, comparator 1 -> Low S=0 R=0 -> Q=0 Q.=1 (no change , C continues to discharge )VC will drop again below V

_{TH}immediately after discharging process is started. S=0 and R=0

will not affect the system (no change state)

The voltage Vc decreases exponentially with a time constant R

_{B}. C toward 0V.This state will

continue until Vc reaches V

_{TL.}

**5- Vc = VTL , comparator 2 -> High S=1 R=0 -> Q=1 Q.=0 (C begins to charge )**

When Vc reaches the threshold of comparator 2 , V

_{TL}, the output of comparator 2 goes high

and then S=1 R=0 causes Q=1 and Q=0.Thus output Vo goes high and Q. goes low , turning off the

transistor.

Capacitor C begins to charge through the series equivalent of R

_{A}and R

_{B}, and its voltage rises

exponentially toward Vcc with a time constant (R

_{A}+R

_{B}).C. This rise continues until Vc reaches V

_{TH}, at

which time the output of comparator 1 goes high , resetting the flip-flop , and the cycle goes on.

Determining the Period T =T

_{H}+ T

_{L}:

For T

_{H}:

From the general solution for step and natural responses :

`X(t) = X`_{F} + [X(t_{o})-X_{F}].e^{-(t-t0)/t}

V_{c}= V_{cc}+[V_{TL}+ Vcc ]

`ε`

^{-τ/t}
or ;

V_{C} = (Final Val-Initial Val) ( 1- e^{-t /RC} ) + shifting

V_{c} = (b-a) ( 1- e^{-t/RC} ) + a

V_{c}= (V_{cc})(1-

`ε`

^{-τ/t}
) +V_{TL}

V_{c}= (V_{cc})(1-

`ε`

^{-τ/t}
) +V_{TL} is equal V_{c}= V_{cc}+[ V_{TL }+ Vcc
]

`ε`

^{-τ/t}
where τ =(R_{A} + R_{B}).C

`Substituting t=T`_{H} V_{C}=V_{TH}=2/3V_{cc}
and V_{TL}=1/3V_{cc} in the equation

V_{c}= (V_{cc})(1- ε

^{-τ/t}

```
) +V
```_{TL}

Vcc=( Vcc - 1/3)(1- ε

^{-τ/t}

```
) + V
```_{cc}

`where τ =(R`_{A}
+ R_{B}).C

ε

^{-τ/t}

`=1/2`

T_{H} = (R_{A}+R_{B}).(C).(ln2)

T_{H} = 0.69(C)(R_{A}+R_{B})

For T

_{L}:

`X(t) = X`_{F} + [ X(t_{o}) - X_{F}].e^{-(t-t0)
/ t}

V_{c}= 0 V+[ V_{TH} -0 ]e^{-τ/t}

V_{c} = V_{TH}. e^{-τ/t} where τ =R_{B}.C

For t=T_{L} VC=V_{TL}=1/3V_{cc} and V_{TH}=2/3V_{cc}

V_{c} = V_{TH}.e^{-τ/t} where τ =R_{B}.C

1/3V_{cc}=2/3V_{cc}. e^{-τ/t}

T_{L} = R_{B}.C.ln2

T_{L} = 0.69R_{B}.C

T = T_{H} + T_{L }

T = 0.69(C)(R_{A}+R_{B}) + 0.69R_{B}.C

T = 0.69.C.(R_{A} + 2R_{B})

Also the

**duty cycle**of the output square wave can be found as:

Duty Cycle=

`T`_{H}/T_{H}+T_{L}

_{ }=

`R`_{A}+R_{B}/R_{A}+2R_{B}

Note that the duty cycle will always be greater than 0.5(50%).It approaches to 0.5 if R

_{A}is

selected much smaller than R

_{B}.

**Most popular**

Questions index C Questions C++ Questions Win32 MFC COM/DCOM DLL Questions

**Compilers & Editors**

Download Visual Studio Download XCode Download Visual Studio Code Android studio install sdk Eclipse installer Best C compilers IDEs

**Development system setup**

Windows media creation tool MSDN subscription Ubuntu virtualbox

**New updated posts**

Why learn C? Calculate weighted average