
555 Timer and Related Theory
ASTABLE
MULTIVIBRATOR USING 555 TIMER ::
A multivibrator is a relaxation oscillator generating nonsinusoidal waveforms.
Basically, the multivibrator is a twostage 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 555timer chip has the following components:
1.
A voltage divider.
2. Two comparators
3. An Rs flipflop
4. An npn 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 FreeRunning 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.
1Initial 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
flipflop as S=0 and R=0 (no change state) which means outputs of flipflop 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(flipflop 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 flipflop , 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^{(tt0)/t}
V_{c}= V_{cc}+[V_{TL}+ Vcc ] ε ^{τ/t}
or ;
V_{C} = (Final ValInitial Val) ( 1 e^{t /RC} ) + shifting
V_{c} = (ba) ( 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^{(tt0)
/ 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}.
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