Oscillator multivibrotor

CONTENTS
 Introduction

 Criterion for oscillations
 Classification of oscillators

 Tuned collector oscillators

 Collpitt and clap oscillator

 Hartley oscillator

 R-C phase shift oscillator

 Wein bridge oscillator

CONTENTS
 Crystal oscillator
 Blocking oscillator

 Others (Negative resistance osc. etc.)

-:OSCILLATORS:-
INTRODUCTION

 An electronic oscillator is an electronic circuit that
produces a repetitive electronic signal, often a sine wave
or a square wave.
 It basically converts the dc signal to a required ac signal.

 An oscillator is an amplifier, which uses a positive

feedback and without any external input signal,
generates an output waveform of a desired frequency.

 An oscillator is basically a waveform generator
which generates an output waveform, which
oscillates with constant amplitude & constant
desired frequency.
 It is basic element of all the ac signal sources and generates
sinusoidal signal of required frequency and for electrical and
electronic measurement.
 Oscillators are generally required to generate carrier frequency
signal needed in modulation of audio and video waves
broadcasting.

OSCILLATORS V/S AMPLIFIER

Amplifier oscillator
1. Negative feedback is 1. Positive feedback is
    applied.     applied.
2. It strengthens the 2. It strengthen the
input signal without     causing charge as
any change in      well waveform.
frequency.
3. Require an external 3. It just require a dc
    wave signal to be      signal.
    applied.

MERITS OF OSCILLATOR

 Portable and cheap in cost.

 An oscillator is a nonrotating device. Consequently,

there is no wear & tear & hence longer life.
 Frequency of oscillation may be conveniently varied.

 Frequency once set remain constant for a considerable

period of time.
 It operates at a very high efficiency since there is no

wastage of energy due to friction.

OSCILLATOR

The factors responsible for the stability of frequency
1. Variation in power supply.
2. Variation in intermediate capacitance.

BARKHAUSAIN Criterion
1. The loop gain Aβ >= 1.
2. The phase shift around the circuit must be 2π or
multiple of 2π.
As the feedback in this case is +ve, if loop gain increases
than gain increases.

TYPES OF OSCILLATION

 Damped oscillation

 Undamped or sustained oscillation

The electrical oscillation whose amplitude goes
on decreasing with time are known as damped
oscillation.
The electrical oscillations whose amplitude
remains constant with time are known as
undamped oscillations.

BASIC BLOCK DIAGRAM OF
TRANSISTORIZED OSCILLATOR

Automatic
Amplifier Frequency OUTPUT
Amplitude
(gain=A) Selector
Control

Feedback
network(β)

 Oscillation circuit : it is basic tank circuit (LC circuit)

which is used to produce frequency of oscillation
f = 1/2π(LC)1/2

 Electronic amplifier : receive dc power from battery

& convert into the ac power for supply to the tank
circuit.

 Feedback network : it is supplied o/p part to tank

circuit to the electronic amplifier.

 Frequency Selector:-
Oscillator must be able to provide
oscillation of any desired frequency , therefore
frequency selector is provided.


Automatic Amplitude
Control:-
If for any reason, the amplitude of
oscillations in the output increases,it may
continue to increase on account of feedback.
Therefore to overcome this problem an
automatic amplitude control unit is provided.

TUNED COLLECTOR OSCILLATOR

1. Circuit Diagram:-

2. INTRODUCTION:-
• It is first oscillator come into the external &
clear from the its name that tuned circuit
connected to the collector of transistor thus is
called tuned collector oscillator.
• The tuned circuit is formed by capacitor C and
transformer primary coil L.

Frequency of oscillation :–
f = 1/2π(LC)1/2

2. CONSTRUCTION:-
 The tuned circuit, constituted by the capacitor C and
transformer primary coil L, forms the load
impedance and determine the frequency of
oscillation.
 The resistor R1, R2 and Re form the dc biasing
circuit of the transistor.
 Capacitors C1 and Ce bypass capacitors for R2 and
Re respectively so that the ac operation of the circuit
is not affected.
 Transistor amplifier provides sufficient gain for
oscillator action to place.

 Feedback voltage provided by secondary coil L
appears across base-emitter junction of transistor.

 As the transistor is connected in CE configuration,
it produces a phase shift of 180° between the
input and output circuit. Another 180° phase shift
is provided by the transformer.

3. WORKING:-
 When give the Vcc supply, a transient current
is flows through the tuned L-C circuit. It is due
to increase of collector current to its quiescent
value.
 This transient current initiates natural
oscillation in the tank circuit. These natural
oscillations induce some voltage into L1 by
mutual induction which cause corresponding
variation in base current. These variation in
base current are amplified β times appear in
the collector circuit.

 A part of this amplified energy is used to meet
the losses that occur in the tank circuit and the
rest is radiated out in the form of elector-
magnetic waves.
 The turn ratio of L1 and L is determined by total
losses.
 The frequency of oscillation i.e. at which
Barkhausen criterion is satisfied differs from the
resonant frequency of the tuned circuit.

TUNED BASE OSCILLATOR

1. Circuit Diagram:-

2.INRODUCTION:-
 When parallel tuned LC circuit is placed
in the base-to-ground circuit, the oscillator
is known as the tuned base oscillator.

Frequency of oscillation –
f = 1/2π(LC)1/2

3. CONSTRUCTION:-

 The dc bias is determined by the resistors
R1,R2 and Re.
 The parallel Re-Ce network in the emitter
circuit is a stabilizing circuit and prevents
degeneration of the signal.
 Cc is the dc blocking capacitor.
 L1 & L2 are the primary & secondary mutually
coupled coils of an RF transformer and
provides the required feedback b/w the
collector & base circuit. The primary coil L1 is

4. WORKING:
 When the Vcc is switched on, the collector
current start increasing.
 The rising collector current, which also flows
through the tickler coil L1, create a varying
magnetic field around L1.
 This varying magnetic field links with the coil
L and therefore, induces a voltage in the tuned
circuit. Because of correct phasing of the coils,
and sufficient gain of amplifier, the oscillations
start building up.

TYPES OF OSCILLATOR
OSCILLATOR

HARMONIC RELAXATION

BLOCKING
R-C L-C OTHER

HARTLEY WEIN’S CRYSTAL
BRIDGE

COLPITTS
NEGATIVE
RC-PHASE RESISTANCE
CLAP
SHIFT

HARMONIC OSCILLATOR
Introduction:-
 The harmonic oscillator produces a sinusoidal
output and energy in this unidirectional. It
means energy is transformed from active to
passive components and consume by them.
 The basic form of an harmonic oscillator is an

electronic amplifier with the output attached to a
narrow-band electronic filter, and the output of
the filter attached to the input of the amplifier.

1. CIRCUIT DIAGRAM:

Vcc

2. CONSTRUCTION:-
 It consists of basically a L-C phase shift
network, known as tank circuit and a
single stage invaries capacitating
amplifier.
 Two series capacitors C1&C2 form
potential divider network.
 The voltage across C1 is feedback
positively.
 The resistors R1 ,R2 , and Re provides the
necessary d.c. bias to keep the transistor
in active region.

 The potential divider network is formed by Re
and Ce along with resistance R1&R2 to provide
stability to the circuit.
 The radio freq. coil RFC at which Vcc is applied
helps in easy flow of current as it allows the dc
current to flow easily and blocks the high
frequency current.
 The output is taken out of this inductor by
transformer coupling.
 The junction of C1 and C2 is grounded.

3. OPERATION:-

 When collector supply voltage Vcc switched on
then a transient current flows through the
tank circuit.
 The current produces an AC voltage acrossC1
and another a.c. voltage across C2.
 Thus,the tank circuit produces a 180° phase
shift
between output collector voltage and feedback
voltage.

 The transistor produces a further 180° phase
shift.
 These capacitors discharged through the coil L.
 The coil again discharges through the
capacitors and the oscillations starts up.

4. HEQUIVALENT CIRCUIT
DIAGRAM:

B C

hie 1k
hfeIb

E
Ib

3 1
Z1 2 Z2

Z3

5. FREQUENCY OSCILLATION:

General Assumptions:-

 hre is too small so the source hreVout is
neglected.
 hoe is small so 1/hoe can be neglected.

FREQUENCY OF OSCILLATION:-

Av = Vo/Vi
= -hfe Ib ZL/hie Ib
= -hfe ZL/hie
ZL = Z2 || [Z1Z3 + hieZ3+ Z1hie] / [Z1+ hie]
= Z1 Z2 Z3+ hieZ2Z3+Z1Z2hie
Z1Z2 + hie(Z1+Z2+Z3)+Z1Z3


β = Vf / Vout
Vf = I1 * [Z1.hie / (Z1+hie)]
Vout = I1 * [Z3+ Z1hie / (Z1+hie)]
β = Z1.hie / (Z1Z3+(Z1+Z3)hie)
According to Barkhausen criterion for
sustained oscillations :- Aβ >= 1
or
Aβ = 1.


1 = - hfeZ1Z2 / {Z1(Z2+Z3)+(Z1+Z2+Z3)hie}
Z1(Z2+Z3)+(Z1+Z2+Z3)hie = - hfeZ1Z2

hie(Z1+Z2+Z3)+(1+hfe)Z1.Z2+Z1.Z3 = 0 --
(1)

Above equation is the general equation of
LC oscillators.

FREQUENCY OF
OSCILLATION:-

Using the Key equation from hartley oscillator-
(1+hfe)Z1Z2+hie(Z1+Z2+Z3)+Z1Z3 = 0
so on substituting the values of
Z1 = 1/jwC1 , Z2 = 1/jwC2, Z3 = jwL
putting the value in the equation
hie(1/jwC1+1/jwC2+jwL)-1/w²(1+hfe)
+jwL(1/jwC1) = 0


    =>  jhie(1/jw²C1+1/jw²C2+wL)
             = (1+hfe)/w²C1C2L/C1
   => jhie(C2+C1+jw2LC1C2/jw²C1C2)
             = 1+hfe/w2C1C2L/C1
      => jhie(C1+C2w²LC1C2/jw²C1C2)
             = 1+hfe/w2C1C2l/C1


Equating imaginary term to be zero-
hie(C1+C2-w2LC2C1/jwC1C2) = 0
hie(-1/wC1-1/wC2+wL) = 0
1/wC1+1/wC2 = wL
1/C1+1/C2 = w2L
C1+C2/C1C2 = w2L


w2 = 1/l(C1+C2/C1C2)
w2 = 1/LCeq
(2πf)2 = 1/LCeq
f2 = 1/4π2LCeq
f=1/2π (LCeq)1/2
Here
1/Ceq=1/C1+1/C2
=> Ceq = C1 .C2 / (C1 +C2 )


Equating real terms to be zero -
L/C1 = 1/w2C1C2(1+hfe)
1+hfe = Lw2C2
1+hfe = (L/LCeq)*C2 (w2 = 1/LCeq)
1+hfe = C2/Ceq
neglecting 1 as compared to hfe
hfe = C2/C1
This is the condition for sustained oscillation.

1. CIRCUIT DIAGRAM:
Vcc

2. CONSTRUCTION:
 Clapp oscillator is just an extension of
that of the colpitt’s oscillator.
 Here one more capacitor is joined in series
with the other two.
 The extension of capacitors with series in
C1 and C2 , remove frequency distortion.

3. OPERATION:
The operation of the clapp
oscillator is just same as that of colpitt’s
oscillator.

4. FREQUENCY OF
OSCILLATION:-
 The frequency of operation of the clapp
oscillator is
f=1/2π (LCeq)1/2
Where
Ceq =(1/C1 +1/C2 +1/C3)-1

1. INTRODUCTION:-
 The Hartley oscillator is an LC
electronic oscillator that derives its
feedback from a
tapped coil in parallel with a capacitor
(tank circuit).
 The tapping is done at the quarter length
so it is not central tapped inductor.
 A Hartley oscillator is essentially any
configuration that uses a pair of series-
connected coils and a single capacitor.

C1 1u
R1 1k R2 1k

R3 1k L1 1m

C3 1u
C2 1u

T1 !NPN

C4 1u
Vcc

C5 1u
2. CIRCUIT DIAGRAM:-

L2 1m L1 1m
NI1
Vout

3. CONSTRUCTION:-

 The construction is similar to that of colpitt
oscillator the change is-
 Capacitor C1&C2 are replaced by a single

capacitor C.
 The coil L is replaced by the coils L1&L2.

 Ce is a bypass capacitor used for stabilization
of operating point (Q).

 Capacitors Cb and Ce are coupling capacitors.

 The tank circuit consists of inductors L1 and L2
and a variable capacitor C. The tank circuit
determines the frequency of oscillations.

 RFC serves two functions first it prevents
radio frequency current from reaching the d.c.
power supply. Second, it prevents the d.c.
supply from short circuiting the a.c. output
voltage.

4. OPERATION:-
 When collector supply voltage Vcc switched
on the capacitor C is charged.
 This capacitor dischargs through the coils
L1&L2.
 The coils again discharges through the
capacitor and the oscillations starts up.
 The voltage across L1 is feedback in to the
circuit and voltage across L2 is output.
 The total phase shift is 360 .

5. H-EQUIVALENT DIAGRAM:-

B C
hie 1k
H fe Ib

E
Ib

Z2 Z1

Z3

6. FREQUENCY OF
OSCILLATIONS:-

General Assumptions:-

 hre is to small so the source hreVout is
neglected.
 hoe is small so 1/hoe can be neglected.


hie(Z1+Z2+Z3)+(1+hfe)Z1Z2+Z1Z3 = 0
Here
Z1 = jw(L1+M), Z2 = jw(L2+M)
Z3 = 1/jwC
Putting these values in above equation and
equating imaginary term


w(L1+L2+2M) = 1/wC
w2C = 1/(L1+L2+2M)
w2 = 1/C(L1+L2+2M)
Here neglecting the mutual inductance and
putting
w = 2πf
(2πf)2 = 1/C(L1+L2+2M)
f = 1/2π(C(L1+L2+2M))1/2


2M+L1+L2 = Leq
So
f = 1/2π(LeqC)1/2


Equating real term equal to zero
w2(1+hfe)(L1+M)(L2+M) = (L1+M)/C
Putting the value of w2
(1+hfe) (L2+M) / LeqC = 1/C
So that Gain
hfe = (L1+M)/(L2+M)

7. ADVANTAGES:-

1.The frequency is varied using a variable
capacitor .
2. The output amplitude remains
constant over the frequency range .
3. The feedback ratio of the tapped
inductor remains constant .

8. DISADVANTAGES:-

1. Harmonic-rich content of the output .
2. It is not suitable for a pure sine wave.

9. APPLICATIONS:-
 The Hartley oscillator is extensively used
on broadcasting bands.

1. CONSTRUCTION :-
 Voltage shunt feedback is used for a
transistor phase shift oscillator .
 Feed back signal is coupled through the
feedback resistor R’ should be such that
when added amplifier stage input
resistance hie it is equal to i.e.,
R’+ hie = R

3. OPERATION :-
 The circuit is set into oscillation by any
random or variation caused in the base current
.
 This variation in base current is amplified in
collector circuit .
 The output of the amplifier is supplied to an
RC feedback network .
 The RC network produces a phase shift of 1800
between output and input voltages.

 So CE amplifier produces a phase reversal of the
input signal ,total phase shift becomes 3600 or 00
which is essential for regeneration or for
sustained oscillation.
 Thus sustained variation in collector current
between saturation and cut-off values are
obtained .
 RC phase shift network is the frequency
determining network.
 The circuit arrangement of a phase shift
oscillator using NPN transistor in CE
configuration.

4. H-EQUIVALENT CIRCUIT :-

R4 1k C3 1u C1 1u C2 1u
+

VG1

R1 1k

R2 1k

R3 1k

5. FREQUENCY OF
OSCILLATION :-

Following assumption are made-
 hre of the transistor is negligibly small so

hre Vout omitted from the circuit
 hoe of the transistor is very small 1/hoe is
much larger than Rc so the effect of hoe can
be neglected.
 Making above assumption and replacing
current source by equivalent voltage
source .


Applying KVL to the three loops, we have

(R+Rc+1/jwC)I1 -RI2
+hfeIbRC=0……………...(1)
-RI1 +(2R+1/jwC) I2 -RIb=0
……………….(2)
0-RI2 +(2R+1/jwC) Ib=0
………………(3)


Substituting 1/wC= Xc
(R+Rc-jXc) (-R) (hfeRc)
(-R) (2R-jwC) (-R) =0
0 (-R) (2R-jwC)

(R+Rc-jXc) [(2R-jXc)2-R2] +R[(-R) (2R-jXc)- hfeRc(-R)] = 0


R3+R2Rc(3+hfe)-(5R)Xc2-RcXc2-6jR2Xc-j4RRcXc+jX3c = 0
Equating the imaginary components of the above
equation to zero we have
or 6R2Xc+4RRcXc-X3c = 0
or Xc = (6R2+4RRc)1/2
XC=1/wC=1/jwfC

FREQUENCY OF OSCILLATION :-

2πfc = 1/(6R2+4RRc)1/2
f = 1/2πRC (6+4Rc/R)1/2 (if Rc=R)

f = 1/2πRC(10)1/2

f is frequency of oscillation


Now equating the real components of equation
to zero we have
R3+R2Rc(3+hfe)-X2c(5R+Rc) = 0
R3+R2Rc(3+hfe)-(6R2+4RRc)(5R+Rc) = 0
29R3-23R2Rc+hfeR2Rc4RR2c = 0
-29R/Rc-23+hfe-4R/Rc = 0
hfe = 23+ 29R/Rc +4 Rc/R


For the loop gain to be greater than unity,
the requirement of the current gain of the
transistor is found to be
hfe > 23+ 29R/Rc + 4R/Rc
If R = Rc, then
hfe > (23+29+4)
hfe > 56

Rs 1k
Rd 1k

T1 2N3369
Vcc

Cs 1u

R 1k
C 1u

R 1k
C 1u
1.CIRCUIT DIAGRAM:-

R 1k
C 1u

2. CONSTRUCTION:-
 The circuit consists of a conventional FET
common source amplifier followed by a
three section RC phase shift network.
 The output of the last section is fed back
to the gate of the FET.
 The circuit uses voltage series feedback.

3. FREQUENCY OF OSCILLATOR
:-

 The basic amplifier shift the input voltage at the
gate by 180° .
 The phase shift network produces additional phase

shift which is a function of frequency .
 At some particular frequency the phase shift is

180° .
 Hence the total phase shift from the gate around

the entire circuit &back to the gate will be exactly
zero.

4. H-EQUIVALENT CIRCUIT:-

C 1u C 1u C 1u

Vi=Vf
Rd 1k
Vf'=Vf
rd 1k

R 1k

R 1k
gmVi R 1k

FREQUENCY OF OSCILLATOR :-

From figure
I1R = Vf ’……(i)
I2R = I1{R+1/jwC}
I2R = Vf’{1+1/jwC}……. (ii)
I3R = (I1+I2)/jwC + I2R
= [Vf’/R +Vf’/R{1+1/jwC}]/jwC
+Vf’{1+1/jwC}
= Vf’/R[R+ (3+1/jwC)/jwC] ……….
(iii)


Vo = (I1+I2+I3)/jwC +I3R ….. (iv)
On substituting the values of I1,I2&I3
using (i), (ii), (iii) in (iv) :-
Vo/Vf ’= 1 – 6j/wCR -5/w2C2R2 + j/w3C3R3 ………(V)

We know
-1/β = Vo/Vf ’ = 1-5A2-j(6A A3) ……(vi)


where A= 1/wCR
For sustained oscillation –Aβ =1
Hence
1-5A2-j(6A A3) = A = A+j0 ……. (vii)
Equating real & imaginary part to 0.
6A - A3 = 0;


A2 = 6
A = (6)1/2
Hence w = 1/(6)1/2 CR
2πf= 1/(6)1/2 CR
f=1/2πRC(6)1/2
Real part
1-5A2 = A
1-(5*6) = A29 = A

Aβ =1
Thus
β=1/29 = hfe

5. APPLICATION :-

 It is well suited to the range of frequencies
from several hertz to several
kilohertz(20Hz to 200KHz)
 For generating different audio-frequencies
 It is not suitable for higher frequency
operations.

6. ADVANTAGE :-
 It is cheap and simple circuit as it contains
resistor and capacitor.
 The output is sinusoidal that is quite distortion
free.
 They have wide frequency range.
 It provides good frequency stability.
 They are particularly suitable for low
frequencies.
 They are much simpler than the wein bridge
oscillators because it does not need negative
feedback and the stabilization arrangement.

7.DISADVANTAGES :-
 The output is small,it is due to smaller feedback.
 The frequency stability is not as good as that of

wein,s bridge oscillator.
 It is difficult for the circuit to start oscillation as

the feedback is usually small.
 It needs high voltages (12V) battery so as to

develop sufficiently large feedback voltage.
 It is essentially a two stage amplifier with an RC

bridge circuit .

 Itis a lead -lag network .
 The phase–shift across the network lags with

increasing frequency and leads with
decreasing frequency.

2. CONSTRUCTION :-
o It is one of the popular type of oscillator used
in audio and sub-audio frequency ranges
(20Hz-20KHz)
o It’s output is free from distortion and it’s
frequency can be varied easily. However the
maximum frequency output of a typical wein
bridge oscillator is only about 1MHz.
o At all other frequency the bridge is off –balance (the
voltage feedback and output voltage do not have the
correct phase relationship for sustained oscillation).

 So bridge circuit can be used as feedback
network for an oscillator provided that the
phase shift through the amplifier is zero .

3. OPERATION :-
• The circuit is set in oscillation by any random
change of in base current of transistor Q1. Base
current is amplified in collector circuit of
transistor with the phase shift of 1800 the
output of transistor Q1 is fed to the base of
second transistor Q1 through capacitor C4 .
• The output signal will be in phase with the
signal input of the base of transistor Q1.
• A part of the output signal at transistor Q2 is
feedback to the input point (AC) of the bridge

•Feedback signal is applied to emitter resistor R4
where it produces degenerative effect (-
ve feedback).
•A part of feedback signal is applied across the

base bias resistance R2 where it produces
regenerative effect (or +ve feedback)
•At the rated frequency,effect of regenerative is

made slightly more than that of generation so as
to obtain sustained oscillation.
• The continuous frequency variation in the

oscillator can be held by varying the two
capacitors C1&C2 simultaneously .
•This capacitor are called variable Air Gang

4. FREQUENCY OF
OSCILLATION:-

For bridge to be balanced :-
Z2Z3 = Z1Z4
Putting value in standard :-
R3[R2/1+jwC2R2] = R4(R1-j/wC1)
R2*R3=R4 (1+jwC2R2)(R1-jwC1)
R2*R3 -R4*R1- (C2/C1 )R2R4 + jwC2R2R1R4 =0
Separating real and imaginary terms we have
R2R3-R4R1- (C2/C1 )R2R4 = 0
C2/C1 = R3/R4-R1/R2
R /wC -wC R R R = 0


w = 1/C1C2R1R2
w = 1/(C1C2R1R2)1/2
F = 1/2π(R1C1R2C2)1/2
If C1=C2=C and R1=R2=R , then
f = 1/2πCR
and
R3 = 2R4
 Thus we see that in a bridge circuit the output
will be in phase with the input only when the
bridge is balanced (at resonant frequency).

5. APPLICATION :-

1. The wein bridge oscillator is a standard
device used for generating a low frequency
in the range of 10Hz to 1MHz.
2. All commercial audio generators make
use of wein bridge oscillator.

6. ADVANTAGE :-
1. It provides a stable low distortion
sinusoidal output over a wide range of
frequency.
2. The frequency range can be selected
simply by using decade resistance boxes.
3.The overall gain is high because of two
transistor.

NEGATIVE RESISTANCE
OSCILLATOR
 Thisis a particular class of oscillator
which uses negative resistance element
such as tuned diode, unijunction
transistor etc.

1. INTRODUCTION:-
• A crystal oscillator is an electronic
circuit that uses the mechanical resonance
of a vibrating crystal of piezoelectric
material (ex. –quartz,rochellesalt) to
create an electrical signal with a very
precise frequency.
• This frequency is commonly used to keep
track of time (as in quartz wristwatches),
to provide a stable clock signal for digital
integrated circuits, and to stabilize
frequencies for radio transmitters.

2. CIRCUIT
DIAGRAM(BASIC):-

U1 HC49S_CY11BS

C1 1u R1 1k

L1 1m

CIRCUIT DIAGRAM (VOLTAGE
SERIES)

(VOLTAGE SERIES)
CONSTRUCTION AND WORKING

 Resistor R1,R2 and Re provide a voltage-
divider stabilized dc bias circuit.
 Capacitor Ce provides ac bypass of emitter
resistor Re .
 Radio frequency coil (RFC) provides for dc
bias while decoupling any ac signal on
power lines form affecting the output
signal.
 The voltage feedback signal form the
collector to the base in maximum when
the crystal impedance is minimum.

CRYSTAL OSCILLATOR (VOLTAGE
SERIES)
 The coupling capacitor Cc has negligible
impedance at the circuit operation frequency but
block any dc between collector to base .
 The resulting circuit frequency of oscillation is set
by the series resonant frequency of the crystal.
 Variation in power supply voltage, transistor
parameter, etc. have no effect on the circuit
operating frequency which is held stabilized by
the crystal
 The circuit frequency stability is set by the crystal
frequency stability, which is good.

CRYSTAL OSCILLATOR
CIRCUIT DIAGRAM ( VOLTAGE
SHUNT)

SHUNT)

 Parallel resonant impedance is of a crystal
of a maximum value, it is connected in
parallel.
 C1 and C2 form a capacitor voltage divider

which returns a portion of the output voltage
to the transistor emitter.
 Transistor NPN combined with R1, R2, RFC

and Re constitutes a common base emitter.
 Capacitor C3 provides an ac short circuit R2

to ensure that the transistor base remain at

(VOLTAGE SHUNT)

 As the voltage increase positively, the
emitter voltage also increases, & since the
base voltage is fixed, the base-emitter
voltage is reduce.
 The reduction in VBE causes collector

current Ic to diminish, & this in turn
causes the collector voltage Vc to increase
positively. Thus, the circuit is applying its
own input, & a state of oscillation exists.

SHUNT)
 The crystal in parallel with C1 & C2 permit
max. voltage feedback form the collector to
emitter when its impedance is maximum,
i.e., at its parallel resonant frequency.
 At other frequencies, the crystal impedance
is low, and so the resultant feedback voltage
is too small to sustain oscillation.
 The oscillation frequency is stabilized at the
parallel resonant frequency of the crystal.

CRYSTAL OSCILLATOR
FREQUENCY OF OSCILLATOR

 fseries= 1/2π (LsCs)1/2

 fShunt = 1/2π (LCt)1/2

 Ct=Cs * Cm/(Cs+Cm)

 There is no effect of temperature on
crystal oscillator.

CRYSTAL OSCILLATOR
ADVANTAGE
 It is very simple circuit as it does need any tank
circuit other than crystal itself.
 Different oscillation frequencies can be had by

simply replacing one crystal with an-other.

The Q factor , which is measure of the quality of
a resonant circuit of a crystal is very high.

The crystal oscillator provides excellent
frequency stability.

CRYSTAL OSCILLATOR
DISADVANTAGE

 The crystal oscillators have a very limited
turning range (or not all). They used for
frequencies exceeding 100KHz.
 The crystal oscillator are fragile and,

therefore, can only be used in low power
circuit.

RELAXATION
OSCILLATOR
A Relaxation Oscillator is an oscillator
in which a capacitor is charged gradually
and then discharged rapidly. It's usually
implemented with a resistor, a capacitor,
and some sort of "threshold" device such
as a neon lamp, diac, uni junction
transistor, or Gunn diode .
 Example :- Blocking oscillators.

BLOCKING OSCILLATOR
A Blocking Oscillator is the
minimal configuration of discrete
electronic components which can produce
a free-running signal, requiring only a
capacitor, transformer, and one
amplifying component.
The name is derived from the fact that
the transistor (or tube) is cut-off or
"blocked" for most of the duty-cycle,
producing periodic pulses.

BLOCKING OSCILLATOR
CIRCUIT DIAGRAM

CONTENTS

 Introduction
 Devices used in Multivibrators
 Types of Multivibrators
 Astable Multivibrators
 Monostable Multivibrators
 Bistable Multivibrators
 Schmitt Trigger

INTRODUCTION
A Multivibrators is an electronic circuit which
generates square wave or other non-sinusoidal
waveforms(i.e. square waves, rectangular
waves, triangular or saw tooth waves, etc.).

A MV has two-stage amplifier with positive
feedback between two stages. It can be seen
that output of one amplifier stage is input to
the second stage.
So in Multivibrators, each amplifier
supplies feedback to the other stage in such a
way that one transistor is driven into
saturation and the other in to cut-off, i.e. when
one transistor is ON the other is in OFF state
or vice versa.

MULTIVIBRATORS:

The two possible states of Multivibrators are :-

First state : Q1 ON and Q2 OFF
Second state: Q2 ON and Q1 OFF

A Multivibrator switches between these two
states. The condition in which a Multivibrator
remains in one state only indefinitely and does
not changes its state until it is triggered by some
external signal is known as stable state.
Otherwise known as quasi-state.

DEVICES USED IN MULTIVIBRATORS

Multivibrators use
(i) Active devices such as electron tubes, BJTs
or FETs.
(ii) Negative resistance devices such as UJT,
tunnel diode.
(iii) OP Amps.

BISTABLE MULTIVIBRATOR
The Bistable MV also known as Two Shot
MV, requires application of two trigger pulses to
return the circuit to its original state.
The first trigger pulse causes the conducting
transistor to be cut-off and the second trigger
pulse causes a transition back to the conducting
state. Because two trigger pulses are required,
therefore Bistable circuit are sometimes called
flip-flop.

BISTABLE MULTIVIBRATOR
In this circuit , both coupling networks provide dc
coupling and no energy storage element is used.

USES
(a) Storage of binary information.
(b) Counting pulses.
(c) Generating of pulse waveform of square waveform.
(d) For frequency division.

COLLECTOR COUPLED BISTABLE
MULTIVIBRATOR
Construction:-
The circuit consist of two identical NPN
transistor Q1 and Q2 with equal collector
resistance RC1 and RC2 and with output of one
supplied to the input of other. The forward bias is
coupled through each resistor R1 and R3.R2, R4
and VBB provides fix bias for the base junction.

WORKING

The multivibrator can be driven from first
stable state to another stable state by applying
either a negative trigger pulse to the base of Q1or
positive trigger pulse to the base of Q2

WORKING
This increase in potential will forward bias
the emitter base junction of Q2, as it is connected
to the collector terminal C1 by R3. as a result
collector current (IC2) of a transistor Q2 increases
and therefore its collector voltage falls.

WORKING

The decreases in the collector voltage
appears across the emitter base junction of Q1
where it further reverse biases the emitter base
junction of transistor Q1 to make the collector
current (IC1) to fall.
After few cycle, Q2 is driven into saturation
state and Q1 is in cut-off state. This is the
second stable state to the multivibrator . The
circuit will remain now in second stable state until
any trigger pulse is given.

Stable state 1:-
Q1 is OFF and Q2 is ON:-
IC2 (sat) =(VCC-VCE (sat))/RC2= VCC/RC2
IB2 (sat)>=IC2 (sat)/bon
I3=(VCC-VB2 (ON))/(Rc1+R3)
I4=(VB2 (ON)-(-VBB))/R4
IB2 (sat)= (VCC-VB2 (ON))/(RC1+R3)-(VB2 (ON)-(-VBB))/R4
VB1(OFF)=-[|VBE (CUTOFF)|-VBE (sat)]
VB1(OFF) = (VCE (sat)R2)/(R1+R2)-VBBR1/(R1+R2)

Stable state 2 :-
Q1 is ON and Q2 is OFF:-
IC1(sat)=(VCC-VCE (sat))/RC1= VCC/RC1
IB1 (sat)>=IC1 (sat)/bon
I1=(VCC-VB1 (ON))/(R1+RC2)
I2=(VB1(sat)-(-VBB))/R2
IB1 (sat)=I1-I2
IB1 (sat)= (VCC-VB1 (ON))/(R1+RC2)-(VB1(sat)-(VBB))/R2
VB2 (off)=[|VBE(cutoff)|-VBE (sat)]
VB2 (off) = (VCE (sat)R4)/(R3+R4)-VBBR3/(R3+R4)

COMMUTATING CAPACITORS
Transition Time:-
It is define as the time interval during which
conduction transfer from ON transistor to the OFF
transistor. Usually, it is desirable that the transition
should be small and the transition should require a
finite amount of trigger energy.

COMMUTATING CAPACITORS
Transition time may be reduced by
introducing the binary capacitances Cm1 andCm2
in shunt with the coupling resistor R1 and R2
respectively. These capacitors speed up the
transition from OFF state to ON state. Hence
these capacitor are known as speed up
capacitor or commutating capacitor or
transpose capacitor.
The main purpose of these capacitors is to
improve the switching characteristics of the
circuit by passing high frequency component of
square wave.

RESOLVING TIME:-

The minimum time interval between two
consecutive trigger pulse is known as resolving
time.

Delay time, D=0.693RC
where, R=>Resistance
C=>Capacitance

TYPES OF MULTIVIBRATOR:-

(c)Self Biased Transistor MV
(d)JFET Bistable MV

NUMERICAL:

Question:-
Design a BMV who has two o/p as 0 and 10v;
given
IcMAX = 25mA, β = 100, VCE(ON) = VBE(ON) = 0 &
voltage of -1v is required to reverse bias the
transistor & VBE(OFF) = -10v ?

Solution:-
Given that:
VCC = 10v
VBE(ON)=VCE(ON) = 0
β = 100
VBB = -1V
VBE(OFF) = -10V

To Find :-

All the parameters which are required for
the circuit
i.e. R1,R2,R3,R4,Rc1,Rc2 .

Calculation:
Q1 is off and Q2 is on:-

IB2 sat>=IC2 sat/βon
=25/100 mA
=0.25 mA

IC2 sat =(VCC-VCE sat)/RC2= VCC/RC2
RC2 =(10*1000)/25
=400 Ω

VB2 off = (VCE satR4)/(R3+R4)-VBBR3/(R3+R4)

VB2 off = R3/(R3+R4)

-10 = R1/(R1+R2)
R2 = -11R1/10

R1=35.9 KΩ

R2 = -11*35.9/10
=39.4 KΩ

BLOCKING OSCILLATOR
WORKING

This voltage applied to base of transistor. Thus Ic is
further increased due to increase in FB to base-emitter
junction. The transistor is quickly driven into
saturation. Now capacitor is charged and –ive charge
on the base of transistor which RB the base-emitter
junction.
Thus the transistor is driven cut-off. The transistor
remains at cut-off, as capacitor now commences
discharge.

MONOSTABLE MULTIVIBRATOR
It is also called a single swing, or delay MV.
In this circuit, one coupling network provides ac
coupling while the other provides dc coupling.
It has only one stable (stand by) state and one
quasi stable state.

MONOSTABLE MULTIVIBRATORS
The circuit remains in its stable state until a
triggering signal causes a transition to the quasi
stable state. Then after a time T, the circuits
return to its stable state.

MONOSTABLE MULTIVIBRATORS
As only one triggering signal is required to
induces a transition from a stable state to quasi-
stable and the circuit returns to its initial stable
automatically after a definite period, it is called
single-shot MV.

MONOSTABLE MULTIVIBRATOR
Application
(1)This MV is employed for generating clean and
sharp pulses from distorted, old pulses.
(2) The falling part of the monostable multivibrator
output is often used for triggering another pulse
generator circuit thus producing a pulse delayed
by a time T with respect to the input pulse.

CONSTRUCTION:-
It consists of two identical transistor Q1 and
Q2 of N-P-N type. Output of Q1 is coupled to base
of Q2 by capacitor C. Q1 is reversed biased due
to power supply VBB. While Q2 will be in ON state,
because base potential required for Q2 is
supplied through VCC by R continuously.

So initially in stable state Q1 is OFF and Q2
is ON .

COLLECTOR-COUPLED MMV
CIRCUIT DIAGRAM:-

CONSTRUCTION:-
Capacitor C charges to VCC through RC1 and base
current of Q2 , then this current stops flowing. So
capacitor C is completely charged to VCC with left
plate positive. Cm is a commutating capacitor or
speed up capacitor which provided to improve the
switching characteristics of the circuit.

WORKING
When the positive trigger is applied to the base of
Q1 transistor through the capacitor C2 the base
voltage of Q1 increases and it starts conducting .
Thus Q1 starts conducting and the potential of
collector of Q1 comes down to ground.
Since charge on capacitor C cannot disappear
instantly, the voltage across the capacitor plates is
maintained.

WORKING
As the capacitor discharges the negative bias is
applied to the base of Q2 and Q2 is cutoff. The
collector of Q2 rises towards VCC and is now
capable of supplying base current of Q1 through its
base resistance R1. thus transistor Q1 remains
tuned On after the positive spike from the
transistor Q1 is removed.

WORKING
Stage 1:- stable state:- In stable state Q1 is off
and Q2 is on.
(A) when Q2 is ON:-
VC2 sat = vCE sat
IC2 sat = (VCC – VCE2 sat )/ RC2 = VCC/ RC2

(VCE2sat = 0)
IB2on >= IC2 sat /bon = (VCC-VBE sat)/R
VB2ON = VBEON = VBE sat

WORKING
(A) when Q1 is OFF:-
VC1 OFF = VCC
VB1OFF = (-VBBR1+VCE satR2)/(R1+R2)
VB1off <= -0.5to -1V neglecting VCE sat
VB1OFF = -VBBR1/(R1+R2)

STAGE 2(A):- Q1 IS ON:-
VC1sat = VCE sat = 0
IC1sat = (VCC-VCE sat)/ RC1 = VCC/RC1
IB1sat >= IC1sat/bon
VB1sat = VBE sat
I1 = (VCC-VBE sat)/(RC2+R1)
I2 = (VBE sat-(-VBB))/R2
IB1sat = I1-I2

WORKING:-
V(t) = Vin+(Vf-Vin)e-t/t
(V (t)-Vin)/(Vf-V in) = e-t/t
(V f-V in)/(V (t)-Vin) = et/t
At t = T
V(t) = Vl (cut in voltage)
V f= VCC and Vin= (VBE sat-VCC+VCE sat)

WORKING:-
(VCC-(VBE sat-VCC+VCE sat))/ (Vl- (VBE sat -VCC+VCEsat))= e-
t/t

At VBE sat=0.7V VCE sat =0.3V
VCC-(0.7-VCC+0.3) / Vl-(0.7-VCC+0.3)=e-t/t
Solving the equation (V l=0.5)
2= et/t
t=t ln2
T=0.693RC

Numerical
Question:-
Design a monostable multivibrator which gives a
pulse duration of 10ms & a height of 10v when it
triggered for transistor to be used ;
Β=125,vce(on)=vbe(on)=0,reverse bias of 1v is
required to turn off the transistor vbe(off)=10v
icmax =20mA

Solution:-
Given that:
Β=125
vce(on)=vbe(on)=0
Vbe(off)=1v
vbb=10v
icmax =20mA
t=10ms

Let us assume that T1 –OFF
T2-ON

IC2 sat=(VCC – VCE2 sat )/ RC2 = VCC/ RC2
=10/20mA=1000/2=500

IB2on>= IC2 sat /bon
=20/125=16mA

(VCC-VBE sat)/R=16mA
R=10/16=62.5kΩ
VB1OFF= -VBBR1/(R1+R2)
1= -10/(R1+R2)

R2=-11R1

In Next Clock Pulse When
T1-ON
T2-OFF

I1=(VCC-VBE sat)/(RC2+R1)

I2=(VBE sat-(-VBB))/R2

IB1sat=I1-I2

= (VCC-VBE sat)/(RC2+R1) -(VBE sat-(-VBB))/R2

0.6mA=10/500+R1-1/r2
R1=33.7KΩ

R2=11*33.7KΩ

T=0.693RC

10ms =.693*62.5*C

C=0.23µF

ASTABLE MULTIVIBRATOR
In an astable MV , both coupling network provide
ac coupling through coupling capacitors.
It has no stable state. The two states had by the
astable MV are quasi-states The astable MV ,
therefore, makes successive transition from one
quasi-state to the other one after a
predetermined time interval, without the aid of an
external triggering signal.

Since its output oscillates in between off and on
states freely, it is called a free-running MV.

Frequency of oscillation;
f=0.7/RC

TYPES OF AMV:-

2) Asymmetrical astable Multivibrator
3) Symmetrical Astable Multivibrator
4) Saturating Astable Multivibrator
5) Non Saturating Astable Multivibrator:-
6) Voltage Controlled Astable Multivibrator:-

COLLECTOR COUPLED SATURATING ASTABLE
MULTIVIBRATOR:-

MULTIVIBRATOR:-

 Resistors Rc1 and Rc2 are the collector circuit
resistors.
 Capacitors Cb1 and Cb2 are the coupling
capacitors.
 Capacitor Cb1 connects the output of transistor
Q1 to the base terminal of the transistor Q2.

MULTIVIBRATOR:-

Resistor RB1 and RB2 provides the ON state base
currents to the transistors Q1 and Q2 respectively
during the saturation region. For a symmetrical
astable MV, we should have RB1= RB2, Cb1= Cb2
and Rc1= Rc2.

WORKING OF THE CIRCUIT:-
Let at time instant t=0, the power supply voltage
Vcc get applied abruptly. Then in a symmetrical
astable multi, due to slight mismatch, let the
current Ic1 flowing in transistor Q1 be slightly more
than the current Ic2 in transistor Q2. Hence the rate
of Vc1 at collector Cb1 of Q1 is greater than of Vc2 at
collector of Q2.

For the transients, the capacitor act as short-
circuit and the voltage across them cannot
change instantaneously. Hence the drop in
collector voltage of Q1 from the initial value Vcc
to Vc1 (Vc1< Vcc) makes the voltage of transistor
Q2 to reduce by the same amount.

This negative increment in the voltage at the
base of transistor Q2 reduce the conduction
current and cause increase in the collector
voltage Vc2 thereby making it move towards
Vcc.

This increase in Vc2 gets transferred through
capacitor Cb1 to the base of the transistor Q1
making its voltage more positive thereby
increasing the conduction in Q1.

Increased conduction in Q1 further reduces the
collector voltage Vc1 which in turn causes further
reduction in voltage of base of transistor
Q2.

Thus a regenerative or positive feedback action
with loop gain greater than unity sets in. As a
result, the above mentioned sequence of
operation occur instantaneously causing the
transistor Q1 to go into saturation and transistor
Q2 to go into OFF region.

Thus when astable MV is switched, we have the
following condition:
(i) Transistor Q1 is in saturation region.
(ii) Transistor Q2 is in OFF region.

LINEAR EQUIVALENT CIRCUIT WITH Q1 IN SATURATION
AND Q2 OFF

CIRCUIT CONDITION FOR TRANSISTOR Q1 IN
SATURATION REGION

VC1(0) =VCE sat
IC1 sat=(VCC-VCE sat)/RC1 = vcc /RC1
(vCEsat=0)
VB1ON >=VBEON =VBESat
IB1ON=(Vcc –VBE Sat )/RB1
IB1ON >=IC1Sat / bON
VB20ff=-VCC+VBEon
VCEoff = VCC-VCEsat

SATURATION REGION

For silicon transistor Q1, the base voltage should
be about 0.7v. With collector –to-emitter saturation
voltage VCE=0.2v,this base voltage of 0.7v will
forward bias both the emitter and the collector
junctions.

SATURATION REGION

On making the simplifying assumption that the
transistor Q1 is ideal .
For large Vcc,Vc1(0)=0v
IC1sat=VCC /RC1
For large VBB ,VB1ON=0v
IB1ON= Vcc /RB1

CIRCUIT CONDITION FOR TRANSISTOR Q 2 IN
OFF REGION:-

VB2OFF (0+)=-VCC+VB2on
IC2(0)=0
VC2 (OFF)=VCC-VCE (sat)=VCC

CIRCUIT BEHAVIOUR IN QUASI STABLE STATE:
CHARGING OF CAPACITOR CB2 (0<T<T1 ):-

The time interval 0<t<t1 , the voltage across
capacitor Cb2 rises from
VB2 OFF(0) towards Vcc. The charging path of capacitor
Cb2 is shown figure.

LINEAR EQUIVALENT CIRCUIT WITH Q2 IN
SATURATION AND Q1 OFF

AND Q1 OFF

VCE sat of transistor Q1 to be negligible .At any time t,
the expression for voltage on capacitor Cb2 or VB2
OFF(t) may be written:-

VB2 OFF (t) = Vcc +[VB2 OFF(0)- Vcc ]e-t/Rc2CB2


During the quasi stable state ,capacitor Cb2
charges towards VBB through resistor RB2.At the
time instant t=t1, the instantaneous base voltage
of Q2 equals VB2on and Q2 enters conduction .
Figure shows the variations of voltageVB2.Thus
we have:-

VB2 ON=VB2(∞)+[VB2 OFF(0)-VB2(∞)]e-t1/RB2CB2
But VB2off(0)=-VCC+VB2 ON
And VB2(∞)= Vcc
Hence on solving for time interval t1 we get :-
t1=RB2CB2In[(Vcc +(VCC-VB2 ON))/ Vcc -VB2 ON]

AND Q1 OFF
We assume that VCE sat<<VCC and VB2on<<VBB. hence
voltage VCE sat and VB2on may be neglected.so
above equation gets reduces as given below:-
t1=0.693RB2Cb2.

BEHAVIOUR AT TIME INSTANT T=T1
At time instant t=t1 transistor Q2 enters into
conduction. The collector voltage of Q2 begins to
fall. This falling collector voltage of Q2 gets
communicated to the base of transistor Q1 by a
capacitor Cb2 consequently the conduction of Q1
reduces resulting in the increase of collector
voltage of Q1

BEHAVIOUR AT TIME INSTANT T=T1
This increase in collector voltage of Q1 is
communicated to the base of Q2 via capacitor Cb2
increasing the conduction of Q2 .This process
continuous and Q2 goes into saturation while Q1
goes OFF instantaneously.

CIRCUIT BEHAVIOUR DURING QUASI ASTABLE
STATE(T1<T<T2):-

During this time period capacitor Cb1 charges from
VB1 OFF(0) towards Vcc analogously. At the time
instant t=t2 the instantaneous base voltage is VB1ON
which brings Q1 into conduction.

CIRCUIT BEHAVIOUR DURING QUASI ASTABLE
STATE(T1<T<T2):-

 The time interval (t2-t1) may then be expressed as:-

 (t2-t1)=RB1Cb1In[((VCC +(VCC-VB1ON))/(Vcc -VB1ON)]

 Assuming that VCE sat<<VCC and VB1ON<< Vcc the

 equation reduces to the following simple form:-

 (t2-t1)=0.693RB1Cb1

TIME PERIOD:-
The time periods T is the sum of periods t1 and (t2-
t1).thus:-
T = 0.693[RB1Cb1+RB2Cb2]

For symmetrical multivibrator, we have
RB1=RB2=RB say
Cb1=Cb2=C say
Then eq. reduces to the following form:-
T=1.38RbCb
f=1/T
f=0.72/RC

from above eqn. we see that the time period of
astable multivibrator is independent of the supply
voltage, temperature and junction voltages.
Duty cycle:
=t1/T=t2/T
= TIME IN WHICH CAPACITOR
CHARGES/
TOTAL TIME PERIOD

NUMERICAL:-

Q. Design a astable multivibrator which
generates a waveform of prf(pulse
repetition frequency) 50khz & duty cycle of
60% of height 10v icmax=20mA,β=100.

Solution:- Given that
f=50KHZ
Duty cycle =60%
Vcc=10V
β =100
Icmax=20mA

Formula to be used
Rc=Vcc(max) / Ic(max)
IB(sat)=Ic(sat) / β
RB1=Vcc-VBE/Ion
Duty cycle=t1/T=t2/t
T=1/f
T=t1+t2
t1=0.693R1C1
t2=0.693R2C2

Rc=Vcc(max)/Ic(max)
= 10/20mA
Rc=500Ω
IB(sat)=IC(sat)/β
IB(sat)=20mA/100
IB(sat)=0.2mA
RB1=Vcc-VBE/Ion
RB1= 10V-0/.2mA=50KΩ
Duty cycle=t1/T

T=1/50K
t1=12 µs
T=t1+t2
1/50K-0.12µs=t2
t2=8µs
t1=0.693R1C1
C1=12 µ/0.693*50K
C1=346pF
C2=8 µ/0.693*50K
C2=230pF

APPLICATION:-
 Amplitude comparator
 Flip flop circuit.
 Squaring circuit
 Signal Regeneration
 FM Demodulation
 Robotics & Radio control
 Frequency Synthesizer

SCHMITT TRIGGER
 It is a device that generates square & rectangular
waveforms.
 It is bistable multivibrator.
 It has two stable state (one high, other low)
 (I) In schmitt trriger base of transistor is kept
open.
 (II) Feedback from output of transistor Q2 to
transistor Q1 is achieved through Re.

OPERATION OF SCHMITT TRIGGER

As input voltage Vin increases from zero(Q1 off, Q2
on) output voltage V0=Vcc-Ic2Rc2, and when input
voltage reachesV1,output suddenly rises to Vcc as
Q2 become off and Q1 remains on,if Vin is
increased beyond V1 circuit remains in stable
state(i.e. Q1 on and Q2 off) but if Vin
decreases,during negative half cycle then when it
reaches to V2 the circuit makes as abrupt change
i.e. Q1 off and Q2 on.
V0=Vin-Ic2Rc2 again.

SCHMITT TRIGGER AS A FLIP FLOP
 When the power supply is switched on for the
first time, R3, R1 and R2 forming a potential
divider across VCC and –VBB.

 Forward biases. Slightly the transistor Q2 and ,
therefore Q2 starts conducting . The transistor Q1
is now reverse biased due to flow of current in
emitter resistor RE from Q2.

Thus the Q1 goes to cut off. As a result the
potential of Q1 collector rises to VCC. This
positive going signal appears across the emitter
base junction of transistor Q2, as it is connected
to terminal C1 by R1, and drives transistor Q2 into
saturation and holds there .
Thus in the initial static condition of the
Schmitt trigger circuit; transistor Q1, is in cut off
and Q2 is in saturation


Now when the input ac signal (say positive
going) is applied to the base of transistor Q1 ,if it
is to sufficient strength to overcome the reverse
bias placed on the base of Q1, due to voltage
drop across emitter resistor RE, the Q1 is forward
biased.

 Now Q1 start conducting , its collector terminal C1
potential drops, this negative going signal
coupled to the base of transistor Q2 via resistor
R1 reduces its forward bias and consequently
emitter current.

 Withreduced emitter current voltage drop across
emitter resistor RE falls and therefore reverse
bias placed on the Q1 due to it decreases and

Q1 starts conducting more. As a result
collector Voltage of Q1 drops further and
therefore Q2 is further driven to cutoff at the end
Q1 goes into saturation and Q2 goes into cutoff.
After half a cycle of the input signal, when the
input signal to the base of transistor Q1 is
negative going Q1 becomes reverse biased, its
collector current drops and therefore its collector
terminal potential rises


As a result the transistor Q2 is forward
biased, it starts conducting again, emitter current
increase voltage drop resistor RE increases and
therefore the Q1 in further reversed biased, at the
end Q2 comes into saturation and Q1 come into
cutoff.

APPLICATION:-
 Squaring circuit
 Amplitude comparator

 Flip flop circuit.

DIFFERENCE BETWEEN SCHMITT
TRIGGER AND BI STABLE MULTI
VIBRATOR:-
SCHMITT TRIGGER BISTABLE
MULTI VIBRATOR
1.It is a square wave 1.It is a square wave or any
generator. non sinusoidal wave
2.V(in) is applied in it. generator.
3.We do not apply 2.V(in) is not applied in it.
external triggering pulse 3.We have to apply external
in it. triggering pulse in it.
4.Re resistance is there 4.There is no Re resistance.
in Schmitt trigger.

SCHMITT TRIGGER BI STABLE
MULTIVIBRATOR
5.Feedback from Q2 is 5. In bi stable it is not like
obtained through that.
resistance R2 6. In bi stable both are
connected.
6.In it output of 1 is
st

connected to base of
2nd transistor but
Output of 2nd is not
connected to base of
1st .

HYSTERESIS IN SCHMITT TRIGGER:-
In the non-inverting configuration, when the input is
higher than a certain chosen threshold, the output is
high; when the input is below a different (lower)
chosen threshold, the output is low; when the input is
between the two, the output retains its value. The
trigger is so named because the output retains its
value until the input changes sufficiently to trigger a
change. This dual threshold action is called
hysteresis, and implies that the Schmitt trigger has
some memory

Fig.:-Schmitt trigger using OP-AMP

FORMULAE:-
 The feedback fraction, b =R2 /(R2 + R1)
The trip point are defined as the two input
voltages where the output changes its states.
 The upper trip point has a value:
UTP= b Vsat
= R2/(R2 + R 1) Vsat
The lower trip point has a value:
LTP= - b Vsat
= - b R2/(R2 +R1 ) Vsat
 Hysteresis
H=UTP-LTP= 2b Vsat

NUMERICAL

Question:-
Derive a Schmitt trigger given that VUTP=5v and VLTP=
-5v
What is hysteresis voltage?

Solution:-
Given that:-
VUTP = 5v
V LTP = -5v
Assume that
Vsat =10v
now by solving equations for VUTP and VLTP we get

10R1/ R1+ R 2 =5

hysteresis voltage =VUTP-VLTP=5+5= 10v

S UBM TTED BY-
I
OSCILLATOR-
 Diggujawal Kumar
 Dilip Kumawat
 Hemant Bhawariya
 Jogendra Ajmera

MULTIVIBRATOR:-
 Bhawani Singh Kanawat
 Bhawna Kaushik
 Deepak Kumar Maru
 Gopal Kumar Roy

Oscillator multivibrotor

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