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

applied.

1. Positive feedback is

applied.

2. It strengthens the

input signal without

any change in

frequency.

2. It strengthen the

causing charge as

well waveform.

3. Require an external

wave signal to be

applied.

3. It just require a dc

signal.

MERITS OF OSCILLATOR

Portable and cheap in cost. An oscillator is a non-rotating 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.

FREQUENCY STABILITY OF 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

OUTPUTAmplifier (gain=A)

Frequency

Selector

AutomaticAmplitude 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 sometimes called a tickler coil.

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

R-C L-C OTHER

HARTLEY

COLPITTS

CLAP

WEIN’S BRIDGE

RC-PHASE SHIFT

CRYSTAL

NEGATIVERESISTANCE

BLOCKING

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.

Z1 Z2

Z3

hie 1k hfeIb

IbB C

E3

21

4. H-EQUIVALENT CIRCUIT DIAGRAM:-

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.


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)/w²C1C2-L/C1

=> jhie(C2+C1+jw2LC1C2/jw²C1C2)

= 1+hfe/w2C1C2-L/C1

=> jhie(C1+C2-w²LC1C2/jw²C1C2)

= 1+hfe/w2C1C2-l/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.


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 1uR1

1k

R2 1

k

C2 1u

R3 1

k

C3 1

u

C4 1

u

C5 1

u

L1 1

m

L1 1

mL2

1m NI1 Vout

T1 !NPN

Vcc

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:-

Z2 Z1

Z3

hie 1k H fe Ib

IbB C

E

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 :-

R1 1

k

R2 1

k

R3 1

k

C1 1u C2 1uC3 1uR4 1k

+

VG1

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-RcXc

2-6jR2Xc-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

1.CIRCUIT DIAGRAM:-

C 1u C 1u C 1u

R 1

k

R 1

k

R 1

k

Cs 1

u

Rs 1

kR

d 1

k

T1

2N

33

69

Vcc

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

R 1

k

R 1

k

R 1

kRd

1k

rd 1

k

gmVi

Vi=Vf Vf'=Vf

FREQUENCY OF OSCILLATOR :-

From figure I1R = Vf ’……(i)I2R = I1R+1/jwCI2R = Vf’1+1/jwC……. (ii)I3R = (I1+I2)/jwC + I2R

= [Vf’/R +Vf’/R1+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 =1Thusβ=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 .

It is a lead -lag network . The phase–shift across the network lags

with increasing frequency and leads with decreasing frequency.

1. CIRCUIT DIAGRAM :-

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 circuit .

•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 capacitor

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

R4/wC1-wC2R2R1R4 = 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

This is 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):-

R1 1kC1 1u

L1 1m

U1 HC49S_CY11BS

CRYSTAL OSCILLATOR CIRCUIT DIAGRAM (VOLTAGE SERIES)

CRYSTAL OSCILLATOR (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)CONSTRUCTION AND WORKING

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)

CRYSTAL OSCILLATOR (VOLTAGE SHUNT)CONSTRUCTION AND WORKING

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 a fixed voltage level.


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.


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 OSCILLATORDISADVANTAGE

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.).

BASIC BLOCK DIAGRAM

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.

TYPES OF MULTIVIBRATORS

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.


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:-

(a)Self Biased Transistor MV(b)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 = 10vVBE(ON)=VCE(ON) = 0β = 100VBB = -1VVBE(OFF) = -10V

To Find :-

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

Calculation:Q1 is off and Q2 is on:-

IB2 sat>=IC2 sat/on

=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

Designed Multivibrator

BLOCKING OSCILLATORWORKING

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 MMVCIRCUIT 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)

Q1 IS ON Q2 IS OFF:-

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

NumericalQuestion:-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=125vce(on)=vbe(on)=0Vbe(off)=1v vbb=10v icmax =20mAt=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=62.5kΩVB1OFF= -VBBR1/(R1+R2) 1= -10/(R1+R2)

R2=-11R1

In Next Clock Pulse WhenT1-ONT2-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

Designed Multivibrator

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:-

1) Asymmetrical astable Multivibrator2) Symmetrical Astable Multivibrator3) Saturating Astable Multivibrator4) Non Saturating Astable Multivibrator:-5) Voltage Controlled Astable Multivibrator:-

COLLECTOR COUPLED SATURATING ASTABLE 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.


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


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.


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 Q2 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.


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

/RB2

CB2

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]


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 haveRB1=RB2=RB sayCb1=Cb2=C say

Then eq. reduces to the following form:-T=1.38RbCb

f=1/Tf=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 CIRCUIT

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.

SCHMITT TRIGGER AS A FLIP FLOPThus 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.

With reduced 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.

SCHMITT TRIGGER BISTABLE

MULTI VIBRATOR1.It is a square wave generator.

2.V(in) is applied in it.

3.We do not apply external triggering pulse in it.

4.Re resistance is there in Schmitt trigger.

1.It is a square wave or any non sinusoidal wave generator.

2.V(in) is not applied in it.

3.We have to apply external triggering pulse in it.

4.There is no Re resistance.

DIFFERENCE BETWEEN SCHMITT DIFFERENCE BETWEEN SCHMITT TRIGGER AND BI STABLE MULTI TRIGGER AND BI STABLE MULTI VIBRATOR:-VIBRATOR:-

SCHMITT TRIGGER BI STABLE MULTIVIBRATOR

5.Feedback from Q2 is obtained through

resistance R2

6.In it output of 1st is connected to base of

2nd transistor but

Output of 2nd is not connected to base of

1st .

5. In bi stable it is not like that.

6. In bi stable both are connected.

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 = 5vV 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

SUBMITTED BY-OSCILLATOR- Diggujawal Kumar Dilip Kumawat Hemant Bhawariya Jogendra Ajmera

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

Oscillator multivibrotor

Education

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