(24) oscillators

7/31/2019 (24) Oscillators

1/52

Oscillators 1

Oscillators


2/52

Oscillators 2

Need of an Oscillator An oscillator circuit is capable of producing ac

voltage of desired frequency and waveshape. To test performance of electronic circuits, it is

called signal generator.

It can produce square, pulse, triangular, orsawtooth waveshape.

High frequency oscillator are used inbroadcasting.

Microwave oven uses an oscillator. Used for induction heating and dielectric

heating.


3/52

Oscillators 3

Types of Oscillators

Sinusoidal or non-sinusoidal.

An oscillator generating square wave or apulse train is called multivibrator :

1. Bistable multivibrator (Flip-Flop Circuit).

2. Monostable multivibrator.

3. Astable multivibrator (Free-running).

Depending upon type of feedback, we have

1. Tuned Circuit (LC) oscillators.

2. RC oscillators, and

3. Crystal oscillators.


4/52

Oscillators 4

In our discussion of negative feedback , we hadassume that the output of the amplifier was feed

backed in opposite phase to that of the input andthat the feed back network did not introduce anyphase shift in the feedback signal so that VF and

VO remains in phase and thus VF and VS were out

of phase . Consequently we can writeVI= VS - VO and AF=A/(1+ A)

Suppose that the feed back network introduces an

phase shift of -1800

then VS and VF will be inphase so it will be equivalent to positive feedback.


5/52

Oscillators 5

Using Positive Feedback The gain with positive feedback is given

as

By making 1A= 0, orA = 1, we getgain as infinity.

This condition (A = 1) is known as

Barkhausen Criterion of oscillations.

It means you get output without any input !


6/52

Oscillators 6

How is it Possible ?

Connecting point x to y, feedback voltagedrives the amplifier.


7/52

Oscillators 7

Building of oscillations

As soon as X get connected to Y, the amplifieramplify the noise signal , a part of which isfeedback to the input. The frequency of oscillation

is the point at which A=1. The amplitude ofoscillation builds up and ultimately limited by thenonlinearities of the device which brings saturationin the output. Because of the nonlinearities in the

output , the output contains the fundamentalfrequency and its harmonics.


8/52

Oscillators 8

(1) IfA < 1, we get decaying ofdampedoscillations.


9/52

Oscillators 9

(2) IfA > 1, we get growing oscillations.


10/52

Oscillators 10

(3) IfA = 1, we get sustained oscillations.In this case, the circuit supplies its own input

signal.


11/52

Oscillators 11

Wherefrom comes the startingvoltage ?

Each resistor is a noise generator. The feedback network is a resonant circuit

giving maximum feedback voltage at

frequency f0, providing phase shift of 0 onlyat this frequency.

The initial loop gainA> 1.

The oscillations build up only at thisfrequency.

After the desired output is reached,Areduces to unity.


12/52

Oscillators 12

RC Oscillators

Two types :

1. RC Phase shift Oscillator.

2. Wein Bridge Oscillator.


13/52

Oscillators 13

RC PSO( Phase Shift Oscillator)

The basic structure of a phase shift RC oscillator isshown here. It consist of a negative gain amplifierwith three section of RC ladder network in thefeedback


14/52

Oscillators 14

The circuit will oscillate at a frequency at

which phase shift of the RC network is 1800

.because this will cause the total phase shiftto 3600 or 00.

The reason for using a three section RC

network is that the three is minimum no ofsections that is capable of producing a 1800phase shift at a finite frequency.

An RC network can produce a phase shiftfrom 0 to 900. Thus in the three section,each section produces a phase shift of 600.


15/52

Oscillators 15

And the feedback factor is =1/29 and the A to begreater then 1 , so we must have gain greater then29.

By using network , we can derive the relationship betweenFeedback voltage and output

VOUT/VIN= 1 - j6/RC 5/(RC)

2

+ j/(RC)

3

puttingImaginary part equals zero we have-6/RC + 1/(RC)3 =0 solving we will get

Frequency as given below

Putting this value of f ,we have feedback factor as=VIN/VOUT = -1/29


16/52

Oscillators 16

RC Phase shift Oscillator

Here the amplifier is realized by n-FET. As we know that FETgives inverted output so already phase shift of 1800 so forsatisfying barkhausen criteria phase shift of 1800 moreneeded by feedback network.


17/52

Oscillators 17

Problem:

It is desired to design a PSO using aFET with gm=5x10

-3A/V, rd=40K, and

the feedback value R=10K.Determine the value of C for oscillatorto operate at 1kHz and the value of RL

for A>>29 to ensure oscillator action.Assume |A|=40.


18/52

Oscillators 18

SOLUTION

We know from small signal analysis

that gain for the above circuit isA=-gm.RL where R

L=RL||rd

So 40=5x10-3.RL => RL =8K

Since RL=RL||rd so we can solve for RL.RL=10K.

Using the formula of we have

C=6.5nF.


19/52

Oscillators 19

THE WIEN BRIDGE OSCILLATOR

The wien bridge oscillator is the standard oscillator circuit

in the range of 5Hz to 1 MHz. It is used in commercialaudio generators and is usually preferred for other lowfrequency application.

This oscillator circuit uses a resonator feedback circuit

called a lead-leg circuit as shown below

Transfer function:V0/VIN=Z2/(Z1+Z2)Where Z2=1/(1+jRC)

Z1=R+1/jRC

V0/VIN=jRC/[1-(RC)2+3jRC)]

(eq.1)


20/52

Oscillators 20

At low frequency series capacitor act as open circuit andhence there is no output . At very high frequency , theshunt capacitance looks shorted and hence there will notbe any output.

In between these limits , the output voltage reaches amaximum value of 1/3(Using eq.1 find maximum value).This frequency is the resonant frequency fr. At thisfrequency the phase angle between the input and output isequals 00. This is shown below


21/52

Oscillators 21

The actual circuit using OPAMP isshown below


22/52

Oscillators 22


23/52

Oscillators 23


24/52

Oscillators 24


25/52

Oscillators 25


26/52

Oscillators 26

Here the resistances R1 and R2 and capacitors C1 and C2 formthe frequency adjustment elements while the resistances R3and R4 form part of the feedback path. The OPAMP output is

connected as the bridge input at point a and c. The bridgecircuit output at point a and d is the input to the OPAMP.

Using feedback concept and barkhausen criteria we have thefrequency of oscillation a given below

By using same value of capacitors and resistors we can have

the following expression for frequency (using eq.1 put phaseshift to be zero, ie real part in denominator should be zero)


27/52

Oscillators 27

This analysis gives the feedback factor as1/3 so gain must be at least greater the3(barkhausen criteria)

As we have gain for the circuit as

1 + R3/R4 = 3

so we have to take R3/R4=2 for oscillation tobuild up.


28/52

Oscillators 28

Solution :


29/52

Oscillators 29

Tuned Oscillator Tuned oscillator employ tuned LC circuit and are favorite

at high frequencies. Figure shown below is tuned oscillatorand next fig shows its equivalent circuit of amplifier whereA , represent the open loop gain of amplifier.


30/52

Oscillators 30

Equivalent circuit:


31/52

Oscillators 31

Note that the input is between terminals 1 and 3and the output is between terminal 2 and 3. Herethe load impedance is given by

ZL=(Z1+Z3)||Z2= [(Z1+Z3).Z2]/(Z1+Z3 +Z2)

From equivalent circuit we have

VO/ZL= - (VO-A.Vi)/RO (applying KCL)

The feedback factor of the circuit

=VI/VO=Z1/(Z1+Z3)


32/52

Oscillators 32

For oscillation to occur we must have

That will give us

AZ1ZL/[(Z3+Z1).(ZL +R0)]=-1

After putting the value of ZL we have the followingexpression

AZ1Z3 /[RO(Z1+Z2+Z3) +Z2(Z1+Z3)]=-1

AL=-1


33/52

Oscillators 33

Assuming Z1, Z2 , Z3 are reactance of thefollowing value

Z1 = jX1 , Z2 = jX2 , Z3 = jX3

We can have the following

AX1.X2=jRO(X1+X2+X3) - X2 (X1+X3)

Since left hand side is real so equating right

hand side imaginary part as zero(X1+X2+X3)= 0

This condition determines the frequency ofoscillation and shows that all the reactance

can not be of same type. It means if twoof them are capacitor then third one will beinductor and if two of them are inductorthen the third one will be capacitor.


34/52

Oscillators 34

Equating the real part we have

AX1.X2= - X2 (X1+X3) since(X1+X2+X3)= 0 ie (X1+X3)= -X2

So we have

AX1.X2= - X2.(-X2)

i.e A= X2/ X1

This gives indication that X2 and X1 must be ofsame type i.e either both of them will becapacitor or will be inductor.

Note: above equation of gain also indicates theminimum gain required for oscillation.


35/52

Oscillators 35

Types of tuned oscillator

There are two types of tuned oscillator(1) COLPITS Oscillator

(2) HARTLEY Oscillator

In Colpits oscillator we have X1 and X2 arecapacitors and X3 is inductor.

In Hartley oscillator we have X1 and X2 areinductor and X3 is capacitor.


36/52

Oscillators 36

COLPITS OSCILLATOROPAMP based colpits oscillator

F f ti


37/52

Oscillators 37

Frequency of operation(X1+X2+X3)= 01/jC1 + 1/jC2+ jL =0L= 1/C

1+ 1/C

2=[1/C

1+1/C

2].1/

2=[1/C1+1/C2].1/L

Calculate the oscillation frequency for C2=750pF,C1=2500pF,

and L=40 Hsol: f0=1.048 MHz.gain of amplifier is A=X2/X1= C1/C2= C1/C2=250/75=10/3Since amplifier gain is

RF/R1=10/3 so taking R1=3K we have Rf=10K


38/52

A colpits oscillator is designed with C =100pF


39/52

Oscillators 39

A colpits oscillator is designed with C1=100pF,and L=42.2mH. The frequency of oscillation is150Khz . Determine C2 and the minimum gainof the amplifier.

Since we have

2=[1/C1+1/C2].1/LUsing the data given we can find

the value of C2=36.4pF.

Gain of amplifier is A=C1/C2=100/36.4


40/52

Oscillators 40

HARTLEY Oscillator


41/52

Oscillators 41

Using (X1+X2+X3)= 0 we have

jL1 +jL2 +1/jC = 0

(L1 +L2) =1/C

2=1/[C.(L1 +L2)]


42/52

Oscillators 42

Problem: A Hartley oscillator is designed with a voltage gain

of 100 and C=13.9pF. (1) Determine theinductances L1 and L2 for f0=950KHz.(2) if f0 raised to 2.05MHz , determine the value ofcapacitor C.

SOL

:Part 1:A=L2/L1=100

L=L1+L2=101L1 2=1/L.C ie L=2.014mH L

1

=2.01/101=19.9 H. L2=1.99mH.:part 2:L.C=1/2. assuming L=101.L1and using f=2.05MHz

SOLVE FOR C =3pF


43/52

Oscillators 43

Solution :


44/52

Oscillators 44

Solution :

Note: here C1 andC2 is different from ourassumption


45/52

Oscillators 45


46/52

Oscillators 46

Crystal Oscillator

A piezoelectric crystal , usually quartz , has theproperty that if electrode are placed on its oppositefaces and a potential is applied across them, the

electric field so set up exert forces on boundcharges in the crystal . As a result the crystalresponds electromechanically i.e it vibrates. Theresonant frequency fr and the Q of the crystaldepends on the crystal depend upon its dimension,

orientation of its surface and mechanical mounting. The resonant frequency of quartz crystal is

extremely stable with respect to temperaturechange and aging.


47/52

Oscillators 47

Fig shown below is symbol , circuit model and reactance asA function of frequency.

Cm(mounting capacitance) = 3.5 pF;

Cs= 0.0235 pF; L= 137 H; R = 15 k

S i d P ll l


48/52

Oscillators 48

Series and ParallelResonance

First, resonance occurs at fs for the seriescombination ofL and Cs.

Above fsthe series branch LCsRhasinductive reactance.

It then resonates at fp, with Cm.

For this parallel resonance, equivalentseries capacitance is Cp.


49/52

Oscillators 49


50/52

Oscillators 50

Normally, Csis much smaller than Cm.

Therefore, Cp is slightly less than Cs.

Hence, the frequency fp is slightly greaterthan fs.

The crystal is inductive only between the

frequencies fs and fp.

The frequency of oscillation must liebetween these frequencies.

Hence the stability.


51/52

Oscillators 51


52/52

Oscillators 52The fo is between 411 kHz and 412 kHz.

(24) oscillators

Documents