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Oscillators 1
Oscillators
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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.
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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.
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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.
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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 !
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Oscillators 6
How is it Possible ?
Connecting point x to y, feedback voltagedrives the amplifier.
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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.
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Oscillators 8
(1) IfA < 1, we get decaying ofdampedoscillations.
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Oscillators 9
(2) IfA > 1, we get growing oscillations.
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Oscillators 10
(3) IfA = 1, we get sustained oscillations.In this case, the circuit supplies its own input
signal.
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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.
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Oscillators 12
RC Oscillators
Two types :
1. RC Phase shift Oscillator.
2. Wein Bridge Oscillator.
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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
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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.
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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
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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.
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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.
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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.
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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)
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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
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Oscillators 21
The actual circuit using OPAMP isshown below
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Oscillators 22
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Oscillators 23
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Oscillators 24
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Oscillators 25
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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)
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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.
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Oscillators 28
Solution :
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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.
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Oscillators 30
Equivalent circuit:
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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)
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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
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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.
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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.
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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.
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Oscillators 36
COLPITS OSCILLATOROPAMP based colpits oscillator
F f ti
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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
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A colpits oscillator is designed with C =100pF
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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
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Oscillators 40
HARTLEY Oscillator
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Oscillators 41
Using (X1+X2+X3)= 0 we have
jL1 +jL2 +1/jC = 0
(L1 +L2) =1/C
2=1/[C.(L1 +L2)]
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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
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Oscillators 43
Solution :
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Oscillators 44
Solution :
Note: here C1 andC2 is different from ourassumption
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Oscillators 45
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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.
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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
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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.
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Oscillators 49
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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.
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Oscillators 51
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Oscillators 52The fo is between 411 kHz and 412 kHz.