these slides contain basic principal of oscillators, and a brief discussion of various type of oscillators

1. UNIT-3
Mohammad Asif Iqbal
Assistant Professor,
Deptt of ECE,
JETGI, Barabanki

2. OSCILLATORS

3. Will you agree if I say, you are an oscillator??
Surprisingly you are!!!
But you can’t generate a sinusoids
Don’t worry! We have something for that…

4. There are two different method of generating
signal
1. The first approach, employs a positive-feedback loop consisting
of an amplifier and an RC or LC frequency-selective network
2. In this approach we generate sine waves utilizing resonance
phenomena and the resulting oscillators are known as linear
oscillators.

5. Basic Principles of Sinusoidal Oscillators
Notethe+vefeedback.
Allthedifferenceis
createdbythis
The gain of this circuit is given by..
where we note the
negative sign in the
denominator

6. The loop gain of this circuit is given by
L(s) ≡ A(s)β (s) ;
And thus the characteristic equation is
1 − L(s) = 0
The Oscillation Criterion known as Barkhausen
criterion
Consider the case when
loop gain A(s)β (s) is
equal to 1 for an
specific frequency f0.
Af = ∞That will leads to
this

7. Af = ∞
This should be
equal to 0
And we know that
input
output

f
A = ∞
And this may be
any finite value
Here we observe that at this particular frequency f0 we
are getting output, without any input. By definition
these circuits are known as oscillator.

8. •Thus the condition for the feedback loop of Fig. to provide sinusoidal
oscillations of frequency ω0 is
L(j ω0) ≡ A (jω0) β (jω0) = 1
•That is, at ω0 the phase of the loop gain should be zero and the magnitude
of the loop gain should be unity. This is known as the Barkhausen
criterion.
•Note that for the circuit to oscillate at one frequency, the oscillation
criterion should be satisfied only at one frequency (i.e.,ω0); otherwise the
resulting waveform will not be a simple sinusoid

9. The Wien-Bridge Oscillator
Amplifier
Frequency selective network
In order to work like
oscillator, this should
follow theBarkhausen
criterion. That is, at ω0
the phase of the loop
gain should be zero and
the magnitude of the
loop gain should be
unity.
For that the loop gain can be easily obtained by multiplying the transfer function
Va(s) ⁄ Vo(s) of the feedback network by the amplifier gain,

10. Now, to follow the first
criterion The loop gain
will be a real number
(i.e., the phase will be
zero) at one frequency,
that will be given by:-
That is, ω0 = 1 ⁄CR
L(s) = A(s)β(s)
Andaccordingtosecond
criterionweshouldsetthe
magnitudeoftheloop
gaintounity.Thiscanbe
achievedbyselecting
R2 ⁄R1 = 2

11. The Phase-Shift Oscillator
Feedback Network. Amplifier
• The circuit will oscillate at the frequency for which the phase shift of the RC network is 180°,
• As the amplifier will introduce an additional phase shift of ±180, so at this frequency the total
phase shift around the loop be 0° or 360°. And that is the first required criterion for the
oscillation.
• For oscillations to be sustained, the value of K should be equal to the inverse of the magnitude of
the RC network transfer function at the frequency of oscillation. That will ensure over loop gain
to be equal to 1 and complete the second criterion of the oscillation.

12. FET-Based Phase-Shift Oscillator
+
+
𝑉𝑓
𝑔 𝑚 𝑉𝐼
Ŕ 𝐷
+
𝑉𝑖
+
𝑉𝑜

13. Ŕ 𝐷
𝑔 𝑚Ŕ 𝐷 𝑉𝑖
+
𝑉𝑖
+
𝑉𝑜
Neglect
R⫺Ŕ 𝐷
𝐼2 𝐼3

14. The mesh equations for the network can be written as
N.B. in writing these equations we
have make the following assumptions:-
𝑍1 =
1
𝑆𝐶
, 𝑍2= R,−𝑔 𝑚Ŕ 𝐷 𝑉𝑖= 𝑉1
𝑉𝑓

16. Putting all these values in the above equation
𝑍1 =
1
𝑆𝐶
, 𝑍2= R,−𝑔 𝑚Ŕ 𝐷 𝑉𝑖= 𝑉1
𝑉𝑓
𝑉𝑓
𝑉1

17. 𝑉𝑓
−𝑔 𝑚Ŕ 𝐷 𝑉𝑖
=
𝑅3
1
𝑆𝐶 3+5
𝑅
𝑆𝐶 2+6
𝑅2
𝑆𝐶
+𝑅3
𝑉𝑓
−𝑔 𝑚Ŕ 𝐷 𝑉𝑖
=
1
1
𝑆𝑅𝐶 3+5
1
𝑆𝑅𝐶 2+6
1
𝑆𝑅𝐶
+1
Put S = jω and assuming that
1
ω 𝑅𝐶
=α, we will get
𝑽 𝒇
𝑽 𝒊
=
−𝒈 𝒎Ŕ 𝑫
(𝟏−𝟓α 𝟐
)−𝒋(α 𝟑
−𝟔α )
AB=

18. We know that loop gain must be real, in order to make phase equal to
zero;
α 𝟑
− 𝟔α = 0
α 𝟐= 𝟔
ω 𝟐 𝑅 𝟐 𝐶 𝟐 = 𝟔
The frequency of oscillaion become
𝑓0=
1
2𝜋𝑅𝐶 6
Putting these value of α in gain equation we get
AB =
𝒈 𝒎Ŕ 𝑫
𝟐𝟗
And we know that for sustaoned oscillation|AB | ≥ 1
𝒈 𝒎Ŕ 𝑫 ≥ 𝟐𝟗

19. • Figure shows two commonly used configurations of LC-tuned oscillators.
• They are known as the Colpitts oscillator (a)and the Hartley oscillator (b).
• Both utilize a parallel LC circuit connected between collector and base with a fraction of the tuned-circuit voltage
fed to the emitter.
• This feedback is achieved by way of a capacitive divider in the Colpitts oscillator and by way of an inductive
divider in the Hartley circuit.
• To focus attention on the oscillator’s structure, the bias details are not shown. In both circuits, the resistor R
models the combination of the losses of the inductors, the load resistance of the oscillator, and the output
resistance of the transistor.
LC-Tuned Oscillators

20. Colpitts oscillator, Frequency of operation
Equivalent circuit of the
Colpitts oscillator of Fig.
(a). To simplify the
analysis, Cμ and rπ are
neglected. We can
consider Cπ to be part of
C2, and we can include ro
in R.
Now lets find out the potential at node C.
𝑉𝑐−𝑉𝜋
𝐿
=
𝑉π−0
𝐶2

21. 𝑉𝑐−𝑉𝜋
𝐿
=
𝑉π−0
𝐶2
Putting the values of L & C and simplifying, we will get
Have another look at the previous circuit
s𝐶2Vπ + 𝑔 𝑚Vπ +
1
𝑅
+ 𝑆𝐶2 (1 + 𝑠2
L𝐶2) Vπ = 0
Applying
KCL at
node C

22. Since Vπ will have no existence once oscillations starts, we can eliminate it. So
eliminating Vπ and rearranging the equation, we get
which is the resonance frequency of the colpitts oscillator. Equating the real
part to zero together with this value of ω, we get.
which has a simple physical interpretation: For
sustained oscillations, the magnitude of the
gain from base to collector (gmR) must be
equal to the inverse of the voltage ratio
provided by the capacitive divider,

23. Some important points
• Remember in colpitts we will have more c.
• It’s a variable, radio frequency oscillator
• The principal is parallel resonance
• It is also known as tapped capacitance oscillator
• The condition for sustained oscillation may be
given as
Advantage
Circuit is economical and small in size due to the
requirement of one inductor
Disadvantage
Inductive tuning offers very high wear and tear
problem

24. Alternate approach for finding the frequency of
oscillation.(finally we got it)
The frequency of oscillation in the circuit
can be obtain by equating 𝑋1+ 𝑋2+ 𝑋3 = 0
1
JωC1
+
1
JωC2
+ JωL= 0
JωL=
J
ω
1
C1
+
1
C2
𝜔2=
1
L
1
C1
+
1
C2
𝜔 =
1
𝐿
C1C2
C1+C2
𝑓0=
1
2𝜋 𝐿
C1C2
C1+C2
=
1
2𝜋 𝐿𝐶 𝑒𝑞

25. Hartley oscillator
• It’s a variable, radio frequency oscillator
• The principal is parallel resonance
• It is also known as tapped inductive oscillator
• As shown in the figure the total inductance is
split in two parts & are connected in series
across a variable capacitor C.
• The condition for sustained oscillation may be
given as
• Feedback is provided through inductor 𝐿1
Advantage
Capacitive tuning, hence offers very low wear and
tear problem .
Disadvantage
Circuit is bulky due to the presence of two inductors

26. Frequency of oscillation.
Again we can use the same technique for
finding the frequency of oscillation by
equating 𝑋1+ 𝑋2+ 𝑋3 = 0
1
JωC
+ Jω𝐿1 + Jω𝐿2 = 0
Jω(𝐿1+ 𝐿1) =
J
ωC
𝜔2=
1
C(𝐿1+ 𝐿1)
𝜔 =
1
C(𝐿1+ 𝐿1)
𝑓0=
1
2𝜋 C(𝐿1+ 𝐿1)

27. Which one is better ??
The total capacitor of the tank circuit
is split into two part, 𝐶1 & 𝐶2, and
connected in series, so that the net
capacitance of tank circuit is
reduced..
And thereby quality factor
Q =
1
𝑅
𝐿
𝐶
, of the tank circuit
increases, hence colpitts is having
better frequency stability when
compared to hartley oscillator

28. Crystal Oscillators
• It’s a fixed frequency radio oscillator
• Operates on the principal of piezoelectric effect
(Piezoelectric Effect is the ability of certain
materials to generate an electric charge in
response to applied mechanical stress)
• The frequency of oscillation generated by the
crystal depends on
• the physical size of the crystal.
• Edge cutting of the crystal
• The mounting position of the crystal.
• The frequency of oscillation produced by the
crystal is independent of time.
• It has a large Quality factor.
• Crystal is simultaneously subjected to series and
parallel resonance.

29. Now some formulas..
From the figure of crystal oscillator it is very much obvious that it
will have two resonance frequency
• Series resonance
• Parallel resonance

30. Series resonance
When these
both are in
resonance
𝑋 𝐿 = 𝑋 𝐶 𝑠
𝜔𝑠L=
1
𝜔 𝑠 𝐶 𝑠
ω 𝑠
2
=
1
L 𝐶 𝑠
ω 𝑠 = 1
𝐿 𝐶 𝑠
Parallel resonance
When L is in
resonance
with 𝐶𝑠 & 𝐶 𝑝
𝑋 𝐿 =𝑋 𝐶 𝑠
𝑋 𝐶 𝑝
𝜔 𝑝L=
𝐶 𝑠+𝐶 𝑝
𝜔 𝑝 𝐶 𝑝 𝐶𝑠
ω 𝑝 =
1
𝐿
𝐶 𝑝 𝐶 𝑠
𝐶 𝑠+𝐶 𝑝

31. By using them, and considering s=jω we can rewrite the impedance
Advantage
• Excellent frequency stability
• Performance is independent of ageing
Disadvantage
• Highly sensitive to temperature, (always enclosed in a constant
temperature chamber)
• It’s a fixed frequency oscillator

32. Important numerical
Q:1-In a Colpitt's oscillator, if the desired frequency is 500 kHz estimate the value of L and C.
Q:2
Q:3

33. THANK YOU!