• In f/b amplifier, if f/b signal is applied in such a way that it is in phase with the i/p signal, then it is called positive feedback.

This positive feedback results in a feedback(A$_{f}$) to be greater than open loop gain(A). It results into instability and operates as an oscillator circuit.

Comparison between oscillator and amplifier

Consider following block diagrams:

• The amplifier is essentially an energy convention device. i.e. a device which gets energy from the d.c source and converts it into an a.c energy at the same frequency as that of input signal.

• This d.c to a.c conversion is controlled by input signal. It means that, if there is no input signal then no energy conversion takes place. Thus no o/p signal.

• On the other hand, oscillator is a device, which produces an o/p signal, without any input signal of any desired frequency.

It keeps producing an output signal, so long as the d.c power is supplied.

Q) Explain block diagram of sinusoidal oscillator.

The block diagram of oscillator is shown in following figure.

• An oscillator consists of an amplifier with gain A & feedback network with gain $\beta$.

• Feedback network is basically phase shifting network.

• Amplifier receives the output of phase shifting network. The amplifier then amplifies it and add 180$^{\circ}$ phase shift.

• This phase shifted o/p of amplifier is applied to the input of feedback(phase shifting network). Feedback networks shift the amplifier o/p through 180$^{\circ}$.

• Thus, due to total 360$^{\circ}$ phase shift, the feedback becomes a positive feedback. This gives rise to the oscillations, if the Barkhausen criterion is satisfied.

Q) State and explain Barkhausen criterion of oscillation.

• An amplifier with positive feedback will work as an oscillator if and only if it satisfies a condition, known as "Barkhausen Criterion".

• The voltage gain of a positive feedback amplifier is given by,

A$_{f}$ = $\frac{A}{1 - A\beta}$

where,

A = Gain of amplifier without feedback(open loop gain)

$\beta$ = Gain of feedback network

• In above above eqn ,term A$\beta$ i.e product of open loop gain and gain of feedback network is known as 'Loop gain'.

• Barkhausen criterion states that,

a) At oscillator frequency, the magnitude of the product of open loop gain of amplifier (A) and feedback factor($\beta$) is equal to greater than unity.

i.e $\mid A\beta \mid\;\;\ge1$

• If the product $A\beta$ is equal to unity, then denominator of above eqn will become zero.

i.e A$_{f}$ increases to infinity and circuit will work as oscillator.

b) Amplifier circuit causes the phase shift of 180$^{\circ}$ between input and output signal. In order to provide positive feedback, feedback network must provide 180$^{\circ}$ phase shift. So the total phase shift of 360$^{\circ}$ or 0$^{\circ}$ is obtained around amplifier output and feedback circuit.

• Above two conditions are called as 'Barkhausen criterion' for oscillator.

Q) Give the classification of oscillator.

Following chart shows classification of oscillator.

• The oscillators having sine waveform at the o/p called as sinusoidal oscillator.

• The oscillator having a square, rectangular or sawtooth waveform are called as non-sinusoidal oscillators.

RC oscillators

• These oscillators use R and C as circuit elements in their feedback network.

• Frequency of oscillation depends on the values of R and C.

These oscillations are useful only for low output frequencies.

• RC oscillators are further classified as:

a) RC phase shift oscillator

b) Wien-bridge oscillator

a) R-C phase shift oscillator

RC phase shift oscillator consists of an amplifier and a phase shifting network(f/b network) made of resistors and capacitors.

• Consider the R-C network as shown in fig below:

• When sinusoidal voltage is applied to the circuit of above fig, there is phase shift of 60$^{\circ}$ between i/p V$_{in}$ and output V$_{out}$.

• If a ladder network consisting of 3 RC sections is built as shown in fig below then, total phase shift is given by 180$^{\circ}$.

• Following fig shows circuit diagram for RC phase shift oscillator.

• In this oscillator, transistor works as common emitter(C$_{e}$) amplifier. C$_{e}$ amplifier provides 180$^{\circ}$ phase shift.

• Three RC sections are used which are called ladder network.

• Total phase shift provided by three RC network is 180$^{\circ}$. Thus the total phase shift in the circuit is 360$^{\circ}$ or 0$^{\circ}$.

• The gain of the amplifier must be sufficient so that,

$\mid A\beta \mid\;\;\ge1$

• Then, Barkhausen criterion is satisfied and sustained(continuous) oscillations are produced.
• The frequency of oscillation is given by,

f$_{0}$ = $\frac{1}{2\pi RC\sqrt{6}}$

Advantages:

• The phase shift oscillator is useful over a wide frequency range, from a few Hz to several hundred Hz.

• Large size transformer or inductance are not required.

Disadvantages:

(i) The gain of transistor must be high to overcome losses in the RC network.

(ii) The frequency of oscillation cannot be varied easily. Since it is difficult to vary the capacitors simultaneously.

Q) For RC phase shift oscillator, the component values are as follows:

R = 8.2 k$\Omega$

C = 0.01 $\mu$F

R$_{1}$ = 1.2 k$\Omega$

R$_{f}$ = 39 k$\Omega$

(i) Determine whether we can get sustained oscillations

(ii) What will be the frequency of oscillations

Given that,

R = 8.2 k$\Omega$ = 8.2 $\times 10^3 \Omega$

C = 0.01 $\mu$F = 0.01 $\times 10^-6$F

R$_{1}$ = 1.2 k$\Omega$ = 1.2 $\times 10^3 \Omega$

R$_{f}$ = 39 k$\Omega$ = 39 $\times 10^3 \Omega$

Solutions:

Frequency of oscillation for phase shift oscillator is,

F = $\frac{1}{2\pi RC \sqrt{6}}$

= $\frac{1}{2\pi \times 8.2 \times 10^3 \times 0.01 \times 10^-6 \times \sqrt{6}}$

= $\frac{1}{1.26 \times 10^-3}$

= 0.7937 $\times 10^3$

F= 793.7 Hz.

Q) A phase shift oscillator has R = 220 K$\Omega$ and C = 500pF. Calculate frequency of sine wave generated by oscillator.

Given that,

R = 220k$\Omega$ = 220 $\times 10^3 \Omega$

C = 500pF = 500 $\times 10^-12$F

Solution:

Frequency of oscillation for phase shift oscillator is,

F = $\frac{1}{2\pi RC\sqrt{6}}$

= $\frac{1}{2 \times \pi \times 220 \times 10^3 \times 500 \times 10^-12 \times \sqrt{6}}$

=0.590 $\times 10^3$Hz

F = 0.590kHz.

Q) The phase shift oscillators uses equal resistance is 1M$\Omega$ and equal capacitance 68pF at what frequency does the circuit oscillates? and also find value of resistance to produce the frequency 800kHz if phase shift oscillator uses 5pF capacitor.

Given that,

(i) R = 1M$\Omega$ = 1 $\times 10^6$

C = 68pF = 68 $\times 10^-12$

Solution:

Frequency of oscillator for phase shift oscillator is

F = $\frac{1}{2\pi RC\sqrt{6}}$

= $\frac{1}{2 \times \pi \times 1 \times 10^6 \times 68 \times 10^-12 \times \sqrt{6}}$

= 955.510 Hz

= 0.955 $\times 10^3$

F = 0.955kHz.

(ii) Given:

R = 800kHz = 800 $\times 10^3$Hz

C = 5pF = 5 $\times 10^-12$F

Solution:

Frequency of oscillator for phase shift oscillator is

F = $\frac{1}{2\pi RC\sqrt{6}}$

= $\frac{1}{2 \times \pi \times 800 \times 10^3 \times 5 \times 10^-12 \times \sqrt{6}}$

= 16.243 $\times 10^3$ Hz

F = 16.243 kHz.

Q) The AC equivalent circuit of crystal has these value:

L = 1H

C = 0.01pF

R = 1000$\Omega$

C$_{m}$ = 20pF

Determine series and parallel resonant frequency

Given,

L = 1H

C = 0.01pF = 0.01 $\times 10^-12$F

R = 1000$\Omega$

C$_{m}$ = 20pF = 20 $\times 10^-12$F

Solution:

(i) Resonant frequency of series RLC circuit is

F$_{s}$ = $\frac{1}{2\pi \sqrt{LC}}$

= $\frac{1}{2 \times \pi \times \sqrt{1 \times 0.01 \times 10^-12}}$

= 1.591 $\times 10^6$Hz

= 1.591 MHz.

(ii) Resonant frequency of parallel RLC circuit is

F$_{p}$ = $\frac{1}{2\pi \sqrt{LC_{eq}}}$

where,

C$_{eq}$ = $\frac{C. C_{m}}{C + C_{m}}$

= $\frac{(0.01 \times 10^-12) \times (20 \times 10^-12)}{(0.01 \times 10^-12) + (20 \times 10^-12)}$

C$_{eq}$ = 9.995 $\times 10^-15$

Now,

F$_{p}$ = $\frac{1}{2 \times \pi \times \sqrt{1 \times 9.995 \times 10^-15}}$

= 1.591 $\times 10^6$Hz

F$_{p}$ = 1.591 MHz.

Crystal oscillator:

1. Crystal oscillator uses crystal material(quartz) in feedback network.
2. Crystal oscillator provide high frequency output over a windband.
3. Crystal shows a property of piezo-electric effect which is used for generation of oscillator.

Symbol of crystal:

Piezo-electric effect:

1. If electric energy is applied to crystal then it gives mechanical vibration(oscillation) at its output and vice versa.

2. This effect is known as Piezo electric effect.

3. This effect is used to produce oscillation in crystal oscillator.

Equivalent circuit of a crystal:

The equivalent circuit of a crystal consists of two resonant circuits:

1. series RLC

2. parallel RLC

Therefore,

Resonant frequency of series RLC circuit is given by,

F$_{s}$ = $\frac{1}{2\pi \sqrt{LC}}$

The resonant frequency of parallel RLC circuit is,

F$_{p}$ = $\frac{1}{2\pi \sqrt{L.C_{eq}}}$

where,

$C_{eq}$ = $\frac{CC'}{C t C'}$

There are two types of crystal oscillator:

1. Pierce crystal oscillator

2. Miller crystal oscillator

Crystal impedance:

1. To obtain the expression for crystal impedance let us use the approximate ac equivalent circuit of figure.

The resistance 'R' in the original equivalent circuit has been neglected.

1. Let X be the reactance offered by the crystal. Then X is dependent on frequency as shown in figure.

(I) If $\omega \lt \omega_{s}$ = the reactance of the crystal i.e 'X' will be negative i.e capacitive.

(II) If $\omega\;=\;\omega_{s}$ = Reactance X = 0

(III) If $\omega_{s} \lt \omega \lt \omega_{p}$ = X will be positive i.e inductive.

(IV) If $\omega \gt \omega_{p}$ = X will be negative i.e capacitive.

(V) If $\omega\;=\;\omega_{p}$ = At $\omega\;=\;\omega_{p}$, X = $\infty$

The variation of jx with $\omega$ as shown in figure.

1) Pierce crystal oscillator:

• If you compare figure with the colpitt's oscillator circuit then you will find that both these circuits are identical except for one change. L$_{3}$ is replaced by the crystal here.

• The crystal is made to operate as an inductance. This is possible of the frequency of oscillation $\omega$ is adjusted between $\omega_{s}$ and $\omega_{p}$ as shown in figure.

• The basic operation of the pierce crystal oscillator is same as that of the colpitt's oscillator.

• RFC is a radio frequency choke which connects the DC supply to the circuit but isolates the DC supply from the high frequency oscillations generated in the tank circuit(feedback network).

2) Miller Crystal Oscillator:

(A)

(B) (1) If the Hartley oscillator circuit is modified by replacing one of the inductors by a crystal then we get the miller crystal oscillator configuration shown in figure.

(2) The crystal acts like an inductor as the frequency of oscillations $\omega$ is adjusted between $\omega_{s}$ and $\omega_{p}$ as shown in figure.

(C) Advantages of Crystal oscillators

1. very high frequency stability

2. very low frequency drift due to change in temperature and other parameters.

3. It is possible to obtain very high, precise and stable frequency of oscillators.

4. The Q is very high.

5. It is possible to obtain frequencies higher than the fundamental frequency by operating the crystal in the overtone mode.

(D) Disadvantages of crystal oscillators:

1. These are suitable for high frequency applications.

2. Crystals of low frequency are not easily available.

(E) Application of crystal oscillators:

1. As a crystal clock in microprocessors

2. In the frequency synthesizers.

3. In the radio and TV transmitters.

4. In special types of receivers.

General application of Oscillators:

1. As a local oscillation in radio receivers.

2. In T.V receivers

3. In signal generators

4. As clock generation for logic circuits

5. Am and FM transmitters

6. In phase lock loops

Comparison of RC, LC and crystal oscillators