OSCILLATOR FUNDAMENTALS

Frequently, electronic circuits require AC signals that can range from a few hertz to
many millions of hertz. Oscillators are usually used to generate these frequencies.
Furnishing this wide range of frequencies requires many different oscillators, and
there are literally hundreds. However, all oscillators operate on the same basic
principles and, if you thoroughly understand these basic principles, you should be able
to analyze the operation of most common oscillators.

What is an Oscillator?

An oscillator is a circuit that generates a repetitive AC signal. As mentioned

previously, the frequency of this AC signal may be a few hertz, a thousand hertz, a
million hertz, or even higher, in the giga-hertz range. The 60 hertz AC signal available
at the normal wall outlet is produced by an AC generator or alternator at the power
station and is then transmitted through the power lines to your home or office. Since
AC can be transmit-ted easily with subsequent low power losses, it is quite suitable for
use in power systems.

The 60 hertz AC signal from the wall outlet is a convenient source of a relatively
constant 60 hertz signal. However, except for supplying operating power (its primary
purpose), this 60 hertz sine wave has few applications in electronic circuits. So, other
means are required to supply AC signals of different frequencies.

Why not use an AC generator? Generators are relatively large and expensive. Also,
they would be satisfactory only for low frequency applications because generator
output frequency depends on the number of poles in the generator field and the speed
of rotation (Figure 3-1). Generator frequency is limited since, at high speeds, the
generator will fly apart. Therefore, the AC generator is not a feasible solution.

Figure 3-1 Generator frequency depends on poles and speed.

Dr. Ahmed El-Shazly |Page1

The electronic generator or "oscillator" is an alternative to the mechanical generator. It
has no moving parts and is capable of producing AC signals ranging from a few hertz
to many millions of hertz. Such an oscillator is shown in Figure 3-2 it operates from
the DC power supply and generates an AC signal. Oscillator output can be a sine
wave, rectangular wave, or a saw tooth, depending on the type of oscillator. The major
requirement of a good oscillator is that the output is uniform, not varying in frequency
or amplitude. In this unit the major concern is with the sinusoidal or sine wave
oscillator.

Figure 3-2:
The electronic oscillator
The Basic Oscillator

When a tank circuit is "excited" by a DC source, it has a tendency to oscillate.

Circulating current flows inside the "tank," producing a back and forth oscillatory
motion. However, resistance of the tank circuit dissipates energy and oscillations are
damped as shown in Figure 3-3.

Figure 3-3: Damped oscillation waveform

For the tank circuit to continue oscillation, lost energy must be replaced. A crude
method of replacing energy is to close the power switch once each cycle, as shown in

Dr. Ahmed El-Shazly |Page2

Figure 3-4. It is important for the switch to close so the energy is replaced at exactly
the instant that reinforces the charge on capacitor C. Replacement energy then has a
reinforcing effect and is in phase with the tank waveform. Reinforcing energy that
"adds to" circuit action, is a type of "positive feedback." Positive feedback is the most
important requisite for oscillation.

Figure 3-4
Replacing tank energy

There is nothing unique about oscillation or positive feedback, especially in high

frequency circuits. Distributed capacitance of these circuits "feeds back" signals that
can cause high frequency circuits to break into oscillation. Of course this is
undesirable, so circuit designers "neutralize" such circuits.

One form of positive feedback is common in public address systems. If the

microphone is placed too close to the loudspeaker, positive feedback results, as shown
in Figure 3-5. Speaker output is fed back to the
microphone, amplified, and again applied to the speaker. This sets up a continual cycle
and the loudspeaker emits a high-pitched acoustical squeal.

Figure 3-5: A form of positive feedback.

Dr. Ahmed El-Shazly |Page3

From this example, it may seem that an oscillator can be created by simply taking a
portion of amplifier output and feeding it back to the input. In some cases this is true,
but not in all. For the circuit to oscillate, the feedback signal must be in phase with the
input signal. Figure 3-6 illustrates this point.

Figure 3-6: Feedback in the common-emitter amplifier

This is a common-emitter amplifier, where a portion of the output signal is fed back to
the input. Remember, in the common-emitter arrangement, output is 180° out of phase
with input. So in this circuit, the feedback signal opposes the input signal, resulting in
reduced input. Such feed-back is degenerative, or negative in nature, and is used in
amplifier circuits to reduce distortion. This circuit will not oscillate.

However, if you place a 180° phase-shift network between the output and input,
feedback will be of the correct phase. Figure 3-7 illustrates this principle. Notice that
the feedback signal is in phase with the input signal, adding to the input. The result is
positive or regenerative feed-back.

Figure 3-7

Common-emitter amplifier with feedback shifted 180

Once the amplifier begins to operate, the input signal can be removed and the circuit
will continue to oscillate. The gain of the amplifier replaces energy lost in the circuit
and positive feedback sustains oscillation. This circuit meets all of the requirements of
an oscillator, except one the frequency of oscillation. This circuit could oscillate on

Dr. Ahmed El-Shazly |Page4

any number of frequencies. Even minor noise pulses could change oscillator
frequency. An oscillator should have a constant output, so a means of setting
frequency is necessary.

Here, the frequency selectivity of the parallel LC network is useful be-cause the
network resonates at a specific frequency, determined by the values of inductance and
capacitance. Also, because the inductor and capacitor are reactive, they can produce
the required 180° phase shift. Therefore, a tank circuit in the positive feedback loop, as
shown in Figure 3-8A, controls the frequency. The tank resonates at its natural
frequency and amplifier gain replaces energy lost in the tank.

Figure 3-8
LC and RC networks set frequency
A similar result is obtained if the regenerative feedback loop contains enough RC
(resistance-capacitance) networks to produce the desired 180° phase shift (Figure 3-
8B). Here, RC time constants determine oscillator frequency and the amplifier
replaces energy lost across the RC networks.

Up to this point, the oscillators were amplifiers with input signals applied to start the
circuit. In actual practice, oscillators must start on their own. This is a natural
phenomenon. When a circuit is first turned on, energy levels do not instantly reach
maximum, but gradually approach it. This produces many noise pulses that can be
phase shifted and fed back to the input, as shown in Figure 3-9. The amplifier steps up
these pulses, which are again supplied to the input. This action continues and
oscillation is underway. Therefore, the oscillator is naturally "self-excited," meaning it
starts on its own.

Dr. Ahmed El-Shazly |Page5

 Figure 3-9

Oscillators are self-starting

Oscillators are electronic circuits that are used to generate electrical waveforms of any
desired frequency, shape, and power. Consider the amplifier with a positive feedback
connection shown in Figure 3-10. From the Figure, we can write the following
relations:

V0 = AVi (3.1)

Vf = BVo (3.2)

Vi = Vs + BV0 (3.3)

Where vs is the external input voltage, Vi the voltage input to the amplifier, Vo the
output voltage, Vf the feedback voltage, and B the feedback factor.

Figure 3-10

Amplifier with positive feedback.

Since the oscillator must produce its own input signal and this condition must occur
continuously, the product of amplifier gain (A) and feedback factor (B) must equal 1,
or a condition of unity. Plug this value into the stage gain formula.

Dr. Ahmed El-Shazly |Page6

When the product of amplifier gain and feedback factor is equal to 1, the denominator
of the gain equation is 0, resulting in infinite stage gain. This may seem impractical,
but it is very practical, even desirable for oscillators. Infinite stage gain implies that a
signal is present at the output, without an input. This is one of the conditions required
for an oscillator. The qualification that AB=1,⎳0 for oscillation actually represents
the following two distinct conditions:

1. Magnitude of total loop gain is unity.

2. Total phase shift around the loop is 0° or 360.

These conditions are called the Barkhausen criteria for oscillations (named after the
originator of these conditions). For oscillations to be generated.

With infinite stage gain you might think that output would continuously increase. This
happens when the oscillator initially starts, but after a few oscillations, the amplifier
saturates and brings the oscillator quickly under control.

When an amplifier is saturated, it is providing maximum gain. This has a damping

effect, returning the feedback product (AB) to one.

Oscillator designers strive for a feedback product (AB) of slightly greater than 1.
When AB is at this point, maximum stability and the cleanest output waveform results.
If the feedback product is too small, the circuit cannot sustain oscillation. If it is too
large, output waveform clipping results. Clipping is desirable in some oscillators but
not in sine wave oscillators.

In every practical oscillator the loop gain is slightly larger than unity [because circuit
components and more importantly, transistors change characteristic. (drift) with age,
temperature, voltage ….etc] say about 5%.

Dr. Ahmed El-Shazly |Page7

Briefly review the following oscillator requirements:

1. An amplifier is necessary to replace circuit losses.

2. Frequency determining components are necessary to set the oscillation
frequency.
3. Positive feedback supplies a regenerative signal to the input to sustain
oscillation.
4. The oscillator must be self-starting.

These fundamentals are common to all oscillators, which will be discussed later with
actual oscillator circuits.

The general principle of oscillator operation

Oscillators are feedback systems which provide their own input signal by positive
feedback. Referring to the general discussion on feedback (p.00) we know that the
gain of a feedback system is given by
⎳
⎳

In general, both (  + ) and | | change with a signal frequency. In a system which

can oscillate, there will be one or more frequencies at which  + = 00. If, at one of
these frequencies, | | = 1, the gain with feedback becomes

Thus, at one particular frequency, the system has infinite gain, such a system only
requires an infinitely small input to produce an output at that frequency. This behavior
can be justified in terms of an initial input resulting from electrical noise but this is not
necessary as analysis by complex frequency techniques shows that a system of this
type must generate a sine wave with an amplitude that will grow exponentially until it
is limited by system saturation. Returning to equation [3.4] we can see that two
conditions must be satisfied for oscillation to occur, the angle, condition, which must
be correct at the required frequency of oscillation, and the gain or maintenance
condition which must be equal to or greater than one at that frequency. A further
problem arises from the maintenance condition as if this is satisfied for all signal
amplitudes, the oscillation will grow until the amplitude is limited by saturation. This
can result in a distorted output waveform (particularly in RC oscillators) which may be
minimised by the use of negative feed circuitry.

Analysis of oscillator systems can be accomplished by a number of methods, the most

appropriate depending on the type of circuit involved and on the information required

Dr. Ahmed El-Shazly |Page8

from the analysis. For RC oscillators, the angle condition and the required gain are
usually obtained by consideration of the feedback network in isolation: a suitable
amplifier is then added to provide the necessary gain. With LC oscillators, it is usually
best to write a set of general equations for the complete circuit and then, on the.
Assumption that oscillation is occurring, the required frequency and maintenance
conditions are extracted. Examples of both methods of analysis are given in the
following sections.

Dr. Ahmed El-Shazly |Page9

 RC OSCILLATORS
Up to this point, LC and crystal oscillators that are commonly used in RF applications
have been discussed. However, in the low and audio frequency ranges, these
oscillators are usually not practical. For example, inductors for the low frequency
range, around 60 Hz, would be large and expensive. And, the practical low limit on
crystals is usually around 50 kHz. So an inexpensive approach to oscillator design in
these low frequency ranges is the RC oscillator.

The RC oscillator uses resistance-capacitance networks to determine oscillator

frequency. This makes the oscillator inexpensive, easy to construct and relatively
stable. There are basically two types of RC oscillators that produce sine wave outputs
the phase-shift oscillator and the Wien bridge oscillator. Many other RC oscillators,
such as the multi-vibrator and Schmitt trigger, produce non-sinusoidal outputs.

The Phase-Shift Oscillator

The phase-shift oscillator, as the name implies, is a conventional amplifier and a phase
shifting RC feedback network. It is typically used in fixed frequency applications. The
collector output signal must be shifted 180° to produce the required regenerative
feedback. The phase-shift oscillator accomplishes this with a series of RC networks
connected in the collector-to-base feedback loop.

Briefly review the basic principles of an RC phase-shift network. Remember that in a

purely capacitive circuit, current leads voltage by 90°. However, in an RC network the
phase difference between current and voltage falls between 0° and 90°, because
resistance now affects the phase relationship. Thus, the phase difference in an RC
circuit is a function of the capacitive-reactance (XC) and resistance of the network. By
carefully selecting the resistance and capacitive values, the amount of phase shift
across an RC network can be controlled.

Resistance does not vary with frequency. However, the capacitor is frequency
sensitive, since its reactance changes with frequency. Therefore, any change in
frequency, changes XC. As XC changes, the phase shift of the RC network also varies.
Hence, the capacitor is the frequency sensitive component in the RC network.

Note: Referring to Fig. 3-11, the amplifier will in general have zero or 180° phase
shift between input and output. If  +  is to be, 00, then feedback angle  will also
have to be 0° or 180° at the frequency of oscillation. In phase-shift oscillators, the
network has an angle of 180° at the required frequency allowing the system to be
completed with a simple inverting amplifier. Suitable phase-shift.

Dr. Ahmed El-Shazly | P a g e 11

 Figure 3-11

The arrangement feedback sinusoidal

The 180° phase-shift can be provided in two simple manners, namely,

• Phase-leading network

• Phase-lagging network.

The RC type phase-lead and lag networks are drawn in Figs. 3.12(a) and (b). From
Fig. 3.12(a), the relationship between output and input voltages is written as

(3.4)
√ ( )

Figure 3-12(a) phase-lead network Figure 3-12(b) phase-lag network

⎳ ⎳
(3-5)
√ √

Where tan 1 = (3-6)

Similarly, from Fig. 13.2(b) the relationship between output and input voltages is

( ) ( )
√

Dr. Ahmed El-Shazly | P a g e 11

 ⎳
=
√

⎳ ⎳
(3-7)
√

Where, tan 2 = CR = f/fH (3-8)

It is clear from Eqns. 3-6 and 3.8 that the maximum values of tan 1 and tan 2 will
occur at  = /2. Hence, one section of either types of RC network will provide only
90° phase shift. Thus, we see that one section of either RC phase-lag or phase-lead
network can provide maximum of 90° phase shift. Considering the loss component,
maximum phase shift of 90° can not be achieved by one section of simple RC network
shown in Figs. 3.12(a) and (b). Hence, in order to get additional 180`' phase shift at
least three sections of RC network are required.

15-2 The circuit connection of the phase-shift oscillator:

We select the phase-shift oscillator Fig. 3-13(a) as a first example because it

exemplifies very simply the principles set forth above. Here a discrete-component
JFET amplifier is followed by three cascaded arrangements of a capacitor C and a
resistor R, the output of the last RC combination being returned to the gate. If the
loading of the phase-shift network on the amplifier can be neglected that is, R >> R L,
the amplifier shifts by 180° the phase of any voltage which appears on the gate, and
the network of resistors and capacitors shifts the phase by an additional amount. At
some frequency the phase shift introduced by the RC network will be precisely 180°,
and at this frequency the total phase shift from the gate around the circuit and back to
the gate will be exactly zero. This particular frequency will be the one at

Figure 3-13 (a) A JFET RC phase-shift oscillator and (b)

its equivalent circuit. (c) An Op-Amp version of
The phase-shift oscillator

Dr. Ahmed El-Shazly | P a g e 12

Which the circuit will oscillate, provided that the magnitude of the amplification is
sufficiently large.

Now, let us cascade three RC network sections, as shown in Figure 3-13(a) or (b).
Since each section of these networks can produce a phase shift of approximately 90°,
by cascading three such sections, we can produce a phase shift of about 180°. Then by
using an amplifier, which can produce another 180° phase shift, we can convert the
cascaded sections into an oscillator called RC phase-shift oscillator (PSO).

Figure 3-14 (a) RC low-pass circuit (b) RC high-pass circuit

Figure 3-15(a) Three-section RC high-pass Circuit, (b) three-section RC low-pass

circuit

RC phase lead network

refer the circuit in Figure (3-15(a))

Figure 3-15(a)

Dr. Ahmed El-Shazly | P a g e 13

In this circuit, equal C and R values are used but the final R may include the input
impedance to the following amplifier. The input impedance to

the network will also be high compared with the amplifier output impedance. The
second network has. Staggered constants which may be used to minimize the required
amplifier gain.

The first circuit can be analyzed by mesh analysis

(3-9)

And Vo = I3R (3-10)

Solving equation [3.9] for I3 by determinants

(3-11)

Now remembering the required conditions for oscillation, the angle of the feedback
expression must be either 0° or 180° (so that A will be 0°). For equation [3-11], the
angle of the numerator is 90°, the angle of the expression can 'only be 0° or 180° if the
angle of the denominator is ±90°. For this to be satisfied, the real terms of equation
(3-11) must be zero, i.e

1 – 6 2C2R2 = 0

6 2C2R2 = 1

(3-12)
√

Dr. Ahmed El-Shazly | P a g e 14

This means that if Vi was applied with a frequency of then the voltage ratio
√
for the network is given by




But at this frequency, from equation (3-12) 2 C2 R2 = 1/6. Therefore

Summarizing the results of this analysis, for the network shown in Fig. 3-15a, the
phase shift will be 180° at a signal frequency of  = 1/CR√ , and at this frequency,
the network gain | | = 1/29.

From which gmRL = 29. Thus, to sustain oscillations, the gain of the JFET amplifier
stage must be at least 29.

If the loading effect of the RC phase-shift network is included (Prob. 15-3), the
frequency of oscillation decreases and the gain of the JFET amplifier stage must be
increased.

The FET in Fig. 3-13(a) may be replaced by an Op-Amp as indicated in Fig. 15-2c.
Because of the virtual ground, the resistance from the input node P to ground is R I = R
and, hence, the phase-shift network in Fig. 3-13c, identical with that in Fig. 3-13a.
Therefore, the oscillation frequency is given by Eq. (15-5). Since the Op-Amp gain Av
= - R1/R and | | must be at least 29, then R2/R must be greater than 29 (by about 5
percent).

It is possible to replace the Op-Amp in Fig 3-13c by a single transistor stage with R2 =
 and R1 = R .

It should be pointed out that it is not always necessary to make use of an amplifier
with transfer gain A > 1 to satisfy the Barkhausen criterion. It is necessary only that
AB > 1. Passive network structures exist for which the transfer function of the
feedback network is greater than unity at some particular frequency. As an oscillator
circuit consisting of a source follower and the RC circuit appropriately connected. As
shown in Example.

Dr. Ahmed El-Shazly | P a g e 15

Transistor Phase Shift Oscillator.
The circuit arrangement of a phase-shift oscillator using N-P-N transistor in CE
configuration is shown in Fig. 3.16. As usual, the voltage divider R1- R2 provides dc
emitter base bias, RE and CE combination provides temperature stability and prevent ac
signal degeneration and collector resistor Rc controls the collector voltage. The
oscillator output voltage is capacitively coupled to the load by Cc.

In case of a transistor phase shift oscillator, the output of the feedback network is
loaded appreciably by the relatively small input resistance (h ie) of the transistor.
Hence, instead of employing voltage-series feedback (as used in case of FET phase
shift oscillator), voltage-shunt feedback is used for a transistor phase shift oscillator,
as shown in Fig, 3-16. In this circuit, the feedback signal is coupled through the
feedback resistor R' in series with the amplifier stage input resistance hie. The value of
R' should be such that when added with amplifier stage input resistance h ie, it is equal
to R i.e., R' + hie = R.

Fig. 3.16
Operation.
The circuit is set into oscillations by any random or variation caused in the base
current, that may be either due to noise inherent in the transistor or minor variation in
voltage of dc power supply. This variation in base current is amplified in collector

Dr. Ahmed El-Shazly | P a g e 16

circuit. The output of the amplifier is supplied to an R-C feedback network. The R-C
network produces a phase shift of 180° between output and input voltages. Since CE
amplifier produces a phase reversal of the input signal, total phase shift becomes 360°
or 0° which is essential for regeneration or for sustained oscillations. The output of
this network is thus in the same phase as the originally assumed input to the amplifier
and is applied to the base terminal of the transistor. Thus sustained variation in
collector current between saturation and cutoff values are obtained. R-C phase shift
network is the frequency determining network, as already explained in case of FET
phase-shift oscillator.

Frequency of Oscillation.
The equivalent circuit for the analysis of a transistor phase shift oscillator (circuit
shown in Fig 3-16) is shown in Fig. 3-17. The equivalent circuit shown in Fig. 3-17 is
simplified if the following assumptions are made.

(i) r0 >> Rc
(ii) hie = r

Making above assumptions placing current source by equivalent voltage source, the
simplified equivalent circuit is shown in Fig

Fig 3-17 (b) replace the dependent current source by its thevenin's equivalent,
and write the equitation for the resulting circuit.

Fig 3-18

1) - hfe IbRc = I1 (Rc + R -jX) – I2 R + I3

Dr. Ahmed El-Shazly | P a g e 17

 2) 0 = - I1R + I2(2R – jx) – I3R
3) 0 = (I1(0) – I2R + I3(2R - jX); let  and k = , Then (3) becomes,
I2 = I3 (2 – jX) and (2) becomes, I1 – I3 (3-2 – j4). Substituting the
expressions of I1 and I2 in (1) and simplifying we get:
–hfeIb k = I3 (I + 3k – (5 + k) 2 – j[(6 + 4k)  - 3]
The loop current gain is I3/ Ib and if this is to be real, then the coefficient of 
2
must be more or  = 6 + 4k =
Thus, f = . At this frequency
√
(4k2 + 23k + 29): For > I then hfe > 4k + 23 +
c) hfe + 4k + 23 + 29/k thus
dhfe / dk = 4 – 29 / k2 = 0, or k = (29/4)1/2 = 2.7
Thus hfe(min) = (4)(2.7) + 23 + 29/207 = 44.5

Variable-Frequency Operation:
The phase-shift oscillator is particularly suited to the range of frequencies from several
hertz to several hundred kilohertz, and so includes the range of audio frequencies. The
frequency of oscillation may be varied by. Changing any of the impedance elements in
the phase-shifting network. For variations of frequency over a large range, the three
capacitors are usually varied simultaneously. Such a variation keeps the input
impedance to the phase-shifting network constant) and also keeps constant the
magnitude of AB. Hence the amplitude of oscillation will not be affected as the
frequency is adjusted. The phase-shift oscillator is operated in class A in order to keep
distortion to a minimum.

Two active phase shifters may be used in place of the passive feedback network to
obtain a sinusoidal oscillator with quadrature outputs, sine, and cosine waveforms.

RC phase shift oscillator using phase-lag network

consider the phase shift network in figure (3-19).

Fig (3-19)

Dr. Ahmed El-Shazly | P a g e 18

Let X = the mesh equations are :
C

V0 = I1 (R – jX) +I2jX + 0 (3-13)

0 = + I1jX + I2 (R – 2jX) + I3jX (3-14)

0 = 0 + I2jX + I3(R – 2jX) (3-15)

Divide by jX and let = - j where  = = RC

= - I1 (I + j) + I2 (3-16)

0 = I1 – I2 (2 + j) + I3 (3-17)

0 = I2 – I3 (2 + j) (3-18)

From (18) I2 = I3 (2 + j)

From (17) I1 = + (2 + j) (2+ j) I3 – I3 = (3 - 2 +4j) I3

From (16) = + (1 + j) (3 - 2 +4j) – (2 + j)

= + 3 -2 + 4 j + j3 - j3 - 42 – 2 - j

= 1 - 52 + j (6 - 3)

Since V/f = jXI3

6 - 3 = 0 or  = √ = RC
Since numerator is real than
√
F=

And A = 1 - 52 = 1 – 5(6) = -29

In practice A > 1. Hence A > 29.

Since A is negative, an inverting OP AMP must be used. The system is that indicated
in Fig. 3-20 with R and C interchanged. However since R2 shunts C to ground then

Dr. Ahmed El-Shazly | P a g e 19

 R2 >> 1/C. If this is not true then a voltage follower must be used as a buffer, as
indicated.

fig :(3-20)

A similar analysis, left for the reader, the circuit shown below in fig. (3-21).

fig (3-21)

The analysis of this network gives and at this frequency

 = 1/16 (3-19)

Fig. 3-22 shows an op-amp circuit using this network to make a complete phase-shift
oscillator.

Fig. 3-22 an op-amp RC phase shift oscillator

Equation [3.19] shows that the network provides the necessary phase condition at

Dr. Ahmed El-Shazly | P a g e 21

 𝟂= 4120 r/s or f = 658 Hz. The network will cause no. appreciable loading of the
amplifier as Zout will be very low. The amplifier input impedance is 22 k and this is
in parallel with the final capacitor. This capacitive reactance at the oscillatory
frequency is j2.4 k and the effect of the 22 k, will be small causing only a slight
shift in the oscillatory frequency. The amplifier gain is 360/22 or 16.4. This is slightly
larger than the required 16 and this will ensure the start of oscillation when the circuit
is . Switched on. The amplitude of oscillation will stabilize between the op-amp
saturation levels with some distortion of waveform at Vol. This being a low impedance
point in the circuit, any external load changes will have little effect. Vo2 however will
be less distorted as the network will have greater attenuation to the higher frequency
harmonics. Excessive loading at this point will either change the frequency or stop the
oscillation.-Phase-shift oscillators provide useful fixed frequency circuits but if tuning
of the frequency is required, the need to adjust three components simultaneously is
inconvenient.

Phase-Shift Oscillator with voltage follower

An example of an op-amp oscillator is the phase-shift oscillator. One configuration of

this oscillator circuit is shown in figure 3-23. The basic amplifier of the circuit is

Fig 3-23 phase-shift oscillator circuit with voltage-follower buffer stages

The op-amp A3, which is connected as an inverting amplifier with its output connected
to a three-stage RC filter. The voltage followers in the circuit eliminate loading effects
between each RC filter stage.

The inverting amplifier introduces a —180 degree phase shift, which means that each
RC network must provide 60 degrees of phase shift to produce the 180 degrees
required of the frequency-sensitive feedback network in order to produce positive
feedback. Note that the inverting terminal of op-amp A3 is at virtual ground; there-
fore, the RC network between op-amps A2 and A3 functions exactly as the other two
RC networks. We assume that the frequency effects of the op-amps themselves occur
at much higher frequencies than the response due to the RC networks. Also, to aid in

Dr. Ahmed El-Shazly | P a g e 21

the analysis, we assume an input signal (v1) exists at one node as shown in the figure.
The transfer function of the first RC network is

( ) (3-20)

Since the RC networks are assumed to be identical, and since there is no loading effect
of one RC stage on another, we have

( ) (3-21)

Where (s) is the feedback transfer function. The amplifier gain A(s) in Equation
(15.33) and (15.34) is actually the magnitude of the gain, or

| | (3-22)

The loop gain is then

T(s) = A(s)(s) = ( ) ( ) (3-23)

From Equation (15.35), the condition for oscillation is that T (j0) = 1 and the phase
of T (j0), must be 180 degrees. When these requirements are satisfied, then v o will
equal (v1) and a separate input signal will not be required.

If we set s = j, Equation (3-23) becomes

( )
( ) ( ) (3-24)
[ ]

To satisfy the condition T (jo) = -1, the imaginary component of Equation (3-24)
must equal zero. Since the numerator is purely imaginary, the denominator must
become purely imaginary, or

[1 - 320R2C2] = 0 Which yields (3-25)

√

where o, is the oscillation frequency. At this frequency, Equation (3-24) becomes

( )( )
√
( ) ( )( ) (3-26)
( )* ( )+
√

Consequently, the condition T (j0) = -1 is satisfied when =8 (3-27)

Dr. Ahmed El-Shazly | P a g e 22

Equation (3-27) implies that if the magnitude of the inverting greater than 8. The
circuit will spontaneously begin oscillating oscillation.

EXAMPLE

Objective: Determine the oscillation frequency and required amplifier gain for a
phase-shift oscillator.

Consider the phase-shift oscillator in Figure 3-23 with parameters C = 0.1 F and R =
1 k.

Solution: From Equation (3.25), the oscillation frequency is

√ √
The minimum amplifier gain magnitude is 8 from Equation 3-27 therefore, the
minimum value of R2 is 8 k.

Comment: Higher oscillation frequencies can easily be obtained by using smaller

capacitor values.

EXERCISE PROBLEM A

Ex: Design the phase-shift oscillator shown in Figure 3-23 to oscillate at f0 = 15 kHz.
Choose appropriate component values. (Ans. For example, C = 0.001 F, R = 6.13k,
R2 = 49 k)

Using Equation (15.36), we can determine the effect of each RC network in the phase-
shift oscillator. At the oscillation frequency co„, the transfer function of each RC
network stage is

√
3-28
√ √

Which can be written in terms of the magnitude and phase, as follows:

⎳
⎳ ⎳ 3-29
√ ⎳ √

Or 1/2 (⎳90 - ⎳30) = 1/2 ⎳60

As required, each RC network introduces a 60 degree phase shift, but they each also
introduce an attenuation factor of (1/2) for which the amplifier must compensate.

Dr. Ahmed El-Shazly | P a g e 23

The two voltage followers in the circuit in Figure 3-23need not be included i n a
practical phase-shift oscillator. Figure shows a phase-shift oscillator without the
voltage-follower buffer stages. The three RC network stages and the inverting
amplifier are still included. The loading effect of each successive RC network

The Wien-Bridge Oscillator

Like the phase-shift oscillator, the Wien bridge uses RC networks. How-ever, in the
Wien bridge oscillator, the RC networks are part of a bridge circuit that produces both
regenerative and degenerative feedback. The result is an excellent sine wave oscillator
that can be used to generate frequencies ranging from 5 Hz to 1 MHz. In the phase-
shift oscillator just discussed, the RC networks produce the desired 180 phase shift
for regenerative feedback. In (he Wien bridge oscillator, the RC networks select the
frequency at which feedback occurs, hut do not shift the phase of the feedback
voltage.

It is easy to understand the bridge oscillator if you understand the regenerative

feedback network.

Figure 3-24

A lead-lag network for the Wien bridge

The circuit shown in Figure 3-24 is referred to as a lead-lag network. It is a simple

band pass filter comprised of a series RC network, R1C1, and a parallel RC network,
R2C2. It is called a lead-lag network because the output phase angle leads for some
frequencies and lags for others. How ever, at the resonant frequency, the phase shift
exactly equals 0. This important characteristic allows the lead-lag network to
determine oscillation frequency in the bridge oscillator.

At low frequencies, series capacitor C1 appears as an open and there is no output. At

very high frequencies, the parallel capacitor, C2, shunts the output to ground and again

Dr. Ahmed El-Shazly | P a g e 24

there is no output. However, at the resonant frequency, the output voltage is
maximum. This is illustrated by the output-versus-frequency curve at the output of the
circuit. Out put is maximum at F0; therefore, the RC network is frequency selective.
On both sides of F0' output decreases significantly. Examine the output phase angle
carefully and notice that, at low frequencies, the phase angle is positive and the circuit
acts like a lead network. At high frequencies, the output phase angle is negative and
the circuit acts like a lag network. At the circuit's resonant frequency, the phase shifts
of the series and parallel circuits cancel, since they are equal but opposite polarity, and
the resultant output is in phase with the input. This is desirable since, when the circuit
is at resonance, 00 phase shift occurs.

Analysis of Wien bridge

The starting point is to investigate the feedback network which may either be used by
itself or as part of a bridge. These alternative arrangements are shown in Fig. 3-25

Fig 3-25

The circuit in fig. 3-25 a may be treated as a series parallel potential divider
network as follows:

Multiplying numerator and denominator by (R2 + 1/jC2)

Multiplying through by (j)2 and dividing by R1R2.

Dr. Ahmed El-Shazly | P a g e 25

 .

+ ( )

As with the phase shift oscillator, the necessary angle co for this expression is either
0° or 180°. Since the numerator imaginary, the condition can only be satisfied if the
denominator also imaginary; this will occur at any frequency when the real parts of the
denominator become zero.

and since j2 = -1,

(3-30)
√

Also, from equation [3-25] the voltage ratio at this frequent is given by

( )

The j cancels, and multiplying through by R1R2C1C2,

(3-31)

In many practical circuits, R1 = R2 = R and C1 = C2 = C. Equations [3-30] and [3-31]

then become respectively

[3-30(I)]

= 1/3 [3-31(I)]

Apply this lead-lag network to a Wien-bridge oscillator. Figure 3-26 active illustrates
a wien-bridge oscillator using an operational amplifier as the device. The lead-lag
network, comprised of RlCl and R2C2, makes up one side of the bridge. A voltage
divider, Ra and R4, is the remaining leg of the bridge. The inverting and non-inverting
inputs of the op amp make it ideal for use in the Wien-bridge oscillator, since

Dr. Ahmed El-Shazly | P a g e 26

regenerative and degenerative feedback are required. The op amp's high gain is also
very useful in offsetting circuit losses

Figure 3-26 An IC Wien-bridge Oscillator

Op amp output is fed back to the bridge input. Regenerative feedback is developed
across the lead-lag network and is applied to the non-inverting input. Therefore,
regenerative feedback is in phase with the output signal.
Degenerative feedback is developed across resistors Ra and R4 and is applied to the
inverting input. Of course, for the circuit to oscillate, regenerative feedback must be
greater than degenerative feedback.

Degenerative feedback remains constant regardless of the frequency, since the

resistance values do not change. However, regenerative feedback depends on the
frequency response of the lead-lag network which is frequency sensitive.

Component values are selected so that, at the desired oscillator frequency, regenerative
feedback is larger than degenerative feedback and oscillation occurs. If however,
oscillator frequency attempts to increase, the reactance of capacitor C 2 will decrease
and shunt more voltage to ground, reducing regenerative feedback. Likewise, a
decrease in frequency increases the reactance of Cl. Less voltage is developed across
the R2C2 network and again regenerative feedback is reduced. Only over a narrow
range of frequencies, set by the lead-lag network, will regenerative feedback be great
enough to sustain oscillation. Thus, the oscillator is forced on frequency by this
network.

The resonant frequency of oscillation is determined by the values of R1' R2' C1, and C2
and can be computed using the formula:

,
√

, for R1 + R2, C1 = C2

Dr. Ahmed El-Shazly | P a g e 27

Oscillator frequency may be varied by changing either the resistance or capacitance in
the lead-lag network. Usually resistors RI and R2 are ganged potentiometers,
permitting frequency variations. The formula shows that an increase in resistance or
capacitance, decreases oscillator frequency. Conversely, a reduction in resistance or
capacitance, increases oscillator frequency.

A wein bridge oscillator can be drawn as shown in figure below 3-27

Figure 3-27

a) A wien bridge oscillator, b) the bridge network

the IC Wien-bridge oscillator is simple to construct and relatively inexpensive. Before

integrated circuits were widely used for electronic design, Wien-bridge oscillators
were assembled from discrete components. Figure 3-28 is such a circuit.

Figure 3-28 A discrete Wien-bridge Oscillator.

FET Wien bridge oscillator is shown in Fig 3-29 The advantages and disadvantages of
Wien bridge oscillators are given below:

Advantages. 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.

Dr. Ahmed El-Shazly | P a g e 28

3. The frequency of oscillation can be easily varied by varying capacitances C1 and C2
simultaneously.

4. The overall gain is high because of two transistors.

Disadvantages. 1. The circuit needs two transistors and a large number of other
components.

2. The maximum frequency output is limited because of amplitude and the phase shift
characteristics of the amplifier.

Figure 3-29
FET Wien Bridge Oscillator

Example 3.1. Design the R-C elements of a Wien bridge oscillator shown in Fig. 3.28
for operation at f = 15 kHz.

Solution: Using values of R1 equal to R2 and C1 equal to C2 and taking each of R1 and
R2 of 200 k,

Required value of C = = 53pF Ans.

We can use R3 = 400 k and R4 = 200 k Ans.

Example 3.2. Find the frequency of oscillations of a Wien bridge oscillator with R =
20k and C = 1,000 pF.

Solution: R = R1 = R2 = 20 k = 20,000 

C = C1 = C2 = 1,000 pF = 1 x 10-9 F

Frequency of oscillation, f= = 7.96 kHz

Dr. Ahmed El-Shazly | P a g e 29

Example 3-3:Determine the frequency of oscillation for the Wien bridge oscillator
circuit shown in Fig. 3-30 and the value of R1 necessary to maintain oscillation.

Fig 3-30 Wien bridge oscillator circuit for example 3.3.

From equation [3.31]

√
Or f = 480 Hz.

The attenuation of the feedback network is given by equation [3-26]

Example 3.4. Design a Wien bridge oscillator for an output frequency of 10 kHz. 741
IC op-amp with Vc =± 10 V may be used.

Solution: The amplifier maximum input current, IB max = 500 nA

Let current through resistor R1 be equal to 100 times IB max

So I1 = 100 x 500 x 10-9 = 50 A

Output voltage, Vout = + (Vcc - 1 V) = + (10 V - 1 V) = ± 9 V

= 18 k

Rf. = 2 R1 , because β = 1/3, A=3=1+Rf/R1

So RI = = 6 k  (use 5.6 k standard value resistor)

Rf = 2 R1 = 2 x 5.6 = 11.2 k (use 12k standard value resistor)

This will make Rf > 2R1 and A > 3 Let R = R1 = 5.6 k

Dr. Ahmed El-Shazly | P a g e 31

From Eq. (3.30 (I)) C = = 2,842 pF

2,700 pF standard capacitor maybe used.

Thus R1 = 5.6k; Rf = 12 k, R = 5.6 k; and C = 2,700 pF Ans.

Example3-5: Determine the frequency of oscillation of the circuit shown in Fig. 3-31
Assume op-amp to be ideal.

Solution: For an ideal op-amp V1 = V2 = Va

At Node 2, or Vout = Va ( ) ( )

At Node =0

or ( )( ) + Va – Vout = 0

or Va = + Va – Vout = 0

or Va (1 + 3jCR = j2 2 C2 R2) = j C R Vout.

or (1 + 3jCR = 2 C2 R2) = j  CR ( )

This is characteristic equation.

Substituting value of R2 and R1 we get

1 + 3 j  CR - 2 C2 R2 = 3 jCR or 2 C2 R2 = 1

or  = = 212.76 radians/second

or = 33.86 Hz Ans.

you can use the expression directly

= = 212.76 rad/sec

Or = 33.86 Hz

Dr. Ahmed El-Shazly | P a g e 31

Amplitude Stabilization
We consider modification of the circuit in Fig.3-31 which serves to stabilize the
amplitude against variations due to fluctuations occasioned by the aging of transistors,
components, etc. One modification consists simply in replacing the resistor R 2 by a
sensistor (a resistor which has a positive thermal coefficient).

The amplitude of oscillation is determined by the extent to which the loop gain is
greater than unity. If the output V0 increases (for any reason), the current in R2
increases and A decreases. The regulation mechanism introduced by the sensistor
operates by automatically changing A so as to keep the return ratio more constant. The
temperature of R2 is determined by the root-mean-square (rms) value of the current
which passes through it. If the rms value of the current changes, then, because of the
thermal lag of the sensistor, the temperature will be determined by the average value
over a large number of cycles of the rms value of the current. An important fact to
keep in mind about the mechanism just described is that, because of the thermal lag of
the sensistor,

Figure 3-31

Zener diodes are used to automatically control the gain of

the oscillator and, hence, stabilize the amplitude of the sinusoid.

the resistance of the sensistor during the course of a single cycle is very nearly
absolutely constant. Therefore, at any fixed amplitude of oscillation, the sensistor
behaves entirely like an ordinary linear resistor.

A thermistor which has a negative temperature coefficient can also be used but it must
replace R1 rather than R2.

Another method of stabilizing the amplitude is indicated in Fig. 3-32 initially, both
Zener diodes are non conducting, and the loop gain is

Dr. Ahmed El-Shazly | P a g e 32

 ( ) ( )

and, hence, oscillations start. Because the loop gain exceeds unity, the amplitude of
these oscillations grows until the peaks exceed the diode breakdown voltage V z. When
this happens, the shunting action of the resistor 6R' reduces the gain and limits the
amplitude to approximately Vz. Distortion can be reduced to approximately 0.5
percent with this circuit.

The two methods of amplitude stabilization described in the foregoing are examples of
automatic gain control (AGC). An active AGC loop may also be used with a FET as a
voltage-controlled resistor.

Another method to control the gain uses a JFET as a voltage-controlled resistor in a

negative feedback path. A JFET operating with a small or zero V Ds is operating in the
ohmic region. As the gate voltage increases, the drain-source resistance increases. If
the JFET is placed in the negative feedback path, automatic gain control can be
achieved because of this voltage-controlled resistance.

A JFET stabilized Wien bridge is shown in Figure 3-32. The gain of the op-amp is
controlled by the components shown in the shaded box, which include the JFET. The
JFET's drain-source resistance depends on the gate voltage. With no output signal. the
gate is at zero volts, causing the drain-source resistance to be at the minimum. With
this condition, the loop gain is greater than I. Oscillations begin and rapidly build to a
large output signal. Negative excursions of the output signal forward-bias D1, causing
capacitor C3 to charge to a negative voltage. This voltage increases the drain-source
resistance of the JFET and reduces the gain (and hence the output). This is classic
negative feedback at work. With the proper selection of components, the gain can be
stabilized at the required level.

FIGURE 3-32 Self-starting Wien-bridge

oscillator using a JFET in the negative feedback loop.

Dr. Ahmed El-Shazly | P a g e 33

Example 3.6: Determine the frequency of oscillation for the Wien-bridge oscillator in
Figure 3-33 Also, verify that oscillations will start and then continue when the output
signal reaches 5.4 V.

Figure 3-33

Solution: For the lead-lag circuit, R4 = R5 = R = 10 k and C1 = C2 = C = 0.001 F.

The resonant frequency is

Initially, the closed-loop gain is




Since Acl > 3, the start-up condition is met.

When the output reaches 5.4 V (4.7 V + 0.7 V), the zeners conduct (their forward
resistance is assumed small, compared to 10 k), and the closed-loop gain is reached.
Thus, oscillation is sustained.



Related Exercise: What change is required in the oscillator in Figure 3-34 to pro-duce
an output with an amplitude of 6.8 V?

Example 3.7

Calculate the minimum and maximum frequencies in Fig 3-34. The two variable
resistors are ganged, which means that they change together and have the same value
for any wiper position.

Dr. Ahmed El-Shazly | P a g e 34

 Figure 3-34

Solution: With Eq. (23-4), the minimum frequency of oscillation is:

The maximum frequency of oscillation is:

Example 3.8: Figure 3-36 shows the lamp resistance of Fig.3-35 versus lamp voltage.
If the lamp voltage is expressed in rms volts, what is the output voltage of the
oscillator?

Solution: In Fig. 3-34 the feedback resistance is 2 k. Therefore, the oscillator output
signal becomes constant when the lamp resistance equals 1 k because this produces a
closed-loop gain of 3. In Fig. 3-35, a lamp resistance of 1 k corresponds to a lamp
voltage of 2 V rms. The lamp current is:


This 2 mA of current flows through the feedback resistance of 2 k, which means that
the output voltage of the oscillator is: Vout = (2 mA)(1 k + 2 k) = 6 V rms

Fig. 3-35

Dr. Ahmed El-Shazly | P a g e 35

42.2.3. Quadrature Oscillators
The quadrature oscillator generates two signals (sine and cosine) that are in and out of
phase by 90°. Figure 3-36 shows the circuit arrangement of a quadrature oscillator.
Output of op-amp Al is labeled as sine and output of A2 as cosine, although the actual
location of sine and cosine is arbitrary. The circuit requires two op-amps and three RC
arrangements.

The op-amp Al works as a non-inverting integrator. The second op-amp A2 is working

as a pure integrator. The feedback circuit consists of a resistance R 3 and a capacitor C3
which is connected at the output of A2 as a voltage divider section. Al and A2
constitute the amplifier stage.

The total phase shift of 360° is achieved in two parts. The op-amp A2,

Figure3-36: Quadrature Oscillator, Al and A2 Dual Op-amp 1C1458/353

as a pure integrator, provides -270° or 90 phase shift and remaining -90° or 270° i
achieved by the s voltage divider R3C3 section and op-amp A1. The total phase shift of
360° (or 0) is achieved only at one frequency fo, called frequency of oscillation.

The oscillation frequency is given by :

where R1C1 = R2C2 = R3C3 = RC

At frequency fo, A = 1 = 1.414

Analysis of the Circuit

With respect to node 1 (3-32)

Dr. Ahmed El-Shazly | P a g e 36

Rearranging Eq. (3.104), we obtain

Vo1 = - sC2R2vo2 (3-33)

Figure 3-37 The quadrature oscillator.

Similarly, at node 2, we find

(3-34)

which yields

Vo2 =(1+ sC3R3)Vo3 (3-35)

Now, at node X, by the virtual short-circuit property of op-amp Al

Vx =Vo3 (3-36)

Then at X, using nodal analysis, we get

= sC1 (Vo1 – Vo3) (3-37)

Rearranging Eq. (3-33) gives

Vo3 = Vo1 (3-38)

Substituting for Vo3 from (3-34) into Eq. (3-31) yields

Vo2 = (1+ sC3R3) * + Vol (3-39)

But from Eq. (3-29) we find

Dr. Ahmed El-Shazly | P a g e 37

 Vo2 = (3-40)

Substituting for Vo2 from Eq. (3-39) into Eq. (3-40), and rearranging, we obtain

= (1 + sC3R3) * + Vo1 (3-41)

Canceling the term Vo1 from both sides and further rearranging yields

- (1 + sC1R1) = (1 + SC3R3)(sC1R1)(sC2R2) (3-42)

Expanding and simplifying, we get

s3C1C2C3R1R2R3 + s2C1C2R1R2 + sC1R1 +1= 0 (3-43)

Substituting for s = j into Eq. (3-39), we find

( j)3C1C2C3R1R2R3 + ( j)2C1C2R1R2 + (j)C1R1 + 1= 0 (3-44)

Equating the imaginary terms to zero yields

(jr)3 C1C2C3R1R2R3 + ( jr )C1R1 =0 (3-45)

from which we have

(3-46)

Choosing C2 = C3 = C and R2 = R3 = R, we obtain the angular frequency of oscillation

as

(3-47)

and the frequency of oscillation as

(3-48)

Note: Notice that in a practical design, we choose C1 = C2 = C3 = C and R1 = R2 = R3

= R.

Design Steps
Step 1: Choice of C and R

Given the frequency of oscillation = 10 kHz, we find from Eq. (3-48)

Dr. Ahmed El-Shazly | P a g e 38

 RC= = 1.5 k (3-49)

Choosing C = 0.01F we find

R= = 1.5 k (3-50)

In actual practice, one of the resistors may be a fixed resistor of 1 k and a

potentiometer of 4.7 k, as shown in Figure 3-39.

Figure 3-38 The fully designed quadrature oscillator.

Example 3.9: Design a quadrature oscillator of frequency fo = 318 Hz using an op-

amp 1458/772.

Solution: Taking C = 0.015 F

R= = 33.33k (say, 33k)

Thus C1 = C2 = C3 = 0.015 F and R1 = R2 = R3 = 33 k

Example 3.10. What are the two requirements for oscillations? Discuss quadrature
oscillator and design it to operate at a frequency of 1.5 kHz.

Solution: Let C = 0.01F

R= = 33.33 k (say, 33k)

Thus C1 = C2 = C3 = 0.01 F and R1 = R2 = R3 =10k

Dr. Ahmed El-Shazly | P a g e 39

 A General Form of Oscillator Configuration

Many oscillator circuits fall into the general form shown in Fig. 3.40a. In the analysis
that follows we assume an active device with extremely high input resistance such as
an Op-Amp or a FET. Figure 3.40b shows the linear equivalent circuit in Fig. 3.40a,
using an amplifier with an open-circuit negative gain —Av, and output resistance Ro.
Clearly, the topology shown in Fig. 3.39 is that of shunt-series feedback.

FIGURE 3 –39

(a) General form of an oscillator circuit.

(b) The equivalent circuit of (a).

The Return Ratio (Feedback Ratio AB)

The value of AB will be obtained by considering the circuit of Fig.3 - to a feedback
amplifier with output taken from terminals 2 and 3 and with input terminals 1 and 3.
The load impedance ZL consists of Z2 in parallel with the series combination of Z1 and
Z3. Then

B= , ZL = (Z1 + Z3)   Z2

Dr. Ahmed El-Shazly | P a g e 41

 *
*

LC-Tunable Oscillators

The oscillators described in the two previous sections are RC-tunable circuits. That is,
the frequency of oscillation is determined by the resistance and capacitance values
used. Often, the frequency obtainable in such circuits is limited to a few hundred
kilohertz. Where higher frequencies of oscillation are required, such as those used in
AM and FM receivers, tuning is accomplished by varying a capacitance or inductance.
In the general oscillator con-figuration shown in Fig. 15-5, making Z1, Z2, and Z3 pure
reactances (either, inductive or capacitive), an LC-tunable oscillator results. If we let
Z1 = jX1, Z2 = jX2, and Z3 = jX3, where X = L, for an inductance and — 1/C for a
capacitance, Eq. (15-9) becomes

For AB to be real

X1 + X2 + X3 = 0 (3.52)

and

From Eq. (3-52) we see that the circuit will oscillate of the series combination of X1,
X2, and X3.

Dr. Ahmed El-Shazly | P a g e 41

Use of Eq. (3-52) in Eq. (3-53) yields

Since AB must be positive and at least unity in magnitude, then X1 and X 2 must have
the same sign (Av is positive). In other words, they must be the same kind of
reactance, either both inductive or both capacitive. Then, from Eq. (3-52), X3 = - (X1 +
X2) must be inductive if X1 and X2 are capacitive, or vice versa. If X1 and X2 are
capacitors and X3 is an inductor, the circuit is called a Colpitts oscillator. If X1 and X2
are inductors and X3 is a capacitor, the circuit is called a Hartley oscillator. In this
latter case, there may be mutual coupling between X1 and X2 (and the above equations
will then not apply).

Transistor versions of the types of LC oscillators described above are possible. As an

example, a transistor Colpitts oscillator is indicated in Fig. 3.40a. Qualitatively, this
circuit operates in the manner described above. However, the detailed analysis of a
transistor oscillator circuit is more difficult, for two fundamental reasons. First, the
low input impedance of the transistor shunts Z1 in Fig. 3.41a, and hence complicates
the expressions for the loop gain given above. Second, if the oscillation frequency is
beyond the audio range, the simple low-frequency model is no longer valid. Under
these circumstances the high-frequency hybrid- model must be used. A transistor
Hartley. Oscillator is shown in Fig. 3.40b.

FIGURE 3- 40: Inductance-capacitance LC oscillators: (a) Colpitts, (b) Hartley

oscillators.

a circuit that can be used for frequencies between 1 and 500 Mhz. This frequency
range is beyond the funity of most op amps. This is why a bipolar transistor or an FET is

Dr. Ahmed El-Shazly | P a g e 42

typically used for the amplifier. With an amplifier and LC tank circuit, we can feed
back a signal with the right amplitude and phase to sustain oscillations.

FIGURE 3.41 Colpitts oscillator.

CE CONNECTION:

Figure 3.41 shows a Colpitts oscillator the voltage-divider bias sets up a quiescent
operating point. The RF choke has a very high inductive reactance, so it appears open
to the ac signal. The circuit has a low-frequency voltage gain of rc gm, where rc is the
ac collector resistance. Because the RF choke appears open to the ac signal, the ac
collector resistance is primarily the ac resistance of the resonant tank circuit. This ac
resistance has a maximum value at resonance. You will encounter many variations of
the Colpitts oscillator. One way to recognize a Colpitts oscillator is by the capacitive
voltage divider formed by C1 and C2. It produces the feedback voltage necessary for
oscillations. In other kinds of oscillators, the feedback voltage is produced by
transformers, inductive voltage dividers, and so on.

AC EQUIVALENT CIRCUIT

Figure 3.42 is a simplified ac equivalent circuit for the Colpitts oscillator. The
circulating or loop current in the tank flows through C 1 in series with C2. Notice that
vout equals the ac voltage across Cl. Also, the feedback voltage of appears

Figure 3 -42 Equivalent circuit of colpitts oscillator

Dr. Ahmed El-Shazly | P a g e 43

across C2. This feedback voltage drives the base and sustains the oscillations
developed across the tank circuit, provided there is enough voltage gain at the
oscillation frequency. Since the emitter is at ac ground, the circuit is a CE connection.

RESONANT FREQUENCY,

Most LC oscillators use tank circuits with a Q greater than 10. Because of this, we can
calculate the approximate resonant frequency as:

√
This is accurate to better than 1 percent when Q is greater than 10.

The capacitance to use in Eq. (23-5) is the equivalent capacitance through which the
circulating current passes. In the Colpitts tank circuit of Fig. 23-16. the circulating
current flows through C1 in series with C2. Therefore, the equivalent capacitance is:

For instance, if C1 and C2 are 100 pF each, you would use 50 pF in Eq. (23-5).

STARTING CONDITION

The required starting condition for any oscillator is AB > 1 at the resonant frequency
of the tank circuit. This is equivalent to A > 1/B. In Fig. 23-16, the out-put voltage
appears across C1 and the feedback voltage appears across C2. The feedback fraction
in this type of oscillator is given by:

For the oscillator to start, the minimum voltage gain is:

What does A equal? This depends on the upper cutoff frequencies of the amplifier.
There are base and collector bypass circuits in a bipolar amplifier. If the cutoff
frequencies of these bypass circuits are greater than the oscillation frequency, A is
approximately equal to rc/r'e. If the cutoff frequencies are lower than the oscillation
frequency, the voltage gain is less than rc/r'e and there is additional phase shift through
the amplifier.

Dr. Ahmed El-Shazly | P a g e 44

OUTPUT VOLTAGE

With light feedback (small B), A is only slightly larger than 1/B, and the operation is
approximately class A. When you first turn on the power, the oscillations build up, and
the signal swings over more and more of the ac load line. With this increased signal
swing, the operation changes from small-signal to large-signal. When this happens, the
voltage gain decreases slightly. With light feedback. the value of AB can decrease to 1
without excessive clipping.

With heavy feedback (large B), the large feedback signal drives the base of Fig. 23-15
into saturation and cutoff. This charges capacitor C3. Producing negative dc clamping
at the base. The negative clamping automatically adjusts the

Value of AB to 1. If the feedback is too heavy, ou ma\ lose some of the output voltage
because of stray power losses.

When you build an oscillator, you can adjust the feedback to maximize the output
voltage. The idea is to use enough feedback to start under all conditions (different
transistors, temperature. voltage, and so on), but not so much that you lose output
signal. Designing reliable high-frequency oscillators is a challenge. Most designers
use computers to model high-frequency oscillators.

COUPLING TO A LOAD

The exact frequency of oscillation depends on the Q of the circuit and is given by:

√
When Q is greater than 10. this equation simplifies to the ideal value given by Eq.
(3.55). If Q is less than 10. the frequency is lower than the ideal value. Furthermore, a
low Q may prevent the oscillator from starting because it may reduce the high-
frequency voltage gain below the starting value of 1/B.

Figure 3.43a shows one way to couple the oscillator signal to the load resistance. If the
load resistance is large, it will load the resonant circuit only slightly and the Q will be
greater than 10. But if the load resistance is small, Q drops

Dr. Ahmed El-Shazly | P a g e 45

 FIGURE 3-43

(a) Capacitor coupling; (b) link coupling.

under 10 and the oscillations may not start. One solution to a small load resistance is
to use a small capacitance C4, one whose Xc is large compared with the load
resistance. This prevents excessive loading of the tank circuit.

Figure 3.43b shows link coupling, another way of coupling the signal to a small load
resistance. Link coupling means using only a few turns on the secondary winding of
an RF transformer. This light coupling ensures that the load resistance will not lower
the Q of the tank circuit to the point at which the oscillator will not start.

Whether capacitive or link coupling is used, the loading effect is kept as small as
possible. In this way, the high Q of the tank ensures an undistorted sinusoidal output
with a reliable start for the oscillations.

CB CONNECTION

When the feedback signal in an oscillator drives the base, a large Miller capacitance
appears across the input. This produces a relatively low cutoff frequency, which
means that the voltage gain may be too low at the desired resonant frequency.

To get a higher cutoff frequency, the feedback signal can be applied to the emitter, as
shown in Fig. 3.44. Capacitor C3 ac-grounds the base, and so the transistor acts like a
common-base (CB) amplifier. A circuit like this can oscillate at higher frequencies

Dr. Ahmed El-Shazly | P a g e 46

because its high-frequency gain is larger than that of a CE oscillator. With link
coupling on the output, the tank is lightly loaded, and the resonant frequency is given
by Eq. (3.58).

The feedback fraction is slightly different in a CB oscillator. The output voltage

appears across C1 and C2 in series, and the feedback voltage appears across C2.
Ideally, the feedback fraction is:

For the oscillations to start, A must be greater than 1/B. As an approximation, this
means that:

Amin = (3.60)

This is an approximation because it ignores the input impedance of the emitter, which
is in parallel with C2.

FIGURE 3.44 CB oscillator can oscillate at higher frequencies than CE oscillator.

FIGURE 3-45 JFET oscillator has less loading effect on tank circuit.

Dr. Ahmed El-Shazly | P a g e 47

FET COLPITTS OSCILLATOR

Figure 3.45 is an example of an FET Colpitts oscillator in which the feedback signal is
applied to the gate. Since the gate has a high input resistance, the loading effect on the
tank circuit is much less than with a bipolar transistor. The feedback fraction for the
circuit is:

The minimum gain needed to start the FET oscillator is:

In an FET oscillator, the low-frequency voltage gain is gmrd. Above the cut-off
frequency of the FET amplifier, the voltage gain decreases. In Eq. (23-13), Amin is the
voltage gain at the oscillation frequency. As a rule, we try to keep the oscillation
frequency lower than the cutoff frequency of the FET amplifier. Otherwise, the
additional phase shift through the amplifier may prevent the oscillator from starting.

Example 3.11

What is the frequency of oscillation in Fig. 3-46? What is the feedback fraction? How
much voltage gain does the circuit need to start oscillating?

- Solution: This is a Colpitts oscillator using the CE connection of a transistor. With

Eq. (3-56), the equivalent capacitance is:

FIGURE 3.46 Example.

Dr. Ahmed El-Shazly | P a g e 48

The inductance is 15 H. With Eq. (3.55), the frequency of oscillation is:

With Eq. (3-57), the feedback fraction is:

To start oscillating, the circuit needs a minimum voltage gain of:

23-5 Other LC Oscillators

The Colpitts oscillator is the most widely used LC oscillator. The capacitive voltage
divider in the resonant circuit is a convenient way to develop the feed-back voltage.
But other kinds of oscillators can also be used.

ARMSTRONG OSCILLATOR

Figure 3.47 is an example of an Armstrong oscillator. In this circuit, the collector

drives an LC resonant tank. The feedback signal is taken from a small secondary
winding and fed back to the base. There is a phase shift of 180° in the transformer.
which means that the phase shift around the loop is zero. If we ignore the loading
effect of the base, the feedback fraction is:

FIGURE 3.47 Armstrong oscillator.

Dr. Ahmed El-Shazly | P a g e 49

Where M is the mutual inductance and L is the primary inductance. For the
Armstrong. oscillator to start, the voltage gain must be greater than I/B.

An Armstrong oscillator uses transformer coupling for the feedback signal. This is
how you can recognize variations of this basic circuit. The small secondary winding is
sometimes called a tickler coil because it feeds back the signal that sustains the
oscillations. The resonant frequency is given by Eq. (3.55), using the L and C shown
in Fig. 23-21.

As a rule, you do not see the Armstrong oscillator used much because many designers
avoid transformers whenever possible.

HARTLEY OSCILLATOR

Figure 3.48 is an example of the Hartley oscillator. When the LC tank is resonant. the
circulating current flows through L1 in series with L2. The equivalent L to use in Eq.
(3.55) is:

L = L1 + L2 (3.64)

In a Hartley oscillator. the feedback voltage is developed by the inductive voltage

divider, L1 and L2. Since the output voltage appears across L1 and the feed-back
voltage appears across L2, the feedback fraction is:

Figure 3 - 48 Hartley oscillator

As usual, this ignores the loading effects of the base. For oscillations to start, the
voltage gain must be greater than 1/B.

Often a Hartley oscillator uses a single tapped inductor instead of two separate
inductors. Another variation sends the feedback signal to the emitter instead of to the
base. Also, you may see an FET used instead of a bipolar transistor. The output signal
can be either capacitively coupled or link-coupled

Dr. Ahmed El-Shazly | P a g e 51

Note:
The coil L1 is inductively coupled to coil L2, the combination functions as an auto-
transformer. However, because of direct connection, the junction of L1 and L2 cannot
be directly grounded. Instead, another capacitor CL is used. The operation of the
circuit is similar to that of the Colpitt's oscillator circuit.

Considering the fact that there exists mutual inductance between coils L 1 and L2
because the coils are wound on the same core, their net effective inductance is
increased by mutual inductance M. So in this case effective inductance is given by the
equation

L = L1 + L2 + 2 M (3.68)

and resonant or oscillation frequency is given by the equation

(3.69) as derived below :

√

Fig.3.49

Circuit for Hartley Oscillator Using CL

Example 3.12: In a transistorized Hartley oscillator, the tank circuit has the
capacitance of 100pF. The value of inductance between the collector and tapping point
is 30 mH and the value of inductance between the rapping point and the transistor base
is 1 X 10-8 H. determine the frequency of oscillations. Neglect the mutual inductance.

Solution: Effective inductance, L = L1 + L2 + 2M = 30 x 10-3 + 1 x 10-8 + 0 ~ 0.03 H

Capacitance, C = 100 pF = 100 x 10-12 F = 1 x 10-10 F

Dr. Ahmed El-Shazly | P a g e 51

CLAPP OSCILLATOR

The Clapp oscillator of Fig. 3.50 is a refinement of the Colpitts oscillator. The
capacitive voltage divider produces the feedback signal as before. An additional
capacitor C3 is in series with the inductor. Since the circulating tank current flows
through C1, C2, and C3 in series, the equivalent capacitance used to calculate the
resonant frequency is:

In a Clapp oscillator, C3 is much smaller than C1 and C2. As a result. C is

approximately equal to C3, and the resonant frequency is given by:

√
Why is this important? Because C1 and C2 are shunted by transistor and stray
capacitances. These extra capacitances alter the values of C1 and C2 slightly. In a
Colpitts oscillator, the resonant frequency therefore depends on the transistor and stray
capacitances. But in a Clapp oscillator, the transistor and stray capacitances have no
effect on C3, so the oscillation frequency is more stable and ac-curate. This is why you
occasionally see the Clapp oscillator used.

FIGURE 3.50 Clapp oscillator.

Example 3.13: If 50 pF is added in series with the 15-H inductor of Fig. 3.50, the
circuit becomes a Clapp oscillator. What is the frequency of oscillation?

Solution : We can calculate the equivalent capacitance with Eq. (3.70):

Dr. Ahmed El-Shazly | P a g e 52

Notice: how the term 1/50 pF swamps the other values because the 50 pF is much
smaller than the other capacitances. The frequency of oscillation is:

CRYSTAL CONTROLLED OSCILLATORS

LC oscillators, such as those just discussed, are widely used. However, in applications
where extreme oscillator stability is required, the LC oscillator is unsatisfactory.
Temperature changes, component aging and load Al fluctuations cause oscillator drift,
which makes the oscillator unstable. When a high degree of stability is required,
crystal oscillators are generally used.

What is meant by a high degree of stability? Suppose an acceptable frequency change

is one part per million. The allowable oscillator drift then would be .0001%. Compare
this with the stability of the LC oscillators just studied, where 1% frequency drift is
common. If an electronic wristwatch had a timing oscillator that drifts 1%, the watch
could either gain or lose 14 minutes each day and still be within oscillator tolerance.
However, if the watch oscillator is stable to within .0001%, the maximum time lost or
gained each day is .09 seconds, or 32 seconds each year. To achieve this accuracy,
electronic watches use crystal oscillators as the basic timing device.

Crystal Characteristics
Crystal material produce piezoelectricity. That is, when mechanical pressure is applied
to a crystal, a difference in potential is developed. Figure 3-51 illustrates this point. In
Figure 3-51A, a normal crystal has charges evenly distributed and is therefore neutral.
If force is applied to the sides of the crystal, as shown in Figure 3-51B, the crystal is
compressed and opposite charges accumulate on the sides . . . a difference in •
potential is developed.

If pressure is applied to the top and bottom, Figure 3-51C, the crystal is stretched and
again opposite polarities appear across the crystal. Thus, if the crystal is alternately
compressed and stretched, an AC voltage can be generated. Therefore, a crystal can
convert mechanical energy to electrical energy.

Crystal microphones use this principle. Rochelle salt, a crystalline material, is

alternately compressed and stretched by sound waves. The salt crystals in the
microphone generate a small voltage corresponding to sound wave variations.

Dr. Ahmed El-Shazly | P a g e 53

 Figure 3-51 Mechanical stress applied to a crystal.

Just the opposite effect occurs if AC voltage is applied to a crystal. The electrical
energy from the voltage source is converted to mechanical energy in the crystal.
Figure 3-52 illustrates this point. The AC input signal causes the crystal to stretch and
compress, which created mechanical vibrations that correspond to the frequency of the
AC signal.

Figure 3-52

Because of their structure, crystals have a natural frequency of vibration. If the

frequency of the applied AC signal matches this natural frequency, the crystal will
stretch and compress a large amount. However, if the frequency of the exciting voltage
is slightly different than the crystal's natural frequency, little vibration is produced.
The crystal, therefore, is extremely frequency selective, making it desirable for filter
circuits. The crystal's mechanical frequency of vibration is extremely constant, which
makes it ideal for oscillator circuits.

Many crystals produce piezoelectricity, but three types are the most useful: Rochelle
salt, tourmaline, and quartz. Rochelle salt has the greatest electrical activity, but it is
also the weakest and fractures easily. Tourmaline has the least electrical activity, yet it
is the strongest of the three. Quartz is a compromise, since it is inexpensive, rugged,

Dr. Ahmed El-Shazly | P a g e 54

and has good electrical activity. Therefore, quartz is the most commonly used crystal
in oscillator circuits.

The natural shape of quartz is a hexagonal prism with pyramids at the ends. Slabs are
cut from the natural crystal, or "mother stone," to obtain a usable crystal. There are
many ways to cut a crystal, all with different names; such as the X cut, Y cut, X-Y cut,
and AT cut. Each cut has a different piezoelectric property. For example the AT cut
has a good temperature coefficient, meaning the frequency changes very little with
temperature changes. The other cuts also have characteristics that are desirable for
specific applications see figure (3-53).

Figure (3-53 a ) Natural quartz crystal, (b) slab, (c) input current is maximum at
resonance

The natural frequency of a crystal is usually determined by its thickness. As shown in

Figure 3-54, the thinner the crystal, the higher its natural frequency. Conversely, the
thicker a crystal, the lower its natural frequency. To obtain a specific frequency, the
crystal slab is ground to the required dimensions.

Figure 3-54

Of course, there are practical limits on just how thin a crystal can be cut, without it
becoming extremely fragile. This places an upper limit on the crystal's natural
frequency, around 50 MHz.

To reach higher frequencies, crystals are mounted in such a way that they vibrate on
"overtones" or harmonics of the fundamental frequency. For example, a 10 MHz
crystal can be mounted so it vibrates at 30 MHz, the third overtone. The same crystal

Dr. Ahmed El-Shazly | P a g e 55

could be mounted to operate on the fifth overtone, 50 MHz. Only odd order overtone
vibrations are possible with crystals.

The formula for the frequency of a crystal is:

f = k/t (3.72)

where K is constant and t is the thickness of the crystal. Since the fundamental

Equivalent Crystal Circuits

A crystal is usually mounted between two metal plates and a spring applies
mechanical pressure on the plates. The metal plates secure the crystal and also
provide electrical contact. The crystal is then placed in a metal casing or holder. The
schematic symbol for a crystal is derived from the way it is mounted and represents
the crystal slab held between two plates as shown in Figure 3-55. The word crystal is
often abbreviated "XTAL" or "Y" on schematics.

Figure 3.55: Schematic symbol for a crystal

The crystal alone looks electrically like a series-resonant circuit, as shown in Figure3-
56. In the series equivalent circuit, inductance (L), represents the crystal mass that
effectively causes vibration; C represents crystal stiffness, which is the equivalent of
capacitance; R is the electrical equivalent of internal resistance caused by friction.
Therefore, at the crystal's natural mechanical resonant frequency, the electrical circuit
is series resonant and offers minimum impedance to current flow. When the circuit's
characteristics are plotted on an impedance-frequency curve, it shows sharp skirts and
minimum impedance at the series-resonant frequency. The sharp skirts indicate the
highly-selective frequency characteristic of the crystal.

Figure 3-56 The crystal wafer is a series-resonant circuit.

Dr. Ahmed El-Shazly | P a g e 56

When the crystal is mounted between metal plates, the equivalent circuit is modified
as shown in Figure 3-57. The metal mounting plates now appear as a capacitor, Cp in
parallel with the series-resonant circuit of the crystal. The value of Cp is relatively high
and, at lower frequencies, does not appreciably affect the series-resonant circuit of the
crystal. However, at frequencies above the crystal's series-resonant frequency, the
inductive reactance of the crystal is greater than the crystal's capacitive reactance, and
the crystal appears inductive. At these higher frequencies, a point is reached where the
inductive reactance of the crystal equals the capacitive reactance of the mounting
plates (XL = Xcp). Here, equivalent circuit is parallel-resonant and impedance is
maximum. The electrical equivalent of the crystal at this frequency is a parallel-tuned
LC circuit. Therefore, a crystal has two resonant frequencies.

Figure 3-57

When mounted, the crystal becomes a parallel-resonant circuit.

At the natural mechanical frequency of the crystal, the crystal is series-resonant and
impedance is minimum. At a slightly higher frequency, the crystal and the capacitance
of its mounting plates form a parallel-resonant

circuit and impedance is maximum. The overall crystal response curve is illustrated in
.Figure 3-58.

Figure 3.58: the crystal response curve

Dr. Ahmed El-Shazly | P a g e 57

As mentioned before, a crystal is highly frequency selective as indicated by the sharp
skirts in the response curve. This is natural, since a crystal has an extremely high Q,
sometimes approaching a Q of 50,000. When Q's of this value are compared with the
Q of an LC circuit, usually 100, it is clear why crystal oscillators are more stable than
normal LC oscillators. You will also find that crystal oscillators use either the series-
resonant or parallel-resonant characteristic.

If we neglect the resistance R, the impedance of the crystal is a reactance jX whose

dependence upon frequency is given by

where 2s = 1/LC is the series resonant frequency (the zero impedance frequency),
and 2p = (1/L) (1/C + 1/Cp) is the parallel resonant frequency (the infinite impedance
frequency). Since Cp >>C, then p s. For the crystal whose parameters are specified
above, the parallel frequency is only three-tenths of 1 percent higher than the series
frequency. For s <  < p, the

FIGURE 3.59

A piezoelectric crystal: (a) symbol, (b) circuit model, (c) the reactance as a
function of frequency assuming R = 0.

Reactance is inductive, and outside the range it is capacitive

CRYSTAL OSCILLATORS

Figure 3.60 a shows a Colpitts crystal oscillator. The capacitive voltage divider
produces the feedback voltage for the base of the transistor. The crystal acts like an
inductor that resonates with C1 and The oscillation frequency is between the series
and parallel resonant frequencies of the crystal.

Dr. Ahmed El-Shazly | P a g e 58

 FIGURE 3.60

Stray capacitances are in parallel with mounting capacitance

Figure 3.61

Crystal oscillators: (a) Colpitts; (b) variation of Colpitts (d) Pierce (c) Clapp

Figure 3.61b is a variation of the Colpitts crystal oscillator. The feedback signal is
applied to the emitter instead of the base. This variation allows the circuit to work at
higher resonant frequencies.

Figure 3.61c is an FET Clapp oscillator. The intention is to improve the frequency
stability by reducing the effect of stray capacitances. Figure 3.61d is a circuit called a
Pierce crystal oscillator. Its main advantage is simplicity.

FIGURE 3.62

A 1-MHz FET crystal, oscillator. (Courtesy of Siliconix Co.)

Dr. Ahmed El-Shazly | P a g e 59

A variety of crystal-oscillator circuits is possible. If a crystal is used for Z1 in the basic
configuration in Fig. 3-61a, a tuned LC combination for Z2, and the capacitance Cgd
between gate and drain for Z3, the resulting circuit is as indicated in Fig. 3-62. From
the theory given in the preceding section, the crystal reactance, as well as that of the
LC network, must be inductive. For the loop gain to be greater than unity, we see from
Eq. (3-13) that X1 cannot be too small. Hence the circuit will oscillate at a frequency
which lies between s and p but close to the parallel-resonance value. Since p  s,
the oscillator frequency is essentially determined by the crystal, and not by the rest of
the circuit.

Exercises
REVIEW QUESTIONS

1. What is an oscillator? How does it differ from an amplifier? What are the essential
parts of an oscillator circuit?

2. What are the Barkhausen conditions of oscillations it electronic systems? What are
their significance? What are the factors which affect the frequency stability of an
oscillator?

3. How an amplifier can be converted into an oscillator? Write the oscillation criteria.

4. Draw the circuit diagram of a tuned--collector oscillator and explain its operation.

5. Sketch the topology for a generalized resonant circuit oscillator, using impedances
Z1, Z2, Z3. At what frequency will the circuit oscillate? Under what conditions the
configuration does reduces to Hartley oscillator?

6. Draw the circuit diagrams of Hartley and Colpitts oscillators and obtain the
expression for the frequency of oscillation.

7. How does the circuit of a Clapp oscillator differ from that of Colpitt's oscillator?
Derive frequency of oscillation of any one of them.

8. Draw the circuit of a Clapp oscillator and derive the expression for its frequency of
oscillation.

9. Draw the circuit diagram of Hartley oscillator and briefly explain, how the
oscillations are maintained in this oscillator.

10. What are the advantages of a crystal oscillator? Draw the equivalent circuit of a
piezoelectric crystal and show how its impedance varies with frequency.

Dr. Ahmed El-Shazly | P a g e 61

11. Why do we use a crystal in an oscillator circuit? Obtain the expression of the
impedance and show its plot w.r.t. the frequency.

12. Explain with figure the working of a typical inductance type high frequency
oscillator. Discuss the limitations in realization of radio-frequency signals using
transistor circuits. Explain how to realize radio-frequency signals using crystal.

13. Explain with the aid of the circuit diagram the working of a transistor RC phase
shift oscillator and derive the condition for sustained oscillations.

14. Explain, with neat circuit diagram, the working of a phase shift oscillator using
three sections of RC network. State the expression for the frequency of oscillations
and minimum gain of the amplifier for sustained oscillations.

15. Point out the names of various sinusoidal oscillators in actual use. Draw the circuit
diagram for a RC phase shift oscillator using JFET as the active device. Deduce
that the frequency of oscillation (f) is equal to in the tank circuit
√
following the Barkhausen criterion.

16. Draw the circuit for Wien bridge oscillator and hence obtain its condition of
sustained oscillations.

17. Draw a Wien bridge oscillator circuit. Why is it mandatory that the minimum gain
of the amplifier be equal to 3?

18. Describe the principle of operation of a Wien bridge oscillator and give the
condition for sustained oscillation.

19. Draw the block diagram of a 'beat frequency oscillator' and explain the working of
each block.

20. Explain the working principle of a tunnel diode oscillator. Give the frequency
range over which it is employed.

21. Write the important factors in choosing a certain type of oscillator. Why do we use
R-C oscillators for audio frequencies and L-C oscillators for RF frequencies? Why
does not an L-C tank circuit produce sustained oscillations? Explain how the low
frequency parasitics can be suppressed in an oscillator.

22. Define the following terms used to specify the performance of an oscillator.

(i) dial resolution, (ii) amplitude stability, (iii) frequency stability, (iv) frequency
range, (v) output impedance.

Dr. Ahmed El-Shazly | P a g e 61

23. What are the main considerations which are to be kept in view while selecting an
oscillator for a particular application?

24. Write short notes on the following:

(i) Colpitt's oscillator.

(ii) Crystal oscillator.

(iii) R-C phase shift oscillator.

(iv) Wien bridge oscillator.

(v) Beat frequency oscillator (BEG).

25. Draw the circuit diagram of an R-C phase shift oscillator and obtain an expression
for its frequency of oscillation.

26. Explain how an op-amp can be used as R-C phase shift oscillator.

27. Derive the expressions for the condition of oscillations and calculate the frequency
of oscillation for R-C phase shift oscillator using op-amp.

28. Show that the circuit given in Fig. 1 will work as an oscillator at if R1
= 2R2.

Fig 1

29. What is Wien bridge? What are its uses? Show how variable frequency oscillator
can be built using an operational amplifier with a bridge. Derive an expression for
the frequency of oscillation of the circuit.

30. What are the two requirements for oscillations ? Discuss quadrature oscillator.

31. Explain the basic principle of operation of crystal oscillator using op-amp with
neat block diagram.

Dr. Ahmed El-Shazly | P a g e 62

x32. Draw the circuit diagram of an a stable multivibrator using op-amp and derive the
expression for its frequency of oscillation. How can the duty cycle be varied?
x33. Draw and explain the operation of a square-wave generator.

x34. Draw the circuit of a square-wave generator using an op-amp. Explain its
operation by drawing the capacitor voltage waveforms,
x35. Draw and explain the operation of a triangular—' wave generator.

x36. How does a sawtooth wave differ from a triangular-wave? How the ramp
generation can be controlled by using (i) bipolar junction transistor (BJT) and (ii)
programmable unijunction transistor (PUT)? Dis-cuss the problems associated
with PUT-controlled sawtooth wave generator.
x
37. Draw and explain the circuit of square- and triangle-wave generators using op-amp
and also draw the wave-forms at various nodes of circuit diagram.
x
38. Draw the diagram of triangular-wave generator using op-amp and find out an
expression for the frequency of oscillation.
x39. Discuss the operation of voltage-controlled oscillator with neat diagrams.

x
40. How does a twin-T oscillator differ from a Wien bridge oscillator? What is a notch
filter?

Draw the circuit diagram of a twin-T oscillator and explain its operation. Give its
advantages and disadvantages in comparison to Wien bridge oscillator.
x
41. Write short notes on the following:

(i) Square-wave generator.

(ii) Sawtooth-wave generator.

(iii) Triangular-wave generator.

(iv) Voltage controlled oscillator.

(v) Twin-T oscillator

42. is State the Barkhausen criterion, that is, the conditions necessary for sinusoidal
oscillations to be sustained.

43. What are the gain margin and phase margin needed to, sustain sinusoidal oscil-
lations?

44. Sketch the phase-shift oscillator using (a) an Op-Amp and (b) a JFET.

Dr. Ahmed El-Shazly | P a g e 63

45. a) Sketch the topology for a generalized resonant-circuit oscillator, using
impedances Z1 , Z2, Z3.

b) At what frequency will the circuit oscillate?

c) Under what conditions does the configuration reduce to a Colpitts oscillator? A

Hartley oscillator?

46. a) Sketch the circuit of a Wien bridge oscillator.

b) Which components determine the frequency of oscillation?

c) Which elements determine the amplitude of oscillation?

47. A) Draw the electrical model of a piezoelectric crystal.

b) Sketch the reactance versus frequency function.

c) Over what portion of the reactance curve do we desire oscillations to take place
when the crystal is used as part of a sinusoidal oscillator? Explain.

48. Sketch a circuit of a crystal-controlled oscillator.

Dr. Ahmed El-Shazly | P a g e 64

 SHORT ANSWER TYPE QUESTIONS WITH Answers

Q. 1. What is an oscillator?

Ans. Oscillator may be defined as a circuit which generates an ac output signal of very
high frequency without requiring any externally applied input signal or it may
be defined as an electronic source of alternating current (or voltage) having
sinusoidal or non-sinusoidal (square, sawtooth or pulse) wave shape. It can also
be defined as a circuit that converts dc energy into very high-frequency ac
energy.

Q. 2. What are the conditions which are to be satisfied for the successful operation of
phase-shift oscillator?

Ans. The phase-shift oscillator will generate sinusoidal waveform only if the gain is
29 and total phase-shift of the circuit is 360°.

Q. 3. Why is quadrature oscillator called by this name?

xxxcx

Ans. Because quadrature oscillator generates two signals (sine and cosine) that are in
and out of phase by 90, it is called quadrature oscillator.
xxQ. 4. Why is square-wave generator also called the astable multivibrator?

Ans. Square-wave generator is also called the free running or a stable multivibrator
because it has two quasi-stable states.
xxx
Q. 5. Give the frequency range of square-wave generator?

Ans. 10 Hz to 10 kHz.

Q. 6. Why are crystal oscillators used in communication transmitters and receivers?

Ans. Crystal oscillators are used in communication transmitters and receivers because
of their greater frequency stability

Q. 7. What is difference between a triangular-wave and a sawtooth wave?

Ans The difference between a triangular-wave and a sawtooth-wave is that in a

triangular-wave the rise time is always equal to its fall time while the sawtooth
waveform has different rise time and fall time.

Q. 8. What is a VCO? Give two applications that require VCO. [P.T.U. 2006-07]

Ans. The oscillator whose output frequency is control-led by varying the magnitude of
the input voltage is called a voltage-controlled oscillator (VCO). Applications

Dr. Ahmed El-Shazly | P a g e 65

 requiring VCO are frequency modulation (FM) and frequency shift keying
(FSK).

Q. 9. How does a twin-T oscillator differ from a Wien bridge oscillator?

Ans. In the Wien bridge oscillator, a bandpass filter is used in the positive feedback
path, but in twin-T oscillator, a notch filter is placed in the negative feedback
path.

Q. 10. Why twin-T oscillator is not a popular circuit like Wien bridge oscillator?

Ans. Twin-T oscillator is not popular like Wien bridge oscillator because it operates
well only at one frequency i.e. it cannot be easily adjusted over a large
frequency range as the Wien bridge oscillator can.

Dr. Ahmed El-Shazly | P a g e 66

PROBLEMS
1. Design a Wien bridge oscillator that will oscillate at 2 kHz.

2. Design a quadrature oscillator to operate at frequency of 2.5 kHz.

3. Consider the figure given below (Fig.1) and ideal op-amp.

Fig- 1

(i) Compute condition of oscillation.

(ii) Find the frequency of oscillation.

4. Given the circuit of Fig. 2. Calculate frequency of oscillation if R 1 = 200 k, R2 =

1.6 M; R3 = 20 k; R4 = 2 k, C1 = 30 F, C2 = 3 F; C3 = 0.3 F.

Fig. 2

5. For the circuit shown in figure 3, calculate frequency of oscillation if R1 = R2 =

22k; R3 = 33 k ; C = 3.3 F

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 Fig. 3

6. Design a phase shift oscillator of frequency 1 kHz.

7. The circuit in Figure 4 is an alternative configuration of a phase-shift oscillator. (a)

Assume that R1 = R2 = R3 = RA1 = RA2 = R and C1 C2 = C3 = C. Show that the
frequency of oscillation is √ (b) Assume equal magnitudes of gain in
each amplifier stage. What is the minimum magnitude of gain required in each
stage to sustain oscillation?

8. Consider the phase-shift oscillator in Figure 5. (a) Derive the expression for the
frequency of oscillation. (b) If R = 5 k, find the values of C and RF that will
produce sustained oscillations at 5 kHz.

9. A Wien-bridge oscillator is shown in Figure 6. (a) Derive the expression for the
frequency of oscillation. (b) What is the condition for sustained oscillations?

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10. Consider the oscillator circuit in Figure 7. (a) Derive the expression for the loop
gain T(s). (b) Determine the expression for the frequency of oscillation. (c) Find
the condition for oscillation.

11. Figure 8 shows a Colpitts oscillator with a BJT. Assume r and ro are both very
large. Derive the expressions for the frequency of oscillation and the condition of
oscillation.

12. Find the loop gain functions T(s) and T (j ), the frequency of oscillation, and the
R2/R1 required for oscillation for the circuit in Figure 9.

13. Repeat Problem 12 for the circuit in Figure 10.

14. Repeat Problem 12 for the circuit in Figure 11.

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 Solved Examples
3-1 Verify Eq. (17-27) for the feedback factor of the phase-shift network of Fig. 17-
25, assuming that this network does not load the amplifier. Prove that the phase
shift of Vf/ Vo is 180° for 2 = 6 and that at this frequency

Ans. let X = . Then, using mash analysis,we have:

V0 = I1(R – jX) – I2R+I3(0)

0 = -I1R + I2(2R-jX) – I3R

0 = I1(0) – I2R + I3(2R – jX) :

I3 is found from I3 = 3 /  where  =

| | | |

= R3[(1-ja)(2-ja)2 – (1-ja) – (2-ja) = R3[1 – 5a2 + j(a3 – 6a)]

| | ,

Hence,

For 180 phase shin, a3-6a = 0 or a2 = 6 and

3-2 (a) For the network of Prob. 17-22, show that the input impedance is given by

(b) Show that the input impedance at the frequency of the oscillator, √ ,is
(0.83 — j2.70)R.

Note that if the frequency is varied by 'varying C, the input impedance remains
constant. However, if the frequency is varied by varying R, the impedance is
varied in Proportion to R.

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Ans. (a) From the mesh equations derived in Frob.17-22 where

| | VoR2 (3-2 –j4)

( √ √ )
(b) For √ * + = R(0.53-j2:7)
√ √

3-3 (a) Draw the equivalent circuit of the transistor phase -shift oscillator shown.
Assume that hoeRc < 0.1 so that the approximate hybrid model may be used to
characterize the low-frequency behavior of the transistor. Let R3 = R — hie.
Neglect the effect of the biasing resistors R1, R3, and Ra,. Imagine the loop
broken at the base between B1 and B2, but in order not to change the loading on
the feedback network, place Ri = hie between B2 and ground.

(b) Find the current loop gain - xf/xi = I3/Ib by writing Kirchhoff's voltage
equations for the meshes of the phase-shift network. Verify that the Barkhausen
condition that the phase of I3/Ib must equal zero leads to the following
expression for the frequency of oscillation:
√

where k  Re / R. The requirement that the magnitude of I3/1b must exceed unity
in order for oscillations to start leads to the inequality hfe > 4k + 23 +

(c) Find the value of k which gives a minimum for h fe. Show that hfe(min) = 44.5. A
transistor with hfe < 44.5 cannot be used in a phase-shift-oscillator
configuration.

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Ans. (a)

(b) Replace the dependent Current source by Its Thevenin's equivalent, and write the
mesh equations for the resulting circuit.

(1) -hfe Ib Rc = Il (Rc + R- jX) = I2 R + I3 (0)

(2) 0 = -I1R + I2 (2R- jX) - I3R
(3) 0 = 11(0) –I2R +I3(2R – jX); Let  = and

K= , Then (3) becomes, I2 = I3 (2 - j) and (2) becomes, Il = I3(3 -2 - j4).
Substituting the expressions of I1 and I2 in (1) and simplifying we get:

- bfe Ibk = I3[I +3k-(5+k)2 –j [(6 + 4k) - 3]]

The loop current gain is I3/Ib and if this in to be real, then the coefficient of  must
be zero or

Thus, . At this frequency

√

(4k2 +23k+29): For > l then hfe > 4k + 23 + 29/k

(c) hfe = 4k + 23 +29/k. Thus

dhfe/dk = 4-19/k2 = 0, or k = (29/4)1/2 = 2.7.

Thus hfe(min) = (4)(2.7) + 23 + 29/2.7 = 44.5

3-4 Design a phase-shift oscillator to operate at a frequency of 5 kHz,. Use a MOSFET

with  = 5.5k. The phase-shift network is not to load down the amplifier.

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 (a) Find the minimum value of the drain-circuit resistance Rd for which the circuit
will oscillate.

(b) Find the product RC.

(C) Choose a reasonable value for R, and find C.

Ans. (a) | | or . For example


 = 55 and rd = 5.5 k .'. Rd = = 6.13k

(b) √ or RC = 1/(2 X 5 X103 √ ) = 12.99s

(c) In order not to load down the emplIfier, this input impedance of the phase shift
network must be high compared to the output impedance of the amplifier. If R is
chosen too large, however, C will be impractically small, e.g. If R = 1M, C=
12.9pF which is of the some order of magnitude as stray wiring capacitance, If we
choose this FET. the amplifier output impedance is = 2.9k.
If we choose R to be, say, 10 times this value, or 30k, then the load of the phase
shift network will be negligible. Then = 430 pF.

3-5 For the FET oscillator shown, find (a) Vf/Vo (b) the frequency of oscillation, (c)
the minimum gain of the source follower required for oscillations.

Ans. (a) Consider the phase shift network redrawn as shown

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 Then, from Eq. (17-27) we see that at 2 = 6m

= ( ) , the phase shift of is 180 or Vo. Applying KVI, around the

outside loop, we have: V'1 = - V'o Vo = Vo + Vo or

(b) From part (a), f =

√

(c) For oscillations to occur -BA>1 or A > = 0.967

3-6 (a) For the feedback network shown, find the transfer function Vf/V0. (b) This
network is used with an OP AMP t form an oscillator. Find the frequency of
oscillation and the minimum amplifier voltage gain.

(c) Draw the network connected to the OP AMP to form the oscillator.

Let X = The meah equations are

Vo = I1(R-jX) +I2jX + 0 (1)

0 = +I1jX +I2(R -2jX) +I3jX (2)

0 = 0 +I2jX +I3(R -zjX) (3)

Divide by jX and let = -j 'where  = RC

= - I1 (I + j) + I2 (4)

0 = I1 – I2 (2+ j) +13 (5)

0 = I2 – I3 (2+ j) (6)

From (6) I2 = I3 (2+ j)

From (5) I1 = + (2+ j) (2+ j) I3 – I3 = (3 - 2 +4J) I3

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From (4) = = + (1+ j) (3 - 2 + 4j) - (2+j)

= +3- 2 + 4j + j32 j3 - 42 – 2 -j

= 1 - 52 + j(6 - 3)

since V'f = -jXI3

Since A is real then 6-3 = 0 or  = √ = RC

√
f =

and A = I -52 = 1-5(6) = -29

in practice A > l . Hence A > 29

Since A is negative, an inverting OP AMP must be used. The system is that indicated
in Fig.17-25 f with R, and C interchanged. However, since R2 shunts C to ground then
R2 >>. 1/C. If this is not true then a voltage follower must be used as a buffer, as
indicated.

3-7 find an expression for the loop gain of a Wien-bridge oscillator.

Ans.

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 =

- AB=

3-8 (a) For the network shown prove that

(b) This network is used with an OP AMP to form an. oscillator. Show that the
frequency of oscillation is f = 1/2RC and that the gain must exceed 3.

(c) Draw the network connected. to the OP AMP

7.9

Ans.

(a) Let X = • Then

V0 = I1 (R- JX) – I2 (-jX)

0 = - I1(-jX) + I2 (R – j2X)

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 I1 = ( )

Vo = I2 [(R –jX)(2+j ) +jX] = I2 [3 R+ j ( –X)]

(b) B = - and – AB > 1 is

Hence RC = or CR = 1 , f = and = l, A >3

(c) Since A must be positive than we must use a non-inverting OP AMP as for the
Wien Bridge oscillator in figure b

3-9 (a) Verify that the transmission characteristic of the OP AMP, configuration
shown is given by

Note the magnitude  Vo/Vi = 1 for all frequencies and that the output leads the
input by a phase  = - 2 arctan RS. If R is varied as C remains constant, this
configuration acts as a constant-amplitude phase shifter. The phase is 0 for R = 0
and -180° for R .

(b) Cascade two identical phase shifters of the type given in part (a). Complete the
loop with an inverting OP AMP. Show at this system will oscillate at the
frequency f = 1 /2RC provided that the amplifier gain exceeds unity (in
magnitude).

(c) Demonstrate that two quadrature sinusoids (sine waves differing in phase by 90)
are obtained.

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Ans. (a) The negative pan is -1 and the positive gain is

( ) where X =

√
| | =1
√

 = -2rc tan RC – atc tan / CR = -2 arc tan CR

(b)

There is 1.80 deg phase shift in the inverting OP AMP. Hence for the loop gain to
equal 1+Jo there must be 180 deg shift through the two, phase shifter, or 90 deg
through each. Hence Z arc tan RC = or RC =1 or f = . Since the gain in
each phase shifter is 1 then the gain in the OP AMP must exceed unity or >1.

(c) Since the phase shift, is 90° In each phase shifter then if Vo1 is a sinusoid then Vo2
is a sinusoid of the same amplitude shifted by 90 deg.

3-10 In the Wien-bridge topology Z1 consists of R, C, and L in series, and Z2 is a

resistor R3. Find the frequency of oscillation and the minimum ratio R1/ R2.

Ans.

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 B=

A=1+

Since AB = 1 then S must be real

Or L = f=
√

at resonance AB = ( )( )

( )

3-12 (a) find an expression for the reactance of a crystal.

(b) Prove that the ratio of the parallel- to series-resonant frequencies is given
approximately by 1 + 1/2C/C dach

(c) If C = 0.04 pF and C' = 2.0 pF, by what percent is the parallel-resonant frequency
greater than the series-resonant frequency?

Ans. In order for the loop gain to be real (equal to unity) the capacitive reactance
must cancel the inductive reactance. Hence,

= 15.92 kHz
√ √

At the resonant frequency the parallel resistance of L and G is very large

compared with R < 10k. If we break the loop at the noninverting terminal
amplifier pain is (1 + 06/5) = 21 and the gain from the output back to the +Input
is . Hence the loop gain is

21 x > 1 or R > k = 476 

(a)

Where and

(b)

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 =

(c) If C = 0.04 pF and C1 = 2pF, then

= *100 = 1%

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