UNIT-4-FEEDBACK AMPLIFIERS AND OSCILLATORS (1).pdf

P.SRIDHAR,AP/EEE,KONGUNADU COLLEGE OF ENGINEERING AND
TECHNOLOGY,TRICHY
4.1
KONGUNADU COLLEGE OF ENGINEERING AND TECHNOLOGY
(AUTONOMOUS)
NAMAKKAL-TRICHY MAIN ROAD, THOTTIAM, TRICHY -621 215
DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
24EE302-ELECTRON DEVICES AND CIRCUITS
UNIT IV
FEEDBACK AMPLIFIERS AND OSCILLATORS
Introduction:
A block diagram of an amplifier with feedback is shown in figure.The output quantity
(either voltage or current) is sampled by a suitable sampler which is of two types, namely,
voltage sampler and current sampler and fed to the feedback network. The output of feedback
network which has a fraction of the output signal is combined with external source signal фs
through a mixer and fed to the basic amplifier. Mixer, also known as comparator, is of two types,
namely, series mixer and shunt mixer.
Advantages of negative feedback:
1. Stabilizes Gain: Negative feedback helps stabilize the gain of amplifiers by reducing the
dependence on various parameters, such as temperature and supply voltage variations. This
results in a more consistent output.
2. Reduces Distortion: By applying negative feedback, nonlinear distortion in amplifiers is
significantly reduced. This leads to a cleaner and more accurate signal output, enhancing the
fidelity of the system.
3. Decreases Noise: Negative feedback effectively reduces the noise present in the output signal.
This is particularly beneficial in sensitive applications where signal integrity is crucial.
4. Increases Bandwidth: The use of negative feedback can increase the bandwidth of amplifiers.
As the gain is reduced, the gain-bandwidth product remains constant, allowing for a wider
frequency response.

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4.2
5. Improves Input Impedance: Negative feedback can increase the input impedance of
amplifiers, making them more compatible with various signal sources and reducing loading
effects.
6. Enhances System Stability: In control systems, negative feedback helps maintain stability by
counteracting changes in output, ensuring that the system responds predictably to input
variations.
7. Better Tolerance to Component Variations: Systems with negative feedback are less
sensitive to variations in component values, leading to improved reliability and performance over
time.
In summary, negative feedback is a powerful tool in electronics and control systems, providing
significant advantages that enhance performance, stability, and reliability across various
applications.
Voltage / Current feedback Amplifiers:
Voltage-Series Feedback
A block diagram of a voltage-series feedback is illustrated in figure 5.2. Here, the input to
the feedback network is in parallel with the output of the amplifier. A fraction of the output
voltage through the feedback network is applied in series with the input voltage of the amplifier.
The shunt connection at the output reduces the output resistance R0. The series connection at the
input increases the input resistance. It is also known as series-shunt amplifier (or) voltage
amplifier. In this case, the amplifier is a true voltage amplifier. The voltage feedback factor is
given by 𝛽 =
𝑉𝑓
𝑉0
.
Figure 5.2 Block diagram of the voltage-series feedback
Input and output resistances
Figure 5.3 shows the voltage-series feedback circuit used to calculate input and output
resistances. Here,
𝑉𝑆 = 𝑉𝑖 + 𝑉
𝑓 = 𝐼𝑖𝑅𝑖 + 𝑉
𝑓
= 𝐼𝑖𝑅𝑖 + 𝛽𝑉0
= 𝐼𝑖𝑅𝑖 + 𝛽𝐴𝐼𝑖𝑅𝑖, 𝑤ℎ𝑒𝑟𝑒, 𝑉0 = 𝐴𝐼𝑖𝑅𝑖

TECHNOLOGY,TRICHY
4.3
Voltage-series feedback circuit for the calculation of input and output resistances
Therefore,
𝑅𝑖𝑓 =
𝑉𝑠
𝐼𝑖
= (1 + 𝐴𝛽)𝑅𝑖
Hence, the input resistance of a voltage-series feedback amplifier is given by
𝑅𝑖𝑓 = (1 + 𝐴𝛽)𝑅𝑖
where Ri is the input resistance of the amplifier without feedback.
For measuring the output resistance, RL is disconnected and Vs is set to zero. Then an
external voltage V is applied across the output terminals and the current I delivered by V is
calculated. Then,𝑅0𝑓 =
𝑉
𝐼
.Due to feedback, input voltage Vf reduces output voltage AVi, which
opposes V. Therefore,
𝐼 =
𝑉 − 𝐴𝑉𝑖
𝑅0
=
𝑉 + 𝛽𝐴𝑉
𝑅0
Therefore,
𝑅0𝑓 =
𝑉
𝐼
=
𝑅0
1 + 𝐴𝛽
Hence, the output resistance of a voltage-series feedback amplifier is given by
𝑅0𝑓 =
𝑅0
1+𝐴𝛽
where 𝑅0 is the output resistance of the amplifier without feedback.
Emitter follower
The Common collector or Emitter follower as shown in figure is an example of voltage-
series feedback since the voltage developed in the output is in series with the input voltage as far
as the base-emitter junction is concerned. This is a single stage RC coupled amplifier without
emitter bypass capacitor across RE. R1 and R2 provide the base bias. The emitter follower

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4.4
inherently exhibits 100% negative feedback since the voltage at the emitter follows the input
voltage.
As the output voltage is taken across 𝑅𝐸 = 𝑅𝐿, the feedback ratio, 𝛽 =
𝑅𝐸
𝑅𝐿
= 1.
Therefore, the overall voltage gain,𝐴𝑓 =
𝐴
1+𝐴
,which is less than unity.
The emitter follower simultaneously increases input resistance and decreases output
resistance.
A common collector amplifier or emitter follower
Characteristics of an emitter follower:
𝐴𝑖 =
𝐼𝑒
𝐼𝑏
=
𝐼𝑏+𝐼𝑐
𝐼𝑏
=
𝐼𝑏+ℎ𝑓𝑒𝐼𝑏
𝐼𝑏
= 1 + ℎ𝑓𝑒
𝑅𝑖 = ℎ𝑖𝑒 + (1 + ℎ𝑓𝑒)𝑅𝐸
= ℎ𝑖𝑒 + (1 + ℎ𝑓𝑒)𝑅𝐿
𝐴𝑉 =
𝐴𝑖𝑅𝐿
𝑅𝑖
=
(1 + ℎ𝑓𝑒)𝑅𝐿
ℎ𝑖𝑒 + (1 + ℎ𝑓𝑒)𝑅𝐿
= 1 −
ℎ𝑖𝑒
𝑅𝑖
𝑅0 =
ℎ𝑖𝑒+𝑅𝑆
1+ℎ𝑓𝑒
---------------------- (5.18)
𝑅0𝑓 = 𝑅0‖𝑅𝐿

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4.5
Voltage-Shunt Feedback
A voltage-shunt feedback is illustrated in figure 5.5. It is called a shunt derived, shunt-fed
feedback connection. Here,a fraction of the output voltage is supplied in parallel with the input
voltage through the feedback network. The feedback signal Ifis proportional to the output voltage
V0.Therefore, the feedback factor is given by𝛽 =
𝐼𝑓
𝑉0
.This type of amplifier is called a trans-
resistance amplifier (or) shunt-shunt feedback.
Block diagram of the voltage-shunt feedback
The voltage-shunt feedback provides a stabilised overall gain and decreases both input
and output resistances by a factor (1+Aβ).
𝑅𝑖𝑓 =
𝑅𝑖
(1 + 𝐴𝛽)
𝑎𝑛𝑑 𝑅0𝑓 =
𝑅0
(1 + 𝐴𝛽)
Common emitter amplifier with voltage-shunt feedback
The collector feedback biased common emitter amplifier as shown in figure 5.6 is an
example of voltage-shunt feedback. Here, a current which is proportional to the output voltage is
feedback to the input. Since 𝑉0 ≫ 𝑉𝑖,the feedback current 𝐼𝑓 ≈
𝑉0
𝑅𝐵
,so that the feedback ratio 𝛽 ≈
1
𝑅𝐵
.
The reduction in input and output resistances occurs due to Miller effect with RB.
Hence,
𝑅𝑖𝑓 = ℎ𝑖𝑒 ‖
𝑅𝐵
1 − 𝐴𝑉
′ , 𝑤ℎ𝑒𝑟𝑒 𝐴𝑉 =
𝑉0
𝑉𝑖
=
−ℎ𝑓𝑒𝑅𝐿
′
ℎ𝑖𝑒
𝑎𝑛𝑑 𝑅𝐿
′
= 𝑅𝐵‖𝑅𝐿

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4.6
CE amplifier with voltage-shunt feedback
𝐴𝑉𝑓 =
𝑉0
𝑉𝑆
= 𝐴𝑉
𝑅𝑖𝑓
𝑅𝑠 + 𝑅𝑖𝑓
𝑅𝑜𝑓 ≈
𝑅𝐵
𝑅𝑆
×
𝑅𝑠+ℎ𝑖𝑒
ℎ𝑓𝑒
𝑅0𝑓
′
= 𝑅𝑜𝑓‖𝑅𝐿
Current-Series Feedback
 A block diagram of a current-series feedback is illustrated in figure .In current-series feedback,
a voltage is developed which is proportional to the output current.
 This is called current feedback even though it is a voltage that subtracts from the input voltage.
Because of the series connection at the input and output, the input and output resistances get
increased.
 This type of amplifier is called transconductance amplifier (or) series-series feedback amplifier.
The transconductance feedback factor or ratio is given by 𝛽 =
𝑉𝑓
𝐼0
.
Block diagram of the current-series feedback
 One of the most common method of applying the current-series feedback is to place a resistor
Re, between the emitter lead of a common emitter amplifier and ground.
 As the common emitter amplifier has a high gain, this is most often used with series negative
feedback so that it can afford to lose some gain. Such a circuit is illustrated in figure.

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4.7
 When Re is properly bypassed with a large capacitor Ce, the output voltage is Vo and the
voltage gain without feedback is A. Resistor Re provides d.c. bias stabilisation, but no a.c.
feedback.
 When the capacitor Ce is removed, an a.c. voltage will be developed across Re due to the
emitter current flowing through Re and this current is approximately equal to the output
collector current.
 This voltage drop across Re will serve to decrease the input voltage between base and
emitter, so that the output voltage will decrease to Vo
’
. The gain of the amplifier with negative
feedback is now Af.
Feedback Ratio (β)
Referring to the figure 5.8(b),it is possible to calculate the approximate value of feedback
ratio β.
𝑉0
′
= 𝐼𝑐𝑅𝐿 𝑎𝑛𝑑 𝑉
𝑒 = 𝐼𝑒𝑅𝑒 = (𝐼𝑐 + 𝐼𝑏)𝑅𝑒 ≈ 𝐼𝑐𝑅𝑒
Therefore,
𝛽 =
𝑉𝑒
𝑉0
′ =
𝐼𝑐𝑅𝑒
𝐼𝑐𝑅𝐿
=
𝑅𝑒
𝑅𝐿
Input Resistance (Rif)
The equation for input resistance Rif and voltage gain Af for the common emitter
amplifier in figure 5.8(a),with an un by passed emitter resistor Re can be calculated using the
circuit of figure 5.8(b).
Input resistance,
𝑅𝑖𝑓 =
𝑉𝑖
𝐼𝑖
=
𝑉𝑖
′
+𝑉𝑒
𝐼𝑖
=
𝑉𝑖
′
𝐼𝑖
+=
𝑉𝑒
𝐼𝑖
But the input impedance without feedback,
𝑅𝑖 =
𝑉𝑖
′
𝐼𝑖
(a) CE amplifier with resistor Re when Ce is removed

TECHNOLOGY,TRICHY
4.8
(b) Simplified diagram for ac quantities
Also, 𝑉
𝑒 = 𝐼𝑒𝑅𝑒 𝑎𝑛𝑑 𝐼𝑖 = 𝐼𝑒 − 𝐼𝐶
Therefore, 𝑅𝑖𝑓 = 𝑅𝑖 +
𝑖𝑒𝑅𝑒
𝑖𝑒−𝑖𝑐
= 𝑅𝑖 +
𝑅𝑒
1−
𝑖𝑐
𝑖𝑒
But we know that
𝑖𝑐
𝐼𝑒
= ℎ𝑓𝑏 =
ℎ𝑓𝑒
1+ℎ𝑓𝑒
Where hfb and hfe are common base and common emitter short circuited current gain,
respectively.
Therefore, 𝑅𝑖𝑓 = 𝑅𝑖 + (1 + ℎ𝑓𝑒)𝑅𝑒
Thus, we find that there is a large decrease in voltage gain due to negative feedback.
Input resistance without feedback, 𝑅𝑖 = ℎ𝑖𝑒
Thus, 𝑅𝑖𝑓 = ℎ𝑖𝑒 + (1 + ℎ𝑓𝑒)𝑅𝑒
If the bias resistance,𝑅 = 𝑅1‖𝑅2 =
𝑅1𝑅2
𝑅1+𝑅2
is considered, the effective input resistance
with feedback is 𝑅𝑖𝑓
′
= 𝑅𝑖𝑓‖𝑅.
Voltage Gain (Af)
We know that
𝐴𝑓 =
𝐴𝑖𝑅𝐿
𝑅𝑖𝑓
𝑎𝑛𝑑 𝐴𝑖 = − ℎ𝑓𝑒
Therefore,

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4.9
𝐴𝑓 =
− ℎ𝑓𝑒𝑅𝐿
ℎ𝑖𝑒 + (1 + ℎ𝑓𝑒)𝑅𝑒
Voltage gain without feedback,
𝐴 =
− ℎ𝑓𝑒𝑅𝐿
ℎ𝑖𝑒
Therefore, we find that there is a large decrease in voltage gain due to negative feedback.
Output resistance (Rof)
The expression for the output resistance Rof looking back into the collector involves Rs,
Re and all the h-parameters. For values of Re in the order of Rs and hie, an approximate
expression for Rof is
𝑅0𝑓 =
1+ℎ𝑓𝑒
ℎ0𝑒
=
1
ℎ0𝑏
This has usually a large value in the range of MΩ, so the overall output resistance 𝑅0𝑓
′
taking the load resistance RL into consideration is approximately RL.
Note: It can be shown that the current-series feedback increases the input resistance but
decreases the output resistance of a feedback amplifier by a factor equal to (1+Aβ).Thus,
𝑅𝑖𝑓 = (1 + 𝐴𝛽)𝑅𝑖 𝑎𝑛𝑑 𝑅0𝑓 = (1 + 𝐴𝛽)𝑅0
Current-Shunt Feedback
A current-shunt feedback is illustrated in figure 5.9. It is called a series-derived, shunt-fed
feedback. The shunt connection at the input reduces the input resistance and the series
connection at the output increases the output resistance. This is true current amplifier also known
as shunt-series feedback amplifier. The current feedback factor is given by 𝛽 =
𝐼𝑓
𝐼0
.
Block diagram of the current-shunt feedback

TECHNOLOGY,TRICHY
4.10
Input and output resistances
Shows the current-shunt feedback circuit used to calculate input and output resistances.
𝐼𝑠 = 𝐼𝑖 + 𝐼𝑓 = 𝐼𝑖 + 𝛽𝐼𝑜
Where, 𝐼0 = 𝐴𝐼𝑖
Therefore, 𝐼𝑠 = 𝐼𝑖 + 𝐴𝛽𝐼𝑖 = 𝐼𝑖(1 + 𝐴𝛽)
𝑅𝑖𝑓 =
𝑉𝑖
𝐼𝑠
Current-series feedback circuit for the calculation of input and output resistances
𝑅𝑖𝑓 =
𝑉𝑖
𝐼𝑖(1 + 𝐴𝛽)
=
𝑅𝑖
1 + 𝐴𝛽
For measuring the output resistance, RL is disconnected and Vs is set to zero. Then
external voltage V is applied across the output terminals and the current I delivered by V is
calculated. Then 𝑅0𝑓 =
𝑉
𝐼
.
With 𝐼𝑠 = 0; 𝐼𝑖 = −𝐼𝑓 = −𝛽𝐼0 = 𝛽𝐼. Since current in the output circuit due to feedback
opposes the current, I, due to applied voltage, V
𝐼 =
𝑉
𝑅0
− 𝛽𝐴𝐼
Simplifying this, 𝑅0𝑓 =
𝑉
𝐼
= 𝑅0(1 + 𝛽𝐴)
Thus, this type of feedback decreases the input resistance and increases the output
resistance, i.e.,𝑅𝑖𝑓 =
𝑅𝑖
1+𝐴𝛽
𝑎𝑛𝑑 𝑅0𝑓 = (1 + 𝛽𝐴)𝑅0
As this type of feedback has the least desirable effects, this connection will not be
considered at all for practical applications.
summarizes the effects of negative feedback on amplifier characteristics according to the
way in which feedback is accomplished.

TECHNOLOGY,TRICHY
4.11
Effect of Negative Feedback on Amplifier Characteristics
Characteristic
Type of feedback
Current-series Voltage-series Voltage-shunt Current-shunt
Voltage gain Decreases Decreases Decreases Decreases
Bandwidth Increases Increases Increases Increases
Harmonic
Distortion
Decreases Decreases Decreases Decreases
Noise Decreases Decreases Decreases Decreases
Input resistance Increases Increases Decreases Decreases
Output
resistance
Increases Decreases Decreases Increases
Series, Shunt feedback Amplifiers:
Positive feedback:
Condition for oscillations:
Mechanism for Start of Oscillations:
 The oscillator circuit is set into oscillations by a random variation caused in the base current
due to noise component or a small variation in the d.c. power supply. The noise components
i.e., extremely small random electrical voltages and currents are always present in any
conductor, tube or transistor.
 Even when no external signal is applied, the ever-present noise will cause some small signal
at the output of the amplifier. When the amplifier is tuned at a particular frequency f0, the
output signal caused by noise signals will be predominantly at f0.
 If a small fraction (β) of the output signal is fed back to the input with proper phase relation,
then this feedback signal will be amplified by the amplifier. If the amplifier has a gain of
more than 1/β, then the output increases and thereby the feedback signal becomes larger.
 This process continues and the output goes on increasing. But as the signal level increases, the
gain of the amplifier decreases and at a particular value of output, the pain of the amplifier is
reduced exactly equal to 1/β. Then the output voltage remains constant at frequency f0, called
frequency of oscillation.
 The essential conditions for maintaining oscillations are:
1. |Aβ| = 1, i.e., the magnitude of loop gain must be unity.
2. The total phase shift around the closed loop is zero or 360 degrees.

TECHNOLOGY,TRICHY
4.12
Practical Considerations:
The condition |Aβ| = 1 gives a single and precise value of Aβ which should be set
throughout the operation of the oscillator circuit. But in practice, as transistor characteristics and
performance of other circuit components change with time, |Aβ| will become greater less than
unity. Hence in all practical circuits |Aβ| should be set greater than unity so that the amplitude of
oscillation will continue to increase without limit but such an increase in amplitude is limited by
the onset of the nonlinearity of operation in the active devices associated with the amplifier as
shown in figure 5.20. In this circuit, Aβ is larger than unity for positive feedback. This onset of
nonlinearity is an essential feature of all practical oscillators.
Block diagram of an oscillator
GENERAL FORM OF AN LC OSCILLATOR
In the general form of oscillator shown in figure, any of the active devices such as
Vacuum tube, Transistor, FET and Operational amplifier may be used in the amplifier section.
Z1, Z2 and Z3 are reactive elements constituting the feedback tank circuit which determines the
frequency of oscillation. Here, Z1 and Z2 serve as an a.c. voltage divider for the output voltage
and feedback signal. Therefore, the voltage across Z1 is the feedback signal. The frequency of
oscillation of the LC oscillator is
𝑓0 =
1
2𝜋√𝐿𝐶
(a) General form of an oscillator and (b) its equivalent circuit

TECHNOLOGY,TRICHY
4.13
The inductive or capacitive reactances are represented by Z1, Z2 and Z3. In figure 5.21,
the output terminals are 2 and 3, and input terminals are 1 and 3. Figure 5.21(b) gives the
equivalent circuit of figure.
Load impedance
Since Z1 and the input resistance hie of the transistor are in parallel, their equivalent
impedance Z1
is given by
1
𝑍1
=
1
𝑍1
+
1
ℎ𝑖𝑒
From this equation, we get
𝑍1
=
𝑍1ℎ𝑖𝑒
𝑍1 + ℎ𝑖𝑒
Now the load impedance ZL between the output terminals 2 and 3 is the equivalent
impedance of Z2 in parallel with the series combination of Z1
and Z3.
Therefore,
1
𝑍𝐿
=
1
𝑍2
+
1
𝑍1 + 𝑍3
=
1
𝑍2
+
1
𝑍1ℎ𝑖𝑒
𝑍1 + ℎ𝑖𝑒
+ 𝑍3
=
1
𝑍2
+
𝑍1 + ℎ𝑖𝑒
𝑍1ℎ𝑖𝑒 + 𝑍1𝑍3 + ℎ𝑖𝑒𝑍3
=
1
𝑍2
+
𝑍1 + ℎ𝑖𝑒
ℎ𝑖𝑒(𝑍1 + 𝑍3) + 𝑍1𝑍3
=
ℎ𝑖𝑒(𝑍1 + 𝑍3) + 𝑍1𝑍3 + 𝑍2(𝑍1 + ℎ𝑖𝑒)
𝑍2[ℎ𝑖𝑒(𝑍1 + 𝑍3) + 𝑍1𝑍3]
=
ℎ𝑖𝑒(𝑍1 + 𝑍2 + 𝑍3) + 𝑍1𝑍2 + 𝑍1𝑍3
𝑍2[ℎ𝑖𝑒(𝑍1 + 𝑍3) + 𝑍1𝑍3]
Therefore,
𝑍𝐿 =
𝑍2[ℎ𝑖𝑒(𝑍1+𝑍3)+𝑍1𝑍3]
ℎ𝑖𝑒(𝑍1+𝑍2+𝑍3)+𝑍1𝑍2+𝑍1𝑍3
Voltage gain without feedback
This is given by,

TECHNOLOGY,TRICHY
4.14
𝐴𝑣𝑒 = −
ℎ𝑓𝑒𝑍𝐿
ℎ𝑖𝑒
Feedback fraction 𝛃
The output voltage between the terminals 3 and 2 in terms of current I1 is given by
𝑉0 = −𝐼1(𝑍1
+ 𝑍3) = −𝐼1 (
𝑍1ℎ𝑖𝑒
𝑍1 + ℎ𝑖𝑒
+ 𝑍3)
= −𝐼1 (
ℎ𝑖𝑒(𝑍1+𝑍3)+𝑍1𝑍3
𝑍1+ℎ𝑖𝑒
)
The voltage fedback to the terminals 3 and 1 is given by
𝑉
𝑓𝑏 = −𝐼1𝑍1
− 𝐼1 (
𝑍1ℎ𝑖𝑒
𝑍1 + ℎ𝑖𝑒
)
Therefore, the feedback ratio is given by
𝛽 =
𝑉
𝑓𝑏
𝑉0
= 𝐼1 (
𝑍1ℎ𝑖𝑒
𝑍1 + ℎ𝑖𝑒
)[
𝑍1 + ℎ𝑖𝑒
ℎ𝑖𝑒(𝑍1 + 𝑍3) + 𝑍1𝑍3
]
𝛽 =
𝑍1+ℎ𝑖𝑒
ℎ𝑖𝑒(𝑍1+𝑍3)+𝑍1𝑍3
Equation for the oscillator
For oscillation, we must have
𝐴𝑣𝑒𝛽 = 1
Substituting the values of Ave and β,we get
(
−ℎ𝑓𝑒𝑍𝐿
ℎ𝑖𝑒
) [
𝑍1ℎ𝑖𝑒
ℎ𝑖𝑒(𝑍1 + 𝑍3) + 𝑍1𝑍3
] = 1
{
ℎ𝑓𝑒𝑍2[ℎ𝑖𝑒(𝑍1 + 𝑍3) + 𝑍1𝑍3]
ℎ𝑖𝑒(𝑍1 + 𝑍2 + 𝑍3) + 𝑍1𝑍2 + 𝑍1𝑍3
} [
𝑍1
ℎ𝑖𝑒(𝑍1 + 𝑍3) + 𝑍1𝑍3
] = −1
ℎ𝑓𝑒𝑍2𝑍1
ℎ𝑖𝑒(𝑍1 + 𝑍2 + 𝑍3) + 𝑍1𝑍2 + 𝑍1𝑍3
= −1
ℎ𝑖𝑒(𝑍1 + 𝑍2 + 𝑍3) + 𝑍1𝑍2 + 𝑍1𝑍3 = −ℎ𝑓𝑒𝑍2𝑍1
ℎ𝑖𝑒(𝑍1 + 𝑍2 + 𝑍3) + 𝑍1𝑍2(1 + ℎ𝑓𝑒) + 𝑍1𝑍3 = 0
This is the general equation for the oscillator.
Phase shift:

TECHNOLOGY,TRICHY
4.15
(i) RC phase shift oscillator using cascade connection of high pass filter
 In this oscillator the required phase shift of 180° in the feedback loop from output to input
is obtained by using R and C components instead of tank circuit.The figure shows the
circuit of RC phase shift oscillator using cascade connection of high pass filter.
 Here, a common emitter amplifier is followed by three sections of RC phase shift network,
the output of the last section being returned to the input.
 In order to make the three RC sections identical, R3 is chosen as R3 = R – Ri, where Ri is
the input impedance of the circuit.
 The phase shift , given by each RC section is ф = tan−1
(
1
𝜔𝐶𝑅
).If R is made zero, then
will become 90°. But making R = 0 is impracticable because if R is zero, then the voltage
across it will become zero.
 Therefore, in practice the value of R is adjusted such that becomes 60°.
(a) RC phase shift oscillator using cascade connection of high pass filter (b) Equivalent
circuit
If the values of R and C are so chosen that, for the given frequency f0, the phase shift
of each RC section is 60°. Thus, such a RC ladder network produces a total phase shift of
180° between its input and output voltages for the given frequency. Therefore, at the specific
frequency f0, the total phase shift from the base of the transistor around the circuit and back to

TECHNOLOGY,TRICHY
4.16
the base will be exactly 360° or 0°, thereby satisfying Barkhausen condition for oscillation.
The frequency of oscillation is given by
𝑓0 =
1
2𝜋𝑅𝐶√6
At this frequency, it is found that the feedback factor of the network is |β|= 1/29. In
order that |Aβ| shall not be less than unity, it is required that the amplifier gain |A| must be
more than 29 for oscillator operation.
Analysis:
From figure 5.24 (b), we can write the following three equations
𝐼1 (𝑅 +
1
𝑗𝜔𝐶
) − 𝐼2𝑅 = 𝐸𝑖
−𝐼1𝑅 + 𝐼2 (2𝑅 +
1
𝑗𝜔𝐶
) − 𝐼3𝑅 = 0
−𝐼2𝑅 + 𝐼3 (2𝑅 +
1
𝑗𝜔𝐶
) = 0
Solving the above simultaneous equations, we get
𝐼3 =
𝐸𝑖
𝑅
[
1
(1−5𝛼2)+𝑗𝛼(𝛼2−6)
] , 𝑤ℎ𝑒𝑟𝑒 𝛼 =
1
𝜔𝐶𝑅
Therefore,
𝐸0 = 𝐸𝑖 [
1
(1 − 5𝛼2) + 𝑗𝛼(𝛼2 − 6)
], 𝑤ℎ𝑒𝑟𝑒 𝐸0 = 𝐼3𝑅
Hence,
𝛽 =
𝐸0
𝐸𝑖
= [
1
(1−5𝛼2)+𝑗𝛼(𝛼2−6)
]
For determining the frequency of oscillation, the phase shift, i.e., the imaginary part
must be equal to zero. Hence,
𝛼(𝛼2
− 6) = 0
From this equation,
𝛼 =
1
𝜔0𝐶𝑅
= √6
Therefore, the frequency of oscillation
𝑓0 =
1
2𝜋𝑅𝐶√6

TECHNOLOGY,TRICHY
4.17
The condition for maintenance of oscillation is obtained by substituting 𝛼 = √6 into
equation Thus, we get,
𝛽 = −
1
29
=
1
29
∟180°
Hence, a transistor with gain <29 cannot be made to oscillate in a circuit and the
frequency of oscillation may be varied by changing any of the impedance element in the
phase shifting network.
(ii) RC phase shift oscillator using cascade connection of low pass filter
The figure shows the RC phase shift oscillator using cascade connection of low pass
filters. There are three RC networks with the output of the last section returned to the input.
In this oscillator, the required phase shift of 180° is obtained by using RC network. In
practice, the resistor R of the last section is adjusted in such a way that the total phase shift
produced by the cascade connection of RC network is exactly equal to 180°. The transistor in
the amplifier circuit gives a phase shift of another 180°. Hence, the total phase shift around
the circuit is 360° i.e., 0°.
(b)
(a) RC phase shift oscillator circuit using cascade connection of low pass filters (b) Its
equivalent circuit
From figure 5.25(b) we can write the following three equations
−
𝐼1
𝑗𝜔𝐶
+ (𝑅 +
2
𝑗𝜔𝐶
) 𝐼2 −
𝐼3
𝑗𝜔𝐶
= 0

TECHNOLOGY,TRICHY
4.18
−
1
𝑗𝜔𝐶
𝐼2 + (𝑅 +
2
𝑗𝜔𝐶
) 𝐼3 −
1
𝑗𝜔𝐶
𝐼4 = 0
−
1
𝑗𝜔𝐶
𝐼3 + (𝑅 +
2
𝑗𝜔𝐶
) 𝐼4 = 0
From equation 5.54, the value of I3 becomes
𝐼3 = 𝐼4(1 + 𝑗𝜔𝑅𝐶)
Substituting equation 5.55 into equation 5.53, we get
−
1
𝑗𝜔𝐶
𝐼2 + (1 + 𝑗𝜔𝑅𝐶) (𝑅 +
2
𝑗𝜔𝐶
) 𝐼4 −
1
𝑗𝜔𝐶
𝐼4 = 0
−𝐼2 + (3𝑗𝜔𝑅𝐶 + 1 − 𝜔2
𝑅2
𝐶2)𝐼4 = 0
Therefore, 𝐼2 = (3𝑗𝜔𝑅𝐶 + 1 − 𝜔2
𝑅2
𝐶2)𝐼4
By substituting the values of I2 and I3 into equation 5.52, we get
𝐼4 (6𝑅 − 𝜔2
𝑅2
𝐶2
+
1
𝑗𝜔𝐶
+ 5𝑗𝜔𝐶𝑅2
) =
𝐼1
𝑗𝜔𝐶
= 0
Therefore,
𝐼4 =
𝐼1
1 − 5𝜔2𝑅2𝐶2 + 𝑗𝜔𝑅𝐶(6 − 𝜔2𝑅2𝐶2)
=
𝐼1
1 − 5𝛼2 + 𝑗𝛼(6 − 𝛼2)
𝑤ℎ𝑒𝑟𝑒 𝛼 = 𝜔𝑅𝐶
Hence
𝛽 =
𝐼4
𝐼1
=
1
1−5𝛼2+𝑗𝛼(6−𝛼2)
To determine the frequency of oscillation, the imaginary part is equated to zero, i.e.,
𝛼(6 − 𝛼2) = 0
Therefore 𝛼 = √6 ; 𝜔𝐶𝑅 = √6
Hence the frequency of oscillation is
𝑓0 =
√6
2𝜋𝑅𝐶
By substituting the values of 𝑓0 into equation 5.56, we get
𝛽 =
1
1 − 30 + 𝑗√6(6 − 6)
= −
1
29

TECHNOLOGY,TRICHY
4.19
Or
|𝛽| =
1
29
Thus, sustained oscillation is obtained by having the gain of transistor amplifier
greater than 29 and frequency of oscillation may be varied by changing the value of
impedance element in the phase shifting network.
WIEN-BRIDGE OSCILLATOR
Figure shows the circuit of a Wien-bridge oscillator. The circuit consists of a two-stage
RC coupled amplifier which provides a phase shift of 360° or 0°. A balanced bridge is used as the
feedback network which has no need to provide any additional phase shift. The feedback network
consists of a lead-lag network (R1 –C1 and R2- C2) and a voltage divider (R3– R4). The lead-lag
network provides a positive feedback to the input of the first stage and the voltage divider
provides a negative feedback to the emitter of Q1.
If the bridge is balanced,
𝑅3
𝑅4
=
𝑅1−𝑗𝑋𝐶1
[
𝑅2(−𝑗𝑋𝐶2)
𝑅2−𝑗𝑋𝐶2
]
---------------------- (5.61)
where XC1 and XC2 are the reactances of the capacitors.
(a)

TECHNOLOGY,TRICHY
4.20
(b)
(a) Wien-bridge oscillator and (b) feedback circuit
Simplifying equation 5.61 and equating the real and imaginary parts on both sides, we get
the frequency of oscillation as,
𝑓0 =
1
2𝜋√𝑅1𝑅2𝐶1𝐶2
=
1
2𝜋𝑅𝐶
, 𝑖𝑓 𝑅1 = 𝑅2 = 𝑅 𝑎𝑛𝑑 𝐶1 = 𝐶2 = 𝐶.
The ratio of R3 to R4 being greater than 2 will provide a sufficient gain for the circuit to
oscillate at the desired frequency. This oscillator is used in commercial audio signal generators.
To determine the gain of Wien bridge oscillator using BJT amplifier:
Assume that 𝑅1 = 𝑅2 = 𝑅 𝑎𝑛𝑑 𝐶1 = 𝐶2 = 𝐶
Then the feedback circuit is as shown figure 5.27(b)
Therefore,
𝑉
𝑓(𝑠) = 𝑉0(𝑠)
𝑅 ‖
1
𝑠𝐶
𝑅 +
1
𝑠𝐶
+ 𝑅 ‖
1
𝑠𝐶
𝑉
𝑓(𝑠) = 𝑉0(𝑠)
𝑅
1 + 𝑠𝑅𝐶
𝑅 +
1
𝑠𝐶
+
𝑅
1 + 𝑠𝑅𝐶
= 𝑉0(𝑠)
𝑠𝑅𝐶
𝑠2𝑅2𝐶2 + 3𝑠𝑅𝐶 + 1
Hence, the feedback factor is
𝛽 =
𝑉
𝑓(𝑠)
𝑉0(𝑠)
=
𝑠𝑅𝐶
𝑠2𝑅2𝐶2 + 3𝑠𝑅𝐶 + 1
We know that Aβ=1
Therefore, the gain of the amplifier,

TECHNOLOGY,TRICHY
4.21
𝐴 =
1
𝛽
=
𝑠2𝑅2𝐶2+3𝑠𝑅𝐶+1
𝑠𝑅𝐶
Substituting 𝑠 = 𝑗𝜔0, where the frequency of oscillation 𝑓0 =
1
2𝜋𝑅𝐶
, 𝑖. 𝑒. , 𝜔0 =
1
𝑅𝐶
, in the
above equation and simplifying, we get A=3. Hence the gain of the Wien bridge oscillator using
BJT amplifier is at least equal to 3 for oscillators to occur.
HARTLEY OSCILLATOR
 In the Hartley oscillator shown in figure Z1 and Z2 are inductors and Z3 is a capacitor. Resistors
R1, R2 and RE provide the necessary d.c. bias to the transistor. CE is a bypass capacitor. CC1 and
CC2 are coupling capacitors.
 The feedback network consisting of inductors L1 and L2, and capacitor C determines the
frequency of the oscillator.
 When the supply voltage +Vcc is switched ON, a transient current is produced in the tank
circuit and consequently, damped harmonic oscillations are set up in the circuit.
 The oscillatory current in the tank circuit produces a.c. voltages across L1 and L2. As terminal 3
is earthed, it is at zero potential.
 If terminal 1 is at a positive potential with respect to 3 at any instant, terminal 2 will be at a
negative potential with respect to 3 at the same instant. Thus, the phase difference between the
terminals 1 and 2 is always 180°.
Hartley oscillator
 In the CE mode, the transistor provides the phase difference of 180° between the input and
output. Therefore, the total phase shift is 360°.
 Thus, at the frequency determined for the tank circuit, the necessary condition for sustained
oscillations is satisfied. If the feedback is adjusted so that the loop gain Aβ= 1, the circuit acts
as an oscillator.
 The frequency of oscillation is 𝑓0 =
1
2𝜋√𝐿𝐶
, where 𝐿 = 𝐿1 + 𝐿2 + 2𝑀, and M is the value of
mutual inductance between coils L1 and L2. The condition for sustained oscillation is
ℎ𝑓𝑒 ≥
𝐿1+𝑀
𝐿2+𝑀
Analysis

TECHNOLOGY,TRICHY
4.22
In the Hartley oscillator, Z1 and Z2 are inductive reactances and Z3, is the capacitive
reactance. Suppose M is the mutual inductance between the inductors, then
𝑍1 = 𝑗𝜔𝐿1 + 𝑗𝜔𝑀
𝑍2 = 𝑗𝜔𝐿2 + 𝑗𝜔𝑀
𝑍3 =
1
𝑗𝜔𝐶
= −
𝑗
𝜔𝐶
Substituting the values in equation 5.34 and simplifying, we get
𝑗𝜔ℎ𝑖𝑒 (𝐿1 + 𝐿2 + 2𝑀 −
1
𝜔2𝐶
) − 𝜔2(𝐿1 + 𝑀)[(𝐿2 + 𝑀)(1 + ℎ𝑓𝑒) −
1
𝜔2𝐶
] = 0
The frequency of oscillation 𝑓0 =
𝜔0
2𝜋
can be determined by equating the imaginary part of
equation 5.37 to zero.
Therefore,
[𝐿1 + 𝐿2 + 2𝑀 −
1
𝜔0
2
𝐶
] = 0
Simplifying this equation, we obtain
𝑓0 =
𝜔0
2𝜋
=
1
2𝜋√(𝐿1+𝐿2+2𝑀)𝐶
The condition for maintenance of oscillation is obtained by substituting equation 5.38into
equation 5.37. Now the imaginary part becomes zero and hence,
[(𝐿2 + 𝑀)(1 + ℎ𝑓𝑒) −
1
𝜔0
2
𝐶
] = 0
Substituting equation 5.38 into the above equation and simplifying, we get
ℎ𝑓𝑒 =
𝐿1 + 𝑀
𝐿2 + 𝑀
COLPITTS OSCILLATOR
 In the Colpitts oscillator shown in figure 5.23, Z1 and Z2 are capacitors and Z3 is an
inductor. The resistors R1, R2 and RE provide the necessary d.c. bias to the transistor.
 CE is a bypass capacitor. CC1 and CC2 are coupling capacitors. The feedback network
consisting of capacitors C1 and C2 and an inductor L determines the frequency of the oscillator.
 When the supply voltage +Vcc is switched ON, a transient current is produced in the tank
circuit and consequently, damped harmonic oscillations are set up in the circuit.
 The oscillatory current in the tank circuit produces a.c. voltages across C1 and C2. As
terminal 3 is earthed, it is at zero potential. If terminal 1 is at a positive potential with respect
to 3 at any instant, terminal 2 will be at a negative potential with respect to 3 at the same
instant.

TECHNOLOGY,TRICHY
4.23
 Thus, the phase difference between the terminals 1 and 2 is always 180°. In the CE mode, the
transistor provides the phase difference of 180° between the input and output. Therefore, the
total phase shift is 360°.
 Thus, at the frequency determined for the tank circuit, the necessary condition for sustained
oscillations is satisfied. If the feedback is adjusted so that the loop gain Aβ= 1, the circuit
acts as an oscillator.
Colpitts oscillator
The frequency of oscillation is
𝑓0 =
1
2𝜋√𝐿𝐶
where
1
𝐶
=
1
𝐶1
+
1
𝐶2
, 𝑖. 𝑒. , 𝐶 =
𝐶1𝐶2
𝐶1+𝐶2
It is widely used in commercial signal generators for frequencies between 1MHz and 500
MHz. It is also used as a local oscillator in super heterodyne radio receiver.
Analysis:
For this oscillator,
𝑍1 =
1
𝑗𝜔𝐶1
= −
𝑗
𝜔𝐶1
𝑍2 =
1
𝑗𝜔𝐶2
= −
𝑗
𝜔𝐶2
𝑍3 = 𝑗𝜔𝐿
Substituting these values in equation 5.37 and simplifying, we get

TECHNOLOGY,TRICHY
4.24
𝑗ℎ𝑖𝑒 (
1
𝜔𝐶1
+
1
𝜔𝐶2
− 𝜔𝐿) + (
1 + ℎ𝑓𝑒
𝜔2𝐶1𝐶2
−
𝐿
𝐶1
) = 0
The frequency of oscillation, 𝑓0 =
𝜔0
2𝜋
, is found by equating the imaginary part of equation
5.42 to zero. Thus, we get
𝑓0 =
𝜔0
2𝜋
=
1
2𝜋
√
𝐶1+𝐶2
𝐿𝐶1𝐶2
Substituting equation 5.43 into equation 5.42 and simplifying, we get the condition for
maintenance of oscillation as
ℎ𝑓𝑒 =
𝐶2
𝐶1
CRYSTAL OSCILLATORS
Shows a crystal-controlled oscillator circuit. Here, it is a Colpitts crystal oscillator in which the
inductor is replaced by the crystal. In this type, a piezo-electric crystal, usually quartz, is used as
a resonant circuit replacing an LC circuit.
The crystal is a thin slice of piezo-electric material, such as quartz, tourmaline and
rochelle salt, which exhibit a property called Piezo-electric effect. The piezo-electric effect
represents the characteristics that the crystal reacts to any mechanical stress by producing an
electric charge; in the reverse effect, an electric field results in mechanical strain.
Colpitts crystal oscillator
Quartz Crystal Construction
 In order to obtain high degree of frequency stability, crystal oscillators are essentially
used. Generally, the crystal is a ground wafer of translucent quartz or tourmaline stone placed
between two metal plates and housed in a stamp sized package.
 There are two different methods of cutting this crystal wafer from the crude quartz. The
method of cutting determines the natural resonant frequency and temperature coefficient of the

TECHNOLOGY,TRICHY
4.25
crystal. When the wafer is cut in such a way that its flat surfaces are perpendicular to its
electrical axis (X-axis), it is called an X-cut crystal as shown in figure.
 When the wafer is cut in such a way that its flat surfaces are perpendicular to its
mechanical axis (Y-axis), it is called Y-cut crystal as shown in figure .
 If an alternating voltage is applied, then the crystal wafer is set into vibration. The
frequency of vibration equal to the resonant frequency of the crystal is determined by its
structural characteristics.
 If the frequency of the applied a.c. voltage is equal to the natural resonant frequency of
the crystal, then the maximum amplitude of vibration will be obtained. In general, the
frequency of vibration is inversely proportional to the thickness of the crystal.
(a) Quartz crystal (b) X-cut (c) Y-cut
Piezoelectric crystal
The frequency of vibration is
𝑓 =
𝑃
2𝑙
√
𝑌
𝜌
where Y is the Young modulus, p is the density of the material and P = 1, 2, 3, ...
The crystal is suitably cut and polished to vibrate at a certain frequency and mounted
between two metal plates as shown in figure. The equivalent circuit of the crystal is shown in
figure. The ratio of Cp to Cs may be several hundred or more so that series resonance frequency
is very close to parallel resonant frequency. The resonant frequency is inversely proportional to
the thickness of the crystal. Resonant frequencies from 0.5 to 30 MHz can be obtained.
The reactance function shown in figure
𝑗𝑋 =
1
𝑗𝜔𝐶𝑝
.
𝜔2
− 𝜔𝑠
2
𝜔2 − 𝜔𝑝
2
neglecting R. Here 𝜔𝑠
2
=
1
𝐿𝐶𝑠
is the serious resonant frequency and

TECHNOLOGY,TRICHY
4.26
𝜔𝑝
2
=
1
𝐿(
1
𝐶𝑠
+
1
𝐶𝑝
)
is the parallel resonant frequency. Since Cp>> Cs, ωp = ωs. For ωs<ω<ωp, the reactance is
inductive and for ω out of the above range, it is capacitive.
A piezo electric crystal: (a) Symbol (b) Electrical equivalent circuit and (c) The reactance
function when R=0
For crystal Hartley oscillator, the capacitors C1 and C2 shown in figure 5.28 are replaced
with inductors L1 and L2 respectively, so that the reactance of the crystal is capacitive. Hence, its
oscillation frequency is
1
2𝜋√(𝐿1+𝐿2)𝐶
.
The advantage of the crystal is its very high Q as a resonant circuit, which results in good
frequency stability for the oscillator. However, since the resonant frequencies of the crystals are
temperature dependent, it is necessary to enclose the crystal in a temperature-controlled oven to
achieve the frequency stability of the order of 1 part in 1010
.
Staff in charge HoD

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