UNIT-IV FEEDBACK AMPLIFIERS AND OSCILLATORS

PART-A
FEEDBACK AMPLIFIERS

1. Define feedback and feedback factor. Define Positive feedback and Negative feedback.

Feedback: The process of injecting a fraction of the output voltage of an amplifier into the input so that it becomes a
part of the input is known as feedback.

Feedback Factor: Feedback factor is defined as the ratio of feedback signal (Voltage/Current) to the amplifier output
which is given
𝑉 as input to the feedback network. Hence, it is also called as feedback ratio and is denoted by β.
i.e., 𝛽 = 𝑓 ; 𝑉 − 𝐹𝑒𝑒𝑑𝑏𝑎𝑐𝑘 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑉 − 𝐴𝑚𝑝𝑙𝑖𝑓𝑖𝑒𝑟 𝑂𝑢𝑡𝑝𝑢𝑡 𝑉𝑜𝑙𝑡𝑎𝑔𝑒
𝑉𝑜 𝑓 𝑂

Positive feedback: If the feedback voltage is in-phase to the input from the source, i.e., feedback signal in-phase with
the original input signal. It is called positive feedback.

Negative feedback: If the feedback voltage is opposite (out of phase) to the input from the source, i.e., feedback signal
opposes the original input signal. It is called negative or degenerative feedback.

Advantages of negative feedback

2. Mention/List the advantages of negative feedback circuits. (Nov/Dec2015), (May/June2016)

➢ In negative feedback amplifiers, the voltage gain of the amplifier remains stable.
➢ High input resistance of a voltage amplifier can be made larger
➢ Low output resistance of a voltage amplified can be lowered
➢ Frequency response improves
➢ Significant improvement in the linearity of operation
➢ The transfer gain of the amplifier with feedback can be stabilized against variation in the h parameters.

3. Write the disadvantages of negative feedback in amplifier circuits and how it can be overcome? (April/May 2015)
The main disadvantage of using negative or degenerative feedback in amplifier is Reduction in Gain.
The required Gain can be attained by increasing the number of amplifier stages

4. What are the effects of a negative feedback?

a) Reduces noise
b) Reduces distortion
c) Reduces gain
d) Increases band width
e) The gain becomes stabilized with respect to changes in the amplifier active device parameters like hfe.
f) The non-linear distortion is reduced there by increasing the signal handling capacity or the dynamic range of the
amplifier.

5. What is the condition required for satisfactory operation of a negative feedback amplifier? (April/May 2019)
The open-loop voltage gain must be much greater than the required closed-loop gain.
𝐴𝑣
Overall Voltage Gain with -ve feedback (Closed-loop Gain), 𝐴𝑣𝑓 = 1+𝛽𝐴𝑣
𝐴𝑣
𝐴𝑣𝑓 = {𝑆𝑖𝑛𝑐𝑒, 𝛽𝐴𝑣 ≫ 1}
𝛽𝐴𝑣
1
Therefore, 𝐴𝑣𝑓 = 𝛽
{Where 𝐴𝑣 is the voltage gain without a feedback and β is the feedback factor is due to negative feedback the gain is reduced
by factor 1 + 𝛽𝐴𝑣}
6. With negative feedback the bandwidth of the amplifier increases- True/False?
True.
Bandwidth of amplifier with feedback is greater than bandwidth of amplifier without feedback.

Voltage / current, Series, Shunt feedback

7. Mention the four connections in Feedback.

a. Voltage series feedback.
b. Voltage shunts feedback.
c. Current series feedback.
d. Current shunt feedback.

8. Explain the voltage series feedback.

In this case, the feedback voltage is derived from the output voltage and fed in series with input signal. The input of the
amplifier and the feedback network are in series is also known as series parallel in parallel, hence this configuration is
also known as series parallel feedback network.

9. Explain the voltage shunt feedback.

The input of amplifier and the feedback network are in parallel and known as parallel –parallel feedback network.
This type of feedback to the ideal current to voltage converter, a circulating having very low input impedance and
very low output impedance.

10. Explain the current series feedback.

When the feedback voltage derived from the load current and is fed in series with the input signal, the feedback is said
to be current series feedback, the inputs of the amplifier and the feedback network are in series and the output are also
in series. This configuration is also called as series-series feedback configuration.

11. Explain the current shunt feedback.

When the feedback voltage is derived from the load current and a fed in parallel with the input signal, the feedback is
said to be current shunt feedback. Here in the inputs of the amplifier and the feedback network are in parallel and the
outputs are in series. This configuration is also known as parallel series feedback.

12. Which is the most commonly used feedback arrangement in cascaded amplifier and why? (Nov/Dec-2013-R13)
A voltage series feedback s commonly used in cascaded amplifiers. Since, it has high input impedance and low output
impedance that is needed for cascaded amplifiers.

Positive feedback (Oscillators)

13. What is Oscillator?
Oscillator is an electronic device which generates electrical oscillations (i.e., repeated waveforms) of required frequency.
It is used for converting DC energy into AC energy of the desired frequency.
{An oscillator is a circuit which generates an alternating voltage without any input signal. Instead of external input
signal, it uses feedback path through which it provides its own input signal.
It is used for converting DC energy into AC energy of the desired frequency.}

14. What are sustained Oscillations?

Electrical oscillations in which amplitude does not change with time are called sustained oscillations. It is called as un-
damped oscillations.

15. What is frequency of Oscillations?

The frequency at which circuit satisfies both the Barkhausen conditions i.e. |𝐴𝛽| = 1 and ∠𝐴𝛽 = 0° or 360°
simultaneously is called frequency of oscillations

16. Classify the various oscillators based on the output waveforms, circuit components, operating frequencies and
feedback used.
According to the nature of waveform generated.
1. Sinusoidal or Harmonic Oscillators
2. Non-sinusoidal or Relaxation oscillators
 Based on circuit components. (Nov/Dec 2017)
According to the frequency determining networks,
1. RC oscillators (Phase-shift Oscillator and Wien Bridge Oscillator)
2. LC oscillators (Hartley Oscillator and Colpitts Oscillator)
3. Crystal oscillators

According to the frequency of the Generated Signals

1. AFO (Audio Frequency Oscillators) – upto 20 KHz

2. RFO (Radio Frequency Oscillators) – 20 KHz to 30 MHz
3. VHFO (Very High Frequency Oscillators) - 30 MHz to 300 MHz
4. UHFO (Ultra High Frequency Oscillators) - 300 MHz to 3 GHz
5. MFO (Microwave Frequency Oscillators) – above 3 GHz

17. What are the types of sinusoidal oscillator? [or] Mention the different types of sinusoidal oscillator?
a) RC phase shift Oscillator.
b) Wein bridge Oscillator.
c) Hartley Oscillator
d) Colpitts Oscillator
e) Crystal Oscillator

18. Name two low frequency oscillators?

a) RC phase shift oscillator.
b) Wein bridge oscillator.

19. Name three high frequency oscillators?

The high frequency oscillators are
a) Hartley oscillator.
b) Colpitts oscillator.
c) Crystal oscillator

Condition for oscillations

20. Write the conditions for a Oscillator. (OR)

State. Barkhausen criterion (Barkhausen condition) for sustained oscillations. (Nov/Dec-2012,2011,09),
(May/June2016) (Nov/Dec-2016) (May 2017)
The Barkhausen criterion for obtaining sustained oscillations,
1. The feedback voltage must be in-phase with the input, i.e., total phase-shift around the closed-loop must be 0° or
360°, and
2. Magnitude of the loop gain must be unity i.e., |𝐴𝛽| = 1
Where, A – Open loop Gain of the system & β – Feedback ratio.

Phase Shift and Wien bridge oscillator (RC oscillators)

21. Why an RC phase shift oscillator is called so?

An RC network products 180o phase shift. Hence it is called RC phase shift oscillator.

22. List the advantages of phase shift oscillator. (May/June-2012)

• The phase shift oscillator does not required conductance or transformers.
• It is suitable for the low frequency range i.e., from a few hertz to several 100 kHz. The upper frequency is limited
because the impedance of RC network may become so small that it loads the amplifier heavily.

23. Write the disadvantages of Phase shift oscillator.

1. It is necessary to change the C or R in all the three RC networks simultaneously for changing the frequency of
oscillations. This is practically difficult.
2. It is not suitable for high frequencies.
24. Which oscillator uses both positive and negative feedback?
Wien bridge oscillator.

Hartley and Colpitts oscillators. (LC oscillators)

25. Distinguish between LC and RC oscillator.

LC Oscillator RC Oscillator

It operates at high frequencies It operates at low frequencies

It is suitable for RF only It is suitable for AF only
The frequency is constant.
Frequency is variable It is known as fixed frequency oscillator.

26. Write the main drawback of LC oscillators.

1. The frequency stability is not very good.
2. They are too bulky and expensive and cannot be used to generate low frequencies.

27. What is the advantage of a colpitts oscillator compared to a phase shift oscillator? (Nov/Dec 2015)
ii) The advantage of colpitts oscillator is the frequency of oscillation is very high.
iii) We can vary the frequency of oscillation.

Crystal oscillators.

28. What is piezo electric effect? (May/June-2013)

The piezo electric crystal exhibits a property, that is, if a mechanical stress is applied across one face, an electrical potential is
developed across the opposite face. The inverse is also true. This phenomenon is called piezo-electric effect.

29. Why Quartz crystal is commonly used in crystal oscillator?

Quartz crystals are generally used in crystal oscillator because of their great mechanical strength, simplicity of manufacture and
abeyance to the piezo electric effect accurately.

30. What are the advantages of crystal oscillators? (NOV/DEC 2012)

The advantages of crystal oscillators are
a) Excellent frequency stability.
b) High frequency of operation
c) Automatic amplitude control.
d) It is suitable for only low power circuits
e) Large amplitude of vibrations may crack the crystal.
f) It large in frequency is only possible replacing the crystal with another one by different frequency.

31. An oscillator operating at 1 MHz has a stability of 1 in 104. What will be the minimum value of frequency generated?
(April/May 2019)
The typical frequency stability of oscillators that do not use CRYSTAL is about 1 in 104.
The minimum value of frequency generated might be 100KHZ or lower than 1MHZ for the oscillator
operating at 1MHZ.
{If the crystal is used, the frequency stability can be improved to better than 1 in 106, which gives a ±1 Hz
variation in the output of a 1 MHz oscillator.}

32. How does an oscillator differ from an amplifier? (or) Differentiate oscillator & amplifier. [Nov/Dec 2013] [Nov/Dec 2016]

S.No. Oscillators Amplifiers

1 They are self-generating circuits. They are not self-generating circuits.
They generate waveforms like sine, square and triangular They need a signal at the input and they just increase
waveforms of their own without having input signal. the level of the input waveform.
2 It has infinite gain It has finite gain
3 Oscillator uses positive feedback. Amplifier uses negative feedback.
33. Compare RC Phase-Shift Oscillators and Wien Bridge Oscillator.

34. Classification of Oscillators

 PART-B
Advantages of negative feedback & positive feedback

1. What is meant by feedback? What are the types of feedback and effects of negative feedback?
(May/June-2012) (Nov/Dec 2017)

Negative feedback
If β is negative, the voltage feedback subtracts from the input yielding a lower output and reduced voltage
gain. Hence this feedback is known as negative feedback.

Positive feedback
If the phase of the voltage feedback is such as to increase the input, then β is positive and the result is
positive feedback.

Increase Stability:
The voltage gain due to a negative feedback is given by

𝐴𝑣
𝐴𝑣𝑓 = … … … … … . . (1)
1+𝛽𝐴𝑣

Where 𝐴𝑣 is the voltage gain without a feedback and β is the feedback factor is due to negative feedback
the gain is reduced by factor 1 + 𝛽𝐴𝑣

If 𝛽𝐴 >> 1 then 𝐴 𝐴𝑣 1
𝑣 𝑣𝑓 = =
𝛽𝐴𝑣 𝛽
Hence the gain of the amplifier with feedback has been stabilized against such problems as ageing of a transistor or
a transistor being re-placed by a transistor with a different value of β.

Sensitivity of transfer gain:

The fractional change in amplification with feedback divided by the fractional change without feedback is
called the sensitivity of the transfer gain

From equ 1 (1+𝛽𝐴𝑣)−𝐴𝑣𝛽 1

𝑑𝐴𝑣𝑓 = =
𝑑𝐴𝑣 (1+𝛽𝐴𝑣).2 (1+𝛽𝐴𝑣).2

𝑑𝐴𝑣𝑓 1
=
𝑑𝐴𝑣 ((1+𝛽𝐴𝑣)2)

𝑑𝐴𝑣
𝑑𝐴𝑣𝑓 =
(1+𝛽𝐴𝑣)2

Dividing both side by 𝐴𝑣𝑓

𝑑𝐴𝑣𝑓 𝑑𝐴𝑣
𝐴𝑣𝑓
=
((1+𝛽𝐴𝑣)2).𝐴𝑣𝑓
 𝐴𝑣
Instead of 𝐴𝑣𝑓 𝑠𝑢𝑏 in above equation
1+𝛽𝐴𝑣

𝑑𝐴𝑣𝑓 𝑑𝐴𝑣
= 𝐴
((1+𝛽𝐴 )2).( 𝑣 )
𝐴𝑣𝑓 𝑣 1+𝛽𝐴𝑣

𝑑𝐴𝑣
=
𝐴𝑣(1+𝛽𝐴𝑣)

Taking absolute value of the resultant equation we get

𝑑𝐴𝑣𝑓 1 𝑑𝐴𝑣
= | | … … … … … . .3
𝐴𝑣𝑓 |1+𝛽𝐴𝑣| 𝐴𝑣

|𝑑𝐴𝑣𝑓|
𝐴𝑣𝑓 1
Sensitivity= 𝑑𝐴 = |1+𝛽𝐴 4
| 𝑣| 𝑣|...............................
𝐴𝑣

The densitivity is reciprocal of sensitivity. Hence

𝐷 = 1 + 𝐴𝑣β......................... 5
Frequency distortion
From equ 1 we find that for a negative feedback amplifier having 𝐴𝑣β>> 1 the gain withfeedback is
𝐴𝑣𝑓=1/β. If the feedback network does not contain any reactive elements the gain is not function of frequency.
Reduction in noise
There are many sources of noise is an amplifier. If the noise present at the output is N and the amplifier
gain is A. then the noise present in the amplifier with negative feedback is

𝑁
N1 = .
1+𝛽𝐴𝑣

Reduction in distortion
Let us assume that the distortion in the absence of feedback is D. Because the effect of feedback the
distortion present at the input is equal to

𝐷
𝐷𝑓 = 1+𝛽𝐴
𝑣

Bandwidth
If the bandwidth of an amplifier without feedback is given by

Bwf=BW(1+β𝐴𝑣)

In curve a source the frequency response of an amplifier without feedback when a negative feedback is
introduced the gain of the amplifier decreases.
 Frequency response of an amplifier with and without feedback

Obtain curve C. from fig we can observe that there is decrease in the lower cutoff frequency and increase in upper
cutoff frequency hence the bandwidth increases. Therefore β increases Bandwidth also increases Loop Gain
A loop gain is used to describe the product of voltage gain 𝐴𝑣 and feedback factor β. The amount of
feedback introduced into an amplifier may be expressed in decibels according to the following definition.
F=feedback in db
𝐴𝑣𝑓
= 20 log
𝐴𝑣
1
= 20 log
1 + 𝛽𝐴𝑣

2. Advantages of Negative feedback in amplifiers. (Nov/Dec 2018)

The advantages of negative feedback in amplifiers are listed as follows.
1. The negative feedback amplifiers, the voltage gain of an amplifier remains stable.
2. It reduces the non-linear distortion produced in large signal amplifiers.
3. It improves the frequency response of the amplifier.
4. It increases the stability of the circuit.
5. Negative feedback increases the input impedance and decreases the output impedance of the amplifier.
6. It decreases the noise voltage in the amplifier.
7. Negative feedback amplifier is less sensitive to variations in amplifier parameters.
8. It increases the amplifier bandwidth.
9. The input and output impedances of feedback amplifier can be adjusted to desired value.
10. It has less phase, amplitude and frequency distortion.
11. Amplifier with negative feedback operates linearly.
12. Operating point of amplifier can be stabilized.
3. With proper mathematical derivation, proven that bandwidth increases in a negative feedback amplifier.
(April/May 2019)

The negative feedback increases amplifier bandwidth which can be proven mathematically as
below
TYPES OF NEGATIVE FEEDBACK AMPLIFIER

4. Explain the various types of feedback amplifier (May 2017)

(OR)
With a neat block diagram, explain the operation of Current Shunt Feedback Amplifier.
(OR)
Determine Rif, Rof, Av, Avf for the following feedback amplifier
A. Voltage series feedback amplifier (Series-Shunt feedback amplifier) (Nov/Dec 2016)
(May 2017)
B. Current Series Feedback Amplifier (Shunt-Series feedback amplifier)
C. Current Shunt Feedback Amplifier (Series-Series feedback amplifier) (May 2017)
D. Voltage Shunt Feedback Amplifier (Shunt-Shunt feedback amplifier)
(OR)
Discuss the effect of voltage series feedback and derive the expression for input resistance, output
resistance and voltage gain.
(OR)
Discuss about the following feedback configurations of amplifiers and obtain the feedback factor and
closed loop gain. (April/May 2018-R13)
A. Shunt-Shunt Feed Back
B. Series-Series Feed Back
C. Shunt-Series Feed Back
D. Series-Shunt Feed Back

Feedback amplifier, the output signal sampled may be either voltage or current and sampled signal can be
mixed either is series or in shunt with the input

The four types of amplifiers, they are

➢ Voltage series feedback amplifier (Series-Shunt feedback amplifier) (Nov/Dec 2016) (May 2017)
➢ Current Series Feedback Amplifier (Shunt-Series feedback amplifier)
➢ Current Shunt Feedback Amplifier (Series-Series feedback amplifier) (May 2017)
➢ Voltage Shunt Feedback Amplifier (Shunt-Shunt feedback amplifier)

(A) VOLTAGE SERIES AMPLIFIER:

With proper mathematical derivation, proven that output resistance reduces in a negative feedback
amplifier. Assume a series shunt feedback scheme. (April/May 2019)

• Ri – input resistance
• Rs – source resistance
• RL – load resistance
• RO – output resistance
• AV – voltage gain
• Ri >> RS then Vi = Vs
• RL >> Ro then Vo = AVVi = Av Vs
• Amplifier provides a voltage output proportional to the voltage input
• The proportionality factor does not depend on magnitudes of the source an load resistance
• Hence it is called voltage amplifier

Feedback Topology

Input resistance

Step 1: equivalent circuit

Step 2: obtain expression for VS

Applying KVL to the input side we get,
V S – Ii – V f = 0 ⸫ Vs = Ii Ri + Vf = Ii Ri + βVo
⸫ Vf = βVo

Step 3: obtain expression for Vo in terms of Ii

The output voltage Vo is given as
𝐴𝑉 𝑉𝑖𝑅𝐿 𝐴𝑣𝑅𝐿
𝑉𝑜 = = 𝐴𝑉 𝑉𝑖 𝑤ℎ𝑒𝑟𝑒, 𝐴𝑣 =
𝑅𝑂 + 𝑅𝐿 𝑅𝑜 + 𝑅𝐿
Vo = Av Ii Ri ⸫ Vi = Ii Ri
 Step 4: obtain expression for Rif
Substituting value of V0 from above equation we get
VS = Ii Ri + βAv Ii Ri ⸫ Rif = Vs / Ii = Ri + βAv Ri
Rif = Ri (1+βAv)

Output Resistance
Step 1: Equivalent circuit

Step 2: obtain expression for I in terms of V

Applying KVL to the output side we get
𝑉−𝐴𝑉𝑉𝑖
Av Vi + IR0 –V = 0 ⸫𝐼=
𝑅0

The input voltage is given as

Vi = - Vf = -βV ⸫Vs = 0
Substituting the Vi from above equation we get
𝑉 + 𝐴𝑣𝛽𝑉 = 𝑉(1 + 𝛽𝐴𝑉)
𝐼=
𝑅0 𝑅0
Step 3: obtain expression for Rof
𝑉 𝑅𝑜
𝑅𝑜𝑓 = 𝑅𝑜𝑓 =
𝐼 (1 + 𝛽𝐴𝑣)

Step 4: obtain expression for R of’

( 𝑅𝑂 ) 𝑋 𝑅
𝑅𝑂𝑓 𝑋 𝑅𝐿 1+𝛽𝐴 𝐿
R’of = Rof || RL = 𝑅𝑂𝑓 + 𝑅𝐿 = 𝑅𝑂 𝑉
+ 𝑅𝐿
(1+𝛽𝐴 ) 𝑉

RORL RORL
= =
RO+RL(1+βAV) RO+RL+βAVRL

Dividing numerator and denominator by (Ro + RL)

 𝑅 𝑜𝑅 𝐿
𝑅′ = 𝑅𝑜 + 𝑅𝐿 ⸫𝑅′ =
𝑅 𝑜𝑅 𝐿 𝑎𝑛𝑑 𝐴 = 𝐴𝑣𝑅𝐿

𝑜𝑓 𝛽𝐴𝑣𝑅𝐿
1+
𝑜 𝑅𝑜 + 𝑅𝐿 𝑣 𝑅𝑜 + 𝑅𝐿
𝑅𝑜 + 𝑅𝐿
′
′ 𝑅𝑜
𝑅𝑜𝑓 =
1 + 𝛽𝐴𝑣

(B)CURRENT SERIES AMPLIFIER:

• Ri – input resistance
• Rs – source resistance
• RL – load resistance
• RO – output resistance
• AI – current gain

• Rs >> Ri and Ii = Is
• Ro >> RL IL = AI Ii
• Amplifier provides a current output proportional to the current input
• The proportionality factor does not independent on source and load resistance
• Hence it is called current amplifier
Feedback Topology

Input Resistance
Step 1: equivalent circuit
 Step 2: obtain expression for VS
Applying KVL to the input side we get,
VS – Ii Ri – Vf = 0 ⸫ Vs = Ii Ri + Vf = Ii Ri + βIo
⸫ Vf = βIo

Step 3: obtain expression for Io in terms of Vi

The output current Io is given by

𝐺𝑚𝑉𝑖𝑅𝑜
𝐼 = = 𝐺 𝑉 where 𝐺 = 𝐺𝑚𝑅𝑜
𝑜 𝑅𝑜+𝑅𝐿 𝑀 𝑖 𝑀 𝑅𝑜+𝑅𝐿

Step 4: obtain expression for Rif

Substituting value of Io from above equation

VS = Ii Ri + β GM Vi = Ii Ri + β GM Ii Ri {Since, Vi = Ii Ri}

Rif = Vs / Ii = Ri (1+ β GM)

Output Resistance

Step 1: equivalent circuit

Step 2: obtain expression for I in terms of V

Applying KVL to the output node we get

𝑉
𝐼= − 𝐺𝑚𝑉𝑖
𝑅𝑜

The input voltage is given as Vi = -Vf = - β Io = β I ⸫ Io = - I

Substituting value of Vi from above equation we get

𝑉 𝑉
𝐼= − 𝐺𝑚 β I = 𝐼 + 𝐺𝑚βI = I(1 + 𝐺𝑚β)
𝑅𝑜 𝑅𝑜
 Step 3: obtain expression for Rof
𝑉
𝑅𝑜𝑓 = = 𝑅𝑜(1 + 𝐺𝑚β)
𝐼
𝑅′ = 𝑅 ||𝑅 = 𝑅𝑜𝑓 𝑋 𝑅𝐿
𝑜𝑓 𝑜𝑓 𝐿
𝑅𝑜𝑓 + 𝑅𝐿

𝑅𝑜(1 + β𝐺𝑚)𝑅𝐿 𝑅𝑜𝑅𝐿(1 + β𝐺𝑚)

= =
𝑅𝑜 (1 + ) + 𝑅𝐿 𝑅𝑜 + 𝑅𝐿 + β𝐺𝑚 𝑅𝑜
β𝐺𝑚

Dividing numerator and denominator by Ro + RL we get

𝑅𝐿𝑅𝑜(1 + β𝐺𝑚)
𝑅𝑜 + 𝑅𝐿
𝑅𝑓𝑜 =
β𝐺𝑚𝑅𝑜
1 + 𝑅𝑜 + 𝑅𝐿
𝑅𝑜′ (1 + β𝐺 ) 𝑅 𝑅 𝐺 𝑅
𝑅′ = 𝑚 ⸫𝑅′ = 𝑜 𝐿 𝑎𝑛𝑑 𝐺 = 𝑚 𝑜
𝑜𝑓
1+ 𝑜 𝑅𝑜 + 𝑅𝐿 𝑀 𝑅 𝑜 + 𝑅𝐿
β𝐺𝑚

(C) VOLTAGE SHUNT AMPLIFIER

• Ri << Rs and Ro << Rs ’

• Since Ri << Rs’
• Ii = Is and Ro << RL ,
Vo = Rm Is
• Where Rm = Vo / Is is
the transfer or mutual
resistance

Feedback Topology
Input Resistance

Step 1: Equivalent Circuit

Step 2: obtain expression for Is

Applying KCL at input node we get

Is = Ii + If = Ii + β Vo ⸫ If = β V o

Step 3: obtain expression for Rif

The output voltage Vo is given by

𝑅𝑚𝐼𝑖𝑅𝑜
𝑉 = = 𝑅 𝐼 where 𝑅 = 𝑅𝑚𝑅𝑜
𝑜 𝑅𝑜+𝑅𝐿 𝑀 𝑖 𝑀 𝑅𝑜+𝑅𝐿

Step 4: obtain expression for Rif

Substituting value of Vo from above equation we get

Is = Ii + β RM Ii = Ii (1+ β RM)

The input resistance with feedback Rif is given by

𝑅𝑖𝑓 =
𝑉𝑖
=
𝑉𝑖 ⸫ 𝑅 = 𝑉𝑖
𝐼𝑠 𝐼𝑖(1+β𝑅𝑀) 𝑖 𝐼𝑖

𝑅𝑖
⸫ 𝑅𝑖𝑓 =
(1+β𝑅𝑀)
Output Resistance

Step 1: Equivalent Circuit

Step 2: obtain expression for I in terms of V

Applying KVL to the output side we get

𝑉−𝑅𝑚𝐼𝑖
Rm Ii + I Ro – V =0 ⸫𝐼 =
𝑅𝑜

The input current is given as

Ii = - If = - β V

Substituting Ii in above equation we get

𝑉+𝑅𝑚 β V 𝑉(1+𝑅𝑚β)
𝐼= =
𝑅𝑜 𝑅𝑜

Step 4: obtain expression for 𝑹𝒐𝒇

′

𝑅 𝑜 𝑋 𝑅𝐿
𝑅′ = 𝑅 ||𝑅 = 𝑅𝑜𝑓 𝑋 𝑅𝐿 = 1 + 𝑅𝑚β 𝑅𝑜𝑅𝐿
𝑜𝑓 𝑜𝑓 𝐿 𝑅𝑜 =
𝑅𝑜𝑓 + 𝑅𝐿 + 𝑅𝐿 𝑅𝑜 + 𝑅𝐿(1 + 𝑅𝑚β)
1 + 𝑅𝑚β
Dividing numerator and denominator by (Ro + RL) we get

𝑅 𝑜𝑅 𝐿
′ 𝑅 𝑜 + 𝑅𝐿
𝑅𝑓𝑜 =
β𝑅 𝑅
1+ 𝑚 𝐿
𝑅𝑜 + 𝑅𝐿
𝑅𝘍𝑜 𝑅𝐿 𝑋 𝑅𝑜𝑓 𝑅 𝑅
′
𝑅𝑓𝑜 = where 𝑅′ = 𝑎𝑛𝑑 𝑅 = 𝑚 𝐿
1+β𝑅𝑀 𝑜 𝑀 (𝑅𝑜+𝑅𝐿)
𝑅𝐿+𝑅𝑜𝑓
(D) CURRENT SHUNT AMPLIFIER:

• Ri << Rs and Ro <<

Rs ’
• Since Ri << Rs’
• Ii = Is and Ro <<
RL , Vo = Rm Is

Feedback Topology

Input Resistance
Step 1: Equivalent Circuit

Step 2: obtain expression for Is

Applying KCL to the input node we get

Is = Ii + If = Ii + β Io ⸫ If = β Io

Step 3: obtain expression for Io in terms of Ii

𝐼 =
𝐴𝑖𝐼𝑖𝑅𝑜 = 𝐴𝐼 where 𝐴 = 𝐴𝑖𝑅𝑜
𝑜 𝑅𝑜+𝑅𝐿 𝐼 𝑖 𝐼 𝑅𝑜+𝑅𝐿

Step 4: obtain expression for Rif

Substituting value of Io in above equation we get

Is = Ii + β AI Ii = Ii (1+ β AI)

The input resistance with feedback is given as

 𝑉𝑖 𝑉𝑖
𝑅 = =
𝑖𝑓
𝐼𝑠 𝐼𝑖(1 + β 𝐴I)
𝑅𝑖
𝑅𝑖𝑓 =
(1 + β𝐴𝐼)
Output Resistance
Step 1: Equivalent Circuit

Step 2: obtain expression for I in terms of V

Applying KCL to the output node we get

𝑉
𝐼= − 𝐴𝑖𝐼𝑖
𝑅𝑜

The input current is given as

Ii = - If = - β Io ⸫ Is = 0

Ii = β I ⸫ I = - Io

Substituting value of Ii in above equation we get

𝑉
𝐼= −𝐴 βI ⸫ 𝑉 = 𝐼 + 𝐴 β = I (1 + β𝐴 )
𝑅𝑜 𝑖 𝑅𝑜 𝑖 𝑖

Step 3: obtain expression for Rof

𝑅′ = 𝑅 ||𝑅 = 𝑅𝑜𝑓 𝑋 𝑅𝐿
𝑜𝑓 𝑜𝑓 𝐿
𝑅𝑜𝑓 + 𝑅𝐿

𝑅𝑜(1 + β𝐴𝑖)𝑅𝐿 𝑅𝑜𝑅𝐿(1 + β𝐴𝑖)

= ⸫ =
𝑅𝑜(1 + β𝐴𝑖) + 𝑅𝐿 𝑅𝑜 + 𝑅𝐿 + β𝐴𝑖𝑅𝑜

Dividing numerator and denominator by (Ro + RL) we get

 𝑅𝑜𝑅𝐿(1 + β𝐴𝑖)
′ 𝑅𝑜 + 𝑅𝐿
𝑅𝑜𝑓 =
β𝐴𝑖𝑅𝑜
1+
𝑅𝑜 + 𝑅𝐿
𝑅′ (1 + β𝐴 )
′ 𝑜 𝑖
𝑅𝑜𝑓 =
(1 + β𝐴𝐼)

𝑅 ′ 𝑅𝑜𝑅𝐿 𝑎𝑛𝑑 𝐴 = 𝐴𝑖𝑅𝑜

𝑜 = 𝑅𝑜+𝑅𝐿 𝐼 𝑅𝑜+𝑅𝐿

OSCILLATORS:

5. Explain the construction and working of the following oscillators and derive the expression for
frequency of oscillation. Also, write about advantages and disadvantages.
A. Phase-Shift Oscillator (RC type Oscillator)
B. Wein Bridge Oscillator (RC type Oscillator)
C. Hartley Oscillator (LC type Oscillator)
D. Colpitts Oscillator (LC type Oscillator)
E. Crystal Oscillator

(A) RC Phase Shift Oscillator:

Explain the construction and working of RC Phase-Shift oscillator and derive the expression for frequency
of oscillation.

• It consists of an amplifier and feedback network consisting of resistors and capacitors.

• An amplifier can be BJT, FET or operational amplifier.

Analysis of RC circuit:
• In this circuit output is taken across resistor R.

1
• The capacitive reactance XC is given by 𝑋𝐶 =
2𝜋𝑓𝐶
Ω where f is frequency of the input.
• The total impedance of the circuit is,
1
𝑍 = 𝑅 − 𝑗𝑋𝐶 = 𝑅 − 𝑗 ( ) Ω
2𝜋𝑓𝐶
 = |𝑍| < −Ф0 Ω
• The current ‘I’ flowing in the circuit is,
𝑉 𝑖 < 00 𝑉𝑖 < 0 0 𝑉𝑖
𝐼= = = | | < +Ф0 𝐴
𝑍 |𝑍| < −Ф0 𝑍

𝑋𝐶
|𝑍| =√𝑅2 + 𝑋𝐶2 𝑎𝑛𝑑 Ф = tan−1
𝑅
• In this equation the current ‘I’ leads input voltage by angle Ф
• The output voltage is drop across R hence VO=VR=IR
• The output voltage is in phase with current hence it leads input voltage by angle Ф
• Thus, RC circuit introduces a phase shift Ф between input and output which depends on R, C and frequency f.

RC Feedback Network for phase shift oscillator:

• In RC phase shift oscillator, amplifier introduces a phase shift of 1800
• Thus, the feedback network must introduce a phase shift of 1800 to satisfy Barkhausen condition.
• The RC feedback network consists of three RC sections, with each RC section contributing 600 phase-shift.
• Hence in RC phase shift oscillator, the feedback network consists of three RC sections are shown in fig.
• In all the three sections, resistance values and capacitance values are same so that at a particular frequency,
each section produces precisely 600 phase-shift. This is the operating frequency of oscillator.

Transistorized RC phase shift oscillator:

• The RC phase shift oscillator uses BJT amplifier stage which is single stage amplifier in common emitter
configuration.
• A phase shift network has three RC sections
• The output of CE amplifier is connected as input to the RC phase shifting network
• The output of RC phase shifting network is connected as input to the amplifier
• Due to common emitter amplifier it introduces a phase shift of 1800 between its input and output
• The RC phase shift network contributes further 1800 phase shift so that phase shift around a loop is 3600
 • From the fig. neglecting R1 and R2 we can write hie= input impedance of amplifier stage
• Thus, to have all three resistance values in three RC section equal, resistance in the last section is selected
as R3 so that R3+hie=R
R3 + hie = R i.e R3 = R- hie ------------- eq. 1
• If R1 and R2 are not neglected then, R3 = R- [R1 || R2 || hie] ----- eq. 2
• When gain A of the amplifier stage and feedback factor β are adjusted to give |Aβ| = 1, then the circuit
works as an oscillator, satisfying both Barkhausen condition.

Derivation for frequency of oscillation:

• Replacing the transistor by its approximate h-parameter model, the equivalent circuit of RC
phase shift oscillator is shown in fig.

• It is known that R = hie + R3 and replace current source by equivalent voltage source.
𝑅𝐶
• The ratio of resistance RC to R is K. =𝐾
𝑅
• The modified equivalent circuit is shown below
• Applying KVL to the three loops
1
𝐼1𝑅𝐶 − 𝐼 − 𝑅(𝐼1 − 𝐼2) − ℎ𝑓𝑒𝐼𝑏𝑅𝑐 = 0 𝑎𝑛𝑑 𝑢𝑠𝑒 𝑅𝐶 = 𝑘 𝑅
𝑗𝜔𝐶 1
1
⸫ 𝐼1 [𝑘𝑅 + 𝑅 + ] + 𝐼2 𝑅 = ℎ𝑓𝑒 𝐼𝑏 𝑘 𝑅 --------- eq. 3
𝑗𝜔𝐶
1 1
− 𝐼 − 𝑅(𝐼2 − 𝐼1 ) − 𝑅(𝐼2 − 𝐼3 ) = 0 𝑖. 𝑒 𝐼1 𝑅 − 𝐼2 (2𝑅 +
𝑗𝜔𝐶 2
) + 𝐼3 𝑅 = 0 ----- eq. 4
𝑗𝜔𝐶
1 1
− 𝐼 − 𝐼3 𝑅 − 𝑅(𝐼3 − 𝐼2 ) = 0 𝑖. 𝑒 𝐼2 𝑅 − 𝐼3 (2𝑅 +
𝑗𝜔𝐶 3
) = 0 ----- eq. 5
𝑗𝜔𝐶
• Using jω = s and Cramers’s rule
1
−(𝑘 + 1)𝑅 − +𝑅 0
𝑠𝐶 |
| 1
𝐷= 𝑅 −2𝑅 − 𝑠𝐶 𝑅
| |
1
0 𝑅 −2𝑅 −
𝑠𝐶
• Solving the determinant, we get,
𝑠3𝐶3𝑅3(3𝑘+1)+𝑠2𝐶2𝑅2(4𝑘+6)+𝑠𝑅𝐶(5+𝑘)+1
𝐷 = −{ } ------- eq. 6
𝑠3𝐶3

• To find I3, find D3 as,

1
−(𝑘 + 1)𝑅 − 𝑅 ℎ𝑓𝑒 𝐼𝑏𝑘𝑅
𝑠𝐶
1
𝐷3 = | 𝑅 −2𝑅 − 0 | = 𝑘𝑅3ℎ𝑓𝑒𝐼𝑏 --------------------eq. 7
𝑠𝐶
0 𝑅 0

𝐷3 −𝑘𝑅3ℎ𝑓𝑒𝐼𝑏𝑠3𝐶3
𝐼3 = = eq. 8
𝐷 𝑠3𝐶3𝑅3(3𝑘+1)+𝑠2𝐶2𝑅2(4𝑘+6)+𝑠𝑅𝐶(5+𝑘)+1------------------------

I3 = Output current of the feedback circuit

Ib = Input current of the amplifier
IC = hfe Ib = input current of the feedback circuit
 Output of the feedback circuit I3 I3
β = = =
Input to feedback circuit IC hfeIb

Output of the amplifier I3

A= = = h fe
Input to the amplifier Ib
𝐴𝛽 = ℎ 𝑋 3 = 3
𝐼 𝐼 ------- eq. 9
𝑓𝑒 ℎ𝑓𝑒𝐼𝑏 𝐼𝑏

From equation 8 and 9,

−𝑘𝑅3ℎ𝑓𝑒𝑠3𝐶3
𝐴𝛽 = 𝑠3𝐶3𝑅3(3𝑘+1)+𝑠2𝐶2𝑅2(4𝑘+6)+𝑠𝑅𝐶(5+𝑘)+1
− − − − − − 𝒆𝒒. 𝟏𝟎

Using s = jω s2 = j2 ω2 = -ω2, s3 = j3 ω3 = -jω3 and separating the real and imaginary

part we get,
+𝑗𝜔3𝑘𝑅3𝐶3ℎ𝑓𝑒
𝐴𝛽 =
[1 − 4𝑘𝜔2𝐶2𝑅2 − 6𝜔2𝐶2𝑅2] − 𝑗𝜔[3𝑘𝜔2𝑅3𝐶3 + 𝜔 2𝑅 3𝐶 3 − 5𝑅𝐶 − 𝑘𝑅𝐶

Dividing numerator and denominator by j ω3 R3 C3 and replacing -1/j = +j

𝑘ℎ𝑓𝑒
𝐴𝛽 = 1 4𝑘 6 5 𝑘
−𝑗 { 3 3 3 − 𝜔𝑅𝐶 − 𝜔𝑅𝐶} − {3𝑘 + 1 − − }
𝜔𝑅𝐶 𝜔𝑅2𝐶2 𝜔 2 𝑅 2 𝐶 2
Replacing 1/ωRC by α for simplicity
𝑘ℎ𝑓𝑒
𝐴𝛽 = ------- eq. 11
[−3𝑘−1+5𝛼2+𝑘𝛼2]−𝑗[𝛼3−4𝑘𝛼−6𝛼]

To satisfy Barkhausen criterion, <Aβ = 00 hence imaginary part of the denominator term
must be 0

⸫ α3 – 4kα - 6α = 0 i.e. α ( α2 – 4 k – 6 ) = 0

⸫ α2 = 4 k + 6 (α ≠ 0 ) i.e. 𝛼 = √4𝑘 + 6 ------ eq. 12

1 1
⸫ 1/ωRC = √4𝑘 + 6 i.e. 𝜔 = 𝑅𝐶√4𝑘+6 i.e. 𝑓 = 2𝜋√4𝑘+6

This is the required frequency of oscillations.

Substituting 𝛼 = √4𝑘 + 6 in equation 11 we get,

𝑘ℎ𝑓𝑒 𝑘ℎ𝑓𝑒
𝐴𝛽 = = 2
−3𝑘 − 1 + (4𝑘 + 6)(5 + 𝑘) 4𝑘 + 23𝑘 + 29
 𝑘ℎ𝑓𝑒
But |𝐴𝛽| = 1 𝑖. 𝑒 | |=1
4𝑘2+23𝑘+29
29
⸫ ℎ𝑓𝑒 = 4𝑘 + 23𝑘 +
𝑘

This is the required hfe for the oscillations.

Minimum value of hfe:

• For satisfying Aβ = 1, the expression for the value of hfe of the transistor used in RC phase
shift oscillator is given by,
hfe ≥ 4 k +23 + 29 where k = 𝑅𝐶
𝑘 𝑅
• For minimum h fe, find k for minimum h fe from the expression 𝑑ℎ𝑓𝑒 = 0
𝑑𝑘
29 29 29
⸫ 𝑑 [4𝑘 + 23 + ] = 0 i.e. 4 − =0 i.e. 𝑘2 =
𝑑𝑘 𝑘 𝑘2 4
k = 2.6925 for minimum hfe
using in the expression of hfe,
hfe (min) = 4 X 2.6925 + 23 + 29
= 44.54
2.6925
Thus for the circuit to oscillate, the transistor must be selected with hfe greater than 44.54
Advantages:

• The circuit is simple to design

• Can produce output over audio frequency range
• Produces sinusoidal output waveform
• It is fixed frequency oscillator

Disadvantages:

• To vary the frequency, values of R and C of all three sections are to be varied simultaneously which is
practically difficult. Hence frequency cannot be varied
• Frequency stability is poor due to changes in the values of various components due to effect temperature,
aging etc.
(B) WIEN BRIDGE OSCILLATOR: (RC Oscillator)
Explain the working of Wien Bridge Oscillator. Derive the expression for frequency of oscillation and
condition for maintenance of oscillation.
(OR)
Design an oscillator to operate at a frequency of 10 KHz which gives an extremely pure sine
wave output, good frequency stability and highly stabilized amplitude. Discuss the operation
of this oscillator as an audio signal generator.

Construction and operation - (Wien Bridge Oscillator Circuit)

✓ Two stage amplifiers (non-inverting) and feedback network are used in Wien Bridge Oscillator.
✓ Both amplifier and feedback network does not introduce any phase shift i.e. 0° phase-shift around the loop in Wien
Bridge Oscillator.
✓ R1 & C1 in series and R2 & C2 in parallel are frequency sensitive arms.
✓ The output of Amplifier is applied as input to Feedback Network (Vin) between 1 and 3.
✓ The output of Feedback Network (Vf) taken between 2 and 4 is given as input to amplifier.
✓ This Feedback Network is also known as Lead-Lag Network.
 Derive the expression for frequency of oscillation:

Analysis for frequency of oscillation:

1 1 + 𝑗𝜔𝑅1𝐶1
𝑍1 = 𝑅1 + ⥤ 𝑍1 = (1)
𝑗𝜔𝐶1 𝑗𝜔𝐶1
1
1 𝑅2 × 𝑅2
⃦ 𝑗𝜔𝐶2
𝑍2 = 𝑅2 𝑗𝜔𝐶2 ⇒ 𝑍2 = 1
⇒ 𝑍2 = (2)
𝑅2 + 1 + 𝑗𝜔𝑅2𝐶2
𝑗𝜔𝐶2

𝑉𝑓 𝑉𝑖𝑛
𝛽= (3) 𝐼= (4)
𝑉𝑖𝑛 𝑍1 + 𝑍2
Sub (6) in (3) 𝑉𝑓 = 𝐼 𝑍2 (5)

𝑍2 𝑍2
⥤ 𝛽= (7) 𝑆𝑢𝑏 (4)𝑖𝑛 (5) ⥤ 𝑉𝑓 = 𝑉𝑖𝑛 (6)
𝑍1+𝑍2
𝑍1 + 𝑍2

𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 (1) & (2) 𝑖𝑛 (7)

𝑅2
1 + 𝑗𝜔𝑅2𝐶2
𝛽= (8)
1 + 𝑗𝜔𝑅1𝐶1 𝑅2
+
𝑗𝜔𝐶1 1 + 𝑗𝜔𝑅2𝐶2
Simplify the equation (8),

𝑗𝜔𝑅2𝐶1
𝛽= (9)
(1 − 𝜔 2 𝑅1 𝑅2 𝐶1 𝐶2 ) + 𝑗𝜔(𝑅1𝐶1 + 𝑅2𝐶2 + 𝑅2𝐶1)
Rationalizing and Simplifying the equation (9),

𝜔2𝑅2𝐶1(𝑅1𝐶1 + 𝑅2𝐶2 + 𝑅2𝐶1) + 𝑗𝜔𝐶1𝑅2 (1 − 𝜔2𝑅1𝑅2𝐶1𝐶2)

𝛽= (10)
(1 − 𝜔2𝑅 1𝑅 2𝐶1𝐶2)2 + 𝜔2(𝑅 1𝐶 1 + 𝑅 2𝐶2 + 𝑅 2𝐶1 )2
To have zero phase shift, imaginary part of above equation must be zero.

(1 − 𝜔2𝑅1𝑅2𝐶1𝐶2) = 0
𝜔 (𝜔2𝑅1𝑅2𝐶1𝐶2) = 0 but 𝜔 can not be zero. So,
1
𝜔2𝑅1𝑅2𝐶1𝐶2 = 0 ⇒ 𝜔2 =
𝑅1𝑅2𝐶1𝐶2
 1
⇒𝜔= (11)
√𝑅1𝑅2𝐶1𝐶2

𝟏
𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑊𝑖𝑒𝑛 𝐵𝑟𝑖𝑑𝑔𝑒 𝑂𝑠𝑐𝑖𝑙𝑙𝑎𝑡𝑜𝑟, 𝒇 = Hz (12)
𝟐𝝅√𝑹𝟏𝑹𝟐𝑪𝟏𝑪𝟐

𝐼𝑛 𝑝𝑟𝑎𝑡𝑖𝑐𝑒, 𝑅1 = 𝑅2 = 𝑅 and 𝐶1 = 𝐶2 = 𝐶 hence,

𝟏
𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑊𝑖𝑒𝑛 𝐵𝑟𝑖𝑑𝑔𝑒 𝑂𝑠𝑐𝑖𝑙𝑙𝑎𝑡𝑜𝑟, 𝒇= 𝑯𝒛
𝟐𝝅 𝑹𝑪
Derive the condition for maintenance of oscillation:
1
Case (1): 𝐼𝑓 𝑅1 = 𝑅2 = 𝑅 and 𝐶1 = 𝐶2 = 𝐶 𝑡ℎ𝑒𝑛 𝑢se 𝜔 = 𝐻𝑧 in (10),
𝑅𝐶

we get the magnitude of the feedback network as,

3 3 𝟏 𝟏
𝖰= = = ⇒ 𝖰=
1 9 𝟑 𝟑
0+ (3𝑅𝐶)2
𝑅2𝐶2
As |𝐴𝛽| ≥ 1 ℎ𝑒𝑛𝑐𝑒 |𝐴| ≥ 3 𝑓𝑜𝑟 𝑊𝑖𝑒𝑛 𝐵𝑟𝑖𝑑𝑔𝑒 𝑂𝑠𝑐𝑖𝑙𝑙𝑎𝑡𝑜𝑟.

Thus, the gain of amplifier stage must be at least 3 to ensure sustained oscillations in Wien Bridge
Oscillator.

Case (2): If 𝑅1 ≠ 𝑅2 and 𝐶1 ≠ 𝐶2 then use 𝜔 = 1

𝑖𝑛 (10) 𝑡ℎ𝑒𝑛
√𝑅1𝑅2𝐶1𝐶2

𝑹𝟐𝑪𝟏 𝑹𝟏𝑪𝟏 + 𝑹𝟐𝑪𝟐 + 𝑹𝟐𝑪𝟏𝟐

𝖰= ⇒∴𝑨≥ {∵ |𝐴𝛽| ≥ 1}
𝑹𝟏𝑪𝟏 + 𝑹𝟐𝑪𝟐 + 𝑹𝟐𝑪𝟏 𝑹𝟐𝑪𝟏

LC OSCILLATORS:
Outline the LC tuned Oscillator and deduce expression for amplifier Gain, feedback Gain
and necessary condition for LC Oscillator in general.
 Analysis of Amplifier stage

Applying KVL,
𝐴𝑉𝑉𝑖
𝐼=− and 𝑉 = 𝐼𝑍
𝑅𝑜+𝑍𝐿 𝑂 𝐿

𝑽𝒐 𝑨𝑽𝒁𝑳
𝑨= =−
𝑽𝒊 𝑹𝒐 + 𝒁𝑳

Ro – Output impedance of the amplifier stage. A – Gain of amplifier stage.

As, I=0 due to infinite input impedance, Z1 and Z3 appear in series and the
combination in parallel with Z2 as shown in figure.
Analysis of feedback stage
By voltage division in parallel circuit,
𝑉𝑜𝑍1
𝑉=
𝑓 𝑍1+𝑍3
𝑉𝑓 𝑍1
i.e. 𝛽 = =
𝑉𝑜 𝑍1+𝑍3

But as feedback network introduces 180°

phase-shift, use negative sign
𝒁𝟏
𝖰=−
𝒁𝟏 + 𝒁𝟑
Types of LC Oscillators:

(C) Hartley Oscillator:

Explain the working of Hartley Oscillator. Derive the expression for frequency of oscillation and condition
for maintenance of oscillation.

Circuit diagram

Construction:
• The Hartley oscillator circuit using BJT as an active device.
• The resistances R1, R2 and RE are biasing resistors
• The RFC is radio frequency chock whose reactance value is very high and high frequency and can be
treated as open circuit. While for d.c operation, it is shorted hence does not cause problems for d.c
operation.
• Due to RFC, the isolation between a.c and d.c operation is achieved. The C1 and C2 are coupling capacitors
while CE is the emitter bypass capacitor. The CE amplifier provides phase shift of 1800.
• In the feedback circuit, as the centre of L1 and L2 is grounded, it provides additional phase shift of 1800.
This satisfies Barkhausen condition. In this oscillator, X1 = ωL1,
X2 = ωL2, X3 = -1/ωC
Analysis:

• For LC oscillator, X1+X2+X3=0

1
⸫ 𝜔𝐿1 + 𝜔𝐿2 − =0
𝜔𝐶
1
i.e 𝜔(𝐿1 + 𝐿2 ) =
𝜔𝑐
1 1
⸫ 𝜔= i.e 𝑓 =
√(𝐿1+ 2𝜋√(𝐿1+ 𝐿2)𝐶
𝐿2)𝐶

• The inductance L1+L2 is equivalent inductance denoted as Leq. To satisfy |Aβ| = 1, then hfe of
the BJT used must be L1/L2.
𝐿1
ℎ𝑓𝑒 =
𝐿2
• Practically L1 and L2 are wound on a single core and there exists a mutual inductance M
between them.
In this case, 𝐿𝑒𝑞 = 𝐿1 + 𝐿2 + 2𝑀
1 𝐿1+𝑀
𝑓= and ℎ𝑓𝑒 =
2𝜋√𝐿𝑒𝑞𝐶 𝐿2+𝑀
• If capacitor C is kept variable, frequency can be varied over wide range.

Derivation of frequency of Oscillations

• The output current is collector current which is hfe Ib, where Ib is base current. Assuming coupling
capacitors shorted the capacitor C gets connected between collector and base.
• As emitter is grounded for a.c analysis, L1 is between emitter and base while L2 is between emitter and
collector.
• hie is the input impedance of the transistor. The output current is Ib while input current is hfe Ib. Convert
current source to voltage source.
𝑉𝑂 = ℎ𝑓𝑒𝐼𝑏𝑗𝑋𝐿2 = ℎ𝑓𝑒𝐼𝑏𝑗𝜔𝐿2

• Total current I is,

−𝑉𝑜
𝐼=
[𝑋𝐿2 + 𝑋𝐶] + [𝑋𝐿1||ℎ𝑖𝑒]
• Negative sign is because direction of I is opposite to the polarities of Vo
1 −𝜔2𝐿2𝐶 + 1
𝑋𝐿2 + 𝑋𝐶 = 𝑗𝜔𝐿2 + = 𝑗𝜔𝐶
𝑗𝜔𝐶
𝑗𝜔𝐿1ℎ𝑖𝑒
𝑋𝐿1 ||ℎ𝑖𝑒 =
𝑗𝜔𝐿1 + ℎ𝑖𝑒
 −ℎ𝑓𝑒𝐼𝑏𝑗𝜔𝐿2
⸫𝐼 =
−𝜔2𝐿1𝐶 + 1 𝑗𝜔𝐿1ℎ𝑖𝑒
+
𝑗𝜔𝐶 𝑗𝜔𝐿1 + ℎ𝑖𝑒
• Using current division rule for parallel elements,
𝑗𝜔𝐿1
𝐼𝑏 = 𝐼 𝑋
𝑗𝜔𝐿1 + ℎ𝑖𝑒

𝐼𝑏 = −ℎ𝑓𝑒𝐼𝑏𝑗𝜔𝐿2 𝑋 𝑗𝜔𝐿1
−𝜔2𝐿 𝐶 + 1 𝑗𝜔𝐿 ℎ 𝑗𝜔𝐿 + ℎ
2 1 𝑖𝑒 1 𝑖𝑒
+
𝑗𝜔𝐶 𝑗𝜔𝐿1 + ℎ𝑖𝑒

𝑗𝜔3ℎ𝑓𝑒𝐶𝐿1𝐿2
⸫
1= 𝐿 𝐶ℎ (𝐿 +𝐿 )+𝑗𝜔𝐿1 +ℎ𝑖𝑒
−𝑗𝜔3𝐿1 2 𝑖𝑒 1 2

𝑗𝜔3ℎ𝑓𝑒𝐶𝐿1𝐿2
⸫ 1= 2
[ℎ 𝑖𝑒 −𝜔 𝐶ℎ 𝑖𝑒 (𝐿1 +𝐿2 )]+𝑗𝜔𝐿1 (1−𝜔2𝐿2 𝐶)
• Rationalizing R.H.S of the above equation,

𝜔4ℎ𝑓𝑒𝐿21𝐿2𝐶(1 − 𝜔2𝐿2𝐶) + 𝑗𝜔3ℎ𝑓𝑒𝐿1𝐿2𝐶[ℎ𝑖𝑒 − 𝜔2𝐶ℎ𝑖𝑒(𝐿1 + 𝐿2)]

1=
[ℎ𝑖𝑒 − 𝜔2𝐶ℎ𝑖𝑒(𝐿1 + 𝐿2)]2 + 𝜔2𝐿21(1 − 𝜔2𝐿2𝐶)2

• Imaginary part of R.H.S of above equation must be Zero

1
⸫ 1 − 𝜔3𝐶(𝐿1 + 𝐿2 ) = 0 𝑖. 𝑒 𝜔 = (𝜔3ℎ𝑓𝑒 ℎ𝑖𝑒 𝐿1 𝐿2 𝐶 ≠ 0)
√𝐶(𝐿1+𝐿2)

1 1
𝑓= =
2𝜋√𝐶(𝐿1 + 𝐿2) 2𝜋√𝐶𝐿𝑒𝑞
1
• Equating magnitude of both sides of the equation and using 𝜔 = 𝑤𝑒 𝑔𝑒𝑡
√𝐶(𝐿1+𝐿2)
𝐿1
ℎ𝑓𝑒 = ℎ𝑓𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑓𝑜𝑟 𝑜𝑠𝑐𝑖𝑙𝑙𝑎𝑡𝑖𝑜𝑛
𝐿2
• In practice, L1 and L2 may be wound on a single core so that there exists a mutual
inductance between them denoted as M.
• In such a case, the mutual inductance is considered while determining the equivalent
inductance Leq, Leq = L1+L2+2M
• If L1 and L2 are assisting each other, then sign of 2M is positive while if L1 and L2 are in
series opposition then sign of 2M is negative.
Advantage:
• The frequency can be easily varied by variable capacitor
• The output amplitude remains constant over the frequency range
• The feedback ratio of L1 and L2 remains constant
• It can be operated over wide range of frequency

Disadvantage:
• The output is rich in harmonics hence not suitable for pure sine wave requirement
• Poor frequency stability

Applications:
• Used as local oscillators in TV and radio receivers
• In function generators
• In radio frequency sources

(D) COLPITTS OSCILLATOR:

Explain the working of Colpitts Oscillator. Derive the expression for frequency of oscillation and condition
for maintenance of oscillation.
(OR)
With a neat circuit diagram deduce the necessary condition for oscillations and expression for oscillation
frequency in the case of Colpitts Oscillator.

Construction:

• It uses two capacitive resistances and one inductive reactance in its feedback network.
• The amplifier stage uses BJT in common emitter configuration providing 180o phase shift. The resistance
R1, R2 and RE are the biasing resistors.
• The RFC is radio frequency choke providing insulation between AC and DC operations. The CC1 and CC2
are coupling capacitors. In the feedback circuit, as the center C1 and C2 are grounded, it provides additional
phase shift of 1800, satisfying Barkhausen angle condition.
 −1 −1
• In this oscillator 𝑋1 = 𝑋2 = 𝑋3 = 𝜔𝐿
𝜔𝐶1 𝜔𝐶2
• For LC oscillator, X1+X2+X3=0
1 1 1 1 1
⸫− − + 𝜔𝐿 = 0 i.e 𝜔𝐿 = [ + ]
𝜔𝐶1 𝜔𝐶2 𝜔 𝐶1 𝐶2
1
⸫𝜔2 = where 𝐶1𝐶2
= 𝐶𝑒𝑞
𝐿[ 𝐶1𝐶2 ] 𝐶1+𝐶2
𝐶1+𝐶2
1
⸫ 𝜔= i.e 𝑓 =
1 𝐶 𝐶
𝑎𝑛𝑑 𝐶𝑒𝑞 = 𝐶 1+𝐶2
√𝐿𝐶𝑒𝑞 2𝜋 √𝐿𝐶𝑒𝑞 1 2

• To satisfy magnitude condition of Barkhausen criterion, the hfe of BJT used is given by
𝐶2
ℎ𝑓𝑒 =
𝐶1

Derivation of Frequency of oscillations

• The equivalent circuit and simplified equivalent circuit.

−𝑗ℎ𝑓𝑒𝐼𝑏 1 𝑗
𝑉0 = ℎ𝑓𝑒𝐼𝑏𝑋𝐶2 = ….. 𝑋𝐶2 = =−
𝜔𝐶2 𝑗𝜔𝐶2 𝜔𝐶2
• The total current drawn I is,
−𝑉0
𝐼=
[𝑋𝐶2 + 𝑋𝐿] + [𝑋𝐶1||ℎ𝑖𝑒]
−𝑗 −𝑗(1 − 𝜔2𝐿𝐶2)
𝑋𝐶2 + 𝑋𝐿 = + 𝑗𝜔𝐿 =
𝜔𝐶2 𝜔𝐶2
𝑗
− 𝑋ℎ
𝜔𝐶1 𝑖𝑒 −𝑗ℎ𝑖𝑒
𝑋𝐶1||ℎ𝑖𝑒 = 𝑗 = −𝑗 + 𝜔𝐶 ℎ
− 𝜔𝐶 + ℎ𝑒𝑖 1 𝑖𝑒
1
𝑗ℎ𝑓𝑒𝐼𝑏
− [− ]
𝐼= 𝜔𝐶2
−𝑗(1 − 𝜔2𝐿𝐶2) −𝑗ℎ𝑖𝑒
−
𝜔𝐶2 −𝑗 + 𝜔𝐶1ℎ𝑖𝑒
• Using current division rule for parallel elements
 −𝑗
𝜔𝐶1 −𝑗𝐼
𝐼𝑏 = 𝐼 𝑋 =
−𝑗
+ ℎ 𝑖𝑒 −𝑗 + 𝜔𝐶1ℎ𝑖𝑒
𝜔𝐶1
𝑗ℎ𝑓𝑒𝐼𝑏
𝐼 = −𝑗 [ 𝜔𝐶2 1
][ ]
𝑏 2
−𝑗(1 − 𝜔 𝐿𝐶2) −𝑗ℎ𝑖𝑒 −𝑗 + 𝜔𝐶1ℎ𝑖𝑒
𝜔𝐶2 −𝑗 +
𝜔𝐶1ℎ𝑖𝑒
−ℎ𝑓𝑒 1
1= --------------
(1−𝜔2𝐿𝐶2)+𝑗𝜔ℎ𝑖𝑒[𝐶1+𝐶2−𝜔2𝐿𝐶1𝐶2]

• To have imaginary part of above equation

𝐶1+𝐶2
zero 1
2
C1+C2-ω LC C = 0 i.e 𝜔 =
2
=
1 2 𝐶 𝐶
𝐿𝐶1𝐶2 𝐿[ 1 2 ]
𝐶1+𝐶2

1 1 𝐶 1𝐶 2
𝜔= 𝑎𝑛𝑑 𝑓 = 𝑤ℎ𝑒𝑟𝑒 𝐶𝑒𝑞 =
√𝐿𝐶𝑒𝑞 2𝜋√𝐿𝐶𝑒𝑞 𝐶1 + 𝐶2

• Substituting ω in equation 1 and equating magnitudes of both sides

𝐶2
ℎ𝑓𝑒 =
𝐶1
Advantages:
• Pure output waveform
• Good stability at high frequency
• Improved performance at high frequency
• Wide range of frequency
• Simple construction
•
Disadvantages:
• Difficult to adjust the feedback
• Poor isolation
Applications:
• Its main application is high frequency function generators.
 (E) CRYSTAL OSCILLATOR:
Describe and explain the operation of the crystal oscillator.
(OR)
Can you use Piezo-Electric effect for electric oscillators? If so, explain a
component with such characteristics. Also draw a circuit for the same.
• The crystals are either naturally occurring or synthetically manufactured, exhibiting the piezoelectric effect
• The piezoelectric effect means under the influence of mechanical pressure, the voltage gets generated
across the opposite faces of the crystal
• If the mechanical force is applied in such a way to force the crystal to vibrate the a.c voltage gets generated
across it.
• Every crystal has its own resonating frequency depending on its cut. So under the influence of the
mechanical vibrations, the crystal generates an electrical signal of very constant frequency
• The crystal has a greater stability in holding the constant frequency. The crystal oscillators are preferred
when greater frequency stability is stability
• Quartz is a compromise between the piezoelectric activity of Rochelle salt and the strength of the
tourmaline.
• Quartz is inexpensive and easily available in nature hence very commonly used in the crystal oscillators.

Constructional Details:

• The natural shape of quartz is a hexagonal prism. But for its practical use, it is cut to the rectangular slab.
This slab is then mounted between the two metal plates.

• The metal plates are called holding plates, as they hold the crystal slab in between them.

A.C. Equivalent circuit:

CM – Mounting Capacitance (due to two metal plates
separated by dielectric like crystal slab).
R – Resistance (internal friction loss during vibration)
L – Inductance (indication of inertia of mass of crystal)
C – Capacitor (stiffness during vibrating)

• RLC forms a resonating circuit. The expression for the resonating frequency fr is,
1 𝑄2
𝑓𝑟 = 2𝜋√𝐿𝐶 √1+𝑄2 where Q = Quality factor of crystal

𝜔𝐿
𝑄=
𝑅
 2
• The Q factor of the crystal is very high, typically 20,000. Value of Q up to 106 also can be achieved. Hence √ 𝑄
1+𝑄2
1
factor approaches to unity and we get the resonating frequency as 𝑓𝑟 = 2𝜋√𝐿𝐶
• The crystal frequency is in fact inversely proportional to the thickness of the crystal.
• f α 1 where t = Thickness
𝑡
• So to have very frequencies, thickness of the crystal should be very small
• The crystal has two resonating frequencies, series resonant frequency and parallel resonant frequency.

Applications
• Watches
• Communication transmitters and receivers

Series and Parallel resonance:

• Series Resonance frequency

1
𝑓𝑠 =
2𝜋√𝐿𝐶

• Parallel Resonance frequency

1
𝑓𝑃 =
2𝜋√𝐿𝐶𝑒𝑞
•
• If we neglect the resistance R, the impedance of the crystal is a reactance jX which depends on the frequency as,

𝑗 𝜔2 − 𝜔𝑠2 Where, ωs = Series resonant

𝑗𝑋 = − frequency
𝜔𝐶𝑀 𝜔2 − 𝜔𝑝2

• Reactance against frequency is shown in fig.

Crystal Stability:

i. Temperature stability
ii. Long term stability
iii. Short term stability

Types of Crystal Oscillator:

1. Pierce Crystal Oscillator:
2. Miller Crystal Oscillator:

Pierce Crystal Oscillator: Miller Crystal Oscillator:

 Comparison between Crystal and LC Oscillator:

Solved Problems
1. In a Hartley oscillator, if L1=0.2mH, L2=0.3mH and C=0.003µF. Calculate the frequency of oscillations.
[MAY 2012]
Given: L1=0.2mH, L2=0.3mH, C=0.003µF
To find frequency of oscillations f=1/(2π√[(L1+L2) C)] by substituting f=129.949KHz

2. In a RC phase shift oscillator if R1=R2=R3=200KΩ and C1=C2=C3=100PF. Find the frequency of

oscillation? (Apr/May 2018)
Solution:
The frequency of an RC phase shift oscillator is given by
1 1
Fo = Fo =
2𝜋𝑅𝐶√6 2𝜋×200×103×100×10−12×√6 Fo = 3.248KHZ

3. In a phase shift oscillator, R1=R2=R3=1 MΩ and C1=C2=C3=68 pF. At what frequency does the circuit
oscillate. (Nov/Dec 2018)
Given that,
For a phase shift oscillator, Resistance, R1 = R2 = R3 = 1 MΩ; Capacitor, C1 = C2 = C3 = 68 pF
Frequency, f = ?
1
Frequency of phase shift oscillator is given by, 𝑓=
2𝜋𝑅𝐶√6
1
Substituting corresponding values in above equation,𝑓 = = 955.9 𝐻𝑧 frequency, f = 955.9 Hz
2𝜋 𝑋 1𝑋106 𝑋 68 𝑋 √6
4. A wien bridge oscillator is used for operation at 10KHz. If the value of the resistor R is 100Kohms, what
is the value of C required?
Solution:
Given: F = 10KHZ, R = 100KΩ, C= ?
The frequency of oscillation is
F= 1 C= 1 C= 1
2𝜋𝑅𝐶 2𝜋𝑅𝐹 2𝜋×100×103×10×103

C = 1.591× 𝟏𝟎−𝟏𝟎 F

5. An amplifier has a c urrent gain of 240 and input impedance of 15 k Ω without feedback. If negative
current feedback (mi = 0.015) is applied, what will be the input impedance of the amplifier? (Nov/Dec
2017)

6. Design a Wien bridge oscillator circuit to oscillate at a frequency of 20KHz. (Nov/Dec2015)

Solution:
1
f= f = 20 kHz, Let C = 0.01𝜇𝐹
2𝜋𝑅𝑐
1 1 1
f=
2𝜋𝑅𝑐
, 𝑅 = 2𝜋𝑓𝐶 = 2×𝜋×20000×0.01×10−6 = 80ohms.

7. A 1 mH inductor is available. Find the capacitor values of a colpitt’s oscillator so that f=1 MHz and
feedback fraction=0.25 (Nov/Dec 2018)
Solution:
Given that,
For a Colpitts oscillator,
Inductance, L = 1 mH
Resonant frequency, f0 = 1 MHz
Feedback factor, β = 0.25
The resonant frequency of Colpitts oscillator is given by,
1
𝑓0 = − − − −(1)
2𝜋√𝐿𝐶𝑒𝑞
𝐶1𝐶2
Where,𝐶𝑒𝑞 =
𝐶1+𝐶2
From equation (1),
1
𝐶𝑒𝑞 = 2 2 − − − −(2)
4𝜋 𝑓0 𝐿
𝐶
Given feedback factor, 𝛽 = 1 = 0.25
𝐶2
C2 = 4C1
Substituting the given specifications in equation (2)
 1
𝐶𝑒𝑞 =
4𝜋2(106)2𝑋 10−3
𝐶1𝐶2
= 2.533 𝑋 10−11
𝐶1 + 𝐶2
4 𝐶12
= 2.53 𝑋 10−11
5𝐶1
C1 = 3.166 X 10-11 = 31.66 pF
From C2 = 4C1,
C2 = 4 X (3.166 X 10-11)

C2 = 126.65 pF

8. The overall gain of a multistage amplifier is 140. When negative voltage feedback is applied the gain is
reduced to 17.5 find the fraction of the output that is feedback to the input. (Nov/Dec 2018)
Given that,
For a multistage feedback amplifier,
Overall gain, AV = 140
Feedback gain, Avf = 17.5
Feedback fraction, β = ?
Voltage gain of negative feedback amplifier is defined as,
𝐴𝑣 40
𝐴𝑣𝑓 = 17.5 =
1 + 𝐴𝑣 𝛽 1 + 140𝛽
17.5 + 2450 β = 140
1
𝛽 = = 0.05
20 β = 0.05

9. In colpitts oscillator C1 = 1nF and C2 = 100nF. If the frequency of oscillation is 1 kHz find the value of
inductor. Also find the minimum gain required for obtaining sustained oscillations. (May / Jun 2016)
Given data:

C1 = 1nF, C2 = 100nF, Frequency of oscillation f = 100 kHz.

Formulae used:

1 𝐶1+𝐶2 𝐶1
𝑓= √ , 𝐴𝑉 =
2𝑛 𝐿1𝐶1𝐶2 𝐶2

𝐶1+𝐶2 101×10−6
Frequency of oscillations 𝐿= =
4𝑛2𝑓𝑟2𝐶1𝐶2 4𝑛2×(10×1000)2×100×10−12

101×106 101
= = × 10−5 = 25.634 × 10−5𝐻 = 256.34𝜇𝐹
4𝑛2×(100000)2 3.99

𝐶1 1
𝐴𝑉 > = = 0.01𝑛𝐹
𝐶2 100
10. Design a RC phase Shift Oscillator to generate 5KHz sine wave with 20 V peak to Peak amplitude.
Assume hfe=𝖰 = 𝟏𝟓𝟎, 𝑪 = 𝟏. 𝟓𝒏𝑭, hre=1.2KΩ(Nov.Dec 2016)
1 1 1
𝑓= ; 5 × 103 = 𝑅=
2𝜋𝑅𝑐√6 2𝜋×1.5×10−9√6×𝑅 2𝜋×1.5×10−9×√6×5×103

𝑅 = 8.67 𝑘 𝛺
11. In Colpitts Oscillator, the desired frequency is 500 KHz. Find the value of L. Assume C= 1000pF.
(Apr/May 2018)

12. When negative voltage feedback is applied to an amplifier of gain 100, the overall gain falls to 50.
Calculate the fraction of the output voltage fedback. If this fraction is maintained, calculate the value of
the amplifier gain required if the overall stage gain is to be 75. (Nov/Dec 2017)
13. In Colpitts oscillator, C1 = C2 =C and L=100 X 10-6 H. The frequency of oscillation is 500 KHz.
Determine the value of C. (Apr/May 2018)

14. An amplifier in required with a voltage gain of 100 which does not vary by more that 1%. If it is to use
negative feedback with a basic amplifier the voltage gain of which vary by 20%, find the minimum
voltage gain required and the feedback factor. (Nov/Dec 2018)
Solution:
Closed loop voltage gain of amplifier, Af is defined as,
𝐴𝑚
𝐴𝑓 = − − − − − (1)
1 + 𝐴𝑚 𝛽
100 = 𝐴𝑚
1 + 𝐴𝑚 𝛽
𝐴𝑚 = 100 + 100 𝐴𝑚 𝛽 − − − − − (2)
Since, feedback voltage gain, Af does not vary more than 1% and amplifier gain varies by 20% equation (1) can
be written as,
0.8 𝐴𝑚
99 =
1 + 0.8 𝐴𝑚 𝛽
0.8 𝐴𝑚 = 99 + 79.2 𝐴𝑚 𝛽 − − − − − (3)
Multiplying equation (1) with 0.792 or both sides,
0.792 𝐴𝑀 = 79.2 + 79.2 𝐴𝑚 𝛽 − − − − − (4)
Subtracting equation (3) and (4),
19.8
0.008 Am = 19.8; = Am = 2475
0.008
𝐴𝑚
Substituting Am in equation (2),
2475 = 100 + 100 X 2475 X β
2475 − 100
𝛽= β = 0.0096
2475 𝑋 100

⸫ Feedback factor, β = 0.0096 and minimum voltage gain Am = 2475 V.