syllabus - wordpress.comunit 4| principle of positive feedback p a g e | 8 2018-19 6+4 − 1 =0 1...

Unit 4| Principle of Positive feedback P a g e | 1

2018-19

Syllabus: Principle of Positive feedback, Concept of Stability in electronics circuits, Barkhausen criteria

for oscillation, Principle of operation of RC Phase Shift, Wien Bridge, Colpitt’s, Hartley, Crystal

oscillators.

Principle of operation of Transistorized Astable, Bistable and Monostable multivibrator.

FEEDBACK AMPLIFIER: Feedback amplifier is a closed loop network in which some part of output signal is fed back to

the input. If A is the gain of amplifier and β is the gain of feedback network, then the feedback

amplifier will be as shown below:

Therefore �� ≠ ��

CLOSED LOOP GAIN (��) WITH POSITIVE FEEDBACK: In positive feedback amplifier, feedback signal is added with input.

�� + �� = ��

�� = �� − �� --- (1)

Gain of feedback network: � = ��/��

Gain of Amplifier: � = ��/��

Overall closed loop gain of feedback amplifier �� is,

�� =��

��=

��

�� − ��=

��/��

��/�� − ��/��=

�

1 −��

��

=�

1 −��

��×

��

�� =�

1 − ��

Or

�� =��

1 − ��


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BARKHAUSEN CRITERIA:

State the conditions under which a feedback amplifier works as an oscillator. (W-14/3m) State Barkhausen criteria of oscillation. (W-16/2m) (S-16/3m) Explain Barkhausen criteria of oscillation. (S-15/3m) (S-16/3m)

Two conditions are required for sustained oscillations as follows,

1. The loop gain of the circuit must be unity or slightly greater than unity, |��| ≥ 1

2. The total phase shift around the closed loop must be zero or 360°, ∠� + ∠� = 0° �� 360°

These two conditions for sustained oscillations are called as Barkhausen Criteria.

TYPES OF SINUSOIDAL OSCILLATORS 1. RC Oscillator

a. RC Phase Shift Oscillator

b. Wein Bridge Oscillator

2. LC Oscillator

a. Colpitt’s Oscillator

b. Hartley Oscillator

3. Crystal Oscillator

Explain Barkhausen Criteria to generate Oscillations. With the help of neat sketch explain working of RC Phase shift oscillator. (W-15/7m) With the help of neat sketch explain working of RC Phase shift oscillator. Why can’t it be used for RF frequency range? (S-14/7m) Draw and explain working of RC Phase shift oscillator (S-15/4m) (W-13/8m)

RC PHASE SHIFT OSCILLATOR (USING BJT):

C C C

R R R

Q

R1 RC

RECE

C0 VO

Feed Back Signal

VCC


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Circuit Explanation:

Resistor R1 provides biasing to the transistor, RC is the collector resistor, RE & CE are emitter

resistor and capacitor which provide thermal stability. CO is the coupling capacitor. The

capacitances C and resistances R form the feedback network. Such that each R-C pair produces

60° of phase shift. Thus, total phase shift produced by three R-C pairs sums to 180°

(60°+60°+60°).

Working:

Consider the circuit is set into oscillation by any random variation caused in base current. This

variation is amplified in collector circuit. The output of amplifier is supplied to an RC feedback

network, which produces a phase shift of 180° between output and input voltages. Since the

CE amplifier produces phase reversal of the input signal, total phase becomes 360° or 0°, which

is one of the condition for oscillation. The RC network attenuates the amplifier output. When

the necessary phase shift of 180° is obtained, this network attenuates the output voltage by a

factor of (1/29). This means that the amplifier must have a voltage gain of 29 or more.

Therefore, when amplifier gain is 29 (i.e. A = 29) and the feedback gain of RC network is 1/29

(i.e. β = 1/29) then the loop gain will be Aβ = 29 × 1/29 =1 (Unity). This is another condition

for sustained oscillation.

The frequency of oscillation depends upon the values of C & R in the feedback network and is

given by,

� =1

2�√6��

At high frequencies, resistors look like inductors or capacitors so the above equation that

decides oscillation in RC oscillators no longer apply. In other words, because of those parasitic

components, it gets more and more difficult to make a stable RC oscillator as frequencies go

up. Thus, RC phase shift oscillator cannot be used for RF frequency range.

Advantages:

1. It is a simple low-cost circuit as it uses only RC components.

2. It provides good frequency stability.

3. This circuit is much simple than Wein-Bridge oscillator circuit.

4. Output is sinusoidal, which is very much distortion free.

5. Availability of wide range of frequencies.

Disadvantages:

1. Output is small due to smaller feedback,

2. It is difficult for the circuit to start oscillation as the feedback is usually small,

3. Frequency stability is not as good as that of Wein-Bridge Oscillator,

4. It needs high voltage (12V) battery so as to develop sufficient large feedback voltage.


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RC PHASE SHIFT OSCILLATOR (USING FET)

C C C

R R R

RD

RSCS

C0 VOFeedback Signal

VCC

The frequency of oscillation depends upon the values of C & R in the feedback network and is

given by,

� =1

2�√6��

Draw and explain the working of Wein Bridge oscillator. Also derive expression for frequency of oscillation. (W-14/7m) (W-16/6m) (S-16/8m) (S-17/7m) Derive an expression for frequency of oscillation and gain of wein bridge oscillator. (S-14/7m)

WEIN BRIDGE OSCILLATOR


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It is the most popular, distortion free oscillator. Its maximum frequency output is only up to

1MHz. It contains two transistors, each producing a phase shift of 180° & thus producing a total

phase shift of 360° or 0°.

Working:

It is a two-stage amplifier with an RC bridge circuit, which consists of a series RC circuit

connected with a parallel RC of the same component values producing a phase delay depending

upon the frequency. RC bridge circuit is a lead-lag network such that the phase shift across the

network lags with increasing frequency and leads with decreasing frequency. By adding Wein

bridge feedback network, oscillator becomes sensitive to a signal of only one particular

frequency called resonating frequency �. This particular frequency is that at which it is

balanced and for which the phase shift is 0° and the feedback gain is 1/3 of its input. Thus, to

achieve the criterion for oscillation, the gain of amplifier must be equal to 3 or slightly greater

than 3.

In the bridge circuit, ��, ��, �� in series with �� & �� in parallel with �� form the four arms of

the bridge.

�� = �� = � & �� = �� = �

Then the frequency of oscillation is given by,

� =1

2��

Where � depends upon the values of C & R in the feedback network.

The RC network consists of a series RC circuit connected to a parallel RC forming basically a

High Pass Filter connected to a Low Pass Filter producing a Band Pass Filter with a high Q

factor at the resonating frequency �.

The circuit is set in oscillation by a random change in base current of transistor ��, that may

be due to noise. This variation in base current is amplified in collector circuit of transistor ��

but with a phase-shift of 180°. The output of transistor �� is fed to the base of second transistor

�� through capacitor ��. The transistor �� will again amplify and invert the phase by 180°.

Thus, total phase shift between input of transistor �� and output of transistor �� is now 360°,

which is a criterion for sustained oscillation. The output of transistor �� is fed back to the

bridge circuit (point A-C).

The frequency of oscillation can be varied by varying the two capacitors �� and ��

simultaneously, this can be achieved using variable air-gang capacitors. We can change the

frequency range of the oscillator by using different values of resistors �� and ��.

Advantages:

1. It provides stable low distortion sinusoidal output over a wide range of frequency.

2. The frequency range can be selected simply by using decade resistance boxes.

3. The frequency of oscillation can be easily varied by varying capacitances �� & ��

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.


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

characteristics of the amplifier.

FREQUENCY OF OSCILLATION � IN RC PHASE SHIFT OSCILLATOR Hybrid model of fb nw in RC phase shift oscillator is as follows:

hfeibRc

RCC

R

C C

R R

Applying KVL in loop 1:*

−�� − �� − �� − �� + �� = 0

−��(�� + �� + �) + �� = ℎ��

��(�� + � � + �) − �� = −�� ---(1)

Applying KVL in loop 2:

−�� + �� − �� − �� + �� = 0

�� − ��(2� + ��) + �� = 0

−�� + ��(�� + � �) − �� = � ---(2)

Applying KVL in loop 3:

−�� + �� − �� − �� = 0

�� − ��(2� + ��) = 0

−�� + ��(�� + � �) = � ---(3)

Equations 1, 2 & 3 can be written in matrix form as

�

�� + �� + � −� 0−� 2� + �� −�0 −� 2� + ��

��

��

��

��

�= �−ℎ��

00

�

Current �� can be found using Cramer’s Rule

�� =��

�

�� = �� + �� + � −� −ℎ��

−� 2� + �� 00 −� 0

�

�� = (�� + �� + �) × [0]− (−�)[0]+ �−ℎ��[��]

* −ℎ�� is shown with inverted polarity because it is coming from the output of transistor which is

180° phase shifted.


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�� = −ℎ��

� = �

�� + �� + � −� 0−� 2� + �� −�0 −� 2� + ��

�

� = (�� + �� + �)[(2� + ��)� − ��]+ �[−�(2� + ��)]

� = (�� + �� + �)[4�� + 4�� + �� − ��]− ��(2� + ��)

� = (4�� + 4�� + 4��) + (4�� + 4�� + 4��) + ��

�� + �� + ��

��− ��

− �� − �� − 2�� − ��

� = 3�� + 6�� + �� + 4�� + 5�� + ��

�� + ��

�� =��

�=

−ℎ��

3�� + 6�� + �� + 4�� + 5�� + ��

�� + ��

Put �� = −�/�� & separate real & imaginary terms,

�� =−ℎ��

3�� + 6�� −�

�� + �� + 4� �−�

�� + 5� �−�

��

+ �−�

��

�� + �−�

��

�� =−ℎ��

3�� − ��6��

�� + �� − ��4��

�� −5�

(�� )� −��

(�� )� + ��1

(�� )��

�� =−ℎ��

�3�� + �� −5�

(�� )� −��

(�� )��− ��6��

�� +4��

�� −1

� ��

According to Barkhausen Criteria �� = 1

� = ℎ�� & � =��

ℎ��

∴ ℎ�� ×��

ℎ��= 1

��

��= 1

Putting value of �� in above equation

−ℎ��

�3�� + �� −5�

(�� )� −��

(�� )��

��

− ��6��

�� +4��

�� −1

� ��

��

= 1

−ℎ��

��

+ �0⏟��

= �3�� + �� −5�

(�� )�−

��

(�� )��

��

− ��6��

��+

4��

��−

1

� ��

� ��

Equating Imaginary terms on both the sides,

6��

��+

4��

��−

1

� ��= 0


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6�� + 4�� −1

� ��= 0

1

� ��= 6�� + 4��

� �� =1

6�� + 4��

� � =1

��(6�� + 4��)=

1

�� 6 +4��

�� =

1

�� 6 +4��

��

� = �1

�� 6 +4��

� �=

1

�� 6 +4��

�

� =1

2�� 6 +4��

�

��

�= �

� =1

2��√6 + 4�

This is the expression for frequency of oscillation in RC Phase shift oscillator


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FREQUENCY OF OSCILLATION � IN WEIN-BRIDGE OSCILLATOR The general bridge is as follows:

Its balance condition is

��

��=

��

��

The balancing condition for Wein bridge oscillator will be,

(�� + ��)

��

�� + ��

=��

��

�� + 1/ ��

�� × 1/�� + 1/��

�=

��

��

��

��=

(�� + 1/ ��)(�� + 1/��)

(�� × 1/��)

��(�� × 1/��)

= ��(��

+ 1/ ��)(��

+ 1/��)

��

��= �� +

��

��+

��

��+

1

��×

1

��

��

��= �� +

��

��+

��

��−

��

� ��

��

��× �� = �� × �� +

��

��× �� +

��

��× �� −

��

� ��× ��

�� = �� + �� + �� −��

�

�� − �� − �� = �� −��

�

Separating real and imaginary terms,

Real term:

�� − �� − �� = 0

�� − �� = ��

(�� − ��)�� = ��

(�� − ��)

��=

��

��

��

��−

��

��=

��

��

�� −��

�= 0


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�� − �� = 0

� �� − 1 = 0

� � =1

��

� =1

��

� =1

2��

�� = �� = � & �� = �� = �

� =1

2�√�� =

1

2��

This is the expression for frequency of oscillation in Wein-Bridge oscillator.

Gain of wein-bridge (Feedback Network) � is given by

� =��

��

Using voltage division formula,

�� =��

�� + ��× ��

� =�

�� + ��

�

(�� + ��) + ��

�� + ��

=�

�� × 1/�� + 1/��

�

(�� + 1/ ��) + �� × 1/�� + 1/��

�

� =

��

×1

�� +1

��

�� +1

��+

��

�1

�� +1

��

�

=

��

×��

�� × �� + 1�� × �� + 1

��+

��

��

�� × �� + 1�

� =

�� × �� + 1

�� × �� + 1

��+ �

�� × �� + 1�

� =��

(�� × �� + 1)×

(��)(�� × �� + 1)

(�� × �� + 1)(�� × �� + 1) + �� × ��

� =��

(�� × �� + 1)(�� × �� + 1) + �� × ��

� =��

−� �� + �� + �� + 1 + ��

� =��

(1 − � ��) + ��(�� + �� + ��)


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� =��

(1 − � ��) + ��(�� + �� + ��)

×(1 − � ��) − ��(�� + �� + ��)

(1 − � ��) − ��(�� + �� + ��)

� =��[(1 − � ��) − ��(�� + �� + ��)]

(1 − � ��)� − [��(�� + �� + ��)]�

� =[��(1 − � ��) + � ��(�� + �� + ��)]

(1 − � ��)� + � �(�� + �� + ��)�

� =��(1 − � ��) − � ��(�� + �� + ��)

(1 − � ��)� + � �(�� + �� + ��)�

As �� = �� = �, �� = �� = � and � =�

�� =

�

��

� =�

1��

�� 1 −1

�� −1

�� (�� + �� + ��)

�1 −1

��

+1

�� (�� + �� + ��)�

� =�(1 − 1) −

1�� (3��)

(1 − 1)� +1

�� (3��)�=

0 − 3

0 − 3�=

3

9=

1

3

HARTLEY OSCILLATOR

Q

R1 RC

RECE

VCC

R2 C L2

L1

CO

Cin

It consists of two inductors �� & �� and a capacitor �. The output of amplifier is applied across

�� and the voltage across inductor �� forms the feedback voltage. The coil �� is inductively

coupled to coil ��. The mutual inductance exists between coils �� & �� because they are wound

on same core, so their net effective inductance is increased by mutual inductance M.

�� = �� + � & ��

� = �� + �


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Therefore, effective inductance � is

� = �� + ��

�

� = �� + �� + 2�

The resonant frequency f is given by,

� =1

2��[�� + �� + 2�]

� =1

2�√��

Explain working of Colpitt's oscillator and derive expression for frequency of oscillation. (S-16/7m) Draw the circuit diagram of Colpitt’s oscillator and explain its working. (W-14/7m)

COLPITT’S OSCILLATOR

Q

R1 RC

RECE

VCC

R2

C1

L

CO

Cin

C2

The two capacitors �� & �� form the potential divider, which is used for providing the feedback

voltage. Parallel combination of �� & �� along with �� & �� provide stabilized self-bias.

Transistor itself produces a phase shift of 180° and the other phase shift of 180° is provided by

the LC feedback. Thus the total phase shift of 360° is obtained, which is essential condition for

oscillation. The frequency is determined by the tank circuit and can be varied by the gang

tuning of two capacitors �� & ��, which are ganged together.


� =1

2�√��

� ℎ��, � =��

�� + ��


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FREQUENCY OF OSCILLATION � The equivalent circuit of Hartley and Colpitt’s oscillators is

hie hfeIB

Z2 Z1

Z3

213

IBIC

I1

I1

I1

Total load impedance between output terminals 1 & 2 is,

�� = ��||[�� + (��||ℎ��)]

�� = ��|| �� + ��ℎ��

�� + ℎ��

�� = ��|| ��(�� + ℎ��) + ��ℎ��

�� + ℎ��

�� = ��|| �� + ��ℎ�� + ��ℎ��

�� + ℎ��

�� = ��|| �� + ℎ��(�� + ��)

�� + ℎ��

1

��=

1

��+ �

�� + ℎ��

�� + ℎ��(�� + ��)�

1

��=

�� + ��(�� + ��) + (�� + ��)��

�� + ℎ��(�� + ��)�

1

��=

��(�� + �� + ��) + �� + ��

�� + ℎ��(�� + ��)�

�� =�� + ��(�� + ��)�

��(�� + �� + ��) + �� + ��

The voltage gain of CE amplifier without feedback is,

�� =��

��=

�−ℎ��

1 + ℎ��

��=

�−ℎ��

1 + ℎ��

ℎ�� (�� = ℎ��)

�� ℎ�� ≪ 1

�� =−ℎ��

ℎ��


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This is approximate formula for ��

The feedback gain � is given by,

� =��

��

From the equivalent circuit, feedback voltage �� is the voltage at point (3) with respect to (2),

�� = ��

�� + ��

From the equivalent circuit, output voltage �� is the voltage at point (1) with respect to (2),

�� = �� + ��ℎ��

�� + ℎ��

�� = ��(�� + ℎ��) + ��ℎ��

�� + ℎ��

�� = �� + ��ℎ�� + ��ℎ��

�� + ℎ��

�� = �� + (�� + ��)��

�� + ��

� =�

��ℎ�� + ℎ��

��

�� + (�� + ��)ℎ��

�� + ℎ��

� =��

�� + (�� + ��)��

The condition for oscillation is

�� = 1

�−ℎ��

ℎ��

��ℎ��

�� + (�� + ��)ℎ�� = 1

⎝

⎜⎛

−ℎ�� + ℎ��(�� + ��)�

ℎ��(�� + �� + ��) + �� + ��

ℎ��

⎠

⎟⎞

��ℎ��

�� + (�� + ��)ℎ�� = 1

−ℎ��

ℎ��(�� + �� + ��) + �� + ��= 1

−ℎ�� = ℎ��(�� + �� + ��) + �� + ��

ℎ��(�� + �� + ��) + �� + �� + �� = 0

ℎ��(�� + �� + ��) + ��1 + ℎ��+ �� = 0 --- (1)

This is the generalized equation for Hartley and Colpitt’s Oscillators

For Hartley Oscillator

�� = �� + ��


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�� = �� + ��

�� =1

��= −

�

��

Putting these values in equation (1),

ℎ�� (�� + ��) + (�� + ��) + �−�

�� + (�� + ��)(�� + ��)�1 + ℎ��

+ (�� + ��) �−�

�� = 0

ℎ�� + �� + �� + �� −��

� ��+ (�� + ��)(�� + ��)�1 + ℎ��

+ (�� + ��) �−�

�� = 0

��ℎ�� + �� + 2� −1

� ��+ ��(�� + �)��(�� + �)�1 + ℎ��+ ��(�� + �) �−

�

�� = 0

��ℎ�� + �� + 2� −1

� ��+ �� (�� + �)(�� + �)�1 + ℎ��+

(�� + �)

�= 0

�� ℎ�� + �� + 2� −1

� ��

� ��

− �� (�� + �)(�� + �)�1 + ℎ��−(�� + �)

��

� ��

= 0

Equating imaginary part with zero,

� ℎ�� + �� + 2� −1

� �� = 0

�� + �� + 2� −1

� ��= 0

1

� ��= �� + �� + 2�

� � =1

�(�� + �� + 2�)

� =1

��(�� + �� + 2�)

� =�

��(�� + �� + ��)

�� = �� + �� + ��

� =�

��√��

This gives frequency of oscillation in Hartley Oscillator

For Colpitt’s Oscillator

�� =1

��= −

�

� ��


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�� =1

��= −

�

� ��

�� = ��

Putting these values in equation (1),

ℎ�� −�

� ��+ �−

�

� ��+ (��)� + �−

�

� ��−

�

� ��1 + ℎ��+ �−

�

� ��(��) = 0

ℎ�� −�

� ��−

�

� ��+ ��−

1

� ��1 + ℎ��+

�

��= 0

�� ℎ�� −1

� ��−

1

� ��+ ��

� ��

− �1

� ��1 + ℎ��−

�

��

� ��

= 0

Equating imaginary part with zero,

� ℎ�� −1

� ��−

1

� ��+ �� = 0

−1

� ��−

1

� ��+ � = 0

1

� ��+

1

� ��= �

1

� � �1

��+

1

��= �

1

� � �� + ��

��= �

� � ��

�� + ��=

1

�

� � =1

� ��

�� + ��

� =1

� � ��

�� + ��

� =�

��

�� + ��

�� = ��

�� + ��

� =�

��√��

This gives frequency of oscillation in Colpitt’s Oscillator

Explain crystal oscillator. Derive expression of resonant frequencies. (S-15/7m) Explain the principle of working of crystal oscillator. (S-16/3m) Explain the working of crystal oscillator with the help of neat sketch. What are its advantages over other types of filter oscillator? (W-15/7m) Explain in brief crystal oscillator. (W-16/4m)


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CRYSTAL OSCILLATOR

L2 L1

Q

R1

R2CE

RE

C1

VCC

In the above circuit feedback is given through mutually coupled �� & ��. The fundamental

frequency of �� (tank circuit) should be equal to that of crystal. The tank circuit excites the

crystal oscillator, which starts oscillations. This oscillation is fed to the base of transistor,

which amplifies the oscillation with unity gain. As the frequency of crystal oscillator is

independent of temperature, the circuit operates at constant frequency.

In order to use crystal oscillator in a circuit, it must be connected such that its low impedance

in the series resonant operating mode or high impedance in parallel resonant operating mode

is selected.

The circuit has two resonant frequencies i.e. series resonant frequency and parallel resonant

frequency. When crystal is made to operate at series resonant frequency (i.e. � � �� ), the

reactance of its series arm (in equivalent circuit) becomes zero. Thus,

�� +1

��= 0

�� −�

��= 0

�� −1

��= 0

�� =1

��

�� =

1

��

(2�� )� =1

��

�� =1

2�√��

This is Series Resonant Frequency.

At parallel resonant frequency (i.e. �� )*, the reactance of the equivalent circuit becomes

zero. Thus,

�� +1

��+

1

��= 0

* Parallel resonant frequency is greater than series resonant frequency.

�� −1

��−

1

��= 0


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�� =1

��+

1

��

�� =

1

��+

1

��=

1

��

1

�+

1

��

�� =

1

��

� + ��

��

�2��

=1

��

� + ��

��

�� =1

2��

� + ��

�� =1

2��

This is Parallel Resonant Frequency ��

� ℎ��, �� =�. ��

� + ��

QUALITY FACTOR FOR CRYSTAL OSCILLATOR (QS) For Series Resonant Circuit,

�� =� ��

�=

2��

�

For Parallel Resonant Circuit,

�� =� � �

�=

2��

�

AC EQUIVALENT CIRCUIT FOR CRYSTAL OSCILLATOR

L

R

C

CM

CrystalOscillator


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Determine the series resonant & parallel resonant frequency and quality factor for quality crystal. Given � = �� , � = ��, � = �. �� & �� = �. � �� (S-14/6m) (W-14/4m)

Solution: Hint:

Series Resonant Frequency ��

�� =1

2�√��=

1

2�√150 × 10� � × 0.025 × 10� ��= 82.18��

Parallel Resonant Frequency ��

�� =1

2��

� + ��

Quality Factor for Series Resonant Circuit,

�� =� ��

�=

2��

�

Quality Factor for Parallel Resonant Circuit,

�� =� � �

�=

2��

�

A crystal has L = 0.33H, C=0.065 PF and CM = 1pF with R=5.5kΩ. Find (i) Series resonant frequency (ii) Parallel resonant frequency. (iii) By what percentage does the parallel resonant frequency exceed the series resonant frequency? (iv) Find the Q factor of the crystal. (S-17/5m)

The ac equivalent of a crystal has values L = 1H, C = 0.01 pF, R = 1000 Ω & CM = 20pF. Calculate fs, fp & Q of the crystal. By what percentage does fp exceeds is fs? (W-16/6m)

A Hartley Oscillator is designed with �� = � ��, �� = �� and a variable capacitance. If the frequency of oscillation is varied between �. � �� & �. � ��. Find the range of capacitance values. (S-15/6m)

A Hartley Oscillator is designed with �� = � ��, �� = �� and a variable capacitance. If the frequency of oscillation is varied between �� & �� . Find the range of capacitance values. (W-13/6m) (W-15/6m)

Solution: �� & �� .

� = �� + �� + 2�

�� = 0

∴ � = �� + ��


�� =1

2��


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�� =1

2��

A transistor Colpitt’s oscillator has �� = �� , �� = �� & � = ��. Find the frequency of oscillation and �� needed. (W-16/5m)

For the circuit shown in following figure. Calculate the frequency of oscillation. Also find �� for required gain condition if �� = ��.

R2

150KC2

0.0015μF

R1

100K

C1

0.001μF

R3

R4

VO+

-

Solution: Hint: Wein bridge Oscillator

� =1

2��

��

��−

��

��=

��

��

Explain with the help of waveform working of Astable multivibrator. (W-13/7m) (W-14/7m) (S-14/6m) (S-17/6m) Explain the working of astable multivibrator with its circuit diagram (S-15/6m)

TRANSISTORIZED ASTABLE MULTIVIBRATOR

Q2Q1

VCC

R1 R2 R3 R4

C1 C2A BC

Output

+ +L M J K


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Circuit Operation: When VCC is applied, collector currents start flowing in Q1 and Q2. In addition, the coupling

capacitors C1 and C2 also start charging up with the shown polarity. As the characteristics of no

two transistors (i.e. β, VBE) are exactly alike, therefore, one transistor, say Q1 will conduct more

rapidly than the other. This will drain the charge of capacitor C2, by this time the positive


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polarity of capacitor C1 will start the transistor Q2 and a base current will start flowing through

it. The collector current of Q2 is constituted through R3 and R4, which charges the capacitor C2

with the shown polarity. By this time, capacitor C1 discharges completely and charges with

reverse polarity. This reverse polarity switches the transistor Q2 OFF and the charge of

Capacitor C2 switches the transistor Q1 ON. The switching time of transistors depends on the

value of capacitor.

The period of time during which Q1 remains at saturation and Q2 at cut off is determined by C2

and R3.

ON or OFF TIME: The time for which either transistor remains ON or OFF is given by:

ON time for Q1 (or OFF time for Q2) is

�� = 0.694 × ��

OFF time for Q1 (or ON time for Q2) is

�� = 0.694 × ��

Total time period of the square wave is

� = �� + �� = 0.694(�� + ��)

As �� = �� = � and �� = �� = �,

� = 0.694(�� + ��) ≈ 1.4��

Frequency of the square wave will be � =�

�

Note: Numericals will be based on only the above formula.

Explain the working of transistorized monostable multivibrator circuit. (W-13/7m) (W-15/6m) (W-16/5m) Explain working of Monostable Multivibrator with its circuit diagram and waveforms. (S-16/7m)

TRANSISTORIZED MONOSTABLE MULTIVIBRATOR

Q2Q1

VCC

R1 R2 R4

C1A BC

Output

+

C2

R3

R5

Input PulseVBB


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Circuit Operation: With the circuit arrangement shown, Q1 is at cut off and Q2 is at saturation. This is the stable

state for the circuit and it will continue to stay in this state until a triggering pulse

is applied at C2. When a positive pulse of short duration and sufficient magnitude is applied to

the base of Q1 through C2, the transistor Q1 starts conducting and positive potential is

established at its collector. The positive potential at the collector of Q1 is coupled to the base of

Q2 through capacitor C1. This decreases the forward bias on Q2 and its collector current

decreases. The increasing negative potential on the collector of Q2 is applied to the base of Q1

through R3. This further increases the forward bias on Q1 and hence its collector current. With

these set of actions taking place, Q1 is quickly driven to saturation and Q2 to cut off.

With Q1 at saturation and Q2 at cut off, the circuit will come back to the stable stage (i.e. Q2 at

saturation and Q1 at cut off) after some time. The capacitor C1 sends a voltage to the base of Q2

to make it more positive. This goes on until a point is reached when forward bias is re-

established on Q2 and collector current starts to flow in Q2. The step by step events occur and

Q2 is quickly driven to saturation and Q1 to cut off. This is the stable state for the circuit and it

remains in this condition until another pulse causes the circuit to switch over the states.


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TRANSISTORIZED BISTABLE MULTIVIBRATOR


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Circuit Operation: When VCC is applied, one transistor (say ��) will start conducting slightly faster than other (say

��) as no two transistors have identical characteristics. This will drive �� to saturation and ��

to cut off. Assume that Q1 is turned ON and Q2 is cut OFF. This will reduce the forward bias on

�� and cause a decrease in its collector current and an increase in collector voltage. The rising

collector voltage is coupled to the base of �� where it forward biases the base-emitter junction

of ��. This will cause an increase in its collector current and decrease in collector voltage. The

decreasing collector voltage is applied to the base of �� where it further reverse biases the

base-emitter junction of �� to decrease its collector current. With these set of actions taking

place, �� is quickly driven to saturation and �� to cut off. The circuit will now remain stable in

this state until a positive triggering pulse at �� (or a negative triggering pulse at Q1) is applied.


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