GNITC Analog Circuits Lab Manual

Department of Electronics & Communication Engineering
(18ES0EC27)Analog Circuits Laboratory
II B.Tech-ECE II-Semester (2020-21)
LABORATORY MANUAL
(Student Copy)
SCHOOL OF ENGINEERING & TECHNOLOGY
GURU NANAK INSTITUTIONS TECHNICAL CAMPUS
(Affiliated to JNUTH, Approved by AICTE)
Ibrahimpatnam, RR Dist-501506
www.gniindia.org

LABORATORY MANUAL
GURU NANAK INSTITUTIONS TECHNICAL CAMPUS
(Affiliated to JNUTH, Approved by AICTE)
Ibrahimpatnam, RR Dist-501506
www.gniindia.org
Document No:
GNITC/ECE/AC Regulation:
Prepared by: Authorized by
HOD-ECE

GURU NANAK INSTITUITONS TECHNICAL CAMPUS
LABORATORY MANUAL
CONTENT PAGE
S.No Description Page No
1 Syllabus I
2 PEO’s & PO’s II
3 List of Experiments III
4 Experiment Planner IV- V

Syllabus
(18ES0EC27) ANALOG CIRCUITS LABORATORY
II B.Tech II semester, ECE branch.
(For Laboratory Examination – Minimum of 12 experiments)
PART A:
1. Common Emitter Amplifier
2. Common Base Amplifier
3. Common Source Amplifier
4. Two stage RC Coupled Amplifier
5. Current Shunt and Voltage Feedback Amplifier
6. Cascode Amplifier
7. Wien Bridge Oscillator using Transistors
8. RC Phase Shift Oscillator using Transistors
9. Hartley and Colpitt’s Oscillator
PART B:
1. Inverting and Non-inverting Amplifiers using Op Amps.
2. Comparators using Op Amp.
3. Integrator Circuit using IC 741.
4. Differentiator circuit using Op Amp.
5. To plot the frequency response of Ist order LPF.
6. To plot the frequency response of Ist order HPF
I

PROGRAM EDUCATIONAL OBJECTIVES (PEOs):
PEO1: Produce Industry ready graduates having the ability to apply knowledge across the
disciplines and in emerging areas of Electronics and Communication Engineering for
higher studies, employability and handle the realistic problems.
PEO2: Graduates shall have good communication skills, possess ethical conduct, sense of
responsibility to serve the society and protect the environment.
PEO3: Graduates shall have soft skills, managerial skills, leadership qualities and understand the
need for lifelong learning for a successful professional career.
PROGRAM OUTCOMES (POs) :
Engineering Graduates will be able to:
1. Engineering knowledge: Apply the knowledge of mathematics, science, engineering fundamentals,
and an engineering specialization to the solution of complex engineering problems.
2. Problem analysis: Identify, formulate, review research literature, and analyze complex engineering
problems reaching substantiated conclusions using first principles of mathematics, natural sciences, and
engineering sciences.
3. Design/development of solutions: Design solutions for complex engineering problems and design
system components or processes that meet the specified needs with appropriate consideration for the
public health and safety, and the cultural, societal, and environmental considerations.
4. Conduct investigations of complex problems: Use research-based knowledge and research methods
including design of experiments, analysis and interpretation of data, and synthesis of the information to
provide valid conclusions.
5. Modern tool usage: Create, select, and apply appropriate techniques, resources, and modern
engineering and IT tools including prediction and modeling to complex engineering activities with an
understanding of the limitations.
6. The engineer and society: Apply reasoning informed by the contextual knowledge to assess societal,
health, safety, legal and cultural issues and the consequent responsibilities relevant to the professional
engineering practice.
7. Environment and sustainability: Understand the impact of the professional engineering solutions in
societal and environmental contexts, and demonstrate the knowledge of, and need for sustainable
development.

8. Ethics: Apply ethical principles and commit to professional ethics and responsibilities and norms of the
engineering practice.
9. Individual and team work: Function effectively as an individual, and as a member or leader in diverse
teams, and in multidisciplinary settings.
10. Communication: Communicate effectively on complex engineering activities with the engineering
community and with society at large, such as, being able to comprehend and write effective reports and
design documentation, make effective presentations, and give and receive clear instructions.
11. Project management and finance: Demonstrate knowledge and understanding of the engineering and
management principles and apply these to one’s own work, as a member and leader in a team, to manage
projects and in multidisciplinary environments.
12. Life-long learning: Recognize the need for, and have the preparation and ability to engage in
independent and life-long learning in the broadest context of technological change..
PSOs:
PSO1: Solve, Design and analyze circuits in the field of Communication Engineering
PSO2: Design and implement Circuits in the field of Digital Signal Processing, Embedded Systems and
Antenna using various modern software tools
MAPPING OF PROGRAM EDUCATIONAL OBJECTIVES (PEO’S) AND PROGRAM
OUTCOMES (PO) FOR ELECTRONICS AND COMMUNICATION ENGINEERING
Program
Educational
Objectives
Program Outcomes
PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12 PSO1 PSO2
PEO1 3 3 3 3 3 2 1 1 1 2 2 2 3 3
PEO2 1 1 1 1 1 3 3 3 2 3 1 1 1 2
PEO3 1 1 1 1 2 1 1 2 3 3 3 3 2 2
II

LABORATORY MANUAL
LIST OF EXPERIMENTS
SNo Name of Experiment Page No.
PART-A
1
Common Emitter Amplifier Part-A
1
2 Common Base Amplifier 5
3 Common Source Amplifier 9
4
Two stage RC Coupled Amplifier 14
5 Current Shunt and Voltage Feedback Amplifier 18
6 Cascode Amplifier 22
7 Wien Bridge Oscillator using Transistors 25
8 RC Phase Shift Oscillator using Transistors 28
9 Hartley and Colpitt’s Oscillator 32
PART-B
10
Inverting and Non-inverting Amplifiers using Op Amps. Part-B
6
11 Comparators using Op Amp.
12
12 Integrator Circuit using IC 741 15
13 Differentiator circuit using Op Amp. 18
14 To plot the frequency response of Ist order LPF. 21
15 To plot the frequency response of Ist order HPF 25
III

(18ES0EC27) Analog Circuits Laboratory
LABORATORY MANUAL
Lab Experiment Planner
Batches: Total Number of Experiments:
Total Number of Batches: 12 (with a maximum of ‘3’ students/batch)
2019-20
Experiment/
Section
Exp-1
Exp-2
Exp-3
Exp-4
Exp-5
Exp-6
Exp-7
Exp-8
Exp-9
Exp-10
Exp-11
Exp-12
Exp-13
Section-1
Section-2
Section-3
Section-4
Section-5
Section-6
Section-7
IV

II B.Tech -ECE II-Semester (2020-21)
Date of Completion (experiment wise)
Experiment/
Section
Exp-1
Exp-2
Exp-3
Exp-4
Exp-5
Exp-6
Exp-7
Exp-8
Exp-9
Exp-10
Exp-11
Exp-12
Exp-13
Section-1
Section-2
Section-3
Section-4
Section-5
Section-6
V

Department of Electronics and Communication Engineering Analog Electronics Lab
GNITC
EXPERIMENT – 1
COMMON EMITTER AMPLIFIER
AIM: To determine the maximum gain, 3dB gain, lower and upper cutoff frequencies and
bandwidth of CE amplifier by performing the AC analysis
APPARATUS REQUIRED:
S.NO Name of the component quantity
1. Regulated power supply (12V) 1
2. Function generator 1
3. CRO 1
4. Transistor (BC 107) 1
5. Resistors (10KΩ,4.7 KΩ,1 KΩ) 1
6. Resistor (100 KΩ) 2
7. Capacitors (100 µF, 10 µF) 1,2
8. Bread Board 1
9 Connecting wires 10
THEORY:
The practical circuit of CE amplifier is shown in the figure. It consists of different circuit
components. The functions of these components are as follows:
1. Biasing Circuit: The resistances R1, R2 and RE form the voltage divider biasing circuit
for the CE amplifier. It sets the proper operating point for the CE amplifier.
2. Input capacitor C1: This capacitor couples the signal to the transistor. It blocks any dc
component present in the signal and passes only ac signal for amplification. Because of this,
biasing conditions are maintained constant.
3. Emitter Bypass Capacitor CE: An emitter bypass capacitor CE is connected in parallel
with the emitter resistance, RE to provide a low reactance path to the amplified ac signal. If it
is not inserted, the amplified ac signal passing through RE will cause a voltage drop across it.
This will reduce the output voltage, reducing the gain of the amplifier.
4. Output Coupling Capacitor C2: The coupling capacitor C2 couples the output of the
amplifier to the load or to the next stage of the amplifier. It blocks DC and passes only AC
part of the amplified signal.

GNITC
OPERATION:
When positive half of the signal is applied, the voltage between base and emitter (Vbe) is
increased because it is already positive with respect to ground. So forward bias is increased
i.e., the base current is increased. Due to transistor action, the collector current IC is increased
2times. When this current flows through RC the drop IC RC increases considerably. As a
consequence of this, the voltage between collector and emitter (Vce) decreases. In this way,
amplified voltage appears across RC). Therefore the positive going input signal appears as a
negative going output signal i.e., there is a phase shift of 180° between the input and output.
PROCEDURE :
1. Connect the circuit as shown in the CIRCUIT DIAGRAM.
2. Connect the signal generator output to input terminals of the circuit and CH-I of dual
trace CRO.
3. Connect the output terminal of the circuit CH-II of the dual trace CRO.
4. Set the power supply voltage to 12V and connect to the circuit.
5. Set the signal generator output sine wave of 1000 Hz at 40 mV constant. (Vp-p)
6. Vary the function generator frequency from 100 Hz to 1MHz (as per in the given
tabular form) and note the corresponding output voltage.
7. Calculation the gain AV =Vo/Vi .
8. Plot the graph frequency verses gain (dB) on a semi log sheet.
CIRCUIT DIAGRAM:
Fig1: Common Emitter Amplifier CIRCUIT
DIAGRAM MODEL GRAPH

GNITC
Precautions
1. Connections must be given very carefully.
2. Readings should be noted without any parallax error.
3. The applied voltage, current should not exceed the maximum rating of the given
transistor.
Tabulations:
Input AC voltage, VI= 0.1mV (peak-peak)
S.No Frequency
(HZ)
Output Voltage (Vo) Gain in decibels
(Volts-p-p) AV=20 log (Vo/ VI)
1 50 2 33.97
2 100 2.2 34.8
3 500 2.4 35.56
4 1K 2.4 35.56
5 5K 2.4 35.56
6 10K 2.4 35.56
7 100K 2.2 34.8
8 500K 1 27.95
9 1M 0.6 23.52
OBSERVATIONS: practical
. Maximum gain (Av) = 35.56 dB
. Lower cut-off frequency (Fl) = 40kHz
. Upper cut-off frequency (FH) = 200kHz
. Band width (B.W) = (FH – FL) =199.9kHz

GNITC
RESULT: The voltage gain and frequency response of the CE amplifier are obtained. Also
gain bandwidth product of the amplifier is calculated.
Discussion/Viva Questions:
1. What is the equation for voltage gain?
Ans:
2. What is cut off frequency? What is lower 3dB and upper 3dB cut off frequency?
Ans: In electronics, cutoff frequency or corner frequency is the frequency either above or
below which the power output of a circuit, such as a line, amplifier, or electronic filter has
fallen to a given proportion of the power in the pass band. Most frequently this proportion is
one half the pass band power, also referred to as the 3 dB point since a fall of 3 dB
corresponds approximately to half power. As a voltage ratio this is a fall to of the pass band
voltage
3. What are the applications of CE amplifier?
Ans: Low frequency voltage amplifier, radio frequency circuits and low-noise amplifiers
4. What is active region?
Ans: The active region of a transistor is when the transistor has sufficient base current to turn
the transistor on and for a larger current to flow from emitter to collector. This is the region
where the transistor is on and fully operating. In this region JE in forward bias and JC in
reverse bias and transistor works as an amplifier
5. What is Bandwidth of an amplifier?
Ans: Bandwidth is the difference between the upper and lower frequencies in a continuous
set of frequencies. It is typically measured in hertz, and may sometimes refer to pass band
bandwidth, sometimes to baseband bandwidth, depending on context. Pass band bandwidth is
the difference between the upper and lower cutoff frequencies of, for example, a band pass
filter, a communication channel, or a signal spectrum. In case of a low-pass filter or baseband
signal, the bandwidth is equal to its upper cutoff frequency.

GNITC Page 5
EXPERIMENT – 2
COMMON BASE AMPLIFIER
AIM: Obtain the frequency response characteristics of CB amplifier
APPARATUS REQUIRED:
S.N
O Name of the component quantity
1. Regulated power supply 1No.
2. Function generator 1No.
3. CRO 1No.
4. Transistor (BC 107) 1No.
5. Resistors(100Ω,1KΩ,4.7KΩ,10KΩ) 4 No.
6. Capacitors(1 µF,10 µF) 2Nos.
7. Bread Board 1No.
8. Connecting wires As required
THEORY:
In Common Base Amplifier Circuit Base terminal is common to both the input and output
terminals. In this Circuit input is applied between emitter and base and the output is taken
from collector and the base. As we know, the emitter current is greater than any other current
in the transistor, being the sum of base and collector currents i.e. IE= IB+ IC In the CE and
CC amplifier configurations, the signal source was connected to the base lead of the
transistor, thus handling the least current possible. Because the input current exceeds all
other currents in the circuit, including the output current, the current gain of this amplifier is
actually less than 1 (notice how Rload is connected to the collector, thus carrying slightly
less current than the signal source). In other words, it attenuates current rather than
amplifying it. With common-emitter and common-collector amplifier configurations, the
transistor parameter most closely associated with gain was β. In the common-base circuit, we
follow another basic transistor parameter: the ratio between collector current and emitter
current, which is a fraction always less than 1. This fractional value for any transistor is
called the alpha ratio, or α ratio.( α= IC/IE) Since it obviously can't boost signal current, it
only seems reasonable to expect it to boost signal voltage.
Operation: The positive going Pulse of input Source increases the emitter voltage. As the
base voltage is Constant, the forward bias of emitter base junction reduces. This reduces IB,
reducing IC and hence the drop across RC since VO=VCC - IC RC, the reduction in IC
results in an increase in VO. Therefore, we can Say that positive going input produces
positive going output and similarly negative going input produces negative going output and
there is no phase shift between input and output in a common base Amplifier.

GNITC Page 6
PROCEDURE:
1. Connect the circuit on the bread board as shown in fig
2. Set Vin = 200mV at 1 KHz.
3. Increase Vin till undistorted waveform is seen on the CRO.
4. Measure the input voltage Vin.
5. Keeping the input voltage constant at this value, vary the frequency from dc to 1
MHz in convenient steps and measure the Vout at each frequency
6. Find the voltage gain
7. Plot AV VS frequency on a semi-log sheet.
8. Expected Graphs:
CIRCUIT DIAGRAM
MODEL GRAPH

GNITC Page 7
Tabulations:
Input AC voltage, VI=200mV (peak-peak)
1. Maximum gain (Av) =24.34dB
2. Lower cut-off frequency (Fl) =500HZ
3. Upper cut-off frequency (FH) = 400KHZ
4. Band width (B.W) = (FH – FL) =399.5kHz
5. GainbandwidthproductfT =9.723MHz
S.No Frequency (HZ)
Output Voltage (Vo) Gain in decibels
(Volts-p-p) AV=20 log (Vo/ VI)
1 50 0.6 9.54
2 100 0.8 12.04
3 500 2.4 21.58
4 1K 3 23.52
5 5K 3.3 24.34
6 10K 3.3 24.34
7 100K 3.3 24.34
8 500K 2.2 20.82
9 1MHz 1.4 16.9

GNITC Page 8
Precautions
transistor.
RESULT: The voltage gain and frequency response of the CB amplifier are obtained. Also gain
bandwidth product of the amplifier is calculated.
1. What is transistor?
Ans: A transistor is a semiconductor device used to amplify and switch electronic signals and
electrical power. It is composed of semiconductor material with at least three terminals for
connection to an external circuit. The term transistor was coined by John R. Pierce as a
portmanteau of the term "transfer resistor".
2. Write the relation between and ?
Ans:
3. Define (alpha)? What is the range of ?
Ans: The important parameter is the common-base current gain, . The common-base
current gain is approximately the gain of current from emitter to collector in the forward-
active region. This ratio usually has a value close to unity; between 0.98 and 0.998.
4. Why is less thanunity?
Ans: It is less than unity due to recombination of charge carriers as they cross the base
region.
5. Input and output impedance equations for CB configuration?
Ans: hib = VBE / IE, 1 / hoe = VCE / IC

GNITC Page 9
EXPERIMENT –3
COMMON SOURCE AMPLIFIER
AIM:Obtain the frequency response characteristics of CS amplifier
APPARATUS REQUIRED:
S.No Name Of The Component/Equipment Qty
1 Field Effect Transistor(BFW10) 1
2 Capacitors(designed values) 3
3 Resistors(designed values) 4
4 Function Generator 1
5 Cathode Ray Oscilloscope 1
6 Regulated Power Supply 1
THEORY
In common source amplifier circuit source terminal is made common to the other two
terminals. In common source amplifier circuit input is applied between gate and source
and output is taken from drain and source. The coupling capacitors C1 and C2 are used to
isolate the D.C biasing from the applied ac signal, and acts as short circuit for the ac
analysis. The high frequency characteristics of the FET amplifier are determined by the
inter electrode and wiring capacitance.
The CS amplifier which provides good voltage amplification is most frequently used. In
cascade amplifier input impedance of the second stage acts as shunt across output of first
stage and Rd is shunted by Ci. Since the reactance decreases with increasing frequencies,
the output impedance will be low at high frequencies, this will result in decreasing the
gain at high frequencies
PROCEDURE
1. Connect the circuit as per the CIRCUIT DIAGRAM as shown in Fig.1.
2. Apply supply voltage, VDD of 12V.
3. Now feed an AC signal 40mV at the input of the amplifier with different
frequencies ranging from 100HZ to 100 MHZ and measure the amplifier output
voltage.
4. Now calculate the gain in decibels at various input signal frequencies.
5. Draw a graph with frequency on X- axis and gain in dB on Y- axis. From graph
calculate bandwidth.

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CIRCUIT DIAGRAM
Fig: Common Source Amplifier CIRCUIT
DIAGRAM
MODEL GRAPH

GNITC Page 11
Tabular Form
ac Input voltage VI = 40mV (peak-peak)
S.No Frequency (Hz)
Output Voltage(Vo) Gain in decibels
(Volts-p-p) AV=20 log (Vo /
Vi)
1 50 0.08 13.06
2 100 0.21 14.4
3 500 0.21 14.4
4 1K 0.21 14.4
5 5K 0.21 14.4
6 10K 0.22 14.8
7 100K 0.22 14.8
8 500K 0.1 8.78
9 1M 0.07 4.86
3. Upper cut-off frequency (FH) =250KHZ
4. Band width (B.W) = (FH – FL) =249.96KHz

GNITC Page 12
Precautions
1. Connections must be made very carefully.
3. The applied voltage and current should not exceed the maximum
ratings of the given transistor.

13
RESULT: The voltage gain and frequency response of the CS amplifier are obtained. Also gain
bandwidth product of the amplifier is calculated.
1. Why FET is called as unipolar device?
Ans: FETs are unipolar transistors as they involve single-carrier-type operation.
2. Why the common-source (CS) amplifier may be viewed as a transconductance
amplifier or as a voltage amplifier?
Ans: As a transconductance amplifier, the input voltage is seen as modulating the current
going to the load. As a voltage amplifier, input voltage modulates the amount of current
flowing through the FET, changing the voltage across the output resistance according to
Ohm's law. However, the FET device's output resistance typically is not high enough for a
reasonable transconductance amplifier (ideally infinite), nor low enough for a decent voltage
amplifier (ideally zero).
3. What are the characteristics of JFET source amplifier?
Ans: At low frequencies and using a simplified hybrid-pi model, the following small-
signal characteristics can be derived.
4. What is the impedance of
FET?
Ans:

14
EXPERIMENT –4
CASCODE AMPLIFIER
Aim: Obtain the frequency response characteristics of Cascode amplifier
S.NO Name of the component quantity
1. Regulated power supply (12V) 1
2. Function generator 1
3. CRO 1
4. Transistor (BC 107 or 2N2222) 2
5. Resistors (desired values) 7
7. Capacitors (10 µF, 100µF) 3,1
8. Bread Board 1
THEROY:
Cascode amplifier is a special case of cascade amplifier. Cascode amplifier is a two stage
amplifier, which comprises a CE amplifier driving a CB amplifier. The CE amplifier has with
significant current and voltage gain moderate input and output impedance. The high input
impedance is desirable while the high output impedance possess some problems. The higher
the output impedance the less current can be drawn from an amplifier without a significant
drop in output voltage .The CE is used most often for voltage amplification .It can provide a
large output voltage swing. In multistage system this output becomes the input of the next
stage of the system .The emitter resistor amplifier is similar to the CE amplifier but has lower
voltage gain and higher input impedance .Because of the feedback present in this amplifier,
internal noise generated by the transistor is almost eliminated. The CB amplifier has low
input impedance and relatively high output impedance. These properties are less desirable for
signal amplification. If the base is bypassed to ground with a capacitor, the amplifier has high
voltage gain but the current gain is less than unity .Thus if the source driving the amplifier
has a low impedance, and the load is drawing little current, the CB can be used as a voltage
amplifier .If the driving source has a higher impedance, we can offset this undesirable effect
by using a CE amplifier to drive the CB amplifier. Thus the overall input impedance is high.
PROCEDURE:
1. Connect 1Vp-p, 100Hz Sine wave signal at the input (between points Vin and gnd) of
amplifier of board and observe the same on oscilloscope CH I.
2. Observe the output waveform between points Vout and gnd on oscilloscope CH II.
3. Increase the input frequency value and observe the output waveform amplitude on
oscilloscope
4. Measure the maximum amplitude of the output sine wave and the frequency range for
which the output wave amplitude is 3dB down the maximum amplitude.(this will give two
valves of frequency Fl and fH, the lower 3dB frequency and higher 3dB frequency
respectively) as shown in fig.
5. Calculate Bandwidth of Cascode amplifier using Eq.2

15
CIRCUIT DIAGRAM
Tabular Form
Input AC voltage, Vi = _ 40mV (peak-peak)
S.No Frequency (Hz)
Output voltage Vo
(Volts) (peak-
peak)
Gain in decibels
AV=20 log (Vo / Vi)
1 50 1.6 32.04
2 100 3.2 38.06
3 500 4 40
4 1K 4 40
5 5K 4 40
6 10K 4 40
7 100K 4 40
8 500K 2.5 35.9
9 1M 1.6 32.04
1. Maximum gain (Av) =40dB

16
MODEL GRAPH

DEPARTMENT OF ELECTRONICS AND COMMUNICATION
ENGINEERING
GURU NANAK INSTITUTIONS TECHNICAL CAMPUS.
ANALOG CIRCUITS LABORATORY MANUAL
Precautions
transistor.
Results: Obtained the frequency response characteristics of CS amplifier
1. What is an amplifier?
Ans The device that amplifies the amplitude of the input signal is called the amplifier. An
amplifier may be defined as a device that increases the current, voltage or power of an input
signal with the help of a transistor by furnishing the additional power from a separate source
of supply.
2.What do you mean by operating point?
Ans The zero signal values of IC and VCEare known as the operating point. It is called
operating point beacue the variations of IC and VCE take place about this point when signal is
applied. It is also known as the quiescent or Q-point.
3.Define what is transistor biasing?
Ans The proper flow of zero signal collector current and the maintenance of proper collector
emitter voltage during the passage of signal is called the transistor biasing.
4.What is faithful amplification?
Ans The process of raising the strength of a weak signal without any change in its general
shape is referred to as faithful amplification.
5.Define what is meant by small signal amplifier?
Ans:When the input signal is quite weak and produces less small fluctuations in the output
current in comparison to its quiescent value, the amplifier is called the small signal or voltage
amplifier.

ENGINEERING
EXPERIMENT –5
RC PHASE SHIFT OSCILLATOR
Aim:
To design RC phase shift oscillator for given specifications. Compare the theoretical and
practical values.
APPARATUS REQUIRED
S.NO NAME OF THE
EQUIPMENT
RANGE QUANTITY
(NO.S)
1 Transistor BC107 1
2 Resistors 47kΩ, 10kΩ,2.2kΩ,680Ω one from each
3 Resistor 4.7kΩ 3
3 Capacitors 10μF 3
4 Capacitor 0.01 μF 3
5 CRO 1
6 RPS (0 – 30V) 1
7 Bread Board Bread Board 1
THEORY
An oscillator is an electronic circuit for generating an AC signal voltage with a DC supply as the
only input requirement. The frequency of the generated signal is decided by the circuit elements
used. An oscillator requires an amplifier, a frequency selective network and a positive feedback
from the output to the input.
The Barkhausen criterion for sustained oscillation is Aβ = 1 where A is the gain of the amplifier
and β is the feedback factor (gain).The unity gain means signal is in phase. ( If the signal is 1800
out of phase and gain will be -1). RC-Phase shift Oscillator has a CE amplifier followed by three
sections of RC phase shift feed-back Networks. The output of the last stage is return to the input
of the amplifier. The values of R and C are chosen such that the phase shift of each RC section is
60º.Thus The RC ladder network produces a total phase shift of 180º between its input and
output voltage for the given frequency. Since CE Amplifier produces 180 º phases shift. The total
phase shift from the base of the transistor around the circuit and back to the base will be exactly
360º or 0º. This satisfies the Barkhausen condition for sustaining oscillations and total loop gain
of this circuit is greater than or equal to 1, this condition used to generate the sinusoidal
oscillations.
PROCEDURE:
1. Connections are made as per CIRCUIT DIAGRAM Fig. 1.
2. Switch on the power supply.
3. Connect the CRO at the output of the circuit and apply supply voltage of VCC
=12v.
4. Compare the simulation frequency and practical frequency values.
5. Plot the graph for amplitude versus frequency.

ENGINEERING
CIRCUIT DIAGRAM
Fig. RC Phase Shift Oscillator CIRCUIT DIAGRAM
MODEL GRAPH
Tabular Form
Theoretical
S.NO Amplitude(v) p-p Frequency (KHZ)
1 1.4 1.3

ENGINEERING
Practical
S.NO Amplitude(v) p-p Frequency (KHZ)
1 1.4 0.9
Precautions
1. Connections must be done very carefully.
3. The applied Voltage and current should not exceed the maximum ratings of the
given transistor.
Result: Frequency of oscillations of RC phase shift oscillator is calculated.

ENGINEERING
1. What is RC phase shift oscillator ?
Ans: A phase-shift oscillator is a simple electronic oscillator circuit. It consists of an inverting
amplifier element such as a transistor or op amp, with its output fed back into its input through
an filter consisting of a network of resistors and capacitors. The feddback network 'shifts' the
phase of the amplifier output by 180 degrees at the oscillation frequency, to give positive
feedback.[1] Phase shift oscillators are mostly used at lower frequencies, often in the audio
frequency range as audio oscillators.
2. What are the advantages of RC phase shift oscillators?
Ans Advantages:
i. It is best suited for generating fixed frequency signals in the audio frequency range.
ii. It requires no transformer or inductor, hence less bulky.
iii. Simple Circuit.
*Pure sine wave output is possible.
3. What are the disadvantages of RC phase shift oscillators?
Ans Disadvantages:
i) It requires a high β transistor to overcome losses in the network.
ii) These oscillators are not suitable for high frequency operation.
iii) Frequency of oscillation can not be changed easily. To change the frequency of oscillation,
the three capacitor or resistors should be changed simultaneously. This is inconvenient.
4..Define piezoelectric effect.
Ans:The piezo electric Crystals exhibit a property that if a mechanical stress is applied across
one face the electric potential is developed across opposite face and viceversa. This phenomenon
is called piezo electric effect.
5.What are the applications of RC phase shift oscillator?
Ans:RC phase shift oscillators are used for musical instruments, oscillators, voice synthesis, and
GPS units. They work at all audio frequency

ENGINEERING
EXPERIMENT –6
(a) COLPITTS OSCILLATOR.
AIM:
To design and set up a Colpitts oscillator using BJT and to observe the sinusoidal output
waveform
Apparatus
S.No Name Range / Value Quantity
1. DC Regulated Power Supply (0-30V) 1
2. Resistors 100K ,10K , 470 K Each 1
4. Resistors 4.7 K 1
5. Capacitors 10 F ,100 F, 0.01 F 2,1,1
6. Decade Inductance Box -- 2
7. Decade Capacitance Box -- 1
8. CRO -- 1
THEORY:
A Colpitts oscillator is the electrical dual of a Hartley oscillator, where the feedback signal
is taken from an "inductive" voltage divider consisting of two coils in series (or a tapped
coil). Fig. 1 shows the common-base Colpitts circuit. L and the series combination of C1 and
C2 form the parallel resonant tank circuit which determines the frequency of the oscillator.
The voltage across C2 is applied to the base-emitter junction of the transistor, as feedback to
create oscillations. Fig. shows the common-collector version. Here the voltage across C1
provides feedback. The frequency of oscillation is approximately the resonant frequency of
the LC circuit, which is the series combination of the two capacitors in parallel with the
inductor
PROCEDURE
1. Connect the CIRCUIT DIAGRAM as shown in the figure.
2. Switch on the power supply.
3. Connect the output terminals to CRO.
4. Adjust the capacitances until a sinusoidal wave form is observed on the CRO.
5. Measure the time period of the sinusoidal wave form (T) and determine the
Frequency (1/T).
6. Repeat the above steps for different values of L, C1 &C2.
7. Tabulate the readings and compare with theoretical values

ENGINEERING
CIRCUIT DIAGRAM :
MODEL GRAPH

ENGINEERING
TABULATION:
S.NO. L (uH)
C (
F)
Practical
frequenc
y (Hz)
Theoretica
l
Frequency
(Hz)
C1 C2
1
100 0.1 0.1 277.7K 275K
RESULT: the Hartley oscillator was designed and its output waveform was verified.
1. How to obtained colpitt’s oscillator form basic form of LC oscillator?
Ans:Using X1 and X2 as capacitors and X3 as inductors, colpitt’s oscillator from basic form of
LC oscillator is obtained
2. Write down the advantages, disadvantages and applications of colpitt’s
oscillator. Ans: Advantages:
a) Simple construction.
b) It is possible to obtain oscillations at very high frequencies.
3. Write down the disadvantages and applications of colpitt’s
oscillator. Ans: Disadvantages:
a) It is difficult to adjust the feedback as it demands change in capacitor values.
b) Poor frequency stability.
4. What are applications of colpitt’s
oscillator? Ans: Application: a) As a high
frequency generator
5. What is an oscillator?
Ans: An oscillator is a circuit which basically acts as a generator, generating the output signal
which oscillates with constant amplitude and constant desired frequency.

ENGINEERING
EXPERIMENT –9
(b ) HARTLEY OSCILLATOR.
AIM: To design and set up a Hartley oscillator using BJT and to observe the sinusoidal output
waveform.
APPARATUS:
1. D.C Regulated Power Supply (0 – 30V) 1
2. Resistors 100KΩ, 1KΩ, 10kΩ,
47KΩ
Each 1
3. Capacitors 0.01µF,10 µF, 100µF 2
4. Decade Capacitance Box -- 2
5. Decade Inductance Box -- 1
6. CRO -- 1
THEORY:
The Hartley oscillator is an electronic oscillator circuit in which the oscillation frequency is
determined by a tuned circuit consisting of capacitors and inductors, that is, an LC oscillator. The
Hartley oscillator is distinguished by a tank circuit consisting of two series-connected coils (or,
often, a tapped coil) in parallel with a capacitor, with an amplifier between the relatively high
impedance across the entire LC tank and the relatively low voltage/high current point between
the coils. The Hartley oscillator is the dual of the Colpitts oscillator which uses a voltage divider
made of two capacitors rather than two inductors. Although there is no requirement for there to
be mutual coupling between the two coil segments, the circuit is usually implemented using a
tapped coil, with the feedback taken from the tap, as shown here. The optimal tapping point (or
ratio of coil inductances) depends on the amplifying device used, which may be a bipolar
junction transistor.
PROCEDURE:
1.Connect the circuit as shown in the figure.
2. Connect the O / P of the oscillator to the C.R.O.
3. Adjust the Capacitance and Inductance Boxes until a sinusoidal signal is observed in the
CRO. 4.Determine the frequency of the wave form.
5.Determine the frequency by varying the capacitance in convenient steps.
6.Tabulate the readings and compare the readings with the theoretical values

ENGINEERING
CIRCUIT DIAGRAM :
MODEL GRAPH
TABULATION:

ENGINEERING
Capacitance
C ( F )
Inductance
( m H ) Theoretical
Frequency (Hz)
Practical
Frequency (Hz)
L 1 L 2
0.1uF 10 40 2.25K 2.27K
RESULT: The Hartley oscillator was designed and its output waveform was verified
1.How to obtain Hartley oscillator from the basic form of LC oscillator
Ans Using X1 and X2 as inductors and X3 as capacitor, Hartley oscillator from basic form of
LC oscillator is obtained.
2. Write down the advantages, disadvantages and applications of Hartley
oscillator. Ans: Advantages:
a) It is easy to tune
b) It can operate over a wide frequency typically from few Hz and several MHz.
c) It is easy to change the frequency by means of a variable capacitor.
3.Write down the disadvantages of Hartley oscillator.
Ans Disadvantages:
a) Poor frequency stability.
4. Write down applications of Hartley oscillator.
Ans Applications:
a) it is used as local oscillator in radio and TV receivers.
b) In the function generator.
c) In RF sources
5. Write down the general applications of oscillators.
Ans a) As a local oscillator in radio receivers.
b) In T.V receivers.
c) In signal generators.
d) As clock generation for logic circuits.
e) AM and FM transmitters.
f) In phase lock loops

ENGINEERING
EXPERIMENT –7
WIEN BRIDGE OSCILLATOR
AIM : Design and generate a sine wave for different RC values (Wien Bridge oscillator).
APPARATUS REQUIRED
S.NO Name of the equipment Range Quantity
1 Transistor BC547 2
2 Resistors 47kohms,10k,2k,680ohms 1 each
3 Resistor 4.7kohms 3
4 Capacitors 1microF,22microF 1 each
5 Capacitor 0.01microF 3
6 CRO 1
7 RPS (0-30V) 1
8 Bread Board 1
THEORY
An oscillator is an electronic circuit for generating an AC signal voltage with a DC
supply as the only input requirement. The frequency of the generated signal is decided by the
circuit elements used. An oscillator requires an amplifier, a frequency selective network and a
positive feedback from the output to the input.
The Barkhausen criterion for sustained oscillation is Aβ = 1 where A is the gain of the
amplifier and β is the feedback factor (gain).The unity gain means signal is in phase. ( If the
signal is 1800 out of phase and gain will be -1).
A Wien bridge oscillator is a type of electronic oscillator that generates sine waves. It
can generate a large range of frequencies. The oscillator is based on a bridge circuit originally
developed by Max Wien in 1891 for the measurement of impedances. The bridge comprises
four resistors and two capacitors. The oscillator can also be viewed as a positive gain
amplifier combined with a bandpass filter that provides positive feedback. Automatic gain
control, intentional non-linearity and incidental non-linearity limit the output amplitude in
various implementations of the oscillator.
The circuit shown to the right depicts a common implementation of the oscillator,
with automatic gain control, using modern components. Under the condition that R1=R2=R
and C1=C2=C, the frequency of oscillation is given by: and the condition of stable oscillation
is given by and the condition of stable oscillation is given by

ENGINEERING
PROCEDURE
1. Identify the pin details of BC107 Transistor (or equivalent silicon Transistor such as
BC108/547) and test it using a millimetre. Set up the circuit on breadboard as shown
in figure.
2. A 12V Supply Voltage is given by using Regulated power supply and output is taken
from collector of the Transistor.
3. By using CRO the output time period and voltage are noted.
4. Plot all the readings curves on a single graph sheet
CIRCUIT DIAGRAM
OUT PUT WAVE FORM

ENGINEERING
TABULAR FORM
S. No. R C
Theoretical
Frequency
(kHz)
Practical
Frequency
(kHz) Amplitude
(volts-p-p)
% Error
1 1KΩ 0.1µF 1.59 1.0 2.0 37
2 1KΩ 0.01µF 15.91 13.88 2.0 12
Precautions
The applied voltage, current should not exceed the maximum rating of the given transistor
RESULT : Frequency of oscillations of Wien Bridge oscillator is calculated.

ENGINEERING
1. Mention two essential conditions for a circuit to maintain oscillations[
Ans:The conditions for oscillator to produce oscillation are given by Barkhausan
criterion. They are:
i) The total phase shift produced by the circuit should be 360o
or 0o
ii) The Magnitude of loop gain must be greater than or equal to 1 (ie)|Aβ|≥1
2.. Differentiate oscillator from amplifier.
Ans:Oscillators
1. They are self-generating circuits. They generate waveforms like sine, square and triangular
waveforms of their own. Without having input signal.
2. It have infinite gain
3. Oscillator uses positive feedback.
Amplifiers
1. They are not self-generating circuits. They need a signal at the input and they just increase
the level of the input waveform.
2. It have finite gain
3. Amplifier uses negative feedback.
3.State Barkhausen criterion for sustained oscillation. What will happen to the
oscillation if the magnitude of the loop gain is greater than unity?
The conditions for oscillator to produce oscillation are given by Barkhausan criterion. They
are :i) The total phase shift produced by the circuit should be 360o
or 0o
ii) The Magnitude of loop gain must be greater than or equal to 1 (ie)|Aβ|≥1
In practice loop gain is kept slightily greater than unity to ensure that oscillator work even if
there is a slight change in the circuit parameters.
4..Why an LC tank circuit does not produce sustained oscillations.
Ans:We know that the inductor coil has some resistance and dielectric material of the
capacitor has some leakage.so small part of the originally imparted energy is used to
overcome these losses. As a result, the amplitude of oscillating current goes on decreasing
and becomes zero when all energy is consumed as losses. So a LC tank circuit does not
produce sustained oscillations.
5. What is the necessary condition for a Wien bridge oscillator circuit to have sustained
oscillations?
AnsThen for oscillations to occur in a Wien Bridge Oscillator circuit the following conditions
must apply.
ü With no input signal the Wien Bridge Oscillator produces output oscillations.
ü The Wien Bridge Oscillator can produce a large range of frequencies.
ü The Voltage gain of the amplifier must be at least 3.

ENGINEERING
EXPERIMENT –8
TWO STAGE RC COUPLED AMPLIFIER
Aim:Todeterminethephaserelationshipbetweentheinputandoutputvoltagesby performing the
transient analysis..
APPARATUS REQUIRED
1. D.C Regulated power
supply
(0 – 30V) 1
2. Transistors BC 107 2
3. Resistors 10KΩ, 100KΩ, 4.7KΩ Each 2
4. Resistors 1KΩ 3
5. Capacitors 10µF 5
6. Function Generator -- 1
7. CRO -- 1
THEORY
An amplifier is the basic building block of most electronic systems. Just as one brick
does not make a house, a single-stage amplifier is not sufficient to build a practical electronic
system. The gain of the single stage is not sufficient for practical applications. The voltage
level of a signal can be raised to the desired level if we use more than one stage. When a
number of amplifier stages are used in succession (one after the other) it is called a multistage
amplifier or a cascade amplifier. Much higher gains can be obtained from the multi-stage
amplifiers.
In a multi-stage amplifier, the output of one stage makes the input of the next stage.
We must use a suitable coupling network between two stages so that a minimum loss of
voltage occurs when the signal passes through this network to the next stage. Also, the dc
voltage at the output of one stage should not be permitted to go to the input of the next. If it
does, the biasing conditions of the next stage are disturbed.
Figure shows how to couple two stages of amplifiers using RC coupling scheme. This
is the most widely used method. In this scheme, the signal developed across the collector
resistor RC (R2)of the first stage is coupled to the base of the second stage through the
capacitor CC .(C2) The coupling capacitor blocks the dc voltage of the first stage from
reaching the base of the second stage. In this way, the dc biasing of the next stage is not
interfered with. For this reason, the capacitor CC (C2)is also called a blocking capacitor.

ENGINEERING
As the number of stages increases, the gain increases and the bandwidth decreases. RC
coupling scheme finds applications in almost all audio small-signal amplifiers used in record
players, tape recorders, public-address systems, radio receivers, television receivers,etc.
PROCEDURE
1. Connect the circuit as shown in the figure.
2. Switch on the power supply and the Function generator.
3. Apply a 5mV sinusoidal signal at the I/P.
4. Vary the frequency in convenient steps and note down the O/P voltage.
5. Tabulate the readings and calculate the gain in dB.
6. Plot a graph between gain and frequency.
7. Determine the band width.
.
CIRCUIT DIAGRAM
MODEL GRAPH

ENGINEERING
Tabular Form
Input AC voltage, Vi
=100mv peak to peak
S.No Frequency (Hz)
Output voltage Vo Gain in decibels
(mv) (peak-peak) AV=20 log (Vo / Vi)
1 10 0.8 18.06
2 50 2.4 27.60
3 100 4.4 32.86
4 200 6 35.56
5 300 6.4 36.13
6 500 6.8 36.65
7 700 6.8 36.65
8 1K 6.8 36.65
9 2K 7.2 37.14
10 3K 7.2 37.14
11 7K 7.2 37.14
12 10K 7.2 37.14
13 50K 7.2 37.14
14 100K 7.2 37.14
15 200K 4.8 33.62
16 300K 3.2 30.10
17 500K 2 26.02
18 700K 1.2 21.58
19 1M 0.8 18.08

ENGINEERING
Precautions
1. Connections must be done very carefully.
The applied voltage and current should not exceed the maximum ratings of the given
transistor
RESULT: The voltage gain and frequency response of the Two stage RC coupled amplifier is
. obtained.
1. What is difference between Amplifier and Attenuator?
Both are linear systems but Amplifier's gain is more than unity (+ve dB), Attenuator gain is
less than unity (-ve dB)
2. What are the advantages over single stage amplifier?
In single stage amplifier if we try to get more gain, bandwidth will be decreased viceversa. So
get more bandwidth and gain combination we generally use multistage amplifier. but
multistage amp bandwidth is less than single stage.
3.what are the classifications of Multistage amplifiers?
Based on Active device used: BJT and FET
Based on type of coupling: RC coupled, Transformer coupled and Direct coupled amplifiers.
4. Define cut off frequency?
It is the frequency at which the gain is 70.07% of it's maximum or 3dB lesser than maximum
5. what are the formulas for low and high cut off frequencies of multistage amplifiers?
where Fl, Fh represents low and high cut off freq for multistage
amp fl,fh represents low and high cut off freq for single stage amp

ENGINEERING
EXPERIMENT-9
(a) : CURRENT SHUNT FEED BACK AMPLIFIER
AIM: -
1. Study the concept of feedback in amplifiers.
2. Study the characteristics of current shunt feedback amplifier.
3. Identify all the formulae will need in this experiment.
OBJECTIVE:
1. To simulate the Current Shunt Feedback Amplifier in Multisim and study the transient
and frequency response.
2. To determine the maximum gain, 3dB gain, lower and upper cutoff frequencies and
bandwidth of Current Shunt Feedback Amplifier by performing the AC analysis.
3. To determine the effect of feedback on gain and bandwidth and compare with Multisim
results.
REQUIREMENTS:
1. Transistor – 2n2222(2)
2. Resistors – as per circuit diagram
3. Capacitors – as per circuit diagram
4. RPS – 0-30V.
5. CRO.
6. Breadboard.
7. Connecting wires and Probes.
CIRCUIT DIAGRAM:
Current shunt feedback Amplifier circuit diagram
VCC
12V
R3
33k?
C1 R2
4.7k?
XSC1
10µF
R6
10k?
C3
CRO
Q1 10µF
R5 C4 A
B
Ext Trig
+
_
+ +
A B
_ _
1k?
10µF
BC107BP
XFG1 R4
5.6k?
R8 C6
2.2k? 0.001µF
R1
C5
1.8k?
100µF

GurunanakInstituteOfTechnology Page 45
THEORY:
Feedback plays a very important role in electronic circuits and the basic parameters,
such as input impedance, output impedance, current and voltage gain and bandwidth, may
be altered considerably by the use of feedback for a given amplifier.
A portion of the output signal is taken from the output of the amplifier and is
combined with the normal input signal and thereby the feedback is accomplished.
There are two types of feedback. They are i) Positive feedback and ii) Negative
feedback. Negative feedback helps to increase the bandwidth, decrease gain, distortion, and
noise, modify input and output resistances as desired.
A current shunt feedback amplifier circuit is illustrated in the figure. 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 a
true current amplifier.
PROCEDURE:
1. Connect the circuit as per the circuit diagram.
2. Apply the input signal.
3. Vary the frequency conveniently and note down the output voltage.
4. Plot the curve between gain and resonant frequency.
5. Calculate the gain.
6. Calculate the resonant frequency and compare it with the theoretical value.
RESULT: -
1. Frequency response of Current shunt amplifier is plotted.
2. Gain = 28 dB (maximum).
3. Bandwidth= fH--fL = 500K Hz.
(b) : VOLTAGE SERIES FEED BACK AMPLIFIER

AIM: To design a voltage series feedback amplifier with following specifications and to
study the frequency response of amplifier, calculate voltage gain and bandwidth from
the response.
CIRCUIT DIAGRAM:
Fig: 6.a Voltage series feedback Amplifier circuit diagram
PROCEDURE:
1. Switch ON the computer and open the multisim software.
2. Check whether the icons of the instruments are activated and enable.
3. Now connect the circuit using the designed values of each and every component.
4. Connect the function generator with sine wave of 50 mV p-p as input at the
input of terminals of the circuit.
5. Connect the Cathode Ray Oscilloscope (CRO) to the out put terminals of the circuit.
6. Go to simulation button click it for simulation process.
7. From the CRO note the following values
a. Input voltage Vi = 2mV
b. Output voltage V0 = 0.15V
c. Voltage gain AV = V0/Vi = 75

To study the frequency response click the AC analysis, so that a screen displays the
following options
d. Start frequency
e. Stop frequency
f. Vertical scale
8. Assign the proper values for start frequency, stop frequency and vertical scale
according to the circuit requirements and observe the frequency response.
9. From the frequency response calculate the
a. maximum gain AVmax = 75
b. lower cutoff frequency (f1) at AVmax - 3dB (decibel scale) value = 1.7KHz
c. Higher cutoff frequency (f2) at AVmax - 3dB (decibel scale) value = 492.7KHz
RESULT: -
1. Frequency response of Voltage Series Feed Back amplifier is plotted.
2. Gain = 37.5 dB (maximum).
3. Bandwidth= fH--fL = 491K Hz.

Department of Electronics and Communication Engineering
1
INTRODUCTION
IC 741 (Operational Amplifier):
General Description:
The IC 741 is a high performance monolithic operational amplifier constructed using the
planer epitaxial process. High common mode voltage range and absence of latch-up tendencies
make the IC 741 ideal for use as voltage follower. The high gain and wide range of operating
voltage provide superior performance in integrator, summing amplifier and general feedback
applications.
Block Diagram of Op-Amp:
Pin Configuration:
Features:
1. No frequency compensation required.
2. Short circuit protection
3. Offset voltage null capability
4. Large common mode and differential voltage ranges
5. Low power consumption
6. No latch-up
Specifications:

2
1. Voltage gain A = α typically 2, 00,000
2. I/P resistance RL = α Ω, practically 2MΩ
3. O/P resistance R =0, practically 75Ω
4. Bandwidth = α Hz. It can be operated at any frequency
5. Common mode rejection ratio = α (Ability of op amp to reject noise voltage)
6. Slew rate + α V/μsec (Rate of change of O/P voltage)
7. When V1 = V2, VD=0
8. Input offset voltage (Rs ≤ 10KΩ) max 6 mv
9. Input offset current = max 200nA
10 .Input bias current: 500nA
11. Input capacitance: typical value 1.4pF
12. Offset voltage adjustment range: ± 15mV
13. Input voltage range: ± 13V
14. Supply voltage rejection ratio: 150 μV/V
15. Output voltage swing: + 13V and – 13V for RL > 2KΩ
16. Output short-circuit current: 25mA
17supply current: 28mA
18. Power consumption: 85mW
Applications:
1. AC and DC amplifiers
2. Active filters
3. Oscillators
4. Comparators
5. Regulators

3
IC 555 (TIMER):
Description:
The operation of SE/NE 555 timer directly depends on its internal function. The three
equal resistors R1, R2, R3 serve as internal voltage divider for the source voltage. Thus one-third
of the source voltage VCC appears across each resistor.
Comparator is basically an Op amp which changes state when one of its inputs exceeds
the reference voltage. The reference voltage for the lower comparator is +1/3 VCC. If a trigger
pulse applied at the negative input of this comparator drops below +1/3 VCC, it causes a change
in state. The upper comparator is referenced at voltage +2/3 VCC. The output of each comparator
is fed to the input terminals of a flip flop.
The flip-flop used in the SE/NE 555 timer IC is a bistable multivibrator. This flip flop
changes states according to the voltage value of its input. Thus if the voltage at the threshold
terminal rises above +2/3 VCC, it causes upper comparator to cause flip-flop to change its states.
On the other hand, if the trigger voltage falls below +1/3 VCC, it causes lower comparator to
change its states. Thus the output of the flip flop is controlled by the voltages of the two
comparators. A change in state occurs when the threshold voltage rises above +2/3 VCC or when
the trigger voltage drops below +1/3 Vcc.
The output of the flip-flop is used to drive the discharge transistor and the output stage. A
high or positive flip-flop output turns on both the discharge transistor and the output stage. The
discharge transistor becomes conductive and behaves as a low resistance short circuit to ground.
The output stage behaves similarly. When the flip-flop output assumes the low or zero states
reverse action takes place i.e., the discharge transistor behaves as an open circuit or positive VCC
state. Thus the operational state o the discharge transistor and the output stage depends on the
voltage applied to the threshold and the trigger input terminals.
Block Diagram of IC 555:

4
Pin Configuration:
Function of Various Pins of 555 IC:
Pin (1) of 555 is the ground terminal; all the voltages are measured with respect to this pin.
Pin (2) of 555 is the trigger terminal, If the voltage at this terminal is held greater than
one-third of VCC, the output remains low. A negative going pulse from Vcc to less than Vec/3
triggers the output to go High. The amplitude of the pulse should be able to make the comparator
(inside the IC) change its state. However the width of the negative going pulse must not be
greater than the width of the expected output pulse.
Pin (3) is the output terminal of IC 555. There are 2 possible output states. In the low output
state, the output resistance appearing at pin (3) is very low (approximately 10 Ω). As a result the
output current will goes to zero , if the load is connected from Pin (3) to ground , sink a current I
Sink (depending upon load) if the load is connected from Pin (3) to ground, and sinks zero current
if the load is connected between +VCC and Pin (3).
Pin (4) is the Reset terminal. When unused it is connected to +Vcc. drives below 0.4V, the output
is immediately forced to low state. over-ride command signals at Pin (2) of the IC.
Pin (5) is the Control Voltage terminal. This can be used to alter the reference levels at which the
time comparators change state. A resistor connected from Pin (5) to ground can do the job.
Normally 0.01μF capacitor is connected from Pin (5) to ground. This capacitor bypasses supply
noise and does not allow it affect the threshold voltages.
Pin (6) is the threshold terminal. In both astable as well as monostable modes, a capacitor is
connected from Pin (6) to ground. Pin (6) monitors the voltage across the capacitor when it
charges from the supply and forces the already high O/p to Low when the capacitor reaches +2/3
VCC.
Pin (7) is the discharge terminal. It presents an almost open circuit when the output is high and
allows the capacitor charge from the supply through an external resistor and presents an almost
short circuit when the output is low.
Pin (8) is the +Vcc terminal. 555 can operate at any supply voltage from +3 to +18V.

5
Features of 555 IC:
The load can be connected to o/p in two ways i.e. between pin 3 & ground 1 or between pin 3
& VCC (supply)555 can be reset by applying negative pulse, otherwise reset can be connected to
+Vcc to avoid false triggering.
1. An external voltage effect threshold and trigger voltages.
2. Timing from micro seconds through hours.
3. Monostable and bistable operation.
4. Adjustable duty cycle
5. Output compatible with CMOS, DTL, TTL
6. High current output sink or source 200mA
7. High temperature stability
8. Trigger and reset inputs are logic compatible
Specifications:
1. Operating temperature : SE555- -55o
C to 125o
C ,NE555---0 o
to 70 o
2. Supply voltage : +5V to +18V
3. Timing : μSec to Hours
4. Sink current : 200mA
5. Temperature stability : 50 PPM/o
C change in temp or 0-005% /o
C.
APPLICATIONS;
1. Monostable and Astable Multivibrators
2. dc-ac converters
3. Digital logic probes
4. Waveform sgenerators
5. Analog frequency meters
6. Tachometers
7. Temperature measurement and control Infrared transmitters

6
EXPERIMENT-1
INVERTING AND NON-INVERTING AMPLIFIER
USING OP-AMPS
AIM: To design and verify the operation of an inverting and non-inverting amplifier using
Op-Amp.
APPARATUS:
1. IC 741--1
2. Resistors (1KΩ---2, 10 KΩ—1)
3. Function generator
4. Regulated power supply
5. Cathode Ray Oscilloscope
6. Bread board
7. Connecting wires and CRO probes
A) INVERTING AMPLIFIER:
CIRCUIT DIAGRAM:
THEORY:
The input signal is applied through a series input resistor R1 to the inverting input. Also,
the output is fed back through Rf to the same input. The non-inverting input is grounded.
An expression for the output voltage of the inverting amplifier is written as

7
The –ve sign indicates inversion. The closed-loop gain of the inverting amplifier is, thus
The input & output impedances of an inverting amplifier are
The output impedance of both the non-inverting and inverting amplifier configurations is
very low; in fact, it is almost zero in practical cases. Because of this near zero output impedance,
any load impedance connected to the op-amp output can vary greatly and not change the output
voltage at all
PROCEDURE:
1. Connections are made as per the circuit diagram.
2. Apply input voltage as sine wave with 5 Vp-p and 1KHz to 2nd
pin of op-amp through R1.
3. Adjust the 10V positive voltage to the 7 th pin and 10V negative voltage to the 4th
pin of
op-amp.
4. Connect the CRO probe to 6th
pin of op-amp.
5. Observe, draw both input wave and output waves from CRO, calculate gain and compare
with theoretical values.
MODEL WAVEFORMS:

8
B) NON-INVERTING AMPLIFIER:
CIRCUIT DIAGRAM:
THEROY:
The input signal is applied to the non-inverting (+) input. The output is applied back to
the inverting (-) input through the feedback circuit (closed loop) formed by the input resistor R1
and the feedback resistor Rf. This creates –ve feedback as follows. Resistors R1 and Rf form a
voltage-divider circuit, which reduces VO and connects the reduced voltage Vf to the inverting
input. The feedback is expressed as

9
The difference of the input voltage, Vin and the feedback voltage, Vf is the differential input of
the op amp. This differential voltage is amplified by the gain of the op-amp and produces an
output voltage expressed as
The closed-loop gain of the non-inverting amplifier is, thus
Notice that the closed loop gain is
• Independent of open-loop gain of op-amp
• set by selecting values of R1 and Rf
An expression for the input impedance of a non-inverting amplifier can be written as
Where AOL = open-loop voltage gain of op-amp
Zin = internal input impedance of op-amp (without feedback)
β = attenuation of the feedback circuit
The output impedance can be written as
PROCEDURE:
2. Apply input voltage as sine wave with 5 Vp-p and 1KHz to 2nd
pin of op-amp through R1.
3. Adjust the 10V positive voltage to the 7 th pin and 10V negative voltage to the 4th
pin of
op-amp.
4. Connect the CRO probe to 6th
pin of op-amp.
5. Observe, draw both input wave and output waves from CRO, calculate gain and compare
with theoretical values.

10
MODEL WAVEFORMS:
PRECAUTIONS:
1. Make null adjustment before applying the input signal and power supplies.
2. Maintain proper Vcc levels.
3. Check the connections before giving the power supply. Readings should be taken
carefully
RESULT:
The design and verification of IC 741 Op-Amp as inverting and non-inverting amplifiers
are done and waveforms are plotted.
Input to Inverting amplifier =5Vp-p, 1 KHz sine
Output of Inverting amplifier = -50Vp-p, 1 KHz sine
Gain of Inverting amplifier =-10
Input to Non Inverting amplifier = 5Vp-p, 1 KHz sine
Output of Non Inverting amplifier = 55Vp-p, 1 KHz sine
Gain of Non Inverting amplifier =11

11
VIVA QUESTIONS:
1. Why OP-AMP called operational Amplifier?
2. What are ideal characteristics of op amp?
3. Why OP-AMP called direct coupled high differential circuit?
4. What are differential gain and common-mode gain of a differential amplifier?
5. Why does an op-amp have high CMRR?

12
EXPERIMENT-2
COMARATORS USING OP-AMP
AIM: To study the applications of IC 741 as comparator.
APPARATUS:
1. IC 741 --- 1
2. Resistors (1KΩ)—4
5. Cathode Ray oscilloscope
6. Bread board
CIRCUIT DIAGRAM:
THEORY:
A comparator is a circuit which compares a signal voltage applied at one input of an op-
amp with a known reference voltage at the other input. It is basically an open loop op-amp with
output ±Vsat as in the ideal transfer characteristics. It is clear that the change in the output state
takes place with an increment in input Vi of only 2mv. This is the uncertainty region where
output cannot be directly defined.
There are basically 2 types of comparators.
1. Non inverting comparator
2. Inverting comparator

13
The applications of comparator are zero crossing detectors, window detector, and time
marker generator and phase meter.
MODEL GRAPH:
PROCEDURE:
2. Select the sine wave of 5V peak to peak, 1KHz frequency.
3. Apply the reference voltage 4V and trace the input and output wave forms.
4. Superimpose input and output waveforms and measure sine wave amplitude with
reference to Vref.
5. Observe input and outputs using CRO. Slowly increase Vref voltage and observe the
change in saturation voltage.
PRECAUTIONS:
carefully
RESULT:
The operation of IC 741 Op-Amp as comparator is studied and waveforms are plotted.
Input to the comparator = 10V, 1 KHz sine; Vref= 4V DC

14
Output of comparator = 20V, 1 kHz Square
VIVA QUESTIONS:
1. What are the applications of comparator?
2. What is a differential amplifier?
3. What are the advantages of using a voltage follower amplifier?
4. Why open-loop op-amp configurations are not used in linear applications?
5. List the parameters that should be considered for ac and dc applications.

15
EXPERIMENT-3
INTEGRATOR CIRCUIT USING IC 741
AIM: To design and verify the operation of an integrator for a given input.
APPARATUS:
1. IC 741 ---1
2. Capacitors - 0.1μf
3. Resistors – 1KΩ, 10kΩ
7. Bread board
THEORY:
In an integrator circuit, the output voltage is integral of the input signal.
At low frequencies the gain becomes infinite, so the capacitor is fully charged and
behaves like an open circuit. The gain of an integrator at low frequency can be limited by
connecting a resistor in shunt with capacitor.
CIRCUIT DIAGRAMS:
PROCEDURES:

16
1. Connect the circuit as per the diagram shown in figure
2. Apply a square wave/sine input of 4V(p-p) at 10KHz to 2nd
pin of op-amp
3. Observe the output at pin 6.
4. Draw input and output waveforms in a graph.
MODEL WAVEFORMS:

17
OBSERVATIONS:
Input –Square
wave Output - Triangular
Amplitude
(V)
Time
period
(ms)
Amplitude
(V)
Time
period
(ms)
4V 10 ms 1V 0.1 ms
Input –Sine wave Output - Cosine
Amplitude
(V)
Time
period
(ms)
Amplitude
(V)
Time
period
(ms)
4V 0.1 ms -1V 0.1 ms
PRECAUTIONS:
carefully
RESULT:
The operation of IC 741 Op-Amp as integrator is verified and waveforms are plotted.
Input to the Integrator =4Vp-p, 10 KHz Square
Output of the Integrator = 1Vp-p, 10 KHz Triangular
Input to the Integrator =4Vp-p, 10 KHz Sine
Output of the Integrator = -1Vp-p, 10 KHz Cosine
VIVA QUESTIONS:
1. What are the applications of integrators?
2. Op-amp is used mostly as an integrator than a differentiator. Why?
3. What are the problems of ideal integrator?
4. What is the need for Rf in the circuit of integrator?
5. How would you eliminate the high frequency noise in integrator?

18
EXPERIMENT-4
DIFFERENTIATOR CIRCUIT USING IC 741
AIM: To design and verify the operation of a differentiator for a given input.
APPARATUS:
1. IC 741 ---1
2. Capacitors – (0.1μf, 0.01μf)--1
3. Resistors – (1KΩ, 10kΩ)--1
7. Bread board
THEORY:
In the differentiator circuit the output voltage is the differentiation of the input
voltage
The input impedance of this circuit decreases with increase in frequency, thereby making
the circuit sensitive to high frequency noise. At high frequencies circuit may become unstable.
CIRCUIT DIAGRAMS:

19
PROCEDURES:
1. Connect the circuit as per the diagram shown in Fig
2. Apply a square wave/sine input of 4V(p-p) at 100Hz TO 2nd
pin of op-amp
3. Observe the output at pin 6
4. Draw the input and output waveforms in a graph
MODEL GRAPH:

20
OBSERVATIONS:
Input –Square
wave Output - Spikes
Amplitude
(V)
Time
period
(ms)
Amplitude
(V)
Time
period
(ms)
4V 10 ms 0.52V 10 ms
Input –Sine wave Output - Cosine
Amplitude
(V)
Time
period
(ms)
Amplitude
(V)
Time
period
(ms)
4V 10 ms 0.18 V 10 ms
PRECAUTIONS:
carefully
RESULT: The operation of IC 741 Op-Amp as differentiator is verified and waveforms are
plotted.
Input to the Integrator =4Vp-p, 100 Hz Square
Output of the Integrator = 0.52Vp-p, 100 Hz Spikes
Input to the Integrator =4Vp-p, 100 Hz Sine
Output of the Integrator = 0.18Vp-p, 100 Hz Cosine
VIVA QUESTIONS:
1. What are the applications of Differentiator?
2. What is the effect of C1 on the output of a differentiator?
3. What are the problems of ideal differentiator?
4. What is the condition for good differentiation?
5. What is the advantage of IC?

21
EXPERIMENT-5
To Plot the Frequency Response of 1st
Order LPF
AIM: To study Op-Amp as first order LPF to obtain frequency response.
APPARATUS:
1. IC 741.
2. Resistors 2.2KΩ, 1KΩ--2,
3. Capacitors (0.1μf) ----1
7. Bread board
8. Connecting wires and CRO Probes
CIRCUIT DIAGRAM:
A) LOW PASS FILTER:

22
THEORY:
LOW PASS FILTER:
A LPF allows frequencies from 0 to higher cut of frequency, fH. At fH the gain is 0.707
Amax, and after fH gain decreases at a constant rate with an increase in frequency. The gain
decreases 20dB each time the frequency is increased by 10. Hence the rate at which the gain rolls
off after fH is 20dB/decade or 6 dB/ octave, where octave signifies a two fold increase in
frequency. The frequency f=fH is called the cut off frequency because the gain of the filter at this
frequency is down by 3 dB from 0 Hz. Other equivalent terms for cut-off frequency are -3dB
frequency, break frequency, or corner frequency.
PROCEDURE:
LOW PASS FILTER:
1. Connections are made as per the circuit diagram shown in Figure.
2. Apply sinusoidal wave of constant amplitude as the input such that op-amp does not go
into saturation.
3. Vary the input frequency and note down the output amplitude at each step
4. Plot the frequency response in a semi log graph.
OBSERVATIONS:
LOW PASS FILTER:
Input voltage Vin = 5V
Frequency
(Hz)
Output
Voltage (V)
Voltage
Gain
AV=Vo/Vi
Gain
in dB
100 7.6 1.52 3.6
200 7.6 1.52 3.6
300 7.6 1.52 3.6
400 7.6 1.52 3.6
500 7.6 1.52 3.6
1K 6.4 1.28 2.14
2K 4.6 0.92 -0.26
3K 3.4 0.68 -3.3
4K 2.6 0.52 -4.8
5K 2 0.48 -6.3
10K 2 0.8 -13.97

23
11K 0.8 0.16 -15.9
12K 0.6 0.12 -18.4
13K 0.4 0.08 -21.93
14K 0.2 0.04 -27.9
MODEL GRAPH:
A) High Pass Filter B) Low Pass Filter
PRECAUTIONS:
carefully
RESULT:
The frequency response of Low pass filter is plotted using IC741 Op-Amp.
Input to the Low pass filter =5Vp-p, 1 KHz Sine
Gain of the Low pass filter = 2.54dB

24
VIVA QUESTIONS:
1. What is the mean by passive filter?
2. What is mean by active filter?
3. Define pass band and stop band of filters?
4. Give the speciality of a Butterworth filter.
5. What are the applications of active filters?

25
EXPERIMENT-6
To Plot the Frequency Response of 1st
Order HPF
AIM: To study Op-Amp as first order HPF to obtain frequency response.
APPARATUS:
9. IC 741.
10. Resistors 2.2KΩ, 1KΩ--2,
11. Capacitors (0.1μf) ----1
15. Bread board
CIRCUIT DIAGRAM:
HIGH PASS FILTER:
THEORY:
HIGH PASS FILTER:

26
The frequency at which the magnitude of the gain is 0.707 times the maximum value of
gain is called low cut off frequency. Obviously, all frequencies higher than fL are pass band
frequencies with the highest frequency determined by the closed –loop band width all of the op-
amp.
PROCEDURE:
HIGH PASS FILTER:
1. Connections are made as per the circuit diagrams shown in Figure.
2. Apply sinusoidal wave of constant amplitude as the input such that op-amp does not go
into saturation.
3. Vary the input frequency and note down the output amplitude at each step.
4. Plot the frequency response as shown in Figure.
OBSERVATIONS:
HIGH PASS FILTER:
Input voltage Vin = 5V
Frequency
(Hz)
Output
Voltage (V)
Voltage
Gain
AV=Vo/Vi
Gain
in dB
100 0.5 0.1 -20
200 1.2 0.2 -12.34
300 1.6 0.3 -9.89
400 2 0.4 -7.9
500 2.3 0.46 -6.7
1K 4 0.8 -1.9
2K 5 1 0
3K 5.4 1.08 0.6
4K 5.4 1.08 0.6
5K 5.4 1.08 0.6
10K 5.4 1.08 0.6
11K 5.4 1.08 0.6
12K 5.4 1.08 0.6
13K 5.4 1.08 0.6
14K 5.4 1.08 0.6

27
MODEL GRAPH:
B) High Pass Filter B) Low Pass Filter
PRECAUTIONS:
carefully
RESULT:
The frequency response of Low pass filter and High pass filter is plotted using IC741 Op-
Amp.
Input to the High pass filter =5Vp-p, 1 KHz Sine
Gain of the High pass filter = 0.4 dB
VIVA QUESTIONS:
2. What is the mean by passive filter?
2. What is mean by active filter?
3. Define pass band and stop band of filters?
4. Give the speciality of a Butterworth filter.

28
5. What are the applications of active filters?
ADDITIONAL EXPERIMENTS
EXPERIMENT-7
IC 741 WAVEFORM GENERATORS-SINE, SQUARE AND
TRIANGULAR WAVES
AIM: To verify op-Amp as function generator that produces various specific waveforms over a
wide range of frequencies.
APPARATUS:
1. IC 741-(2).
2. Resistors 470KΩ-1, 1KΩ-3, 10KΩ-2, 100KΩ-2, 1MΩ-1,
3. Capacitors(0.1μF-3, 0.01μF-2)
5. Cathode Ray Oscilloscope
7. Breadboard
THEORY:
Function generator is a signal generator that produces various specific waveforms for test
purposes over a wide range of frequencies. In laboratory type function generator generally one of
the functions (sine, square & triangle) is generated using dedicated chips or standard circuits and
converts it in to required signal.
This consists of
1. Sine wave generator Using IC 741.
2. Square wave generator (Astable Multivibrator using IC 741)
3. Active integrator using IC 741

29
CIRCUIT DIAGRAM:
SINE WAVE GENERATOR:
The sine wave generator circuit is shown in fig.1. The operational amplifier (IC 741) used
in the circuit is provided with a positive feed back through two 47KΩ resistors and a 0.047µF
capacitor. This positive feed back provides a fraction of output signal in phase(00
or 3600
) with
the input at the non-inverting terminal (pin No.3) of Op-Amp 741. Once the loop gain (Aβ) of
the circuit equals to 1 the circuit produce oscillates. The frequency of the oscillations can be
controlled by varying the feed back network components. However a negative feed back is also
provided to the Op-Amp 741 to improve the stability of the circuit.
SQUARE WAVE GENERATOR
In comparison to sine wave oscillations, square wave signals are generated when the Op-
Amp is forced to operate in saturated region. That is the output of the Op-Amp is forced to swing
between +Vsat & -Vsat, resulting in square wave output. The circuit arrangement of a square
wave generator using IC 741 is shown in fig.2.

30
Fig: Square & Triangular Wave Generators
TRIANGULAR WAVE GENERATOR:
The circuit arrangement of a triangular wave generator is shown in Fig.2. A square wave
from the square wave generator is fed to the integrator. The RC time constant of the integrator
has been chosen in such a way that it is very small value compared to the time period of the
incoming square wave. For the basic operation of integrator, it is known that the output of the
integrator for a given square wave input is a triangle wave.

31
Model waveforms:
PROCEDURE:
1. Connect trainer kit to the 230V AC mains and switch on the supply.
2. Observe the output of the sine wave generator. If signal is not coming or distorted in shape
adjust the gain trim pot provided on the kit until a good signal is obtained. Measure the signal
frequency using Oscilloscope.
3. Observe the output of the square wave generator and measure the output signal frequency.
4. Observe the output of the Integrator (triangular wave generator) by varying the input signal
frequency (square wave is internally connected to the circuit).
5. Measure the frequency of the triangular wave using CRO.

32
PRECAUTIONS:
carefully
RESULT: Op-Amp as function generator that produces various specific waveforms over a wide
range of frequencies is verified
Output frequency of sine wave generator circuit = 653.59Hz
Output frequency of square wave generator circuit = 0.04MHz
Output frequency of triangular wave generator circuit = 0.04MHz
VIVA QUESTIONS:
1. Why do we call sine to square wave converter as zero crossing detector?
2. What are the different ways of generating Sinusoidal waves?
3. What are different ways of generating square wave voltage waveforms?
4. Give Barkhausen criterion for oscillation.
5. What is the duty cycle of a pulse waveform?

33
EXPERIMENT-8
SCHMITT TRIGGER CIRCUITS- USING IC 741
AIM: To design the Schmitt trigger circuit using IC 741.
APPARATUS:
1. IC 741
2. Resistors 1KΩ
6. Bread board
THEORY:
The circuit shows an inverting comparator with positive feed back. This circuit converts
orbitrary wave forms to a square wave or pulse. The circuit is known as the Schmitt trigger (or)
squaring circuit. The input voltage Vin changes the state of the output Vo every time it exceeds
certain voltage levels called the upper threshold voltage Vut and lower threshold voltage
Vlt.When Vo= - Vsat, the voltage across R1 is referred to as lower threshold voltage, Vlt. When
Vo=+Vsat, the voltage across R1 is referred to as upper threshold voltage Vut.The comparator with
positive feed back is said to exhibit hysterisis, a dead band condition.
CIRCUIT DIAGRAMS:

34
PROCEDURE:
1. Connect the circuit as shown in Figure.
2. Apply an orbitrary waveform (sine/triangular) of peak voltage greater than UTP to the
input of a Schmitt trigger.
3. Observe the output at pin6 of the IC 741 Schmitt trigger circuit by varying the input and
note down the readings as shown in Tables.
4. Find the upper and lower threshold voltages (Vutp, VLtp) from the output wave form.
MODEL GRAFH:
Fig : (a) Schmitt trigger input wave form (b) Schmitt trigger output wave form
OBSERVATIONS:
Theoritical Practical
VLTP 5V 4.5V
VUTP 5V 5V
VH 0V 0.5V

35
PRECAUTIONS:
carefully
RESULT: The Schmitt trigger circuit using IC 741 and IC 555 is designed.
Input voltage to Schmitt trigger = 10Vp-p, 1 KHz Sine
Output voltage of Schmitt trigger = 16Vp-p, 1KHz Square
VIVA QUESTIONS:
1. What is the other name for Schmitt trigger circuit?
2. In Schmitt trigger which type of feed back is used?
3. What is mean by hysteresis?
4. Define the threshold points in a Schmitt trigger circuit.
5. Give two applications of Schmitt Trigger circuit

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