HARDWARE-ECA LAB

1. CE AND CS AMPLIFIER

AIM:

1.To construct common emitter and common source amplifier.

2.To verify their voltage gain, current gain, input resistance theoretically and

practically.

3.To obtain the frequency response of common emitter and common source

amplifier and to find the band width.

COMPONENTS REQUIRED:

S.NO Name Range Quality

1

2

3

4

5

6

7

Regulated Power Supply

Function Generator

Cathode Ray Oscilloscope

Transistor

FET

Capacitors

Resistors

(0-20)V

-------

20MHZ

BC547

BFW11

10µf/63v

2.2kΩ,33kΩ,8.2kΩ,1kΩ47kΩ,820Ω3.3kΩ,4.7kΩ

1

1

1

1

1

3

1

THEORY:

Common emitter amplifier:

The transistor connected in CE configuration provides high gain compared to

other configuration . This is generally a self biased circuit R1 and R2 provides the voltage

divider network that provides necessary forward bias to base emitter junction. The

emitter resistor Re provides stabilization since the emitter junction is reverse biased the

transistor is in active region and thus act as an amplifier. Since the emitter junction is in

forward bias input impedance of CE amplifier is low and similarly collector junction is

reverse biased so output impedance is very high.

Common source amplifier:

The most frequently used biasing circuit for FET amplifier is the self bias

arrangement Rg, Rd and Rs is the short circuited by a capacitor .Cs, C1,C2 are the

coupling capacitors which blocks the dc and allows the ac signal to pass through it .Cs is

the bypass capacitor which stabilizes the gain. The low frequency range is governed by

the coupling and bypass capacitor. Therefore voltage gain and phase angle are constant in

the mid frequency range region, there is no effect of input or output coupling capacitors.

In high frequency range the response is effected by inter electrode capacitance which

play an important role.

PROCEDURE:

Common emitter amplifier:

1. connect the circuit on bread board as per the circuit diagram.

2. Set the function generator input voltage to 20mv.

3. Connect the CRO at the output varying input frequencies from 50Hz to 1MHz.

Observe the output voltage in CRO without changing input voltage.

4. Tabulate the frequencies and corresponding output voltages.

5. Plot the graph for frequency and gain in db.

Common source amplifier:

1. Connect the circuit on bread board as per the circuit diagram.


3. Connect the CRO at the output, varying input frequencies from 50Hz to 1MHz.

Observe the output voltage in CRO without changing input voltage.



CIRCUIT DIAGRAM FOR CE AMPLIFIER :

CIRCUIT DIAGRAM FOR CS AMPLIFIER :

TABULAR COLUMN FOR CE AMLIFIER:

Vi = 20mv

Frequency(Hz) Output voltage(V) Gain=Vo/Vi Gain (db)

EXPECTED GRAPH:

TABULAR COLUMN FOR CS AMLIFIER:

Vi = 50mv

Frequency(Hz) Output voltage(V) Gain=Vo/Vi Gain (db)

EXPECTED GRAPH:

CALCULATIONS:

For CE Amplifier:

Lower half power frequency fL = ____________ Hz.

Higher half power frequency fH = ____________ KHz.

Band width = fH - fL = ____________ KHz.

Maximum gain =

For CS Amplifier:

Lower half power frequency fL = ____________ Hz.

Higher half power frequency fH = ____________ MHz.

Band width = fH - fL = ____________ MHz.

Maximum gain =

PRECAUTIONS:

1. Avoid wrong connections.

2. Avoid loose connections.

3. Check the connections once or twice.

RESULT: The frequency response of CE and CS are obtained and Maximum gain

and bandwidth are also determined.

QUESTIONS:

1. Does phase reversal exist in CE amplifier?

2. Does phase reversal affect amplification?

3. What is signification of operating point?

4. Which components are responsible for fixing operating point?

5. Calculate with the help of observed in this experiment?

6. In which transistor configuration current gain is less than unity.

2. CURRENT SHUNT FEED BACK AMPLIFIER

AIM:To obtain the frequency response of current shunt amplifier with and without

feedback and to calculate the maximum voltage gain and bandwidth.



1

2

3

4

5

6


Function Generator


Transistor

Capacitors

Resistors

(0-20)V

(0-1)MHz

20MHZ

BC547

10µF/63v100µF

2.2kΩ,33kΩ,8.2kΩ,1kΩ

4.7kΩ

1

1

1

1

113

1

THEORY:

Current shunt feedback is also known as derived shunt fed feedback amplifier. In

this circuit, the feedback network picks up a part of the output current and produces a

feedback voltage in parallel with the input signal voltage as shown in figure.since the

feedback shunts the input,so input impedence is reduced with feedback whereas output

impedence is increased because of feedback network being in series with the output.This

type of amplifier is called current amplifier.

Figure shows a CE amplifier circuit in which the resistor Rf acts as feedback

resistor.If the input signal increases Isincreases and as a consequence the feedback current

If also increases . As a result Ii=Is-If is smaller than it would be without feedback. This is

the nature of negative feedback.

Salient features of current shunt feedback amplifiers:

Gain reduces its value with feedback it is given by,Avf = A/1+Aß

Where , Avf is gain with feedback

Av is gain with out feedback

ß is feedback network

. Input impedence decreases i.e,Rif = Ri/1+A ß

where Rif is input resistance with feedback

Ri is resistance without feedback

Output impedence increases i.e,Rof =Ro(1+_Aß)

where Rof is output impedence with feedback

Ro is output impedence without feedback

Bandwidth increases i.e, Bwf=Bw(1+Aß)

Where Bwf is bandwidth with feedback.

Bw is bandwidth without feedbackl

PROCEDURE:

1. Connect the circuit on breadboard as per the circuit diagram for amplifier with out

feed back.


3. Connect the CRO at the output varying input frequencies from 50Hz to

1MHz.Observe the output voltage in CRO without changing input voltage.


5. Calculate voltage gain and gain in dB for noted values.


7. Repeat the same procedure for the circuit with feedback of Current shunt

feedback Amplifier.

CIRCUIT DIAGRAM :

EXPECTED GRAPH:

TABULAR COLUMN:

a)With feedback: Vi=100mv

Frequency(Hz) Output voltazge Gain=Vo/Vi(v) Gain (db)

b)Without feedback: Vi=20mv

Frequency(Hz) Outputvoltage(V) Gain=Vo/Vi Gain (db)

CALCULATIONS:

a)With feedback:

Lower cutoff frequency fL = _________ Hz.

Upper cutoff frequency fH = _________ Hz.

Band width BW = fH - fL = _________ Hz.

Maximum gain = _____________ .

a)Without feedback:

Lower cutoff frequency fL = _________ Hz.

Upper cutoff frequency fH = _________ Hz.

Band width BW = fH - fL = _________ Hz.

Maximum gain = _____________ .

PRECAUTIONS:




RESULT:

The frequency response of current shunt feedback with feedback and without

feedback is determined. Maximum gain and bandwidth are also determined.

3.TUNED RF CLASS C AMPLIFIER

AIM:1.To obtain the frequency response of class c amplifier

2.To calculate resonant frequency, inductive reactance, capacitive reactance

3.To calculate band width and Q of the coil

4.calculate gain of the tuned RF amplifier



1

2

3

4

5

Tuned RF amplifier trainer

Function Generator


Multi meter

Patch cards

--------

(0-1)MHz

20MHZ

---------

----------

1

1

1

1

1 SET

THEORY:The power amplifier is said to be class c amplifier if the Q poi8nt and in put

signal are selected such that the out put signal is obtained per less than half

cycle per a full input cycle. For this operation the Q point is shifted below X-

axis. Due to such a selection of Q point transistor remains active per less than

half cycle. Hence only that much part is reproduced at the out put. For

remaining cycle of the in put cycle the transistor remains cut off and no

signal is reproduced at the out put. The angle of the collector current flow is

less than 180.

The class c operation is not suitable for audio frequency

power amplifier. The class c amplifiers are used in tuned circuits in

communication areas and in RF circuits with tuned RLC loads. As used in

tuned circuits class c power amplifiers are called tuned amplifiers. These are

also used in mixers or converter circuits used in radio receivers and wire less

communication systems.

PROCEDURE:

1.connect the circuit as per the circuit diagram.

2.switch ON the power supply.

3.Connent the signal generator with sine wave and an amplitude of 10Volts

and connect it to input of the amplifier.

4.connect the in put terminals of the CRO to the out put of the amplifier.

5.Vary the frequency in steps from 10Hz to 1 MHz and observe the wave

form on CRO to note the out put voltage.

6.Calculate the gain in dB for the respective frequencies.

7.Plot the graph for frequency vs gain in dB.

8.calculate inductive and capacitive reactance, band width and Q of the coil.

CALCULATIONS:

Parallel resonance fr=1/2π√LC

Inductive reactance XL=2πfrL

capacitive reactance =1/2πfrC

band width ∆f=fH - fL

Q factor=fr /∆f

CIRCUIT DIAGRAM:

EXPECTED GRAPH:

TABULAR COLUMN:

Vi=10 V

Frequency(KHz)

Output voltage(Vo) Gain=Vo/Vi Gain (db)

PRECAUTIONS:

1.connect patch cards properly.

2.increase the in put frequency slowly.

RESULT:

Hence the resonant frequency, inductive and capacitive reactance, gain and band

width and Q factor of coil are calculated for the tuned RF class c amplifier

QUESTIONS:

4. SERIES VOLTAGE REGULATOR

AIM:To study and design the series voltage regulated power supply.

EQUIPMENT REQUIRED:

S.No Name of Apparatus Range Quantity

1

2

3

4

Series voltage regulator Power supply

Ammeter

Multimeter

Patch cords

--------

(0-50)mA

---------

----------

1

1

1

1

THEORY:The heart of any voltage regulator circuit is a control element is connected in

series with the load, the regulator circuit is called series voltage regulator.

The unregulated input voltage Vin, rise to provide the load current . But part of

the current is taken by the control element , to maintain the constant voltage across the

load. If there is any change in the load voltage, the sampling circuit provides a feedback

signal to the comparator circuit. The comparator circuit compares the feedback signal

with the reference voltage and generates a control signal which decides the amount of

voltage required to be seriesed to keep the load voltage constant.

PROCEDURE:1. Switch on the main power supply.

2. Observe the unregulated voltage to the output of the rectifier.

3. Connect this voltage to the input of the series regulator circuit.

4. Keep the load resistance 1KΩ constant.

5. Observe the output voltage across the load resistor, Vo= Vz - Vbe

and also observe the IR , IZ , Is, and Ic.

6. Compare practical values with theoretical values.

7. By changing the load across resistance observe the output voltage and current.TABULAR COLUMN FOR LINE REGULATION:

Vi(V) Vo(V) IL(mA)

TABULAR COLUMN FOR LOAD REGULATION:RL(Kohm) Vo(V) IL(mA)

CIRCUIT DIAGRAM:

CALCULATIONS :

1. Output voltage VO = VZ - VBE =

2. Voltage across series resistor Rs = VS = Vin - V0 =

3. Current through Rs = Is = VRS/RS =

4. Load current IL = VO/RL =

5. Base current IB = IL /ß=

6. Zener current IZ = IR – IB =

PRECAUTIONS:



3. Check the connections once or twice

RESULT:Hence we studied and designed the series regulated power supply, and load and

line regulations are obtained. Line regulation and load regulation graphs are plotted.

QUESTIONS:

1.why it is called series transistor regulator?A.2.what is the advantage of transistor used in the circuit?A.3.How the output voltage is regulated for input voltage functions?

A.

5. SHUNT VOLTAGE REGULATOR

AIM:

To study and design the shunt voltage regulator.

EQUIPMENT REQUIRED:


1

2

3

4

Shunt voltage regulator Power

supply

Ammeter

Multimeter

Patch cords

--------

(0-100)mA

(0-20)mA

(0-20)V

----------

1

1

1

2

1

THEORY:

The heart of any voltage regulator circuit is a control element is connected in

shunt with the load, the regulator circuit is called shunt voltage regulator.

The unregulated input voltage Vin, rise to provide the load current . But part of

the current is taken by the control element , to maintain the constant voltage across the

load. If there is any change in the load voltage, the sampling circuit provides a feedback

signal to the comparator circuit. The comparator circuit compares the feedback signal

with the reference voltage and generates a control signal which decides the amount of

current required to be shunted to keep the load voltage constant. For example, if load

voltage increases then comparator circuit decides the control signal based on the feedback

information, which draws increased shunt current Ish value. Due to this , the load current

IL, decreases and hence the load voltage decreases to its normal. Thus control element

maintains the constant output voltage by shunting the current, hence the regulator circuit

is called voltage shunt regulator circuit.

PROCEDURE:

1. Switch on the main power supply.

2. Observe the unregulated voltage to the output of the rectifier.

3. Connect this voltage to the input of the shunt regulator circuit.

4. Keep the load resistance 1KΩ constant.

5. Observe the output voltage across the load resistor, Vo= Vz + Vbe

6. and also observe the IL ,Is and Ic.

7. Compare practical values with theoretical values.

8. For load regulation put the load resistance constant, and vary the input

voltage and note down the load current and output voltage.

9. For line regulation put the input voltage constant, and vary the load

resistance and note down the load current and output voltage.

10. Draw the load regulation and line regulation in graph.

CIRCUIT DIAGRAM:

TABULAR COLUMN FOR LINE REGULATION:

RL = 1Kohm

Vi(V) Vo(V) IL(mA)

TABULAR COLUMN FOR LOAD REGULATION:

Vi=12VRL(Kohm) Vo(V) IL(mA)

CALCULATIONS :A. Output voltage VO = VZ+ VBE =

b. Voltage across shunt resistor Rs = VS = Vin - VZ =

c. Current through Rs = Is = VRS/RS = VIN – V0/RS =

d. Load current IL = VO/RL =

e. Collecter current Ic = IS – Il =

PRECAUTIONS:1. Avoid wrong connections.



RESULT:

Hence we studied and designed the shunt voltage regulated power supply, and

load and line regulations are obtained. Line regulation and load regulation graphs are

plotted.

QUESTIONS:

1.why it is called shunt transistor regulator?

2.How the output voltage is regulated for input voltage functions?

3.what are the advantage of IC voltage regulators over transistor voltage regulators ?

6. RC PHASE SHIFT OSCILLATOR

AIM: To study the operation of RC phase shift oscillator and finding phase shift at each

point of the RC net work.



1

2

3

4

5

Transistor

Capacitors

Resistors

Variable resistor


BC547

0.001µf 10µf

10KΩ100KΩ22KΩ1KΩ

100KΩ

20V

1

32

3211

1

1

6 Cathode Ray Oscilloscope 20MHz 1

THEORY:

The circuit consists of an amplifier and a phase shift network. The phase shift

network is connected between output and input of the amplifier. In general a single stage

amplifier produces a phase shift of 180˚. An additional of 180˚ to satisfy barkhausen

criterion is provided by the feedback network consist of three identical RC section. Each

section produces a phase shift of 60˚

Fo=1/2ΠRC√(6+_4K) ,where K=Rc/R

So at frequency of the netphase shift provided by the feedback network equals

180˚.When the circuit is energised by switching on supply, the circuit starts oscillating.

The oscillations may start due to the minor variations in dc supply or the inherent noise in

the BJT. However ,to start the oscillations the loop gain must be greater than unity using

classical network analysis, it can be found that the short circuit current gain of the

transistor should be greater than 44.5(i.e,hfe >44.5) in order to satisfy this loopgain

condition(i.e,Aß>1)

The RC phase shift oscillator is particularly useful for generating signals in the audio

frequency range.

CIRCUIT DIAGRAM:

PROCEDURE:

1. Connect the circuit as per the circuit diagram.

2. The output wave form is observed by connecting CRO at first capacitor and its

time period is noted.

3. Next connect the CRO to the first stage of RC network and observe the

waveform.It is observed ther is a phase shift in waveform.

4. Repeat the same for next two stages RC network.

5. Calculate the phase shift for each stage theoretically and practically.

CALCULATIONS:

ft = frequencyof RC phace shift oscillator = _____1_________

2πRC√6

=

PRECAUTIONS:




RESULT:

The analysis of RC phase shift oscillator is observed by using multisim7 software and

phase shift between output and different RC sections are verified.

QUESTIONS:

1. what is the purpose of RC phase shift oscillator?

2. List the applications of RC phase shift network?

3. what is the range of frequencies generated by RC phase shift oscillator?

4. what is the phase shift offered by each RC section of the network?

7. VARIABLE REGULATED POWER SUPPLY

AIM: To study variable regulated power supply circuit.

To obtain different regulated D.C supplies from an unregulated power supply

APPARATUS: S.No Name of Apparatus Range Quantity

1

2

3

Bread board Trainer System

Digital Multimeter

Diodes

-------

-------

1N4007

-------

-------

04

4

5

6

Capacitors

IC Regulator

Resistors

2200µF0.27 µF10 µF

LM7805

100Ω220 Ω330 Ω510 Ω

010101

01

01010101

THEORY:

Regulated power supply circuit consist mainly three sections:

1. Rectifier Circuit: it converts an A.C. signal into pulsating D.C signal.

2. Filter Circuit: it converts pulsating D.C signal into pure D.C signal.

3. Regulator Circuit: it converts unregulated D.C voltage into regulated

D.C signal.

In this circuit , Bridge circuit is used as rectifier circuit and capacitor

is used as filter circuit.

In general, from voltage regulator circuit we get a fixed d.c voltage value. With the

help of potential divider circuit we are getting different voltages.

Formula for finding theoretical value of D.C voltage is,

PROCEDURE: 1. Connect the circuit as per circuit diagram .

2. Apply the 230V, 50Hz A.C supply at the input.

3. Observe the d.c voltage obtained because of the resistance value 220Ω.

4. Repeat the above step for different values of resistors 330 Ω and 510

Ω.

5. Calculate theoretically output d.c voltage obtained because of resistor

220 Ω.

6. Similarly find out theoretical value of d.c output voltage for resistors

330 Ω and 510 Ω.

7. compare theoretical and practical values of D.C output voltages.

CIRCUIT DIAGRAM:

TABULAR COLUMN:Resistance

(R2 in Ohms)Output Voltage in Volts

Theoretical value Practical value

RESULT:

Different values of D.C voltages are obtained from Variable Regulated

Power Supply with the help potential divider network.

QUESTIONS:

1. What are the various sections in the RPS?

2. Which device is used as rectifier?

3. Which components are used in filter circuits?

8.CRYSTAL OSCILLAIORAIM:

To study and determine the frequency of oscillations of a crystal oscillator.

APPARATUS: S.No Name of Apparatus Range Quantity

1

2

3

4

5

6

CRO

Transistorised power supply

FETs

Capacitors

Inductance

Resistors

-------

-------

BFW10

0.01µF 10 µF

1mH

10MΩ10 KΩ

01

01

02

0101

01

0201

7

8

9

Potentio meter

Digital frequency counter

Crystal

10 KΩ

-----

3.3 MHz

01

01

01

CIRCUIT DIAGRAM:

THEORY: A crystal oscillator is basically a tuned circuit oscillator using a piezo

electric as a resonant tank circuit. The quartz crystal provides a high degree of

frequency stability. When an a.c voltage is applied, the crystal vibrates at the

frequency of the applied voltage. These vibrations have a natural resonant frequency.

The crystal has two resonant frequencies. First the inductance L resonates with the

capacitance C at a series branch LCR has inductive reactance. At a frequency

slightly higher than fs, the series branch LCR has a parallel resonance with

capacitance Cin. This frequency is called parallel resonant frequency fp. above this

frequency the crystal offers capacitive reactance. Between fs and fp the crystal offers

capacitive reactance. Between fs and fp the crystal behaves as an inductor.

If the crystal is used in place of an inductor in oscillator circuit,

the frequency of oscillation must lie in between fs and fp. This fact gives rise to great

frequency stability of a crystal oscillator.

PROCEDURE:

1. Connect the circuit as per the circuit diagram.

2. Switch on the power supply.

3. Connect the CRO at the output of the circuit.

4. Observe the wave form on CRO and compare it with the value marked on

the crystal.

5. Check yhe crystal frequency with the help of frequency counter.

OBSERVATIONS:

Frequency of the oscillations:

Time period T of the signal available at the output =

The frequency of oscillation f = 1/T =

Amplitude =

RESULT:

1. Stability of frequency of oscillations observed on the CRO.

2. Practical values are compared with the theoretical values.

QUESTIONS:

1. What is piezo electric effect?

2. What are the characteristics of a crystal?

3. How the frequency stability is achieved?

4. What are the applications of crystal oscillator?

5. List different types of crystal materials?

6. What are the three axes of a crystal?

7. How the vibration depends on crystal dimensions?

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