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ELECTRICAL AND ELECTRONICS ENGINEERING

L BOR TORY M NU L

(FOR ME)

DEPARTMENT OF ELECTRONICS AND COMMUNICATION

ENGINEERING

MALLA REDDY COLLEGE OF ENGINEERING AND

TECHNOLOGY

(Sponsored by CMR Educational Society)

(Affiliated to JNTU, Hyderabad)

Secunderabad-14.

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LIST OF EXPERIMENTS

SECTION-A: ELECTRICAL ENGINEERING

S.NO: EXPERIMENT NAME PAGE NO:

1. SWINBURNE’S TEST ON D.C SHUNT MACHINE 2-5

2. OC & SC TEST ON 1-PHASE TRANSFORMER 6-11

3. BRAKE TEST ON 3-PHASE INDUCTION MOTOR 12-16

4. REGULATION OF ALTERNATOR BY SYNCHRONOUS IMEDANCE 17-19

METHOD

5. SPEED CONTROL OF D.C SHUNT MOTOR 20-23

6. BRAKE TEST ON D.C SHUNT MACHINE 24-26

7. MAGNETISATON CHARACTERISTICS OF D.C SHUNT GENERATOR 27-29

SECTION-B: ELECTRONICS ENGINEERING

S.NO: EXPERIMENT NAME PAGE NO:

1. TRANSISTOR CE CHARACTERISTICS 31-34

2. FULL WAVE RECTIFIER WITH AND WITHOUT FILTER 35-38

3. FREAQUENCY RESPONSE OF CE AMPLIFIER 39-42

4. RC PHASE SHIFT OSCILLATOR 43-45

5. CLASS- A POWER AMPLIFIER 46-47

6. MICROPROCESSOR 49-56

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EEE LAB MANUAL MRCET

1

SECTION-A

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1. SWINBURNE’S TEST ON D.C SHUNT MACHINE

AIM: To perform Swinburne’s test on the given D.C machine and predetermine theefficiency at any desired load both as motor and as generator.

NAME PLATE DETAILS:

S.NO Characteristic D.C Motor

1 Voltage 220V

2 Current 20A

3 Speed 1500rpm

4 Power 5HP

APPARATUS REQUIRED:

S.NO Description Type Range Quantity

1 Ammeter MC 0-2A,0-1A,0-10A 3no

2 Voltmeter MC 0-30V,0-300V 2no3 Rheostat WW 370 Ω /2A 1no4 Tachometer Digital 0-10000rpm 1no

CIRCUIT DIAGRAM:

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CIRCUIT DIAGRAM TO FIND ARMATURE RESISTANCE:

THEORY:

This test is to find out the efficiency of the machine .It is a simple indirect method in

which losses are determined separately and from their knowledge, efficiency at anydesired load can be predetermined. The only test needed is no-load test. This test cannot

be performed on DC series motor. The machine is run as a no load shunt motor at rated

speed and with a rated terminal voltage .However, this test is applicable to thosemachines in which flux is practically constant.

The constant losses in a dc shunt machine= Wc = stray losses (magnetic & mechanical

losses) +shunt field copper losses.

PROCEDURE1) Make all the connections are as per the circuit diagram.

2) Keep the field rheostat in minimum resistance position.

3) Excite the motor with 220V, DC supply by closing the DPST switch and start theMotor by moving the handle of 3-point starter from OFF to ON position.

4) By adjusting the rheostat in motor field bring the speed of the motor to its rated value. Note down the readings of Ammeter and Voltmeter at no load condition.

5) The necessary calculations to find efficiency of machine as motor & generator at any

given value of armature current is done.

TO FIND ARMATURE RESISTANCE (Ra):1) Connect the circuit per the circuit diagram

2) Keep the rheostat in maximum position.

3) Now excite the motor terminals by 30V supply by closing DPST switch.

4) Note down the readings of Ammeter and voltmeter.

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MODEL CALCULATIONS:

Constant losses(Wc)= No load input – No load armature copper los= VIL0 – I

2aoRa where Ra is the armature resistance

And Iao=IL-Ish

For motor:IL= Ia+Ish

Input = VIL

Cu losses = Ia2

RaTotal losses =No load losses or constant losses(Wc)+ cu losses

%Efficiency ( η) = (Output/Input)*100Output = input-total losses

For generator:Ia= IL +Ish

Input = VIL

Cu losses = Ia

2

RaTotal losses =No load losses or contant losses(Wc) + cu losses

%Efficiency ( η) = (Output / Input)*100Output = input - total losses

TABULAR COLOUMN:

S.NO

Voltmeter

reading

V Volts

Ammeter

Reading

ILO in Amps

Ammeter

reading

Ishin Amps

Speed in

RPM

ARMATURE RESISTANCE (Ra):

S.No Voltage Current

CALCULATION TABLE:As a Motor:

S.NO IL ( A) Ia=(IL-Ish)A W=Ia2Ra in

watts

Total

losses

%Efficiency

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As a Generator:

S.NO IL (A) Ia=(IL+Ish)

in A

W=Ia Ra

in watts

Total

losses

%Efficiency

MODEL GRAPH:

PRECAUTIONS:1. We should start the motor under no load

2. Take the reading without parallax error.

3. The connections must be tight.4. If voltmeter gives ding then interchange voltmeter terminal connecting of

voltmeter.

RESULT:

QUESTIONS:1. Why the magnetic losses calculated by this method are less than the actual value?

2. Is it applied to D.C series machines?3. Comment on the efficiency determined by this method.

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2. OC & SC TESTS ON 1 – PHASE TRANSFORMER

AIM:To conduct Open circuit and Short circuit tests on 1-phase transformer to pre-

determine the efficiency, regulation and equivalent parameters.

NAME PLATE DETAILS:

Voltage Ratio 220/110V

Full load Current 13.6A

KVA RATING 3KVA

APPARATUS REQUIRED:


1 Ammeter MI0-20A

0-5A2no

2 Voltmeter MI0-150V

0-300V 2no

3 WattmeterLPF

UPF

2A,!50V

20A,300V2no

4Auto

transformer- 230/0-270V 1no

CIRCUIT DIAGRAM:

OPEN CIRCUIT TEST:

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SHORT CIRCUIT TEST:

THEORY: Transformer is a device which transforms the energy from one circuit to other circuit

without change of frequency.

The performance of any transformer can be determined by conducting tests. The OC andSC tests are conducted on transformer to find the efficiency and regulation of the

transformer at any desired load current.

OC TEST:The objectives of OC test are

1. To find out the constant losses or iron losses of the transformer.

2. To find out the no load equivalent parameters.

SC TEST:

The objectives of SC test are1. To find out the variable losses or copper losses of the transformer.

2. To find out the short circuit equivalent parameters.

By calculating the losses and equivalent parameters from the above tests the efficiencyand regulation can be calculated at any desired power factor.

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PROCEDURE (OC TEST):

1. Connections are made as per the circuit diagram.2. Initially variac should be kept in its minimum position.

3. Close the DPST switch.

4. By varying Auto transformer bring the voltage to rated voltage.5. When the voltage in the voltmeter is equal to the rated voltage of HV winding notedown all the readings of the meters.

6. After taking all the readings bring the variac to its minimum position.

7. Now switch off the supply by opening the DPST switch.

PROCEDURE (SC TEST):

1. Connections are made as per the circuit diagram.

2. Short the LV side and connect the meters on HV side.3. Before taking the single phase, 230 V, 50 Hz supply the variac should be in minimum

position.

4. Now close the DPST switch so that the supply is given to the transformer.5. By varying the variac when the ammeter shows the rated current

(i.e. 13. 6A) then note down all the readings.

6. Bring the variac to minimum position after taking the readings and switch off the

supply.

CALCULATIONS:

a) Calculation of equivalent circuit parameters:OC Test:

cosфo = (Wo/Vo*Io)Iw = Io cosфo =Im

= Im

sinфo =

R o = Vo/Iw =

Xm = Vo/Im =

K= V2/V1 =SC Test:

R e1= Wsc2 /Vsc =

Ze1 = Vsc/Isc =

Xe1 = 12−Re12 =

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‘-‘ O.C TEST OBSERVATIONS:

S.NO V0(VOLTS) I0(AMPS) W0(WATTS)

S.C TEST OBSERVATIONS:

S.NO VSC(VOLTS) ISC(AMPS) WSC(WATTS)

TBLER COLUMN:

S.NO % OF LOAD EFFICIENCY

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TABULATION:

LAGGING POWER FACTOR LEADING POWER FACTOR

SNO PF %REG SNO PF %REG

1 0.3 0.3

2 0.4 0.4

3 0.5 0.54 0.6 0.6

5 0.7 0.7

6 0.8 0.8

7 0.9 0.9

8 UNITY UNITY

MODEL GRAPHS:

1. EFFICIENCY VS OUTPUT

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2. EFFICIENCY VS POWER FACTOR

RESULT:

QUESTIONS:1) What is a transformer?

2) Draw the equivalent circuit of transformer?

3) What is the efficiency and regulation of transformer?

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3. BRAKE TEST ON 3 -PHASE INDUCTION MOTOR

AIM: To perform Brake test on 3- phase induction motor to determine performance

characteristics.

NAME PLATE DETAILS:

Voltage 415V

Current 7.46A

Power 5HP

Speed 1500rpm

Frequency 50Hz

APPARATUS REQUIRED:

S.NO Description Type Range Quantity1 Ammeter MI 0-20A 1

2 Voltmeter MI0-150V

0-300V2

3 WattmeterLPF

UPF

1/2A,!50V

5/10A,300V2

CIRCUIT DIAGRAM:

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THEORY: It is the direct method to find the efficiency of the induction motor. In this test the breakig

of the rotor is done with the help of the belt which surrounds the pulley by using spring balances. When the braking power is increased by tightening the springs then the line

current is increased.

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

2. Observing precautions close the TPST switch.

3. Apply he rated voltage to the stator windings of 3 Phase induction motor with the helpof starter.

4. Note down the readings of all meters on no-load.

5. Load the induction motor in steps using the brake-drum arrangement. At each step note

down the readings of all meters up to full load of the motor.6. Gradually release the load and switch OFF the supply.

7. Using thread, measure the circumference of the brake-drum when motor is at rest.

MODEL CALUCLATIONS:Force on pulley (F)=(S1-S2)*9.81*W ( N)

Torque (T) =Force(F)* Re. N-m

Output power of the motor = (S1-S2)*Re*9.81*W(watts)W=2πN/60. S1, S2 = weights on the pulley.

Re= Effective radius of the pulley.W=motor speed in rad/sec.

If V is the terminal voltage IL is the line current and cosΦ is the power factor.Power in put =V*IL* cosΦ watts.Ef ficiency (%η)=(w(S1-S2)*Re*9.81/V*I

L* cosΦ)*100

%slip=Ns – N/Ns

Where Ns is the synchronous speed and N is the speed of the motor.

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TABLER COLUMN:

MODEL GRAPHS:

1. Output Vs Efficiency2. Output Vs Torque

3. Slip Vs Torque

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RESULT:

QUESTIONS:1) What is motor?

2) Why test is conducted on motor?3) What is break test and what is the disadvantage of break test?

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4. REGULATION OF ALTERNATOR BY SYNCHRONOUS IMPEDANCE

METHOD

AIM: To predetermine the regulation of 3-phase alternator by using synchronous impedancemethod by conducting O.C and S.C tests.

NAME PLATE DETAILS:

D.C MOTOR ALTERNATOR

Voltage 220V 415V

Current 20A 7.2A

Power 5HP 3KVA

Speed 1500rpm 1500rpm

Frequency - 50Hz

APPARATUS REQUIRED:

S.NO Description Type Range Quantity1 Ammeter MI 0-20A 0-2A 2

2 Voltmeter MI0-30V0-300V

2

3 Rheostat WW 370 Ω /2A 1

4 D.C supply0-220V

1

5 Tachometer Digital 0-10000rpm 1

CIRCUIT DIAGRAM:

OC &SC TEST:

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TO FIND ARMATURE RESISTANCE:

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RESULT:

QUESTIONS:

1) What is alternator and what is regulation?

2) How we can determine the regulation by using synchronous impedance method?

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5. SPEED CONTROL OF DC SHUNT MOTOR

AIM: To control the speed of a DC shunt motor using the following methods i). field flux control

ii). Armature voltage control

APPARATUS:

S.No Equipment Type Range Qty

1 DC shunt motor DC 220V 1

2 Rheostat Wire wound 370Ω/2.5A 23 Ammeter MC (0-2)A 1

4 Voltmeter MC (0-220)V 1

5 Tachometer Digital (0-30000)rpm 1

NAME PLATE DETAILS FOR DC SHUNT MOTOR:

Rated Voltage: 220VRated Speed: 1500rpmField current: 1.2A

CIRCUIT DIAGRAM:

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THEORY:

The term speed control means intentional sped variation, carried out manually or automatically.DC motors are most suitable for wide range speed control and are therefore indispensible for

many adjustable speed drives. The speed of a motor is given by

Wm = (Vt-IaRa)/Kaф Where Ka – armature constant = PZ/2∏A And ф is the flux per pole. Hence it follows that, for a DC motor, there are basically three methods of speed control andthese are:

1. Variation of resistance in armature circuit

2. Variation of the field flux

3. Variation of armature terminal voltage.

Variation of the field flux: This method of speed control, also called as flux weakening method or filed current control

method gives speeds above the base speed only. Base speed is nothing but the rated speed of themachine. This is one of the simplest and economical methods and is, therefore extensively used

in modern electric drives. Under steady state running conditions, if the field circuit resistance isincreased, the field current and hence the field flux are reduced. Since the rotor speed cannot

change suddenly due to inertia, a decrease in field current causes a reduction of counter emf. As

a result of it, more current flows through armature. The percentage increase in armature currentis much more than the percentage decrease in the field current. In view of this, the

electromagnetic torque is increased and this being more than the load torque, the motor gets

accelerated. The disadvantages of this method are:

a. The armature may get over heated at higher speeds, because the increased armaturecurrent results in more ohmic losses whereas cooling by ventilation doesnot improve

proportionally. b. If the field flux is weakened considerably, the speed becomes very highand due to these

changes; the motor operation may become unstable.

Variation of armature terminal voltage: If the voltage applied to the armature changes the speed changes directly with it. Using this

method, speeds below rated speeds are attained.

PROCEDURE:

1. Make the connections as per the circuit diagram.

2. Keep the field rheostat in minimum position and armature in maximum position and

close the DPST switch.3. Bring the motor to rated speed using field rheostat.

4. At this point take a note of the voltmeter and tachometer readings.5. Now start varying the armature rheostat in steps and for each step note down the

voltmeter and tachometer readings. Take 10 to 15 such readings and bring back the

armature rheostat to initial position.6. Bring back the field rheostat to initial position. Using the armature rheostat set the motor

at rated speed.

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7. Vary the field rheostat in steps and for each step note down the ammeter and tachometer

readings. Take 10 to 15 such readings.

8. Set both the rheostats to their initial positions and open the DPST switch.

TABULAR COLUMN:

Armature voltage control:

S.No Va(Armature

Voltage(volts))

N

(Speed(rpm))

Field Flux Control:

S.No If(Field

current(amperes))

N

(Speed(rpm))

MODEL GRAPHS:

Armature Voltage Control:

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Field Flux Control:

RESULT:

QUESTIONS:

1. What are the three methods of speed control?2. What is the main purpose of speed control?

3. Why is three point starters not used in this circuit?

4. What is the method opted to get speeds above rated speed?

5. How is flux per pole related to the speed of the machine?

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6. BRAKE TEST ON D.C SHUNT MOTOR

AIM:To conduct the brake test on a given D.C shunt motor and to draw its performance curves.

NAME PLATE DETAILS:

S.NO Characteristic D.C Motor

1 Voltage 220V

2 Current 20A

3 Speed 1500rpm

4 Power 5HP

APPARATUS REQUIRED:


1 Ammeter MC 0-20A 1no2 Voltmeter MC 0-300V 1no

3 Rheostat WW 370 Ω /2A 1no4 Tachometer Digital 0-10000rpm 1no

CIRCUIT DIAGRAM:

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THEORY:

This test is direct test to find the efficiency of the DC shunt motor. In this test the motor directly

loaded by connecting brakes which are with pulley and motor is subjected to rated load so theentire power is wasted. The belt around the water cooled pulley has its ends attached to spring

balances s1and s2.The belt tightening hand wheels h1 and h2 help in adjusting the load on the

pulley so that the load on the motor can be varied.

PROCEDURE:

1. All the connections are as per the circuit diagram.2. 220V, DC supply is given to the motor by closing DPST switch.

3. Move the 3-point starter handle form ‘OFF’ to ‘ON’ position slowly and motorStarts running.

4. Vary the field rheostat until the motor reaches its rated Speed and take voltmeter and ammeterreadings.

5. Apply the land by break drum pulley and for each applications of load the

Corresponding Voltmeter (V), Ammeter (I), spring forces S1 & S2 and Speed (N)

Readings are noted.6. Calculate output & efficiency for each reading.

7. Note down all the readings in the tabular form carefully.

8. Remove the load slowly and keep the rheostat as starting position and switch‘OFF’ the supply by using DPST switch.

MODEL CALUCLATIONS:

TABLER COLUMN:

S.

NO Voltage

(V) Current

(A) Input

=VI

watts

Forces

inKG

S1 S2

Net

forces

F =S1~S2

in kg

Torque(T)

=F*Re*9.81

(N-M)

Speed

in

RPM(N)

O/p=

2πNT/60

(Watts)

%Efficiencyη=output/input

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GRAPH:The graph is drawn between

a) Output in Watts Vs Speed(N) in RPM b) Output in Watts Vs Torque (T) in N-m

c) Output in Watts Vs Current (I) in A

d) Output in Watts Vs Efficiency (%η) By taking output in Watts on X axis and speed, Torque, current, Efficiency onY- axis .

MODEL GRAPH:

PRECAUTIONS:

1. Initially 3- point starter should be kept at ‘OFF’ position and later it must be varied slowly anduniformly from ‘OFF’ to ‘ON’ position. 2. The field regulator must be kept at its minimum output position.

3. The brake drum of the motor should filled with cold water.

4. The motor should be started without load.

RESULT:

QUESTIONS:

1 .Why a 3-point starter is used for starting a D.C shunt motor?2. If a 3-point starter is not available, how can a D.C motor be started?

3. Explain the function of overload release coil in 3-point starter.

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7. MAGNETIZATION or OPEN CIRCUIT CHARACTERISTICS OF D.C

SHUNT GENERATOR

AIM: To obtain the no load characteristics of a DC shunt generator and to determine the critical

field resistance.

NAME PLATE DETAILS:

S.NO Characteristic D.C Motor D.C Generator

1 Voltage 220V 220V

2 Current 13.6A 20A

3 Speed 1500rpm 1500rpm

4 Power 5HP 3KW

APPARATUS REQUIRED:


1 Voltmeters MC 0-300V 2no

2 Ammeters MC 0-2A 1no

3 Rheostats WW 370ohm/2A 2no

4 Tachometers Digital 0-10000rpm 1no

CIRCUIT DIAGRAM:

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THEORY:Magnetization curve is relation between the magnetizing forces and the flux density B. this is

also expressed as a relation between the field current and the induced emf , in a D.C machine.Varying the field current and noting corresponding values of induced emf can determine this.

For a self-excited machine the theoretical shape of the magnetization

Curve is as shown in the figure. The induced emf corresponding to residual magnetism existswhen the field current is zero. Hence the curve starts, a little above the origin on y-axis. The fieldresistance line R sh is a straight-line passing through the origin.

PROCEDURE:1) All the connections are as per the circuit diagram.

2) 220V, DC supply is given to the motor by closing DPST switch.

3) Move the 3-point starter handle form ‘OFF’ to ‘ON’ position slowly and motor starts running.4) Adjust the speed of the motor to rated value by the adjusting the field rheostat ofmotor.

5) By using field rheostat vary the field current of generator.

6) By varying the field current in steps note down all the readings of generated voltage atconstant speed.

7) Now the field current & field rheostat of motor is removed slowly and the power is switched

OFF.

TO FIND CRITICAL FIELD RESISTANCE:

1) Draw the shunt field resistance line

2) Draw tangent to the OCC3) The slope of this tangent gives the Rc

Critical field resistance, Rc=Eg/ IF =

TABULAR COLUMN:

Residual Voltage = Speed=

SNO If (A) E (V)

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MODEL GRAPH:Draw the graph between generated voltage at no load and field current. By taking

Generated voltage Eg in volts on Y axis and field current If in amps on X-axis.

PRECAUTIONS:-1) The rheostat is connected such that minimum resistance is included in field circuit of motor.

2) The rheostat is connected such that maximum resistance is included in field circuit of

generator.3) Starter handle is moved slowly.

RESULT:

QUESTIONS:1.What is meant by critical field resistance?

2. What is meant by critical speed?3. Residual magnetism is necessary for self excited generators or not.

4.Why this test is conducted at constant speed.

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SECTION-B

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1. TRASISTOR CE CHARACTERISTICS (INPUT AND OUTPUT)

AIM:1. To draw the input and output characteristics of transistor connected in CE

configuration.

2. To find β of the given transistor and also its input and output Resistances.

APPARATUS: NAME QUANTITYTransistor, BC107 -1No.

Regulated power supply (0-30V) -1No.

Voltmeter (0-20V) - 2No.

Ammeters (0-20mA) -1No.Ammeters (0-200μA) -1No.Resistor, 100Ω -1NoResistor, 1KΩ -1No.

Bread board -1NoConnecting wires Required number

CIRCUIT DIAGRAM:

THEORY:

In common emitter configuration, input voltage is applied between base and emitter terminals

and out put is taken across the collector and emitter terminals. Therefore the emitter terminal is

common to both input and output.

The input characteristics resemble that of a forward biased diode curve. This is expectedsince the Base-Emitter junction of the transistor is forward biased. As compared to CB

arrangement IB increases less rapidly with VBE. Therefore input resistance of CE circuit is

higher than that of CB circuit.

The output characteristics are drawn between Ic and VCE at constant IB. the

collector current varies with VCE upto few volts only. After this the collector current becomesalmost constant, and independent of VCE. The value of VCE up to which the collector current

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changes with V CE is known as Knee voltage. The transistor always operated in the region

above Knee voltage, IC is always constant and is approximately equal to IB.The current

amplification factor of CE configuration is given by

β = ΔIC/ΔIB

Input Resistance, r i = ∆VBE /∆IB (μA) at Constant VCEOutput Résistance, r o = ∆VCE /∆IC at Constant IB (μA)

PROCEDURE:

A) INPUT CHARECTERSTICS:

1. Connect the circuit as per the circuit diagram.2. For plotting the input characteristics the output voltage VCE is kept constant at 1V and

for different values of VBB , note down the values of IB and VBE

3.Repeat the above step by keeping VCE at 2V and 4V and tabulate all the readings.4. plot the graph between VBE and IB for constant VCE

B) OUTPUT CHARACTERSTICS:1. Connect the circuit as per the circuit diagram. 2. for plotting the output characteristics the input current IB is kept constant at 50μA and

for different values of VCC note down the values of IC and VCE 3. Repeat the above step by keeping IB at 75 μA and 100 μA and tabulate the all the

readings. 4. plot the graph between VCE and IC for constant IB.

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MODEL GRAPHS:

A) INPUT CHARACTERISTICS:

B) OUTPUT CHARACTERSITICS:

OBSERVATIONS:

A) INPUT CHARACTERISTICS:

VBB

VCE = 1V VCE = 2V VCE = 4V

VBE(V) IB(μA) VBE(V) IB(μA) VBE(V) IB(μA)

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B) OUTPUT CHAREACTARISTICS:

S.NO

IB = 50 μA IB = 75 μA IB = 100 μA

VCE(V) IC(mA) VCE(V) IC(mA) VCE(V) IC(mA)

PRECAUTIONS: 1. The supply voltage should not exceed the rating of the transistor2. Meters should be connected properly according to their polarities

RESULT:

QUESTIONS:

1. What is the range of β for the transistor? 2. What are the input and output impedances of CE configuration?3. Identify various regions in the output characteristics?

4. What is the relation between α and β? 5. Define current gain in CE configuration?6. Why CE configuration is preferred for amplification?

7. What is the phase relation between input and output?

8. Draw diagram of CE configuration for PNP transistor?

9. What is the power gain of CE configuration?10. What are the applications of CE configuration

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2. FULL-WAVE RECTIFIER WITH AND WITHOUT FILTER

AIM: To Examine the input and output waveforms of Full Wave Rectifier and also calculate

its load regulation and ripple factor.

1. with Filter2. without Filter

APPARATUS: NAME QUANTITY Digital Multimeter - 1No.

Transformer (6V-0-6V) - 1No.

Diode, 1N4007 - 2No.

Capacitor 100μf/470 μf - 1No.Decade Resistance Box - 1No.

Breadboard -1No

CRO and CRO probes -1No

Connecting wires Required number CIRCUIT DIAGRAM:

A) FULL WAVE RECTIFIER WITHOUT FILTER:

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B) FULL WAVE RECTIFIER WITH FILTER:

MODEL WAVEFORMS:

A) INPUT WAVEFORM

B) OUTPUT WAVEFORM WITHOUT FILTER:

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C) OUTPUT WAVEFORM WITHOUT FILTER:

THEORY:The circuit of a center-tapped full wave rectifier uses two diodes D1&D2. During

positive half cycle of secondary voltage (input voltage), the diode D1 is forward biased and

D2is reverse biased. So the diode D1 conducts and current flows through load resistor R L.

During negative half cycle, diode D2 becomes forward biased and D1 reverse biased. Now, D2 conducts and current flows through the load resistor R L in the same direction. There is

a continuous current flow through the load resistor R L, during both the half cycles and will getunidirectional current as show in the model graph. The difference between full wave and half

wave rectification is that a full wave rectifier allows unidirectional (one way) current to the

load during the entire 360 degrees of the input signal and half-wave rectifier allows this onlyduring one half cycle (180 degree).

THEORITICAL CALCULATIONS:Vrms = Vm/ √2 Vm =Vrms√2

Vdc=2Vm/П(i)Without filter:

Ripple factor, r = √ (Vrms/ Vdc )2 -1 = 0.812(ii)With filter:

Ripple factor, r = 1/ (4√3 f C R L)

PROCEDURE: 1. Connections are made as per the circuit diagram.

2. Connect the ac mains to the primary side of the transformer and the secondary side tothe rectifier.

3. Measure the ac voltage at the input side of the rectifier.

4. Measure both ac and dc voltages at the output side the rectifier.5. Find the theoretical value of the dc voltage by using the formula Vdc=2Vm/П 6. Connect the filter capacitor across the load resistor and measure the values of Vac and

Vdc at the output.

7. The theoretical values of Ripple factors with and without capacitor are calculated.

8. From the values of Vac and Vdc practical values of Ripple factors are calculated. The practical values are compared with theoretical values.

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PRECAUTIONS:

1. The primary and secondary side of the transformer should be carefully identified.

2. The polarities of all the diodes should be carefully identified.

RESULT:

QUESTIONS:

1. Define regulation of the full wave rectifier?

2. Define peak inverse voltage (PIV)? And write its value for Full-wave rectifier?

3. If one of the diode is changed in its polarities what wave form would you get?4. Does the process of rectification alter the frequency of the waveform?

5. What is ripple factor of the Full-wave rectifier?

6.

What is the necessity of the transformer in the rectifier circuit?7. What are the applications of a rectifier?

8. What is meant by ripple and define Ripple factor?

9. Explain how capacitor helps to improve the ripple factor?

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3. FREQUENCY RESPONSE OF CE AMPLIFIER

AIM: 1. To Measure the voltage gain of a CE amplifier

2. To draw the frequency response curve of the CE amplifier

APPARATUS: NAME QUANTITYTransistor BC107 -1No.

Regulated power Supply (0-30V) -1No.

Function Generator -1No.CRO -1No.

Resistors [33KΩ, 3.3KΩ, 330Ω, -1No.Each1.5KΩ, 1KΩ, 2.2KΩ, 4.7KΩ]

Capacitors (10µF -2No100µF ) -1No.

Bread Board -1No

Connecting Wires Required number

CIRCUIT DIAGRAM:

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THEORY:

The CE amplifier provides high gain &wide frequency response. The emitter lead iscommon to both input & output circuits and is grounded. The emitter-base circuit is forward

biased. The collector current is controlled by the base current rather than emitter current. When

a transistor is biased in active region it acts like an amplifier. The input signal is applied to baseterminal of the transistor and amplifier output is taken across collector terminal. A very smallchange in base current produces a much larger change in collector current. When positive half-

cycle is fed to the input circuit, it opposes the forward bias of the circuit which causes the

collector current to decrease; it decreases the voltage more negative. Thus when input cyclevaries through a negative half-cycle, increases the forward bias of the circuit, which causes the

collector current to increases thus the output signal is common emitter amplifier is in out of

phase with the input signal. An amplified output signal is obtained when this fluctuating

collector current flows through a collector resistor Rc.

The capacitor across the collector resistor Rc will act as a bypass capacitor. This will improve

high frequency response of amplifier.

PROCEDURE:

1. Connect the circuit as shown in circuit diagram

2. Apply the input of 20mV peak-to-peak and 1 KHz frequency using Function Generator3. Measure the Output Voltage Vo (p-p) for various load resistors.

4. Tabulate the readings in the tabular form.

5. The voltage gain can be calculated by using the expression , Av= (V0/Vi)6. For plotting the frequency response the input voltage is kept Constant at 20mV peak-to-

peak and the frequency is varied from 100Hz to 1MHz Using function generator

7. Note down the value of output voltage for each frequency.

8. All the readings are tabulated and voltage gain in dB is calculated by Using The expressionAv=20 log10 (V0/Vi)

9. A graph is drawn by taking frequency on x-axis and gain in dB on y-axis On Semi-log

graph.10.The band width of the amplifier is calculated from the graph Using the expression,

Bandwidth, BW=f 2-f 1

Where f 1 lower cut-off frequency of CE amplifier, and

Where f 2 upper cut-off frequency of CE amplifier11. The bandwidth product of the amplifier is calculated using the Expression

Gain Bandwidth product=3-dBmidband gain X Bandwidth

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MODELWAVE FORMS:

A) INPUT WAVE FORM:

B) OUTPUT WAVE FORM

FREQUENCY RESPONSE

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OBSERVATIONS:

INPUT VOLTAGE, Vi =20MV

LOAD

RESISTANCE(KΩ) OUTPUT

VOLTAGE (V0)

GAIN

AV=(V0/Vi)

GAIN IN dB

Av=20log10

(V0/Vi)

FREQUENCY RESPONSE: Vi=20mv

Frequency in KHZ OUTPUT

VOLTAGE(Vo)

GAIN IN

dB=20log10(vo/vi)

RESULT:

QUESTIONS:

1. What is phase difference between input and output waveforms of CE amplifier?

2. What type of biasing is used in the given circuit?

3. If the given transistor is replaced by a p-n-p, can we get output or not?

4. What is effect of emitter-bypass capacitor on frequency response?

5. What is the effect of coupling capacitor?6. What is region of the transistor so that it is operated as an amplifier?

7. How does transistor acts as an amplifier?8. Draw the h-parameter model of CE amplifier?

9. What type of transistor configuration is used in intermediate stages of a multistageamplifier?

10. What is early effect?

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4. RC PHASE SHIFT OSCILLATOR

AIM: To conduct experiment on RC phase shift oscillator. Assume R 1 = 100k, R 2 = 22K, R C = 4 K, RE =1K & VCC = 12V.

APPARATUS:Name Quantity

Resistors(100k,20k,4k,1k) 4no

Transistors(2N2222A) 1no

CRO 1no

RPS 1no

Bread Board 1no

Capacitors(1uF,1ouF,100uF) 3no

DESIGN PROCEDURE:

a) Let R = 10K

)tan(1962.0

10

4466102

1

462

1

dard sSelect nF nF

K

K K K

C

R

R K When

K R f C

C

r

nF C K R 1;10

1.97

429

23)

noscillatio sustained for K K

hfeb

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CIRCUIT DIAGRAM:

Q2

2N2222A

R1

100kohm

R2

22kohm

R3

4kohm

R4

1kohm

12V

VCC

C1

10uF

C2

10uF

A

B

T G

XSC1

C3

100uF

C4

1.0nF

C5

1.0nF

C6

1.0nF

R5

10kOhm_5%

R6

10kOhm_5%

R7

10kOhm_5%

4

6 7

VCC

10

11

0

9

12

PROCEDURE: Rig up the circuit using multisim software and verify the results usingOscilloscope.

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RESULT:

QUESTIONS:

1.

What is an Oscillator circuit?2. What is the main difference between an amplifier and an oscillator?3. State Barkhausen criterion for oscillation.

4. State the factors on which oscillators can be classified.

5. Give the expression for the frequency of oscillation and the minimum gain required forsustained oscillations of the RC phase shift oscillator.

6. Why three RC networks are needed for a phase shift oscillator? Can it be two or four?

7. What are the merits and demerits of phase shift oscillator?

8. At low frequency which oscillators are found to be more suitable?9. What are the two important RC oscillators?

10. What makes Quartz produce stable oscillations?

11. What are the factors which contribute to change in frequency in oscillators?

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5. CLASS A POWER AMPLIFIER

AIM: To study the operation of Class A, Class B, Class AB and Class C power amplifiers.

EQUIPMENT:1. Class/A/B/C/AB amplifier trainer.

2. Function generator.

3. C.R.O

4. Connecting patch cords.

CIRCUIT DIAGRAM:

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PROCEDURE:1. Connect the circuit as shown in the circuit diagram, and get the circuit verified by your

Instructor.2. Connect the signal generator with sine wave at 1KHz and keep the

Amplitude at .5V (peak-to-peak)

3. Connect the C.R.O across the output terminals.4. Now switch ON the trainer and see that the supply LED glows.5. Keep the potentiometer at minimum position, observe and record

the waveform from the C.R.O.

6. Slowly varying the potentiometer, observe the outputs for theClass A/B/AB/C amplifiers as shown in fig.

CLASS A:

RESULT:

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MICROPROCESSOR

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INTRODUCTION TO DEVELOPMENT TOOLS

EDITOR

An editor is a program, which allows you to create a file containing the assembly

language statements for your program. As you type in your program, the editor stores the ASCII

codes for the letters and numbers in successive RAM locations. When you have typed in all ofyour programs, you then save the file on a floppy of hard disk. This file is called source file. The

next step is to process the source file with an assembler. In the MASM /TASM assembler, you

should give your source file name the extension, .ASM

ASSEMBLER

An assembler program is used to translate the assembly language mnemonics forinstructions to the corresponding binary codes. When you run the assembler, it reads the source

file of your program the disk, where you saved it after editing on the first pass through the source

program the assembler determines the displacement of named data items, the offset of labels and

pails this information in a symbol table. On the second pass through the source program, the

assembler produces the binary code for each instruction and inserts the offset etc that iscalculated during the first pass. The assembler generates two files on floppy or hard disk. The

first file called the object file is given the extension. OBJ. The object file contains the binarycodes for the instructions and information about the addresses of the instructions. The second file

generated by the assembler is called assembler list file. The list file contains your assembly

language statements, the binary codes for each instructions and the offset for each instruction. InMASM/TASM assembler, MASM/TASM source file name ASM is used to assemble the file.

Edit source file name LST is used to view the list file, which is generated, when you assemble

the file.

LINKER

A linker is a program used to join several object files into one large object file and

convert to an exe file. The linker produces a link file, which contains the binary codes for all thecombined modules. The linker however doesn’t assign absolute addresses to the program, itassigns is said to be relocatable because it can be put anywhere in memory to be run. In

MASM/TASM, LINK/TLINK source filename is used to link the file.

DEBUGGERA debugger is a program which allows you to load your object code program into system

memory, execute the program and troubleshoot are debug it the debugger allows you to look at

the contents of registers and memory locations after your program runs. It allows you to changethe contents of register and memory locations after your program runs. It allows you to change

the contents of register and memory locations and return the program. A debugger also allows

you to set a break point at any point in the program. If you inset a breakpoint the debugger willrun the program upto the instruction where the breakpoint is set and stop execution. You canthen examine register and memory contents to see whether the results are correct at that point. In

MASM/TASM, td filename is issued to debug the file.

DEBUGGER FUNCTIONS:

1. Debugger allows to look at the contents of registers and memory locations.2. We can extend 8-bit register to 16-bit register which the help of extended register option.

3. Debugger allows to set breakpoints at any point with the program.

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4. The debugger will run the program upto the instruction where the breakpoint is set and

then stop execution of program. At this point, we can examine registry and memory

contents at that point.5. With the help of dump we can view register contents.

6. we can trace the program step by step with the help of F7.

7.

We can execute the program completely at a time using F8.

DEBUGGER COMMANDS

ASSEMBLE: To write assembly language program from the given address.

A starting address

Eg: a 100

Starts program at an offset of 100.

DUMP: To see the specified memory contents

D memory location first address last address(While displays the set of values stored in the specified range, which is given above)

Eg: d 0100 0105

Display the contents of memory locations from 100 to 105(including).

ENTER : To enter data into the specified memory locations(s).

E memory location data data data data data …

Eg: e 1200 10 20 30 40 …. Enters the above values starting from memory locations 1200 to 1203, by loading 10 into

1200,20 into 1201 and soon.

GO: To execute the program

G: one instruction executes (address specified by IP)

G address : executes from current IP to the address specifiedG first address last addresses : executes a set of instructions specified between the given

addresses.

MOVE: Moves a set of data from source location to destination locationM first address last address destination address

Eg: m100 104 200

Transfers block of data (from 100 to 104) to destination address 200.

QUIT : To exit from the debugger.

Q

REGISTER: Shows the contents of Registers

R register name

Eg: r ax

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Shows the contents of register.

TRACE: To trace the program instruction by instruction.

T = 0100 : traces only the current instruction. (Instruction specified by IP)T = 0100 02 : Traces instructions from 100 to 101, here the second argument specifies

the number of instructions to be traced.

UNASSEMBLE: To unassembled the program.

Shows the opcodes along with the assembly language program.

U 100 : unassembled 32 instructions starting from 100th

location.

U 0100 0109 : unassebles the lines from 100 to 104

Using Macro – Assembler – Linker – Debugger(MASM, LINK)

UDT-1.11. Open an MSDOS window.2. Set the PATH so that the MASM programs are available

The MASM programs are on the C drive; set the path so that DOS can find them. This only

needs to be done once each time you open an MSDOS prompt.

set PATH=%PATH%;C:\MASM\BIN3. Use a Text Editor to Edit the .ASM FileUDT-1.2

Create your file using one of the following programs:

notepad proj.asmwordpad proj.asm

edit proj.asm

UDT-1.3

4. Compile the source code to create an object module.

tasm/z/zi proj.asm

The /z switch causes MASM to display the lines that generate compilation errors. The /zi switchenables information needed by the debugger to be included in the .OBJ file. Note that you should

use "real mode" assembler, MASM.EXE. Do not use the "protected mode" assembler

MASM32.EXE for the assignments that will be given in class

UDT-1.4

5. Run Linker LINK.EXE- generate .EXE file from the .OBJ file

tlink/v proj6. Run the ProgramYour final program (if there were no errors in the previous step) will have an .EXE ending. To

just run it, type: proj

If you want to use the debugger to examine the instructions, registers, etc., type:

td projThis brings up the regular full-screen version of the Turbo debugger.

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UDT 1.5

1. Tracing the Program's Execution

The Turbo debugger first starts, a Module Window which displays the . Executable lines of

program code, marked with a bullet in the left column of the window. You can set breakpointsor step to any of these lines of code. An arrow in the first column of the window indicates the

location of the instruction pointer. This always points to the next statement to be executed. Toexecute just that instruction use one of the two methods listed under the Run menu item:o Trace into (can use F7 key): executes one instruction; traces "into" procedures.

o Step over (can use F8 key): executes one instruction; skips (does not trace into) procedures.

Hitting either of these executes the instruction, and moves the arrow to the next instruction. Aseach instruction executes, the effects might be visible in the Registers Window and Watches

Window

UDT 1.6

o Trace into (can use F7 key): executes one instruction; traces "into" procedures.

o Step over (can use F8 key): executes one instruction; skips (does not trace into) procedures.

Hitting either of these executes the instruction, and moves the arrow to the next instruction. Aseach instruction executes, the effects might be visible in the Registers Window and Watches

Window

UDT 1.7

2. Setting and Removing Breakpoints

To set a breakpoint, position the cursor on the desired line of source code and press F2. The linecontaining the breakpoint will turn red. Pressing F2 again removes the breakpoint. To execute all

instructions from the current instruction pointer up to the next encountered breakpoint, choose

Run (can use F9 key) from the Run menu item.

UDT 1.8

3. Examining RegistersAnother window, the Registers Window, can be opened to examine the current value of the CPU

registers and flags. The View menu can be used to open this Registers Window. The registersand flags might change as each instruction is executedUDT 1.9

4. Examining Memory

To examine memory, you will need to open an Inspector window. An Inspector window shows

the contents of a data structure (or simple variable) in the program you are debugging. It alsoallows you to modify the contents of the data structure or variable. To open an Inspector

window, place the cursor on what you want to inspect and press CTRL-I. After you've examined

the data item, press the ESC key to remove the Inspector window

UDT 1.10

5. Viewing the Program's OutputOutput written to the screen by a program is not immediately visible, since the main purpose of

using a debugger is to examine the internal operation of the program. To observe what the user

would see, press ALT-F5. The entire screen will change to a user-view showing the program'sinput and output (and possibly that of previous programs as well). Press any key to return to the

debugger screen.

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ADDITION, SUBSTRACTION AND MULTIPLICATION

OF TWO NUMBERS

AIM: To Write and execute an Assembly language Program (ALP) to 8086 processor to add,

subtract and multiply two 16 bit unsigned numbers. Store the result in extra segment.

SOFTWARE REQUIRED:

1. Pc with windows(95/98/XP/NT/2000)

2. MASM SOFTWARE

PROGRAMS:

(A) 16-BIT ADDITION

.MODEL TINY

DATA SEGMENT

NUM1 DW 23C5H

NUM2 DW 6789HDATA ENDS

EXTRA SEGMENTRESULT DW 0000H

EXTRA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA,ES:EXTRASTART: MOV AX,DATA

MOV DS,AX

MOV AX,EXTRAMOV ES,AX

SUB AX,AXMOV AX, NUM1

MOV BX, NUM2ADD AX,BX

MOV DI,OFFSET RESULT

MOV [DI],AXMOV RESULT,AX

INT 3H

CODE ENDSEND START

RESULT:

INPUT: NUM1 23C5H

NUM2 6789H

OUTPUT: RESULT 8B4EH

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(B) 16-BIT SUBTRACTION

.MODEL TINY

DATA SEGMENT

NUM1 DW 23C5H NUM2 DW 1789HRESULT DW 0000H

DATA ENDS


EXTRA ENDS

CODE SEGMENTASSUME CS:CODE,DS:DATA,ES:EXTRA

START: MOV AX,DATA

MOV DS,AXMOV AX,EXTRA

MOV ES,AX

SUB AX,AX

MOV AX, NUM1MOV BX, NUM2

SUB AX,BX

MOV DI,OFFSET RESULTMOV [DI],AX

MOV RESULT,AX

INT 3H

CODE ENDSEND START

RESULT:

INPUT: NUM1 23C5H

NUM2 1789H

OUTPUT: RESULT 0C3CH

(C) 16-BIT MULTIPICATION

.MODEL TINY

DATA SEGMENT NUM1 DW 23C5H

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NUM2 DW 1789H

DATA ENDS


EXTRA ENDS

CODE SEGMENTASSUME CS:CODE,DS:DATA,ES:EXTRASTART: MOV AX,DATA

MOV DS,AX

MOV AX,EXTRAMOV ES,AX

SUB AX,AX

MOV AX, NUM1

MOV BX, NUM2MUL AX,BX

MOV DI,OFFSET RESULT

MOV [DI],AXMOV RESULT,AX

INT 3H

CODE ENDS

END START

RESULT:

INPUT: NUM1 23C5H NUM2 1789H

OUTPUT: DX:AX 0349D76DH

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DIVISION OF TWO NUMBERS

AIM: To Write and execute an Assembly language Program (ALP) to 8086 processor to divide a

32 bit unsigned number by a 16 bit unsigned numbers. Store the result in stack segment.

SOFTWARE REQIRED:

1. Pc with windows(95/98/XP/NT/2000)2. MASM SOFTWARE

PROGRAMS:

MODEL SMALL DATA SEGMENT

NUM1 DD 4096H

NUM2 DW 1024H

DATA ENDSSTACK SEGMENT

RESULT DW ?

STACK ENDS

CODE SEGMENTASSUME CS:CODE,DS:DATA,SS:STACK

START: MOV AX,DATAMOV DS,AX

MOV AX,STACK

MOV SS,AX

SUB AX,AXMOV SI,OFFSET NUM1

MOV AX,[SI]

INC SIINC SI

MOV DX,[SI]MOV BX, NUM2

DIV BXLEA BP,RESULT

MOV [BP],AX

INC BPMOV [BP],DX

INT 3H

CODE ENDSEND START

RESULT: INPUT: NUM1 00004096H NUM2 : 1024H

OUTPUT: AX: 0004H DX:0000H

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EEE MECH222 (110) II - I