Document No:
MLRIT/EC/LAB
MANUAL/EEE
VERSION 2.1
Date of Issue
01-July-2009
Date of Revision
29-December-
2014
Compiled by
G. Karthik Reddy
Verified by
Authorized by
HOD(ECE)
1

institute of technology.
ELECTRICAL AND ELECTRONICS ENGINEERING LAB
DEPARTMENT OF
ELECTRICAL AND ELECTRONICS ENGINEERING
ACADEMIC YEAR 2014-2015
II B.Tech MECH I-SEMESTER & II B.Tech AERO II-SEMESTER.
PREFACE
The significance of the Electrical and Electronics Engineering Lab is renowned in the various fields of engineering applications. For an Mechanical Engineer, it is obligatory to have the practical ideas about the Electrical and Electronics Engineering. By this perspective we have introduced a Laboratory manual cum Observation for Electrical and
Electronics Engineering Lab.
The manual uses the plan, cogent and simple language to explain the fundamental aspects of Electrical and Electronics Engineering in practical. The manual prepared very carefully with our level best. It gives all the steps in executing an experiment.
GUIDELINES TO WRITE YOUR OBSERVATION BOOK
1. Experiment Title, Aim, Apparatus, Procedure should be on right side.
2. Circuit diagrams, Model graphs, Observations table, Calculations table should be left side.
3. Theoretical and model calculations can be any side as per your convenience.
4. Result should always be in the ending.
5. You all are advised to leave sufficient no of pages between experiments for theoretical or model calculations purpose.
2

Institute of technology.
DO’S AND DON’TS IN THE LAB
DO’S:-
1. Proper dress has to be maintained while entering in the Lab. (Boys Tuck in and shoes, girls with apron)
2. All students should come to the Lab with necessary tools. (Cutting Pliers 6”,
Insulation remover and phase tester)
3. Students should carry observation notes and record completed in all aspects.
4. Correct specifications of the equipment have to be mentioned in the circuit diagram.
5. Student should be aware of operating equipment.
6. Students should be at their concerned experiment table, unnecessary moment is restricted.
7. Student should follow the indent procedure to receive and deposit the equipment from the Lab Store Room.
8. After completing the connections Students should verify the circuits by the Lab
Instructor.
9. The reading must be shown to the Lecturer In-Charge for verification.
10. Students must ensure that all switches are in the OFF position, all the connections are removed.
11. All patch cords and stools should be placed at their original positions.
DON’Ts:-
1. Don‟t come late to the Lab.
2. Don‟t enter into the Lab with Golden rings, bracelets and bangles.
3. Don‟t make or remove the connections with power ON.
4. Don‟t switch ON the supply without verifying by the Staff Member.
5. Don‟t switch OFF the machine with load.
6. Don‟t leave the lab without the permission of the Lecturer In-Charge.
3

institute of technology.
The following experiments are required to be conducted as compulsory experiments.
1.
Magnetization characteristics of d.c.shunt generator
2 . Swinburne‟s Test on DC shunt machine (Predetermination of efficiency of a given DC
Shunt machine working as motor and generator).
3.
Brake test on 3-phase Induction motor (performance characteristics).
4 . OC & SC tests on Single-phase transformer (Predetermination of efficiency and regulation at given power factors).
5 . Regulation of alternator by synchronous impedance method.In addition to the above four experiments, any one of the experiments from the following list is required to be conducted
6.
Load Test on a Single Phase transformer
1. PN junction diode characteristics.
2. Transistor CE characteristics.
3. Zener diode characteristics.
4. Rectifiers with filters
5.
Rectifiers without filters
6.
Study of CRO.
4

Institute of technology
Expt. No: 1
AIM: a) To determine the critical field resistance (R c
) and critical speed (N c
) at 1400 r.p.m. b) To determine the critical field resistance (R c
) and critical speed (N c
) at 1450 r.p.m. c) To determine the critical field resistance (R c
) and critical speed (N c
) at 1500 r.p.m. d) To determine the critical field resistance (R c
) and critical speed (N c
) at 1550 r.p.m
NAME PLATE DETAILS:
Specifications
Speed power
Current
Voltage
EQUIPMENTS REQUIRED:
S.No Equipments Type
DC Motor
1500 rpm
5HP
20A
230V
DC Generator
1500 rpm
3KVA
13.6A
230V
Range Quantity
1.
2.
3.
4.
Voltmeter
Ammeter
Rheostats
Tachometer
M.C
M.C
Wire wound
Digital
0-300v
0-1/2A
300Ω/1.5A
0-300)rpm
1 No
1 No
1 No
1 No
THEORY :
Open circuit characteristics or magnetization curve is the graph between the generated emf
(E g
) and field current (I f
) of a dc shunt generator. For field current is equal to zero there will be residual voltage of 10 to 12v because of the residual magnetism present in the machine. If this is absent then the machine cannot build up voltage. To obtain residual magnetism the machine is separately excited by a dc source. We can get critical field resistance (R c
) and critical speed (N c
) from open circuit characteristics curve.
Critical field résistance: It is the maximum value of resistance of field winding of D.C. shunt generator at critical speed..
Critical speed: It is the maximum speed of D.C. shunt generator at critical field resistance beyond which generator fails to generate voltage.
CIRCUIT DIAGRAM:
5

PROCEDURE:
1.) Connections are made as per the circuit diagram and supply is given by closing the
DPST switch
2.) Motor is started with the help of three point starter and bring the motor to rated speed by varying the motor field rheostat from minimum resistance position
3.) At rated speed by varying the generator field rheostat from maximum resistance position note down the readings of open circuit voltage and field current up to the rated voltage of D.C.shunt generator.
4.) Switch off the supply and draw the graph between V oc and I f
and hence it is the magnetization characteristics curve.
PRECAUTIONS:
1.) Loose connections must be avoided
2.) Always check the connections before u switch on the supply
3.) Motor field rheostat is kept at minimum resistance position and generator field rheostat is kept at maximum resistance position
4.) Experiment must be done in the presence of concern lab instructor.
TABULAR COLUMN:
S.No
1
2
3
4
5
V
OC
(Volts) I f
(amps)
6

CALCULATIONS:
N
1
=minimum speed i.e.,1400 r.p.m
R min
=V oc
(max.)/I f
(max.)
R max
= V oc
(min.)/I f
(min.)
Where R max
= critical field resistance
N
1
/N c
= R min
/ R max
Hence critical speed and critical field resistance can be found from the above calculations.
RESULT:
VIVA QUESTIONS
1.) What is the Basic Principle in dc generator?
2.) In DC Machine which Induced emf is suitable?
3.) The direction of induced emf and hence current in a conductor can be determined by?
4.) Mechanical Energy Converted to Electrical energy is called as?
5.) Write the EMF equation for D.C. Generator?
7

Institute of technology
Expt. No: 2
AIM: To perform Swine burn‟s test on the given DC machine and a) Determine the efficiency of D.C. shunt machine at 1500 rpm.
b) Determine the efficiency of D.C. shunt machine at 1450 rpm.
c) Determine the efficiency of D.C. shunt machine at 1400 rpm.
d) Determine the efficiency of D.C. shunt machine at 1550 rpm.
NAME PLATE DETAILS
SWINBURNE’S TEST
Specifications DC meter
Speed 1500 rpm power
Current
5HP
20A
Voltage 230V
EQUIPMENT REQUIRED:
S.No
1
DESCRIPTION RANGE
Voltmeter (0-300)V
2 Ammeter
3
4
Rheostats
Tachometer
CIRCUIT DIAGRAM:
(0-10)A &(0-1/2
A)
300/1.5 A
Digital
TYPE
M.C
M.C
QUANTITY
1 No
3 No
Variable type 1 No
Rotatory 1 No
8

THEORY:
Swine burn‟s Test is to determine the efficiency of a given DC shunt motor when working as both motor as well as generator.
It is simple indirect method in which losses are measured separately and the efficiency at any desired load can be predetermined. This test applicable to those machines in which flux is practically constant i.e. shunt and compound wound machines. The no load power input to armature consist iron losses in core, friction loss, windage loss and armature copper loss. It is convenient and economical because power required to test a large machine is small i.e. only no load power. But no account is taken the change in iron losses from no load to full load due to armature reaction flux is distorted which increases the iron losses in some cases by as 50%
Principle of DC Motor: It starts that rotating conductor is placed in between the magnetic fields the conductors experiences force and rotate the rotor, armature winding, yoke, eye bolt and frame test.
PROCEDURE:
1.
Make connections as per the circuit diagram.
2.
Show the connections to the lab instructor.
3.
Keeping the rheostat at minimum, Start the motor with the help of starter, and by adjusting field rheostat bring the motor to rated speed.i.e.1500 rpm.
4.
Note down all the meter readings at no load.
5.
Do necessary calculations and find out the efficiency of the Machine as a motor and as a generator.
OBSERVATIONS :
V
L
R a
S.No
1
2
I a
I f
I
L
CALCULATIONS:
1.
Load input voltage = V
L
I
L
2.
Load armature copper losses W i
=I a
3.
Constant losses or no-load losses W
2
R a c
= (V
LO
I
LO
) - Ia o
2
R a where W c
= no load input – no load copper losses.
9

Efficiency as a motor:
I
L
= Assumed load current at any load i.e., 5 amp or 10 amp or 15 amp.
V
L
= rated voltage= 230V.
Motor input power = V
L
I
L
Motor armature copper losses (W cu
) = I a
2
R a
= (I
L
–I
F
)
2
Ra
Total losses= iron losses (W i
) + copper losses(W cu
)
Efficiency of motor= output power/input power.
Where, input power = V
L
I
L
And output power = input power – power losses.
Or input power = output power + power losses.
Efficiency as generator:
I
L
= Assumed load current at any load i.e,5 amp or 10 amp or 15 amp.
V
L
= rated voltage= 230V.
Generator output power = V
L
I
L
armature copper losses (W cu
) = I a
2
R a
= (I
L
+I
F
)
2
Ra
Total losses= iron losses(Wi) + copper losses(Wcu)
Efficiency of Generator = output power/input power.
Where, output power = V
L
I
L
And output power = input power + power losses.
Or input power = output power - power losses.
RESULT:
VIVA QUESTIONS
1.
Which of the following is the unit of Torque ?
2.
When the starting resistance of a D.C. motor is used generally?
3.
Armature Reaction is related to?
4.
Brushes are made up of?
5.
The function of the commutator?
10

Institute of technology
Expt. No: 3
Brake Test on 3-Phase Induction Motor
AIM : a) To conduct a brake test on the given 3-¢ Slip ring Induction motor and to draw its performance Characteristics curves at load current of 1,2,3,4,5 amps. b) To conduct a brake test on the given 3-Slip ring Induction motor and to draw its performance Characteristics at load current of 1.2, 2.2, 3.2, 4.2, 5.2 amps. c) To conduct a brake test on the given 3-Slip ring Induction motor and to draw its performance Characteristics at load current of 2.8, 3.8, 4.8, 5.8, 6.8 amps. d) To conduct a brake test on the given 3-Slip ring Induction motor and to draw its performance Characteristics at load current of 1.5, 2, 25, 3, 3.5, 4, 4.5, 5 amps.
NAME PLATE DETAILS : (to be noted from the machine)
APPARATUS REQUIRED :
S.NO DESCRIPTION RANGE TYPE QTY
1
2
Ammeter
Voltmeter
(0-10)A
(0-600)V
MI
MI
1
1
3
4
5
Watt meters 10A,600v,UPF
Autotransformer 600v,3-¢
Tachometer ( 0-3000)
Unity variable digital
2
1
1
CIRCUIT DIAGRAM:
11

PRECAUTIONS :
1. There should not be any load on the motor initially.
2. The brake drum should be filled with water to cool it.
3. If the wattmeter shows negative deflection, reverse either pressure coil or current coil and take that reading as negative.
4. The rotor external resistance should be kept at max resistance position initially.
THEORY :
As a general rule, conversion of electrical energy to mechanical energy takes place in to the rotating part on electrical motor. In DC motors, electrical power is conduct directly to the armature, i.e, rotating part through brushes and commutator. Hence, in this sense, a DC motor can be called as „conduction motor‟. However, in AC motors, rotor does not receive power by conduction but by induction in exactly the same way as secondary of a two winding T/F receives its power from the primary. So, these motors are known as Induction motors. In fact an induction motor can be taken as rotating T/F, i.e, one in which primary winding is stationary and but the secondary is free.
The starting torque of the Induction motor can be increase by improving its p.f by adding external resistance in the rotor circuit from the stator connected rheostat, the rheostat resistance being progressively cut out as the motor gathers speed. Addition of external resistance increases the rotor impedance and so reduces the rotor current. At first, the effect of improved p.f predominates the current-decreasing effect of impedance. So, starting torque is increased. At time of starting, external resistance is kept at maximum resistance position and after a certain time, the effect of increased impedance predominates the effect of improved p.f and so the torque starts decreasing. By this during running period the rotor resistance being progressively cut-out as the motor attains its speed. In this way, it is possible to get good starting torque as well as good running torque.
12

PROCEDURE :
1) Give all the connections as per the circuit diagram.
2) Switch –ON the supply and slowly vary the auto transformer up to the rated voltage carefully.
3) At this voltage no load is applied.
4) Note the no-load readings of ammeter, voltmeter, wattmeters, speed & loads s
1 and s
2
.
5) Gradually increase the load on the motor by tightening the belt or rotating the wheels of a pulley and note the readings of voltmeter, wattmeters, spring balances and speeds at different load currents .
6) Remove the load completely & Switch-Off the power.
GRAPH : Graphs are drawn between the following a) speed vs torque b) efficiency vs output power c) load current vs output power d) slip vs torque.
OBSERVATIONS :
S.No Voltmeter reading
Ammeter reading
Spring Balance
S
1
S
2
Torque Speed
CALCULATIONS:
Input power= (W
1
+W
2
)*M.F
Where M.F= multiplying factor.
Torque = force * distance =F*r = mgr =((S
1
Output power =(2ΠNT/60)watts
-S
2
)9.81*0.1)NM
Efficiency=output power/input power.
Slip (S)=(N
S
-N r
)/N s
Where, W
1
+W
2
= wattmeter readings, T is the torque and N is the speed
And N s
= synchronous speed and N r
= rotor speed.
Also N s
=(120f s
)/p where p= number of magnetic poles and f s
=supply frequency
13

Model graphs:
Graph : Graphs are drawn between the following a) speed vs torque b) efficiency vs output power c) load current vs output power d) slip vs torque.
RESULT
VIVA QUESTIONS
1.
Which motor is suitable for Electric Traction?
2.
Which of the following are Iron Losses?
3.
DC Motors in Starters are used to?
4.
Swinburne‟s test is conducted on DC Shunt Machine at?
5.
Armature windings are made up of?
14

Institute of technology
Expt. No: 4
OC & SC TESTS ON 1- PHASE TRANSFORMER
Aim : a) To conduct OC & SC tests on 1-phase transformer and to calculate efficiency and regulation at unity power factor. b) To conduct OC & SC tests on 1-phase transformer and to calculate efficiency and regulation at 0.7 power factor. c) To conduct OC & SC tests on 1-phase transformer and to calculate efficiency and regulation at 0.6 power factor. d) To conduct OC & SC tests on 1-phase transformer and to calculate efficiency and regulation at 0.8 power factor.
Name plate details :
1-PHASE TRANSFORMER
Power rating 2KVA
Input Voltage 110V
Output voltage
Input current
220V
18.2A
Output Current 9.1A
Apparatus required :
S.No
1
2
3
4
DESCRIPTION
Voltmeter
Ammeter
Wattmeter
Auto Transformer
RANGE
(0-150)V
(0-2)A &(0-
10)A
L.P.F & U.P.F
230 V/0-
270V,8A
TYPE
M.I
M.I
Analog
QUANTITY
1 No
1 No
1 No
Wire Wound 1 No
OPEN CIRCUIT TEST:-
15

SHORT CIRCUIT TEST:
Theory :
Transformer is a static device which transfers electrical power from one circuit to another circuit either by step up or step down the voltage with corresponding decrease or increase in the current, without changing the frequency.
OC Test
The main aim of this test is to determine the Iron losses & No- load current of the T/F which are helpful in finding Ro & Xo.In this test generally supply will be given to primary
16

and secondary kept open. Since secondary is opened a small current (magnetizing current will flow and it will be 5 to 10% of full load current. The wattmeter connected in primary will give directly the Iron losses (core losses).
SC Test:
The main aim of this test is to determine the full load copper losses which is helpful in finding the R
01
, X
01
, Z
01
, efficiency and regulation of the Transformer. Generally low voltage side will be short circuited and supply will be given to high voltage side & it will be of 5-10% of the rated voltage. The wattmeter connected in primary will give directly the full load copper losses of the Transformer.
OC Test :
1) Give connections as per the circuit diagram.
2) Switch-ON the supply and apply rated voltage i.e., 110 volts to the primary winding by using the 1- phase auto transformer.
3) At 110 volts of voltmeter reading wattmeter reading represents the iron losses or core losses of a 1-phase transformer. Note the readings of Ammeter, Voltmeter &
Wattmeter at 110 volts.
SC Test :
1) Give connections as per the circuit diagram.
2) Switch-ON the supply and vary the Dimmerstat till rated full load current flows through transformer.
3) At rated current Note the readings of Ammeter, Voltmeter & Wattmeter .At rated current wattmeter reading represents the copper losses or winding losses.
Observations from Open Circuit Test :
S.No.
1
Voltmeter
2
Observations from Short Circuit Test :
Voltmeter S.No.
1
2
17
Ammeter
Ammeter
Wattmeter
Wattmeter

Calculations:
S.No. load Input power
Output power
Cu loss Iron loss Efficiency
1
2
3
4 full
1/4
1/2
3/4
1) Efficiency=output power/input power where input power = V i
I i cos ¢
2) output power = input power + iron loss + copper loss
3) Regulation=(V r
cos ¢+/-V x sin¢)/V
1
Where V r
=I
1
R
01
and V x
=I
1
X
01 i.e.,V
1
= primary voltage and
V r
=I
1
R
01
= primary resistive drop. And V x
=I
1
X
01
= primary reactive drop+ for power factor lagging and – for power factor leading.
Where cos ¢=W oc
/(V oc
* I oc
) from open circuit test.
Z
01
=V sc
/I sc
, R
01
=W sc
/I sc
2
, and (Z
01
)
2
= (R
01
)
2
+ (X
01
)
2
Precautions:
1) The Dimmer stat should be kept at minimum O/P position initially.
2) In OC test, rated voltage should be applied to the Primary of the Transformer.
3) In SC test, the Dimmer stat should be varied up to the rated load current only.
4) The Dimmer stat should be varied slowly & uniformly.
Result:
Viva Questions:
1.
Induced emf in a coil is directly proportional to?
2.
The transformer rating is?
3.
The transformer transforms?
4.
The main purpose of using Core in transformer is to?
5.
The transformer cores are laminated in order to?
18

Institute of technology
Expt. No: 5
Aim : a) To conduct Load test on the given 1-Φ Transformer at 20Watts b) To conduct Load test on the given 1-Φ Transformer at 40Watts c) To conduct Load test on the given 1-Φ Transformer at 60Watts d) To conduct Load test on the given 1-Φ Transformer at 100Watts
Name plate details :
Apparatus required :
S.
No Apparatus Range Type Qty
1 Voltmeters 0-150V, 0-75V M.I 1, 1 No
2 Ammeters 0-2A, 0-15A M.I 1 No
3 Wattmeter 2A , 150V, 60 0 W LPF 1 No
1 .
5A , 1 50V, 600W UPF 1 No
4 Auto T/F 230V/0-230V
5 Load (0-5/10)A
1 No
1 No
Procedure:
1.
Connections are made as per the circuit diagram.
19

2.
By varying the Auto transformer, rated voltage is applied to the input side of the transformer and should be maintained constant throughout the experiment.
3.
By varying the load in steps, readings of ammeter, voltmeter, and wattmeter are noted down in each step.
4.
Efficiency and Regulations are calculated in each step and tabulated.
5.
Graphs are drawn Output vs Efficiency
Calculations table:
Rated Secondary Voltage, V
2
= 230V
20

Precautions :
1.
The Dimmer stat should be kept at minimum O/P position initially
2.
Rated voltage should be maintained on the Primary of the Transformer
3.
The Dimmer stat should be varied slowly & uniformly
Results:
21

Institute of technology
Expt. No: 6
REGULATION OF 3-PHASE ALTERNATOR
: a) To conduct OC test & SC test on the given 3-Alternator and to determine its regulation by synchronous impedance method at 0.7 power factor. b) To conduct OC test & SC test on the given 3-Alternator and to determine its regulation by synchronous impedance method at 0.6 power factor. c) To conduct OC test & SC test on the given 3-Alternator and to determine its regulation by synchronous impedance method at 0.8 power factor. d) To conduct OC test & SC test on the given 3-Alternator and to determine its regulation by synchronous impedance method at 0.9 power factor.
Nameplate details :
DESCRPTION D.C MOTOR 3-ALTERNATOR
3 KVA Capacity
Voltage
Current
Speed
Excitation
5 H.P
220V
19A
1500 Rpm
220V, 1.5A
415V
4.2A
1500 Rpm
220V, 1.4A
3.
4.
Apparatus required :
S.No. Equipment
1. ammeter
2. voltmeter ammeter rheostats type mc mi mi variable range
(0-1/2)A
(0-10)A
quantity
300 ohms/1.5A
1
(0-300/600)V 1
1
2
22

5.
1.
2.
3.
4.
Tabular column:
For O.C.Test:
Si.no. Voc
1.
2.
3.
4.
5.
For S.C.Test:
Si.no. Isc
If
If
CIRCUIT DIAGRAM: FOR O.C & S.C. TEST:
23

Precautions: a.
Operate the 3-point starter slowly & uniformely. b.
Keep the speed of the prime mover to its rated value throughout the experiment. c.
In OC test, there should not be any load on Alternator. d.
In SC test, the SC current should not exceed its rated value.
Theory:
Alternator is a machine, which converts mechanical energy to electrical energy. Regulation of an Alternator can be calculated by synchronous impedance method. In OC test the terminals of the alternator are kept opened and a voltmeter is connected. Keeping speed constant, a relation b/w field current & open circuit voltage are obtained. In SC test, the terminals are short circuited with a suitable ammeter & a relation b/w field current & short circuit Current are obtained.
Voltage regulation:
It is defined as the rise in terminal voltage of an isolated
Machine when full load is thrown off w.r.t voltage on the full load, when speed & excitation remaining constant.
24

Procedure:
OC test:
1) Give all connections as per the circuit diagram.
2) Switch-ON the supply & by varying the starter, prime mover speed is adjusted to rated.
3) Now keeping the field current at zero, note the induced emf in armature duo to residual Magnetism.
4) By slowly varying the potential divider, field current is increased& corresponding emf Induced is noted up to above 20% of rated voltage.
SC test:
1) Give all connections as per the circuit diagram.
2) Switch-ON the supply & by varying the starter, prime mover speed is adjusted to rated.
3) By slowly varying the potential divider, field current is increased & corresponding short Circuit current is noted up to rated value.
4) Measure Ra value using multimeter.
Graph:
1) A graph is drawn b/w If and V which is known as OC curve, by taking If on X-axis and
V on Y-axis.
2) A graph is drawn b/w If and ISC which is known as SC curve, by Taking If on X-axis and ISCV on Y-axis.
Model graphs:
25

Model calculations:
Now, Synchronous Impedance (Z s
) = V oc
Also, (Z s
)
2
= (X
S
)
2
+ (R a
)
2
/I sc
Where R a
= 5 ohms
% Regulation = [(E o
-V) / V] 100
Where,
E o
=√(Vcos¢+IRa) 2
+/- √(Vsin¢+IX s
)
2 i.e., + for power factor lagging and – for power factor leading.
I=5 amps, cos¢=0.9, and V= 415/1.732=240volts.
RESULT
26

Experiment No:1
AIM:
1.
Plot the Volt-Ampere characteristics of a given P-N junction diode 1N4007 with
1kΩ resistance.
2.
Find the cut-in voltage and dynamic forward-bias resistance of a given P-N junction diode 1N4007 with
1.5 kΩ
3.
Find the cut-in voltage and static forward-bias resistance of a given P-N junction diode 1N4007 with
2 kΩ
4.
Plot the Volt-Ampere characteristics of a given P-N junction diode OA79 with
560Ω
5.
Find the cut-in voltage and dynamic forward-bias resistance of a given P-N junction diode OA79 with 470Ω
Equipment required:
Bread board-----1No.
DC power supply (0-20V) -----1No.
Digital DC Voltmeter (0-20V) -----1No.
Digital DC Ammeter ( 0-200mA)-----1No.
Components required:
Silicon Diode 1N 4007
Germanium Diode OA79
Resistor
THEORY:-
A p-n junction diode conducts only in one direction. The V-I characteristics of the diode are curve between voltage across the diode and current through the diode.
When external voltage is zero, circuit is open and the potential barrier does not allow the current to flow. Therefore, the circuit current is zero. When P-type (Anode is connected to +ve terminal and n- type (cathode) is connected to –ve terminal of the supply voltage, is known as forward bias. The potential barrier is reduced when diode is in the forward biased condition. At some forward voltage, the potential barrier altogether eliminated and current starts flowing through the diode and also in the circuit. The diode is said to be in ON state. The current increases with increasing forward voltage. When N-type (cathode) is connected to +ve terminal and P-type
(Anode) is connected to the –ve terminal of the supply voltage is known as reverse bias and the potential barrier across the junction increases. Therefore, the junction resistance becomes very high and a very small current (reverse saturation current)
27

flows in the circuit. The diode is said to be in OFF state. The reverse bias current is due to minority charge carriers.
Circuit diagrams:
1 Forward bias
R
(0-200mA)
+
A
Diode
(1- 20V)
+
V
(0-20V)
Forwad bias
Fig.1.1
2 Reverse bias
R (0-200mA)
+
A
V
(0-20V)
+
(1-20V)
Reverse bias
Fig.1.2
Procedure:
1. Construct the circuit as shown in the Fig1.1.Use 1N 4007 diode and forward
bias it.
2. Increase the power supply voltage gradually in steps and note the voltmeter and
Ammeter readings in Table -1.
.
4. Reverse-bias the diode, construct the circuit as shown in Fig:1.2. Use 1N 4007.
5. Increase the power supply voltage in convenient steps and note down the micro
ammeter current readings.
28

Observations:
1.
Forward Bias
V
F
(Volts)
I
F
(mA)
Table -1
2.
Reverse Bias
V
R
(Volts)
I
R
(μA)
Table -2
Model Graph:
Draw a Graph with volatage on X- axis and Current on Y-axis for pn jn. diode.
Forward bias & Reverse bias is ploted in fig
Characteristics:
Fig.
From graph calculate:
Dynamic forward Resistance : r
AC
=∆V
F
/∆I
F
Result:
Cut-in Voltage for 1N4007 = Volts
29

Dynamic Forward Resistance = Ω
Viva Questions:
1.
Explain how to obtain dynamic Resistance from the P-N junction diode characteristics?
2.
Define cut-in voltage. Also mention Cut-in voltage of Germanium diode & Silicon diode?
3.
What is breakdown voltage?
4.
Explain the phenomena of Half effect?
5.
What is Diffusion Capacitance of P-N junction diode?
6.
What is transition capacitance of P-N junction diode?
7.
Name Donor Impurity & Acceptor Impurity?
8.
The forbidden energy gap for Silicon diode at room temperature?
9.
What are the Applications of Half Effect?
10.
Explain the effect of the temperature on the diode?
11.
Explain V-I characteristics of P-N junction diode?
30

Experiment No.2
AIM:
1.
Plot the Volt-Ampere characteristics of a given Zener diode BZX 5.1
with
1kΩ
2.
Find the cut-in voltage , dynamic forward-bias resistance and breakdown voltage of a given Zener diode BZX 5.1
with
1.5 kΩ
3.
Plot the Volt-Ampere characteristics of a given Zener diode BZX 9.1
with 560Ω
4.
Find the cut-in voltage , dynamic forward-bias resistance and breakdown voltage of a given Zener diode BZX 9.1
with 470Ω
5.
Find the cut-in voltage , static forward-bias resistance and breakdown voltage of a given Zener diode BZX 12.1
with
1kΩ
Equipment required:
Bread board-----1No.
DC power supply (0-20V) -----1No.
Digital DC Voltmeter ( 0-20V)-----1No.
Digital DC Ammeter ( 0-200mA)-----1No.
Components required:
Zener Diode
Resistor
Theory:-
A zener diode is heavily doped p-n junction diode, specially made to operate in the break down region. A p-n junction diode normally does not conduct when reverse biased. But if the reverse bias is increased, at a particular voltage it starts conducting heavily. This voltage is called Break down Voltage. High current through the diode can permanently damage the device To avoid high current, we connect a resistor in series with zener diode. Once the diode starts conducting it maintains almost constant voltage across the terminals what ever may be the current through it, i.e., it has very low dynamic resistance. It is used in voltage
Circuit diagrams:
1. Forward bias
31

R
(1-20V)
(0-200mA)
+
A
Diode
+
V
(0-20V)
Forwad bias
Fig.2.1
2. Reverse bias
R (0-200mA)
+
A
Diode
+
V
(0-20V) (1-20V)
Reverse bias
Fig.2.2
Procedure:
1. Construct the circuit as shown in the Fig 2.1.Use BZX 11.5V diode and
forward bias it.
2. Increase the power supply voltage gradually in steps and note the voltmeter and
Ammeter readings in Table -3.
3.
Reverse-bias the diode, construct the circuit as shown in Fig 2.2.
4.
Increase the power supply voltage in convenient steps and note down the
ammeter and voltmeter readings in Table -4.
Observations:
1.
Forward Bias
32

V
F
(Volts)
I
F
(mA)
Table -3
2.
Reverse Bias
V
R
(Volts)
I
R
(mA)
Table -4
Model Graph:
Draw a Graph with voltage on X- axis and Current on Y-axis for both
Forward bias and Reverse bias is plotted in fig
Fig.
Result:
Cut-in Voltage for BZX 11.5V = Volts
Dynamic Forward Resistance = Ω
Break down Voltage = Volts
Viva Questions:
33

!. Explain the Zener Breakdown?
P-N junction diode & Zener diode?
3. What are the applications of Zener diode?
4. Explain Avalanche breakdown?
5. Explain V-I characteristics of Zener diode junction diode?
34

Experiment No.3
AIM:
1.
Examine the input and out put wave forms of a half wave rectifier without and with
Capacitor filter C=10µF for various loads. Also find ripple factor.
2.
Examine the input and out put wave forms of a half wave rectifier without and with
Capacitor filter C=470µF for various loads. Also find ripple factor.
3.
Examine the input and out put wave forms of a half wave rectifier without and with
4.
5.
Capacitor filter C=10µF for various loads. Also find percentage of regulation.
Examine the input and out put wave forms of a half wave rectifier without and with
Capacitor filter C=470µF for various loads. Also find percentage of regulation .
Examine the input and out put wave forms of a half wave rectifier without and with
Capacitor filter C=100µF for various loads. Also find percentage of regulation .
Equipment required:
Bread board-----1No
C.R.O-----1No
AC Power supply12V-0-12V-----1No
Digital DCVoltmeter ( 0-20V)----- 1No
Digital AC Voltmeter ( 0-20V)-----1No
Components required:
Diode 1N 4007-----1No
Resistors 390Ω,1.2kΩ,4.7kΩ,10kΩ-----1No Each.
Capacitor 10µF, 470µF-------1No Each.
THEORY:
In Half Wave Rectification, When AC supply is applied at the input, only
Positive Half Cycle appears across the load whereas, the negative Half Cycle is suppressed. How this can be explained as follows: During positive half-cycle of the input voltage, the diode D1 is in forward bias and conducts through the load resistor RL. Hence the current produces an output voltage across the load resistor
RL, which has the same shape as the +ve half cycle of the input voltage. During the negative half-cycle of the input voltage, the diode is reverse biased and there is no current through the circuit. i.e., the voltage across RL is zero. The net result is that only the +ve half cycle of the input voltage appears across the load. The average value of the half wave rectified o/p voltage is the value measured on dc voltmeter. For practical circuits, transformer coupling is usually provided for two reasons.
1. The voltage can be stepped-up or stepped-down, as needed.
2. The ac source is electrically isolated from the rectifier. Thus preventing shock hazards in the secondary circuit.
The efficiency of the Half Wave Rectifier is 40.6%
35

Circuit diagrams:
230V,50Hz
AC
Step down
Transformer
Diode
12V
+
Primary
Side
Secondary
Side
0V
RL
+
1
V Vo
1'
Fig 3.1:Half Wave Rectifier Without Filter
230V, 50Hz
AC
Step down
Transformer
Diode
+
Primary
Side
Secondary y
Side
RL
1
C
+
V Vo
CRO
+ Ch1 + -
CRO
+ Ch1 + -
0V 1'
Fig 3.2:Half Wave Rectifier With Filter
Procedure:
1. Construct the circuit as shown in Fig3.1Use the diode 1N 4007 and load
Resistance, R
L.
2. Observe the voltage across the secondary of the transformer and across the out put
terminals 1-1
‟
by using CRO.
3. Remove the Load resistance from 1-1
‟ and note down no-load voltage,V
NL
4. Vary the load R
L
in convenient steps and note the ac voltage and dc voltage
across the load.
5. Calculate ripple factor and regulation for different loads.
6.Construct the circuit as shown in fig3.2Use the diode 1N 4007,Capacitor and load
resistance RL.Observe the voltage across the secondary of the transformer and
across the out put terminals 1-1
‟
by using CRO.
7. Remove the Load resistance from 1-1
‟ and note down no-load voltage,V
NL
8. Vary the load R
L
in convenient steps and note the ac voltage and dc voltage
across the load.
9. Calculate ripple factor and regulation for different loads.
10. Tabulate the readings as in Table 5&6.
36

Observations:
1
2
3
4
S.NO R
L
(Ω)
V
NL
= Volts
V
FL
=V
DC
(Volts) V
AC
(Volts) Ripple
Factor=V
AC
/V
DC
%Regulation
Table.5 Half Wave Rectifier Without Filter
V
NL
= Volts
S.NO R
L
(Ω) V
FL
=V
DC
(Volts) V
AC
(Volts) Ripple %Regulation
Factor=V
AC
/V
DC
1
2
3
4
Table. 6 Half Wave Rectifier With Filter
Where
(V
NL
-V
FL
)
%Regulation = ----------------- ×100
V
FL
Model Graph:
Input wave form: Output wave form without Filter:
Output wave form with Filter:
37

Result:
Ripple factor without filter at 1.2KΩ =
Ripple factor with capacitor filter at 1.2KΩ =
Regulation without filter at 1.2KΩ =
Regulation with capacitor filter at 1.2KΩ =
Viva Questions:
1.
What is Rectifier?
2.
Classify the Rectifiers?
3.
Explain the operation of HWR through Waveforms
4.
Define Average value?
5.
Define rms value ?
6.
Define peak inverse voltage (PIV)?
7.
What is ripple factor in HWR? Also mention Theoretical ripple factor value of
HWR?
8.
What is filter?
9.
Explain the operation of HWR with capacitor filter
10.
Explain the operation of HWR with inductor filter?
11. What is TUF?
12. What is efficiency of the Half Wave Rectifier?
38

Experment No.4
AIM:
1.
Examine the input and out put wave forms of a full wave rectifier without and with
Capacitor filter C=10µF for various loads. Also find ripple factor.
2.
Examine the input and out put wave forms of a full wave rectifier without and with
Capacitor filter C=470µF for various loads. Also find ripple factor.
3.
Examine the input and out put wave forms of a Full wave rectifier without and with
Capacitor filter C=10µF for various loads. Also find percentage of regulation.
4.
Examine the input and out put wave forms of a half wave rectifier without and with
Capacitor filter C=470µF for various loads. Also find percentage of regulation .
5.
Examine the input and out put wave forms of a half wave rectifier without and with
Capacitor filter C=100µF for various loads. Also find percentage of regulation .
Equipment required:
Bread board-----1No
C.R.O-----1No
AC Power supply12V-0-12V-----1No
Digital DCVoltmeter ( 0-20V)----- 1No
Digital AC Voltmeter ( 0-20V)-----1No
Components required:
Diode 1N 4007-----2No‟s.
Resistors 390Ω,1.2kΩ,4.7kΩ,10kΩ
Capacitor 10µF, 470µF
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 RL. During negative half cycle, diode D2 becomes forward biased and
D1 reverse biased. Now, D2 conducts and current flows through the load resistor
RL in the same direction. There is a continuous current flow through the load resistor RL, during both the half cycles and will get unidirectional 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 only during one half cycle (180 degree).
39

Circuit diagrams:
Center taped
Step down
Transformer
12V
Diode
230V,50Hz
AC
+
Primary side
Secondary
Side
0V
1
RL
+
V Vo
CRO
+ Ch1 + -
12V
Diode
1'
230V, 50Hz
AC
Fig 4.1:Full wave Rectifier Without Filter
Center taped
Step down
Transformer
12V
Diode
1
+ Primary
Side
Secondary
SIde
0V
RL
C +
V Vo
CRO
+ Ch1 + -
12V 1'
Diode
Fig 4.2:Full wave Rectifier With Filter
Procedure:
1. Construct the circuit as shown in Fig 4.1Use the diodes 1N 4007 and load
resistance R
L.
2.Observe the voltage across the secondary of the transformer and across the out
put terminals 1-1
‟
by using CRO.
3. Remove the Load resistance from 1-1
‟ and note down no-load voltage, V
NL
4. Vary the load R
L
in convenient steps and note the ac voltage and dc voltage
across the load.
5. Calculate ripple factor and regulation for different loads.
6.Construct the circuit as shown in Fig4.2Use the diodes 1N 4007,Capacitor and
load resistance RL.Observe the voltage across the secondary of the transformer
and across the out put terminals 1-1
‟
by using CRO.
7. Remove the Load resistance from 1-1
‟ and note down no-load voltage, V
NL
8. Vary the load R
L
in convenient steps and note the ac voltage and dc voltage
across the load.
40

1
2
3
4
9. Calculate ripple factor and regulation for different loads.
10. Tabulate the readings as in Table 7&8.
Observations:
S.NO R
L
(Ω)
V
NL
= Volts
V
FL
=V
DC
(Volts) V
AC
(Volts) Ripple
Factor=V
AC
/V
DC
%Regulation
Table.7 FULL Wave Rectifier Without Filter
V
NL
= Volts
S.NO R
L
(Ω)
V
FL
=V
DC
(Volts) V
AC
(Volts) Ripple
Factor=V
AC
/V
DC
%Regulation
1
2
3
4
Table.8 FULL Wave Rectifier With Filter
Where
(V
NL
-V
FL
)
%Regulation = ----------------- ×100
V
FL
Model Graph:
Input wave form: Output wave form without Filter:
Output wave form with Filter:
41

Result:
Ripple factor without filter at 1.2KΩ =
Ripple factor with capacitor filter at 1.2KΩ =
Regulation without filter at 1.2KΩ =
Regulation with capacitor filter at 1.2KΩ =
Viva Questions:
1.
Explain the operation of FWR through Waveforms?
2.
What is the Peak current, average current and r.m.s current ?
3 . What is efficiency of the Full Wave Rectifier?
4. What is TUF in Center tapped FWR ?
5. What is ripple factor in FWR?
6. What is the general expression for ripple factor in FWR?
7. Explain the operation of FWR with capacitor filter.
8. Give the ripple factor value of FWR?
9. Explain the operation of FWR with inductor filter .
10. Give the ripple factor value in FWR with inductor filter?
42

Experiment no:5
AIM:
1.
Plot a family of output and input characteristics of a given transistor BC 107
(NPN) connected in common base configuration. For input characteristics set V
CB
= 0v, V
CB
= 5v, V
CB
=10v . For Output characteristics I
E
= 1.1mA, 2.1mA and
3.1mA.
2.
Find h-parameters h ib,
h rb, h ob,
and h fb
of a given transistor BC 107 (NPN) connected in Common base configuration. For input characteristics set V
CB
= 0v,
V
CB
=3v, V
CB
=5v . For Output characteristics I
E
= 1.5mA, 2.5mA and 3.5mA.
3.
Plot a family of output and input characteristics of a given BC557 (PNP) transistor connected in common base configuration. . For input characteristics set V
CB
= 0v,
V
CB
=3v, V
CB
=5v . For Output characteristics I
E
= 1.5mA, 2.5mA and 3.5mA
4.
Find h-parameters h ib,
h rb, h ob,
and h fb
of a given transistor BC557 (PNP ) connected in Common base configuration. For input characteristics set V
CB
= 2v, V
CB
=7v,
V
CB
=14v . For Output characteristics I
E
= 1.5mA, 2.5mA and 3.5mA.
5.
Plot a family of output and input characteristics of a given BC557 (PNP) transistor connected in common base configuration. . For input characteristics set V
CB
= 10v,
V
CB
=12v, V
CB
=18v.
For Output characteristics I
E
= 3mA, 4mA and 7mA.
Equipment required:
Bread board-----1No
DC power supplies (0-20V) -----2No‟s.
Digital DC Voltmeters (0-20V) –2No‟s
Digital DC Ammeters (0-200mA) –2No‟s.
Components required:
Transistor
Resistors 1kΩ-----2No‟s.
THEORY:
A transistor is a three terminal active device. T he terminals are emitter, base, collector. In CB configuration, the base is common to both input (emitter) and
43

output (collector). For normal operation, the E-B junction is forward biased and C-
B junction is reverse biased.
So, V
EB
 f
1
(V , I )
CB E
and I
C
 f
2
(V , I )
CB E
.
With an increasing the reverse collector voltage, the space-charge width at the output junction increases and the effective base width „W‟ decreases. This phenomenon is known as “Early effect”. Then, there will be less chance for recombination within the base region. With increase of charge gradient with in the base region, the current of minority carriers injected across the emitter junction increases. The current amplification factor of CB configuration is given by,
α= ΔIC/ ΔIE
Circuit diagrams:
R
E
(0-20)V
0-200mA
+
A
I
E
(0-20V)
+
V
V
EB
NPN
V
+
V
CB
0-20V
R
C
(0-20)V)
R
E
( 0-20)V
Fig:5.1 Transistor CB Input characteristics
(0-200mA)
+
A
I
E
NPN
I
C
+
A
0-200mA
R
C
V
+
V
CB
0-20V
(0-20)V
Fig:5.2 Transistor CB Output characteristics
For plotting input characteristics:
44

1.To construct the circuit as shown in the Fig5.1.
2.Fix V
CB
=open=0V,by varying the out put power supply and note that input power
supply should be in minimum during V
CB
kept constant.
3.Now by varying input power supply ,change V
EB in convenient steps and note
down the emitter current I
E at each step.
4.Repeat steps 2 and 3 for different constant values of V
CB
=5V and 10V.
5.Tabulate the readings a in the Table-9.
Observations:
V
CB
=
V
EB
VOLTS
I
E mA
V
CB
V
=
EB
VOLTS
I
E mA
V
EB
VOLTS
V
CB
=
I
E mA
Table-9
For plotting output characteristics:
6. Set I
E to 1.1 mA by varying the input power supply and note that output power
45

supply should be in minimum during I
E
kept constant.
7.Now by varying tout put power supply ,change V
CB
in convenient steps and note
down the collector current I
C at each voltage step.
8.Repeat steps 6 and 7 for constant values of I
E
equal to 2.1mA,3.1 mA and 4.1mA.
9.Tabulate the readings as in Table-10.
Observations:
V
CB
VOLTS
I
E
=
I
E mA
V
CB
VOLTS
I
E
=
I
E mA
V
CB
VOLTS
I
E
=
I
E mA
Table-10
Model Graph:
1.Plot the family of input characteristics by taking emitter voltage V
EB
on X-
axis and Emitter current I
E
on Y-axis for constant values of output voltage V
CB
.
The model set of input characteristic curves are shown below in Fig 5.3.
2.
Plot the family of output characteristics by taking collector voltage V
CB
on X-
axis and Collector current I
C on Y-axis for constant values of input current I
E
.
The model set of output characteristic curves are shown below in Fig 5.4.
INPUT CHARACTERISTICS OUTPUTCHARACTERISTICS
46

Fig 5.3 Fig5.4
Viva Questions:
1.
Draw the input and output Characteristics of CB Configuration
2.
What is base width modulation?
3
Define “ α”?
4. What is the typical value of “ α” ?
α “
6. Explain the input Characteristics?
7. Explain the output Characteristics?
8. What is the general expression for collector current?
9. What is I
CBO?
47

Experiment no: 6
AIM:
1.
Plot a family of output and input characteristics of a given transistor BC 107
(NPN) connected in common emitter configuration. For input characteristics set
V
CE
= 5v, V
CE
=10v, V
CE
=15v. For Output characteristics I
B
= 10µA, 15µA and
20µA.
2.
Find h-parameters h ie,
h re, h oe,
and h fe
of a given transistor BC 107 (NPN) connected in common emitter configuration. For input characteristics set V
CE
= 0v, V
CE
=5v,
V
CE
=10v . For Output characteristics I
B
= 15µA, 20µA and 25µA .
3.
Plot a family of output and input characteristics of a given transistor BC557 (PNP) connected in common base configuration. For input characteristics set V
CE
= 0v,
V
CE
=3.5v, V
CE
=5.5v.
For Output characteristics I
B
= 25µA, 30µA and 35µA
4.
Find h-parameters h ie,
h re, h oe,
and h fe
of a given transistor BC557 (PNP) connected in common emitter configuration For input characteristics set V
CE
= 0v,
V
CE
=3v, V
CE
=5v.
For Output characteristics I
B
= 5µA, 10µA and 25µA
5.
Plot a family of output and input characteristics of a given transistor BC 107
(NPN) connected in common emitter configuration. For input characteristics set
V
CE
= 5v, V
CE
=15v, V
CE
=20v . For Output characteristics I
B
= 10µA, 15µA and
20µA
Equipment required:
Bread board-----1No
DC power supplies (0-20V) -----2No‟s.
Digital DC Voltmeters (0-20V) –2No‟s
Digital DC Ammeters (0-200mA) –2No‟s.
Components required:
Transistor BC
Resistors 1kΩand 10kΩ-----1No each.
THEORY:
A transistor is a three terminal device. The terminals are emitter, base, collector. 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 expected since 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 unto few
48

volts only. After this the collector current becomes almost constant, and independent of VCE. The value of VCE up to which the collector current 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
Circuit diagrams:
R
C
R
B
(0-20V)
0-200mA
+
A
I
B
NPN
0-20V
+
V V
BE
V
CE
V
+
0-20V
(0-20V)
R
B
Fig:6.1 Transistor CE Input characteristics
( 0-20V)
0-200mA
+
A
Ib
NPN
0-200mA R
C
A
+
Ic
V
CE
V
+
0-20V
(0- 20V)
Fig:6.2 Transistor CE Output characteristics
For plotting input characteristics:
1.To construct the circuit as shown in the Fig6.1.
2.Fix V
CE
=open=0V,by varying the out put power supply and note that input power
49

supply should be in minimum during V
CE
kept constant.
3.Now by varying input power supply ,change V
BE in convenient steps and note
down the base current I
B
for each step.
4.Repeat steps 2 and 3 for different constant values of V
CE
=5V and 10V.
5.Tabulate the readings a in the Table-11.
Observations:
V
BE
V
CE
= VOLTS
I
B
VOLTS mA
V
BE
V
CE
= VOLTS
VOLTS
I
B mA
V
BE
V
CE
= VOLTS
VOLTS
I
B mA
Table-11
For plotting output characteristics:
6. Set I
B to 0.1 mA by varying the input power supply and note that output power
supply should be in minimum during I
B
kept constant.
7.Now by varying the output power supply ,change V
CE
in convenient steps and
note down the collecter current I
C at each voltage step.
8.Repeat steps 6 and 7 for constant values of I
B
=0.3mA and 0.5 mA .
9.Tabulate the readings as in Table-12.
50

Observations:
V
CE
VOLTS
I
B
= µA
I
C mA
V
CE
VOLTS
I
B
= µA
I
C mA
V
CE
VOLTS
I
B
= µA
I
C mA
Table-12
Model Graph:
1.Plot the family of input characteristics by taking base voltage V
BE
on X-axis and base current I
B
on Y-axis for constant values of output voltage V
CE
. The model set of input characteristic curves are shown below in Fig 6.3.
2. Plot the family of output characteristics by taking collector voltage V
CE
on X-axis and Collector current I
C on Y-axis for constant values of input current I
B
. The model set of output characteristic curves are shown below in Fig 6.4.
INPUT CHARACTERISTICS: OUTPUT CHARACTERSITICS:
Fig :6.3 Fig:6.4
51

Viva Questions:
1.
Draw the input and output Characteristics of CE Configuration
2.
What is base width modulation?
3.
Define “ ß ”?
4. What is the typical value of “ ß ”
?
ß “
6. Explain the input Characteristics?
7. Explain the output Characteristics?
8. What is the general expression for collector current?
9. What is I
CEO?
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