1 experiment no. : 1 date: characteristics of pn diode...

1

EXPERIMENT NO. : 1 DATE:

CHARACTERISTICS OF PN DIODE

AIM

To plot the forward and reverse VI characteristics of a PN diode and to calculate the

following parameters 1. Forward resistance

2. Reverse resistance 3. Cut in voltage

APPARATUS REQUIRED

EQUIPMENT RANGE

Power supply (0-10)V

Ammeter (0-30)mA, (0-100)µA

Volt meter (0-1)V, (0-10)V

Resistors 1K

Diode 1N 4001

Bread board & wires

-

THEORY

In a piece of semiconductor material, if one half is doped by P-type impurity and the other half is doped by N-type impurity, a PN junction is formed. The plane dividing the two halves or zones is called PN junction. The N-type material has high concentration of free electrons, while

P-type material has high concentration of holes. Therefore, at the junction there is a tendency for the free electrons to diffuse over the P-side and holes the N-side. This process is called diffusion.

As the free electron moves across the junction from N-type to P-type, the donor irons become positively charged. Hence a positive charge is built on the N-side of the junction. The free

electrons that cross the junction uncover the negative acceptor ions by filling in the holes. Therefore, a net negative charge is established on the P-side of the junction. This net negative

charge on the P-side prevents further diffusion of electron in to the P-side. Similarly, the net positive charge on the N-side repels the holes crossing from P-side to N-side. Thus a barrier is set-up near the junction which prevents further movement of charge carriers. As the consequence

of the induced electric field across the depletion layer, an electrostatic potential difference is established between P and N region, which is called the potential barrier, junction barrier,

diffusion potential, or contact potential, VO. The magnitude of the contact potential VO varies

with doping levels and temperature. VO is 0.3 V for germanium and 0.72 V for silicon. Forward bias

When positive terminal of the battery is connected to the P-type and negative terminal to

the N-type of the PN diode, the bias is known as forward bias. Under the forward bias condition, the applied positive potential repels the holes in P-type region so that the holes move towards the junction and the applied negative potential repels the electron in the N-type region and the

electron move towards the junction. Eventually, when the applied potential is more than the internal barrier potential the depletion region and internal potential barrier disappear. A feature worth to be noted in the forward characteristics is the cut in or threshold voltage VR below which

the current is very small. It is 0.3 V for GE and 0.7 V for Si respectively. At the cut in voltage, the potential is overcome and the current through the junction starts to increase rapidly.

2

SYMBOL

A K

PN diode

PIN DIAGRAM

AK

1N 4001

CONSTRUCTION

P N

A K

CIRCUIT DIAGRAM

FORWARD BIAS

1 K

( 0 - 30 ) mA

+ -

1N 4001

( 0 - 1 ) V

A

V

+ -

R

( 0 - 10 ) V

REVERSE BIAS

1 K + -

( 0 - 100 ) uA

A

1N 4001

( 0 - 10 ) V

V

+ -

R

( 0 - 10 ) V

3

Reverse bias

When the negative terminal of the battery is connected to the P-type and positive terminal

of the battery is connected to the N-type of the PN junction, the bias applied is known as reverse bias. Under applied reverse bias, holes which form the majority carriers or the P-side moves

towards the negative terminal of the battery and electron which form the majority carrier of the N-side are attracted towards the positive terminal of the battery. Hence, the width of the depletion region which is depleted of the mobile charge carriers increases. Thus the electric field produced

by applied reverse bias, is in the same direction as the electric field of the potential barrier. Hence, the resultant potential barrier is increased which prevents the flow of majority carriers in both the

directions. Therefore, theoretically no current should flow in the external circuit. But in practice, a very small current of the order of a few microamperes flows under reverse bias. Electron forming covalent bonds of the semiconductor atoms in the P and N-type regions

may absorb sufficient energy from heat and light to cause breaking of some covalent bonds. Hence electron-hole pairs are continually produced in both the regions. Under the reverse bias

condition, the thermally generated holes in the P-region are attracted towards the negative terminal of the battery and the electrons in the N-region are attracted towards the positive terminal of the battery. Consequently, the minority carriers, electron on the P-region and holes in the N-

region, wander over to the junction and flow towards their majority carrier side giving rise to a small reverse current. This current is known as reverse saturation current, IO. The magnitude of

reverse current depends upon the junction temperature because the major source of minority carriers is thermally broken covalent bonds. For large applied reverse bias, the free electrons from the N-type moving towards the

positive terminal of the battery acquire sufficient energy to move with high velocity to dislodge valence electron from semiconductor atoms in the crystal. These newly liberated electrons, in turn,

acquire sufficient energy to dislodge other parent electrons. Thus, a large number of free electrons are formed which is commonly called as an avalanche of free electrons. This leads to the breakdown of the junction leading to very large reverse current. The reverse voltage at which the junction breakdown occurs is known as breakdown voltage, VBD.

PN diode applications

An ideal PN diode is a two terminal polarity sensitive device that has zero resistance when it is forward biased and infinite resistance when it is reverse biased. Due to this characteristic the

diode finds number of applications as given below. 1. Rectifier

2. Switch 3. Clamper 4. Clipper

5. Demodulation detector circuits PROCEDURE Forward bias

1. Connect the circuit as per the circuit diagram.

2. Vary the power supply voltage in such a way that the readings are taken in steps of 0.1 V in the voltmeter.

3. Note down the corresponding ammeter readings.

4. Plot the graph: V against I 5. Calculate the forward resistance RF = ∆VF / ∆IF

4

TABULATION

FORWARD BIAS

Sl. No.

SUPPLY VOLTAGE

VF (V)

IF (mA)

1

2

3

4

5

6

7

8

9

10

REVERSE BIAS

Sl.

No.

SUPPLY

VOLTAGE

VR

(V)

IR

(µA)

1

2

3

4

5

6

7

8

9

10

5

PROCEDURE

Reverse bias

1. Connect the circuit as per the circuit diagram. 2. Vary the power supply voltage in such a way that the readings are taken in steps of 0.5

V in the voltmeter.

3. Note down the corresponding ammeter readings. 4. Plot the graph: V against I

5. Calculate the forward resistance RR = ∆VR / ∆IR

VIVA-VOCE QUESTIONS

1. Explain how a potential barrier is created within a PN diode.

2. Define peak inverse voltage of a diode.

3. Mention the characteristics of an ideal diode.

4. What are the sources of reverse current in a diode?

5. What are the transition and diffusion capacitances?

6

MODEL GRAPH

I mA

V VOLTS

IuA

CALCULATION

Forward resistance RF = ∆VF / ∆IF

=

Reverse resistance RR = ∆VR / ∆IR =

7

VIVA-VOCE QUESTIONS

6. Write the diode equation.

7. Why silicon devices are more popular than germanium devices?

8. What do you mean by valance electron?

9. Give the barrier potential for silicon and germanium.

10. Write the other name of PN diode.

RESULT

Thus the characteristics of PN diode were plotted and the following results were obtained.

Forward resistance RF

Reverse resistance RR

Cut In voltage

9


CHARACTERISTICS OF ZENER DIODE

AIM

To plot the forward and reverse VI characteristics of a zener diode and to calculate the

following parameters 1. Forward resistance

2. Reverse resistance 3. Break down voltage

APPARATUS REQUIRED

EQUIPMENT RANGE


Ammeter (0-30)mA, (0-100)mA

Volt meter (0-1)V, (0-10)V

Resistors 1K

Diode SZ 5.6

Bread board & wires

-

THEORY

Zener diode is heavily doped PN diode. When the reverse voltage reaches breakdown

voltage in normal PN diode the current through the junction and the power dissipated at the junction will be high. Such an operation is destructive and the diode gets damaged. Diodes can be

designed with adequate power dissipation capabilities to operate in the breakdown region. One such diode is known as Zener diode.

From the VI characteristics of the Zener diode, it is found that the operation of Zener diode

is same as that of ordinary PN diode under forward biased condition. Under reverse biased condition, breakdown of the junction occurs. The breakdown voltage depends upon the amount of

doping. If the diode is heavily doped, depletion layer will be thin and, consequently, breakdown occurs at lower reverse voltage and the breakdown voltage is sharp. The sharp increasing current under breakdown conditions is due to the following two mechanisms.

1. Avalanche breakdown 2. Zener breakdown

Avalanche breakdown

As the applied reverse bias increases, the field across the junction increases

correspondingly, thermally generated carriers while traversing the junction acquire a large amount of kinetic energy from this field. As a result the velocity of these carriers increases. These

electrons disrupt covalent bond by colliding with immobile ions and create new electron-hole pairs. These new carriers again generating energy from the field and collide with other immobile ions thereby generating further electron-hole pairs. This process is cumulative in nature and results in

generation of an avalanche of charge carriers within a short time. This mechanism of carrier generation is known as avalanche multiplication. This process results in flow of large amount of

current at the same value of reverse bias.

10

SYMBOL

A K

Zener diode

PIN DIAGRAM

AK

SZ 5.6

CONSTRUCTION

P N

A K

Heavily doped

CIRCUIT DIAGRAM FORWARD BIAS

A

SZ 5.6

V(0 - 30)V

(0-30)mA

(0-1)V

-

-+

+

1K

R

REVERSE BIAS

(0-100)mA

A

SZ 5.6

V(0 - 30)V

R

1K

(0-10)V

+

+

-

-

11

Zener breakdown

When the PN regions are heavily doped, direct rupture of covalent bonds takes place

because of strong electric fields, at the junction of PN diode. The new electron-hole pair so created increases the reverse current in a reverse biased PN diode. The increase in current takes

place at a constant value of reverse bias typically below 6V for heavily doped diodes. As a result of heavy doping of P and N regions, the depletion regions, the depletion region width becomes very small and for applied voltage of 6V or less, the field across the depletion region becomes very

high, of the order of 107 V/m, making conditions suitable for Zener breakdown. For lightly doped diodes, Zener breakdown voltage becomes high and breakdown is predominantly by Avalanche

multiplication. Though Zener breakdown occurs for lower breakdown voltage and Avalanche breakdown occurs for higher breakdown voltage, such diodes are normally called Zener diodes. Application

Zener diode can be used as a voltage regulator.

PROCEDURE

Forward Bias

1. Connect the circuit as per the circuit diagram. 2. Vary the power supply voltage in such a way that the readings are taken in steps of

0.1 V in the voltmeter.

3. Note down the corresponding ammeter readings. 4. Plot the graph: V against I

5. Calculate the forward resistance RF = ∆VF / ∆IF Reverse Bias


2. Vary the power supply voltage in steps of 0.5 V. 3. Note down the corresponding ammeter readings. 4. Plot the graph: V against I

5. Calculate the forward resistance RR = ∆VR / ∆IR

12

TABULATION

FORWARD BIAS

Sl. No.

SUPPLY VOLTAGE

VF (V)

IF (mA)

1

2

3

4

5

6

7

8

9

10

REVERSE BIAS

Sl.

No.

SUPPLY

VOLTAGE

VR

(V)

IR

(mA)

1

2

3

4

5

6

7

8

9

10

13

VIVA-VOCE QUESTIONS

1. How Zener diode is formed?

2. What is Avalanche breakdown?

3. Mention few differences between PN diode and Zener diode.

4. What are the general applications of Zener diodes?

5. What is Zener breakdown?

14

MODEL GRAPH

I mA

V VOLTS

CALCULATION

Forward resistance RF = ∆VF / ∆IF

=

Reverse resistance RR = ∆VR / ∆IR

=

15

RESULT

Thus the characteristics of Zener diode were plotted and the following results were obtained.

Forward resistance RF

Reverse resistance RR

Breakdown voltage

17


CHARACTERISTICS OF CE TRANSISTOR

AIM

To plot the input and output characteristics of an NPN transistor in common emitter

configuration and to find its 1. Input resistance

2. Output resistance APPARATUS REQUIRED

EQUIPMENT RANGE


Ammeter (0-500)µA,(0-10)mA

Voltmeter (0-5)V, (0-10)V

Resistors 500Ω, 1K

Bread board &

wires

-

Transistor BC 107 THEORY

Transistor can be connected in a circuit in any of the three different configurations namely; common emitter, common base, and common collector. Common emitter (CE) configuration is the most frequently used configuration because it provides voltage, current, power gain more than

unity. The name CE is because the emitter of the transistor is common to the input and output

circuits. Input is applied across the base and emitter, and the output is taken across the collector and emitter. CE configuration is also called grounded emitter configuration.

Input Resistance

Input resistance can be calculated from the input characteristic curves. It is given by the

ratio of small change in base to emitter voltage to corresponding base current; keeping collector to emitter voltage constant. Ri = ∆VBE / ∆IB; keeping VCE constant. Output Resistance

Output resistance can be calculated from the output characteristic curves. It is given by the ratio of small change in collector to emitter voltage to corresponding collector current; keeping base current constant.

Ro = ∆VCE / ∆IC; keeping IB constant. Application

The CE configuration is used for audio frequency circuits.

18

SYMBOL

B

E

C

NPN transistor

PIN DIAGRAM

E

E

B

C

B

C

BC 107

CONSTRUCTION

NE CP N

B

CIRCUIT DIAGRAM

(0 - 30)V

(0 - 30)V

E

C

BBC 107

VVBE

(0 - 5)V +

-

A

(0 - 500)uA

IB

+ -

A

(0 - 10)mA

IC

+-

V

+

-

(0 - 10)V

VCE

1K

500 ohmR1

R2

19

PROCEDURE

Input Characteristics

1. Rig up the circuit as per circuit diagram. 2. Set VCE and vary VBE insteps of 0.1 V and note down the corresponding IB. Repeat the

above procedure for various values of VCE. 3. Plot the graph: VBE vs IB for constant values of VCE.

4. Find the input resistance Ri = ∆VBE / ∆IB; keeping VCE constant. Output Characteristics

1. Rig up the circuit as per circuit diagram.

2. Set IB and vary VCE insteps of 1 V and note down the corresponding IC. Repeat the above procedure for various values of IB.

3. Plot the graph: VCE vs IC for constant values of IB. 4. Find the output resistance Ro = ∆VCE / ∆IC; keeping IB constant.

VIVA-VOCE QUESTIONS

1. What is the importance of a CE amplifier?

2. Mention the regions of operation of a transistor.

3. In which region of operation a transistor acts as an amplifier.

4. What is the use of CC amplifier stage?

5. What is an emitter follower and why it is called so

20

TABULATION

INPUT CHARACTERISTICS

SL. NO.

VCE =

VCE =

VCE =

VBE

(V)

IB

(µA)

VBE

(V)

IB

(µA)

VBE

(V)

IB

(µA)

1

2

3

4

5

6

7

8

9

10

OUTPUT CHARACTERISTICS

SL. NO.

IB =

IB =

IB =

VCE (V)

IC

(mA) VCE (V)

IC

(mA) VCE (V)

IC

(mA)

1

2

3

4

5

6

7

8

9

10

21

VIVA-VOCE QUESTIONS

6. What is application of CB amplifier?

7. What the arrow in the symbol of transistor indicates.

8. Why the emitter of a transistor is highly doped.

9. Why the area of collector of transistors is largest.

10. Define transistor.

22

MODEL GRAPH


IB

uA

VBE Volts

MODEL GRAPH


IC

mA

VCE Volts

CALCULATION

Input resistance RI = ∆VBE / ∆IB

Output resistance Ro = ∆VCE / ∆IC

23

RESULT

Thus the characteristics of CE transistor were plotted and the following results were obtained.

Input resistance RI

Output resistance Ro

25

EXPERIMENT NO. : 4

DATE:

CHARACTERISTICS OF CB TRANSISTOR

AIM

To plot the input and output characteristics of an NPN transistor in common base configuration and to find its

1. Input resistance 2. Output resistance

APPARATUS REQUIRED

EQUIPMENT RANGE

Power supply (0-5)V, (0-30)V

Ammeter (0-100)mA

Voltmeter (0-5)V, (0-10)V

Resistors 1K, 500 ohm

Bread board & wires

-

Transistor BC 107

THEORY

In common base configuration, the base of the transistor is common to the input and output circuits. Input voltage is applied between emitter and base. Output is taken across collector and base. CB configuration is also called as grounded base configuration. In this set up, an increase

in emitter current causes an increase in collector current. Collector current is given by the

expression IC = IE – IB. since IB is in the order of micro amperes. Collector current is almost same as the emitter current in spite of the variations in the collector – base junction. Another important expression for a common base transistor configuration is IC = α IE + ICBO where ICBO is the current

flowing through the collector circuit when the collector – base junction is reverse biased and emitter - base junction is open circuited.

Transistor offers medium input resistance and very high output resistance when it is in CB configuration. It provides almost unit current gain and high voltage gain.

Input Resistance

Input characteristics are plotted between the emitter current IE and emitter to base voltage

VBE for a constant value of collector to base voltage VCB. The reciprocal of the slope of the curve s gives the value of dynamic input resistance Ri and is given by the expression Ri = ∆VBE/∆IE.

Output Resistance

Output characteristics are plotted between the collector current IC and collector to base

voltage VCB for a constant value of emitter current IE. The output resistance can be obtained from the curves and is given by the expression Ro = ∆VCB/∆IC keeping IE constant. Ro has rather high values since the curves are almost flat.

Application

The CB configuration is used for high frequency circuits.

26

SYMBOL

B

E

C

NPN transistor

PIN DIAGRAM

E

E

B

C

B

C

BC 107

CONSTRUCTION

NE CP N

B

CIRCUIT DIAGRAM

(0 - 30)V

E C

B

BC 107

IE

V

(0 - 5)V

VEB

A

+

+

(0 - 100)mA

-

-

(0 - 30)V

A

IC

(0 - 100)mA

V

(0 - 10)V

VCB

+

+

-

-

1K500 ohm

R1 R2

27

PROCEDURE

Input Characteristics

1. Rig up the circuit as per circuit diagram. 2. Set VCB and vary VEB insteps of 0.1 V and note down the corresponding IE. Repeat the

above procedure for various values of VCB. 3. Plot the graph: VEB vs IE for constant values of VCB.

4. Find the input resistance Ri = ∆VEB / ∆IE; keeping VCB constant. Output Characteristics


2. Set IE and vary VCB insteps of 1 V and note down the corresponding IC. Repeat the above procedure for various values of IE.

3. Plot the graph: VCB vs IC for constant values of IB.

4. Find the output resistance Ro = ∆VCB / ∆IC; keeping IE constant.

VIVA-VOCE QUESTIONS

1. What s the current amplification factor for CB stage?

2. Mention the characteristics of CB configuration.

3. Mention the application of CB amplifier?

4. Which configuration is good as a constant current source? Why?

5. What is collector power dissipation of transistor?

28

TABULATION


SL. NO.

VCB =

VCB =

VCB =

VEB

(V)

IE

(mA)

VEB

(V)

IE

(mA)

VEB

(V)

IE

(mA)

1

2

3

4

5

6

7

8

9

10


SL. NO.

IE =

IE =

IE =

VCB (V)

IC

(mA) VCB (V)

IC

(mA) VCB (V)

IC

(mA)

1

2

3

4

5

6

7

8

9

10

29

VIVA-VOCE QUESTIONS

6. What is a bipolar junction transistor?

7. List the difference between NPN and PNP transistor.

8. What are the different BJT configurations?

9. Explain about early effect.

10. List the two types of bread down in transistors.

11. Draw the pin diagram of BC 107 transistor.

12. What is meant by base-width modulation?

13. Explain about punch through.

14. Give the application of CC configuration.

30

MODEL GRAPH


IE

mA

VBE Volts


IC

mA

VCB Volts

CALCULATION

Input resistance RI = ∆VEB / ∆IE

Output resistance Ro = ∆VCB/∆IC

31

RESULT

Thus the characteristics of CE transistor were plotted and the following results were

obtained.

Input resistance RI

Output resistance Ro

33

EXPERIMENT NO. : 5

DATE:

CHARACTERISTICS OF JFET

AIM

To plot the transfer and drain characteristics of JFET and to find the following parameters 1. Drain resistance 2. Mutual conductance

3. Amplification factor APPARATUS REQUIRED

EQUIPMENT RANGE


Ammeter (0-30)µA

Voltmeter (0-30)V

Bread board & wires -

Transistor BFW 10

THEORY Junction field effect transistor (JFET) is a unipolar device since its function depends only

upon one type of carrier. JFET has high input impedance unlike BJT. JFETs are two types, N-channel and P-channel. An N-channel JFET is an N-type silicon bar with a P-type semiconductor is embedded on both sides of the bar. P-type semiconductor

forms the gate and the ends of the N-type bar are source and drain. The P-type regions are externally shorted. The gate of an N-channel JFET is connected to negative potential with respect

to source. The drain is connected to positive with respect to the source. Drain Resistance

It is defined as the ratio of change in drain to source voltage to the change in drain current at an operating point, when gate to source voltage remains constant.

Rd = ∆VDS / ∆ID with VGS constant Mutual Conductance

It is defined as the ratio of change in drain current to the change in gate to source voltage at an operating point, when drain to source voltage remains constant.

Gm = ∆ID / ∆VGS with VDS constant. Amplification Factor

It is defined as the product of drain resistance and mutual conductance. µ = Rd * Gm

34

SYMBOL

G

S

D

N - channel JFET

PIN DIAGRAM

S

D

G

Schield

BFW 10

CONSTRUCTION

N Type

P

P

DS

G

CIRCUIT DIAGRAM

G

D

S

BFW 10

A

V V

(0 - 30)V(0 - 30)V

(0 - 30)mA

(0 - 5)V

VGS VDS

(0 - 10)V

ID

-

+

+

+-

-

35

Application

1. FET is used as a buffer in measuring instruments, receivers since it has high input impedance and low output impedance.

2. FETs are used in RF amplifiers in FM tuners and communication equipment for the low noise level.

3. Since the input capacitance is low, FETs are used in cascade amplifiers in measuring and

test equipments. 4. Since the device is voltage controlled, it is used as a voltage variable resistor in operational

amplifiers and tone controls. 5. FET are used in mixer circuits in FM and TV receivers, and communication equipment

because inter modulation distortion is low.

6. As the coupling capacitor is small, FETs are used in low frequency amplifiers in hearing aids and inductive transducers.

7. FETs are used in digital circuits in computers, LSD and memory circuits because of its small size.

PROCEDURE

Drain Characteristics


2. Set VGS and vary VDS insteps of 0.5 V and note down the corresponding ID. Repeat the above procedure for various values of VGS.

3. Plot the graph: VDS vs ID for constant values of VGS. 4. Find the drain resistance Rd = ∆VDS / ∆ID with VGS constant

Transfer Characteristics

1. Rig up the circuit as per circuit diagram. 2. Set VDS and vary VGS insteps of 0.5 V and note down the corresponding ID. Repeat the

above procedure for various values of VDS.

3. Plot the graph: VGS vs ID for constant values of VDS. 4. Find the Trans conductance. Gm = ∆ID / ∆VGS with VDS constant

36

TABULATION

DRAIN CHARACTERISTICS

SL. NO.

VGS =

VGS =

VGS =

VDS

(V)

ID

(mA)

VDS

(V)

ID

(mA)

VDS

(V)

ID

(mA)

1

2

3

4

5

6

7

8

9

10

TRANSFER CHARACTERISTICS

VDS = _________

SL. No.

VGS (V)

ID (mA)

1

2

3

4

5

MODEL GRAPH


37

ID

mA

V DS Volts

VIVA-VOCE QUESTIONS

1. Why a FET is said to be a voltage controlled device?

2. What is the ohmic region in FET output characteristics?

3. What is pinch off voltage?

4. Give the advantage of FET over BJT.

5. Prove µ = Rd * Gm

6. What does the arrow in the gate of symbol of FET indicate?

7. Why FET is called so?

8. How a FET function as a VVR.

38

9. Mention the types of JFET.

10. Draw the small signal equivalent circuit of JFET

MODEL GRAPH


VGSVolts

mA

ID

CALCULATION

Drain resistance Rd = ∆VDS / ∆ID

=

Mutual conductance Gm = ∆ID / ∆VGS

=

39

Amplification factor µ = Rd * Gm

=

40

RESULT

Thus the characteristics of JFET were plotted and the following results were obtained.

Drain resistance Rd

Mutual conductance Gm

Amplification factor µ

41


CHARACTERISTICS OF MOSFET

AIM

To plot the transfer and drain characteristics of n-channel MOSFET and to find the following

parameters 1. Drain resistance 2. Mutual conductance

3. Amplification factor APPARATUS REQUIRED

EQUIPMENT RANGE


Ammeter (0-100)mA

Voltmeter (0-30)V

Bread board & wires -

Transistor BFW 96

THEORY

MOSFETs are three terminal devices having a source, gate and drain. MOSFET is the

abbreviation of metal oxide semiconductor field effect transistor. It uses a thin layer of silicon dioxide as an insulator between gate and channel. It is also known as insulated gate field effect transistor (IGFET). There are two kinds of MOSFET, depletion and enhancement type.

An N-channel depletion type MOSFET is shown in figure. A heavily doped two N-type wells are doped on P-type substrate to form source and drain. Between these two N-type wells a lightly

doped N-type material forms a channel. A thin layer of Sio2 which is an insulating material is fabricated on the surface above the channel and gate terminal is attached to it. Source and drain terminals are attached to heavily doped N-type material with metal contacts.

A positive voltage VDS is applied at the drain with respect to source to establish drain current. When a negative voltage VGS is applied at the gate with respect to the source, electrons

under gate get repelled causing the channel effectively thinner. This reduces the current through the channel. If the magnitude of VGS is increased, the drain current decreases. If it is increased further drain current stops.

If a positive voltage is applied at the gate, electrons get attracted in to the channel and drain current increases. A depletion type MOSFET can operate in depletion and enhancement modes.

Application

MOSFETs are widely used in VLSI circuits.

42

SYMBOL

G

D

S

N - channel MOSFET

PIN DIAGRAM

S

D

G

Schield

BFW 96

CONSTRUCTION

N+ N+N

S D G

P- type substrate

CIRCUIT DIAGRAM

A

D

VG

VS

BFW 96

(0 - 30)V

(0 - 30)V

(0 - 30)V

(0 - 100)mA

ID

(0 - 30)V

VDS

VGS

+

+

-

+

-

-

1 K

1 K

R

R

43

Drain Resistance

It is defined as the ratio of change in drain to source voltage to the change in drain current

at an operating point, when gate to source voltage remains constant. Rd = ∆VDS / ∆ID with VGS constant Mutual Conductance

It is defined as the ratio of change in drain current to the change in gate to source voltage at

an operating point, when drain to source voltage remains constant. Gm = ∆ID / ∆VGS with VDS constant. Amplification Factor

It is defined as the product of drain resistance and mutual conductance.

µ = Rd * Gm

PROCEDURE Drain Characteristics


2. Set VGS = 0V and vary VDS insteps of 0.5 V and note down the corresponding ID. Repeat the above procedure for VGS = -1V,-2V,-3V , for the depletion mode

3. Repeat the above procedure for VGS = 1V, 2V, 3V, for the enhancement mode.

4. Plot the graph: VDS vs ID for constant values of VGS. 5. Find the drain resistance Rd = ∆VDS / ∆ID with VGS constant

Transfer Characteristics

1. Rig up the circuit as per circuit diagram. 2. Set VDS and vary VGS insteps of 0.5 V and note down the corresponding ID. Repeat the

above procedure for various values of VDS. 3. Plot the graph: VGS vs ID for constant values of VDS. 4. Find the Trans conductance. Gm = ∆ID / ∆VGS with VDS constant

44

TABULATION

DRAIN CHARACTERISTICS – DEPLETION MODE

SL. NO.

VGS =

VGS =

VGS =

VDS

(V) ID

(mA) VDS

(V) ID

(mA) VDS

(V) ID

(mA)

1

2

3

4

5

6

7

8

DRAIN CHARACTERISTICS – ENHANCEMENT MODE

SL.

NO.

VGS =

VGS =

VGS =

VDS

(V)

ID

(mA)

VDS

(V)

ID

(mA)

VDS

(V)

ID

(mA)

1

2

3

4

5

6

7

8


VDS = _________

SL. No.

VGS (V)

ID (mA)

1

2

3

4

5

45

VIVA-VOCE QUESTIONS

1. What is a unipolar device?

2. What are the types of FET?

3. How a MOSFET is formed?

4. Mention the modes of MOSFET.

5. Explain enhancement mode.

46

MODEL GRAPH


ID

mA

V DS Volts


ID

mA

VGS Volts

CALCULATION

Drain resistance Rd = ∆VDS / ∆ID

=

Mutual conductance Gm = ∆ID / ∆VGS

=

Amplification factor µ = Rd * Gm =

47

RESULT

Thus the characteristics of MOSFET were plotted and the following results were obtained.

Drain resistance Rd

Mutual conductance Gm

Amplification factor µ

49


CHARACTERISTICS OF UJT

AIM

To plot the VI characteristics of a uni-junction transistor (UJT) and to measure its intrinsic

stand off ratio. APPARATUS REQUIRED

EQUIPMENT RANGE


Ammeter (0-30)mA

Voltmeter (0-10)V

Resistors 1K

Bread board &

wires

-

Transistor 2N 2646 THEORY

A uni-junction transistor consists of a bar of highly doped N-type semiconductor to which a

heavily doped P-type rod is attached. Ohmic contacts are made at opposite ends of the N-type bar, which are called base1 (B1) and base2 (B2) of the transistor. P-type rod is called the emitter.

The inter-base resistance RBB of N-type silicon bar appears as two resistors RB1 and RB2. The intrinsic stand off ratio is given by the expression η = RB1 / RBB with IE = 0

Due to the applied voltage at B2 of the transistor, a positive voltage gets developed across RB1 and it is equal to ηVBB.

If VE is less than the voltage across RB1, diode becomes reverse biased. When VE increases, forward current flows through the emitter to B1 region. If VE is raised further, a sudden reduction of RB1 occurs. This happens because the

increase in current reduces RB1. Reduction in RB1 causes the increase in current through it. This further reduces RB1 and so forth. In other words, a regenerative action takes place at a particular

value of VE, called peak voltage which is expressed as VP = ηVBB + VD Where VBB is the base supply voltage and VD is the junction voltage drop. After a particular value of VE called valley point, emitter current increases with VE similar to that of an ordinary forward

biased diode.

As explained above, when VE is rises, forward resistance across the junction decreases and the junction behaves as a short circuit. Then current through the E-B1 junction increases and hence voltage across the junction decreases. This is equivalent to a negative resistance across

the junction. This continues up to a voltage called valley voltage after which junction behaves as an ordinary diode.

Application

UJT can be employed in a variety of applications like, saw-tooth wave generator, pulse

generator, switching, timing, and phase control circuits.

50

SYMBOL

E

B2

B1UJT

PIN DIAGRAM

E

B1

B2

EB2

B12N 2646

CONSTRUCTION

P

N

E

B2

B1

CIRCUIT DIAGRAM

EA

V V

B2

B1(0 - 30) V (0 - 30) V

VB1B2

(0 - 10) V(0 - 10) V

VEB1

+

+ +

-

- -

R

1 K

IE

(0 - 30) mA

51

PROCEDURE

1. Rig up the circuit as per the circuit diagram.

2. Keeping VBB = 0, vary VBE from 0V to 5V in steps of 0.5V. 3. Take the voltmeter and ammeter reading s at the input side and enter it in tabular column. 4. Plot the VI characteristics with IE along X-axis and VBE1 along Y-axis.

Calculate the intrinsic stand off ratio from the graph.

VIVA-VOCE QUESTIONS

1. Draw the electrical equivalent circuit of UJT.

2. Define negative resistance.

3. Give two applications of UJT.

4. Comment on UJT when VBB = 0.

5. Why RB1 is greater than RB2?

52

TABULATION

VB1B2 = 5V

Sl. No.

VEB1

(v) IE

(mA)

1

2

3

4

5

6

7

8

9

10

MODEL GRAPH

Volts

VEB1

VP

VV

IP IV IE mA

CALCULATION

Intrinsic stand off ratio η = (VP – VD) / VBB

=

53

RESULT

Thus the characteristic of UJT was plotted and the following result was obtained.

Intrinsic stand off ratio η

55


CHARACTERISTICS OF SCR

AIM

To plot the VI characteristics of a silicon control rectifier (SCR) and to measure it’s holding

current and latching current. APPARATUS REQUIRED

EQUIPMENT RANGE



Voltmeter (0-30)V

Resistors 1.5K,5K

Bread board &

wires

-

SCR 2P4M THEORY

Silicon control rectifier is a four layer PNPN device. It has three terminals namely,

anode (A), cathode (K), gate (G). Keeping gate open, if forward voltage is applied across SCR, it will remain in OFF state. If applied voltage exceeds the break over voltage, it will turn ON and

heavy current will flow through it. The break over voltage can be reduced considerably, if a small voltage is applied at the gate. As the gate current increases, the break over voltage decreases. Once the SCR is fired, gate loses control over the current through the device. Even if gate

current is disconnected, anode current cannot be brought back to zero. To turn OFF the SCR, anode current should be made less than the holding current.

Application

1. Relay control 2. Motor control

3. Phase control 4. Heater control 5. Battery chargers

6. Inverters 7. Regulated power supplies

8. Static switches

56

SYMBOL

SCR

K

A

G

PIN DIAGRAM

K A G2P4M

CONSTRUCTION

P1

N1

P2

N2

G

A

K

57

PROCEDURE

1. Rig up the circuit as per the circuit diagram. 2. Switch ON the anode supply keeping at low voltage.

3. Set the gate current and watch the triggering of SCR by varying the anode voltage. 4. Repeated the above steps for various values of gate current. 5. Plot the VI characteristics with IA along Y-axis and VAK along X-axis.

6. Calculate the holding current and latching current.

VIVA-VOCE QUESTIONS

1. Define holding current and latch current.

2. Why an SCR called a thyristor?

3. Give the applications of SCR.

4. Why an SCR is called so?

5. Draw the transistor analogy of the SCR.

58

CIRCUIT DIAGRAM

A

KG

A-

V

+

-

A+-

(0 - 30) V

(0 - 30) V+

(0 - 5) mA

IGR

1.5 K

(0 - 30) mA

IA

VAK

(0 - 30) V

2P4M

TABULATION

SL. NO.

IG = 2.5 mA

IG = 3 mA

IG = 3.5 mA

VAK (V)

IA

(mA) VAK (V)

IA

(mA) VAK (V)

IA

(mA)

1

2

3

4

5

6

7

MODEL GRAPH

IA

mA

IH

VH

VAK Volts

59

RESULT

Thus the characteristic of SCR was plotted and the following result was obtained.

Holding current IH

Latching current IL

61


CHARACTERISTICS OF DIAC - ( DIODE A.C. SWITCH )

AIM

To plot the VI characteristics of a Diode A.C switch.

APPARATUS REQUIRED

EQUIPMENT RANGE


Ammeter (0-10)mA

Voltmeter (0-100)V

Resistors 1.5K

Bread board &

wires

-

Diac DB3 THEORY

Diac is a three layer, two terminal semiconductor device. MT1 and MT2 are the two main terminals which are interchangeable. It acts as a bidirectional Avalanche diode. It does not have

any control terminal. It has two junctions J1 and J2. Though the Diac resembles a bipolar transistor,

the central layer is free from any connection with the terminals. From the characteristic of a Diac shown in figure it acts as a switch in both directions. As

the doping level at the two ends of the device is the same, the Diac has identical characteristics for

both positive and negative half of an a.c. cycle. During the positive half cycle, MT1 is positive with respect to MT2 where as MT2 is positive with respect to MT1 in the negative half cycle. At voltages

less than the breakover voltage, a very small amount of current called the leakage current flows through the device and the device remains in OFF state. When the voltage level reaches the breakover voltage, the device starts conducting and it exhibits negative resistance characteristics,

i.e. the current flowing in the device starts increasing and the voltage across it starts decreasing.

Application The Diac is not a control device. It is used as triggering device in Triac.

PRICEDURE

1. Rig up the circuit as per the circuit diagram. 2. Vary the power supply in regular steps and note down the corresponding current and

voltage of the diac. 3. Reverse the diac terminal.

4. Repeat the procedure 2. 5. Plot the graph: voltage Vs current.

62

SYMBOL

MT1 MT2

PIN DIAGRAM

MT2MT1 DB3

CONSTRUCTION

P

P

N

N

N

MT2

MT1

CIRCUIT DIAGRAM

MT1

MT2

A

V

R

1.5 K

-

+

-

+

(0 - 30)V

(0 - 30)mA

(0 - 100)V

63

VIVA-VOCE QUESTIONS

1. What is meant by DIAC?

2. Draw the construction of DIAC.

3. Mention some applications of DIAC.

4. The DIAC is a thyristor? Why?

5. Draw the symbol of DIAC.

6. Compare DIAC with SCR.

64

TABULATION

SL. NO.

FORWARD BIAS

REVERSE BIAS

Voltage

(V) Current

(mA) Voltage

(V) Current

(mA)

1

2

3

4

5

6

7

8

9

10

MODEL GRAPH

ON

ON

OFF

OFF

Forward current

IBO

VBO Forward voltage

Reverse voltage

Reversecurrent

VBO

IBO

65

RESULT

Thus the characteristic of DIAC was plotted.

67


CHARACTERISTICS OF TRIAC - ( TRIODE A.C. SWITCH )

AIM

To plot the VI characteristics of a Triode A.C switch.

APPARATUS REQUIRED

EQUIPMENT RANGE



Voltmeter (0-30)V

Resistors 1.5K,5K

Bread board &

wires

-

Triac BT 136 THEORY

Triac is a three terminal semiconductor switching device which can control alternating

current in a load. Its three terminals are MT1, MT2 and the gate (G). The basic structure and circuit symbol of a Triac are shown in figure.

Triac is equivalent to two SCRs connected in parallel but in the reverse direction. So, a Triac will act a switch for both directions. The characteristics of a Triac are shown in figure. Like SCR, a Triac also starts conducting only when the breakover voltage is reached. Earlier to that the

leakage current which is very small in magnitude flows through the device and therefore remains in the OFF state. When the device, starts conducting, allows very heavy amount of current to flow

through it. The high inrush of current must be limited using external resistance, or it may otherwise damage the device.

During the positive half cycle MT1 is positive with respect to MT2, where as MT2 is positive with respect to MT1 during negative half cycle. A Triac is a bidirectional device and can be

triggered either by a positive or by a negative gate signal. By applying proper signal at the gate, the breakover voltage, i.e. firing angle of the device can be changed; thus phase control process can be achieved.

Application

Triac is used for illumination control, temperature control, liquid level control, motor speed control and as static switch to turn a.c. power ON and OFF. Nowadays, the diac-triac pairs are increasingly being replaced by a single component unit known as quadrac. Its main limitation in

comparison to SCR is its low power handling capacity.

68

SYMBOL

MT1

G

MT2

TRIAC

PIN DIAGRAM

MT1

MT2

G

BT 136

CONSTRUCTION

N N

N

N

P

P

MT1 G

MT2

69

PROCEDURE

1. Set up the circuit after verifying the condition of the components. Switch on the anode

power supply at a low voltage. 2. Switch on gate dc supply and adjust it to make gate current 1mA. 3. Increase the anode voltage. At certain anode voltage it drops to a very low value indicating

that the triac has been triggered. The voltage just prior to this is the break over voltage. Allow the anode current to increase further. Note down the values and draw the VI

characteristics in the first quadrant of the graph. Repeat this step for gate current of 1.5 mA. 4. Reverse the polarity of the triac, dc supplies and meters. Repeat the above steps and draw

the VI characteristics in the third quadrant.

VIVA-VOCE QUESTIONS

1. Define TRIAC.

2. Draw the symbol of TRIAC.

3. List the applications of TRIAC.

4. What is meant by quadrac?

5. Draw the pin diagram of TRIAC.

6. Compare SCR with TRIAC.

7. List the advantage of TRIAC over SCR.

70

CIRCUIT DIAGRAM

MT1

G MT2

A

A

V

(0 - 30)V

(0 - 30)V

R1

1.5 K

+ -

IG

(0 - 5)mA

+

-

(0 - 30)V

R2

5 K

- +

(0 - 10)mA

BT 136

TABULATION

MT1 Positive

SL.

NO.

IG = 1 mA

IG = 1.5 mA

VMT1:MT2

(V)

IMT1

(mA)

VMT1:MT2

(V)

IMT1

(mA)

1

2

3

4

5

6

7

8

9

10

71

VIVA-VOCE QUESTIONS

8. Why TRIAC is called as thyristor?

9. What is the function of gate terminal in SCR and TRIAC?

10. Compare the DIAC with TRIAC.

72

TABULATION

MT1 Negative

SL. NO.

IG = -1 mA

IG = -1.5 mA

VMT1:MT2

(V)

IMT1

(mA)

VMT1:MT2

(V)

IMT1

(mA)

1

2

3

4

5

6

7

8

9

10

MODEL GRAPH

73

ON

ON

OFF

OFF-VH

VH

IH

IH

MT1 +

IA

- IA

VAK- VAK

MT1 -

74

RESULT

Thus the characteristic of TRIAC was plotted.

75


VERIFICATION OF KIRCHHOFF’S LAW

AIM

To verify the Kirchoff’s current and voltage laws for a given circuit. APPARATUS REQUIRED

EQUIPMENT RANGE


Ammeter (0-30)mA

Voltmeter (0-10)V

Resistors 680Ω,820Ω,1K

Bread board & wires

-

THEORY Kirchhoff’s Current Law (KCL)

It states that, at any node the sum of the currents entering a node is equal to the sum of the currents leaving the node. Kirchhoff’s Voltage Law (KVL)

It states that, any closed path in any network, the algebraic sum of the emfs is equal to the

algebraic sum of the IR drops. PROCEDURE

1. Connections are given as per the circuit diagram.

2. Input supply is given to the circuit. 3. For various values of input supply the corresponding meter readings are noted.

4. Tabulate the readings. 5. Verify the Kirchhoff’s current law and voltage law.

VIVA-VOCE QUESTIONS

1. What is positive ion?

2. What is negative ion?

3. Define charge.

76

CIRCUIT DIAGRAM

KIRCHHOFF’S CURRENT LAW

A+ -

I1 R1

680

A

A

I2 R2 + -

820

(0-30)V

(0-30)mA (0-30)mA

(0-30)mA

+

-

1K

R3

I3

TABULATION

SL.

NO.

SUPPLY

VOLTAGE

AMMETER READINGS BY KCL I1=I2+I3

(mA) I1

(mA) I2

(mA) I3

(mA)

1

2

3

4

5

6

7

8

9

10

77

VIVA-VOCE QUESTIONS

4. What is meant by static charge?

5. Define current.

6. Define potential difference.

7. Define power.

8. Give the relationship between power, voltage and current.

9. What are called passive elements?

10. What are called active elements?

11. Define resistance.

12. Define conductance.

13. What will be the equivalent resistance when it connected in series?

14. What will be the equivalent resistance when it connected in parallel?

15. State ohm’s law.

16. Define resistivity.

78

CIRCUIT DIAGRAM

KIRCHHOFF’S VOLTAGE LAW

(0-30)V

(0-10)V

V1

(0-10)V

V2

(0-10)V

V3

R1

680R2

820R3

1k

+ - + - -+

TABULATION

SL. NO.

SUPPLY

VOLTAGE

VOLTMETER READINGS

BY KVL VS=V1+V2+V3

(V) V1 (V) V2 (V) V3 (V)

1

2

3

4

5

6

7

8

9

10

79

VIVA-VOCE QUESTIONS

17. What is the internal resistance of an ideal voltage source?

18. What is the internal resistance of an ideal current source?

19. State Kirchhoff’s current law.

20. State Kirchhoff’s voltage law.

THEORETICAL VERIFICATION - KCL

THEORETICAL VERIFICATION - KVL

RESULT

Thus the Kirchoff’s current and voltage laws were verified for the given circuit.

81


VERIFICATION OF THEVENIN’S THEOREM

AIM

To verify Thevenin’s theorem for a given circuit. APPARATUS REQUIRED

EQUIPMENT RANGE


Ammeter (0-30)mA

Voltmeter (0-30)V

Resistors 1K,10K,DRB

Bread board & wires

-

THEORY Statement

Any linear bilateral network containing one or more voltage sources can be replaced by a single voltage source whose value is equal to open circuit voltage at output terminal with a series

Thevenin’s resistance. The Thevenin’s resistance is equal to the effective resistance looking back from the output terminal by removing the load resistance. PROCEDURE

To Find (IL)

1. Connect the ammeter in series with load. 2. Give the input supply.

3. Note the ammeter reading. To Find (VTH)

1. Remove the load resistance. 2. Connect the voltmeter across open circuited terminals.

3. Give the input supply. 4. Note the voltmeter reading.

To Find (RTH)

1. Short circuit the source voltage.

2. Remove the load resistance. 3. Connect the Ohm meter across open circuited terminals

4. Note the Ohm meter reading.

82

CIRCUIT DIAGRAM

ACTUAL CIRCUIT

25V

1K

10K

1K

10KRL

1K

CIRCUIT TO FIND ACTUAL LOAD CURRENT (IL)

25V

1K 1K

10K

(0 - 30)mA

+ -

10KRL

1K

IL

A

CIRCUIT TO FIND THEVENIN’S VOLTAGE (VTH)

25V

1K

10K

1K

10K VTH

+

-

V

CIRCUIT TO FIND THEVENIN’S RESISTANCE (RTH)

1K

10K

1K

10KOhm

MeterRTH

+

-

83

VIVA-VOCE QUESTIONS

1. What is meant by short circuit?

2. What is meant by open circuit?

3. What is the voltage across the short circuit?

4. What is the current across the open circuit?

5. Define mesh.

6. Define loop.

7. Name the basic circuit law upon which the mesh analysis is based?

8. Name the basic circuit law upon which the nodal analysis is based?

9. State Thevenin’s theorem.

10. List the application of Thevenin’s theorem.

84

THEVENIN’S EQUAVELENT CIRCUIT

VTH

RTH

RL

1K

CIRCUIT TO FIND LOAD CURRENT OF THEVENIN’S CIRCUIT (IL

’)

VTH

(0 - 30)mA

IL'

RTH

RL

1K

A

85

THEORETICAL VERIFICATION

RESULT

Thus the Thevenin’s circuit is verified for a given circuit and the following results were

obtained.

Thevenin’s Resistance RTH

Thevenin’s Voltage VTH

Load Current of Actual Ckt IL

Load Current of Thevenin’s Ckt IL’

87


VERIFICATION OF NORTON’S THEOREM

AIM

To verify the Norton’s theorem for a given circuit. APPARATUS REQUIRED

EQUIPMENT RANGE


Ammeter (0-30)mA

Voltmeter (0-30)V


Bread board & wires

-

THEORY Statement

Any linear bilateral network containing one or more generators can be replaced by an equivalent circuit consisting of current source (INOR) in parallel with admittance (YNOR). The INOR is

the short circuit current flowing through the output terminals and YNOR is the admittance measured across the output terminals with all the sources replaced by its internal impedance.

PROCEDURE

To Find (VL)

1. Connect the voltmeter across the load. 2. Give the input supply.

3. Note the voltmeter reading. To Find (IN)

1. Remove the load resistance. 2. Connect the ammeter across short circuited output terminals.

3. Give the input supply. 4. Note the ammeter reading.

To Find (RN)

1. Short circuit the source voltage.

2. Remove the load resistance. 3. Connect the Ohm meter across open circuited terminals

4. Note the Ohm meter reading.

88

CIRCUIT DIAGRAM

ACTUAL CIRCUIT

25V

1K

10K

1K

10K RL

1K

CIRCUIT TO FIND ACTUAL LOAD VOLTAGE (VL)

25V

1K

10K

1K

10KRL

1KVL

(0-30)V

+

-

V

CIRCUIT TO FIND NORTON’S CURRENT (IN)

25V

1K

10K

1K

10KIN

(0-30)mA

+

-

A

CIRCUIT TO FIND NORTON’S RESISTANCE (RN)

RN

Ohm

Meter

1K

10K

1K

10K

+

-

89

VIVA-VOCE QUESTIONS

1. Define inductance.

2. Give the formula for energy stored in the inductor.

3. Write the formula for equivalent inductance when the inductors are connected in series.

4. Write the formula for equivalent inductance when the inductors are connected in parallel.

5. Define capacitance.

6. Give the formula for energy stored in the capacitor.

7. Give the charge – voltage relationship in a capacitor.

8. What is current voltage relation for capacitor?

9. State Norton’s theorem.

10. Give the application of Norton’s theorem.

90

NORTON’S EQUAVELENT CIRCUIT

IN RN RL

1K

CIRCUIT TO FIND LOAD VOLTAGE OF NORTON’S CIRCUIT (VL

’)

VN

RN

RL

1K

VL'

(0-30)V

+

-

V

Where

VN = IN * RN

91


RESULT

Thus the Norton’s circuit is verified for a given circuit and the following results were

obtained.

Norton’s Resistance RN

Norton’s Voltage VN

Load Voltage of Actual Ckt VL

Load Voltage of Norton’s Ckt VL’

93


VERIFICATION OF SUPER POSITION THEOREM

AIM

To verify the super position theorem for a given circuit. APPARATUS REQUIRED

EQUIPMENT RANGE


Ammeter (0-10)mA

Resistors 1K

Bread board & wires

-

THEORY Statement

In linear bilateral network containing more than one generator, current flowing through any branch is the algebraic sum of current flowing through that branch when generators are

considered one at a time replacing other generators by their internal impedance.

PROCEDURE


2. Switch on the DC power supplies (V1, V2) and note down the corresponding ammeter reading (IL).

3. Replace the power supply (V2) by short circuit.

4. Switch on the DC power supply (V1) and note down the ammeter reading (I1). 5. Connect back the power supply (V2).

6. Replace the power supply (V1) by short circuit. 7. Switch on the DC power supply (V2) and note down the ammeter reading (I2). 8. Verify the condition: IL = I1 + I2

VIVA-VOCE QUESTIONS

1. Write the formula for equivalent capacitance when the capacitors are connected in series.

2. Write the formula for equivalent capacitance when the capacitors are connected in parallel.

3. When an inductor makes its presence known in a circuit?

94

CIRCUIT DIAGRAM

ACTUAL CIRCUIT

20 V

V1

(0-10)mAA

10V

+

-

IL

1K

1K

1K

V2

IL =________

CIRCUIT TO FIND CURRENT DUE TO VOLTAGE SOURCE V1 (I1)

20 V

V1

I1 A (0-10)mA

+

-

1K

1K

1K

I1 =________

CIRCUIT TO FIND CURRENT DUE TO VOLTAGE SOURCE V2 (I2)

1K

I2

+

-

10V

A (0-10)mA

1K

1K

V2

I2 =________

95

VIVA-VOCE QUESTIONS

4. By what factor is the inductance of a coil increased when the number of turns is doubled?

5. Distinguish between an inductor and inductance.

6. When a capacitor makes its presence known in circuit? Explain.

7. What are lumped elements?

8. Define network.

9. State superposition theorem.

10. Where we can use superposition theorem.


RESULT

Thus the Norton’s circuit is verified for a given circuit and the following results were

obtained.

Actual Current IL

Current Due To V1 (I1)

Current Due To V2 (I2)

IL = I1 + I2

97


VERIFICATION OF MAXIMUM POWER TRANSFER THEOREM

AIM

To verify the Maximum power transfer theorem for a given circuit.

APPARATUS REQUIRED

EQUIPMENT RANGE


Ammeter (0-30)mA


Ohm meter -

Bread board & wires

-

THEORY Statement

In any linear bilateral network maximum power will be transferred from the voltage source to the load when the load impedance is equal to source impedance.

PROCEDURE

1. Connections are made as per circuit diagram. 2. Switch on the power supply.

3. Vary the resistance using DRB. 4. Note down the ammeter readings (IL). 5. Tabulate the readings and calculate the power.

VIVA-VOCE QUESTIONS

1. What are discrete elements?

2. Describe about voltage divider rule.

3. Describe about current divider rule.

98

CIRCUIT DIAGRAM

ACTUAL CIRCUIT

A

25V

1K 1K (0-30)mA

IL

10K10K

RL

+ -

CIRCUIT TO FIND SOURCE RESISTANCE (RS)

RS

1K

10K

1K

10KOhm

Meter

+

-

TABULATION

SL.

NO.

RL

(Ω)

IL

(mA)

P = IL2 RL

(mW)

1

2

3

4

5

6

7

8

9

10

99

VIVA-VOCE QUESTIONS

4. Distinguish between node and independent node.

5. Distinguish between a mesh and a loop of a circuit.

6. Which has the greater capacitance, two equal capacitors in series or in parallel?

7. What is meant by source transformation? Explain.

8. What is superposition? Explain.

9. State maximum power transfer theorem.

10. What is the significance of maximum power transfer theorem?


RESULT

Thus the maximum power transfer thermo was verified and for the given circuit maximum power was transferred if RS __________ = RL = _____________.

101


VERIFICATION OF RECIPROCITY THEOREM

AIM To verify the reciprocity theorem for a given circuit.

APPARATUS REQUIRED

EQUIPMENT RANGE


Ammeter (0-30)mA

Resistors 100 ,470 ,940

Bread board & wires

-

THEORY

Statement

In any linear bilateral network the ratio of voltage to current response, in any element to the

input is constant even the position of the input and output are interchanged. PROCEDURE

1. Connections are given as per the circuit diagram.

2. Note down the ammeter reading and find the ratio of output current to the input voltage. 3. Interchange the position of ammeter and voltage source.

4. Note down the ammeter reading and find the ratio of output current and input voltage. 5. Compare this value with the value obtained in step.2.

VIVA-VOCE QUESTIONS

1. What is called linear network?

2. What is called bilateral network?

3. State reciprocity theorem.

4. Mention the applications of reciprocity theorem.

102

CIRCUIT DIAGRAM

CIRCUIT - I

A

(0 -30)V

940 100

100470

(0 - 30)mA

+

-

IT IL

CIRCUIT – II

940 100

100470

IT IL

-

+

A(0 -30)V

(0 - 30)mA

TABULATION

CIRCUIT – I

CIRCUIT - II

VOLTAGE

V

CURRENT

I

Z = V/I VOLTAGE

V

CURRENT

I

Z = V/I

103


RESULT

Thus the reciprocity theorem was verified for a given circuit.

105


SERIES RESONANCE CIRCUIT

AIM

To design and set up a series resonant circuit and to measure its resonant frequency, bandwidth and quality factor.

APPARATUS REQUIRED

EQUIPMENT RANGE

AFO (0-2)MHz

Capacitor 0.01 uF

Resistors 5.6K

DIB 56 mH

Bread board &

wires

-

THEORY

A circuit is said to be in resonance, when the applied voltage V and the resulting current I is

inphase. It means, at a particular frequency called resonance frequency the capacitive reactance XC equals the inductive reactance XL and the circuit behaves as a purely resistive network. i.e.

2 foL = 1/2 foC. So, fo = 1/2 LC.

Since the impedance is minimum at resonance frequency, the current through the network

will be maximum. An ideal series resonance circuit acts as an ON switch for the supply at resonant frequency and acts as an OFF switch for all other frequencies. That is, the circuit accepts signals of a frequency only. Hence a series resonant circuit is also called as an acceptor circuit.

The graph of current versus frequency is plotted. At the resonant frequency fo, the current is maximum (IO). Two points at which the current is 0.707 of its maximum are indicated. These points are called half power points. The frequencies corresponding to half power points are f1 and f2. The difference between these frequencies is called the bandwidth of the circuit. i.e. bandwidth =

f2 – f1.

Quality factor of a series resonant circuit is given by the expression QO = OL/R = 1/ OCR.

Quality factor can also be expressed as the ratio of resonant frequency to the bandwidth. Hence

QO = fO/B.W. PROCEDURE

1. Set up the circuit on bread board after testing the components using a multimeter.

2. Set the signal generator output at 2V peak to peak sine wave and feed it to the input of the circuit and observe the input and output of the circuit across the resistor simultaneously on

the CRO screen. 3. Vary the input from Hz range to higher KHz range and note the frequency of the signal and

corresponding voltages across the resistor. Enter it in the tabular column and draw the

graph with frequency along X-axis and current along Y-axis. 4. Mark fo and half power points on the graph. Measure the bandwidth from the graph and

calculate Q of the circuit using the formula Q = fo/BW.

106

DESIGN

1. Resonant frequency is given by the expression: fo = 1/2 LC

2. Let fo be 5 KHz

3. To avoid the over loading of the signal generator take R = 10 times of the output impedance of the function generator. Take R = 5.6 K .

4. Take C = 0.01 F, then L = 56 mH.

CIRCUIT DIAGRAM

R L

56 mH

C

5.6 K0.01 uF

VIN

2 V

VO

TABULATION

FREQ (Hz)

Vo (mV)

I = VO/R Amps

FREQ (Hz)

Vo (mV)

I = VO/R Amps

FREQ (Hz)

Vo (mV)

I = VO/R Amps

10

20

30

40

50

60

70

80

90

100

200

300

400

500

600

700

800

900

1k

2k

3k

4k

5k

6k

7k

8k

9k

10k

20k

30k

40k

50k

60k

70k

80k

90k

100k

MODEL GRAPH

I

Ampere

IO

0.707 IO

f1 f2f0 f

Hz

107

VIVA-VOCE QUESTIONS

1. Define resonance.

2. Why the series RLC resonant circuit is called acceptor circuit?

3. What change can be observed on the graph of RLC series circuit, if the resistance in the

circuit is increased?

4. Why the cut off frequency points are referred to as half power points?

5. State the application of series resonant circuit.

6. How does the coil resistance affect the performance of the series resonant circuit?

RESULT

Thus a series resonant circuit was designed, implemented and the following results were obtained.

Resonant frequency fc

Bandwidth BW

Quality factor Q

109


PARALLEL RESONANCE CIRCUIT

AIM

To design and set up a parallel resonant circuit and to measure its resonant frequency, bandwidth and quality factor.

APPARATUS REQUIRED

EQUIPMENT RANGE

AFO (0-2)MHz

Capacitor 0.01 uF

Resistors 5.6K,100

DIB 56 mH

Bread board &

wires

-

THEORY

A circuit is said to be in resonance, when the applied voltage V and the resulting current I is inphase. It means, at a particular frequency called resonance frequency the capacitive reactance XC equals the inductive reactance XL and the circuit behaves as a purely resistive network. i.e.

2 foL = 1/2 foC. So, fo = 1/2 LC.

In the parallel resonance circuit, the current through the inductor IL and that through the

capacitor IC are in opposite phase and it will cancel each other at the resonance frequency fo. So, the impedance offered by an idea parallel resonant circuit is infinite at fo. It means that an ideal

parallel resonant circuit behaves as an OFF switch at resonance frequency. For all other frequencies, the impedance becomes small and the circuit acts as an ON switch. That is, the parallel LC circuit rejects supply at the frequency fo. Hence the circuit is also called as rejecter.

The graph of current versus frequency is plotted. At the resonant frequency fo, the current is maximum (IO). Two points at which the current is 0.707 of its maximum are indicated. These points are called half power points. The frequencies corresponding to half power points are f1 and f2. The difference between these frequencies is called the bandwidth of the circuit. i.e. bandwidth =

f2 – f1.

Quality factor of a parallel resonant circuit is given by the expression QO = R/ OL = OCR.

Quality factor can also be expressed as the ratio of resonant frequency to the bandwidth. Hence

QO = fO/B.W.

PROCEDURE

1. Set up the circuit on bread board after testing the components using a multimeter.

2. Set the signal generator output at 2V peak to peak sine wave and feed it to the input of the circuit and observe the input and output of the circuit across the resistor simultaneously on

the CRO screen. 3. Vary the input from Hz range to higher KHz range and note the frequency of the signal and

corresponding voltages across the resistor. Enter it in the tabular column and draw the

graph with frequency along X-axis and current along Y-axis. 4. Mark fo and half power points on the graph. Measure the bandwidth from the graph and

calculate Q of the circuit using the formula Q = fo/BW.

110

DESIGN

1. Resonant frequency is given by the expression: fo = 1/2 LC

2. Let fo be 5 KHz

3. To avoid the over loading of the signal generator take R = 10 times of the output impedance of the function generator. Take R = 5.6 K .

4. Take C = 0.01 F, then L = 56 mH.

5. Take RL = 100 to limit the current through the inductor.

CIRCUIT DIAGRAM

VIN

2 V

R

RL

56 mHL

5.6 K

100

C

0.01 uF

TABULATION

FREQ

(Hz)

Vo

(mV)

I = VO/R

Amps

FREQ

(Hz)

Vo

(mV)

I = VO/R

Amps

FREQ

(Hz)

Vo

(mV)

I = VO/R

Amps

10

20

30

40

50

60

70

80

90

100

200

300

400

500

600

700

800

900

1k

2k

3k

4k

5k

6k

7k

8k

9k

10k

20k

30k

40k

50k

60k

70k

80k

90k

100k

MODEL GRAPH

I

Ampere

f

Hz

IO

0.707 IO

f0f1 f2

111

VIVA-VOCE QUESTIONS

1. What is called parallel resonance circuit?

2. Why the parallel RLC resonant circuit is called rejecter circuit?

3. Define cut off frequency.

4. Define band width.

5. What are the applications of parallel LC circuits?

6. Define Q factor.

7. Compare series and parallel resonance circuits.

RESULT

Thus a parallel resonant circuit was designed, implemented and the following results were obtained.

Resonant frequency fc

Bandwidth BW

Quality factor Q

113


AUGMENTED EXPERIMENT

CHARACTERISTICS OF LED

AIM

To plot the VI characteristics of a Light Emitting Diode (LED). APPARATUS REQUIRED

EQUIPMENT RANGE


Ammeter (0-100)mA

Voltmeter (0-10)V

Resistors 100Ω

Bread board & wires

-

LED -

THEORY

A PN junction diode, emits light on forward biasing, is known as light emitting diode. The emitted light may be either in the visible range, and the intensity of light depends on the applied

potential. In a PN junction charge carrier recombination takes place when the electrons cross from

the N-layer to the P-layer. The electrons are in conduction band on the P-side while the holes are in the valance ban on the P-side the conduction band has a higher energy level compared to the valance band and so when the electrons recombines with a hole the difference in energy is given

out in the form of heat or light. In case of silicon or germanium, the energy dissipation in the form of heat, where in the case of gallium- arsenide and gallium-phosphide, it is in the form of heat. But

this light is in the invisible region and so these materials cannot used in the manufacture of LED. Hence gallium-arsenide phosphide which emits light in the visible region is used to manufacture an LED.

PROCEDURE

1. Connect the circuit as per the circuit diagram. 2. Vary the power supply voltage in such a way that the readings are taken in steps of 0.1 V in

the voltmeter. 3. Note down the corresponding ammeter readings.

4. Plot the graph: V against I

114

CIRCUIT DIAGRAM

(0-15)V

100

R

(0-100)mA

-

LED

V

(0-10)V

A

+ -

+

TABULATION

Sl.

No.

SUPPLY

VOLTAGE

VF

(V)

IF

(mA)

1

2

3

4

5

6

7

8

9

10

MODEL GRAPH

VF Volts

IF

mA

115

VIVA-VOCE QUESTIONS

1. What is a LED?

2. How light is emitted from LED?

3. Name some semiconductor material used to form LED.

4. What is the advantage of LED?

5. Name some applications of the LED.

RESULT

Thus the VI characteristic of a light emitting diode was studied and the graph was plotted.

116

RESISTOR COLOUR CODING

COLOR 1st

DIGIT 2nd

DIGIT MULTIPLIER TOLERANCE

BLACK 0 0 1 Silver +/-10%; Gold +/-5%

BROWN 1 1 10 Silver +/-10%; Gold +/-5%

RED 2 2 100 Silver +/-10%; Gold +/-5%

ORANGE 3 3 1,000 Silver +/-10%; Gold +/-5%

YELLOW 4 4 10,000 Silver +/-10%; Gold +/-5%

GREEN 5 5 100,000 Silver +/-10%; Gold +/-5%

BLUE 6 6 1,000,000 Silver +/-10%; Gold +/-5%

VIOLET 7 7 10,000,000 Silver +/-10%; Gold +/-5%

GRAY 8 8 100,000,000 Silver +/-10%; Gold +/-5%

WHITE 9 9 1,000,000,000 Silver +/-10%; Gold +/-5%

RESISTANCE = (1ST DIGIT X 10 + 2ND DIGIT) X MULTIPLIER

1 experiment no. : 1 date: characteristics of pn diode...

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