Ece-III-Analog Electronic Ckts [10es32]-Notes

University Syllabus

PART A

UNIT 1:

Diode Circuits: Diode Resistance, Diode equivalent circuits, Transition and diffusion capacitance, Reverse

recovery time, Load line analysis, Rectifiers, Clippers and clampers. (Chapter 1.6 to 1.14, 2.1 to 2.9)

6 Hours

UNIT 2:

Transistor Biasing: Operating point, Fixed bias circuits, Emitter stabilized biased circuits, Voltage divider

biased, DC bias with voltage feedback, Miscellaneous bias configurations, Design operations, Transistor

switching networks, PNP transistors, Bias stabilization. (Chapter 4.1 to 4.12)

7 Hours

UNIT 3:

Transistor at Low Frequencies: BJT transistor modeling, Hybrid equivalent model, CE Fixed bias

configuration, Voltage divider bias, Emitter follower, CB configuration, Collector feedback configuration,

Hybrid equivalent model. (Chapter 5.1 to 5.3, 5.5 to 5.17)

7 Hours

UNIT 4: Transistor Frequency Response: General frequency considerations, low frequency response, Miller effect

capacitance, High frequency response, multistage frequency effects. (Chapter 9.1 to 9.5, 9.6, 9.8,

9.9)

6 Hours

PART B

UNIT 5:

(a) General Amplifiers: Cascade connections, Cascode connections, Darlington connections. (Chapter 5.19

to 5.27)

3 Hours (b) Feedback Amplifier: Feedback concept, Feedback connections type, Practical feedback circuits.

(Chapter 14.1 to 14.4) 3

Hours

UNIT 6:

Sub Code 10ES32 IA Marks 25

Hrs/ Week 04 Exam Hours 03

Total Hrs. 52 Exam Marks 100

Analog Electronic Circuits 10ES32

SJBIT/ECE Page 2

Power Amplifiers: Definitions and amplifier types, series fed class A amplifier, Transformer coupled Class

A amplifiers, Class B amplifier operations, Class B amplifier circuits, Amplifier distortions. (Chapter 12.1 to

12.9)

7 Hours

UNIT 7: Oscillators: Oscillator operation, Phase shift Oscillator, Wienbridge Oscillator, Tuned Oscillator circuits,

Crystal Oscillator. (Chapter 14.5 to 14.11) (BJT version only)

6 Hours

UNIT 8: FET Amplifiers: FET small signal model, Biasing of FET, Common drain common gate configurations,

MOSFETs, FET amplifier networks. (Chapter 8.1 to 8.13)

7 Hours

TEXT BOOK:

1. Electronic Devices and Circuit Theory, Robert L. Boylestad and Louis Nashelsky, PHI/Pearson Eduication. 9TH Edition.

REFERENCE BOOKS:

1. Integrated Electronics, Jacob Millman & Christos C. Halkias, Tata - McGraw Hill, 1991 Edition 2. Electronic Devices and Circuits, David A. Bell, PHI, 4th Edition, 2004

3 Analog electronics circuits: A simplified approach U B mahadevaswamy, pearson education 9 th edition.

Question Paper Pattern: Student should answer FIVE full questions out of 8 questions to be set each

carrying 20 marks, selecting at least TWO questions from each part.


SJBIT/ECE Page 3

INDEX SHEET

SL.NO TOPIC PAGE NO.

1 University syllabus 04

PART A

UNIT 1: Diode Circuits

1.1 Diode Resistance 07

1.2 Diode equivalent circuits 08

1.3 Transition and diffusion capacitance 09

1.4 Reverse recovery time 11

1.5 Load line analysis 14

1.6 Clippers and clampers 20

UNIT - 2: Transistor Biasing

2.1 Bipolar Transistor 57

2.2 Bipolar Stability 59

2.3 Fixed with Emitter 63

2.4 Voltage divider biased 64

UNIT - 3: Transistor at Low Frequencies

3.1 AC Analysis BJT transistor modeling 87

3.2 Hybrid equivalent model 90

3.3 CE Fixed bias configuration 95

3.4 Voltage divider bias 96

3.5 Emitter follower 98

3.6 CB configuration 100

UNIT - 4: Transistor Frequency Response

4.1 General frequency Response 112

4.2 Low frequency response 113

4.3 Miller effect capacitance 118

4.4 High frequency response 120

PART B UNIT - 5: a) General Amplifiers

5.1 Amplifier Basics 132

5.2 Classification of Amplifier 133

5.3 Multistage Amplifier 134

5.4 RC couples Amplifier 140

UNIT - 5: b) Feedback Amplifiers

5.5 Feedback concept 140

5.6 Feedback connections type 142

UNIT - 6: Power Amplifiers

6.1 Definitions and amplifier types 180

6.2 Cass A amplifier 180

6.3 Transformer coupled Class A amplifiers 186

6.4 Class B amplifier operations 193

UNIT - 7: Oscillators


SJBIT/ECE Page 4

7.1 Oscillator operation 209

7.2 Phase shift Oscillator 209

7.3 Wienbridge Oscillator 210

7.4 Tuned Oscillator circuits 211

7.5 Crystal Oscillator 212

UNIT - 8: FET Amplifiers

8.1 FET small signal model 222

8.2 Biasing FET 226


SJBIT/ECE Page 5

Unit: 1 Hrs: 6

Diode Circuits: Diode Resistance, Diode equivalent circuits, Transition and diffusion capacitance, Reverse

recovery time, Load line analysis, Rectifiers, Clippers and clampers.

Recommended readings:

TEXT BOOK: 1. Electronic Devices and Circuit Theory, Robert L. Boylestad and Louis Nashelsky, PHI/Pearson

Eduication. 9TH Edition.

REFERENCE BOOKS: 1. Integrated Electronics, Jacob Millman & Christos C. Halkias, Tata - McGraw Hill, 1991 Edition

2. Electronic Devices and Circuits, David A. Bell, PHI, 4th Edition, 2004


SJBIT/ECE Page 6

1.1 DIODE RESITANCE

As the operating point of a diode moves from one region to another, the resistance of the diode will also

change due to the nonlinear shape of the diode characteristic curve.

The type of applied voltage or signal will define the resistance level of interest.

Three different levels will be introduced.

DC or Static Resistance

The application of a dc voltage to a circuit containing a semiconductor diode will result in an

operating point on the characteristic curve that will not change with time.

The resistance of the diode at the operating point is simply the quotient of the corresponding levels of

VD and ID.

The dc resistance levels at the knee and below will be greater than the resistance levels obtained for the

vertical rise section of the characteristics.

The resistance levels in the reverse bias region will be high.

In general, the lower the current through a diode the higher the dc resistance level

AC Resistance

It is used to find the diode resistance when the small signal ac input voltage is applied across the

diode.

VD

ID

DC resistance Rd= VD / ID


SJBIT/ECE Page 7

For small signal ac voltage, ID & VD changes around Q point which is fixed by large signal DC

voltage

The ac resistances is determined by

Drawing a tangent line at Q point

Then find the change in voltage and the current.

The ratio of this change in the voltage and the current is called ac resistance.

Average Resistance

It is used to find the diode resistance when the large signal ac input voltage is applied across the

diode.

For large signal, there is no Q point and limits of operation is large due large swing in current and

voltage.

Average resistance is ratio of change voltage to the change in current between two extreme points.

The average resistances is determined by

Drawing a straight line between two extreme voltages on characteristic curve

Then finding the difference in voltages and respective currents between the two

points.

1.2 Equivalent Circuits of Diode

Order of simplification:

Diode Characteristic Curve Non linear graph

Ac resistance rd= Vd / Id

Q Id

Vd


SJBIT/ECE Page 8

Piecewise Linear Equivalent Circuit approximate in to two lines, one horizontal and other with

slope 1/r

Simplified Equivalent Circuit approximate in to two lines, one horizontal and other one vertical

Ideal Diode with zero voltage across diode during forward bias and zero current through diode during

reverse bias

1.3 Diode Capacitance

Ideal Equivalent circuit

Ideal diode

rave = 0, Vk =0

Simplified Equivalent Circuit

Ideal diode

rave = 0


SJBIT/ECE Page 9

Two types of Diode Capacitance :

Transition Capacitance

Diffusion capacitance.

Effect of capacitance:

It is stray capacitance and has very low value

Diode becomes frequency sensitive, mainly at very high frequency.

At high frequency, Xc becomes low enough to introduce a low reactance shorting path.

Transition capacitance (CT):

Predominant effect in reverse bias condition.

Also called as Depletion region Capacitance or space charge capacitance

Basic capacitance eqn = A/d where

= permittivity of dielectric between the two plates

A =Area and d = distance between the plates.

Depletion region behaves like dielectric between two charged plates.

Depletion width d increases with increase in reverse bias.

So CT decreases as reverse bias increases.

Application Ex Schottky diode, varactor (Varicap) diodes.

CT is present in forward bias also, but is effect is neglected by the presence of larger CD

Diffusion capacitance (CD):

Predominant effect in forward bias condition.

Also called as Storage Capacitance.

Depends on rate at which charge is injected in to the PN region. (outside the depletion).

So as current increases, CD increases.

However increase in current reduces resistance. This helps in high frequency operation as T =

RC


SJBIT/ECE Page 10

1.4 Reverse Recovery Time:-

Denoted by trr.

In forward bias, large number of free electrons in P region and holes in N region during conduction.

This results in minority carriers in each region

Sudden changing to reverse bias results into large reverse current due to large minority carriers. ( I

reverse = I forward)

Stays for initial storage time ts

After movement of minority carriers top other region Ir decreases to Is within time tt.

trr = ts +tt

Important in high speed switching applications

Normal value few nanosec to 1us . Very low trr of picosecs are also available

-25 -20 -15 -10 -5 0 0.25 0.5

5

10

15

CT

Cpf

CT + CD ~ CD

CT

1uF

CD

1

D1

1N5402


SJBIT/ECE Page 11

DIODE SPECIFICATIONS

Data provided by manufactures.

Must be included data :-

VF at specified temp and IF

IF max at specified temp.

IR at specified voltage and temp.

PIV or BR or PRV at specified temp

PD max = VDID

Capacitance levels

Reverse recovery time .. trr

Operating temp Range

Additional Data depends on application :-

Frequency range

Noise Level.

Switching time.

Thermal resistance levels

Peak repetitive values.

ID

t

IF

t1

ts

tt

Diode is reverse biased

Ir

Is

Desired response.

trr = ts + tt


SJBIT/ECE Page 12

Diode Notation:- A & K Depends on application, manufactures, current/voltage rating

Testing can be done by using

DMM (Digital Multimeter)- with diode checking function

ohm meter

Curve Tracer

With diode checking :- Internal meter voltage is used

In one direction it shows 0.7 V as diode is forward biased

In other direction it is around 2.5V (depends on Vbattery)

With Ohm meter

One direction low resistance (RF)

Other direction it shows high resistance (RR)

With curve tracer

Characteristic curve of the diode is displayed.

Vertical axis is 1mA/div(Can be adjusted)

Horizontal axis is 100mV/div (can be adjusted)

Expensive and looks more complex.


SJBIT/ECE Page 13

Load line Analysis:-

Load line - defined by the network

Characteristic curve defined by device

V1 = VD+IDR

ID = V1/R at VD =0V

VD = V1/R at ID =0A

1. In any given circuit, check biasing of diode.

2. During forward bias (i.e diode is ON) replace diode by short for ideal diodes or with 0.7V

3. During reverse bias (i.e diode is OFF) replace diode with open circuit.

4. Do the ckt analysis and find the output voltage.

In a circuit, diode can be in Series, Parallel or Series and Parallel

Answers :-

Load line Analysis:-

R11k

+ V110V

D1

1N5402

10mA

V1

Q

Load line

Characteristic


SJBIT/ECE Page 14

5.Determine the current I for each of the configurations of fig 2.150 using the approximate equivalent model

for the diode.

a) I = 0 mA; diode reverse-biased.

(b)V in loop of 20 = 20 V 0.7 V = 19.3 V (Kirchhoffs voltage law)

I = 19.3/20 = 0.965 A

(c)I = 10v/10 = 1 A; center branch open is open as one diode is forward biased and the other one is reverse

biased.


SJBIT/ECE Page 15

6) Determine Vo and Id for the networks of fig.2.151

a)Diode forward-biased,

Kirchhoffs voltage law (CW):

5 V + 0.7 V Vo = 0

So Vo = 4.3 V

IR = ID = 4.3V/2.2K = 1.955 mA

(b)Diode forward-biased,

ID = (8-0.7)/ (1.2k+4.7k) = 1.24 mA

Vo = ID* 4.7 k + VD

= (1.24 mA)(4.7 k ) + 0.7 V = 6.53 V

7) Determine the level of Vo for each network of fig.2.152

a)Vo = (Vdc-VD1-VD2) = (20 V 1 V)

= (19 V) = 9.5 V

b) I = (10-(-2)-0.7)/ (1.2+4.7)k

= (11.3/5.9) = 1.915 mA


SJBIT/ECE Page 16

V = IR = (1.915 mA)(4.7 k ) = 9 V

Vo = V 2 V = 9 V 2 V = 7 V

8) Determine Vo and Id for the networks of fig.2.153.

a) Determine the Thevenin equivalent circuit for the 10mA source and 2.2 k resistor.

ETh = IR = (10 mA)(2.2 k ) = 22 V and RTh = 2.2k

So ID =22/(2.2+1.2) = 6.26 mA

Vo = ID(1.2 k ) = (6.26 mA)(1.2 k ) = 7.51 V

(b)Diode forward-biased,

ID = 20-(-5)-0.7 /6.8k= 3.57 mA

Kirchhoffs voltage law (CW):

Vo 0.7 V + 5 V = 0

Vo = 4.3 V

9) Determine Vo1 and Vo2 for the networks of fig.2.154.

(a)Vo1 = 12 V 0.7 V = 11.3 V

Vo2 = 0.3 V


SJBIT/ECE Page 17

(b) Vo1 = 10 V + 0.3 V + 0.7 V = 9 V

I = 9V/(1.2+3.3)k = 2 mA,

Vo2 = (2 mA)(3.3 k ) = 6.6 V

10) Determine Vo and Id for the netwoks of Fig.2.155.

(a) Both diodes forward-biased

(b) IR = (20-0.7)/4.7K = 4.106 mA

Assuming identical diodes:

ID = 4.106/2 = 2.05 mA,Vo = 20 V 0.7 V = 19.3 V

(b)Right diode forward-biased:

ID =15-(-5)-0.7 /2.2K =20/2.2 8.77 mA

Vo = 15 V 0.7 V = 14.3 V


SJBIT/ECE Page 18

11) Determine Vo and I for the networks of Fig 2.156.

(a)Both diodes forward-biased

IR = (20-0.7)/4.7K = 4.106 mA

Assuming identical diodes:

ID = 4.106/2 = 2.05 mA

Vo = 20 V 0.7 V = 19.3 V

(b)Right diode forward-biased:

ID =15-(-5)-0.7 /2.2K

=20-.7/2.2 = 8.77 mA

Vo = 15 V 0.7 V = 14.3 V


SJBIT/ECE Page 19

12) Determine Vo1 and Vo2 and I for the network of Fig.2.157.

Both diodes forward-biased:

Vsi = 0.7 V, Vge = 0.3 V

Vo1 = 20-0.7=19.3V Vo2 = 0.3V

Current through 1k resistor (I1)

= (20-0.7)I1 k 19.3/1k = 19.3 mA

Current through 0.7k resistor (I2)

13) Determine Vo and Id for the network of fig.2.158.

Both diodes are forward biased and parallel (in series with 2k ).

Thevinins eqt circuit for this is 2k//2k with 0.7 V in series = 1kohm in series with 0.7V

So current through load resistor

= (10-0.7)/ (1+2)k = 3.1mA

ID = 3.1/ 2 = 1.55 m

Vo = 3.1mA* 2 K = 6.2V


SJBIT/ECE Page 20

1.5 Clippers

A clipper is a circuit that is used to eliminate a portion of an input signal. There are two basic types of clippers: series

clippers and shunt/parallel clippers. As shown in Figure 4-1, the series clipper contains a diode that is in series with

the load. The shunt clipper contains a diode that is in parallel with the load.

FIGURE 4-1 Basic clippers.

The series clipper is a familiar circuit. The half-wave rectifier is nothing more than a series clipper. When the diode in

the series clipper is conducting, the load waveform follows the input waveform. When the diode is not conducting, the

output is approximately 0 V or fixed dc voltage which is connected in parallel. (Figure 4.2). The direction of the

diode determines the polarity of the output waveform. If the diode symbol (in the schematic diagram) points toward the

source, the circuit is a positive series clipper, meaning that it clips the positive alternation of the input. If the diode

symbol points toward the load, the circuit is a negative series clipper, meaning that it clips the negative alternation of

the input (Figure 4.11). With this di

Ideally, a series clipper has an output of when the diode is conducting (ignoring the voltage across the diode).

When the diode is not conducting, the input voltage is dropped across the diode, and .

Unlike a series clipper, a shunt clipper provides an output when the diode is not conducting. For example, refer to

Figure 4-1. When the diode is off (not conducting), the component acts as an open. When this is the case, and

form a voltage divider, and the output from the circuit is found using

When the diode in the circuit is on (conducting), it shorts out the load. In this case, the circuit ideally has an output of

. Again, this relationship ignores the voltage across the diode. In practice, the output from the circuit is

generally assumed to equal 0.7 V, depending upon whether the circuit is a positive shunt clipper or a negative shunt

clipper. The direction of the diode determines whether the circuit is a positive or negative shunt clipper. The series

current-limiting resistor ( ) is included to prevent the conducting diode from shorting out the source.

A biased clipper is a shunt clipper that uses a dc voltage source to bias the diode. A biased clipper is shown in Figure

4-2. (Several more are shown in Figures 4.9 and 4.10). The biasing voltage ( ) determines the voltage at which the

diode begins conducting. The diode in the biased clipper turns on when the load voltage reaches a value of .

In practice, the dc biasing voltage is usually set using a potentiometer and a dc supply voltage, as shown in Figure 4.10.


SJBIT/ECE Page 21

FIGURE 4-2 A biased clipper.

Clippers are used in a variety of systems, most commonly to perform one of two functions:

1. Altering the shape of a waveform

2. Protecting circuits from transients

The first application is apparent in the operation of half-wave rectifiers. As you know, these circuits are series clippers

that change an alternating voltage into a pulsating dc waveform. A transient is an abrupt current or voltage spike of

extremely short duration. Left unprotected, many circuits can be damaged by transients. Clippers can be used to protect

sensitive circuits from the effects of transients, as illustrated in Figure 4.12.

Staff:- KRS

Session (Aug 08 Dec08)

TE Department

PESIT, Bangalore

1

Syllabus:- Analog Electronic Circuit

Unit I /c:- Diode Applications - Clippers

OFF

ON,

Diode condition

Vo = VR =0V

Vo= Vin - Vd

Output voltage

IVinI < I0.7IVFor all values of Vin

Reverse Biased

IVinI > I0.7IVForward biased

Negative Cycle

Positive Cycle


SJBIT/ECE Page 22

Staff:- KRS


TE Department

PESIT, Bangalore

4



Vo

Vin0V+5V

-4.3V

Staff:- KRS


TE Department

PESIT, Bangalore

5



OFF

ON,

Diode condition

Vo = VR =0V

Vo= Vin Vdc+ Vd

= 0 for +ive cycle

=-(Vin + 1.3V) in ive

cycle

Output voltage

Vin >2V -0.7Reverse Biased

For all values of Vin

Vin


SJBIT/ECE Page 23

Staff:- KRS


TE Department

PESIT, Bangalore

6



OFF

ON,

Diode condition

Vo = VR =0V

Vo= Vin- (Vdc+0.7)

Output voltage

I Vin I<

Vdc+0.7

For all values

of Vin

Reverse

Biased

I Vin I > Vdc+0.7V

Forward biased

Negative Cycle

Positive Cycle

Staff:- KRS


TE Department

PESIT, Bangalore

7



Vo

Vin-5V

- 2.3V


SJBIT/ECE Page 24

Staff:- KRS


TE Department

PESIT, Bangalore

8



OFF

ON,

Diode condition

Vo = Vdc =2V

Vo= Vin- 0.7V

Output voltage

Vin> Vdc-0.7Reverse

Biased


Vin< Vdc-0.7V

Forward biased

Negative Cycle

Positive Cycle

Staff:- KRS


TE Department

PESIT, Bangalore

9



Vo

Vin-5V

- 4.3V

2V


SJBIT/ECE Page 25

Staff:- KRS


TE Department

PESIT, Bangalore

11



Vo

Vin

0V

+5V

4.93V

Staff: - KRS Session (Aug 08 Dec08)

TE Department PESIT, Bangalore

10

Syllabus: - Analog Electronic Circuit

Unit I /c: - Diode Applications - Clippers

OFF

ON,

Diode condition

Vo = VR =0V

Vo= Vin - Vd

Output voltage


Vin < 0.7V Reverse Biased

Vin > 0.7V Forward biased

Negative Cycle

Positive Cycle


SJBIT/ECE Page 26

Staff:- KRS


TE Department

PESIT, Bangalore

12



OFF

ON,

Diode condition

Vo = Vin.

Vo= Vd =0.7V

Output voltage

For all

values of Vin

Vin 0.7VForward biased

Negative Cycle

Positive Cycle

Staff:- KRS


TE Department

PESIT, Bangalore

13



Vo

Vin

0.7V

-5V

-5V


SJBIT/ECE Page 27

Staff:- KRS


TE Department

PESIT, Bangalore

15



Vo

Vin

- 0.7V

+5V

-5V

Staff:- KRS


TE Department

PESIT, Bangalore

16



OFF

ON,

Diode condition

Vo = Vin

Vo= Vdc + Vd== 2+0.7 =2.7V

Output voltage


Vin < Vdc +0.7VReverse Biased

Vin > Vdc +0.7VForward biased

Negative Cycle

Positive Cycle


SJBIT/ECE Page 28

Staff:- KRS


TE Department

PESIT, Bangalore

17



Vo

Vin

2.7V

+5V

4.93V

Staff:- KRS


TE Department

PESIT, Bangalore

18



OFF

ON,

Diode

Vo = Vin.

Vo= -Vdc+Vd ==

-2 + 0.7 = -1.3V

Output voltage

IVinI > I(Vdc-0.7V)IReverse Biased

IVinI < I(Vdc-0. 7V)IFor all

values of Vin

Forw ard

biased

Negative CyclePositive

Cycle


SJBIT/ECE Page 29

Staff:- KRS


TE Department

PESIT, Bangalore

19



Vo

Vin

-1.3V

-5V

5V

Staff:- KRS


TE Department

PESIT, Bangalore

20



OFF

ON,

Diode

Vo = Vin.

Vo= Vdc- Vd == 2 -0.7

=1.3V

Output voltage


Vin > Vdc -0.7VReverse Biased

Vin < Vdc -0.7VForward biased

Negative CyclePositive Cycle


SJBIT/ECE Page 30

Staff:- KRS


TE Department

PESIT, Bangalore

21



Vo

Vin

+2.7V

-5V

-5V

Staff:- KRS


TE Department

PESIT, Bangalore

22



OFF

ON,

Diode

Vo = Vin.

Vo= -(Vdc+Vd)

== -2.7

Output voltage

IVinI< I(Vdc+0.7V) IFor all values of Vin

Reverse Biased

IVinI> I(Vdc+0.7V) IForward

biased



SJBIT/ECE Page 31

Staff:- KRS


TE Department

PESIT, Bangalore

23



Vo

Vin

-2.7V+5V

5V

Staff:- KRS


TE Department

PESIT, Bangalore

24



OFF

ON,

Diode

Vo = Vin.

Vo= -(Vdc+Vd)

== -2.7

Output voltage

IVinI< I(Vdc+0.7V) IFor all values of Vin

Reverse Biased

IVinI> I(Vdc+0.7V) IForward

biased



SJBIT/ECE Page 32

Staff:- KRS


TE Department

PESIT, Bangalore

25



Vo

Vin

-2.7V

2.7V

Clampers (DC Restorers)

A clamper is a circuit that is designed to shift a waveform above or below a dc reference voltage without altering the

shape of the waveform. This results in a change in the dc average of the waveform. Both of these statements are

illustrated in Figure 4-3. (The clamper has changed the dc average of the input waveform from 0 V to +5 V without

altering its shape.)

FIGURE 4-3 A clamper with its input and (ideal) output waveforms.

There are two basic types of clampers:

A positive clamper shifts its input waveform in a positive direction, so that it lies above a dc reference voltage.

For example, the positive clamper in Figure 4-3 shifts the input waveform so that it lies above 0 V (the dc

reference voltage).

A negative clamper shifts its input waveform in a negative direction, so that it lies below a dc reference

voltage.


SJBIT/ECE Page 33

Both types of clampers, along with their input and output waveforms, are shown in Figure. The direction of the diode

determines whether the circuit is a positive or negative clamper.

Clamper operation is based on the concept of switching time constants. The capacitor charges through the diode and

discharges through the load. As a result, the circuit has two time constants:

For the charge cycle, and (where is the resistance of the diode)

For the discharge cycle, and (where is the resistance of the load)

Since is normally much greater than , the capacitor charges much more quickly than it discharges. As a result,

the input waveform is shifted as illustrated in Figure 4.16.

A biased clamper allows a waveform to be shifted above (or below) a dc reference other than 0 V. Several examples

of biased clampers are shown in Figure 4-4.

FIGURE Several biased clampers.

The circuit in Figure (a) uses a dc supply voltage (V) and a potentiometer to set the potential at the cathode of . By

varying the setting of , the dc reference voltage for the circuit can be varied between approximately 0 V and the

value of the dc supply voltage.

The zener clamper in Figure (b) uses a zener diode to set the dc reference voltage for the circuit. The dc reference

voltage for this circuit is approximately equal to . Note that zener clampers are limited to two varieties:

Negative clampers with positive dc reference voltages

Positive clampers with negative dc reference voltages


SJBIT/ECE Page 34

Staff:- KRS


TE Department

PESIT, Bangalore

26


Unit I /c:- Diode Applications - Clampers

First positive cycle:-

Diode is reverse biased and Vo= 0V.

First negative cycle:-

Diode is forward biased and capacitor is charging with very low time constant. At

negative peak, Vc=Vm.-Vdc After peak diode becomes reverse biased as

Vc>Vin.

Vo = Vin+Vc

Subsequent positive and negative cycles :- Time constant of Capacitor

discharge is very high.(=C*100k). In each negative cycle, Vc charges to max. value.

In both cycles Vo= Vin + Vc


SJBIT/ECE Page 35

Staff:- KRS


TE Department

PESIT, Bangalore

27



Positive clamper with waveform in negative side. Swing level decreases with increase in voltage. Swing level is max at

Vdc =0V. Swing level can be varied from 0V to Vm


SJBIT/ECE Page 36

Staff:- KRS


TE Department

PESIT, Bangalore

28




Diode is reverse biased and Vo= 0V.

First negative cycle:-

Diode is forward biased and capacitor is charging with very low time constant. At

negative peak, Vc=Vm.+Vdc After peak diode becomes reverse biased as

Vc>Vin.

Vo = Vin+Vc

Subsequent positive and negative cycles :- Time constant of Capacitor

discharge is very high.(=C*100k). In each negative cycle, Vc charges to max. value.

In both cycles Vo= Vin + Vc (Vin is +ivefor positve cycle and ive for ve cycle)


SJBIT/ECE Page 37

Staff:- KRS


TE Department

PESIT, Bangalore

29



Positive clamper with waveform in positive side. Swing level increases with increase in voltage. Swing level is min at Vdc

=0V. Swing level can be varied from Vm to Vm+Vdc.


SJBIT/ECE Page 38

Staff:- KRS


TE Department

PESIT, Bangalore

30




Diode is forward biased and capacitor is charging with very low time constant. At positive peak, Vc=Vm.-Vdc After peak,

diode becomes reverse biased as Vc>Vin.

Vo = Vin-Vc

Subsequent negative and positive cycles :- Time constant of Capacitor discharge is very high.(=C*100k). In each

positive cycle, Vc charges to max. value. In both cycles Vo= Vin Vc. (Vin is +ive

for postive cycle and ive for ve cycle)


SJBIT/ECE Page 39

Staff:- KRS


TE Department

PESIT, Bangalore

31



Negative clamper with waveform in positive side. Swing level decreases with increase in voltage. Swing level is max

Vdc =0V. Swing level can be varied from 0 to Vm


SJBIT/ECE Page 40

Staff:- KRS


TE Department

PESIT, Bangalore

32




Diode is forward biased and capacitor is charging with very low time constant. At positive peak, Vc=Vm.-Vdc After peak,

diode becomes reverse biased as Vc>Vin.

Vo = Vin-Vc

Subsequent negative and positive cycles :- Time constant of Capacitor discharge is very high.(=C*100k). In each

positive cycle, Vc charges to max. value. In both cycles Vo= Vin Vc. (Vin is +ive

for positive cycle and ive for ve cycle)


SJBIT/ECE Page 41

Staff:- KRS


TE Department

PESIT, Bangalore

33



Negative clamper with waveform in negative side only. Swing level increases with increase in voltage. Swing level

is min Vdc =0V. Swing level can be varied from -Vm to -Vm +VDC).

Troubleshooting Diode Circuits

Because diodes are so common in the electronics industry, it is important to be able to troubleshoot and repair

systems that employ diodes.

Diode defects include:

Anode-to-cathode short.

Anode-to-cathode open.

Low front-to-back ratio.

Out-of-tolerance parameters.

Tests that can performed on diodes to check for their operation are:

Voltage measurements.

Ohmmeter tests.

Diode testers.

Instruments that used to measure the healthiness of diode are

Digital multimeter in diode mode

Ohm-meter ( multimeter in resistance mode)

Curve tracer


SJBIT/ECE Page 42

Question paper with Solutions

Q. 1 What is the origin of diffusion capacitance. Obtain an expression for the diffusion capacitance in terms of current in a p-n diode.

(Jan 2004(6), July 2004 (6), Jan 2007 (7), July 2007 (5) ) Jan 2009 (7)

Sol.: In forward biased condition, the width of the depletion region decreases and holes from p side get diffused in 'n' side while

electrons from 'n' side move into the p-side the applied voltage increases, concentration of injected charged particles increases.

This rate of change of the injected charge with applied voltage is defined as capacitance called diffusion capaacitance.


SJBIT/ECE Page 43

Q 2) Draw a double diode clipper which limits at two independent levels and explain its working. (Jan 2004(6), July 2004 (8), July 2005 (6), Jan 2007(6))


SJBIT/ECE Page 44

Q. 3 Define the terms P.I.V and regulation as applied to rectifiers.

(July 2004 (4), Jan 2008 (4) , Jan 2009

Sol.: i) Peak Inverse Voltage (PIV) :

When the diode is not conducting, the reverse voltage gets applied across the diode. The peak value of such

voltage decides the peak universe voltage i.e. PIV rating of a diode.

Regulation of the output voltage:

As the load current changes, load voltage changes. Practically load voltage should remain constant So

concept of regulation is to study the effect of change in load current on the load voltage.

Q 4) Draw the piece-wise linear volt-ampere characteristics of a p-n diode. Give the circuit model for the

ON state and OFF state. Jan./Feb. 2005, July 2007 (10).

Another way to analyse the diode circuits is to approximate the V-I characteristics of a diode using only

straight lines i.e. linear relationships. In such approximation, the diode forward resistance is neglected and

the diode is assumed to conduct instantaneously when applied forward biased voltage Vo is equal to cut-in

voltage Vy' And then it is assumed that current increases instantaneously giving straight line nahlre of V-I

characteristics. While in reverse biased condition


SJBIT/ECE Page 45

when Vo < 0, the diode does not conduct at all.

Hence when diode forward resistance is assumed zero, the circuit model of diode is as shown in the

Fig. 1 (a). In reverse biased, the diode is open circuit as shown in the Fig. 1 (b). As the diode conducts at Vo

=Vy' the V-I characteristics with straight lines is as shown in the Fig. 1 (c). As the method models the diode

with the pieces of straight lines, the name given to such approximation is piecewise-linear method. The

characteristics of diode shown in the Fig. 1 (c) are called the piecewise linear diode characteristics.

Open circuit

For the clipping circuit shown incharacteristic. Assume ideal diode.

150 volts.

the following figure, obtain its transfer

The input varies linearly from 0 to(7)

Q 5) Sketch and explain the circuit of a double ended clipper using ideal p-n diodes which limit the output

between 10 V. (6) (July 2005(6) July 2007(10),

July 2008 (10))

Vin = Vim sin w t

During positive half cycle, the diode D} becomes forward biased and conducts, only when Vin is greater

than battery voltage Vl' So as long as Vin is less that V1 both the diodes are reverse biased and output

follows input. When D1 conducts, D2 is OFF and hence the output is constant at V1 volts. This is shown in

the Fig. 2


SJBIT/ECE Page 46

In case of negative half cycle, as long as Viis greater than V2' the diodes D1 and D2 both remain reverse

biased and the output follows input. Once input goes below V 2 then the diode D2 conducts and output

remains constant equal to V2' This is shown in the Fig. 3 (a) and (b).


SJBIT/ECE Page 47

Q 6) Draw the bridge rectifier with capacitor filter and explain.

(July 2005(10), june 2008)


SJBIT/ECE Page 48


SJBIT/ECE Page 49

Q 7) Explain the working of a full wove voltage doubler circuit. Jan./Feb. - 2004.Jan-

2006,July2008

Q 8) Design a power supply usinfS a FWR with capacitance filter to given an output voltage of 10V at 10mA from a 220

Hz, 50 Hz supply. The ripple factor must be less than 0.01. (Jan 2004(10))


SJBIT/ECE Page 50

Q 9) For the clipping circuit shown in characteristic. Assume ideal diode.150 volts.the following figure, obtain its transferThe input

varies linearly from 0 to 7 Jan 2005 (10) July 2007 (10) Jan 2009 (10)


SJBIT/ECE Page 51


SJBIT/ECE Page 52

Q 10) Design a full wave' rectifier with a capacitor filter to meet the following specifications.

DC output voltage = 15 volts, Load resistance = 1 kD. RMS ripple voltage on capacitor = < 1% of DC output

voltage. Assume the AC supply voltage as 230 Volts, 50 Hz. (8)


SJBIT/ECE Page 53

Jan 2005(10)


SJBIT/ECE Page 54


SJBIT/ECE Page 55

Recommended Questions

1. What do you understand by diffusion Capacitance? (Jan /Feb 2004, 6 marks) 2. Draw a doubl diode clipper, which limits at two independent levels and explain its operation

(Jan /Feb 2004, 6 marks)

3. what is the origin of diffusion capacitance? (July/ Aub 2004 6 marks) 4. Draw a double diode clipper which limits two independent levels and explain its workin?

(July/ Aub 2004 8 marks) 5. Draw a simple clamping circuit and explain its working?

(July/ Aub 2004 6 marks) 6. Define the terms P.I.V and regulation as applied to rectifiers

(July/ Aub 2004 4marks) 7. Explain the validity of the piecewise linear approximation of the diode model

(July/ Aub 2004 4 marks) 8. Draw the piece-wise linear volt-ampere characteristics of a p-n diode. Give the circuit model for the

ON state and OFF state.

9. Sketch and explain the circuit of a double ended clipper using ideal p-n diodes which limit the output between +/- 10V (July / Aug 2005 6 marks)

10. Draw the circuit diagram ofa bridge rectifier. Plot its input and output waveforms. (July / Aug 2005- 10 Marks)

11. Explain diffusion capacitance? (Jan/Feb 2007, 6 marks) 12. Draw and explain a double diode clipper circuit, which limits the output at two independent levels?


SJBIT/ECE Page 56

Unit: 2 Hrs: 7

Transistor Biasing: Operating point, Fixed bias circuits, Emitter stabilized biased circuits, Voltage divider

biased, DC bias with voltage feedback, Miscellaneous bias configurations, Design operations, Transistor

switching networks, PNP transistors, Bias stabilization.


TEXT BOOK:


REFERENCE BOOKS:



SJBIT/ECE Page 57

2.1 Bipolar transistor biasing

DC Biasing is a static operation. It deals with setting a fixed level of the current which should flow through

the transistor with a devised fixed voltage drop across the transistor junctions.

Bias establishes the dc operating point for proper linear operation of an amplifier.

The proper flow of zero signal collector current and the maintenance of proper collector emitter voltage

during the passage of signal is known as transistor.

The purpose of dc biasing of a transistor is to obtain certain dc collector current at a certain dc collector

voltage. These values of current and voltage define the point at which transistor operates. This point is

known as operating point.

The transistor functions most linearly when it is constrained to operate in its active region. Once an operating

point Q is established, time-varying excursions of the input signal should cause an output signal of the same

wave form. If the output is not the one desired, i.e., the o/p does not suit the required conditions, the

operating point is unsatisfactory and should be relocated on the output characteristics.

Now, about choosing the operating point, we should note that the transistor cannot be operated everywhere in

the active region even if we have the liberty to choose the external circuit parameters. This is because of the

various transistor ratings which limit the range of operation. These ratings are maximum collector dissipation

Pcmax, maximum collector voltage V cmax, and maximum collector current Icmax & maximum emitter to base

voltage VEBmax.

Requirements upon biasing circuit

The operating point of a device, also known as bias point or quiescent point (or simply Q-point), is the DC

voltage and/or current which, when applied to a device, causes it to operate in a certain desired fashion.


SJBIT/ECE Page 58

1. For analog circuit operation, the Q-point is placed so the transistor stays in active mode (does not shift to operation in the saturation region or cut-off region) when input is applied. For digital

operation, the Q-point is placed so the transistor does the contrary - switches from "on" to "off" state.

Often, Q-point is established near the center of active region of transistor characteristic to allow

similar signal swings in positive and negative directions.

2. Q-point should be stable. In particular, it should be insensitive to variations in transistor parameters (for example, should not shift if transistor is replaced by another of the same type), variations in

temperature, variations in power supply voltage and so forth.

3. The circuit must be practical: easily implemented and cost-effective.

Note on temperature dependence

At constant current, the voltage across the emitter-base junction VBE of a bipolar transistor decreases about 2

mV for each 1C rise in temperature. Oppositely, if VBE: is held constant and the temperature rises, IC

increases, also increasing the power consumed in the transistor, tending to further increase its temperature.

Unless steps are taken to control this positive feedback of increased temperature increased current increased temperature, thermal runaway ensues. An electrical approach to avoid thermal runaway is to use

negative feedback, as described in conjunction with some of the circuits below. A different approach is to use

heat sinks that carry away the extra heat.

Comparison between various configurations for a given transistor sample


SJBIT/ECE Page 59

2.2 BIAS STABILITY-

There are two reasons for the operating point to shift. Firstly, the transistor parameters such as , VBE are not the same for every transistor, even of the same type. Secondly, the

transistor parameters (,IC0 , VBE ) are functions of temperature. It is therefore, very important that biasing network be so designed that operating point should be independent of

transistor parameter variations.

The techniques normally used to do so maybe classified into-

1.Stabilization techniques

2. Compensation techniques

STABILITY FACTOR-

As Ic is a function of ICO , VBE, & , it is convenient to introduce three partial derivatives of IC w.r.t these variables. These are called stability factors S,S&S and defined as follows: S = (Ic / ICO ) = (1+ )[ (1+(Rb/Re))/(1+ +(Rb/Re))] S = (Ic / VBE ) = -/Re [1+ +(Rb/Re)] S = (Ic / ) (Ic1/1) [ (1+(Rb/Re))/(1+ 2+(Rb/Re))]

Types of bias circuit

The following discussion treats five common biasing circuits used with bipolar transistors:

1. Fixed bias 2. Collector-to-base bias 3. Fixed bias with emitter resistor 4. Voltage divider bias 5. Emitter bias


SJBIT/ECE Page 60

Fixed bias (base bias)

Fixed bias (Base bias)

This form of biasing is also called base bias. In the example image on the right, the single power source (for

example, a battery) is used for both collector and base of transistor, although separate batteries can also be

used.

In the given circuit,

VCC = IBRB + Vbe

Therefore,

IB = (VCC - Vbe)/RB

For a given transistor, Vbe does not vary significantly during use. As VCC is of fixed value, on selection of

RB, the base current IB is fixed. Therefore this type is called fixed bias type of circuit.

Also for given circuit,

VCC = ICRC + Vce

Therefore,

Vce = VCC - ICRC

From this equation we can obtain Vce. Since IC = IB, we can obtain IC as well. In this manner, operating point given as (VCE,IC) can be set for given transistor.

Merits:

It is simple to shift the operating point anywhere in the active region by merely changing the base

resistor (RB).

Very few number of components are required.


SJBIT/ECE Page 61

Demerits:

The collector current does not remain constant with variation in temperature or power supply voltage.

Therefore the operating point is unstable.

When the transistor is replaced with another one, considerable change in the value of can be expected. Due to this change the operating point will shift.

Usage:

Due to the above inherent drawbacks, fixed bias is rarely used in linear circuits, ie. those circuits which use

the transistor as a current source. Instead it is often used in circuits where transistor is used as a switch.

Collector-to-base bias

Collector-to-base bias

In this form of biasing, the base resistor RB is connected to the collector instead of connecting it to the battery

VCC. That means this circuit employs negative feedback to stabilize the operating point.

From Kirchhoff's voltage law, the voltage across the base resistor is

VRb = VCC - (IC + Ib)RC - Vbe.

From Ohm's law, the base current is

Ib = VRb / Rb.

The way feedback controls the bias point is as follows. If Vbe is held constant and temperature increases,

collector current increases. However, a larger IC causes the voltage drop across resistor RC to increase, which

in turn reduces the voltage VRb across the base resistor. A lower base-resistor voltage drop reduces the base

current, which results in less collector current, so increase in collector current with temperature is opposed,

and operating point is kept stable.

For the given circuit,

IB = (VCC - Vbe) / (RB+RC).


SJBIT/ECE Page 62

Merits:

Circuit stabilizes the operating point against variations in temperature and (ie. replacement of transistor)

Demerits:

In this circuit, to keep IC independent of the following condition must be met:

which is approximately the case if

RC >> RB.

As -value is fixed for a given transistor, this relation can be satisfied either by keeping RC fairly large, or making RB very low.

If RC is of large value, high VCC is necessary. This increases cost as well as precautions

necessary while handling.

If RB is low, the reverse bias of the collector-base is small, which limits the range of collector

voltage swing that leaves the transistor in active mode.

The resistor RB causes an ac feedback, reducing the voltage gain of the amplifier. This undesirable

effect is a trade-off for greater Q-point stability.

Usage: The feedback also decreases the input impedance of the amplifier as seen from the base, which can

be advantageous. Due to the gain reduction from feedback, this biasing form is used only when the trade-off

for stability is warranted.

2.3 Fixed bias with emitter resistor

Fixed bias with emitter resistor


SJBIT/ECE Page 63

The fixed bias circuit is modified by attaching an external resistor to the emitter. This resistor introduces

negative feedback that stabilizes the Q-point. From Kirchhoff's voltage law, the voltage across the base

resistor is

VRb = VCC - IeRe - Vbe.

From Ohm's law, the base current is

Ib = VRb / Rb.

The way feedback controls the bias point is as follows. If Vbe is held constant and temperature increases,

emitter current increases. However, a larger Ie increases the emitter voltage Ve = IeRe, which in turn reduces

the voltage VRb across the base resistor. A lower base-resistor voltage drop reduces the base current, which

results in less collector current because Ic = IB. Collector current and emitter current are related by Ic = Ie with 1, so increase in emitter current with temperature is opposed, and operating point is kept stable.

Similarly, if the transistor is replaced by another, there may be a change in IC (corresponding to change in -value, for example). By similar process as above, the change is negated and operating point kept stable.


IB = (VCC - Vbe)/(RB + (+1)RE).

Merits:

The circuit has the tendency to stabilize operating point against changes in temperature and -value.

Demerits:



( + 1 )RE >> RB.

As -value is fixed for a given transistor, this relation can be satisfied either by keeping RE very large, or making RB very low.

If RE is of large value, high VCC is necessary. This increases cost as well as precautions


If RB is low, a separate low voltage supply should be used in the base circuit. Using two

supplies of different voltages is impractical.

In addition to the above, RE causes ac feedback which reduces the voltage gain of the amplifier.


SJBIT/ECE Page 64

Usage:

The feedback also increases the input impedance of the amplifier as seen from the base, which can be

advantageous. Due to the above disadvantages, this type of biasing circuit is used only with careful

consideration of the trade-offs involved.

2.4 Voltage divider bias

Voltage divider bias

The voltage divider is formed using external resistors R1 and R2. The voltage across R2 forward biases the

emitter junction. By proper selection of resistors R1 and R2, the operating point of the transistor can be made

independent of . In this circuit, the voltage divider holds the base voltage fixed independent of base current provided the divider current is large compared to the base current. However, even with a fixed base voltage,

collector current varies with temperature (for example) so an emitter resistor is added to stabilize the Q-point,

similar to the above circuits with emitter resistor.

In this circuit the base voltage is given by:

voltage across

provided .

Also



SJBIT/ECE Page 65

Merits:

Unlike above circuits, only one dc supply is necessary.

Operating point is almost independent of variation. Operating point stabilized against shift in temperature.

Demerits:



where R1 || R2 denotes the equivalent resistance of R1 and R2 connected in parallel.

As -value is fixed for a given transistor, this relation can be satisfied either by keeping RE fairly large, or making R1||R2 very low.

If RE is of large value, high VCC is necessary. This increases cost as well as precautions


If R1 || R2 is low, either R1 is low, or R2 is low, or both are low. A low R1 raises VB closer to

VC, reducing the available swing in collector voltage, and limiting how large RC can be made

without driving the transistor out of active mode. A low R2 lowers Vbe, reducing the allowed

collector current. Lowering both resistor values draws more current from the power supply

and lowers the input resistance of the amplifier as seen from the base.

AC as well as DC feedback is caused by RE, which reduces the AC voltage gain of the amplifier. A

method to avoid AC feedback while retaining DC feedback is discussed below.

Usage:

The circuit's stability and merits as above make it widely used for linear circuits.

Voltage divider with AC bypass capacitor


SJBIT/ECE Page 66

Voltage divider with capacitor

The standard voltage divider circuit discussed above faces a drawback - AC feedback caused by resistor RE

reduces the gain. This can be avoided by placing a capacitor (CE) in parallel with RE, as shown in circuit

diagram.

This capacitor is usually chosen to have a low enough reactance at the signal frequencies of interest such that

RE is essentially shorted at AC, thus grounding the emitter. Feedback is therefore only present at DC to

stabilize the operating point. Of course, any AC advantages of feedback are lost.

Of course, this idea can be used to shunt only a portion of RE, thereby retaining some AC feedback.

Emitter bias


SJBIT/ECE Page 67

When a split supply (dual power supply) is available, this biasing circuit is the most effective. The negative

supply VEE is used to forward-bias the emitter junction through RE. The positive supply VCC is used to

reverse-bias the collector junction. Only three resistors are necessary.

We know that,

VB - VE = Vbe

If RB is small enough, base voltage will be approximately zero. Therefore emitter current is,

IE = (VEE - Vbe)/RE

The operating point is independent of if RE >> RB/

Merit:

Good stability of operating point similar to voltage divider bias.

Demerit:

This type can only be used when a split (dual) power supply is available.

Stability factors

S (ICO) = IC / IC0

S (VBE) = IC / VBE

S () = IC /

Networks that are quite stable and relatively insensitive to temperature variations have

low stability factors.

The higher the stability factor, the more sensitive is the network to variations in that

parameter.

S( ICO)

Analyze S( ICO) for emitter bias configuration

fixed bias configuration Voltage divider configuration

For the emitter bias configuration, S( ICO) = ( + 1) [ 1 + RB / RE] / [( + 1) + RB / RE] If

RB / RE >> ( + 1) , then

S( ICO) = ( + 1)


SJBIT/ECE Page 68

For RB / RE 10R2, the voltage VB will remain fairly constant for changing levels of IC. VBE = VB VE, as IC increases, VE increases, since VB is constant, VBE drops making IB to fall, which will try to offset the increases level of IC.

S(VBE)


SJBIT/ECE Page 69

S(VBE) = IC / VBE

For an emitter bias circuit, S(VBE) = - / [ RB + ( + 1)RE]

If RE =0 in the above equation, we get S(VBE) for a fixed bias circuit as,

S(VBE) = - / RB. For an emitter bias,

S(VBE) = - / [ RB + ( + 1)RE] can be rewritten as,

S(VBE) = - (/RE )/ [RB/RE + ( + 1)]

If ( + 1)>> RB/RE, then

S(VBE) = - (/RE )/ ( + 1) = - 1/ RE The larger the RE, lower the S(VBE) and more stable is the system.

Total effect of all the three parameters on IC can be written as,

IC = S(ICO) ICO + S(VBE) VBE + S() General conclusion: The ratio RB / RE or Rth / RE should be as small as possible considering all aspects of

design.


SJBIT/ECE Page 70

Question paper with Solutions

Q 1) Desigh a self bias transistor circuit for a stability of s< 5. The give date is as follows: Jan 2004(8)


SJBIT/ECE Page 71


SJBIT/ECE Page 72


SJBIT/ECE Page 73

Q 2) For a self bias circuit, derive an expression for the stability factor s.

July 2004(8)


SJBIT/ECE Page 74


SJBIT/ECE Page 75


SJBIT/ECE Page 76


SJBIT/ECE Page 77

Q 4.

Jan 2005 (10) JAN2009(10)


SJBIT/ECE Page 78

Q 5) In the circuit of Fig. 9 given below, Vcc = 10V, Rc = 1.5 kn, ICQ = 2 mA, VCE = 5V, VBE = 0.7 V, 0 = 50 and stability factor S ~ 5. Find R] and R2.

July 2005 (9),July2009(9)


SJBIT/ECE Page 79


SJBIT/ECE Page 80

Q. 6


SJBIT/ECE Page 81

July 2006 (10)


SJBIT/ECE Page 82


SJBIT/ECE Page 83


SJBIT/ECE Page 84

Recommended Questions

1. What are the causes of instability in a transistor? Explain them in brief.(Jan/Feb 2006, 5 marks) 2. Discuss the causes for bias instability in a transistor(July / Aug 2005, 5 marks) 3. What is meant by biasing of a transistor? List the different types of transistor biasing circuits. 4. What to do you mean by operating point of a transistor? Draw the output characteristic of transistor

with various limits of operation and explain it.

5. Differentiate the active region, saturating region and cut off region of a transistor with the requirement of biasing.

6. Analyze the fixed bias circuit operation and derive the expression for operating point (Iceq, Vceq) Vce max and Ic max

7. Analyse the Emitter bias circuit operation and derive the expression for operating point (Iceq, Vceq) Vce max and Ic max

8. What are the different areas of operation in the BJT Characteristic curve? And explain them. 9. Analyse the voltage divider bias circuit operation and derive the expression for operating point

(Iceq, Vceq) Vce max and Ic max (using both approximate and exact method)

10. List out the various types of biasing circuits and compare their merits and demerits. 11. What do you understand of designing the transistor bias circuit? List the parameters to be

calculated and list the parameters required to design.

12. What is meant by transistor switching circuit? Explain with the required biasing. 13. What do you mean by stabilization? 14. Give the essential requirements of stabilization 15. Differentiate between saturation, linear region & cutoff region of transistor operation & show this

in the characteristic curve

16. Explain the fixed bias of transistor with circuit diagram and output equations 17. What is meant by biasing of a transistor? List the different types of transistor biasing circuits 18. a) Draw the transistor amplifier with the fixed bias circuit using the given component values

Input coupling Capacitor C1 =10uF, RB= 240Kohm, RC = 22Kohm VCC= +12V Output coupling

capacitor C2 = 10uF, Beta = 50 Input signal is ac signal

b)Determining the following for the fixed bias transistor configuration

i)Ibq & Icq ii) Vceq iii) VB & Vc iv) VBC v)Ve

Recalculate for B =100 and compare the results

19. a) Draw the transistor amplifier with the Emitter bias circuit using the given component values Input coupling Capacitor C1 =10uF, RB= 510Kohm, RC = 2.4Kohm, RE= 1.5K ohm VCC= +20V

Output coupling capacitor C2 = 10uF, Beta = 100 Input signal is ac signal

b)Determining the following for the Emitter bias transistor configuration



20. a) Draw the transistor amplifier with the voltage divider bias circuit using the given component values

Input coupling Capacitor C1 =10uF, R1= 62Kohm, R2=9.1 kohm, RC = 3.9 k ohm, RE= 0.68 k

ohm VCC= +16V Output coupling capacitor C2 = 10uF, Beta = 80 Input signal is ac signal

b)Determining the following for the voltage divider bias transistor configuration



SJBIT/ECE Page 85


21. Explain the concept of Load line in case of transistors and thus discuss the biasing techniques applied to NPN transistors

22. What do you mean by bias stabilization? 23. Define stability factor. Find the relationship between stability factor and Ib? What is its ideal

value?

24. Give the essential requirements of stabilization of transistor

25. Design the transistor inverter with Rb & RC , Vcc=5V to operate with saturation current of 8mA, B=100. Use level of Ib equal to 120% Ibmax and standard resistor values.

26. Write short notes on Relay driver circuit using transistor.


SJBIT/ECE Page 86

Unit: 3 Hrs: 6

Transistor at Low Frequencies: BJT transistor modeling, Hybrid equivalent model, CE Fixed bias

configuration, Voltage divider bias, Emitter follower, CB configuration, Collector feedback configuration,

Hybrid equivalent model.


TEXT BOOK:


REFERENCE BOOKS:



SJBIT/ECE Page 87

3.1 AC Analysis of BJT transistors

Two types of analyses are usually used depending on the voltage and currents of the input ac signal

relative to the bias voltages and currents. They are small-signal analysis and large-signal analysis. In ac

analysis of BJT amplifier is done using small signal analysis,

Amplification in the AC domain

The transistor can be employed as an amplifying device. That is, the output sinusoidal signal is greater than

the input signal or the ac input power is greater than ac input power.

How the ac power output can be greater than the input ac power?

Conservation- output power of a system cannot be larger than its input and the efficiency cannot be greater

than 1.

The input dc plays an important role in the amplification and contributes in increasing its level to the ac

domain where the conversion will become as =Po(ac)/Pi(dc)

The superposition theorem is applicable for the analysis and design of the dc & ac components of a BJT

network. It permits the separation of the analysis of the dc & ac responses of the system.

In other words, one can make a complete dc analysis of a system before considering the ac response.

Once the dc analysis is complete, the ac response can be determined by doing a complete ac analysis.

Important Parameters for the ac analysis

Zi, Zo, Av, Ai are important parameters for the analysis of the AC characteristics of a

transistor circuit.

Using equivalent circuit

Zi = Vi/Ii where Ii= (Vs-Vi)/Rsense

Where Rsense is very low value resistor

used to measure input current


SJBIT/ECE Page 88

Zo= Vo/Io where Io= (V-Vo)/Rsense

Where Rsense is very low value resistor

used to measure output current

Av = Vo/Vi

AVNL = Vo/Vi with RL = infinite

AVNL > AVLoad

Ai = Io/Ii

It also can be calculated as

Ai = -AvZi/RL

Phase Relationship

The phase relationship between input and output signal depends on the amplifier

Common Emitter : 180 degrees Common - Base : 0 degrees

Common Collector: 0 degrees

AC analysis using equivalent circuit:-

Schematic symbol for the device can be replaced by this equivalent circuit. Basic methods of circuit analysis

are applied.

DC levels are important to determine the Q-point. Once determined, the DC level can be ignored in the AC

analysis of the network.

Coupling capacitors & bypass capacitor are chosen, to have a very small reactance at the frequency of

applications.

The AC equivalent of a network is obtained by:

Setting all DC sources to zero & replacing them by a short-circuit equivalent.

Replacing all capacitors by a short-circuit equivalent.

Removing all elements bypassed by short-circuit equivalent.

Redrawing the network.


SJBIT/ECE Page 89

The first step in modeling the ac behavior of the transistor is to determine its ac equivalent circuit and

use it to replace the transistor circuit symbol in the schematic. Normal circuit analysis is then performed.

To explain the transistor operation during small signal analysis, one of three models are usually used: the re

model, the hybrid model, and the hybrid equivalent model. The re model is a reduced version of the hybrid model which is exclusively used for high frequency analysis.

Disadvantage

Re model- It fails to account the output impedance level of device and feedback effect from output to input.

Hybrid equivalent model-It is limited to specified operating condition in order to obtain accurate result.

A device model is a combination of properly chosen circuit elements that best approximates the

actual behavior of the device under specific operating conditions.

The subsequent figures shows an example of how a typical CE circuit is usually converted to its ac

equivalent circuit. This is achieve by setting all DC sources as ground potential (or ac ground) and capacitors

as ac shorts and with small signal ac modeling of a transistor circuit

Short out capacitors

Set Vdc to ac gnd

Re-arranging


SJBIT/ECE Page 90

3.2 The Hybrid Model The hybrid model is used for high frequency modeling of the transistor. We will apply this to frequency analysis discussions later on.

The re Model This model is more suitable for when transistor circuit is used at dc and low frequencies (e.g. audio). Its the same as the hybrid model except that the high frequency components are not included

Transistor Models

In this session, we will only be looking re model, and hybrid equivalent model..

The re transistor model

Common Base PNP Configuration

Transistor is replaced by a single diode between E & B, and control current source between B & C.

Collector current Ic is controlled by the level of emitter current Ie.

For the ac response the diode can be replaced by its equivalent ac resistance.


SJBIT/ECE Page 91

The ac resistance of a diode can be determined by the equation;

where ID is the dc current through the diode at the Q-point.

Input impedance is relatively small and output impedance quite high.

Input impedance ranges from a few to max 50 . Typical values are in the M .

The common-base characteristics

Voltage Gain

EI

mVre

26

CBi reZ

CBZo

re

RA

re

R

rI

RI

V

VA

rI

ZI

ZIV

RI

RI

RIV

LV

L

ee

Le

i

O

V

ee

ie

iii

Le

LC

Loo

:gain voltage

: ageinput volt

)(

: tageoutput vol


SJBIT/ECE Page 92

Current Gain

The fact that the polarity of the Vo as determined by the current IC is the same as defined by figure below.

It reveals that Vo and Vi are in phase for the common-base configuration.

Approximate model for a common-base npn transistor configuration

Example 1: For a common-base configuration in figure below with IE=4mA, =0.98 and AC signal of 2mV

is applied between the base and emitter terminal:

a) Determine the Zi b) Calculate Av if RL=0.56k

c) Find Zo and Ai

Solution:

1i

e

e

e

C

i

oi

A

I

I

I

I

I

IA

e

b b

c

ec I I

IcI

e

common-base re equivalent cct

re

5.64

26

I

26mr Za)

E

ei

m

m

43.845.6

)56.0(98.0

r

R b)

e

Lv

kA

98.0

Zc) o

i

oi

I

IA


SJBIT/ECE Page 93

Example 2: For a common-base configuration in previous example with Ie=0.5mA, =0.98 and AC signal of

10mV is applied, determine:

a) Zi b) Vo if RL=1.2k c) Av d)Ai e) Ib

Common Emitter NPN Configuration

Base and emitter are input terminals.

Collector and emitter are output terminals.

Substitute re equivalent circuit

Current through diode

Input impedance

The output graph

205.0

10 Za)

:Solution

i

m

m

I

V

e

i

88mV5

(1.2k)0.98(0.5m)

RIRI b) LeLcoV8.58

10

588A c) v

m

m

V

V

i

o

98.0A d) i

A

m

m

I

10

)98.01(5.0

)1(5.0

I-I

I-I e)

ee

ceb

bc II

bbe

bbbce

III

IIIII

)1(

ei

ei

b

ebi

eb

eei

b

be

i

ii

rZ

rZ

I

rIZ

rI

rIV

I

V

I

VZ

; 1an greater thusually

)1(

)1( that so

)1(

:ageinput volt

:impedanceinput


SJBIT/ECE Page 94

Output impedance Zo

Voltage Gain

Current Gain

re model for common-emitter

re model for common-emitter

Example 3: Given =120 and IE(dc)=3.2mA for a common- emitter configuration with ro= , determine:

a) Zi b)Av if a load of 2 k is applied c) Ai with the 2 k load

Example 4: Using the npn common-emitter configuration, determine the following if =80, IE(dc)=2 mA and

ro=40 k .

a) Zi b) Ai if RL =1.2k c) Av if RL=1.2k

bI

c

e

bIi=I

b

re model for the C-E transistor configuration

re

ro

e

0AbI

c

e

bI

i=I

b

re

ro

e

Vs=0V

= 0A

impedance)high cct,(open Z

the thusignored is r if

o

o

oo rZ

e

LV

eb

Lb

i

o

V

eb

iii

Lb

Lco

Loo

r

RA

rI

RI

V

VA

rI

ZIV

RI

RIV

RIV

that so

:ageinput volt

: tageoutput vol

i

b

b

b

C

i

oi

A

I

I

I

I

I

IA

975)125.8(120

125.82.3

26

I

26mr a)

E

e

ei rZ

m

m

:Solution

15.246125.8

2

r

Rb)A

e

Lv

k

120I

IA c)

i

oi

bI

cbIi=I

b

re model for the C-E transistor configuration

re

ro

e

RL

Io

krZ

m

m

ei 04.1)13(80

132

26

I

26mr a)

E

e

:Solution

67.77

)80(2.140

40)(

)(

ib)

(cont)Solution

kk

k

Rr

r

I

Rr

Ir

A

Rr

IrI

I

I

I

IA

Lo

o

b

Lo

bo

i

Lo

boL

b

L

i

o


SJBIT/ECE Page 95

Common Collector Configuration

For the CC configuration, the model defined for the common-emitter configuration is normally applied rather

than defining a model for the common-collector configuration.

3.3 The re Transistor Model for the CE Fixed Biased Configuration

and ro are given in spec sheet; and re is determined from dc analysis

Ac analysis

Input impedance, Zi =Vi/Ii

From the figure, it is clear that, Ii = IRB+IB = Vi/RB+Vi/re = Vi(1/RB +1/re) Zi = Vi/Ii= (1/RB +1/re) i.e Zi = RB// re

6.8913

402.1vc)

kk

r

rRA

e

oL


SJBIT/ECE Page 96

Output impedance Zo=Vo/Io

By keeping Vi=0, Ii=0 So IB =0 i.e open in output side. Output current Io= Vo(1/ro+1/Rc)

So, Output impedance Zo = Vo/Io = Rc//ro Rc when ro>>Rc or ro>10Rc

Voltage gain Av=Vo/Vi

Vo= -IB(Rc//ro) = -(Vi/ re )(Rc//ro) by replacing IB= Vi/ re = -Vi Rc//ro /re

Av= Vo/Vi =-( Rc//ro)/re -Rc/re

-ive sign indicates the 1800 phase shift between input & output voltage signal

Current gain Ai =Io/Ii

Ii =Vi/Zi

Io=Vo/RL

Ai= Io/Ii = (Vo/RL)/(Vi/Zi)= -AvZi/RL

3.4 Ac analysis for Voltage divider circuit

Example

It is similar to that of fixed bias circuit with RB is replaced by R1//R2

So

Zi = R1//R2 //re here R1//R2 is comparitely smaller value than that of RB in fixed n\bias. So it may not be possible to ignore R1//R2 in calculation of Zi. So Zi with voltage divider is lesser than that of fixed bias

Zo =Rc//roRc Same as that of fixed bias. Av =-(Rc//ro)/re Rc/re Same as that of fixed bias

Hybrid Equivalent Model The hybrid parameters: hie, hre, hfe, hoe are developed and used to model the transistor.

These parameters can be found in a specification sheet for a transistor.

hi = input resistance hr = reverse transfer voltage ratio (Vi/Vo) hf = forward transfer current ratio (Io/Ii)


SJBIT/ECE Page 97

ho = output conductance

General h-Parameters for any

Transistor Configuration

Vi =f1 (Ii, Vo) and Io =f2((Ii, Vo)

Vi = h11Ii+h12Vo

Io=h21Ii+h22Vo

Where

h11 = Vi/Ii with Vo=0 ie short circuit input resistance, unit , & designated as hi h12 = Vi/Vo with Ii=0 ie open circuit reverse transfer voltage ratio, unitless, & designated as hr

h21 = Io/Ii with Vo=0 ie short circuit forward transfer current ratio, unitless, & designated as hf

h22 = Vo/Io with Ii=0 ie open circuit output conductance, unit Siemens & designated as ho

So hi,hr,hf & ho are called hybrid parameters

By placing second subscript as b for CB, c for CC and e for CE, we can get hybrid parameters for each

configuration.

Simplified General h-Parameter Model

The above model can be simplified based on these approximations:

hr =0 therefore hrVo = 0 and 1/ho =

Common-Emitter h-Parameters

ac fe

hie =25mV/IBQ =hfeIBQ/IEQ

hfe =ac


SJBIT/ECE Page 98

Common-Base h-Parameters

hib =25mV/IEQ

hib=-ac = -1

3.5 Common-Emitter (CE) Fixed-Bias Configuration

The input (Vi) is applied to the base and the output (Vo) is from the collector.

The Common-Emitter is characterized as having high input impedance and low output

impedance with a high voltage and current gain.

Determine hfe, hie, and hoe:

hfe and hoe: look in the specification sheet for the transistor or test the transistor using

a curve tracer.

hie: calculate hie using DC analysis:

hie =25mV/IBQ =hfe25mV/IEQ

Input impedance Zi = RB//hie hie if RB > 10hie

Output impedance Zo= Rc//(1/ho) Rc if 10Rc


SJBIT/ECE Page 99

Voltage gain hie

hfeRc

hie

hohfeRcAv

)/1//(

if 1/ho>10Rc

Current gain Ai hfe if RB>10hie & 1/ho>10Rc

Or Ai = -AvZi/Rc

Phase Relationship

The phase relationship between input and output is 180 degrees. The negative sign used in

the voltage gain formulas indicates the inversion.

CE Voltage-Divider Bias Configuration

Input impedance Zi = R1//R2//hie

Output impedance Zo= Rc//(1/ho) Rc if 10Rc 10Rc


SJBIT/ECE Page 100

Current gain Ai hfe if RB>10hie & 1/ho>10Rc

Or Ai = -AvZi/Rc Phase Relationship

A CE amplifier configuration will always have a phase relationship between input and

output is 180 degrees. This is independent of the DC bias.

CE Emitter-Bias Configuration

Unbypassed RE

Input impedance Zi = RB//hie+(1+hfeRE)

Output impedance Zo= Rc//(1/ho) Rc if 10Rc 10Rc, 1+hfe hfe

Current Gain Ai =

ieCoe

oefe

hRRRh

hRRh

IiIo)2//1(()1(

1)2//1(/ = -AvZi/Rc

For Emitter follower circuit

Zi = RB//(1+)(re+RE) = RB// RE if RE > 10re & 10RE


SJBIT/ECE Page 101

3.6 For CB amplifier

Zi = RE//re =re ( hib in hybrid eq ckt)

Zo = Rc//ro=Rc

AV = Rc/re ( Rc/hib)

Effect of load resistance and source impedance:-

AvNL >AVL>AVs

Both load resistance & source impedance reduces the gain

If load resistance is very low compare to Zo, gain reduces drastically

If source impedance is very high compare to Zi gain reduces.

If we consider output equivalent circuit as voltage source with value AvNL , inseries with output

impedance Zo we can find the reduction factor as below

If AVNL is the no load gain, AVL is the gain with load RL, AVs is the gain with load and source resitance

rs

Then

RoR

R

rsZi

ZiAV

RoR

RAVAVs

rsZi

ZiAVAV

L

L

LN

L

LL

LNL

Analog Electronic Circuits 10

Ece-III-Analog Electronic Ckts [10es32]-Notes

Documents

Transcript of Ece-III-Analog Electronic Ckts [10es32]-Notes