Analog Electronics

 Analog Electronics First Edition Authors Dr. R. Thirumamagal Dr. M. Karunakaran Iterative International Publishers Title of the Book: Analog Electronics First Edition - 2022 Copyright 2022 © Authors Dr. R. Thirumamagal, Assistant Professor, Department of Physics, Ananda College, Devakottai. Dr. M. Karunakaran, Assistant Professor, PG & Research Department of Physics, Alagappa Government Arts College, Karaikudi. No part of this book may be reproduced or transmitted in any form by any means, electronic or mechanical, including photocopy, recording or any information storage and retrieval system, without permission in writing from the copyright owners. Disclaimer The authors are responsible for the contents published in this book. The publisher or editors don’t take any responsibility for the same in any manner. Errors, if any, are purely unintentional and readers are requested to communicate such errors to the editors or publishers to avoid discrepancies in future. ISBN: 978-81-960738-4-8 MRP Rs.450/- Publisher, Printed at & Distribution by: Selfypage Developers Pvt Ltd., Pushpagiri Complex, Beside SBI Housing Board, K.M. Road Chikkamagaluru, Karnataka. Tel.: +91-8861518868 E-mail:[email protected] IMPRINT: I I P Iterative International Publishers ii  Preface This book has been written for B.Sc. Physics and Electronics students at various Indian Universities. The book covers the Alagappa University, Karaikudi's recommended course curriculum. We were inspired to write the book by the students' enthusiasm for the topic of "Analog Electronics." With a modern perspective, proper attention has been taken in dealing with the issue. At the end of each unit, a huge number of questions and problems have been provided. Students should make an effort to address them effectively in order to have a better comprehension of the subject. It is clear and easy to read the text. Both regular students and top students should like this book. The students' exam performance would improve, while at the same time, their intellectual interest would increase. This book is divided into five sections. The first unit discusses semiconductor diodes and regulated power supply. The transistor and its biasing are discussed in detail in the next unit, which includes detailed experiments. The third section discusses several types of amplifiers. Oscillators are discussed in detail in the fourth unit. The amplifier and its features are discussed in the final unit. We would warmly accept and highly value any useful comments you have on how to make the book better. iii  Acknowledgement At every step of our activity, we praise the Almighty for his favours. Words cannot describe how grateful we are to Dr. A. Pethalakshmi, Principal of Alagappa Government Arts College, Karaikudi, for her support and encouragement during this incredible trip. We are grateful to Rev. Fr. Jesuraj K. Christy, Secretary of Ananda College, and Rev. Dr. S. John Vasanthakumar, Principal of Ananda College, Devakottai for their kind assistance and useful direction in the successful completion of this book. We would also like to thank Dr. R. Chandramohan, Former Principal, Sree Sevugan Annamalai College, Devakottai, and Research Advisor Vidhyaagiri College of Arts and Science College, Puduvayal, as well as Dr. A. Ayesha Mariam, Assistant Professor, PG and Research Department of Physics, Kadhir Mohaideen College, Adhirampattinam, for their constant encouragement. We would like to express our deepest appreciation to students from Ananda College and Alagappa Government Arts College, as well as Research Scholars from the Thin Film and Nano Science Research Laboratory. We are grateful to our friends and colleagues who pushed us to start working on the project, stick with it, and finally publish it. Finally, we'd want to thank our individual families for their love and support. They all kept us going, and none of this would have been possible without them. Dr. R. Thirumamagal Dr. M. Karunakaran iv  Contents Unit 1 Semiconductor Diodes and Regulated Power 1 - 48 Supplies 1.1 Electrical Conductivity of the Materials 1 1.1.1 What is a Semiconductor? 1 1.1.2 Properties of Semiconductor 1 1.1.3 Commonly Used Semiconductors 1 1.1.4 Energy Band Description of Semiconductor 3 1.1.5 Effect of Temperature on Semiconductors 3 1.1.6 Intrinsic Semiconductor 5 1.1.7 Extrinsic Semiconductor 6 1.1.8 Charges on n-type and p-type Semiconductor 10 1.1.9 PN Junction Diode 11 1.1.10 Formation of PN Junction 12 1.1.11 Properties of PN Junction 13 1.1.12 Biasing a PN Junction 14 1.1.13 Forward Biasing 14 1.1.14 Reverse Biasing 14 1.1.15 Current Flow in a Forward Biased PN junction 15 1.1.16 V-I Characteristics of PN Junction 16 1.1.17 Crystal Diode Rectifier 18 1.1.18 Types of Rectifier 18 1.1.19 Half Wave Rectifier 19 1.1.20 Full Wave Rectifier 19 1.1.21 Construction and working principle of Half Wave 19 Rectifier 1.1.22 Efficiency of a half Wave Rectifier 21 1.1.23 Construction and working principle of Full wave 23 Rectifier 1.1.24 Bridge Rectifier 25 1.1.25 Efficiency of Full Wave Rectifier 28 1.1.26 What is a Filter Circuit? 31 1.1.27 Types of Filter Circuits 32 1.1.28 Capacitor Filter 32 1.1.29 Choke Input Filter (LC- Filter) 34 1.1.30 Capacitor input Filter or π-Filter 35 1.1.31 What is a Zener Diode? 36 1.1.32 V-I Chracteristics of Zener diode (voltage- 37 Current Characteristics) v 1.1.33 Zener Diode as a Voltage Regulator 38 Solved Problems 41 Review Questions 48 Unit 2 Transistors and Biasing 49- 98 2.1 Transistor 49 2.1.1 Necessity for Transistors? 49 2.1.2 Constructional Details of a Transistor 49 2.1.3 The Symbols of PNP and NPN Transistors are as 51 Shown Below 2.1.4 Types of Transistor 51 2.1.5 N-P-N Transistor 51 2.1.6 6 P-N-P Transistor 52 2.1.7 The Following Points can be noted in Each Type 52 of Transistors 2.1.8 Working principle of Emitter 52 2.1.9 Working Principle of Collector 53 2.1.10 Working principle of Base 54 2.1.11 Transistor Action 54 2.1.12 Working of NPN Transistor 54 2.1.13 Working of PNP Transistor 55 2.1.14 Transistor Characteristics 56 2.1.15 Transistor Connections (Configurations) 56 2.1.16 Common Base (CB) Configuration of Transistor 56 2.1.17 Characteristics of Common Base Configuration 57 (CB mode) 2.1.18 Common Emitter Connection (CE Mode) 59 2.1.19 characteristics of Common Emitter Configuration 60 (CE mode) 2.1.20 Common Collector Configuration (CC mode) 63 2.1.21 Comparison of Transistor Configuration (CB, CE 67 and CC) 2.1.22 Transistor as an amplifier in CE arrangement 68 2.1.23 DC Load Line Analysis 69 2.1.24 Operating Point 70 2.1.25 Biasing 72 2.1.26 What is Transistor Biasing? 74 2.1.27 Zero Signal Collector Current 75 2.1.28 Stability Factor 75 2.1.29 Methods of Transistor Biasing 75 vi 2.1.30 Base Resistor Method 75 2.1.31 Voltage Divider Bias Method 77 2.1.32 Biasing with Collector Feedback Resistor 80 2.1.33 FET 81 2.1.34 Junction Field Effect Transistor 81 2.1.35 N-Channel FET 82 2.1.36 P-Channel JFET 83 2.1.37 Construction 83 2.1.38 Parameters of JFET 85 Solved Problems 87 Review Questions 97 Unit 3 Amplifiers 99 - 138 3.1 Amplification 99 3.1.1 Single-stage Transistor Amplifier 99 3.1.2 Working of a single stage Amplifier 100 3.1.3 Practical circuit of a Transistor Amplifier 100 3.1.4 Biasing Circuit 100 3.1.5 Phase reversal 103 3.1.6 DC & AC Equivalent circuits 103 3.1.7 Voltage Gain 106 3.1.8 Classification of Amplifiers 107 3.1.9 Input Impedance 109 3.1.10 Multistage Amplifier 110 3.1.11 Purpose of Coupling Device 110 3.1.12 Types of Coupling 111 3.1.13 Role of Capacitors in Amplifiers 112 3.1.14 Amplifier Consideration 113 3.1.15 RC Coupled Amplifier 113 3.1.16 Transformer Coupled Amplifier 116 3.1.17 Direct Coupled Amplifier 119 3.1.18 Comparisons of Different Types of Coupling 120 3.1.19 Comparison of Different Types of Amplifiers 121 3.1.20 CB Amplifier 121 3.1.21 CE Amplifier 123 3.1.22 CC Amplifier 125 3.1.23 Comparison between CB, CE and CC Amplifiers 126 Solved Problems 127 Review Questions 136 vii UNIT 4 Oscillators 139-170 4.1 Power Amplifier 139 4.1.1 Power Transistor (Transistor Audio Power 139 Amplifier) 4.1.2 Difference between Voltage and Power 140 Amplifiers 4.1.3 The Characteristics of a Power Amplifier are as 141 follows 4.1.4 Performance quantities of Power Amplifier 141 4.1.5 Expression for Collector Efficiency 142 4.1.6 Classification of Power Amplifiers 143 4.1.7 Transformer Coupled Class a Power Amplifier 143 4.1.8 Push-Pull Amplifiers 147 4.1.9 Amplifier Feedback 149 4.1.10 Feedback 149 4.1.11 Principle of Feedback Amplifier 149 4.1.12 Types of Feedback 151 4.1.13 Positive Feedback 151 4.1.14 Negative Feedback 152 4.1.15 Current Gain of Negative Feedback Amplifier 153 4.1.16 Voltage Gain of Negative Feedback Amplifier 155 4.1.17 Oscillators 156 4.1.18 Sinusoidal Oscillator 156 4.1.19 Types of Sinusoidal Oscillators 157 4.1.20 Damped Oscillations 157 4.1.21 Undamped Oscillations 157 4.1.22 Barkhausen Criterion 158 4.1.23 Essentials of Transistor Oscillator 158 4.1.24 Different Types of Transistor Oscillators 159 4.1.25 Colpitt’s Oscillator 159 4.1.26 Circuit Operation 160 4.1.27 Feedback Circuit 160 4.1.28 Feedback Fraction (MV) 161 4.1.29 Hartley Oscillator 161 4.1.30 Phase Shift Oscillator 163 Solved Problems 165 Review Questions 169 viii UNIT 5 Operational Amplifier 171-201 5.1 Operational Amplifier 171 5.1.1 Block Diagram of Operational Amplifier 171 5.1.2 The Following Points may be noted about Op- 172 Amps 5.1.3 Schematic Symbol of Operational Amplifier 172 5.1.4 Differential Amplifier 173 5.1.5 Basic circuit of Differential Amplifier 173 5.1.6 Operation of Differential Amplifier 174 5.1.7 Common Mode and Differential Mode Signals 176 5.1.8 Common-Mode Rejection Ratio (CMRR) 176 5.1.9 A.C. Analysis of Operational Amplifier 176 5.1.10 Bandwidth of an Operational Amplifier 178 5.1.11 Slew Rate 178 5.1.12 Op-Amp as Inverting Amplifier 180 5.1.13 Voltage Gain 180 5.1.14 Op-Amp as Non-Inverting Amplifier 182 5.1.15 Voltage Gain 182 5.1.16 Voltage Follower 183 5.1.17 Summing Amplifier 184 5.1.18 Op -Amp as a Subtractor 187 5.1.19 Op-Amp as an Integrator 187 5.1.20 Op-Amp as a Differentiator 189 5.1.21 Comparator 190 5.1.22 Signal Generator 191 5.1.23 Phase Shift Signal Generator 192 5.1.24 Hartley Signal Generator 193 5.1.25 Square Wave Generator 194 5.1.26 Triangular Wave Generator 196 Solved Problems 197 Review Questions 201 ix x UNIT 1 Semiconductor Diodes and Regulated Power Supplies 1.1 Electrical Conductivity of the Materials As the energy in the system increases, electrons leave the valence band and enter the conduction band. Semiconductors and insulators are further distinguished by the relative band gap. In semiconductors, the band gap is small, allowing electrons to populate the conduction band. In insulators, it is large, making it difficult for electrons to flow through the conduction band. 1.1.1 What is a Semiconductor? A semiconductor is a substance which has resistivity in between conductors and insulators, e.g. germanium, silicon, selenium, carbon etc. It has a resistivity of (10-4 to 0.5Ώ). 1.1.2 Properties of Semiconductor 1. The resistivity of a semiconductor is less than an insulator but more than a conductor. 2. It has negative temperature co-efficient of resistance. That means the resistance of a semiconductor decreases with increase in temperature and vice-versa. 3. When a suitable metallic impurity is added to a semiconductor, its current conducting properties change appreciably. 1.1.3 Commonly Used Semiconductors The two most frequently used semiconductors are (i) germanium (Ge) and (ii) silicon (Si). It is because the energy required to break their co-valent bonds is very small; being 0.7 eV for Ge and 1.1 eV for Si. 1  UNIT 1: Semiconductor Diodes and Regulated Power Supplies (i) Germanium(Ge) Atomic number of germanium is 32. So it has 32 protons and 32 electrons. Two electrons are in the first orbit, eight electrons are in the second, eighteen electrons in the third and and four electrons in the outer or valence orbit. This is shown in fig.1 (i). It is clear that germanium atom has four valence electrons i.e. it is a tetravalent element.Fig.1 (ii) shows how the various germanium atoms are held through co-valent co valent bonds. Fig.1 (i) Fig.1 (ii) (ii) Silicon(Si) Atomic number of silicon is 14. So it has 14 protons and 14 electrons. Fig.2 (i) Fig.2 (ii) Two electrons are in the first orbit, eight electrons lectrons are in the second orbit and four electrons in the third orbit. This is shown in fig.2(i).It is clear that silicon atom has four valence electrons i.e. it is a tetravalent element.Fig.2(ii) shows how various silicon atoms are held through co- co valent bonds. 2  UNIT 1: Semiconductor Diodes and Regulated Power Supplies 1.1.4 Energy Band Description of Semiconductor A semiconductor can be defined much more comprehensively on the basis of energy bands as under: A semiconductor is a substance which has almost filled valence band and nearly empty conduction band withw a very small energy gap (nearly equal to 1 eV ) separating the two. Fig.3 (i) and 3 (ii) shows the energy band diagrams of germanium and silicon respectively. Fig.3(i) Fig.3(ii) ig.3(ii) As the forbidden gap is very small; being 0.7 eV for Ge and 1.1 eV for Si, therefore, relatively small energy is needed by their valence electrons to cross over to the conduction band. Even at room temperature, some of the valence electrons may acquire acquire sufficient energy to enter into the conduction band and thus become free electrons. However, at this temperature, the number of free electrons available is very small. Hence, at room temperature, a piece of Ge or Si is neither a good conductor nor an insulator. For this reason, such substances are called semiconductors. 1.1.5 Effect of Temperature on Semiconductors The electrical conductivity of a semiconductor changes appreciably with temperature variations. (i) At absolute zero At absolute zeroo temperature, all the electrons are tightly held by the semiconductor atoms. The inner orbit electrons are bound whereas the valence electrons are engaged in co-valent co bonding. At this temperature, 3  UNIT 1: Semiconductor Diodes and Regulated Power Supplies the co-valent valent bonds are very strong and there are no free electrons. Therefore, the semiconductor behaves as a perfect insulator. This is shown in fig.4(i). In terms of energy band description, description the valence band is filled and there is a large energy gap between valence band and conduction band. Therefore, no valence ence electrons can reach the conduction band to become free electron. Fig.4(i) Fig.4(ii) Hence, the semiconductor behaves as an insulator due to the non- non availability of free electrons. ctrons. (ii) Above absolute zero When the temperature is raised, some of the co-valent co valent bonds in the semiconductor break due to the thermal energy supplied. The breaking of bonds set those electrons free which are engaged in the formation of these bonds. These free electrons constitute a tiny electric current if potential difference is applied across the semiconductor. This is shown in fig.5(i). Fig.5 (i) Fig.5 (ii) This shows that the resistance of a semiconductor semiconductor decreases with the increase in temperature i.e. it has negative temperature co-efficient co of resistance. Fig.5(ii) shows the energy band diagram. 4  UNIT 1: Semiconductor Diodes and Regulated Power Supplies As the temperature is raised, some of the valence electrons acquire acqui sufficient energy to enter into the conduction band and thus become free electrons. Under the influence of electric field, these free electrons will constitute electric current. It may be noted that each time a valence electron enters into the conduction on band, a hole is created in the valence band. 1.1.6 Intrinsic Semiconductor A semiconductor in an extremely pure form is known as an intrinsic semiconductor. In an intrinsic semiconductor, even at room temperature, hole-electron electron pairs are created. Whenn electric field is applied across an intrinsic semiconductor, the current conduction take place by free electrons and holes as shown in fig.1. The free electrons are produced due to the breaking up of some co-valent co bonds by thermal energy. At the same time time holes are created in the covalent-bonds. Under the influence of electric field, conduction takes place by both free electrons and holes. Fig.1 Therefore, the total current inside the semiconductor is the sum of currents due ue to free electrons and holes. It can be noted that current in the external wire is fully due to electrons. Referring to fig.1, holes being positively charged move towards the negative terminal of supply. As the holes reach the terminal B, electrons enter enter the semiconductor crystal near the terminal and combine with holes, thus cancelling them. At 5  UNIT 1: Semiconductor Diodes and Regulated Power Supplies the same time, the loosely held electrons near the positive terminal A are attracted away from their atoms into the positive terminal. This creates new holes near the positive terminal which again drift towards the negative terminal. 1.1.7 Extrinsic Semiconductor The conducting properties of an intrinsic semiconductor can be increased by adding small amount of suitable impurities to it. It is then called impurity or extrinsic semiconductor. The process of adding impurities to a semiconductor is known as doping. Generally for 108 two types atoms of semiconductor, one impurity atom is added. Depending upon the type of impurity added, extrinsic semiconductor are classified into two types, such as: (i) n-type semiconductor (ii) p-type semiconductor N-type Semiconductor When a small amount of pentavalent impurity is added to a pure semiconductor, it is known as n-type semiconductor. The addition of pentavalent impurity provides a large number of free electrons in the semiconductor crystal. Typical examples of pentavalent impurities are arsenic (As), and antimony (Sb). Such impurities which produce n-type semiconductor are known as donor impurities as they donate free electrons to the semiconductor crystal. To explain the formation of n-type semiconductor, consider a pure germanium crystal. We know that germanium atom has four valence electrons. When a small amount of pentavalent impurity like arsenic is added to germanium crystal, a large number of free electrons become available in the crystal. The reason is explained below. 6  UNIT 1: Semiconductor Diodes and Regulated Power Supplies Arsenic is pentavalent i.e. its atom has five valence electrons. An arsenic atom fits in the germanium crystal in such a way that its four valence electrons form covalent bonds with four germanium atoms. The fifth valence electron of arsenic atom finds no place in co-valent co valent bonds and thus remains free as shown in fig.2. Fig.2 Fig.3 Therefore, for each arsenic atom added, one free electron will be available in the germanium crystal. Fig.3 shows the energy band description of n- n type semiconductor. The addition of pentavalent impurity has produced a number of conduction ion band electrons i.e. free electrons. The four valence electrons of pentavalent atom form co-valent co valent bonds with four neighbouring germanium atoms. The fifth left over valence electron of the pentavalent atom cannot be accommodated in the valence band and travels to the conduction band. The following two points may be noted carefully: (i) Many new free electrons are produced by the addition of pentavalent impurity. (ii) Thermal energy of room temperature still generated a few hole- hole electron pairs. However, the number number of free electrons provided by the pentavalent impurity far exceeds the number of holes. Hence it is called n-type type semiconductor (n stands for negative) N-type Conductivity The current conduction in an n-type n type semiconductor is predominantly by free electrons trons i.e. negative charges and is called n-type n type or electron type conductivity. 7  UNIT 1: Semiconductor Diodes and Regulated Power Supplies When a potential difference is applied across the n-type n type semiconductor, the free electrons, donated by the impurity in the crystal, will be directed towards the positive terminal, termin constituting electric current. Fig.4 As the current flow through the crystal is by free electrons which are carriers of negative charge, therefore, this type of conductivity is called negative or n-type type conductivity. P-type Semiconductor When a small amount of trivalent impurity is added to a pure semiconductor, it is called p-type p semiconductor. The addition of trivalent impurity provides a large number of holes in the semiconductor. Typical example of trivalent impurities are gallium (Ga) and indium indium (In).Such impurities which produce p-type p type semiconductor are known as acceptor impurities as the holes created can accept electrons. To explain the formation of p-type type semiconductor, consider a pure germanium crystal. When a small amount of trivalent impurity like gallium is added to germanium crystal, there exist a large number of holes in the crystal. The reason is explained below: Gallium is trivalent i.e. its atom as three valence electrons. Each atom of gallium fits into the germanium crystal but but now only three co-valent co bonds can be formed. It is because three valence electrons of gallium atom can form only three single co-valent co valent bonds with three germanium atoms as shown in fig.5. 8  UNIT 1: Semiconductor Diodes and Regulated Power Supplies Fig.5 Fig.6 In the fourth co-valent valent bond only germanium atom contributes one valence electron while gallium has no valence electron to contribute as its three valence electrons are already engaged in the co-valent co valent bonds with neighbouring germanium atoms. In other word, fourth bond is incomplete; being short of one electron. This missing electron is called a hole. Therefore, for each gallium atom added, one hole is created.Fig.6 shows the energy band diagram diagram of p-type semiconductor. Thee addition of trivalent impurity has produced a large number of holes. However, there are a few conduction band electrons due to thermal energy at room temperature. But the number of holes far exceeds the number of conduction band electrons. It is due to the the predominance of holes over free electrons that it is called p-type p type semiconductor (p stands for positive). P-type Conductivity The current conduction in p-type p type semiconductor is predominantly by holes i.e. Positive charges and is called p-type p or hole-type type conductivity. Fig.7 9  UNIT 1: Semiconductor Diodes and Regulated Power Supplies When a potential difference is applied to the p-type semiconductor, the holes, donated by the impurity, are shifted from one co-valent bond to another. As the holes are positively charged, therefore, they are directed towards the negative terminal, constituting hole current. 1.1.8 Charges on n-type and p-type Semiconductor In n-type semiconductor, current conduction is due to excess of electrons whereas in a p-type semiconductor, conduction is due to holes. One may think that n-type material has a net negative charge and p-type has a net positive charge. But this conclusion is wrong. It is true that n- type semiconductor has excess of electrons but these extra electrons were supplied by the atoms of donor impurity and each atom of donor impurity is electrically neutral. When the impurity atom is added, the term excess electrons refers to an excess with regard to the number of electrons needed to fill the co-valent bonds in the semiconductor crystal. The extra electrons are free electrons and increase the conductivity of the semiconductor. This situation with regard to p-type semiconductor is also similar. Hence, n-type as well as p-type semiconductor is electrically neutral. Majority and Minority Carriers Due to the impurity, n-type semiconductor has a large number of free electrons whereas p-type semiconductor has a large number of holes. However, at room temperature, some of the co-valent bonds break, thus releasing equal number of free electrons and holes. So,an n-type semiconductor has its share of electron-hole pairs due to breaking of co-valent bonds and in addition has a much larger quantity of free electrons due to the effect of impurity. Consequently, an n-type semiconductor has a large number of free electrons and a small number of holes as shown in fig.8 10  UNIT 1: Semiconductor Diodes and Regulated Power Supplies Fig.8 The free electrons in this case are known as majority carriers, since the majority portion of current in n-type n type material is by the flow of free electrons. And the holes are known know as minority carriers. Similarly, in a p- p type semiconductor, holes outnumber the free free electrons as shown in fig 9. Therefore, holes are the majority carriers and free electrons are the minority carriers. Fig 9 1.1.9 PN Junction Diode When a p-type semiconductor miconductor is suitably joined to an n-type n semiconductor, the contact surface is called PN junction. Most semiconductor devices contain one or more PN junctions. The PN junction is of great importance as it is the main control element for the semiconductor devices. Symbol 11  UNIT 1: Semiconductor Diodes and Regulated Power Supplies 1.1.10 Formation of PN Junction In actual practice, a PN junction will not be formed if a p-type type block is just brought in contactt with n-type n block. In fact, PN junction is fabricated by special techniques. One common method of makingmakin PN junction is known as alloying. In this method, a small block of indium (trivalent impurity) is placed on an n-type type germanium slab as shown in fig.1 (i). The system is then heated to a temperature of about 500o C. The indium and some of the germanium melt to form a small puddle of molten germanium-indium germanium indium mixture as shown in fig.1 (ii). Fig.1(i) Fig.1(ii) The temperature is then lowered and puddle begins to solidify. Under proper conditions, the atoms atoms of indium impurity will be suitably adjusted in the germanium slab to form a single crystal. The addition of indium overcomes the excess of electrons in the n-type n type germanium to such an extent that it creates a p--type region. As the process goes on, the remaining molten mixture becomes increasingly rich in indium. When all the germanium has been redeposited, the remaining material appears as indium button which is frozen on the outer surface of the crystallized portion as shown in fig.1(iii).This button serves erves as a suitable base for soldering on leads. Fig.1(iii) 12  UNIT 1: Semiconductor Diodes and Regulated Power Supplies 1.1.11 Properties of PN Junction At the instant of PN junction formation, the free electrons near the junction in the n region begin to diffuse across the junction into the p region where they combine with holes near the junction. As a result n region loses free electrons and this creates a layer of positive charges (pentavalent ions) near the junction. As the electrons move across the junction, the p region loses holes as the electrons and holes hol combine. The result is that there is a layer of negative charges (trivalent ions) near the junction. These two layer of positive and negative charges form the depletion region or depletion layer. The term depletion is due to the fact that near the junction, tion, the region is depleted i.e emptied of charge carriers (free electrons and holes) due to diffusion across the junction. The depletion layer is formed very quickly and is very thin as compared to the n region and the p region. Once PN junction is formed rmed and depletion layer is created, the diffusion of free electrons stops. In other words, the depletion layer acts as a barrier to the further movement of free electrons across the junction. The positive and negative charges set up an electric field which whi acts as a barrier to the free electrons in the n region. This is shown in fig.2. Fig.2 There exist a potential difference across the depletion layer known as barrier potential (V0). The typical barrier potential is approximately: For silicon, V0= 0.7 0 V, for germanium, V0= 0.3 V. 13  UNIT 1: Semiconductor Diodes and Regulated Power Supplies 1.1.12 Biasing a PN Junction In electronics, the term bias refers to the use of d.c. voltage to establish certain operating conditions for an electronic device. In case of a pn junction, there are following two bias conditions: conditions: 1. Forward biasing 2. Reverse biasing 1.1.13 Forward Biasing When external d.c. voltage applied to the junction is in such a direction that it cancels the potential barrier, thus permitting current flow, it is called forward biasing. To apply forward bias, the positive terminal of the battery is connected to p-type type and negative terminal isis connected to n-type n of the PN junction as shown in fig.3. The applied forward potential establishes an electric field which acts against the field due to potential barrier. b Therefore, the resultant field is weakened and the barrier height is reduced at the junction as shown in fig.3. Fig.3 As potential barrier voltage is very small (0.1 to 0.3 V), therefore, a small forward voltage is sufficient to completely eliminate eliminate the barrier. Once the barrier is eliminated by the forward voltage, junction resistance becomes almost zero and a low resistance path is established for the entire circuit. Therefore, current flows in the circuit. This is called forward current. 1.1.14 Reverse Biasing When the external d.c. voltage applied to the junction is in such a direction that potential barrier is increased, it is called reverse biasing. To apply reverse bias, the positive terminal of the battery is connected to n-type n and negative terminal to p-type type of the PN junction as shown in fig.4. 14  UNIT 1: Semiconductor Diodes and Regulated Power Supplies The applied reverse voltage establishes an electric field which acts in the same direction as the field due to potential barrier. Therefore, the resultant field at the junction is strengthened and the barrier height is increased as shown in fig.4. Fig.4 The increased potential barrier prevents the flow of charge carriers across the junction. Thus, a high resistance path is established for the entire circuit and hence the current does not flow. flow 1.1.15 Current rent Flow in a Forward Biased PN junction Fig .5 shows a forward biased PN junction. Fig.5 Under the influence of forward voltage, the free electrons in n-type n move towards the junction, leaving behind positively charged atoms. However, more ore electrons arrive from the negative terminal of the battery and enter the n-region region to take up their places. As the free electrons reach the junction, they become valence electron. As valence electron, they move through the holes in the p-region. p region. The valence electron move towards left in the p-region p region which is equivalent to holes 15  UNIT 1: Semiconductor Diodes and Regulated Power Supplies moving to right. When the valence electron reaches the left end of the crystal, they flow into the positive terminal of the battery. Summery 1. The free electrons from the negative negative terminal continue to pour into the n-region region while the free electrons in the n-region n region move towards the junction. 2. The electrons travel through the n-region n region as free electrons i.e current in n-region region is by free electrons. 3. When these electrons reach the junction, junction, they combine with holes and become valence electrons. 4. The electrons travel through p-region p region as valence electrons i.e. current in the p-region region is by holes. 5. When these valence electrons reach the left end of the crystal, they flow into the positive terminal term of the battery. 1.1.16 V-II Characteristics of PN Junction Volt-Ampere or V-I characteristics haracteristics of a PN junction is the curve between voltage across the junction and the circuit current. Usually voltage is taken across x-axis axis and current along y-axis. y Fig.6 shows the circuit arrangement for determining the V-I V characteristics of a PN junction. The characteristics can be explained under three conditions namely zero external voltage, forward bias and reverse bias. Fig.6 (i) Zero External Voltage When the external voltage is zero, i.e. circuit is open atK, K, the potential barrier at the junction does not permit current flow. Therefore, circuit current is zero as indicated by point O in fig.7. 16  UNIT 1: Semiconductor Diodes and Regulated Power Supplies (ii) Forward Bias With forward bias to the PN junction i.e. p-type type is connected to positive terminal and n-type type is connected to negative terminal, the potential barrier is reduced. At some forward voltage (0.7 V for Si and 0.3 V for Ge), the potential barrier is altogether eliminated and current starts flowing in the circuit. Fig.7 From now onwards, the current increases with the increase in forward voltage. Thus a rising curve OB is obtained with forward bias as shown in fig.7. From the forward characteristics, it is seen that at first (i.e. region OA), the current rrent increase very slowly and curve is non-linear. non linear. It is because the external applied voltage is used to overcome the potential barrier. However, once the external applied voltage exceeds the potential barrier voltage, the pn junction behaves like an ordinary ordinary conductor. Therefore, current rises very sharply with increase in voltage (region AB). The curve is almost linear. (iii) Reverse Bias With reverse bias to the PN junction i.e. p-type p type connected to negative terminal and n-type type connected to positive terminal, potential barrier at the junction is increased. Therefore, the junction resistance becomes very high and practically no current flows through the circuit. However, in practice, a very small current (of the order of µA) µA) flows in the circuit with reverse bias as shown 17  UNIT 1: Semiconductor Diodes and Regulated Power Supplies in fig.8.This is called reverse saturation satur current (Is)) and is due to the minority carriers. Fig.8 To the free electrons in p-type p and holes in n-type, type, the applied reverse bias appears as forward bias. Therefore, a small current flows in the reverse direction. If the reverse voltage is increased increased continuously, the kinetic energy of minority carriers may become high enough to knock out electrons from the semiconductor atom. At this stage breakdown of the junction occurs. This is characterized by a sudden rise of reverse current and a sudden sudde fall of the resistance of barrier region. This may destroy the junction permanently. 1.1.17 Crystal Diode Rectifier The electric power available for transmission is usually AC supply, but most of the electronic devices operate on DC supply. So, we need to convert AC supply into DC. The circuit used for converting AC into DC is known as rectifier. Most of the rectifier circuit which uses crystal diode along with some resistors for this conversion are known known as crystal diode rectifier. 1.1.18 Types of Rectifier ectifier • Half Wave Rectifier • Full Wave Rectifier 18  UNIT 1: Semiconductor Diodes and Regulated Power Supplies 1.1.19 Half Wave Rectifier The half wave rectifier conducts current only during the positive half cycles of the input supply voltage. The negative half cycles of the a.c. supply are suppressed. So no current current is conducted and hence no voltage appears across the load. So current always flows in one direction (i.e. d.c.) through the load. 1.1.20 Full Wave Rectifier In full wave rectifier, current flows through the load in the same direction ( i.e. d.c.) for both the half cycles of input a.c. supply voltage. 1.1.21 Construction and working principle of Half Wave Rectifier The half wave rectifier conducts current only during the positive half cycles of the input supply voltage. voltage The negative half cycles of the the a.c. supply are suppressed. So no current is conducted and hence no voltage appears across the load. So current always flows in one direction (i.e. d.c.) through the load. Circuit Connections of Half wave Rectifier Fig (1) Fig(2) Fig(1) shows the circuit connection of a half wave rectifier . Fig(2) shows the input and output waveform of a half wave rectifier . In this circuit a single crystal diode is used as a half wave rectifier . 19  UNIT 1: Semiconductor Diodes and Regulated Power Supplies The a.c. supply is applied in series with the diode D and load resistance RL. The a.c. supply is normally applied through a transformer. It gives two advantages. First it allows us to step up or step down the a.c. input voltage as required. Secondly, the transformer isolates the rectifier circuit from power line and thus reduces the risk of electric shock. Operation of Half Wave Rectifier During the positive half cycles of input supply voltage, end A becomes positive with respect to end B. This makes the diode D forward biased and hence it conducts current. During negative half cycles, end A becomes negative with respect to end B. So the diode D is reverse biased and hence conducts no current. So current flows through the diode only during the positive half cycles of the input a.c. voltage. Hence current flows through the load RL always in the same direction and d.c. output is obtained across RL. In half wave rectifier, output across the load is pulsating. So a filter circuit is used to smoothen the output. Applications of Half wave rectifier • For rectification applications • For signal demodulation applications • For signal peak applications Advantages of Half wave rectifier 1. The main advantage of half-wave rectifiers is in their simplicity. As they don’t require as many components, they are simpler and cheaper to setup and construct. 2. Simple (lower number of components), cheaper up front cost (as there is less equipment. Although there is a higher cost over time due to increased power 20  UNIT 1: Semiconductor Diodes and Regulated Power Supplies Disadvantages of Half Wave Rectifier 1. The output is pulsating, so a filter circuit is required. 2. The a.c. supply delivers power only half the time. So output is low 3. The output frequency of a half wave rectifier is equal to the input frequency i.e 50 Hz. fout = fin 1.1.22 Efficiency of a half Wave Rectifier The ratio of d.c. power output to the applied input a.c. power is known as rectifier efficiency. Rectifier efficiency Let the input supply voltage which appears across the secondary winding v = Vm sinθ Diode resistance = rf Load resistance = RL The diode conducts only during the positive half cycles of the a.c. supply. D.C. Power Output The output current is pulsating direct current. Iav=Idc 21  UNIT 1: Semiconductor Diodes and Regulated Power Supplies D.C. power A.C. Power input Rectifier Efficiency η will be maximum if rf is negligible as compared to RL Hence maximum efficiency = 40.6% 22  UNIT 1: Semiconductor Diodes and Regulated Power Supplies This means in Half wave rectifier, a maximum of 40.6% of a.c. power is converted into d.c. power. Ripple Factor In half-wave wave rectification, hence, Hence it is clear that a.c. component exceeds the d.c. component in the output of a half wave rectifier .It results more pulsation in the output. So half wave rectifier is ineffective for conversion of a.c into d.c. 1.1.23 Construction and working principle of Full wave Rectifier In full wave rectifier. Current flows through the load in the same direction (i.e. d.c.) for both the half cycles of input a.c. supply voltage. There are two types of circuits commonly used us for full-wave wave rectifications: 1. Centre-tap tap full wave rectifier 2. Full wave bridge rectifier Centre-tap Full wave Rectifier Fig (1) shows the circuit diagram of a Centre-tap Cent tap full wave rectifier and Fig(2) shows the input and output waveform of a center-tap center full wave rectifier . 23  UNIT 1: Semiconductor Diodes and Regulated Power Supplies Fig.1 Fig. 2 The circuit uses two diodes D1 and D2. A centre-tap tap secondary winding AB is connected with the two diodes such that each diode uses one half- half cycle of input a.c. voltage. That means diode D1 utilises the upper half of secondary winding for rectification and diode D2 uses the lower half. Operation During the positive half cycle of secondary voltage, the end A of the secondary winding becomes positive and end B negative. So diode D1 is forwardd biased and diode D2 is reverse biased. Hence, diode D1 conducts and diode D2 does not. The current flows through diode D1, load resistance RL and the upper half of the secondary winding OA. This is shown by the dotted arrows. During the negative half cycle of secondary voltage, the end A of the secondary winding becomes negative and end B positive. So diode D2 is forward biased and D1 is reverse biased. Hence D2 conducts while D1 does not. The conventional current flows through diode D2, load resistance resistanc RL and the lower half of the secondary winding OB as shown by the solid arrows. As we can see that current in the load RL flows in the same direction for both the half cycles of input supply voltage. So d.c. is obtained across RL. Peak Inverse Voltage Let Vm is the maximum voltage across the half secondary winding. Fig(2) Fig( shows the circuit at the instant secondary voltage reaches its maximum 24  UNIT 1: Semiconductor Diodes and Regulated Power Supplies value Vm in the positive direction. At this instant diode D1 is conducting and D2is not conducting. So whole of the secondary voltage appears across the non-conducting conducting diode. Hence the peak inverse voltage is twice the maximum voltage across the half secondary winding. PIV = 2 Vm Advantages • The ripple frequency is two times the input frequency • Efficiency is higher er • The large DC power output • Ripple factor is less • Higher output voltage • Higher transformer utilization factor • Utilizes both halves of the AC waveform Disadvantages • It is difficult to locate the centre tap on secondary winding. • The d.c. output is low as each diode utilizes only half of the secondary voltage. • The diodes must have high peak inverse voltage. 1.1.24 Bridge Rectifier Fig. 3 Fig. 4 Fig(3) shows the circuit connection of a full wave bridge rectifier and Fig(4) shows the input and output waveform of full-wave wave bridge rectifier. 25  UNIT 1: Semiconductor Diodes and Regulated Power Supplies The full wave bridge rectifier circuit contains four diodes D1, D2,D3 and D4, connected to form a bridge as shown in Fig(3). The a.c. supply to be rectified rectified is applied to the diagonally opposite ends of the bridge through the transformer. Between other two ends of the bridge, the load resistance RL is connected. Operation During the positive half-cycle half of secondary voltage, the end P of the secondary winding nding becomes positive and end Q negative. This makes diodes D1 and D3 forward biased while D2 and D4 are reverse biased. Hence only diodes D1 and D3 conducts. These two diodes will be in series through the load RL as shown in Fig. 5. 5 The conventional current urrent flows through load RL is shown by the dotted arrows. It may be seen that current flows from A to B through the load RL. RL Fig.5 During the negative half cycle of secondary voltage, end P becomes negative and end Q positive. This makes diodes D2 and D4and forward biased and diodes D1 and D3 are reverse biased. Hence only diodes D2 and D4 conduct.. These two diodes will be in series through the load RL as shown in Fig. 6. Fig.6 26  UNIT 1: Semiconductor Diodes and Regulated Power Supplies The conventional current flow through load RL is shown by the solid sol arrows. It may be seen that again the current flows from A to B through the load i.e. in the same direction as for the positive half-cycle. half cycle. Therefore, d.c. output is obtained across load RL. Peak Inverse Voltage The peak inverse voltage (PIV) of each diode is equal to the maximum secondary voltage of transformer i.e. Vm. Suppose during positive half cycle of input a.c., end P of secondary is positive and end Q negative. Under such conditions, diodes D1 and D3 are forward biased while diodes D2 and D4 are reverse biased. Since the diodes are considered ideal, diodes D1 and D3 can be replaced by wires as shown in Fig. 7 (i). This circuit is the same as shown in Fig. 7 (ii). Fig.7 (i) Fig.7 (ii) From the Fig. 7 (ii), it is clear that two reverse biased diodes (i.e., D2 and D4) and the secondary of transformer are in parallel. Hence PIV of each diode (D2 and D4) is equal to the maximum voltage (Vm) across the secondary. Similarly, during the next half cycle, D2 and D4 are forward biased while D1 and D3 will be reverse biased. It is easy to see that reverse voltage across D1 and D3 is equal to Vm. Hence, PIV = Vm Advantages 1. There is no need of centre taped transformer in full-wave full wave bridge rectifier. 2. The outputt is twice that of centre-tap centre tap circuit for the same secondary voltage. 3. The PIV is half that of the centre-tap centre circuit. 27  UNIT 1: Semiconductor Diodes and Regulated Power Supplies Disadvantages 1. It needs four diodes: diodes Since during each half cycle two diodes that conducts are in series so voltage drop in the internal resistance r of the rectifying unit is twice as in the centre-tap centre tap circuit. This is objectionable when secondary voltage is small. Output Frequency of Bridge Rectifier The output frequency of a full wave rectifier is double the input frequency. Fig. 8 As we know, a wave has a complete cycle when it repeats the same pattern. In Fig. 8 (i), the input a.c. completes onecycle from 0° – 360°. However, the full-wave wave rectified wave completes 2 cycles in this period As shown in Fig.8(ii). Therefore, output frequency frequency is twice the input frequency i.e.fout= 2fin 1.1.25 Efficiency of Full Wave Rectifier Fig.9 shows the full wave rectification process. Fig.9 28  UNIT 1: Semiconductor Diodes and Regulated Power Supplies Let v = Vmsinθθ (the a.c. voltage to be rectified) rf = diode resistance RL= Load resistance i= instantaneous current Efficiency η d.c. output power The output current is pulsating direct current. Therefore, in order to find the d.c. power, average current has ha to be found out. Let Idc= Iav =average current 29  UNIT 1: Semiconductor Diodes and Regulated Power Supplies So d.c.power output, a.c. input power The a.c. input power is given by: For a full-wave wave rectified wave, Efficiency The efficiency will be maximum if rf is negligiblee as compared to RL So Maximum Efficiency = 81.2% 30  UNIT 1: Semiconductor Diodes and Regulated Power Supplies This is double the efficiency effic of a half wave rectifier. Therefore, a full wave rectifier is twice as effective as a half-wave half rectifier. Ripple Factor The output of a rectifier consists of a d.c. component component and an a.c. component, which is also known as ripple. The a.c. component is undesirable and accounts for the pulsations in the rectifier output. The effectiveness of a rectifier depends upon the magnitude of a.c. component in the output. i.e., the smaller smaller this component, the more effective is the rectifier. The ratio of r.m.s. value of a.c. component to the d.c. component in the rectifier output is known as ripple factor i.e. Ripple factor = r.m.s. value of a.c component / value of d.c. component = Iac /Idc In full-wave wave rectification, Ripple Factor = 0.48 It is clear that d.c. component exceeds the a.c. component in the output of a full wave rectifier .This results in lesser pulsation in the output of a full wave rectifier as compared to a half wave rectifier . Therefore, full-wave full rectification is invariably used for conversion rectification. 1.1.26 What is a Filter Circuit? A rectifier is actually required to produce pure d.c. supply for using at various places in the electronics electronics circuits. However, the output of a rectifier 31  UNIT 1: Semiconductor Diodes and Regulated Power Supplies is pulsating. That means it contains both a.c. component and d.c. component. If such a pulsating d.c. is applied in an electronics circuit, it will produce a hum. So the a.c. component in the pulsating rectifier output is undesirable and must be kept away from the load. To do so, a filter circuit is used which removes the a.c. component and allows only the d.c. component. A filter circuit is a device which removes the a.c. component of rectifier output and allows only d.c. component to reach the load. A filter circuit is installed between the rectifier and the load as shown in figure given below Rectifier Output Pure D.C. Output A filter circuit is generally a combination of inductors(L) and capacitors(C).The filtering action of L and C depends upon the basic electrical principles. A capacitor C passes a.c. readily but does not pass d.c. at all. On the other hand, an inductor L opposes a.c. but allows d.c. to pass through it. Hence a suitable network of L and C can effectively remove the a.c. component and allows only the d.c. component to reach the load. 1.1.27 Types of Filter Circuits The three most commonly used filter circuits are: 1. Capacitor Filter 2. Choke input Filter 3. Capacitor input filter or π-Filter 1.1.28 Capacitor Filter The circuit diagram of a typical capacitor filter is shown in Fig.2(i).Its input and output waveform are shown in fig.2(ii) and Fig.2(iii) respectively. 32  UNIT 1: Semiconductor Diodes and Regulated Power Supplies Fig.2(i) Fig.2(ii) Fig.2(iii) The capacitor filter circuit consists of a capacitor C placed across the rectifier output in parallel with load resistance RL. The pulsating d.c. output of the rectifier is applied across the capacitor. As the rectifier output voltage increases, it charges the capacitor and also supplies current to the load. At the end of quarter cycle i.e. at point A in fig.2(iii), the capacitor is charged to the peak value of the rectifier voltage i.e. Vm. As the rectifier voltage now starts to decrease, the capacitor discharges through the load and voltage across it decreases. So voltage across RL also decreases. This is shown by the line AB in fig.2(iii). The voltage across the load will decrease only slightly because immediately the next voltage peak comes and recharges the capacitor. This process is repeated again and again and the output voltage waveform becomes ABCDEFG as shown in the fig.2(iii).We can see that very little ripple is left in the output. The output voltage is also higher as it remains substantially near the peak value of rectifier output voltage. 33  UNIT 1: Semiconductor Diodes and Regulated Power Supplies Advantages of Capacitor Filter 1. Capacitor filter circuits are extremely popular because of its low cost. 2. These filters are of very small size. 3. It has a little weight. 4. It has good characteristics. For small load currents up to 50mA this type of filter is preferred. It is commonly used in transistor radio battery eliminators. 1.1.29 Choke Input Filter (LC- Filter) Fig.3(i) shows a typical choke input filter circuit. The rectifier output which is applied as input to the choke input filter is shown in fig.3(ii) and the output of this filter circuit is shown in Fig.3(iii). Fig.3(i) Fig.3(ii) Fig.3(iii) The circuit of a choke input filter consists of a choke L connected in series with the rectifier output and a filter capacitor C, which is connected across the load resistance RL. Here in this fig.3(i) only a single filter section is shown. But normally several identical sections are used to reduce the pulsations as effectively as possible. The pulsating output of the rectifier is applied across the terminal 1 and 2 of the filter circuit. This pulsating output contains both a.c. and d.c. component. 34  UNIT 1: Semiconductor Diodes and Regulated Power Supplies As we know, the choke L offers a high resistance to the passage of a.c. component and passes the d.c. component readily. So most of the a.c. component appears across the choke L, while all the d.c. component passes through the choke L on its way to the load. This results in the reduced pulsation at terminal 3 as most of the a.c. component are blocked by the choke L now. At terminal 3, the rectifier output contains d.c. component and the remaining part of a.c. component which are managed to pass through the choke L. Now, the filter capacitor by passes the a.c. component but opposes the d.c. component to flow through it. Therefore, only the d.c. component reaches the load RL. 1.1.30 Capacitor input Filter or π-Filter Fig.4(i) shows the circuit diagram of a typical capacitor input filter or π– filter. Fig.4(ii) shows the rectifier output, which is applied as the filter input and the filter output wave forms. Fig.4(i) Fig.4(ii) As we can see in Fig.4(i) the shape of the circuit diagram of this filter circuit looks like π, hence it is also known as π-filter. In this circuit a filter capacitor C1 is connected across the rectifier output. A choke L is connected in series and another filter capacitor C2 is connected across the load. 35  UNIT 1: Semiconductor Diodes and Regulated Power Supplies Here only one filter section is shown, but most often several identical sections are used to improve the smoothing action. The pulsating output from the rectifier is applied across the input terminals 1 and 2 of the filter. The filtering action of the three components i.e. C1,L and C2 of the filter is described as below. The filter capacitor C1 offers low reactance to the a.c. component of the rectifier output while it offers infinite reactance to the d.c. component. Therefore, capacitor C1 bypasses an appreciable amount of a.c. component while the d.c. component continues its journey to the choke L. The choke L offers a high reactance to the a.c. component while it offers almost zero reactance to the d.c. component. Therefore, it allows the d.c. component to flow through it, while the unbypassed a.c component is blocked. The filter capacitor C2 bypasses the a.c. component which the chock L has failed to block. Hence, only the d.c. component appears across the load RL. 1.1.31 What is a Zener Diode? When the reverse bias voltage on a normal crystal diode is increased, a point known as Knee point is reached at a voltage known as breakdown voltage, when the reverse current increases sharply to a high value. This breakdown voltage is also known as Zener Voltage and the sharp increase in current is known as Zener current. The breakdown voltage of a diode depends on the amount of doping. In a heavily doped diode, the depletion layer is thin, so the breakdown of the junction occurs at a lower reverse bias voltage. When an ordinary crystal diode is properly doped so as to have a sharp breakdown voltage. It is known as zener diode. This particular voltage is known as zener breakdown voltage. Fig(1) shows the symbol of a zener diode. Fig. (1) 36  UNIT 1: Semiconductor Diodes and Regulated Power Supplies A zener diode is always used in reverse biased state. It has sharp breakdown voltage known as Zener voltage VZ. When forward biased ,it behaves just like ordinary crystal diode. Fig(2) shows the characteristics of a zener diode. 1.1.32 V-I Chracteristics of Zener diode (voltage-Current Characteristics) A Zener diode like an ordinary diode except that it is properly doped. So as to have sharp breakdown voltage. A zener diode is always reverse biased. When forward biased its characteristics are just those ordinary diode. The zener diode operated in this will have a relatively constant voltage across it. This permits the zener diode to be used as a voltage regulator. When a PN junction diode is reverse biased, the depletion layer becomes wider. If this reverse biased voltage across the diode is increased continually, the depletion layer becomes more and more wider. At the same time, there will be a constant reverse saturation current due to minority carriers. After certain reverse voltage across the junction, the minority carriers get sufficient kinetic energy due to the strong electric field. Free electrons with sufficient kinetic energy collide with stationary ions of the depletion layer and knock out more free electrons. These newly created electrons also get sufficient kinetic energy due to the same electric field, and they create more free electrons by collision cumulatively. Due to this commutative phenomenon very soon huge electrons will be created in the depletion layer, and the entire diode will become conductive. This type of breakdown in depletion layer is called Zener breakdown 37  UNIT 1: Semiconductor Diodes and Regulated Power Supplies 1.1.33 Zener Diode as a Voltage Regulator A zener diode can be used as a voltage stabilizer or voltage regulator to provide a constant voltage from a source whose voltage may vary over a particular range. The circuit connection is shown in fig(5). The zener diode of zener voltage VZ is connected reversely across the load resistance RL across which constant output voltage EO is required. The series resistance R is used to absorb the output voltage fluctuations, so as to maintain constant output voltage across RL . When the circuit is properly designed, the output voltage EO remains constant even though the input voltage Ei and load resistance RL may vary over a wide range. Fig. 5 38  UNIT 1: Semiconductor Diodes and Regulated Power Supplies Case 1: Ei Variable and RL constant Fig.6 Suppose the input voltage Ei increases, since the zener is in the breakdown battery of voltage VZ and the region, the zener diode is equivalent to a battery output voltage remains constant at VZ ( EO =VZ). The excess voltage is dropped across R. This will cause an increase in the value of total current I. The zener will conduct the increase of current in I, while the load current remains constant. Hence the output voltage remains constant irrespective of the change in input voltage Ei Summary (Case 1) Suppose Ei increases. Since zener is in breakdown region, VZ remains constant EO = VZ So EO remains constant Excess voltage drops across R. So I increases, since IL remains constant, IZ increases So the excess current is conducted by the zener diode and EO remains constant irrespective of the change in Ei 39  UNIT 1: Semiconductor Diodes and Regulated Power Supplies Case 2: Ei Constant and RL Variable Now suppose the input voltage volta Ei is constant but RL decreases. Since the zener diode is in breakdown region, voltage across it will remain constant at VZ.As the output voltage EO is equal to the zener voltage, So EO will also remain constant at VZ. When RL decreases, in order to maintain EO constant, current through the load resistance IL will increase. Since Ei is constant, total current I is also constant. So the increase in load current IL will come from a decrease in zener current IZ . Peak inverse voltage It is the maximum reverse voltage that a diode can with stand without destroying the junction. 40  UNIT 1: Semiconductor Diodes and Regulated Power Supplies Solved Problems 1. An a.c. voltage of peak value 20 V is connected in series with a silicon diode and load resistance of 500 Ω.. If the forward resistance of diode is 10 Ω, find: (i) peak current through diode (ii) peak output voltage What will be these values if the diode is assumed to be ideal? Solution: Peak input voltage = 20 V Forward resistance, rf = 10 Ω Load resistance, RL= 500 Ω Potential barrier voltage, V0 = 0.7 V The diode will conduct during the positive half-cycles half cycles of a.c. input voltage only. The equivalent circuit is shown in Fig.1(ii) (i). The peak current through the diode will occur at the instant when the input voltage reaches positive peak i.e. Vin = VF = 20 V. (ii) Peak output voltage: 41  UNIT 1: Semiconductor Diodes and Regulated Power Supplies (iii) Ideal Diode Case: 2. Find the current through the diode in the circuit shown in Fig. 2(i). Assume the diode to be ideal. Fig. 2 Solution: We shall use Thevenin’s theorem to find current in the diode. Referring to Fig. 2(i), Fig. 2 (ii) shows Thevenin’s equivalent circuit. Since the diode is ideal, it has zero resistance 42  UNIT 1: Semiconductor Diodes and Regulated Power Supplies 3. Calculate the current through 48 Ω resistor in the circuit shown in Fig. 3 (i). Assume the diodes odes to be of silicon and forward resistance of each diode is 1 Ω. Fig. 3 Solution: Diodes D1 and D3 are forward biased while diodes D2 and D4 are reverse biased. We can, therefore, consider the branches containing diodes D2 and D4 as “open”. Replacing diodes D1 and D3 by their equivalent circuits and making the branches containing diodes D2 and D4 open, we get the circuit shown in Fig. 3 (ii). As we know for a silicon diode, the barrier voltage is 0.7 V. 4. Determine the current I in the circuit shown shown in Fig. 4 (i). Assume the diodes to be of silicon and forward resistance of diodes to be zero. Fig. 4 43  UNIT 1: Semiconductor Diodes and Regulated Power Supplies Solution: The conditions of the problem suggest that diode D1 is forward biased and diode D2 is reverse biased. We can, therefore, consider the branch br containing diode D2 as open as shown in Fig. 4 (ii). Further, diode D1 can be replaced by its simplified equivalent circuit. 5. Find the voltage VA in the circuit shown in Fig. 5 (i). Use simplified model. Fig. 5 Solution: It appears that whenn the applied voltage is switched on, both the diodes will turn “on”. But that is not so. When voltage is applied, germanium diode (V0 = 0.3 V) will turn on first and a level of 0.3V is maintained across the parallel circuit. The silicon diode never gets the opportunity to have 0.7 V across it and, therefore, remains in open-circuit open circuit state as shown in Fig.5(ii). 6. Determine current through each diode in the circuit circuit shown in Fig. 7 (i). Use simplified model. Assume diodes to be similar. 44  UNIT 1: Semiconductor Diodes and Regulated Power Supplies Fig.7 Solution: The applied voltage forward biases each diode so that they conduct current in the same direction. Fig. 7 (ii) shows the equivalent circuit using simplified model. Referring to Fig. 7 (ii), 7. Determine the currents I1, I2 and I3 for the network shown in Fig. 8(i). Use simplified model for the diodes. Fig. 8 Solution: As we can see in Fig. 8 (i) both diodes D1 and D2 are forward biased. Using simplified model for the diodes, the circuit shown in Fig. 8(i) becomes the one shown in Fig. 8 (ii). Applying Kirchhoff’s voltage law to loop ABCDA in Fig. 8 (ii), we have, 45  UNIT 1: Semiconductor Diodes and Regulated Power Supplies 8. Determine if the diode (ideal) in Fig. 9 (i) is forward biased or reverse biased. Fig. 9 Solution: Let us assume that diode in Fig.9 (i) is OFF i.e. it is reverse biased. The circuit then becomes as shown in Fig. 9(ii). Referring to Fig. 9 (ii), we have, Now V1 – V2 = 2V is enough voltage to make the diode forward biased. Therefore, our initial assumption was wrong, and diode is forward biased. 9. Determine the state of diode diode for the circuit shown in Fig. 10 (i) and find ID and VD . Assume simplified model for the diode. Fig. 10 46  UNIT 1: Semiconductor Diodes and Regulated Power Supplies Solution: Let us assume that the diode is ON. Therefore, we can replace the diode with a 0.7V battery as shown in Fig. 10 (ii). Referring to Fig.10 Fig. (ii), we have, Since the diode current is negative, the diode must be OFF and the true value of diode current is ID =0 mA. Hence our initial assumption was wrong. In order to analyses the circuit properly, we should replace the diode in Fig. 10 (i) with an open circuit as shown in Fig.10 (iii). Fig.10 (iii) The voltage VD across the diode is: We know that 0.7V is required to turn ON the diode. Since VD is only 0.4V, the answer confirms that the diode is OFF. 47  UNIT 1: Semiconductor Diodes and Regulated Power Supplies Review Questions Short Answer Questions 1. Conductivity of Germanium is more than silicon. Why? 2. Define ripple factor. 3. What are the majority and minority carriers charge of n type and p type semiconductors? 4. Define intrinsic semiconductor .give an example? 5. Write a short note on breakdown mechanism of zener diode. 6. Write few lines about the rectification using crystal diode. 7. Define peak inverse voltage. 8. What is extrinsic semiconductor? 9. What is half wave rectifier and full wave rectifier? 10. Mention the type of filter circuit. 11. Why collector region is greater than emitter region? 12. Write the action of a filter circuit. 13. What is semiconductor diode? draw it’s symbol 14. What is rectifier? 15. What is meant by zener break down voltage? Big Questions 16. What is filter circuit? What are its types, explain its types. 17. Define zener break down, explain its characteristics. 18. Obtain an expression for the efficiency of half wave rectifier. 19. Obtain an expression for the efficiency of full wave rectifier. 20. Describe the working of zener diode working as regulator. 21. Explain the volt-ampere characteristics of zener diode. 22. Discuss the working principle of bridge rectifier. 23. Give the details of ripple factor. 24. Explain the working of a regulated power supply using zener diode with a necessary circuit. 25. Explain the working of bridge rectifier. 26. Discuss the action of a full-wave rectifier. 27. How P-N type semiconductors are formed? Give an example. 28. Explain the voltage-current characteristics of PN junction diode. 29. Construct the circuit diagram of zener diode. Explain its characteristics. 30. Draw and explain the circuit diagram of full wave rectifier using diodes. 48 UNIT 2 Transistors and Biasing 2.1 Transistor Two PN junctions diodes are connected together which make a new component called Transistor. When a third doped element is added to a crystal diode in such way that two PN-junction are formed, the resulting device is known as transistor. A Transistor is a three terminal semiconductor device that regulates current or voltage flow and acts as a switch or gate for signals. The only difference between an NPN and PNP is the direction of the arrow on the emitter. The arrow on an NPN points out, and on the PNP it points in. 2.1.1 Necessity for Transistors? Suppose that you have a FM receiver which grabs the signal you want. The received signal will obviously be weak due to the disturbances it would face during its journey. Now if this signal is read as it is, you cannot get a fair output. Hence we need to amplify the signal. Amplification means increasing the signal strength. This is just an instance. Amplification is needed wherever the signal strength has to be increased. This is done by a transistor. A transistor also acts as a switch to choose between available options. It also regulates the incoming current and voltage of the signals. 2.1.2 Constructional Details of a Transistor The Transistor is a three terminal solid state device which is formed by connecting two diodes back to back. Hence it has got two PN junctions. Three terminals are drawn out of the three semiconductor materials present in it. This type of connection offers two types of transistors. They are PNP and NPN which means an N-type material between two Ptypes 49  UNIT 2: Transistors and Biasing and the other is a P-type material between two N-types respectively. The construction of transistors is as shown in the following figure which explains the idea discussed above. The three terminals drawn from the transistor indicate Emitter, Base and Collector terminals. They have their functionality as discussed below. Emitter 1. The left hand side of the above shown structure can be understood as Emitter. 2. This has a moderate size and is heavily doped as its main function is to supply a number of majority carriers, i.e. either electrons or holes. 3. As this emits electrons, it is called as an Emitter. 4. This is simply indicated with the letter E. Base 1. The middle material in the above figure is the Base. 2. This is thin and lightly doped. 3. Its main function is to pass the majority carriers from the emitter to the collector. 4. This is indicated by the letter B. Collector 1. The right side material in the above figure can be understood as a Collector. 2. Its name implies its function of collecting the carriers. 3. This is a bit larger in size than emitter and base. It is moderately doped. 4. This is indicated by the letter C. 50  UNIT 2: Transistors and Biasing 2.1.3 The Symbols of PNP and NPN Transistors are as Shown Below Transistors are fundamentally three-terminal devices. On a bi-polar junction transistor (BJT), those pins are labeled collector (C), base (B), and emitter(E).The circuit symbols for both the NPN and PNP Bipolar Junction Transistor (BJT) are below: The only difference between an NPN and PNP is the direction of the arrow on the emitter. The arrow on an NPN points out, and on the PNP it points in. The arrow-head in the above figures indicated the emitter of a transistor. As the collector of a transistor has to dissipate much greater power, it is made large. Due to the specific functions of emitter and collector, they are not interchangeable. Hence the terminals are always to be kept in mind while using a transistor. In a Practical transistor, there is a notch present near the emitter lead for identification. The PNP and NPN transistors can be differentiated using a Multimeter. The following figure shows how different practical transistors look like. 2.1.4 Types of Transistor 1. NPN Transistor 2. PNP Transistor 2.1.5 N-P-N Transistor An n.p.n transistor is composed of two n-type semiconductors separated by a thin section of p-type. 51  UNIT 2: Transistors and Biasing 2.1.6 P-N-P Transistor A p.n.p transistor is formed by two p-type semiconductors separated by a thin section of n-type. 2.1.7 The Following Points can be noted in Each Type of Transistors 1. These are two P-N junctions so, a transistor may be considered as a combination of two diodes connected back to back. 2. There are three terminals, one taken from each type of semiconductor. 3. The middle layer is very thin. This is the most important factor in the function of a transistor. 4. A transistor (npn or pnp) has three sections of doped semiconductors. The section on one side is called emitter and the section on the opposite side is called collector. 5. The middle section is called the base and it forms two junctions between the emitter and collector. 2.1.8 Working principle of Emitter The section on one side that supplies charge carriers (electrons or holes) is called the emitter. The emitter is always forward biased w.r.t. base. so that it can supply a large number of majority carriers. 52  UNIT 2: Transistors and Biasing In fig. 2(i), the emitter (p-type) of PNP transistor is forward biased and supplies holes to its junction with the base. Fig.2(i) Similarly in fig.2(ii), the emitter (n-type) of npn transistor has a forward bias and supplies free electrons to its junction with the base. Fig.2(ii) 2.1.9 Working Principle of Collector The section on the other side that collects the charges is called the collector. The collector is always reverse biased. Its function is to remove charges from its junction with the base. In fig.2(i),the collector (p-type) of pnp transistor has a reverse bias and receives hole charges that flow in the output circuit. Similarly in fig.2 (ii), the collector (n-type) of npn transistor has reverse bias and receives electrons. 53  UNIT 2: Transistors and Biasing 2.1.10 Working principle of Base The middle section which forms two pn junctions between the emitter and collector is called the base. The base-emitter junction is forward biased and allows low resistance for the emitter circuit. The base-collector junction is reverse biased and provides high resistance in the collector circuit. 2.1.11 Transistor Action 1. Before discussing transistor action, it is important to keep in mind the following facts about the transistor: 2. The transistor has three regions, namely; emitter, base and collector. 3. The base is much thinner than the emitter while the collector is wider than both. 4. During transistor operation much heat is produced at collector junction. Hence the collector is made larger to dissipate heat. 5. The emitter is heavily doped. So, that it can inject a large number of charge carriers in to the base. The base is lightly doped and very thin; it passes the charge carriers to the collector. The collector is moderately doped. 6. The emitter–base junction is always forward biased where as collector base junction is always reverse biased. 7. The resistance of emitter is very small as compared to collector. 8. The transistor has two pn junctions that mean it is like two diodes. The junction between emitter and base may be called emitter-base diode or simply the emitter diode. The junction between base and collector may be called collector-base diode or simply collector diode. 9. The emitter diode is always forward biased and the collector diode is always reverse biased. The resistance of emitter diode is very small as compared to collector diode. Therefore, forward bias applied to the emitter diode is generally very small where as reverse bias on the collector diode is much higher. 2.1.12 Working of NPN Transistor The npn transistor with forward bias to emitter-base junction and reverse bias to collector base junction. 54  UNIT 2: Transistors and Biasing The forward bias causes the electrons in the emitter, to flow toward the base. This constitute emitter current IE. As these electrons flow through the p-type base. They tend to combine with the holes. In base only few electrons less than 5%combine with holes to constitute base current IB. The remaining electrons cross over in to the collector region to constitute collector current IC. In this way, the entire emitter current flow in collector circuit. IE= IB + IC 2.1.13 Working of PNP Transistor The forward bias causes the holes in the p-type emitter to flow towards the base. This constitutes the emitter current IE. As these holes cross into n- type base only few holes combine with the electrons. 55  UNIT 2: Transistors and Biasing Their main cross into collector region to constitute collector current. In this way, almost entire emitter current flows in the collector circuit. It may be noted the current conduction within pnp transistor is by holes. However, in the external connecting wires the current is still by electrons. 2.1.14 Transistor Characteristics Transistor Characteristics are the plots which represent the relationships between the current and the voltages of a transistor in a particular configuration. By considering the transistor configuration circuits to be analogous to two-port networks, they can be analyzed using the characteristic-curves which can be of the following types. 1. Input Characteristics: These describe the changes in input current with the variation in the values of input voltage keeping the output voltage constant. 2. Output Characteristics: This is a plot of output current versus output voltage with constant input current. 3. Current Transfer Characteristics: This characteristics curve shows the variation of output current in accordance with the input current, keeping output voltage constant. 2.1.15 Transistor Connections (Configurations) A transistor can be connected in a circuit in the following three ways. 1. Common Base Configuration 2. Common Emitter Configuration 3. Common Collector Configuration 2.1.16 Common Base (CB) Configuration of Transistor In CB Configuration, the base terminal of the transistor will be common between the input and the output terminals as shown by Figure 1. This configuration offers low input impedance, high output impedance, high resistance gain and high voltage gain. In this circuit, input is applied between emitter and base and output is taken from collector and base. 56  UNIT 2: Transistors and Biasing Base Current amplification factor The ratio of change in collector current to the change in emitter current at constant collector base voltage VCB is known as current amplification factor Collector Current With the idea above, let us try to draw some expression for collector current. Along with the emitter current flowing, there is some amount of base current IB which flows through the base terminal due to electron hole recombination. As collector-base junction is reverse biased, there is another current which is flown due to minority charge carriers. This is the leakage current which can be understood as Ileakage. This is due to minority charge carriers and hence very small. The emitter current that reaches the collector terminal is IE IC = α IE + Ileakage 2.1.17 Characteristics of Common Base Configuration (CB mode) 57  UNIT 2: Transistors and Biasing The complete electrical behavior of a transistor described by various current and voltages. These relationships displayed graphically and curves obtainedd are known as characteristics of transistor. Input characteristics It is a curve between emitter current IE and emitter base voltage VEB at constant collector base voltage VCB. The emitter current increases rapidly with small increase in emitter base voltage. It means that input resistance is very small. The emitter current is almost independent of collector base voltage. This leads to the conclusion that emitter current is almost independent of collector voltage. Input Resistance Output Characteristics stics It is the curve between collector current Ic and collector base voltage VCB at constant emitter curren IE. The collector current IC varies with VCB only at very low voltages (<1volt). The transistor is never operated in this region. When the value of VCB is raised above1-2volt, the collect or current becomes constant as indicated by straight horizontal curves. Emitter 58  UNIT 2: Transistors and Biasing current flows entirely to the collector terminal. The transistor is always operated in this region. Output Resistance Current Transfer Characteristics for CB Configuration of Transistor Figure shows the current transfer characteristics for CB configuration which illustrates the variation of IC with the IE keeping VCB as a constant. The resulting current gain has a value less than than 1 and can be mathematically expressed as 2.1.18 Common Emitter Connection (CE Mode) In this circuit, input is applied between base and emitter and output is taken from the collector and emitter. Here emitter of the transistor is common to both input ut and output circuits. Fig 9: Common Emitter (CE) Configuration 59  UNIT 2: Transistors and Biasing Base Current Amplification Factor To ratio of change in collector current to the change in base current is known as base current amplification factor Relation between β and α in CE Configuration Now Substitute equation (3) in (1) Dividing the numerator and denominator of RHS of above equation by ∆IE 60  UNIT 2: Transistors and Biasing 2.1.19 Characteristics of Common Emitter Configuration (CE mode) Fig 10: Transistor circuit in CE mode. Input Characteristics It is the curve between base current IB and Base emitter voltage VBE at constant collector emitter voltage VCE. The characteristics resemble that of a forward biased diode curve. This is expected since the base emitter section of transistor is a diode and it is forward biased. As compared to CB arrangement, IB increases less rapidly with VBE therefore, input resistance of a CE circuit is higher than that of CB circuit. Input Resistance It is the ratio of change in base emitter voltage ∆VBE to the change in base current ∆IB at constant ∆VCE 61  UNIT 2: Transistors and Biasing Output Characteristics It is the curve between collector current and collector emitter voltage at constant base current. Keeping the base current IB fixed at some value say,5 µA, note the collector current IC for various values of VCE. Then plot the reading on a graph. The test can be repeated for various values. The collector current IC varies with VCE for VCE between 0 and 1 volt only. After this collector current becomes almost constant. The transistor is always operated in this region above knee voltage. The value of VCE up to which collector current IC changes with VCE is called the knee voltage. Above the knee voltage, IC is almost constant, however a small increase in IC with increasing VCE is caused by the collector depletion layer getting wider. For any value of VCE above knee voltage, the collector current IC is approximately equal to β X IB Output Resistance It is the ratio of change in collector emitter voltage to the change in collector current at constant IB Current Transfer Characteristics This characteristic of CE configuration shows the variation of IC with IB keeping VCE as a constant. This can be mathematically given by 62  UNIT 2: Transistors and Biasing This ratio is referred to as common-emitter current gain and is always greater than 1. 2.1.20 Common Collector Configuration (CC mode) The configuration in which the collector is common between emitter and base is known as CC configuration. In CC configuration, the input circuit is connected between collector and base and the output is taken from the collector and emitter. The collector is common to both the input and output circuit and hence the name common collector connection or common collector configuration. Current Amplifier Factor (γ) The current amplification factor is defined as the ratio of the output current to the input current. In common collector configuration, the output current is emitter current IE, whereas the input current is base current IB. Thus, the ratio of change in emitter current to the change in base current is known as the current amplification factor. It is expressed by the relation. 63  UNIT 2: Transistors and Biasing Relation Between γ and α The ϒ is the current amplification factor of common collector configuration and the α is current amplification factor of common base connection. and, Substituting the value of ∆IB in above first equation, we get, The above relation shows that the value of Y is nearly equal to β. This circuit is mainly used for amplification because of this arrangement input resistance is high, and output resistance is very low. The voltage gain of the resistance is very low. This circuit arrangement is mainly used for impedance matching. 64  UNIT 2: Transistors and Biasing Collector Current We know that, Characteristics of Common Collector Configuration (CC mode) Input Characteristics The input characteristic of the common collector configuration is drawn between collector base voltage VCE and base current IB at constant emitter current voltage VCE. The value of the output voltage VCE changes with respect to the input voltage VBC and IB With the help of these values input characteristic curve is drawn. The input characteristic curve is shown below. 65  UNIT 2: Transistors and Biasing Input resistance Input resistance is defined as the ratio of change in input voltage or base voltage (VBC) to the corresponding change in input current or base current (IB), with the output voltage or emitter voltage (VEC) kept at constant. Output Characteristics The output characteristic of the common emitter circuit is drawn between the emitter-collector voltage VEC and output current IE at constant input current IB. If the input current IB is zero, then the collector current also becomes zero, and no current flows through the transistor. The transistor operates in active region when the base current increases and reaches to saturation region. The graph is plotted by keeping the base current IB constant and varying the emitter-collector voltage VCE, the values of output current IE are noticed with respect to VCE. By using the VCE and IE at constant IB the output characteristic curve is drawn. Output Resistance Output resistance is defined as the ratio of change in output voltage or emitter voltage (VEC) to the corresponding change in output current or emitter current (IE), with the input current or base current (IB) kept at constant. The output resistance of common collector amplifier is low. 66  UNIT 2: Transistors and Biasing Current amplification factor (γ) ( The current amplification factor is defined as the ratio of change in output current or emitter current IE to the change in input current or base current IB. It is expressed by γ. The current gain of a common collector amplifier is high. 2.1.21 Comparison of Transistor Configuration (CB, CE and CC) S. Characteristics Common Base Common Common No Emitter Collector 1 Input Low Low Very high resistance (about 100 Ώ) (about 750 Ώ) (about 750 K Ώ) 2 Output Very high High Low resistance (about 450 K Ώ) (about 45 K Ώ) (about 50 Ώ) 3 Voltage gain About 150 About 500 Less than one 4 Application For high frequency For audio For impedance application frequency matching application 5 Current gain No (Lessthen1) High Appreciable 67  UNIT 2: Transistors and Biasing 2.1.22 Transistor as an amplifier in CE arrangement Common emitter npn amplifier circuit is shown in fig. Battery VBB is connected in the input circuit in addition to the signal voltage. It always keeps emitter base junction forward biased. biased. Operation During the positive half cycle of the signal, the forward bias across the emitter base junction is increase. Therefore more electrons flow from the emitter to collector via the base. This causes increase in collector current. Due to this, s, the voltage drop across the collector load resistance will be greater. During the negative half cycle of the signal the forward bias across the emitter base junction is decreased. Therefore collector current decrease. This result in the decreased output outp voltage. Hence an amplified output is obtained across the load. In this type of configuration, the current flowing out of the transistor must be equal to the currents flowing into the transistor as the emitter current is given as Ie = Ic + Ib. 68  UNIT 2: Transistors and Biasing As the load resistance (RL) is connected in series with the collector, the current gain of the common emitter transistor configuration is quite large as it is the ratio of Ic/Ib is called Beta. A transistors current gain is given the Greek symbol of Beta, (β). As the emitter current for a common emitter configuration is defined as Ie = Ic + Ib, the ratio of Ic/Ie is called Alpha, given the Greek symbol of α. (Note: that the value of Alpha will always be less than unity) By combining the expressions for both Alpha, α and Beta, β the mathematical relationship between these parameters and therefore the current gain of the transistor can be given as 2.1.23 DC Load Line Analysis When the transistor is given the bias and no signal is applied at its input, the load line drawn at such condition can be understood as DC condition. Here there will be no amplification as the signal is absent. The circuit will be as shown below. 69  UNIT 2: Transistors and Biasing In the transistor circuit analysis, it is generally required to determine the collector current for various collector emitter voltages. It is known as load line method. In the figure, the straight line represents the d.c. load line. To obtain the load line, the two end points of the straight line are to be determined. Let those two points be A and B. To obtain A When collector emitter voltage VCE = 0, the collector current is maximum and is equal to VCC/RC. This gives the maximum value of VCE. This is shown as VCE = VCC − ICRC 0 = VCC − ICRC VCE = VCC / RC To obtain B When the collector current IC = 0, then collector emitter voltage is maximum and will be equal to the VCC. This gives the maximum value of IC. This is shown as VCE = VCC − IC RC = VCC [IC = 0] This gives the point B, which means (OB = VCC) on the collector emitter voltage axis shown in the above figure. Hence we got both the saturation and cutoff point determined and learnt that the load line is a straight line. So, a DC load line can be drawn. 2.1.24 Operating Point I) Saturation Point The point was the load line intersects the IB = IB (sat) curve is called saturation point. At saturation collector base junction no longer remains reverse biased and normal transistor action is lost. When a value for the maximum possible collector current is considered, that point will be present on the Y-axis, which is nothing but the saturation point. 70  UNIT 2: Transistors and Biasing ii) Cut off The point where the load line intersects the IB=0 curve is known as cut off. At this point IB = 0 and only small collector current exists. At cut off the base emitter ter junction no longer remains forward biased and normal action is Lost. The collector emitter voltage is nearly equal to VCC. VCE (cutoff) = VCC. When a value for the maximum possible collector emitter voltage is considered, that point will be present presen on the X-axis, axis, which is the cutoff point. When a line is drawn joining these two points, such a line can be called as Load line.. This is called so as it symbolizes the output at the load. This line, when drawn over the output characteristic curve, makes contact at a point called as Operating point. point iii) Quiescent Point This operating point is also called as quiescent point or simply Q-point. There can be many such intersecting points, but the Q-point Q point is selected in such a way that irrespective of AC signal signal swing, the transistor remains in active region. This can be better understood through the figure below. 71  UNIT 2: Transistors and Biasing The load line has to be drawn in order to obtain the Q-point. A transistor acts as a good amplifier when it is in active region and when it is made to operate at Q-point, faithful amplification is achieved. 2.1.25 Biasing The supply of suitable external dc voltage is called as biasing. Either forward or reverse biasing is done to the emitter and collector junctions of the transistor. These biasing methods make the transistor circuit to work in four kinds of regions such as Active region, Saturation region, Cutoff region and Inverse active region seldomused. This is understood by having a look at the following table. Emitter Junction Collector Junction Region of Operation Forward biased Forward biased Saturation region Forward biased Reverse biased Active region Reverse biased Forward biased Inverse active region Reverse biased Reverse biased Cutoff region Among these regions, Inverse active region, which is just the inverse of active region, is not suitable for any applications and hence not used. Active region This is the region in which transistors have many applications. This is also called as linear region. A transistor while in this region, acts better as an Amplifier. 72  UNIT 2: Transistors and Biasing This region lies between saturation and cutoff. The transistor operates in active region when the emitter junction is forward biased and collector junction is reverse biased. In the active state, collector current is β times the base current, i.e., IC=βIB Where, IC = Collector current β = current amplification factor IB = base current Saturation Region This is the region in which transistor tends to behave as a closed switch. The transistor has the effect of its collector and Emitter being shorted. The collector and Emitter currents are maximum in this mode of operation. The figure below shows a transistor working in saturation region. The transistor operates in saturation region when both the emitter and collector junctions are forward biased. As it is understood that, in the saturation region the transistor tends to behave as a closed switch, we can say that, IC=IE Where IC = collector current and IE = emitter current. 73  UNIT 2: Transistors and Biasing Cutoff region This is the region in which transistor tends to behave as an open switch. The transistor has the effect of its collector and base being opened. The collector, emitter and base currents are all zero in this mode of operation. The following figure shows a transistor working in cutoff region. The transistor operates in cutoff region when both the emitter and collector junctions are reverse biased. As in cutoff region, the collector current, emitter current and base currents are nil, we can write as IC=IE=IB=0 Where IC = collector current, IE = emitter current, and IB = base current. 2.1.26 What is Transistor Biasing? In order to achieve faithful amplification, to ensure that input circuit remains forward biased and output circuit remains reverse biased during all parts of the signal. This is knows as transistor biasing. 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 biasing. Transistor Biasing is the process of setting a transistor’s DC operating voltage or current conditions to the correct level so that any AC input signal can be amplified correctly by the transistor. Transistors are one of the most widely used semiconductor devices which are used for a wide variety of applications, including amplification and switching. However, to achieve these functions satisfactorily, a transistor must be supplied with a certain amount of current and/or voltage. 74  UNIT 2: Transistors and Biasing In order to understand transistor biasing, first we have to understand faithful amplification first .The process of raising the strength of a weak signal without changing its shape is known as faithful amplification . For faithful amplification, a transistor must satisfy the following conditions: 1. Proper zero signal collector current. 2. Minimum proper base-emitter voltage (VBE) at any instant. 3. Minimum proper collector-emitter voltage (VCE) at any instant. 2.1.27 Zero Signal Collector Current When no signal is applied, the input circuit is forward biased by the batter VBB. Therefore, dc collector current IC flows in the collector circuit. This is called zero signal collector current. 2.1.28 Stability Factor The rate of change of collector current IC with respect to the collector leakage current ICO at constant and IB is called stability factor. 2.1.29 Methods of Transistor Biasing The biasing in transistor circuits is done by using two DC sources VBB and VCC. It is economical to minimize the DC source to one supply instead of two which also makes the circuit simple. The commonly used methods of transistor biasing are 1. Base Resistor method 2. Collector to Base bias 3. Biasing with Collector feedback resistor 4. Voltage-divider bias All of these methods have the same basic principle of obtaining the required value of IB and IC from VCC in the zero signal conditions. 2.1.30 Base Resistor Method In this method, a resistor RB of high resistance is connected in base, as the name implies. The required zero signal base current is provided by 75  UNIT 2: Transistors and Biasing VCC which flows through RB. The base emitter junction is forward biased, as base is positive with respect to emitter. The required value of zero signal base current and hence the collector current (as IC = βIB) can be made to flow by selecting the proper value of base resistor RB. Hence the value of RB is to be known. The figure below shows how a base resistor method of biasing circuit looks like. Let IC be the required zero signal collector current. Therefore, Considering the closed circuit from VCC, base, emitter and ground, while applying the Kirchhoff’s voltage law, we get, Since VBE is quite small compared to VCC Hence 76  UNIT 2: Transistors and Biasing We know that VCC is a fixed known quantity and IB is chosen at some suitable value. As RB can be found directly, this method is called as fixed bias method. Stability Factor The rate of change of collector current IC with respect to the collector leakage current ICO at constant and IB is called stability factor. Here, IB is independent of IC so that, Therefore, S = +1 Due to large value of “s” in a fixed bias. It has poor thermal stability. Advantages • The biasing circuit is very simple as only one resistance RB is required. • Biasing conditions can easily be set and the calculations are simple. Disadvantage • This methods provides poor stabilization • The stability factor is very high. Therefore there are strong chances of thermal runway. 2.1.31 Voltage Divider Bias Method Among all the methods of providing biasing and stabilization, the voltage divider bias method is the most prominent one. Here, two resistors R1 and R2 are employed, which are connected to VCC and provide biasing. The resistor RE employed in the emitter provides stabilization. The name voltage divider comes from the voltage divider formed by R1 and R2. The voltage drop across R2 forward biases the base-emitter 77  UNIT 2: Transistors and Biasing junction. This causes the base current and hence collector current flow in the zero signal ignal conditions. The figure below shows the circuit of voltage divider bias method. Suppose that the current flowing through resistance R1 is I1. As base current IB is very small, therefore, it can be assumed with reasonable accuracy that current flowing flow through R2 is also I1. Now let us try to derive the expressions for collector current and collector voltage. Collector Current, IC From the circuit, it is evident that, voltage across resistance R2, Apply Kirchhoff’s voltage law to the base circuit, ci 78  UNIT 2: Transistors and Biasing since, IE _~ IC From the above expression. IC is practically independent of VBE {V2 >>VBE}. Thus IC is independent of transistor parameters and hence good stabilization ensured. Collector Emitter Voltage (VCE) ‟s law to the collector side, Apply kirchoff‟s Stability Factor In this circuit excellent stabilization is provided by RE If IC increases, V2 independent of IC. Hence VBE decreases. This in turn IB decreases. So that IC restored to original dividing ing numerator and denominator of R.H.S by RE If the ratio R0/RE is very small, then R0/RE can be neglected as compared to 1 and the stability factor becomes S = 1 this leads to maximum possible thermal stability. 79  UNIT 2: Transistors and Biasing 2.1.32 Biasing with Collector Feedback Resistor In this method, the base resistor RB has its one end connected to base and the other to the collector as its name implies. In this circuit, the zero signal base current is determined by VCB but not by VCC. It is clear that VCB forward biases the base-emitter junction and hence base current IB flows through RB. This causes the zero signal collector current to flow in the circuit. The below figure shows the biasing with collector feedback resistor circuit. Circuit Analysis The required value of RB needed to give the zero signal current IC can be determined as follows. Alternatively 80  UNIT 2: Transistors and Biasing Stability Factor Therefore this method provides better thermal stability than fixed bias. Advantages • It is simple method as it requires only one resistance resistance RB. • This circuit provides some stabilization of operating point Disadvantage This circuit does not provide good stabilization because stability factor is fairly high. This circuit provides a negative feedback which reduces the gain of the amplifier. 2.1.33 FET A Field Effect Transistor (FET) is a three-terminal three terminal semiconductor device. Its operation is based on a controlled input voltage. 1. Gate 2. Drain 3. Source. By appearance JFET and bipolar transistors are very similar. However, BJT is a current current controlled device and JFET is controlled by input voltage. Most commonly two types of FETs are available. • Junction Field Effect Transistor (JFET) • Metal Oxide Semiconductor FET (MOSFET) 2.1.34 Junction Field Effect Transistor The JFET is abbreviated as Junction Field Effect Transistor. Transistor JFET is just like a normal FET. The types of JFET are n-channel n channel FET and P-channel P FET. A p-type type material is added to the n-type n type substrate in n-channel n FET, whereas an n-type type material is added to the ptype substrate in p-channel FET. A junction field effect transistor is a three terminal semiconductor device in which current conduction is by one type of carrier.i.e.) electron (or) holes. 81  UNIT 2: Transistors and Biasing The functioning of Junction Field Effect Transistor depends upon the flow of majority carriers (electrons or holes) only. Basically, JFETs consist of an N type or P type silicon bar containing PN junctions at the sides. Following are Some Important Points to Remember About FET Gate − By using diffusion or alloying technique, both sides of N type bar are heavily doped to create PN junction. These doped regions are called gate (G). • Source − It is the entry point for majority carriers through which they enter into the semiconductor bar. • Drain − It is the exit point for majority carriers through which they leave the semiconductor bar. • Channel − It is the area of N type material through which majority carriers pass from the source to drain. Schematic Symbol of JFET 2.1.35 N-Channel FET The N-channel FET is the mostly used Field Effect Transistor. For the fabrication of Nchannel FET, a narrow bar of N-type semiconductor is taken on which P-type material is formed by diffusion on the opposite sides. These two sides are joined to draw a single connection for gate terminal. This can be understood from the following figure. These two gate depositions p−type materials form two PN diodes. The area between gates is called as a channel. The majority carriers pass through this channel. Hence the cross sectional form of the FET is understood as the following figure. Ohmic contacts are made at the two 82  UNIT 2: Transistors and Biasing ends of the n-type semiconductor bar, which form the source and the drain. The source and the drain terminals may be interchanged. 2.1.36 P-Channel JFET It has a thin layer of P type material formed on N type substrate. The following figure shows the crystal structure and schematic symbol of an N-channel JFET. The gate is formed on top of the P channel with N type material. At the end of the channel and the gate, lead wires are attached. Rest of the construction details are similar to that of N- channel JFET. 2.1.37 Construction A JFET consist of P-type or N-type silicon bar containing two PNjunction at the sides as shown in figure. The bar forms the conducting channel for the charge carries. If the bar is of N-type, it is called N-channel JFET. The bar is of P-type it is called P-channel JFET. The two PN-junction forming diodes are connected internally and a common terminal called gate is taken out. Other terminals are source and drain taken out from the bar. This JFET has essentially three terminals gate(G), drain (D) and source(S). 83  UNIT 2: Transistors and Biasing Working When a voltage VDS is applied between drain and source terminals and voltage on the gate is zero. The two PN junction at the sides of the bar establish depletion layers. The electron will flow from source to drain though a channel between the depletion layers. The size of these layers determines the width of the channel and hence the current conduction through the bar. When a reverse VGS is applied between the gates and source the width of the depletion layers is increased. This reduces the width of the conducting channel there by increasing the resistance of N – type bar. Consequently the current from source is decreased on the other hand, if the reverse voltage on the gate is decreased the width of the depletion layers also decreases. This increases the width of the conducting channel and hence source to drain current. Output Characteristics of JFET The output characteristics of JFET are drawn between drain current (ID) and drain source voltage (VDS) at constant gate source voltage (VGS) as shown in the following figure. Initially, the drain current (ID) rises rapidly with drain source voltage (VDS) however suddenly becomes constant at a voltage known as pinch-off voltage (VP). Above pinch-off voltage, the channel width becomes so narrow that it allows very small drain current to pass through it. Therefore, drain current (ID) remains constant above pinch-off voltage. 84  UNIT 2: Transistors and Biasing 2.1.38 Parameters of JFET 1. AC drain resistance 2. Tran conductance 3. Amplification factor 1. Ac drain resistance It is the radio of change in drain source voltage to the change in drain current at constant gate source voltage Drain resistance of JFET has a large value because above the pinch of voltage the change in ID is small for the change in VDS. 2. Transcondutance It is the ratio of change in drain current (∆ID) to the change in gate- source voltage (∆VGS) at constant drain source voltage. 3. Amplification factor It is the ratio of change in drain source voltage VDS to the change in gate source voltage (VGS) at constant drain current. 85  UNIT 2: Transistors and Biasing Advantages of JFET • It has very high input impedance • A JFET has a negative temperature co efficient of resistance. This avoids a risk of thermal run away. • It has very high power gain Pinch off voltage It is the minimum drain source voltage at which the drain current essentially becomes constant 86  UNIT 2: Transistors and Biasing Solved Problems 1. In a common base connection, IE = 1mA, IC = 0.95mA. Calculate the value of IB . Solution: 2. In a common base connection, current amplification factor is 0.9. If the emitter current is 1mA, determine the value of base current. Solution: 3. In a common base connection, IC = 0.95 mA and IB = 0.05 mA. Find the value of α. Solution : 4. In a common base connection, the emitter current is 1mA. If the emitter circuit cuit is open, the collector current is 50 µA. Find the total collector current. Given that α = 0.92. Solution : 87  UNIT 2: Transistors and Biasing 5. A transistor employs a 4 kΩk load and VCC = 13V. What is the maximum input signal if β = 100 ? Given Vknee = 1V and a change of 1V in VBE causes auses a change of 5mA in collector current. Solution: Collector supply voltage, VCC = 13 V Knee voltage, Vknee = 1 V Collector load, RC = 4 kΩ ∴ Max. allowed voltage across RC = 13 − 1 = 12 V ∴ Max. allowed collector current, iC =12 V /RC = 12 V/ 4 KΩ K = 3 mA Maximum base current, iB = iC / β = 3 mA / 100 = 30 µAµ Now Collector current / Base voltage (signal voltage) = 5 mA/V ∴ Base voltage (signal voltage) = Collector current / (5 mA/V ) = 3 mA /( 5 mA/V) = 600 mV 6. Fig. 3 (i) shows that a silicon transistor with β = 100 is biased by base resistor method. Draw the d.c. load line and determine the operating point. What is the stability factor ? Solution : Fig. 3 Solution Here, VCC = 6 V, RB = 530 kΩ, k RC = 2 kΩ D.C. load line Referring to Fig. 3 (i), VCE = VCC − ICRC When IC = 0, VCE = VCC = 6 V. This locates the first point B (OB = 6V) of the load line on collector-emitter collector emitter voltage axis as shown in Fig. 3 (ii). 88  UNIT 2: Transistors and Biasing When VCE = 0, IC = VCC/RC = 6V/2 kΩ = 3 mA. This locates the second point A (OA = 3mA) of the load line on the collector current axis. By joining points A and B, d.c. load line AB is constructed as shown in Fig. 3(ii). Operating point Q As it is a silicon transistor, therefore, VBE = 0.7V. .7V. Referring to Fig. 3(i), it is clear that: Fig. 3 (ii) shows the operating point Q on the d.c. load line. Its co- co ordinates are IC = 1mA and VCE = 4V. 7. (i). A germanium transistor is to be operated at zero signal IC = 1m A. If the collector supply sup VCC = 12V, what is the value of RB in the base resistor method? Take β = 100 (ii) If another transistor of the same batch with β = 50 is used, what will be the new value of zero signal IC for the same RB ? Solution: Given, VCC = 12 V, β = 100 As it is a Ge transistor, sistor, therefore, VBE = 0.3 V (i) Zero signal IC = 1 mA 89  UNIT 2: Transistors and Biasing (ii) Now β = 50 8. Calculate the values of three currents in the circuit shown in Fig. 4. Fig. 4 Solution: Applying Kirchhoff‘s voltage law to the base side and taking resistances in i kΩ and currents in mA, we have, 90  UNIT 2: Transistors and Biasing 9. In base bias method, how Q-point Q point is affected by changes in VBE and ICBO. Solution: In addition to being affected by change in β, the Q-point point is also affected by changes in VBE and ICBO in the base bias method. (i) Effect of VBE : The base-emitter-voltage voltage VBE decreases with the increase in temperature (and vice-versa). versa). The expression for IB in base bias method is given by; It is clear that decrease in VBE increases IB . This will shift the Q-point Q (IC = βIB and VCE = VCC – IC RC). The effect of change in VBE is negligible if VCC >> VBE (VCC atleast 10 times greater than VBE). (ii) Effect of ICBO: The reverse leakage current ICBO has the effect of decreasing the net base current and thus increasing the base voltage. voltage. It is because the flow of ICBO creates a voltage drop across RB that adds to the base voltage as shown in Fig. 7. Therefore, change in ICBO shifts the Q-point point of the base bias circuit. Fig. 7 91  UNIT 2: Transistors and Biasing However, in modern transistors, ICBO is usually less than 100 nA and its effect on the bias is negligible if VBB >> ICBO RB. 10. Fig. 8 (i) shows the base resistor transistor circuit. circuit. The device (i.e. transistor) has the characteristics shown in Fig. 8 (ii). Determine VCC, RC and RB . Fig. 8 Solution: rom the d.c load line, VCC = 20V. From 11. What fault is indicated in (i) Fig. 9 (i) and (ii) Fig. 9 (ii)? Fig. 9 92  UNIT 2: Transistors and Biasing Solution: (i) The obvious fault in Fig. 9(i) is that the base is internally open. It is because 3V at the base and 9V at the collector mean that transistor t is in cut-off state. (ii) The obvious fault in Fig. 9 (ii) is that collector is internally open. The voltage at the base is correct. The voltage of 9V appears at the collector because the ‘open’ prevents collector current. 12. Determine how much the Q-point Q point in Fig. 11 will change over a temperature range where β increases from 85 to 100 and VBE ,decreases from 0.7V to 0.6V. Fig. 11 Solution: For β = 85 and VBE = 0.7V As calculated in the above Question.12, IC = 1.73 mA and VCE = 14.6V. 14 For β = 100 and VBE = 0.6V 93  UNIT 2: Transistors and Biasing 13. Fig. 12 shows a silicon transistor biased by collector feedback resistor method. Determine the operating point. Given that β = 100. Fig. 12 Solution: kΩ RC = 1kΩ VCC = 20V, RB = 100 kΩ, Since it is a silicon transistor, nsistor, VBE = 0.7 V. Assuming IB to be in mA and using the relation, 94  UNIT 2: Transistors and Biasing 14. (i) It is required to set the operating point by biasing with collector feedback resistor at IC = 1mA, VCE = 8V. If β = 100, VCC = 12V, VBE = 0.3V, how will you do it? (ii) What will ill be the new operating point if β = 50, all other circuit values remaining the same ? Solution: Given, VCC = 12V, VCE = 8V, IC = 1mA, β = 100, VBE = 0.3V (i) To obtain the required operating point, we should find the value of RB. Now, collector load is (ii) Now β = 50, and other circuit values remain the same. 95  UNIT 2: Transistors and Biasing 15. It is desired to set the operating point at 2V, 1mA by biasing a silicon transistor with collector feedback resistor RB. If β = 100, find the value of RB. Solution: Fig. 13 96  UNIT 2: Transistors and Biasing Review Questions Short Answer Questions 1. What is operating point? give its importance. 2. Give the h-parameter values for common emitter configuration transistor. 3. Draw the input and output characteristics of a transistor in CB mode. 4. Define parameters of a transistor. 5. Write a short note on cut off and saturation points. 6. Explain load line analysis of a CE amplifier. 7. What is stability factor? 8. Mention the essential of biasing circuit. 9. Define current gain of a transistor in cc mode. 10. Define operating point of a transistor. 11. Define voltage gain of a transistor. 12. Mention the essential of a biasing transistor. 13. Define alpha of a transistor. 14. Define beta of a transistor. 15. Give the characteristics of CE amplifier. 16. What do you understand by feedback resistor bias? Big Questions 17. Define input impedance and current gain for a common emitter configuration. 18. Describe the transistor action in detail. 19. Describe the various methods for transistor biasing. 20. Write a note on cut off and saturation points. 21. What is biasing? Describe the voltage divider biasing of a transistor. 22. With the circuit diagram of a transistor CE mode explain its action. 23. Explain the h parameters of a transistor. 24. Explain the transistor action in detail. 25. Draw the circuit diagram of common emitter configuration and explain its input and output characteristics. 26. Draw the circuit diagram of common base configuration and explain its input and output characteristics. 27. Discuss in detail the working of NPN and PNP transistor. 28. Derive the relation between alpha and beta. 29. Explain the base resistor method for biasing. 30. How will you draw the input and output characteristics of a transistor in common emitter configuration. 97  UNIT 2: Transistors and Biasing 31. Draw the circuit diagram of a transistor CE configuration, explain its action. 32. Explain the application of CB amplifier. 33. Obtain the relationship between alpha and beta. 34. Explain the construction and working of JFET with its characteristics. 98 UNIT 3 Amplifiers 3.1 Amplification An electronic signal contains some information which cannot be utilized if doesn’t have proper strength. The process of increasing the signal strength is called as Amplification. Almost all electronic equipment must include some means for amplifying the signals. Uses of amplifiers 1. Medical devices 2. Scientific equipment 3. Automation 4. Military tools 5. Communication devices 6. Household equipment. For practical applications Amplification is done using Multi-stage amplifiers. A number of single-stage amplifiers are cascaded to form a Multi-stage amplifier. 3.1.1 Single-stage Transistor Amplifier When only one transistor with associated circuitry is used for amplifying a weak signal, the circuit is known as single-stage amplifier. A Single stage transistor amplifier has one transistor, bias circuit and other auxiliary components. The following circuit diagram shows how a single stage transistor amplifier looks like. 99  UNIT 3: Amplifiers 3.1.2 Working of a single stage Amplifier When a weak input signal is given to the base of the transistor as shown in the figure, a small amount of base current flows. Due to the transistor action, a larger current flows in the collector of the transistor. (As the collector current is β times of the base current which means IC = βIB). Now, as the collector current increases, the voltage drop across the resistor RC also increases, which is collected as the output. Hence a small input at the base gets amplified as the signal of larger magnitude and strength at the collector output. Hence this transistor acts as an amplifier. 3.1.3 Practical circuit of a Transistor Amplifier The circuit of a practical transistor amplifier is as shown below, which represents a voltage divider biasing circuit. The various prominent circuit elements and their functions are as described below. 3.1.4 Biasing Circuit The resistors R1, R2 and RE form the biasing and stabilization circuit, which helps in establishing a proper operating point. Input capacitor [cin] This capacitor couples the input signal to the base of the transistor. The input capacitor Cin allows AC signal, but isolates the signal source from 100  UNIT 3: Amplifiers R2. If this capacitor is not present, the input signal gets directly applied, which changes the bias at R2. Coupling Capacitor [Cc] This capacitor is present at the end of one stage and connects it to the other stage. As it couples two stages it is called as coupling capacitor. This capacitor blocks DC of one stage to enter the other but allows AC to pass. Hence it is also called as blocking capacitor. Due to the presence of coupling capacitor CC, the output across the resistor RL is free from the collector’s DC voltage. If this is not present, the bias conditions of the next stage will be drastically changed due to the shunting effect of RC, as it would come in parallel to R2 of the next stage. Emitter by-pass capacitor [CE] This capacitor is employed in parallel to the emitter resistor RE. The amplified AC signal is by passed through this. If this is not present, that signal will pass through RE which produces a voltage drop across RE that will feedback the input signal reducing the output voltage. 101  UNIT 3: Amplifiers The load resistor [RL] The resistance RL connected at the output is known as Load resistor. When a number of stages are used, then RL represents the input resistance of the next stage. Voltage gain (Av) It is the ratio of the amplified output voltage to the input voltage. Various circuit currents Let us go through various circuit currents in the complete amplifier circuit. These are already mentioned in the above figure. Base Current When no signal is applied in the base circuit, DC base current IB flows due to biasing circuit. When AC signal is applied, AC base current ib also flows. Therefore, with the application of signal, total base current iB is given by iB=IB+ib Collector current When no signal is applied, a DC collector current IC flows due to biasing circuit. When AC signal is applied, AC collector current ic also flows. Therefore, the total collector current iC is given by iC=IC+ic Where IC=βIB = zero signal collector current ic=βib = collector current due to signal Emitter Current When no signal is applied, a DC emitter current IE flows. With the application of signal, total emitter current iE is given by iE=IE+ie 102  UNIT 3: Amplifiers It should be remembered that IE=IB+IC ie=ib+ic As base current is usually very small, IE≅IC and ie≅ic These are the important considerations for the practical circuit of transistor amplifier. 3.1.5 Phase reversal The Common emitter amplifier is the only configuration that gives a phase inversion or phase rversal of 180o between the input and output signals. This is due to that as the input voltage rises, the current increases through the base. So that the collector current also increases. It tends to turn the transistor ‘on’. This results in the voltage between the collector and emitter terminals falling. In this way, an increase in voltage between the base and emitter gives the voltage drop between the collector and emitter terminals. So the phase of the two signals has been inverted. 3.1.6 DC & AC Equivalent circuits In a transistor amplifier, both dc and ac conditions prevail. The dc sources set up dc currents and voltages whereas the ac sources (ie. Signal) produces fluctuations in the transistor currents and voltages. 103  UNIT 3: Amplifiers Therefore, a simple way to analyze the action of a transistor is to split the analysis into two parts i.e. the dc analysis and an ac analysis. In the dc analysis, we consider all the dc sources at the same time and work out the dc currents and voltages in the circuit. On the other hand, for ac analysis, we consider all the ac sources sources at the same time and work out the ac currents and voltages. By adding the dc and ac currents and voltages, we get the total currents and voltages in the circuit. For example, consider the amplifier circuit as shown in figure. This circuit can be easily analyzed by splitting it into dc equivalent circuit and ac equivalent circuit. Fig 1 (i) DC Equivalent Circuit In the dc equivalent circuit of a transistor amplifier, only dc conditions are to be considered. i.e. it is presumed that no signal is applied. app As direct current cannot flow through a capacitor, therefore all the capacitors look like open circuits in the dc equivalent circuit. It follows, therefore, that in order to draw the equivalent dc circuit, the following two steps are applied to the transistor circuit: a. Reduce all ac sources to zero b. Open all the capacitors. 104  UNIT 3: Amplifiers Fig 2 Applying these two steps to the circuit as shown in figure 1, We get the dc equivalent circuit shown in Figure 2. We can easily calculate the dc currents and voltages from fro this circuit. (ii) AC Equivalent quivalent Circuit In the ac equivalent circuit of a transistor amplifier, only ac conditions are to be considered. Obviously, the dc voltage is not important for a such a circuit and may be considered zero. The capacitors are generally generally used to couple or bypass the ac signal. The designer intentionally selects capacitors that are large enough to appear as short circuits to the ac signal. It follows, therefore that in order to draw the ac equivalent circuit, the following two stepss are applied to the transistor circuit. a. Reduce all dc sources to zero (ie. Vcc = 0) b. Short all the capacitors. Applying these two steps to the circuit shown in figure 1, we get the ac equivalent circuit shown in figure 3. We can easily calculate the ac currents urrents and voltages from this circuit. Fig 3 105  UNIT 3: Amplifiers It may be seen that total current in any branch is the sum of dc and ac currents through that branch. Similarly, the total voltage across any branch is the sum of dc and ac voltages across that branch. 3.1.7 Voltage Gain The basic function of an amplifier is to raise the strength of an ac input signal. The voltage gain of the amplifier is the ratio of ac output voltage to the ac input signal voltage. Therefore, in order to find the voltage gain, we should consider onsider only the ac currents and voltages in the ciruit. For this purpose, we should look at the ac equivalent circuit of transistor amplifier. For facility of reference, the ac equivalent circuit of transistor amplifier is redrawn Fig. 4 Incidentally, ly, power gain is given by 106  UNIT 3: Amplifiers 3.1.8 Classification of Amplifiers Amplifiers are classified according to many considerations. i). Based on Number of Stages Depending upon the number of stages of Amplification, there are Single- stage amplifiers and Multi-stage amplifiers. • Single-stage Amplifiers − This has only one transistor circuit, which is a singlestage amplification. • Multi-stage Amplifiers − This has multiple transistor circuit, which provides multi-stage amplification. ii). Based on its Output Depending upon the parameter that is amplified at the output, there are voltage and power amplifiers. • Voltage Amplifiers − The amplifier circuit that increases the voltage level of the input signal, is called as Voltage amplifier. • Power Amplifiers − The amplifier circuit that increases the power level of the input signal, is called as Power amplifier. iii). Based on the Input Signals Depending upon the magnitude of the input signal applied, they can be categorized as Small signal and large signal amplifiers. • Small signal Amplifiers − When the input signal is so weak so as to produce small fluctuations in the collector current compared to its quiescent value, the amplifier is known as Small signal amplifier. • Large signal amplifiers − When the fluctuations in collector current are large i.e. beyond the linear portion of the characteristics, the amplifier is known as large signal amplifier. iv). Based on the Frequency Range Depending upon the frequency range of the signals being used, there are audio and radio amplifiers. 107  UNIT 3: Amplifiers • Audio Amplifiers − The amplifier circuit that amplifies the signals that lie in the audio frequency range i.e. from 20Hz to 20 KHz frequency range, is called as audio amplifier. • Power Amplifiers − The amplifier circuit that amplifies the signals that lie in a very high frequency range, is called as Power amplifier. v). Based on Biasing Conditions Depending upon their mode of operation, there are class A, class B and class C amplifiers. • Class A amplifier − The biasing conditions in class A power amplifier are such that the collector current flows for the entire AC signal applied. • Class B amplifier − The biasing conditions in class B power amplifier are such that the collector current flows for half-cycle of input AC signal applied. • Class C amplifier − The biasing conditions in class C power amplifier are such that the collector current flows for less than half cycle of input AC signal applied. • Class AB amplifier − The class AB power amplifier is one which is created by combining both class A and class B in order to have all the advantages of both the classes and to minimize the problems they have. vi). Based on the Coupling Method Depending upon the method of coupling one stage to the other, there are RC coupled, Transformer coupled and direct coupled amplifier. • RC Coupled amplifier − A Multi-stage amplifier circuit that is coupled to the next stage using resistor and capacitor (RC) combination can be called as a RC coupled amplifier. • Transformer Coupled amplifier − A Multi-stage amplifier circuit that is coupled to the next stage, with the help of a transformer, can be called as a Transformer coupled amplifier. • Direct Coupled amplifier − A Multi-stage amplifier circuit that is coupled to the next stage directly, can be called as a direct coupled amplifier. 108  UNIT 3: Amplifiers vii). Based on the Transistor Configuration Depending upon the type of transistor configuration, there are CE CB and CC amplifiers. • CE amplifier − The amplifier circuit that is formed using a CE configured transistor combination is called as CE amplifier. • CB amplifier − The amplifier circuit that is formed using a CB configured transistor combination is called as CB amplifier. • CC amplifier − The amplifier circuit that is formed using a CC configured transistor combination is called as CC amplifier. 3.1.9 Input Impedance When one CE amplifier is being used to drive another, the input impedance of the second amplifier will serve as the load resistance of the first. Therefore, in order to calculate the voltage gain (A) of the first amplifier stage correctly, we must calculate the input impedance of the second stage. The input impedance of an amplifier can be found by using the ac equivalent circuit of the amplifier as shown in figure Zin = R1 || R2 || Zin (base) Where, Zin = input impedance of the amplifier Zin (base) = input impedance of transistor base Now, Zin (base) = *βre’ 109  UNIT 3: Amplifiers The input impedance Zin is always less than the input impedance of transistor base Zin (base) 3.1.10 Multistage Amplifier The output of a single state amplifier is usually insufficient, though it is a voltage or power amplifier. Hence they are replaced by Multi-stage transistor amplifiers. In Multi-stage amplifiers, the output of first stage is coupled to the input of next stage using a coupling device. These coupling devices can usually be a capacitor or a transformer. This process of joining two amplifier stages using a coupling device can be called as Cascading. The following figure shows a two-stage amplifier connected in cascade. The overall gain is the product of voltage gain of individual stages. Av = Av1×Av2 = V2 / V1 X Vo / V2 Av = Vo/V1 Where AV = Overall gain, AV1 = Voltage gain of 1st stage, and AV2 = Voltage gain of 2nd stage. If there is n number of stages, the product of voltage gains of those n stages will be the overall gain of that multistage amplifier circuit. 3.1.11 Purpose of Coupling Device The basic purposes of a coupling device are 110  UNIT 3: Amplifiers • To transfer the AC from the output of one stage to the input of next stage. • To block the DC to pass from the output of one stage to the input of next stage, which means to isolate the DC conditions. 3.1.12 Types of Coupling Joining one amplifier stage with the other in cascade, using coupling devices form a Multi-stage amplifier circuit. There are four basic methods of coupling, using these coupling devices such as resistors, capacitors, transformers etc. Let us have an idea about them. 1. Resistance-Capacitance Coupling This is the mostly used method of coupling, formed using simple resistor- capacitor combination. The capacitor which allows AC and blocks DC is the main coupling element used here. The coupling capacitor passes the AC from the output of one stage to the input of its next stage. While blocking the DC components from DC bias voltages to effect the next stage. Let us get into the details of this method of coupling in the coming chapters. 2. Impedance Coupling The coupling network that uses inductance and capacitance as coupling elements can be called as Impedance coupling network. In this impedance coupling method, the impedance of coupling coil depends on its inductance and signal frequency which is jwL. This method is not so popular and is seldom employed. 3. Transformer Coupling The coupling method that uses a transformer as the coupling device can be called as Transformer coupling. There is no capacitor used in this method of coupling because the transformer itself conveys the AC component directly to the base of second stage. 111  UNIT 3: Amplifiers The secondary winding of the transformer provides a base return path and hence there is no need of base resistance. This coupling is popular for its efficiency and its impedance matching and hence it is mostly used. 4. Direct Coupling If the previous amplifier stage is connected to the next amplifier stage directly, it is called as direct coupling. The individual amplifier stage bias conditions are so designed that the stages can be directly connected without DC isolation. The direct coupling method is mostly used when the load is connected in series, with the output terminal of the active circuit element. For example, head-phones, loud speakers etc. 3.1.13 Role of Capacitors in Amplifiers Other than the coupling purpose, there are other purposes for which few capacitors are especially employed in amplifiers. To understand this, let us know about the role of capacitors in Amplifiers. 1. The Input Capacitor Cin The input capacitor Cin present at the initial stage of the amplifier, couples AC signal to the base of the transistor. This capacitor Cin if not present, the signal source will be in parallel to resistor R2 and the bias voltage of the transistor base will be changed. Hence Cin allows, the AC signal from source to flow into input circuit, without affecting the bias conditions. 2. The Emitter By-Pass Capacitor Ce The emitter by-pass capacitor Ce is connected in parallel to the emitter resistor. It offers a low reactance path to the amplified AC signal. In the absence of this capacitor, the voltage developed across RE will feedback to the input side thereby reducing the output voltage. Thus in the presence of Ce the amplified AC will pass through this. 112  UNIT 3: Amplifiers 3. Coupling Capacitor Cc The capacitor CC is the coupling capacitor that connects two stages and prevents DC interference between the stages and controls the operating point from shifting. This is also called as blocking capacitor because it does not allow the DC voltage to pass through it. In the absence of this capacitor, RC will come in parallel with the resistance R1 of the biasing network of the next stage and thereby changing the biasing conditions of the next stage. 3.1.14 Amplifier Consideration For an amplifier circuit, the overall gain of the amplifier is an important consideration. To achieve maximum voltage gain, let us find the most suitable transistor configuration for cascading. 1. CC Amplifier • Its voltage gain is less than unity. • It is not suitable for intermediate stages. 2. CB Amplifier • Its voltage gain is less than unity. • Hence not suitable for cascading. 3. CE Amplifier • Its voltage gain is greater than unity. • Voltage gain is further increased by cascading. The characteristics of CE amplifier are such that, this configuration is very suitable for cascading in amplifier circuits. Hence most of the amplifier circuits use CE configuration. 3.1.15 RC Coupled Amplifier The resistance-capacitance coupling is, in short termed as RC coupling. This is the mostly used coupling technique in amplifiers. 113  UNIT 3: Amplifiers Construction of a Two-Stage RC Coupled Amplifier The two stage amplifier circuit has two transistors, connected in CE configuration and a common power supply VCC is used. The potential divider network R1 and R2 and the resistor RE form the biasing and stabilization network. The emitter by-pass capacitor CE offers a low reactance path to the signal. • The resistor RL is used as a load impedance. • The input capacitor Cin present at the initial stage of the amplifier couples AC signal to the base of the transistor. • The capacitor CC is the coupling capacitor that connects two stages and prevents DC interference between the stages and controls the shift of operating point. Operation of RC Coupled Amplifier When an AC input signal is applied to the base of first transistor, it gets amplified and appears at the collector load RL which is then passed through the coupling capacitor CC to the next stage. This becomes the input of the next stage, whose amplified output again appears across its collector load. Thus the signal is amplified in stage by stage action. The important point that has to be noted here is that the total gain is less than the product of the gains of individual stages. This is because when a second stage is made to follow the first stage, the effective load resistance of the first stage is reduced due to the shunting effect of the input resistance of the second stage. Hence, in a multistage amplifier, only the gain of the last stage remains unchanged. 114  UNIT 3: Amplifiers As we consider a two stage amplifier here, the output phase is same as input. Because the phase reversal is done two times by the two stage CE configured amplifier circuit. Frequency Response of RC Coupled Amplifier Frequency response curve is a graph that indicates the relationship between voltage gain and function of frequency. The frequency response of a RC coupled amplifier is as shown in the following graph. From the above graph, it is understood that the frequency rolls off or decreases for the frequencies below 50Hz and for the frequencies above 20 KHz. whereas the voltage gain for the range of frequencies between 50Hz and 20 KHz is constant. We know that, Xc=1/2πfc It means that the capacitive reactance is inversely proportional to the frequency. Advantages of RC Coupled Amplifier The following are the advantages of RC coupled amplifier. • The frequency response of RC amplifier provides constant gain over a wide frequency range, hence most suitable for audio applications. • The circuit is simple and has lower cost because it employs resistors and capacitors which are cheap. • It becomes more compact with the upgrading technology. 115  UNIT 3: Amplifiers Disadvantages of RC Coupled Amplifier The following are the disadvantages of RC coupled amplifier. • The voltage and power gain are low because of the effective load resistance. • They become noisy with age. • Due to poor impedance matching, power transfer will be low. Applications of RC Coupled Amplifier The following are the applications of RC coupled amplifier. • They have excellent audio fidelity over a wide range of frequency. • Widely used as Voltage amplifiers • Due to poor impedance matching, RC coupling is rarely used in the final stages. 3.1.16 Transformer Coupled Amplifier We have observed that the main drawback of RC coupled amplifier is that the effective load resistance gets reduced. This is because, the input impedance of an amplifier is low, while its output impedance is high. When they are coupled to make a multistage amplifier, the high output impedance of one stage comes in parallel with the low input impedance of next stage. Hence, effective load resistance is decreased. This problem can be overcome by a transformer coupled amplifier. In a transformer-coupled amplifier, the stages of amplifier are coupled using a transformer. Let us go into the constructional and operational details of a transformer coupled amplifier. Construction of Transformer Coupled Amplifier The amplifier circuit in which, the previous stage is connected to the next stage using a coupling transformer, is called as Transformer coupled amplifier. The coupling transformer T1 is used to feed the output of 1st stage to the input of 2nd stage. The collector load is replaced by the primary winding of the transformer. The secondary winding is connected between the potential 116  UNIT 3: Amplifiers divider and the base of 2nd stage, which provides the input to the 2nd stage. Instead of coupling capacitor like in RC coupled amplifier, a transformer is used for coupling any two stages, in the transformer coupled amplifier circuit. The figure below shows the circuit diagram of transformer coupled amplifier. The potential divider network R1 and R2 and the resistor Re together form the biasing and stabilization network. The emitter by-pass capacitor Ce offers a low reactance path to the signal. The resistor RL is used as a load impedance. The input capacitor Cin present at the initial stage of the amplifier couples AC signal to the base of the transistor. The capacitor CC is the coupling capacitor that connects two stages and prevents DC interference between the stages and controls the shift of operating point. Operation of Transformer Coupled Amplifier When an AC signal is applied to the input of the base of the first transistor then it gets amplified by the transistor and appears at the collector to which the primary of the transformer is connected. The transformer which is used as a coupling device in this circuit has the property of impedance changing, which means the low resistance of a 117  UNIT 3: Amplifiers stage (or load) can be reflected as a high load resistance to the previous stage. Hence the voltage at the primary is transferred according to the turns ratio of the secondary winding of the transformer. This transformer coupling provides good impedance matching between the stages of amplifier. The transformer coupled amplifier is generally used for power amplification. Frequency Response of Transformer Coupled Amplifier The figure below shows the frequency response of a transformer coupled amplifier. The gain of the amplifier is constant only for a small range of frequencies. The output voltage is equal to the collector current multiplied by the reactance of primary. At low frequencies, the reactance of primary begins to fall, resulting in decreased gain. At high frequencies, the capacitance between turns of windings acts as a bypass condenser to reduce the output voltage and hence gain. So, the amplification of audio signals will not be proportionate and some distortion will also get introduced, which is called as Frequency distortion. Advantages of Transformer Coupled Amplifier The following are the advantages of a transformer coupled amplifier − • An excellent impedance matching is provided. • Gain achieved is higher. • There will be no power loss in collector and base resistors. • Efficient in operation. 118  UNIT 3: Amplifiers Disadvantages of Transformer Coupled Amplifier The following are the disadvantages of a transformer coupled amplifier − • Though the gain is high, it varies considerably with frequency. Hence a poor frequency response. • Frequency distortion is higher. • Transformers tend to produce hum noise. • Transformers are bulky and costly. Applications The following are the applications of a transformer coupled amplifier − • Mostly used for impedance matching purposes. • Used for Power amplification. • Used in applications where maximum power transfer is needed. 3.1.17 Direct Coupled Amplifier The other type of coupling amplifier is the direct coupled amplifier, which is especially used to amplify lower frequencies, such as amplifying photo- electric current or thermo-couple current or so. As no coupling devices are used, the coupling of the amplifier stages is done directly and hence called as Direct coupled amplifier. Construction The figure below indicates the three stage direct coupled transistor amplifier. The output of first stage transistor T1 is connected to the input of second stage transistor T2. 119  UNIT 3: Amplifiers The transistor in the first stage will be an NPN transistor, while the transistor in the next stage will be a PNP transistor and so on. This is because, the variations in one transistor tend to cancel the variations in the other. The rise in the collector current and the variation in β of one transistor gets cancelled by the decrease in the other. Operation The input signal when applied at the base of transistor T1, it gets amplified due to the transistor action and the amplified output appears at the collector resistor Rc of transistor T1. This output is applied to the base of transistor T2 which further amplifies the signal. In this way, a signal is amplified in a direct coupled amplifier circuit. Advantages The advantages of direct coupled amplifier are as follows. • The circuit arrangement is simple because of minimum use of resistors. • The circuit is of low cost because of the absence of expensive coupling devices. Disadvantages The disadvantages of direct coupled amplifier are as follows. • It cannot be used for amplifying high frequencies. • The operating point is shifted due to temperature variations. Applications The applications of direct coupled amplifier are as follows. • Low frequency amplifications. • Low current amplifications. 3.1.18 Comparisons of Different Types of Coupling Let us try to compare the characteristics of different types of coupling methods discussed till now. 120  UNIT 3: Amplifiers S.No Particular RC Coupling Transformer Direct Coupling Coupling 1 Frequency Excellent in Poor Best response audio frequency range 2 Cost Less More Least 3 Space and Less More Least Weight 4 Impedance Not good Excellent Good matching 5 Use For voltage For Power For amplifying amplification amplification extremely low frequencies 3.1.19 Comparison of Different Types of Amplifiers Any transistor amplifier, uses a transistor to amplify the signals which is connected in one of the three configurations. For an amplifier it is a better state to have a high input impedance, in order to avoid loading effect in Multi-stage circuits and lower output impedance, in order to deliver maximum output to the load. The voltage gain and power gain should also be high to produce a better output. 3.1.20 CB Amplifier The amplifier circuit that is formed using a CB configured transistor combination is called as CB amplifier. Construction The common base amplifier circuit using NPN transistor is as shown below, the input signal being applied at emitter base junction and the output signal being taken from collector base junction. The emitter base junction is forward biased by VEE and collector base junction is reverse biased by VCC. The operating point is adjusted with the help of resistors Re and Rc. Thus the values of Ic, Ib and Icb are decided by VCC, VEE, Re and Rc. 121  UNIT 3: Amplifiers Operation When no input is applied, the quiescent conditions are formed and no output is present. As Vbe is at negative with respect to ground, the forward bias is decreased, for the positive half of the input signal. As a result of this, the base current IB also gets decreased. The below figure shows the CB amplifier with self-bias circuit. As we know that, IC≅IE≅βIB Both the collector current and emitter current get decreased. The voltage drop across RC is VC=ICRC This VC also gets decreased. As ICRC decreases, VCB increases. It is because, VCB=VCC−ICRC 122  UNIT 3: Amplifiers Thus, a positive half cycle output is produced. In CB configuration, a positive input produces a positive output and hence input and output are in phase. So, there is no phase reversal between input and output in a CB amplifier. If CB configuration is considered for amplification, it has low input impedance and high output impedance. The voltage gain is also low compared to CE configuration. Hence CB configured amplifiers are used at high frequency applications. 3.1.21 CE Amplifier The amplifier circuit that is formed using a CE configured transistor combination is called as CE amplifier. Construction The common emitter amplifier circuit using NPN transistor is as shown below, the input signal being applied at emitter base junction and the output signal being taken from collector base junction. The emitter base junction is forward biased by VEE and collector base junction is reverse biased by VCC. The operating point is adjusted with the help of resistors Re and Rc. Thus the values of Ic, Ib and Icb are decided by VCC, VEE, Re and Rc. 123  UNIT 3: Amplifiers Operation When no input is applied, the quiescent conditions are formed and no output is present. When positive half of the signal is being applied, the voltage between base and emitter Vbe is increased because it is already positive with respect to ground. As forward bias increases, the base current too increases accordingly. Since IC = βIB, the collector current increases as well. The following circuit diagram shows a CE amplifier with self-bias circuit. The collector current when flows through RC, the voltage drop increases. VC=ICRC As a consequence of this, the voltage between collector and emitter decreases. Because, VCB=VCC−ICRC Thus, the amplified voltage appears across RC. Therefore, in a CE amplifier, as the positive going signal appears as a negative going signal, it is understood that there is a phase shift of 180o between input and output. CE amplifier has a high input impedance and lower output impedance than CB amplifier. The voltage gain and power gain are also high in CE amplifier and hence this is mostly used in Audio amplifiers. 124  UNIT 3: Amplifiers 3.1.22 CC Amplifier The amplifier circuit that is formed using a CC configured transistor combination is called as CC amplifier. Construction The common collector amplifier circuit using NPN transistor is as shown below, the input signal being applied at base collector junction and the output signal being taken from emitter collector junction. The emitter base junction is forward biased by VEE and collector base junction is reverse biased by VCC. The Q-values of Ib and Ie are adjusted by Rb and Re. Operation When no input is applied, the quiescent conditions are formed and no output is present. When positive half of the signal is being applied, the forward bias is increased because Vbe is positive with respect to collector or ground. With this, the base current IB and the collector current IC are increased. The following circuit diagram shows a CC amplifier with self-bias circuit. 125  UNIT 3: Amplifiers Consequently, the voltage drop across Re i.e. the output voltage is increased. As a result, positive half cycle is obtained. As the input and output are in phase, there is no phase reversal. If CC configuration is considered for amplification, though CC amplifier has better input impedance and lower output impedance than CE amplifier, the voltage gain of CC is very less which limits its applications to impedance matching only. 3.1.23 Comparison between CB, CE and CC Amplifiers Characteristic CE CB CC Input resistance Low (1K to 2K) Very low (30-150 Ω) High(20-500 KΩ) Output resistance Large (≈ 50 K) High (≈ 500 K) Low (50-1000 KΩ) Current gain B high α<1 High (1 + β) Voltage gain High (≈ 1500) High (≈ 1500) Less than one Power gain High (≈ 10,000) High (≈ 7500) Low (250-500) Phase between reversed same same input and output Due to the compatibility and characteristic features, the common-emitter configuration is mostly used in amplifier circuits. 126  UNIT 3: Amplifiers Solved Problems 1. For the transistor amplifier shown in Fig. 2, R1 = 10 kΩ, k R2= 5 kΩ, RC = 1 kΩ, RE = 2 kΩ k and RL= 1 kΩ.. (i) Draw d.c. load line (ii) Determine the operating point (iii) Draw a.c. load line. Assume VBE = 0.7 V. Solution: (i) d.c. load line: To draw d.c. load line, we require two end points i.e., maximum VCE point and maximum IC point. Maximum VCE = VCC = 15 V This locates the point A (OA = 5 mA) of the d.c. load line. Fig. 3 shows the d.c. load line AB. (ii) Operating point Q:: The Voltage across series combination of R1 and R2 is 15 V. Applying voltage age divider theorem, voltage across , R2 = 5 V. 127  UNIT 3: Amplifiers The voltage across R2 (= 5 kΩ k ) is 5 V i.e. V2 = 5 V. ∴ Operating point Q is 8.55 V, 2.15 mA. This is shown on the d.c. load line. (iii) a.c. load line: To draw a.c. load line, we require two end points viz. maximum collector- collector emitter voltage point and maximum collector current point when signal is applied. This locates the point D (OD = 19.25mA) on the iC axis. By joining points C and D, a.c. load line CD is constructed as shown in Fig. 4. Fig. 4 128  UNIT 3: Amplifiers 2. In the transistor amplifier shown in Fig. 5, RC = 10 kΩ, kΩ RL= 30 kΩ and VCC= 20V. The values R1 and R2 are such so as to fix the operating point at 10V, 1mA. Draw the d.c. and a.c. load lines. Assume RE is negligible. Fig. 5 Solution: (i) d.c. load line: For drawing d.c. load line, two end points such as maximum VCE point and maximum IC point are needed. Maximum VCE = 20 V. This locates the point B (OB = 20V) of the d.c. load line on the VCE axis. This locates the point A (OA = 2 mA) on the IC axis. By joining points A and B, the d.c. load line AB is constructed as shown in Fig.6. Fig. 6 129  UNIT 3: Amplifiers (ii) a.c. load line: To draw a.c. load line, we require two end points viz maximum collector- collector emitter voltage point and maximum collector current point when signal is applied. This locates the point C (OC = 2.33 mA) on the iC axis. By joining points C and D, a.c. load line CD is constructed as shown in the above Fig. 6. 3. In the circuit shown in Fig. 7, find the voltage gain. Given that β = 60 and input resistance Rin = 1 kΩ. Fig. 7 Solution: So far as voltage gain of the circuit is concerned, we need only RAC, β and Rin 130  UNIT 3: Amplifiers 4. In the circuit shown in Fig. 8, if RC = 10 kΩ, RL = 10 kΩ, k Rin= 2.5 kΩ, β =100, find the output voltage for an input voltage of 1mV r.m.s. Fig. 8 Solution: 5. In a transistor amplifier, whe whenn the signal changes by 0.02V, the base current changes by 10 µA A and collector current by 1mA. If collector load RC = 5 kΩ and RL= 10 kΩ,, find: (i) current gain (ii) input impedance (iii) a.c. load (iv) voltage gain (v) power gain. Solution: (i) Current gain: 131  UNIT 3: Amplifiers (ii) input impedance (iii) a.c. load (iv) voltage gain (v) Power gain 6. In Fig. 9, the transistor has β = 50. Find the output voltage if input resistance Rin= 0.5 kΩ. Ω. Fig. 9 Solution: β = 50, Rin = 0.5 kΩ 132  UNIT 3: Amplifiers 7. Determine the ac emitter resistance for the transistor circuit shown in Fig. 10. Fig. 10 Solution: 8. For the amplifier circuit shown in Fig. 11, find the voltage gain of the amplifier with (i) CE connected in the circuit (ii) CE removed from the circuit. Fig. 11 133  UNIT 3: Amplifiers Solution: (i) With CE connected: (ii) Without CE : 9. For the circuit shown in Fig. 12, find (i) a.c. emitter resistance (ii) voltage gain (iii) d.c. voltage across both capacitors. Fig. 12 Solution: (i) In order to find a.c. emitter resistance res re′, we shall first find D.C. emitter current IE. To find IE, we proceed as under: 134  UNIT 3: Amplifiers (ii) (iii) The d.c. voltage across input capacitor is equal to the d.c. voltage at the base of the transistor which is V2 = 1V. Therefore, d.c. voltage across Cin is 1V. Similarly, d.c. voltage across CE = d.c voltage at the emitter = VE = 0.3V. 10. (i) A multistage amplifier employs five stages each of which has a power gain of 30. What is the total gain of the amplifier in db? (ii) If a negative feedback of 100 db is employed, find the resultant gain. Solution: Given: Absolute gain of each stage = 30 No. of stages = 5 (i) Power gain of one stage in db = 10 log10 30 = 14.77 ∴ Total power gain = 5 × 14.77 = 73.85 db 11. A certain amplifier has voltage gain of 15 db. If the input signal voltage is 0.8V, what is the output voltage? Solution: 135  UNIT 3: Amplifiers 12. An amplifier rated at 40W output is connected to a 10Ω 10 speaker. (i) Calculate ate the input power required for full power output if the power gain is 25 db. (ii) Calculate the input voltage for rated output if the amplifier voltage gain is 40 db. Solution : (i) (ii) 136  UNIT 3: Amplifiers Review Questions Short Answer Questions 1. What are the characteristics of single stage common base transistor amplifier. 2. Write any two advantages of direct coupled amplifier . 3. What is meant by direct coupling? 4. Mention the merits and demerits of r-c coupled transistor. 5. Distinguish between amplifier and oscillator. 6. What is single stage amplifier? 7. What is multistage amplifier? 8. List the name of couplings and multistage amplifier. 9. Draw the single stage amplifier circuit. 10. Give the basic classification of amplifiers. 11. Define ‘Input impedance’ of a amplifier. 12. What is meant by transformer? 13. What is an amplifier? Mention the advantages of CE based amplifier. 14. What is meant by frequency response of an amplifier? Big Questions 15. Discuss the d.c and a.c equivalent circuit of a transistor. 16. Draw the circuit diagram of RC coupled transistor amplifier, discuss its operation and frequency response curve. 17. Explain the classification of amplifiers. 18. Discuss in brief about phase reversal of single stage amplifier 19. Discuss the circuit of single stage amplifier and explain its components. 20. State the function of each component used in this circuit. 21. Explain D.C load line and Q point. 22. List out the comparison of different types of couplings. 23. Describe the construction and working of direct couple amplifier circuit. 24. Draw the circuit diagram of direct coupled amplifier and explain its function. 25. Draw the circuit diagram of single stage amplifier and explain its working. 26. Explain on transformer coupled amplifier function and mention its merits and demerits 27. Explain the phase reversal of output voltage in amplifier. 137  UNIT 3: Amplifiers 28. Explain on R-C coupled amplifier function and mention its merits and demerits. 29. Briefly explain the classification of amplifier. 30. Explain the working of single stage CE amplifier with a neat diagram. 31. Explain the action of single stage amplifier. 32. With a neat circuit diagram, explain the action of RC coupled amplifier. 33. Explain the action of direct coupled amplifiers with a neat circuit diagram. 138 UNIT 4 Oscillators 4.1 Power Amplifier After the audio signal is converted into electrical signal, it has several voltage amplifications done, after which the power amplification of the amplified signal is done just before the loud speaker stage. This is clearly shown in the below figure. While the voltage amplifier raises the voltage level of the signal, the power amplifier raises the power level of the signal. Besides raising the power level, it can also be said that a power amplifier is a device which converts DC power to AC power and whose action is controlled by the input signal. The DC power is distributed according to the relation, DC power input = AC power output + losses 4.1.1 Power Transistor (Transistor Audio Power Amplifier) A transistor amplifier which raises the power level of the signals that have audio frequency range is known as transistor audio power amplifier For such Power amplification, a normal transistor would not do. A transistor that is manufactured to suit the purpose of power amplification is called as a Power transistor. A Power transistor differs from the other transistors, in the following factors. • It is larger in size, in order to handle large powers. 139  UNIT 4: Oscillators • The collector region of the transistor is made large and a heat sink is placed at the collector-base junction in order to minimize heat generated. • The emitter and base regions of a power transistor are heavily doped. • Due to the low input resistance, it requires low input power. Hence there is a lot of difference in voltage amplification and power amplification. So, let us now try to get into the details to understand the differences between a voltage amplifier and a power amplifier. 4.1.2 Difference between Voltage and Power Amplifiers Voltage Amplifier The function of a voltage amplifier is to raise the voltage level of the signal. A voltage amplifier is designed to achieve maximum voltage amplification. The voltage gain of an amplifier is given by Av = β (Rc/Rin)) The characteristics of a voltage amplifier are as follows – • The base of the transistor should be thin and hence the value of β should be greater than 100. • The resistance of the input resistor Rin should be low when compared to collector load RC. • The collector load RC should be relatively high. To permit high collector load, the voltage amplifiers are always operated at low collector current. • The voltage amplifiers are used for small signal voltages. Power Amplifier The function of a power amplifier is to raise the power level of input signal. It is required to deliver a large amount of power and has to handle large current. 140  UNIT 4: Oscillators 4.1.3 The Characteristics of a Power Amplifier are as follows • The base of transistor is made thicken to handle large currents. The value of β being (β > 100) high. • The size of the transistor is made larger, in order to dissipate more heat, which is produced during transistor operation. • Transformer coupling is used for impedance matching. • Collector resistance is made low. The comparison between voltage and power amplifiers is given below in a tabular form. S.No Particular Voltage Amplifier Power Amplifier 1 β High (>100) Low (5 to 20) 2 RC High (4-10 KΩ) Low (5 to 20 Ω) 3 Coupling Usually R-C Invariably transformer coupling coupling 4 Input voltage Low (a few m V) High (2-4 V) 5 Collector current Low (≈ 1 mA) High (> 100 mA) 6 Power output Low High 7 Output High (≈ 12 K Ω) Low (200 Ω) impendence 4.1.4 Performance quantities of Power Amplifier The primary objective of a power amplifier is to obtain maximum output power. In order to achieve this, the important factors to be considered are collector efficiency, power dissipation capability and distortion. Let us go through them in detail. i) Collector Efficiency This explains how well an amplifier converts DC power to AC power. When the DC supply is given by the battery but no AC signal input is given, the collector output at such a condition is observed as collector efficiency. The collector efficiency is defined as η=average a.c power output / average d.c power input 141  UNIT 4: Oscillators ii) Power Dissipation Capacity Every transistor gets heated up during its operation. As a power transistor handles large currents, it gets more heated up. This heat increases the temperature of the transistor, which alters the operating point of the transistor. So, in order to maintain the operating point stability, the temperature of the transistor has to be kept in permissible limits. For this, the heat produced has to be dissipated. Such a capacity is called as Power dissipation capability. Power dissipation capability can be defined as the ability of a power transistor to dissipate the heat developed in it. Metal cases called heat sinks are used in order to dissipate the heat produced in power transistors. iii) Distortion A transistor is a non-linear device. When compared with the input, there occur few variations in the output. In voltage amplifiers, this problem is not pre-dominant as small currents are used. But in power amplifiers, as large currents are in use, the problem of distortion certainly arises. Distortion is defined as the change of output wave shape from the input wave shape of the amplifier. An amplifier that has lesser distortion, produces a better output and hence considered efficient. 4.1.5 Expression for Collector Efficiency Collector is the main criteria for the power amplifiers. Collector efficiency = η= a.c power output/ d.c power input η = Po / Pdc Where, Pdc = VCCIC Therefore, PO = VCC IC VCC = rms value signal output voltage IC = rms value of output signal current 142  UNIT 4: Oscillators The ac power output PO = [(0.5X0.707) VCE [P-P]] [(0.5X0.707) IC[P-P]] Po = η= η= The greater the collector efficiency the better is the power amplifier. 4.1.6 Classification of Power Amplifiers Amplifiers are generally classified according to their mode of operation. • Class A Power amplifier − When the collector current flows at all times during the full cycle of signal, the power amplifier is known as class A power amplifier. • Class B Power amplifier − When the collector current flows only during the positive half cycle of the input signal, the power amplifier is known as class B power amplifier. • Class C Power amplifier − When the collector current flows for less than half cycle of the input signal, the power amplifier is known as class C power amplifier. There forms another amplifier called Class AB amplifier, if we combine the class A and class B amplifiers so as to utilize the advantages of both. 4.1.7 Transformer Coupled Class a Power Amplifier The construction of class a power amplifier can be understood with the help of below figure. This is similar to the normal amplifier circuit but connected with a transformer in the collector load. Here R1 and R2 provide potential divider arrangement. The resistor Re provides stabilization, Ce is the bypass capacitor and Re to prevent a.c. voltage. The transformer used here is a step-down transformer. 143  UNIT 4: Oscillators The high impedance primary of the transformer is connected to the high impedance collector circuit. The low impedance secondary is connected to the load (generally loud speaker). Transformer Action The transformer used in the collector circuit is for impedance matching. RL is the load connected in the secondary of a transformer. RL’ is the reflected load in the primary of the transformer. The number of turns in the primary are n1 and the secondary are n2. Let V1 and V2 be the primary and secondary voltages and I1 and I2 be the primary and secondary currents respectively. The below figure shows the transformer clearly. 144  UNIT 4: Oscillators Or Hence Here, N1 = n1 and N2 = n2 Therefore But V1 / I1 = RL’ = effective input resistance And V2 / I2 = RL = effective output resistance Therefore, Where, A power amplifier may be matched by taking proper turn ratio in step down transformer. Circuit Operation If the peak value of the collector current due to signal is equal to zero signal collector current, then the maximum a.c.power a.c.power output is obtained. So, in order to achieve complete amplification, the operating point should lie at the center of the load line. The operating point obviously varies when the signal is applied. The collector voltage varies in opposite phase to the collector collector current. The variation of collector voltage appears across the primary of the transformer. 145  UNIT 4: Oscillators Circuit Analysis The power loss in the primary is assumed to be negligible, as its resistance is very small. The input power under dc condition will be Under maximum capacity of class A amplifier, voltage swings from (Vce)max to zero and current from (Ic)max to zero. Hence Therefore, Therefore, Or 146  UNIT 4: Oscillators The efficiency of a class A power amplifier is nearly than 30% whereas it has got improved to 50% by using the transformer coupled class A power amplifier Advantages The advantages of transformer coupled class A power amplifier are as follows. • No loss of signal power in the base or collector resistors. • Excellent impedance matching is achieved. • Gain is high. • DC isolation is provided. Disadvantages The disadvantages of transformer coupled class A power amplifier are as follows. • Low frequency signals are less amplified comparatively. • Hum noise is introduced by transformers. • Transformers are bulky and costly. • Poor frequency response. Applications The applications of transformer coupled class A power amplifier are as follows. • This circuit is where impedance matching is the main criterion. • These are used as driver amplifiers and sometimes as output amplifiers. 4.1.8 Push-Pull Amplifiers The push-pull amplifier is power amplifier and is frequently employed in the output stages of electronic circuits. It is used whenever high output power at high efficiency is required. 147  UNIT 4: Oscillators Circuit Arrangement Two Transistors Tr1 and Tr2 placed back to back are employed. Both transistors are operated in class B operation. i.e. collector current is nearly zero in the absence of signal. The centre tapped secondary of driver transformer T1 supplies equal and opposite voltages to the base circuits of two transistors. The output transformer T2 has the center-tapped primary winding. The supply voltage VCC is connected between the base and this center tap. The loud speaker is connected across the secondary of this transformer. Operation The input signal appears across the secondary AB of driver transformer. Suppose during the first half cycle of the signal, end A becomes positive and end B negative. This will make the base-emitter junction of Tr1 and that of Tr2 forward biased. Therefore the circuit will conduct current due to Tr2 only. In the next half cycle, Tr1 is forward biased and Tr2 is reverse biased. Therefore, Tr1 conducts current. The output transformer combines these two collector currents to a form a sine wave. R’L = [ ]2 RL 148  UNIT 4: Oscillators Advantages 1. The efficiency of the circuit is quite high 2. A high a.c output power is obtained Disadvantages 1. Two Transistors have to be used 2. It requires two equal and opposite voltages 3. The circuit gives more distortion 4. Transformers used are bulky and expensive. 4.1.9 Amplifier Feedback An amplifier circuit simply increases the signal strength. But while amplifying, it just increases the strength of its input signal whether it contains information or some noise along with information. This noise or some disturbance is introduced in the amplifiers because of their strong tendency to introduce hum due to sudden temperature changes or stray electric and magnetic fields. Therefore, every high gain amplifier tends to give noise along with signal in its output, which is very undesirable. The noise level in the amplifier circuits can be considerably reduced by using negative feedback done by injecting a fraction of output in phase opposition to the input signal. 4.1.10 Feedback The process of injecting a fraction of output energy of some device back to the input is known as feedback. It has been found that feedback is very useful in reducing noise and making the amplifier operation stable. 4.1.11 Principle of Feedback Amplifier A feedback amplifier generally consists of two parts. They are the amplifier and the feedback circuit. The feedback circuit usually consists of resistors. The concept of feedback amplifier can be understood from the following figure. 149  UNIT 4: Oscillators From the above figure, the gain of the amplifier is represented as A. the gain of the amplifier is the ratio of output voltage Vo to the input voltage Vi. the feedback network extracts a voltage Vf = β Vo from the output Vo of the amplifier. This voltage is added for positive feedback and subtracted for negative feedback, from the signal voltage Vs. Now, Vi=Vs + Vf = Vs+βVo Vi=Vs−Vf = Vs−βVo The quantity β = Vf/Vo is called as feedback ratio or feedback fraction. Let us consider the case of negative feedback. The output Vo must be equal to the input voltage (Vs - βVo) multiplied by the gain A of the amplifier. Hence, (Vs−βVo)A=Vo Or AVs−AβVo = VoOr AVs = Vo(1+Aβ) Therefore, Vo / Vs=A / 1+Aβ Let Af be the overall gain (gain with the feedback) of the amplifier. 150  UNIT 4: Oscillators This is defined as the ratio of output voltage Vo to the applied signal voltage Vs, i.e., Af=Output voltage / Input signal voltage = Vo / Vs So, from the above two equations, we can understand that, The equation of gain of the feedback amplifier, with negative feedback is given by Af = A/1+Aβ The equation of gain of the feedback amplifier, with positive feedback is given by Af = A / 1−Aβ These are the standard equations to calculate the gain of feedback amplifiers. 4.1.12 Types of Feedback The process of injecting a fraction of output energy of some device back to the input is known as Feedback. It has been found that feedback is very useful in reducing noise and making the amplifier operation stable. Depending upon whether the feedback signal aids or opposes the input signal, there are two types of feedbacks used. 4.1.13 Positive Feedback The feedback in which the feedback energy i.e., either voltage or current is in phase with the input signal and thus aids it is called as Positive feedback. Both the input signal and feedback signal introduces a phase shift of 180o thus making a 360o resultant phase shift around the loop, to be finally in phase with the input signal. 151  UNIT 4: Oscillators Though the positive feedback increases the gain of the amplifier, it has the disadvantages such as • Increasing distortion • Instability It is because of these disadvantages the positive feedback is not recommended for the amplifiers. If the positive feedback is sufficiently large, it leads to oscillations, by which oscillator circuits are formed. 4.1.14 Negative Feedback The feedback in which the feedback energy i.e., either voltage or current is out of phase with the input and thus opposes it, is called as negative feedback. In negative feedback, the amplifier introduces a phase shift of 180o into the circuit while the feedback network is so designed that it produces no phase shift or zero phase shift. Thus the resultant feedback voltage Vf is 180o out of phase with the input signal Vin. 152  UNIT 4: Oscillators Though the gain of negative feedback amplifier is reduced, there are many advantages of negative feedback such as • Stability of gain is improved • Reduction in distortion • Reduction in noise • Increase in input impedance • Decrease in output impedance • Increase in the range of uniform application It is because of these advantages negative feedback is frequently employed in amplifiers. 4.1.15 Current Gain of Negative Feedback Amplifier In this method, a fraction of output current is feedback to the input of the amplifier. In other words, the feedback current (If) is proportional to the output current (Iout) of the amplifier. Fig. shows the principles of negative current feedback. This circuit is called current-shunt feedback circuit. A feedback resistor Rf is connected between input and output of the amplifier. This amplifier has a current gain of Ai without feedback. It means that a current I1 at the input terminals of the amplifier will appear as Ai I1 in the output circuit i.e., Iout = Ai I1. Now a fraction mi of this output current is feedback to the input through Rf. The fact that arrowhead shows the feed current being fed forward is because it is negative feedback. 153  UNIT 4: Oscillators Feedback current, If = mi Iout !" Feedback Fraction, mi = = # " " !" Note that negative current feedback reduces the input current to the amplifier and hence its current gain. In this method, a fraction of output current is fedback to the input of the amplifier. In other words, the arrowhead shows the feed current being fed forward is because it is negative feedback Referring, we have, Iin = I1 + If = I1 + mi Iout But Iout = Ai I1, where Ai is the current gain of the amplifier without feedback. Iin = I1 + mi Ai I1 (ä Iout = Ai I1) Current gain with negative current feedback is This equation looks very much like that for the voltage gain of negative voltage feedback amplifier. The only difference is that we are dealing with current gain rather than the voltage gain. The following points may be noted carefully: 154  UNIT 4: Oscillators • The current gain of the amplifier without feedback is Ai. However, when negative current feedback is applied, the current gain is reduced by a factor (1 + mi Ai). • The feedback fraction (or current attenuation) mi has a value between 0 and 1. • The negative current feedback does not affect the voltage gain of the amplifier. 4.1.16 Voltage Gain of Negative Feedback Amplifier Consider the negative voltage feedback amplifier shown in Fig. The gain of the amplifier without feedback is Av. Negative feedback is then applied by feeding a fraction mv of the output voltage e0 back to amplifier input. Therefore, the actual input to the amplifier is the signal voltage egminus feedback voltage mve0 i.e., Actual input to amplifier = eg - mv e0 The output e0 must be equal to the input voltage eg - mve0 multiplied by gain Av of the amplifier i.e (eg - mv e0) Av = e0 or Aveg - Av mv e0 = e0 or e0 (1+ Av mv) = Aveg $% '( or = $& ) '( *( 155  UNIT 4: Oscillators $% But is the voltage gain of the amplifier with feedback. + Therefore voltage gain with negative feedback is ', Avf = )', -, 4.1.17 Oscillators Many electronic devices require a source of energy at a specific frequency which may range from a few Hz to several MHz. This is achieved by an electronic device called an oscillator. An oscillator is a circuit which produces a continuous, repeated, alternating waveform without any input. Oscillators basically convert unidirectional current flow from a DC source into an alternating waveform which is of the desired frequency, as decided by its circuit components. In radio and television receivers, oscillators are used to generate high frequency wave in the tuning stage. Oscillators are also widely used in radar, electronic computers and other electronic devices. 4.1.18 Sinusoidal Oscillator An electronic device that generates sinusoidal oscillations of desired frequency is known as sinusoidal oscillator. Advantages 1. An oscillator can produce waves from small (20Hz) to extremely high frequencies (>100 MHz). 2. The frequency of oscillations can be easily changed when desired. 3. It has good frequency stability. i.e. once set remains constant for particular period of time. 4. It has very high efficiency 156  UNIT 4: Oscillators 4.1.19 Types of Sinusoidal Oscillators 1. Damped oscillations 2. Undamped oscillations. 4.1.20 Damped Oscillations The electrical oscillations whose amplitude goes on decreasing with time are called damped oscillations. Fig.1(i) shows waveform of damped electrical oscillations. The oscillator which generates these oscillations, losses some energy during each oscillation. As no means are provided to compensate for these losses, Hence the amplitude of the generated wave decreases gradually. It can be noted that the frequency of oscillations remains unchanged since it depends upon the constant of the electrical system. 4.1.21 Undamped Oscillations The electrical oscillations whose amplitude remains constant with time are called undamped oscillations. Fig.1(ii) shows waveform of undamped electrical oscillations. Although the electrical system in which these oscillations are being generated has also losses, but now right amount of energy is being 157  UNIT 4: Oscillators supplied to overcome the losses. Hence, the amplitude of the generated wave remains constant. An oscillator is mainly required to produce undamped oscillations for utilizing in various electronics equipment. 4.1.22 Barkhausen Criterion Barkhausen criterion is that in order to produce continuous undamped oscillations at the output of an amplifier, the positive feedback should be such that: mv Av =1 Once this condition is set in the positive feedback amplifier, continuous undamped oscillations can be obtained at the output immediately after connecting the necessary power supplies. 4.1.23 Essentials of Transistor Oscillator Figure shows the block diagram of an oscillator. Its essential components are: (i) Tank Circuit It consists of inductance (L) connected in parallel with capacitor (C). The frequency of oscillations in the circuit depends upon the values L and C. (ii) Transistor Amplifier The transistor amplifier receives d.c. power from the battery and changes it into a.c. power for supplying to the tank circuit. The oscillations occurring 158  UNIT 4: Oscillators in the tank circuit are applied to the input of the transistor amplifier. Because of the amplifying nature of the transistor, we get increased output of these oscillations. This amplified output of oscillations is due to the d.c. d. power supplied by the battery. The output of the transistor can be supplied to the tank circuit to meet the losses. (iii) Feedback Circuit The feedback circuit supplies a part of collector energy to the tank circuit in correct phase to aid the oscillations oscillations i.e. it provides positive feedback. 4.1.24 Different Types of Transistor Oscillators All oscillators under different names have similar function i.e. they produce continuous undamped output. However, the major difference between these oscillators lies lies in the method by which energy is supplied to the tank circuit to meet the losses. The following are the transistor oscillators commonly used at various places in electronic circuit: 1. Tuned collector oscillator 2. Colpitt’s Oscillator 3. Hartley Oscillator 4. Phase ase shift Oscillator 5. Wien Bridge Oscillator 6. Crystal Oscillator 4.1.25 25 Colpitt’s Oscillator It consists of two capacitors C1 and C2, placed across a common inductor L and the centre of the two capacitors is tapped. The tank circuit is made up of C1, C2 andd L. The frequency of oscillations is determined by the values of C1, C2 and L and is given by: Where, 159  UNIT 4: Oscillators 4.1.26 Circuit Operation When the circuit is turned on, the capacitor C1and C2are charged. The capacitors discharge through L, setting up oscillations of frequency determined by equ.(i). The output voltage of the amplifier appears across C1 and feedback voltage is developed across C2. The voltage across it is 180o out of phase with the voltage developed across C1(Vout) as shown in figure Note that C1-C2-L is also the feedback circuit that produces a phase shift of 180o 4.1.27 Feedback Circuit It can be noted that voltage feedback to the transistor provides positive feedback. A phase shift of 180o is produced by the transistor and a further phase shift of 180o is produced by C1 – C2 voltage divider. 160  UNIT 4: Oscillators In this way, feedback is properly phased to produce continuous undamped oscillations. 4.1.28 Feedback Fraction (MV) The amount of feedback voltage in Colpitt’s oscillator depends upon feedback fraction mv of the circuit. For this circuit, Feedback fraction, Or, 4.1.29 29 Hartley Oscillator The Hartley oscillator is similar to Colpitt’s oscillator with minor modifications. Instead of using tapped capacitors, two inductors L1 and L 2are placed ed across a common capacitor C. The centre of the inductors is tapped as shown in figure. The frequency of oscillations is determined by the values of L1, L2 and C and is given by: …………………………..(i) Where, Here M= mutual inductance between L1 and L2 Circuit Operation When the circuit is turned on, the capacitor is charged. When the capacitor is fully charged, it discharges through coils L1 and L2, setting up oscillations of frequency determined by equ(i). 161  UNIT 4: Oscillators The output voltage of the amplifier appears appe across L1and feedback voltage across L2. The voltage across L2 is 180o out of phase with the voltage developed across L1, as shown in figure It can be noted that voltage feedback to transistor provides positive feedback. A phase shift of 180o is produced pro by the transistorr and a further phase shift o of 180 is produced by L1– L2 voltage divider. In this way, feedback is properly phased to produce continuous undamped oscillations. Feedback Fraction: (MV) In Hartley oscillator, the feedback voltage is across L2 and output voltage is across L1. Feedback Fraction, 162  UNIT 4: Oscillators Or, 4.1.30 30 Phase Shift Oscillator It consists of a conventional single transistor amplifier and a RC phase shift network. The phase shift network consists of three sections of R1C1, R2C2 and R3C3. At some particular frequency f0, the phase shift in each RC section is 600 so that the total phase shift produced by the RC network is 1800. The frequency of oscillation is given by: ……………………………..(i) Where, Circuit Operation When the circuit rcuit is switched on, it produces oscillations of frequency determined by equ.(i) . The output E0 of the amplifier is fed back to RC feedback network. This network produces a phase shift of 1800 and a voltage Ei appears at its output which is applied to the t transistor amplifier. 163  UNIT 4: Oscillators The Feedback Fraction: (Mv) Mv =Ei / E0. The phase shift of 1800 is produced by the transistor amplifier and a further phase shift of 1800 is produced by the RC network. As a result, the phase shift around the entire loop is 3600. Advantages 1. It does not require transformers or inductors. 2. It can be used to produce very low frequencies. 3. The circuit provides good frequency stability. Disadvantages 1. It is difficult for the circuit to start oscillations as the feedback is generally small. 2. The circuit gives small output. 164  UNIT 4: Oscillators Solved Problems 1. The voltage gain of an amplifier without feedback is 3000. Calculate the voltage gain of the amplifier if negative voltage feedback is introduced in the circuit. Given that feedback fraction mv = 0.01. Solution: 2. The overall gain of a multistage amplifier is 140. When negative voltage feedback is applied, the gain is reduced to 17.5. Find the fraction of the output that is fedback to the input. Solution: 3. When negative voltage feedback is applied to an amplifier of gain 100, the overall gain falls to 50. (i) Calculate the fraction of the output voltage fedback. (ii) If this fraction is maintained, calculate the value of the amplifier gain required if the overall stage gain is to be 75. Solution : 165  UNIT 4: Oscillators 4. With a negative voltage feedback, an amplifier gives an output of 10 V with an input of 0.5 V. When feedback is removed, it requires 0.25 V input for the same output. Calculate (i) gain without feedback (ii) feedback fraction mv . Solution: 5. The gain of an amplifier without feedback is 50 whereas with negative voltage feedback, it falls to 25. If due to ageing, the amplifier gain falls to 40, find the percentage reduction in stage gain (i) without feedback and (ii) with negative feedback. Solution: 166  UNIT 4: Oscillators 6. An amplifier has a voltage amplification Av and a fraction mv of its output is fedback in opposition to the input. If mv = 0.1 and Aν = 100, calculate the percentage change in the gain of the system if Aν falls 6 db due to ageing. Solution: 7. An amplifier has a voltage gain of 500 without feedback. If a negative feedback is applied, the gain is reduced to 100. Calculate the fraction of the output fed back. If, due to ageing of components, the gain without feedback ack falls by 20%, calculate the percentage fall in gain with feedback. 167  UNIT 4: Oscillators Solution: Note that without negative feedback, the change in gain is 20%. However, when negative feedback is applied, the change in gain (4.7%) is much less. This shows that negative feedback provides voltage gain stability. 8. An amplifier has an open-loop gain Av = 100,000. A negative feedback of 10 db is applied. Find (i) voltage gain with feedback (ii) value of feedback fraction mv . Solution: 168  UNIT 4: Oscillators Review Questions Short Answer Questions 1. Write down the expression for frequency of oscillations condition for Hartley oscillator. 2. What are the important features of power transistor. 3. How does a Hartley oscillator differ from colpitt’s oscillator. 4. Distinguish between the oscillator and amplifier. 5. What is an emitter follower. 6. What is feed back? 7. What are positive and negative feedback? 8. Differentiate voltage amplifier and power amplifier 9. What is tank circuit? 10. What is oscillator? 11. Write the principle of phase shift oscillator. 12. What is power amplifier? 13. What are the advantages of negative feedback? 14. What is Barkhasan condition for oscillation? 15. What is damped and undamped oscillation? Big Questions 16. On the basis of mode of operation classify the power amplifiers. 17. Explain the principles of negative fees back in amplifiers. 18. Describe the functions of push pull amplifier. 19. Explain the effect of negative feedback on gain stability of an amplifier. 20. Derive an expression for current gain with negative feedback. 21. Discuss in detail the phase shift oscillator. 22. Discuss in detail the Hartley oscillator. 23. Discuss in detail the colpitt’s oscillator. 24. Differentiate voltage amplifier and power amplifier. 25. Describe the positive and negative feedback amplifier with neat diagram. 26. Discuss the transistor audio power amplifier. 27. Sketch and explain the circuit of push — pull amplifier. 28. Mention the difference between voltage and power amplifier. 29. Sketch the circuit diagram of Hartley oscillator and discuss its working. 30. Describe the action of Colpitt oscillator with circuit and also obtain an expression for its frequency of oscillation. 169  UNIT 4: Oscillators 31. Explain the push-pull circuit with neat diagram and mention their advantages and disadvantages. 32. Discuss in detail the phase shift oscillator. 33. Discuss in detail the Hartley oscillator. 170 UNIT 5 Operational Amplifier 5.1 Operational Amplifier An operational amplifier (OP-Amp) is a multi-stage, direct coupled, high gain negative feedback amplifier that has one or more differential amplifiers and its concluded with a level translator and an output stage. A voltage-shunt feedback is provided in an op-amp to obtain a stabilized voltage gain. Initially operational amplifier was used to perform mathematical operations such as addition, subtraction, integration and differentiation. Nowadays , the application of op-amp’s varies from ac and dc signal amplification to use in active filters, oscillators, comparators, voltage regulators, instrumentation and control systems, pulse generators, square wave generators and many more electronic circuits. 5.1.1 Block Diagram of Operational Amplifier An operational amplifier (OP-Amp) is a multistage amplifier and consists of three stages such as: differential amplifier (input stage) followed by a high gain CE amplifier stage and finally class B push-pull emitter follower as the output stage. 171  UNIT 5: Operational Amplifier 5.1.2 The Following Points may be noted about Op-Amps 1. The input stage of an OP-Amp is a differential amplifier (DA) and the output stage is typically a class B push pull emitter follower. 2. The internal stages of an OP-Amp are direct coupled i.e. no coupling capacitors are used. The direct coupling allows the OP- Amp to amplify d.c as well as a.c. signals. 3. An OP-Amp has very high input impedance (ideally infinite) and very low output impedance (ideally zero). The effect of high output impedance is that the amplifier will draw a very small current (ideally zero) from the signal source. The effect of very low output impedance is that the amplifier will provide a constant output voltage independent of current drawn from the source. 4. An OP-Amp has very high open loop voltage gain (ideally infinite); typically more than 200000. 5. The OP-Amps are almost always operated with negative feedback. It is because the open loop voltage gain of these amplifiers is very high and we can sacrifice the gain to achieve the advantages of negative feedback including large bandwidth (BW) and gain stability 5.1.3 Schematic Symbol of Operational Amplifier Figure shows the schematic symbol of an operational amplifier. The basic operational amplifier has five terminals: two high impedance inputs, one called the inverting input, marked with a negative sign (-) and the other one called non-inverting input, marked with a positive sign (+), two terminals for supply voltage +Vs and -Vs and one output terminal. A signal applied to + terminal will appear in the same phase at the output as at the input. 172  UNIT 5: Operational Amplifier 5.1.4 Differential Amplifier The key electronic circuit in an OP-AMP is the differential amplifier. In conventional amplifiers, the signal (generally single input) is applied at the input terminals and amplified output is obtained at the output terminals. A differential amplifier (DA) can accept two input signals and amplifies the difference between these two input signals. But in operational circuit that accepts two signals and amplifies the difference between these two signals such amplifier is called a differential amplifier. Fig 1 Figure shows the block diagram of a differential amplifier. There are two input voltages V1 and V2. This amplifier amplifies the difference between the two input voltages. Therefore, the output voltage is V0 = A (V1 – V2) 5.1.5 Basic circuit of Differential Amplifier Fig.2 (i) shows the basic circuit of a differential amplifier. It consists of two transistors Q1 and Q2 that have identical (ideally) characteristics. They share common positive supply VCC, common emitter resistor RE and common negative supply VEE. 173  UNIT 5: Operational Amplifier 1. The differential amplifier (DA) is a two-input terminal device using at least two transistors. There are two output terminals marked 1 (v out 1) and 2 (v out 2). 2. The transistors Q1 and Q2 are matched so that their characteristics are the same. The collector resistors (RC1 and RC2) are also equal. The equality of the matched circuit components makes the DA circuit arrangement completely symmetrical. 3. We can apply signals to a differential amplifier in the following two ways : a. The signal is applied to one input of DA and the other input is grounded. In this case it is called single-ended input arrangement. b. The signals are applied to both inputs of DA. In this case it is called dual-ended or double-ended input arrangement. 4. We can take output from DA in the following two ways : a. The output can be taken from one of the output terminals and the ground. In this case it is called single-ended output arrangement. b. The output can be taken between the two output terminals i.e. between the collectors of Q1 and Q2). In this case it is called double-ended output arrangement or differential output. 5. Generally the differential amplifier (DA) is operated for single- ended output. 5.1.6 Operation of Differential Amplifier For simplicity, we shall discuss the operation of single-ended input and double-ended output DA. Case – 1: Suppose the signal is applied to input 1 (i.e. base of transistor Q1) and input 2(i.e. base of transistor Q2 ) is grounded as shown in fig.3. 174  UNIT 5: Operational Amplifier The transistor Q1 will act in two ways: as a common emitter amplifier and as a common collector amplifier. As a common emitter amplifier, the input signal to Q1 will appear at output 1 (i.e. collector of Q1) as amplified inverted signal. As a common collector amplifier, the signal appears on the emitter of Q1 in phase with the input and only slightly smaller. Since the emitters of Q1 and Q2 are common, the emitter signal becomes input to Q2. Therefore, Q2 functions as a common base amplifier. As a result, the signal on the emitter Q2 will be amplified and appears on output 2 (i.e. collector of Q2) in phase with the emitter signal and hence in phase with the input signal (signal at input 1). This is shown in Fig.4. Case–2: Now suppose the signal is applied to input 2 (i.e. base of transistor Q2) and input 1 is grounded. Now Q2 acts as a common emitter amplifier and common collector amplifier while Q1 acts as a common base amplifier. Therefore, an inverted and amplified signal appears at output 2 and non- inverted, amplified signal appears at output 1. This is shown in Fig.4. When signal applied to the input of DA produces no phase shift in the output, it is called non-inverting input. In other words, for non-inverting input, the output signal is in phase with the input signal. 175  UNIT 5: Operational Amplifier When the signal applied to the input of DA produces 1800 phase shift, it is called inverting input. In other words, for inverting input, the output signal is 1800 out of phase with the input signal. 5.1.7 Common Mode and Differential Mode Signals The importance of a differential amplifier lies in the fact that the outputs are proportional to the difference between the two input signals. Thus the circuit can be use to amplify the difference between the two input signals or amplify only one input signal simply by grounding the other input. The input signals to a DA are defined as (i) Common mode and Differential mode signals. • Common Mode signals: When the input signals to a DA are in phase and exactly in amplitude, they are called common-mode signals. Therefore the difference between two signals is zero. V1 = V2 • Differential Mode signals: When the input signals to a DA are 180o out of phase and exactly equal in amplitude, they are differential mode signals. The differential mode signals are amplified by the differential amplifier. V1 = -V2 5.1.8 Common-Mode Rejection Ratio (CMRR) A differential amplifier should have high differential voltage gain (ADM) and very low common mode voltage gain (ACM) The ratio (ADM) / (ACM) is called common –mode rejection ratio (CMRR). 5.1.9 A.C. Analysis of Operational Amplifier i). Practical Op-Amp Figure shows the equivalent circuit of a practical OP-Amp. 176  UNIT 5: Operational Amplifier The input voltage difference V2 -V1 appears between the two input terminals and the output voltage is -A(V2 -V1) taken through the output impedance Zout. Since the voltage gain A of a practical OP-Amp OP Amp is very high, an extremely small input voltage (V2 -V1) will produce a large output(Vout). Since the input impedance, Zin is very high, a practical OP-Amp OP has very small input current. Since the output impedance, Zout is very low, the output voltage is practically independent of the value of load connected to OP-Amp. OP ii). Ideal Op-Amp Figure shows the a.c. equivalent circuit of an ideal OP-Amp. OP Amp. The Characteristics of an Ideal Op-Amp Op are 1. It has infinite voltage gain 2. Infinite input impedance 3. Zero output impedance 177  UNIT 5: Operational Amplifier The Consequence of these the properties of an Ideal Op-Amp Amp are 1. Since the voltage gain Av of an ideal OP-AmpAmp is infinite, it means that we can set Vin= 0V. 2. Since the input impedance Zin is infinite, an ideal OP-Amp OP has zero input current. 3. Since the output impedance Zout is zero, thee output voltage does not depends on the value of load connected to OP-Amp. OP 5.1.10 10 Bandwidth of an Operational Amplifier All electronic devices work only over a limited range of frequencies. This range of frequencies is called bandwidth. Every OP-Amp OP Amp has a bandwidth i.e. the range of frequencies over which it will work properly. The bandwidth of an OP-AmpAmp depends upon he closed-loop closed loop gain of the OP- OP Amp circuit. One important parameter is gain-bandwidth gain bandwidth product (GBW). It is defined as: ACL = Closed loop gain at frequency f2 funity = frequency at which the closed loop gain is unity It can be proved that the gain-bandwidth gain bandwidth product of an OP-Amp OP is constant. Since an OP-Amp Amp is capable of operating as d.c. amplifier, its bandwidth is (f2 – 0). The gain-bandwidth gain idth product of an OP-Amp OP is an important parameter because it can be used to find: • The maximum value of ACL at a given value of f2. • The value of f2 for a given value of ACL. 5.1.11 Slew Rate The slew rate of an OP-Amp OP Amp is a measure of how fast the output outpu voltage can change and its measured in volts per microsecond (V/µs). Since frequency is a function of time, the slew rate can be used to determine the maximum operating frequency of the OP-Amp OP Amp as follows: 178  UNIT 5: Operational Amplifier Maximum operating frequency, ./$0 123$ fmax = 4567 Here, Vpk is the peak output voltage. Biasing Op-amp The inputs to an Op-amp require some amount of dc biasing current for the transistors in the differential amplifier. The input bias current is defined as average of the dc base currents. Iin (bias) = (IB1 + IB2)/2 For example, if IB1 µA and IB2 = 75 µA then the input bias current is Iin (bias) = 85 +75 /2 = 80 µA This means that when no signal is applied the inputs of Op-amp will draw a dc current of 80 µA The fact that both transistors in the differential amplifier require an input biasing current leads to the following operating restriction. An op-amp will not work if either of its inputs is open. In figure shown, the non-inverting input is shows to have an open circuit between the op-amp and ground. The open circuit would not allow the dc biasing current required for the operation of the differential amplifier. The transistor associated with the inverting input would work, but not the one associated with the non-inverting input. 179  UNIT 5: Operational Amplifier Since the differential amplifier would network, the overall op-amp op circuit would not work. Thus an input bias current path must always be provided for both op-amp inputs. 5.1.12 Op-Amp Amp as Inverting Amplifier An OP amplifier can be operated as an an inverting amplifier as shown in figure An input signal Vin is applied through input resistor Ri to the minus input (inverting input). The output is fed back to the same inverting input through feedback resistor Rf . The plus input (non-inverting inverting input) input is grounded. The resistor Rf provides the negative feedback. Since the input signal is applied to the inverting input, the output will be inverted i.e. 180o out of phase as compared to the input. Hence, the name inverting amplifier. 5.1.13 Voltage Gain An OP-Amp Amp has infinite input impedance. This means that there is zero current at the inverting input. If there is zero current through the input impedance, then there must be no voltage drop between the inverting and non-inverting inverting inputs. This means that voltage at the inverting input is zero (point A) because the other input is grounded. The 0V at the inverting input terminal is referred to as virtual ground. The point A is said to be at virtual ground because it is at 0V but is not physically connected to the ground. This condition is shown in fig.2 (i). 180  UNIT 5: Operational Amplifier Referring to fig.2 (ii), the current I1 to the inverting input is zero. Therefore, current Iin flowing through Ri entirely flows through feedback resistor Rf . In other words, Now, And Since, Voltage gain, The negative sign indicates that output signal is inverted as compared to input signal. The following points may be noted noted about the inverting amplifier 1. The closed-loop loop voltage gain ACL of an inverting amplifier is the ratio of the feedback resistance Rf to the input resistance Ri .Hence, it is independent of the OP-Amp’s OP Amp’s internal open loop voltage gain. Thus the negative feedback stabilizes the voltage gain. 181  UNIT 5: Operational Amplifier 2. The inverting amplifier can be designed for unity gain if Rf = Ri 3. If Rf is some multiple of Ri , the amplifier gain is constant. Thus the inverting amplifier provides constant voltage gain. 5.1.14 Op-Amp Amp as Non-Inverting Non Amplifier Sometimes, we need to have an output signal of the same polarity as the input signal. In thiss case, the OP-Amp OP Amp is connected as non-inverting non amplifier as shown in fig.3. The input signal is applied to the non-inverting non input (+). Fig.3 The output is applied back to the input through the feedback circuit formed by feedback resistor Rf and input inp resistance Ri . Note that, resistors Rf and Ri form a voltage divider at the inverting input (-).This ).This produces negative feedback in the circuit. Note that Ri is grounded. Since the input signal is applied to the non- non inverting input, the output signal sig is non-inverted inverted i.e. the output signal will be in phase with the input signal. Hence, the name non-inverting non inverting amplifier. 5.1.15 Voltage Gain If we assume that we are not at saturation, the potential at point A, is the same as Vin. Since, the input impedance imp of OP-Amp Amp is very high, all of the current that flows through Rf also flows through Ri . Keeping these things in mind, we have, 182  UNIT 5: Operational Amplifier Voltage across Ri = Vin– 0; Voltage across Rf = Vout – Vin Now Current through Ri = Current through Rf Or Or Or Or Closed-loop loop voltage gain, The following points may be noted about the non-inverting non inverting amplifier: 1. The voltage gain of non-inverting non inverting amplifier also depends on the values of Rf &Ri 2. Thee voltage gain of a non-inverting non inverting amplifier can be made equal to greater than 1. 3. The voltage gain of a non-inverting non inverting amplifier will always be greater than the gain of an equivalent inverting amplifier by a value of 1. 4. The voltage gain is positive as the output output signal is in phase with the input signal. 5.1.16 16 Voltage Follower The voltage follower arrangement is a special case of non-inverting non amplifier where all of the output voltage is fed back to the inverting input as shown in figure 183  UNIT 5: Operational Amplifier It can be noted that Ri and Rf are removed from the non-inverting inverting amplifier and the output of the amplifier is shorted to the inverting input. The voltage gain for the voltage follower is calculated as follows: Thus the closed-loop loop voltage gain of the voltage follower is 1. The most important feature of the voltage follower is that it has a very input impedance. This makes it nearly ideal buffer amplifier to be connected between high-impedance high sources and low-impedance impedance loads. 5.1.17 Summing Amplifier The output voltage age of a summing amplifier is proportional to the negative of the algebraic sum of its input voltages. Hence, the name summing amplifier. A summing amplifier is an inverted OP-Amp OP Amp that can accept two or more inputs. Figure shows a three-input three input summing amplifier. ampl Three voltages V1, V2 and V3 are applied to the inputs and produce currents I1, I2 and I3. 184  UNIT 5: Operational Amplifier The inverting input of the OP-Amp OP Amp is at virtual ground (0 V) and there is no current to the input. So, the three input currents I1, I2 and I3 combine at the he summing point A and form the total current If which goes through Rf as shown in figure When all the three inputs are applied, the output voltage is If R1=R2=R3=R, then, we have, Thus the output voltage is proportional to the algebraic algebraic sum of the input voltages. If Rf =R1=R2=R3=R, then, we have Thus, when the gain of summing amplifier is unity, the output voltage is the algebraic sum of the input voltages. Application of Summing Amplifier There are a number of applications of summing amplifiers. Here we will discuss the following two applications: 1. As averaging amplifier 2. As Subtractor 185  UNIT 5: Operational Amplifier 1. As Averaging Amplifier By using the proper input and feedback resistor values, a summing amplifier can be designed to provide an output voltage that is equal to the average of input voltages. A summing amplifier will act as an averaging amplifier when both of the following conditions are met: • All input resistors (R1, R2 and so on ) are equal in value. • The ratio of any input resistor to the the feedback resistor is equal to the number of input circuits. Figure shows the circuit of averaging amplifier. Here all input resistors are equal in value (3 KΩ). K ). If we take the ratio of kΩ/1 kΩ =3. This is any input resistor to the feedback resistor, we get 3 kΩ/1 equal to the number of inputs to the circuit. Referring to fig.1, fig.1 the output voltage is given by: Now, 186  UNIT 5: Operational Amplifier Note that Vout is equal to the average of the three inputs. The negative sign shows the phase reversal. 5.1.18 Op -Amp as a Subtractor A summing amplifier can be used to provide an output voltage that is equal to the difference of two voltages. voltages. Such a circuit is called a Subtractor and is shown if figure 3 As we can see, this circuit will provide an output voltage that is equal to the difference between V1 and V2. The voltage V1 is applied to a standard inverting amplifier that has unity gain. n. Because of this, the output from the inverting amplifier will be equal to –V1. This output is then applied to the summing amplifier, also having unity gain along with V2. Thus output from second OP-Amp OP Amp is given by: The gain of the second stage in the the Subtractor can be varied to provide an output that is proportional to the difference between the input voltages. 5.1.19 Op-Amp as an Integrator An integrator is a circuit that performs integration of the input signal. The most important application of an integrator is to produce a ramp output voltage. Fig.4 shows the circuit of an OP-Amp OP integrator. 187  UNIT 5: Operational Amplifier Fig.4(i) Fig.4 (ii) When a signal is applied to the input of this this circuit, the output-signal output waveform will be the integration of input-signal input signal waveform. It consists of an OP-Amp, Amp, input resistor R and feedback capacitor C. Circuit Analysis Since point A in fig.4(i) is at virtual ground, the virtual ground equivalent circuit rcuit of operational integrator will be as shown in fig.4 (ii). Because of virtual ground and infinite impedance of the OP-Amp, OP Amp, all of the input current flows through the capacitor i.e. Now Also voltage across capacitor is So, Since, So, 188  UNIT 5: Operational Amplifier Or To find out the output voltage, we integrate both sides of the above equation to get, This equation shows that the output is the integral of the input with an inversion and scale multiplier of 1/RC. Output Voltage If a fixed voltage age is applied to the input of an integrator, the output voltage grows over a period of time, providing a ramp voltage. The output ramp voltage is opposite in polarity to the input voltage and is multiplied by a factor 1//RC. 5.1.20 Op-Amp as a Differentiator Different A differentiator is a circuit that performs differentiation of the input signal. That means, a differentiator produces an output voltage that is proportional to the rate of change of the input voltage. Its important application is to produce a rectangular ngular output from a ramp input. Fig.5 (i) shows the circuit of an OP-Amp Amp differentiator. Fig.5 (i) Fig .5 (ii) 189  UNIT 5: Operational Amplifier It consists of an OP-Amp, Amp, an input capacitor C and feedback resistor R. It can be noted that the placement of the capacitor and resistor differs from the integrator circuit. Circuit Analysis Since point A in fig.5 (i) is at virtual ground, the virtual-ground virtual ground equivalent circuit of the operational differentiator will will be as shown in fig.5 (ii). Because of virtual ground and infinite impedance of OP-Amp, OP Amp, all the input current ic flows through the feedback resistor R. i.e. And Also The above equation shows that output is the differentiation of the inputin with an inversion and scale multiplier of RC. If the input voltage is constant, dvi/dt is zero and the output voltage is zero. The faster the input voltage changes, the larger the magnitude of the output voltage. 5.1.21 Comparator Often we want to compare ompare one voltage to another to see which is larger. In this situation, a comparator may be used. A comparator is an OP-Amp OP circuit without negative feedback and takes advantage of very high open- open loop voltage gain. A comparator has two input voltages and one output voltage. 190  UNIT 5: Operational Amplifier Because of the high open loop voltage gain of an OP-Amp a very small difference voltage between the two inputs drives the amplifier to saturation. This is the key point in the working of comparator. Fig.6 shows the action of a comparator. The input voltages are v1(signal) & v2(reference voltage). Fig.6 If the differential input is positive, the circuit is driven to saturation & output goes to maximum positive value. Reverse happens, when the differential input goes negative i.e. now output is maximum negative. This circuit is called comparator because it compares v1 to v2 to produce a saturated positive or negative output voltage. Comparator Circuits A comparator circuit has the following two characteristics: 1. It uses no feedback so that the voltage gain is equal to the open loop voltage gain of OP-Amp. 2. It is operated in a non-linear mode 5.1.22 Signal Generator Signal Generators are the instrument which provides different output waveforms including sine wave, square wave, triangular wave, rectangular wave etc. It is also called an oscillator, since it produces periodic signals. The signal generator, which produces the periodic signal having a frequency of Audio Frequency (AF) range is called AF signal generator. The range of audio frequencies is 20Hz to 20KHz. Depending upon the different types of waveforms different types of signal generators are used. 191  UNIT 5: Operational Amplifier 5.1.23 Phase Shift Signal Generator In this type of signal generator sine wave is the output waveform. When we use Op-amp for RC phase shift oscillator, it functions as an inverting amplifier. Initially, the input wave has been into the RC network , due to which we get 180o of phase shift. This output of RC fed into the inverting terminal of the op-amp. As we know that the op-amp will produce a 180o of phase shift when function as an inverting amplifier. So we get 360o of phase shift in the output sine wave. This RC phase shift oscillator or signal generator using op-amp provides constant frequency even under the varying load conditions. R1 = R2 = R3 = 100K R4 = R5= 10K R6 = 4.7 K C1 = C2 = C3 = 100PK RC phase shift oscillator provides an accurate sine wave output. Here A is the input wave and D is the output wave. B and D are the phase shift of 90o and 180o. Here the feedback network is offering a phase shift of 180o. We are getting 60o phase shift from each RC network. The remaining 180o phase shift is 192  UNIT 5: Operational Amplifier generated by the op-amp amp in the inverting configuration. Totally we have o 360 phase. Hence the output waveform is sine si wave. The disadvantage of RC phase shift oscillator using op-amp op amp is that cannot be used for high frequency applications. 5.1.24 24 Hartley Signal Generator The Hartley signal generator circuit uses an operational amplifier for generating sine waves. Tank Tank circuit of Hartley uses two inductance coil and one capacitor. The voltage gain of this op-amp op amp oscillator can be expressed by the equation 9: A= 8 9; It provides the good frequency stability. The main advantage of using op- op amp is that the gainn of the oscillator can be individually adjusted using the feedback resistor (Rf) and input resistor (Ri) Frequency of the oscillator is given as F= 4√=> L = L1 + L2 To generate oscillation from this circuit, amplifier gain must be equal to t the ratio of two inductances = Av = = 193  UNIT 5: Operational Amplifier Advantages 1. The output amplitude is not proportional with the variable frequency range and the amplitude remains near constant. 2. Frequency is easily controllable using a trimmer instead of the fixed capacitor in the tank circuit, 3. Well suitable for RF range applications due to stable RF frequency generation 5.1.25 Square Wave Generator A square wave generator is an electronic circuit which generates square wave. This section discusses about op-amp based square wave generators. The circuit diagram of a op-amp based square wave generator is shown in the following figure Observe that in the circuit diagram shown above, the resistor R1 is connected between the inverting input terminal of the op-amp and its output of op-amp. So, the resistor R1 is used in the negative feedback. Similarly, the resistor R2 is connected between the non-inverting input terminal of the op-amp and its output. So, the resistor R2 is used in the positive feedback path. A capacitor C is connected between the inverting input terminal of the op- amp and ground. So, the voltage across capacitor C will be the input voltage at this inverting terminal of op-amp. 194  UNIT 5: Operational Amplifier Similarly, a resistor R3 is connected between the non-inverting input terminal of the op-amp and ground. So, the voltage across resistor R3 will be the input voltage at this non-inverting terminal of the op-amp. The operation of a square wave generator is explained below – • Assume, there is no charge stored in the capacitor initially. Then, the voltage present at the inverting terminal of the op-amp is zero volts. But, there is some offset voltage at non-inverting terminal of op-amp. Due to this, the value present at the output of above circuit will be +Vsat. • Now, the capacitor C starts charging through a resistor R1. The value present at the output of the above circuit will change to −Vsat, when the voltage across the capacitor C reaches just greater than the voltage (positive value) across resistor R3. • The capacitor C starts discharging through a resistor R1, when the output of above circuit is −Vsat. The value present at the output of above circuit will change to +Vsat, when the voltage across capacitor C reaches just less than (more negative) the voltage (negative value) across resistor R3. Thus, the circuit shown in the above diagram will produce a square wave at the output as shown in the following figure – From the above figure we can observe that the output of square wave generator will have one of the two values: +Vsat and −Vsat−Vsat. So, the output remains at one value for some duration and then transitions to another value and remains there for some duration. In this way, it continues. 195  UNIT 5: Operational Amplifier 5.1.26 Triangular Wave Generator A triangular wave generator is an electronic circuit, which generates a triangular wave. The block diagram of a triangular wave generator is shown in the following figure – The block diagram of a triangular wave generator contains mainly two blocks: a square wave generator and an integrator. These two blocks are cascaded. That means, the output of square wave generator is applied as an input of integrator. Note that the integration of a square wave is nothing but a triangular wave. The circuit diagram of an op-amp based triangular wave generator is shown in the following figure- We have already seen the circuit diagrams of a square wave generator and an integrator. Observe that we got the above circuit diagram of an op-amp based triangular wave generator by replacing the blocks with the respective circuit diagrams in the block diagram of a triangular wave generator. 196  UNIT 5: Operational Amplifier Solved Problem 1. The tuned collector oscillator circuit used in the local oscillator of a radio receiver makes use of an LC tuned circuit with L1 = 58.6 µH and C1 = 300 pF. Calculate the frequency of oscillations. 2. Find the capacitance of the capacitor required to build an LC oscillator that uses an inductance of L1 = 1 mH to produce a sine wave of frequency 1 GHz (1 GHz = 1 × 1012 Hz). A 1 mH inductor is available. Choose the capacitor values in a Colpitts oscillator so that f = 1 MHz and mv = 0.25. 197  UNIT 5: Operational Amplifier 3. Calculate the (i) operating frequency and (ii) feedback fraction for Hartley oscillator shown in Fig. 2. The mutual inductance between the coils, M = 20 µH. A 1 pF capacitor is available. Choose the inductor values in a Hartley oscillator so that f = 1 MHz and mv = 0.2. 198  UNIT 5: Operational Amplifier 4. In the phase shift oscillator shown in Fig. 3, R1 = R2 = R3 = 1MΩ and C1 = C2 = C3 = 68 pF. At what frequency does the circuit oscillate ? Fig. 3 199  UNIT 5: Operational Amplifier 5. A phase shift oscillator uses 5 pF capacitors. Find the value of R to produce a frequency of 800 kHz. 6. In the Wien bridge oscillator shown in Fig. 4 , R1 = R2 = 220 kΩ and C1 = C2 = 250 pF. Determine the frequency of oscillations. Fig. 4 200  UNIT 5: Operational Amplifier Review Questions Short Answer Questions 1. What are the characteristics of ideal operational amplifier. 2. Draw the op-amp pin diagram. 3. Draw the op-amp Hartley oscillator. 4. What is op-amp? What is slew rate of an op-amp? 5. Define CMRR of op-amp. 6. What is virtual ground in an op-amp? 7. What is open loop gain of the op-amp. 8. What is closed loop gain of the op-amp. 9. How op-amp produce square wave? 10. What is slew rate? 11. Define CMRR. 12. What is an ideal operational amplifier? 13. Define ‘virtual ground’. Big Questions 14. With a neat diagram explain the working of non-inverting op-amp. Derive an expression for the voltage gain 15. Explain how an op-amp is used as an adder 16. Explain how an op-amp is used as an differentiator 17. Explain how an op-amp is used as an subractor 18. Explain how an op-amp is used as an integrator 19. State the important characteristics of op-amp 20. Define slew rate. Describe how the slew rate of an op-amp can be improved 21. Explain the working of inverting amplifier 22. Draw the circuit diagram of triangular wave generator using op-amp 23. Draw the circuit diagram of square wave generator using op-amp 24. Describe the operation of (a) Differentiator and (b) Integrator using op-amplifier. 25. Mention the ideal characteristics of an op amp. 26. Sketch and explain the inverting and non-inverting circuit using op amp. 201