Fundamentals of Electronics Course Material PDF

ECE - 1051 : Basic Electronics ECE 1051 : BASIC ELECTRONICS TABLE OF CONTENTS Part - I : Analog Electronics Chapter – 1 : Diodes and Ap...

ECE - 1051 : Basic Electronics ECE 1051 : BASIC ELECTRONICS TABLE OF CONTENTS Part - I : Analog Electronics Chapter – 1 : Diodes and Applications Module-1 : PN Junction Diodes 1.1.1 Introduction 5 1.1.2 Concept of PN junction 5 1.1.3 PN junction under bias 6 1.1.4 V-I Characteristics of diode 9 1.1.5 Static and dynamic Resistance of Diode 13 1.1.6 Ideal and Practical diodes 14 1.1.7 Equivalent circuit of diode 15 1.1.8 Break down phenomenon in diodes 18 1.1.9 Zener diode characteristics 19 1.1.10 Diode as a capacitor 20 Module-2 : Applications of Diodes 1.2.1 Introduction 23 1.2.2 Basic Block diagram of DC power supply 24 1.2.3 Half wave rectifier 28 1.2.4 Full Wave Rectifier 31 1.2.5 Full wave bridge rectifier 34 1.2.6 Comparison of Rectifiers 37 1.2.7 Rectifier with filter 38 Module-3 : Voltage Regulators 1.3.1 Zener Voltage Regulator 44 1.3.2 IC Voltage Regulators 51 Module-4 : Special purpose diodes 1.4.1 Light Emitting Diodes 53 1.4.2 Photo diodes and applications 54 1.4.3 Optocoupler 55 1.4.4 Solar Cell 56 Chapter – 2 : BJT and Applications Module-1 : BJT Characteristics 2.1.1 BJT construction and operation 61 2.1.2 BJT configurations 62 Department of Electronics and Communication Engineering, M.I.T. 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Page 1  ECE - 1051 : Basic Electronics 2.1.3 BJT current components 63 2.1.4 BJT characteristics 65 Module-2 : BJT Biasing 2.2.1 DC load line and Ned for Biasing 68 2.2.2 Fixed biasing 70 2.2.3 Voltage divider biasing or self –biasing 73 Module-3 : Transistor as an amplifier 2.3.1 RC coupled Amplifier 77 2.3.2 Frequency Response of an Amplifier 78 2.3.3 Multistage amplifier 79 Module-4 : Transistor as a Switch 2.4.1 Introduction 81 2.4.2 Transistor as LED driver 82 2.4.3 Transistor as Inverter 82 Chapter – 3 : Operational amplifier and applications: Module-1 : Operational Amplifier 3.1.1 Introduction 84 3.1.2 Internal Block diagram of Op-amp 84 3.1.3 Op-Amp Characteristics 88 Module-2 : Linear applications of Operational Amplifier 3.2.1 Inverting amplifier 90 3.2.2 Non-inverting Amplifier 91 3.2.3 Voltage follower 92 3.2.4 Inverting Summing Amplifier 92 3.2.5 Difference amplifier 93 3.2.6 Integrator 94 3.2.7 Differentiator 94 Module-3 : Non-linear applications of Operational Amplifier 3.3.1 Voltage Comparator 96 3.3.2 Square wave generator 96 Part - II : Digital Electronics Chapter – 4 : Number systems & Codes Module-1 : Number Systems 4.1.1 Decimal number system 100 4.1.2 Binary number system 101 4.1.3 Octal number system 101 4.1.4 Hexadecimal number systems 102 Department of Electronics and Communication Engineering, M.I.T. 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Page 2  ECE - 1051 : Basic Electronics 4.1.5 Number system – Arithmetic 103 4.1.6 Complementation of Numbers 105 Module-2 : Codes 4.2.1 Weighted binary codes 108 4.2.2 Non-Weighted codes 110 4.2.3 Self-complimenting codes 110 4.2.4 Error detecting and correcting codes 111 Chapter – 5 : Boolean Algebra & Logic gates Module-1 : Introduction to Boolean Algebra 5.1.1 Boolean algebraic theorems 115 5.1.2 Simplification of Boolean algebraic expressions 118 Module-2 : Logic gates 5.2.1 Logic gates and operation 121 5.2.2 Concept of Universal Logic 126 5.2.3 Classification of digital circuits 128 5.2.4 Implementation of Combinational Circuits 129 Module-3 : Simplification of Boolean functions using K - map 5.3.1 Standard form of Boolean expression s 135 5.3.2 Introduction to K-map 137 5.3.3 Simplification of Boolean expressions using K-map 138 Chapter – 6 : Sequential circuits Module-1 Flip-flops and applications 6.1.1 Introduction to Flip Flops 142 6.1.2 Counters 147 6.1.3 Shift Registers 151 Part - III : Principles of Electronic Communication Chapter – 7 : Fundamentals of Analog communication Module-1 : Amplitude Modulation 7.1.1 Principle of Electronic Communication 156 7.1.2 Need for modulation 158 7.1.3 Amplitude modulation 159 7.1.4 Time domain analysis 160 7.1.5 Spectrum of AM signal 162 7.1.6 Power content of AM 162 7.1.7 Different types of AM signals 165 Department of Electronics and Communication Engineering, M.I.T. 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Page 3  ECE - 1051 : Basic Electronics 7.1.8 Detection of AM signal 166 7.1.9 Super heterodyne principle of AM reception 167 Module-2 : Frequency Modulation 7.2.1 Frequency Modulation 170 7.2.2 Time domain analysis of FM signal 171 7.2.3 Comparison of AM and FM 174 Chapter – 8 : Introduction to digital communication Module-1 : Digitization of Analog signals 8.1.1 Basic principle of Sampling 176 8.1.2 Pulse Amplitude Modulation 179 8.1.3 Pulse Width Modulation 180 8.1.4 Pulse Position Modulation 180 8.1.5 Digital communication system 180 8.1.6 Digital Modulation Techniques 183 Chapter – 9 : Communication Networks Module-1 : Introduction to Communication networks 9.1.1 Introduction to Data communication 186 9.1.2 Introduction to communication networks 187 9.1.3 Types of communication networks 188 9.1.4 Communication Network topology 189 9.1.5 Network protocols and Reference models 190 Chapter – 10 : Introduction to Mobile Communication Module-1 : Introduction to Mobile Communication 10.1.1 Principle of Cellular communication 195 10.1.2 Multiple access Technology 196 10.1.3 Global System for Mobile Communications (GSM) 198 Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 4  ECE - 1051 : Basic Electronics Part –I ANALOG ELECTRONICS Chapter -1: Diodes and Applications The term diode is used to represent a device or element which has two elecrodes. These divices are characterized by the fact that they allow electric current to flow in one direction, and block flow of current in the opposite direction. This unilateral behaviour is predominantly used in swithching and rectification. Diode is in fact, the very first electronic device invented. Initially, for several years vacuum tube version was in use, and were bulky in size, required higher power for operation, and were slow. Today, we have semiconductor diodes, which are very small in size, requires relatively low power and operates at higher speeds. Semiconductor diodes are available in various forms and are used in wide variety of applications. In this unit, we shall look at the operating behavior and charecteristics of semiconductor diodes along with their typical applications such as rectifiers, voltage regulators and some special purpouse applications. Module – 1 : Diodes Learning Outcomes: At the end of this module, students will be able to: 1. Explain the operation of PN junction diode under different biasing conditions. 2. Plot the I-V characteristics of the diode. 3. Define static and dynamic resistance of the diode. 4. Explain the breakdown phenomenon observed in diodes. 5. Describe the working of Zener diode and plot its I-V characteristics. 6. Explain the operation of diode as a capacitor. 1.1.1 Introduction Materials are broadly classified as metals, insulators and semiconductors. A semiconductor like Germanium or Silicon has electrical conductivity lying between conductor and insulator. Semiconductors are the basic materials used in modern electronics. For example, Diodes, Transistors, Solar cells, Light-emitting diodes (LEDs), and integrated circuits. Self Reading: 1. Crystal structure of Germanium and Silicon 2. Intrinsic and extrinsic semiconductors 3. N-type and P-type semiconductors and concept of minority and majority cariiers 4. Diffussion and drift currents 1.1.2 Concept of PN Junction As you know, P-type semiconductor has large number of holes while, N-type semiconductor has large number of free electrons. When P-type and N-type materials are joined together, a gradient of charge carriers densities is created at the junction. This will Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 5  ECE - 1051 : Basic Electronics cause the electrons to move from N-type material to the P-type material and holes to move from N-type material to the P-type material. This process of movement of charge carriers form the region of higher concentration to a lower concentration in the absence of external electric field is called diffusion. Diffusion of charge carriers across the junction will continue until the equilibrium condition is established. Also at the junction, N-type material will have positively charged immobile ions and P-type material have negatively charged immobile ions. Thus the regions on either sides of p–n interface lose their charge neutrality and become charged. For this reason it is caleld space charge region. As the region is devoid or depleted of mobile charge carriers it is also called depletion region and is as shown in Figure. 1.1.1. = P N Figure 1.1.1. Schematic of PN junction The space charge on either sides of the junction causes a potential difference accross the P-N junction and it is called the barrier potential. This is the minimum amount of voltage required to initiate flow of charge carriers across the junction. Doped germanium has a barrier potential of about 0.3 volts where as, doped silicon has a barrier voltage of about 0.7 volts. 1.1.3 P-N junction under bias Application of external voltage across the diode is called biasing. Depending upon the polarity and magnitude of voltage applied, we can have three biasing conditions, as listed below. a) No bias or Zero bias b) Forward bias c) Reverse bias Figure 1.1.2. Circuit symbol of a diode Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 6  ECE - 1051 : Basic Electronics a) Zero Bias: In the absence of any bias voltage, the net flow of charge carriers in any one direction for a semiconductor diode is zero. This occurs because minority carriers (holes) in the N-type material will encounter barrier in the depletion region to cross the junction and move to the P-type region. Same is the case for electrons in P-type material. This results in depletion region with high impedance, and hence no current flows through the diode. The built-in potential varies from 0.3 to 0.7 eV depending upon the type of semiconductor material. A diode operated without any biasing is shown in Figure 1.1.3. Figure. 1.1.3. P-N diode with zero biase. b) Forward Bias: When a negative voltage is applied to the N-type material and a positive voltage is applied to the P-type material, the diode is said to be in a Forward Bias condition. Figure 1.1.4. shows the diode with forward biase. If the external voltage applied is greater than the value of the barrier potential, the carriers start crossing eth junction and hence there will be a forward current. The diode is said to be in the ON condition. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 7  ECE - 1051 : Basic Electronics Figure.1.1.4. Forward biasing of P-N junction diode. [http://www.imagesco.com/articles/photovoltaic/photovoltaic-pg3.html]. c) Reverse Bias: When a positive voltage is applied to the N-type material and a negative voltage is applied to the P-type material , the diode is said to be in a reverse biased condition, as shown in Figure. 1.1.5. The positive voltage applied to the N-type semiconductor attracts electrons towards the positive electrode and hence away from the junction. At the same time, the holes in the P-type semiconductor are attracted towards the negative electrode. This results in widening of depletion layer due to a lack of electrons and holes near the junction and presents a high impedance path for the majority carriers. The height of potential barrier is increased, which prevents the flow of forward current through the diode. However, the applied potential favors the movement of minority carriers across the junction causing flow of current in the reverse direction. This current is called reverse saturation current and is represented by I0 or IS. Figure. 1.1.5. Reverse biasing of P-N junction diode. [http://www.imagesco.com/articles/photovoltaic/photovoltaic-pg3.html]. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 8  ECE - 1051 : Basic Electronics Self test: 1. The arrow direction in the diode symbol indicates a. Direction of electron flow. b. Direction of hole flow (Direction of conventional current) c. Opposite to the direction of hole flow d. None of the above 2. When the diode is forward biased, it is equivalent to a. An OFF switch b. An On switch c. A high resistance d. None of the above 3. The barrier potential voltage of Si diode is a. 0.2 V b. 0.7 V c. 0.8 V d. 1.0 V 5. List the methods available for tesing the diode. How cut-in voltage of diode is measured in practie? 1.1.4 I-V characteristics of diode I-V characteristics of practical diode is shown in Figure 1.1.6. When the forward biased voltage is applied to diode, current is initially zero and then increases sharply after crossing the cut-in voltage.In this case,the diode behaves like a closed switch. Similarly, in reversed biased condition, the diode behaves like an open switch and very small current flows due to minority charge carriers, which is known as reverse saturation current. In the reverse biased condition, beyond a particular reverse voltage, a sudden rise of current will be observed and this voltage is called breakdown voltage. Figure. 1.1.6. I-V characteristics of practical diode [http://www.learningaboutelectronics.com/Articles/Ideal-diode.php] Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 9  ECE - 1051 : Basic Electronics a) Forward biased Characteristic: The application of a forward bias voltage to the junction diode results in the depletion layer becoming very thin and narrow which represents a low resistance path through the junction thereby aiding flow of current through the diode. The point at which this sudden increase in current takes place is called “knee” point and is represented on the static I-V characteristics as shown in Figure 1.1.7. Figure 1.1.7. I-V characteristics of P-N junction diode under forward biased condition. [http://www.electronics-tutorials.ws/diode/diode_3.html] b) Reverse biased Characteristic: In this case, PN junction offers high resistance value and practically zero current flows through the junction diode with an increase in bias voltage. However, a very small leakage current flows through the junction which is in the order of microamperes (μA ) for ordinary rectifier diodes. Figure. 1.1.8. I-V characteristics of P-N junction diode under reversed biased condition. [ http://www.electronics-tutorials.ws/diode/diode_3.html] Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 10  ECE - 1051 : Basic Electronics If the reverse bias voltage VR applied to the diode is increased beyond certain limits there will a large current due to avalanche effect and cause breakdown. This is shown in figure 1.1.8. c) The Diode Current: The current flowing through the diode is given by the following equation.  VD  I D  I 0  eVT  1 (1.1.1)     ID= Diode current I0 = Reverse saturation current. VD= Applied bias voltage (Positive for forward and negative for reverse bias) T VT  11600 Volt equivalent of temperature (T is in degree Kelvin) for Germanium η =1 and for Silicon η =2 For large forward bias, equation 1.1.1 approximates to VD (1.1.2) I D  I0 (e ηVT ) For large reverse bias, equation 1.1.1 approximates to I D   I0 (1.1.3) d) Effect of Temperature on the Reverse current: The reverse saturation current is a temperature dependent parameter. It doubles for every 10o C rise in temperature. Let I01 be the reverse saturation current at temperature T1 and I02 be the reverse saturation current at temperature T2, where T2 > T1. Thus the rise in reverse saturation current can be modelled as I02  I01 2(T2 T1 )/10 (1.1.4) Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 11  ECE - 1051 : Basic Electronics Exercise Problem 1: 1. Sketch I versus V to scale for each of the circuits shown below. Assume that the diodes are ideal and allow V to range from -10 V to +10 V. i + v _ 2kΩ Solution: Diode is ON for v > 0 ; 5 4 3 i (mA) 2 1 0 -10 -5 0 5 10 v (V) 2. Sketch I versus V to scale for each of the circuits shown below. Assume that the diodes are ideal and allow V to range from -10 V to +10 V. Solution: Diode B is ON for v > 0 and R=1kΩ. Diode A is on for v < 0 and R=2kΩ. 10 5 i (mA) 0 -5 -10 -5 0 5 10 v (V) Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 12  ECE - 1051 : Basic Electronics 1.1.5 Static and Dynamic Resistance of Diode There are two types of diode resistance namely, DC and AC resistance. a) DC or Static Resistance The application of a dc voltage to a circuit containing a semiconductor diode will result in an operating point on the characteristic curve that will not change with time. The resistance of the diode at the operating point can be found by the corresponding values of VD and ID as shown in Figure 1.1.9. The equation for the DC resistance can then be written as, V RD  D (1.1.5) ID Note that the DC or static resistance of a diode does not depend on the curve shape, it depends only on the operating point or the values of diode voltage and current. Figure. 1.1.9. Static resistance of a diode [http://www.freewebs.com] b) AC or Dynamic Resistance: To determine the dynamic resistance of a diode, a a tangent is drawn to the curve through the operating point as shown in Figure 1.1.10. The dynamic resistance of the diode is found using the following equation: 1 Vd rd   (1.1.6) slope I d Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 13  ECE - 1051 : Basic Electronics Figure 1.1.10. Dynamic resistance [http://www.freewebs.com] Also, dynamic resistance is found by the derivative of the diode equation; where:  V   I D  I s exp  D   1 , (1.1.7)   VT   dI D d    V D     I s  V   I s exp    1   exp    (1.1.8) dV D dV D    VT     T  V  VT   dV D  VT     (1.1.9) dI D  ID  Is  Since ID>> Is, then dV D VT rd   (1.1.10) dI D ID T For Ge , η=1 ; VT  (1.1.11) 11600 dV D 300 And at room temperature, T=300 K, rd   (1.1.12) dI D 11600 xI D 1.1.6 Ideal and Practical diode In an ideal diode, current flows freely through the device when forward biased, offering no resistance. An ideal diode is simply a P-N junction where the change from P-type to N-type material is assumed to occur instantaneously, also referred to as an abrupt junction. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 14  ECE - 1051 : Basic Electronics The simplified diode model ignores the effect of diode resistance in comparison with values of other elements of the circuit. The voltage drop across the diode is zero. A practical diode does offer some resistance to current flow when forward biased. Since there is some resistance (built-in potential) , there will be some power dissipated when current flows through a forward biased diode. Therefore, there is a practical limit to the amount of current a diode can conduct without damage. A reverse biased diode has very high resistance and excessive reverse bias can cause the diode to damage. 1.1.7 Equivalent circuit of diode: Diode is often replaced by its equivalent circuit during circuit analysis and design. For DC diode model, characteristics of an ideal diode and the modifications that were required due to practical considerations has been considered for following cases: (i) Ideal Diode (ii) Second approximation of diode (iii) Practical diode (i) For an Ideal diode Vγ = 0, RR = ∞ and RF = 0 as shown in Figure 1.1.14. In other words, the ideal diode is a short in the forward bias region and an open in the reverse bias region. A K RF = 0 A K Forward bias RR =  Vγ = 0 A K Figure 1.1.14 Equivalent circuit of Ideal diode. Reverse bias (ii) In second approximation: Vγ ≠ 0, RR = ∞ and RF = 0 as shown in Figure 1.1.11. A K RF = 0 A K Forward bias RR =  Vγ A K Reverse bias Vγ Figure 1.1.11 Equivalent circuit of diode for second approximation. (iii) In Practical diode (silicon) Vγ = 0.7 V, RR < ∞ (typically several MΩ), RF ≈ rd (typically < 50 Ω) as shown in Figure 1.1.12. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 15  ECE - 1051 : Basic Electronics Figure 1.1.12 Equivalent circuit of practical diode. Exercise Problem 1: 1. A Silicon diode has a saturation current of 1pA at 200C. Determine (a) Diode bias voltage when diode current is 3mA (b) Diode bias current when the temperature changes to 1000C, for the same bias voltage. Ans. Given: The diode current ID=3mA, Reverse saturation current I0=1x10-12 A, Temperature T=20°C = 273+20 = 293K The diode is silicon η=2 The equation for the diode current ID is given by  VD  I D  I 0  eVT  1 and VT  T 293   25.25 mV   11600 11600 (a) The diode  bias voltage   I  VD  VT ln 1  D   1.103V  I0  (b) The diode current when the temperature is 1000 oC T 393 VT    32.15 mV 11600 11600 The temperature is raised to 100°C (So the reverse saturation current I0 changes) T2 T1  /10  1001020  I 02  I 01 2  10  2 12   256 pA     1.103  12  ( 2 x 32.15 x10 3 1)  I D  256 x10  e   7.21 mA   2. Find the static and dynamic resistance of a P-N junction germanium diode if Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 16  ECE - 1051 : Basic Electronics the temperature is 27°C and I0=1μA for an applied forward bias of 0.2V. Solution: Given: Applied forward voltage= 0.2 V. Reverse saturation current I0=1x10-6 A. Temperature T=27°C = 273+27 = 300°K. The diode is Ge, η=1. T 300 VT    25.86 mV 11600 11600  0.2   6  (1x 25.86 x10 3 1)  I D  1x10  e   2.28 mA    V Static Resistance:  0.087 K I ηVT 1x25.86 x10 3 Dynamic resistance    11.33 I D  I 0 2.28 x10 3  1x10 6. 3. Determine the dc resistance levels for the diode at (a) ID= 2 mA, (b) ID = 20mA, (c) VD = -10V Solution: From the graph, find corresponding voltage for the mentioned current values. (a) RD= VD/ID= 0.5V/2mA= 250 Ω. (b) RD= VD/ID= 0.8V/20mA= 40 Ω. (c) At VD= -10V, ID= -Is= -1μA (from the curve) and RD= VD/ID= 10V/1μA= 40 Ω Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 17  ECE - 1051 : Basic Electronics 1.1.8 Break down phenomenon in diodes Breakdown voltage is the largest reverse voltage that can be applied without causing an exponential increase in the current in the diode. As long as the current is limited, exceeding the breakdown voltage of a diode does no harm to the diode. Two breakdown mechanisms exist in diode.They are a) Zener breakdown b) Avalanche breakdown a) Zener breakdown: In Zener breakdown , the electric field established due to the reverse voltage capapble of getting the electrons out of their covalent bonds and away from their parent atoms as shown in Figure 1.1.13. Electrons are transferred from the valence to the conduction band. In this situation, the current can still be limited by the limited number of free electrons produced by the applied voltage so it is possible to cause Zener breakdown without damaging the semiconductor. Figure. 1.1.13 Schematic of the zener breakdown mechanisms http://shrdocs.com/presentations/12656/index.html b) Avalanche Breakdown: Avalanche breakdown occurs when the applied voltage is so large that electrons achieve kinetic energy sufficiently high and collide with the silicon atoms and knock off more electrons. These electrons are then also accelerated and subsequently collide with other atoms. Each collision produces more electrons which leads to more collisions etc as shown in Figure. 1.1.14. The current in the semiconductor rapidly increases and the material can quickly be destroyed. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 18  ECE - 1051 : Basic Electronics Figure. 1.1.14. Avalanche breakdown phenomenon http://shrdocs.com/presentations/12656/index.html 1.1.9 Zener diode characteristics: Zener diode (also referred as regulated diode) is a two terminal device that is widely used in voltage regulators. When the reverse bias, applied to the semiconductor, has reached to zener voltage Vz , the current will be dramatically increased while the voltage keeps constant. Circuit symbol of Zener diode is shown in Figure 1.1.15.It is special kind of diode that is heavily doped during manfacturing. This results in a high number of free carriers. These additional current carriers permit reverse current flow when a specific reverse bias voltage is reached. This voltage level is referred to as the avalanche point or Zener point. When forward biased, the Zener diode acts like a regular diode. Zener diodes are rated according to the voltage at which they will "turn ON", or begin to conduct reverse bias current. The regulated values of the zener diode are thus distributed in the range from 3V to several hundreds of volts, whereas the power range is distributed from 200mW to 100W. Figure. 1.1.15. Circuit symbol of Zener diode. Zener Diodes are used in the "REVERSE" bias mode, i.e. the anode is connected to the negative supply. From its I-V characteristics curve as shown in Figure. 1.1.16, it can be said that Zener diode has a region in its reverse bias characteristics of almost a constant voltage regardless of the current flowing through the diode. This voltage across the diode (Zener Voltage, Vz) remains nearly constant even with large changes in current through the diode caused by variations in the supply voltage or load. This ability to control itself can be used to great effect to regulate or stabilise a voltage source against supply or load variations. The diode will continue to regulate until the diode current falls below the minimum Iz value in the reverse breakdown region. When the Zener diode is forward biased it behaves like ordinary diode. In the reverse biased condition, as the reverse voltage is increased beyond the break down voltage of the diode, the current rises sharply with the applied voltage but the voltage across the diode Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 19  ECE - 1051 : Basic Electronics remains constant. This behavior of Zener diode is used to provide a constant reference voltage such as in the case of voltage regulation. Figure 1.1.16. I-V characteristic of Zener diode [http://www.nptel.ac.in]. IZK or IZmin – Minimum current necessary to maintain breakdown. IZM or IZMax – Maximum current that can be safely passed through the zener diode PZM or PZMax – Maximum power dissipation across zener diode PZM = VZ. IZM Equivalent circuit of zener diode in different conditions is shown in Figure 1.1.17. N N N N – + Vγ VZ + – RR ≈  RZ RF P P P P (a) (b) (c) (d) Figure. 1.1.17 (a) Zener diode symbol (b) equivalent circuit in forward biased condition (c) equivalent circuit in reversed biased condition (d) equivalent circuit in breakdown condition 1.1.10 Diode as a capacitor Varactor diode is a special purpose diode that acts as a capacitor and always operates in reverse-bias. It is doped to maximize the inherent capacitance of the depletion region. The Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 20  ECE - 1051 : Basic Electronics depletion region acts as a capacitor dielectric because of its nonconductive characteristic. The p and n regions are conductive and acts as the capacitor plates, as shown in Figure 1.1.18. Figure. 1.1.18. Schematic of varactor diode Capacitance is determined by the parameters of plate area (A), dielectric constant (ε) , and plate separation (d) and is given by Ax C (1.1.13) d As the reverse-bias voltage increases the depletion region widens, effectively increasing the plate separation, thus decreasing the capacitance. Major application of varactors is in tuning circuits. For example very high frequency (VHF), ultra high frequency (UHF), and satellite receivers employ varactors. When used in a parallel resonant circuit, the varactor acts as a variable capacitor. Thus, allowing the resonant frequency to be adjusted by a variable voltage level. They are also used in Frequency Modulation circuits. Summary 1. PN junction diode in forward biased condition behaves like a closed switch and in reversed biased condition behaves like an open switch. 2. The reverse saturation current of a diode doubles for every 100 rise in temperature. 3. There are two breakdown mechanisms in a zener diode: avalanche breakdown and zener breakdown. 4. The zener diode is generally used in reverse breakdown region. 5. A zener diode maintains a nearly constant voltage across its terminals over a specified range of zener currents. 6. Varactor diode operates in reverse biased condition and acts as a variable capacitance. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 21  ECE - 1051 : Basic Electronics Exercise Problems: 1. In each diode circuit shown below, find whether the diodes are forward or reverse biased. 2. Determine the state of diode for the circuit shown below and find ID and VD. Assume practical model for the diode. 3. Calculate the dynamic forward and reverse resistance of a PN junction diode, when the applied voltage is 0.25V for Germanium Diode. I0 = lμA and T = 300 K. (Ans: rf=1.734Ω; rr=390MΩ) 4. A germanium diode has reverse saturation current of 0.19μA. Assuming η =1, find the current in the diode when it is forward biased with 0.3 V at 27oC. (Ans: 19.5mA) 5. The forward current in a Si diode is 15 mA at 27oC. If reverse saturation current is 0.24nA, what is the forward bias voltage? (Ans: 0.93V) 6. A germanium diode carries a current of 10mA when it is forward biased with 0.2V at 27oC. (a) Find reverse saturation current. (b) Find the bias voltage required to get a current of 100mA. (Ans: 4.42μA, 0.259V) Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 22  ECE - 1051 : Basic Electronics Module 2: Application of Diodes In the previous module we have discussed the behaviour and V-I characeteristics of PN junction diode. Diode conducts when it is forward biased and behaves like a closed switch. Whereas, during the reverse bias, it goes off and behaves like an open switch. This unilateral behaviour depending on the polarity of external voltage applied to the diode is used in many circuit applications. One such application is conversion of AC voltage to DC, called rectification. In this module we will study various forms of rectifier circuits and their analysis in detail. Learning Outcomes: At the end of this module, students will be able to: 1. Explain the need for AC to DC conversion 2. Draw and explain the block diagram of a basic dc power supply unit. 3. Discuss the importance of the various components used in the rectifier circuits. 4. Discuss the working of a half wave and full wave rectifier circuits. 5. Analyse the performance of rectifier circuits and compare them. 6. Explain the working of rectifier circuits with capacitor filter. 1.2.1. Introduction Today, we cannot imagine life without electronic products like cell phones, computers, laptops, music systems etc., in our daily lives. Some of these electronic systems work on a constant DC voltage derived from the AC mains, while others use internal batteries, which requires regular charging. In any case, some device which at one end receives AC voltage, which is 230V sinusoidal signal of 50Hz as input and produces a constant DC voltage at its output is required. This module aims at providing an insight into some of the basic circuits which converts AC mains in to a constant DC voltage. Such circuits or devices are called regulated power supplies. A signal obtained from main AC power supply is purely sinusoidal that can be defined in terms of Peak amplitude and Frequency.  Peak amplitude: The maximum amplitude of an alternating signal on either sides measured from its zero value.  Frequency: Number of cycles that passes a given point per second. It is equal to reciprocal of time taken to complete one full cycle. It is mathematically expressed as V (t )  A sin(t ) and plotted as shown in Figure 1.2.1, where Peak amplitude A= 230  2 V, ω= 2πf radians/sec, frequency(f) =50Hz = 1/20ms. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 23  ECE - 1051 : Basic Electronics Figure 1.2.1: AC sinusoidal signal with amplitude A= 230  2 V, f=50 Hz signal Average or DC value: The dc value of a signal V (t ) is the average value of that signal. It is mathematically evaluated as: 1 T  Vav  Vdc    V (t ) d (t ) Time Period  0  1 2  Vav  Vdc    V (t ) d ( wt) (1.2.1) 2  0  The root mean square(RMS) value of the signal V (t ) mathematically evaluated as: 1 T 2  1 2  Vrms    V (t ) d ( t ) or Vrms    V 2 (t ) d (t ) (1.2.2) Time Period  0   2  0  Note 1 : A pure sinusoidal signal has an average value equal to zero. It means the dc value of this signal is zero. 1.2.2. Basic Block Diagram of DC power supply The DC power supply converts the AC sinusoidal signal to a DC signal. The block diagram of a basic DC power supply unit is as shown in Figure 1.2.2. A DC regulated power supply unit consists of the following key components. a) Step down transformer b) Rectifier circuits c) Filter circuit d) Regulator Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 24  ECE - 1051 : Basic Electronics AC sinusoidal signal with Signal with varying amplitude in DC signal reduced amplitude in both one direction, i.e. only ‘+’ or across load ‘+’ and ‘-‘ directions only ‘-’values Filter Step down Rectifier Regulator Transformer circuit circuits Input AC mains LOAD Figure 1.2.2: Basic block diagram of DC power supply a) Step down Transformer: A transformer is used to bring voltage up or down in an AC electrical circuit. A step down transformer consists of two coils of wire called primary and secondary winding placed such that they are not in contact with each other as shown in Figure 1.2.3a. The symbolic representation of the transformer is shown in Figure 1.2.3b. (a) (b) Figure 1.2.3: (a) Transformer core with primary and secondary windings (b) circuit representation of a step down transformer. A step down transformer converts a high voltage low current power to a low voltage high current power. A step down transformer has large number of turns in primary coil compared to the number of turns in the secondary coil. Hence in this case the secondary voltage is less than the primary voltage and equivalently there is rise in secondary current. The ratio of primary voltage to secondary voltage is proportional to the ratio of number of turns in the primary to the number of turns in the secondary. The 155V AC mains applied at the primary of a step down transformer is stepped down by ten times at the secondary as shown in Figure 1.2.4. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 25  ECE - 1051 : Basic Electronics (a) (b) Figure 1.2.4: (a) Primary voltage waveform with peak amplitude of 220V (b) Stepped down version is available at the secondary with peak amplitude of 22V Transformer used in the DC power supply units plays two important roles. First role is, as discussed earlier, it steps down AC voltage to a suitable value and secondly it provides electrical isolation to the low voltage low power components on the secondary side of the transformer. The second equally important role is to provide electrical isolation to the voltage low power components on the other side of the transformer. This also provides some amount of safety to the equipments using such DC supplies. Note 2: The frequency of the primary voltage is equal to the frequency of the secondary voltage of the transformer. b) Rectifier circuit: A rectifier circuit is the heart of a DC power supply. It converts an AC sinusoidal signal that is bidirectional (with both positive and negative amplitudes) into a signal which is unidirectional (either only positive or only negative). Thus rectifier circuit forces the current through the load to flow in only one direction. The rectified output is usually called a pulsating DC voltage. Thus in general the process of converting an AC signal into pulsating DC signal is called rectification and the circuit is called rectifier. Rectification is commonly performed using semiconductor diodes because of its inherent unidirectional conduction property. Figure 1.2.5: Secondary transformer voltage and rectified voltage waveform Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 26  ECE - 1051 : Basic Electronics The rectified output voltage with respect to the input voltage using a single diode is as shown in Figure 1.2.5. The rectifier passes positive half cycle to the output and blocks the negative half cycle. In the case of a full wave rectifier both half cycles will be rectified and available as unidirectional pulses at the output. Note 3: The AC component of a rectified output signal is not equal to zero. That is the rectified pulsating DC signal has both DC and AC components. c) Filter circuit: The pulsating DC signal is not suitable for appliances that require pure DC voltage. Filters can be used to minimize (smooth out) the pulsations or eliminate the AC content from the rectified signal to achieve approximately constant valued (or a DC) signal. A filter contains a capacitor, an energy storing component that can hold the voltage to the peak value of the rectified pulsating DC and then dissipate energy to load when the pulsating DC drops as shown in Figure 1.2.6. Essentially, the filter minimises the ac component present in the output of the rectifier. This increases the DC value at the output. Figure 1.2.6: Illustration of filtering function Note 4: The output of filter circuit is a DC signal with small AC component called ripples d) Regulator Regulation is defined as the ability of a system to provide near constant supply over a wide range of load and power line fluctuating conditions. A circuit that can provide constant DC voltage despite of variations in the mains AC power supply or load variations is called voltage regulator. Self test: 1. List out the appliances or electronic products used in daily life which require DC power supply for their operation. Mention the DC values recemoneded for their operation. 2. List and classify the appliances or electronic products which use DC and AC power supply. 3. Explain the block diagram of a DC power supply. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 27  ECE - 1051 : Basic Electronics 1.2.3 Half Wave Rectifier The half wave rectifier consists of a single semiconductor diode. The secondary voltage of transformer is fed as an input to the rectifier circuit. The output is measured across the load resistance RL. a) Working of an HWR circuit For simplicity it is assumed that the diodes are ideal, and is represented as an open circuit when reversed biased and short circuit when forward biased. The circuit diagram of half wave rectifier and rectified output waveform with respect to secondary voltage waveform is as shown in Figure 1.2.7(a) and (b) respectively. During the positive half cycle of the input waveform, voltage at node A is positive with respect to voltage at node B, which forces the diode to be forward biased and acts as a short. The equivalent circuit for the positive half cycle is as shown in Figure 1.2.8.This results in current flow through the load resistance RL. Hence output voltage is approximately equal to the secondary voltage. Similarly during the negative half cycle, the voltage at node A is negative with respect to node B that forces the diode to be reverse biased and acts as open circuit. The equivalent circuit is as shown in Figure 1.2.9. This results in no current flow through the load. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 28  ECE - 1051 : Basic Electronics Thus the diode conducts only during positive half cycle and hence the circuit is referred as ‘Half Wave Rectifier’. The rectified voltage is pulsating DC, which can be smoothened using filter circuits. Analysis of a HWR circuit Filter performance is measured using certain standard parameters. Few of them will be defined and analyzed for each of the filter circuits. A. Dc voltage Vdc : The average value of the output voltage measured across the load resistor. B. Ripple Factor: Ripple factor γ is defined as the ratio of rms value of AC component to DC component of the signal. Vr   rms (1.2.3) Vdc where Vrrms is the rms value of the ripple (AC component in the pulsating DC) and is given by Vrrms  Vrms 2  Vdc2 (1.2.4) Ripple factor actually measures the amount of AC content present as compared to the DC (or average) content in the pulsating dc. Using 1.2.3 in 1.2.4, we can write 2 Vrms   1 (1.2.5) Vdc2 C. Efficiency: Efficiency (η) is the ratio of the DC output power to AC input power supplied by the secondary of the transformer. dc output power Pdc   ac input power Pac Efficiency signifies the outcome of the rectifier circuit to output DC power in comparison to AC input power. Thus Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 29  ECE - 1051 : Basic Electronics  V2   dc   RL    (1.2.6) 2  Vrms   RL    D. Peak Inverse Voltage (PIV): It is defined as the peak value of input voltage across the reverse biased diode before the diode breaks down. It is essential that the diode used in rectifier circuit should be able to withstand the voltage available across it when it is reversed biased so that it acts as open circuit and does not enter into a breakdown state. Computation of DC Voltage and current for HWR Consider input signal VS Vm Sin ( t ) as the secondary voltage signal applied to the half wave rectifier. Then Vav the average or the DC component of the voltage across the load is can be computed using equation (1.2.1). 1   2 Vav  Vdc   Vm sin (t ) d (t )   0 d (t )  2  0   Vdc  Vm  cos (t )0  Vm (1.2.7) 2  V V I I dc  dc  m  m (1.2.8) RL RL  Computation of ripple factor for HWR RMS voltage at the load resistance can be calculated using 1.2.2 1  Vrms  2  0 (Vm Sin t ) 2 d (t ) (1.2.9) Vm2  1  cos 2 t V Vrms  2 0 ( 2 ) d (t )  m 2 (1.2.10) Using 1.2.10, and 1.2.7 in 1.2.5, the ripple factor is 2  Vm    2    2  1   1  1.21 (1.2.11)  Vm  4    Computation of efficiency for HWR Consider VS Vm Sin ( t ) as the secondary voltage signal applied to the half wave rectifier and using equation 1.2.7 and 1.2.10 in 1.2.6, 2  Vdc2   Vm        2 L R     4  0.406 or 40.6% (1.2.12)  Vrms   Vm  2 2    2  RL    Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 30  ECE - 1051 : Basic Electronics Computation of PIV for HWR For the HWR, during the negative half cycle the diode is reverse biased. The maximum voltage across the reverse biased diode will be equal to the secondary peak voltage. If the secondary voltage is VS Vm Sin ( t ) , the PIV of the diode should be greater than Vm , peak of secondary voltage. Self test 2: 1. Input AC signal of 25V peak value is to be rectifed using HWR. For proper working it is essential to choose the diodes whose PIV rating is (a) 5V (b) 15V (c) 30V (d) both a and b 2. In the circuit of figure 1.2.7 what happens when the diode connection is reversed? Draw the input and output waveforms. Will the values of PIV, ripple factor and efficiency for this circuit change? Solved Example 1. A sinusoidal secondary voltage of peak value 10V and frequency 50Hz is applied to half wave rectifier. If the load resistance is 800Ω. calculate average load current. Given: peak voltage Vm =10V, f =50Hz, RL=800 Ω V Solution: Vdc  m  3.18309 V (here V=volts)  V I dc  dc  3.978mA (here mA=milli Amperes) RL 1.2.4 Full Wave Rectifier Figure 1.2.11: Circuit diagram of a center tapped FWR Figure 1.2.12: waveform at different nodes of a center tapped Full wave rectifier Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 31  ECE - 1051 : Basic Electronics The circuit of full wave rectifier is shown in Figure 1.2.11 with the center tap of transformer grounded. The center tapped transformer consists of two input terminals (nodes) and three output nodes. The extreme nodes are labelled as node A and node B. The center node is usually used as a common reference voltage terminal relative to which all the voltages are measured. Hence the center output node of center tapped transformer is considered to be a common ground. The secondary voltage observed between the extreme end nodes i.e. between node A and node B is a stepped down voltage as shown in Figure 1.2.12(a). The voltages measured between node A and center node (ground) or node B and center node is half in magnitude in comparison to the voltage measured between node A and node B. Also voltage at node B is 180° out of phase with the voltage at node A when measured relative to the center node (ground). All these secondary waveforms are shown in Figure 1.2.12(b) and (c). Working of Center tapped Full wave rectifier circuit During the first half cycle, as shown in Figure 1.2.11, the voltage at node A is positive and voltage at node B is negative measured with respect to ground. Diode D1 gets forward biased and acts as a short whereas diode D2 is reverse biased and acts as open. Load resistor is connected at the junction of cathodes of the two diodes D1 and D2 with respect to ground. Considering ideal diodes, the equivalent circuit of a center tapped full wave rectifier for half cycle when voltage at node A with respect to node B is positive is as shown in Figure 1.2.13 This results in a current flow through upper half secondary windings of transformer, diode D1 and the load RL as shown in Figure 1.2.13. Direction of the current through the load is towards the ground from node C. Hence the output voltage measured at node C with respect Figure 1.2.13: Equivalent circuit of a center Figure 1.2.14: Equivalent circuit of a tapped FWR when node A positive center tapped FWR when node B positive to ground is equal to voltage at node A measured with respect to ground. Similarly, during the second half cycle, as shown in Figure 1.2.12, the voltage at node A is negative and voltage at node B is positive measured with respect to ground. Diode D1 gets reverse biased and acts as a open whereas diode D2 is forward biased and acts as short. The equivalent circuit of a center tapped full wave rectifier for half cycle when voltage at node A Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 32  ECE - 1051 : Basic Electronics with respect to node B is negative is as shown in Figure 1.2.14. This results in a current flow through lower half secondary windings of transformer, diode D2 and the load RL as shown in Figure 1.2.14. Again note that the direction of the current through the load is towards the ground from node C. Hence the output voltage measured at node C with respect to ground is equal to voltage at node B measured with respect to ground. The output waveform observed across load resistor along with voltage waveforms at node A and node B with respect to ground is shown in Figure 1.2.15. Figure 1.2.15: Secondary and output waveforms of center tapped FWR Analysis of Center tap FWR circuit: Computation of PIV for center-tapped FWR Consider Figures 1.2.15 and 1.2.16, the maximum voltage across the reverse biased diode will be twice the peak voltage measured between either of extreme ends and center node of secondary winding of a center tapped transformer. Hence PIV of the full wave rectifier is 2Vm. Computation of Ripple Factor for center-tapped FWR The average or the DC component of the voltage across the load using 1.2.1 is given by V av Vdc   Vm sin (t ) d (t )  m  cos (t )0  m 1  V  2V (1.2.13)   0    V 2V 2I I dc  dc  m  m (1.2.14) RL RL  RMS value of the voltage across the load using 1.2.2 is 1  Vm Im  Vrms  (Vm sin t ) 2 d (t )  and I rms  (1.2.15) 0 2 2 Using equation 1.2.13 and 1.2.15 in 1.2.5, the ripple factor is Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 33  ECE - 1051 : Basic Electronics 2  Vm    2   1    1  0.483 2    (1.2.16)  2Vm  8    Computation of Efficiency for center-tapped full wave rectifier From equation 1.2.6, the efficiency is calculated as 2  Vdc2   2Vm         2 L R   8  2  0.812 or 81.2% (1.2.17)  Vrms   Vm  2       RL   2 Self-test : Choose the correct answer: (T is the time period of the input signal) 1. In center tapped FWR, each diode is forward biased for what duration of the time period? (a) T/2 b) T/4 c) 3T/4 d) T 2. In center tapped FWR, current through the load flows for what duration of the time period? (a) T/2 b) T/4 c) 3T/4 d) T 3. Input AC signal of 25V peak value is to be rectifed using center tapped FWR. For proper working it is essential to choose the diodes whose PIV rating is (a) 5V (b) 15V (c) 30V (d) both a and b 1.2.5 Full Wave Bridge Rectifier The bridge rectifier consists of four diodes in the form of a bridge. two parallel paths can be represented with each path having two diodes and all diodes are directed in the same direction as shown in Figure 1.2.16. The top path consists of diodes D3 and D1 whereas bottom path consists of diodes D2 and D4 respectively. Load resistor is connected between the ends of two parallel paths (i.e. node C and ground). Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 34  ECE - 1051 : Basic Electronics Figure 1.2.16: Bridge rectifier represented in two ways If the secondary signal given to the bridge is as shown in Figure 1.2.17(a), the output voltage measured across load is expected to be as shown in Figure 1.2.17(b) assuming ideal diodes. Figure 1.2.19: Input – output waveforms of a bridge rectifier Figure 1.2.17: input and output waveforms of bridge FWR Working of Bridge FWR circuit During positve half cycle, node A is positive with respect to B, D1 and D2 are forward biased whereas D3 and D4 are reverse biased as shown in Figure 1.2.18. This results in current flow taking a closed path from node A, through D1, R-Load and D2 and from node B through the secondary coil as indicated in Figure 1.2.18. Note that current through the load resistor flows from node C to ground. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 35  ECE - 1051 : Basic Electronics Figure 1.2.18: Equivalent circuit of a Figure 1.2.19: Equivalent circuit of a bridge rectifier when node A is positive bridge rectifier when node B is positive During negative half cycle, node B is positive with respect to A, D3 and D4 are forward biased whereas D1 and D2 are reverse biased as shown in Figure 1.2.19. This results in current flow taking a closed path from node B, through D4, R-Load and D3 and from node A through the secondary coil as indicated in Figure 1.2.19. Note that again current through the load resistor flows from node C to ground. Thus output voltage is unidirectional for both the half cycles. Analysis of Bridge FWR circuit Consider Figures 1.2.18 and 1.2.19, maximum voltage across the reverse biased diode will be secondary peak voltage. Hence PIV rating of the diode should be greater than Vm. The Idc, Vdc, Irms, Vrms, Ripple factor, Efficiency are same as center tapped full wave rectifier. Self-test : Choose the correct answer: (T is the time period of the input signal) 1. In Bridge rectifier, each diode is forward biased for what duration of the time period? (a) T/2 b) T/4 c) 3T/4 d) T 2. In Bridge rectifier, current through the load flows for what duration of the time period? (a) T/2 b) T/4 c) 3T/4 d) T 3. Input AC signal of 25V peak value is to be rectifed using bridge. For proper working it is essential to choose the diodes whose PIV rating is (a) 5V (b) 15V (c) 30V (d) both a and b Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 36  ECE - 1051 : Basic Electronics Solved Example 2. Find the PIV rating of the diode used for proper working of the bridge rectifier when it is supplied with 230V, 50 Hz AC mains through a step down transformer with turns ratio equal to 10. Given: Input AC mains peak voltage =230V, turns ratio=10, Solution: Secondary peak voltage Vm =230/10=23V therefore PIV rating of diode >= Vm=23V 1.2.6 Comparison of Rectifiers Advantages of HWR over FWR  Simple circuit  Single diode  PIV rating is V m Disadvantages of HWR  High ripple factor  Low efficiency Advantages of Center tapped FWR rectifier over HWR  High Efficiency  Low ripple factor Advantages of bridge rectifier over to centre-tap full wave rectifier:  PIV rating is Vm  Centre-tap transformer is not required Disadvantage of bridge rectifier over to centre-tap full wave rectifier:  Need for four diodes Comparison of rectifier circuits based on the performance parametrs is listed in Table 1.2.1, considering the input signal Vi (t )  Vm sin(2  fi t ) Table 1.2.1: Comparison of Rectifiers Parameters of HWR Center-tapped FWR Bridge FWR rectified signal Vdc Vm 2Vm 2Vm    VRMS Vm Vm Vm 2 2 2 PIV ≥Vm ≥2Vm ≥Vm Ripple factor 1.21 0.483 0.483 Efficiency 40.6% 81.2% 81.2% Frequency fo fi 2 fi 2 fi Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 37  ECE - 1051 : Basic Electronics 1.2.7 Rectifier with Filter A capacitor filter with HWR and FWR is as shown Figure 1.2.20. The capacitor allows AC component and blocks DC component. The capacitor filter minimizes the ripple and increases the average value of output voltage. In each of the positive half cycle, the capacitor charges up to the peak value of the transformer secondary voltage, Vin. Capacitor tries to maintain this maximum value when the input drops to zero. The capacitor will discharge through the load resistance slowly until the input voltage again increases to a value greater than the capacitor voltage. The filtered output waveform for both HWR and FWR is as shown in figure 1.2.21. Figure 1.2.20: Filter circuit for (a) half wave rectifier and (b) full wave rectifier Large value of the product “CRL” results in a small ripple factor. Thus increasing C or RL (both) an approximate perfect DC voltage can be obtained. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 38  ECE - 1051 : Basic Electronics Figure 1.2.21: Filtered waveform (a) half wave rectifier and (b) full wave rectifier [floyd] (c) filtered and rectified wave (d) approximation of filtered wave The performance of the filter circuits is measured by ripple factor. The approximate filtered waveform shown in Figure 1.2.21(d). The ripple factor for Capacitor filter for the HWR and FWR is given by equation (1.2.18) and (1.2.19) respectively. 1 r (1.2.18) 2 3 fCRL 1 r (1.2.19) 4 3 fCR L The corresponding dc value for HWR and FWR is given by equation (1.2.20) and (1.2.21) respectively. 2 f CRL Vdc  Vm (1.2.20) 1  2 f CRL 4 f CRL Vdc  Vm (1.2.21) 1  4 f CRL Note 5: Here ‘f’ in equation (1.2.18) to (1.2.21) is the frequency of the input signal Comparison of ripple factor and the output dc voltage achieved from the input secondary Vi (t )  Vm sin(2ft) is listed in Table 1.2.2 Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 39  ECE - 1051 : Basic Electronics Table 1.2.2: Comparison of C-filter with HWR and FWR Parameters of HWR FWR rectified signal Vdc 2 f CRL Vm 4 f CRL Vm 1  2 f CRL 1  4 f CRL Ripple factor 1 1 r r 2 3 fCRL 4 3 fCRL Solved Exercise: 3. A sinusoidal voltage of peak value Vi  20 sin(2 50t ) V is applied to FWR. If the load resistance is 1000Ω. calculate the average and RMS value of load current, efficiency and ripple factor. Find the frequency of the output signal. Given Vm = 20V, f=50Hz, RL = 1000 Ω, Rf=10 Ω Solution: V 2I m I I m  m  20mA , I dc   12.73 mA , I rms  m  14.142 mA RL  2  V2   dc   RL  Efficiency    81.2% 2  Vrms   RL    I Ripple factor =  dc  0.6365 I ac FWR output signal frequency =2 x frequency of the input signal =100Hz. 4. A particular load has to be supplied with 10 mA average current at 5 V dc voltage, with ripple factor not more than 10%. Calculate the value of the filter capacitor that needs to be connected to the output of full wave bridge rectifier. Given: Idc = 10 mA, Vdc = 5 V, r = 0.1, Note: Input is not given but can be assumed appropriately. For example in INDIA, AC mains has Vi(rms) = 230  2 V, f = 50 Hz Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 40  ECE - 1051 : Basic Electronics Solution: V 2I m I I m  m  20mA , I dc   12.73 mA , I rms  m  14.142 mA RL  2  V2   dc   RL  Efficiency    81.2% 2  Vrms   RL    I Ripple factor =  dc  0.6365 I ac FWR output signal frequency =2 x frequency of the input signal =100Hz. 5. A particular load has to be supplied with 10 mA average current at 5 V dc voltage, with ripple factor not more than 10%. Calculate the value of the filter capacitor that needs to be connected to the output of full wave bridge rectifier. Given: Idc = 10 mA, Vdc = 5 V, r = 0.1, Note: Input is not given but can be assumed appropriately. For example in INDIA, AC mains has Vi(rms) = 230  2 V, f = 50 Hz RL = Vdc / Idc = 5 V / 10 mA = 500 Ω 1 1 Solution: r. Hence, C  = 57.74 µF 4 3 fCRL 4 3 frRL 6. For the data provided in problem 4, find the turns ratio required for the transformer. Solution: Using equation (1.2.21), the peak value of the secondary is V (1  4 f CRL ) Vm  dc =5.866 V 4 f CRL By assumption AC mains Vi(rms) = 230  2 to get secondary peak voltage 5.866V a 230  2 step down transformer. The turns ratio = =55 5.886 Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 41  ECE - 1051 : Basic Electronics Summary 1. A pure sinusoidal signal has an average value equal to zero. It means the dc value of this signal is zero. 2. A DC power supply unit consists of : Step down transformer, Rectifier circuits, Filter circuit and Regulator. 3. In a HWR the diode conducts only during positive half cycle and hence the circuit is referred as ‘Half Wave Rectifier’. The rectified voltage is pulsating DC, which can be smoothened using filter circuits. 4. In a FWR the diodes conduct in both half cycles and hence the name Full Wave Rectifier. 5. In a bridge rectifier there are four diodes. 6. A capacitor filter is used in rectifier circuits to remove DC components called as ripples. Exercise: 1. Primary voltage is 120V, 60Hz. Turns ratio is 5:1. This transformer supplies to bridge rectifier employing 4 identical ideal diodes. The load resistance is 1kΩ. Calculate average and rms load voltage, efficiency, ripple factor, PIV rating and the frequency of output waveform. (Ans.: 108V, 120V, 81%, 0.484, 169.7V,120HZ) 2. Repeat this problem for center tapped FWR. Comment on the results comparing it with results of problem 1. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 42  ECE - 1051 : Basic Electronics 3. A half wave rectifier with capacitor filter is supplied from a transformer having peak secondary voltage 20V and frequency 50Hz. The load resistance is 560Ω and capacitor used is 1000μF. Calculate ripple factor and dc output voltage. Draw the filtered output and label peak and dc value. (Ans. 0.0103, 19.65V) 4. Repeat problem 3 for a full wave rectifier 5. A half wave rectifier with capacitor filter has to supply an average voltage of 30V to 900Ω load. Calculate the rms input voltage and value of capacitor needed to get ripple factor of 0.05, assuming f = 50Hz. (Ans. : a) 23V, 128.3μF) 6. Repeat problem 5 for a full wave rectifier. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 43  ECE - 1051 : Basic Electronics Module – 3: Voltage Regulators Learning Outcomes: At the end of this module, students will be able to: 1.Describe the working of Zener as voltage regulator 2.Understand the IC based voltage regulator. 1.3.1 Zener Voltage Regulator Zener diodes are widely used to regulate the voltage. It is connected in parallel with a variable voltage source in such a way that it is reverse biased, zener diode acts as an open circuit until teh voltage reaches the diode's reverse breakdown voltage and the voltage at the output is equal to the voltage applied. Once the voltage goes beyond the breakdown voltage, the voltage across the diode remains constant at the value equal to the breakdown voltage. This behavior of the Zener diode make it a good reference voltage source and one of its major application is in voltage regulators. Figure 1.3.1 is a simple voltage control circuit using a Zener diode. Figure. 1.3.1. Zener Voltage Regulator Circuit [http://rsandas.com/P2_Session_4-4.html]. In Figure 1.3.1, there is a power supply that outputs unregulated voltage of 12-volts. A load, represented by resistor RL , requires a regulated 5-volt source. Using a 5-volt Zener diode as illustrated, this requirement can be met. In the circuits, load drops a portion of the source voltage. Since the Zener Diode is in series with resistor R, Zener diode will drop 5- volts (VZ) and series resistor R will drop the remaining 7 Volts. With the load, R L, connected across the Zener Diode, it will provide a constant 5-volts; regardless of any variation of the power supply. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 44  ECE - 1051 : Basic Electronics If the power supply voltage drops to 10-volts from initial 12 volts as mentioned above. The Zener diode still drops 5-volts (designed voltage) and the series resistor R will drop 5 volts. Again, because the load is connected across the Zener diode, it will produce 5- volts. Therefore, irrespective of change in line voltage, output voltage remains constant of 5V. Self test: 1. If the series resistance increases in an unloaded Zener regulator, the Zener current a. Decreases b. Stays the same c. Increases d. Equals the voltage divided by the resistance 2. In a loaded Zener regulator, which is the largest current? a. Series current b. Zener current c. Load current d. None of these 3. If the load resistance increases in a Zener regulator, the Zener current a. Decreases b. Stays the same c. Increases d. Equals the source voltage divided by series resistance 4. A voltage regulator is a circuit which a. Converts the ac voltage to dc voltage b. Smoothens the ac variation in dc output voltage c. Maintains a constant dc output voltage inspite of the fluctuations in ac input voltage or load current d. None of the above Zener diode used for line and load regulation A zener diode can be used as a shunt voltage regulator (shunt meaning connected in parallel), and voltage regulator being a class of circuit that produces a stable voltage across varying load and input voltages. Consier the circuit for the voltage regulator as shown in Figure 1.3.2. The circuit has to mainatin constant voltage across a load resisitor RL. The circuit holds the voltage across the load RL almost equal to the voltage across zener VZ even after the input Vin and load resistor RL undergo changes. If the unregulated DC voltage Vin rises, the current through R increases. This extra current is directed to the zener diode instead of flowing through the load. The zener diode voltage is virtually unaffected by the increase in this current and load voltage which is same as the diode voltage Vz remains constant. If the load Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 45  ECE - 1051 : Basic Electronics requires more current when RL is decreased, the zener diode can supply the extra current without affecting the load voltage. Figure 1.3.2. Zener voltage regulator Let I be the current through the resister R , and it can be written as I  Iz  IL Vin  Vz I  (1.3.1) R The power dessipated in the diode is Pz = IZ Vz (1.3.2) The selection of R can be done by using the equation, Vin  VZ R (1.3.3) I After substituting the value if I , Vin  VZ R (1.3.4) IZ  IL VZ (i) For Line regulation RL is constant and IL  R is also constant and Vin varies between L Vin(min) to Vin(max) (ii) For Load Regulation, Vin is constant and RL varies between RLmin and RLmax and load VZ VZ current is given by I Lmin  and I Lmax  R Lmax R Lmin When Vin=Vin(min),and IL is constant then Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 46  ECE - 1051 : Basic Electronics Vin(min)  Vz I min  , (1.3.5) R I min  I z(min)  I L (1.3.6) Vin (max)  VZ Similarly when Vin=Vin(max) I max  , (1.3.7) R I max  I Z ( nax)  I L (1.3.8) The selected R must be small enough to permit minimum zener current to ensure that the diode is in its breakdown region. That is R must be small enough to ensure that minimum current IZ(min ) flows under worst condition. This is when Vin falls to its smallest possible value Vin(min) and IL is its largest possible value ILmax (Load Regulation). At the same time R must be selected large enough to ensure that the current through the zener diode should not exceed the maximum zener current Iz(max) so that power desipation in the diode will not exceed Pz. That is the condition when Vin rises to the value of Vin(max) and load current IL to its minimum ILmin Therefore, Vin(min)  VZ Vin(max)  VZ R and R (1.3.9) I zmin  I Lmax I zmax  I Lmin Applications:  As Voltage regulators  As Voltage Limiters  Wave shaping  Protection diode  Fixed reference voltage Example of Zener Regulation with varying input voltage (Line regulation) A Zener diode can be used as a voltage regulator. To illustrate this, let’s use a Zener diode 1N4740A in the circuit as shown in Figure 1.3.3. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 47  ECE - 1051 : Basic Electronics Figure 1.3.3 Zener diode as regulator for line regulation. As VIN changes IZ changes, the limitations on the input voltage variation (VIN(.min) and VIN(.max) ) are set by the minimum and maximum current levels (IZK and IZM ) with which the Zener diode can operate. The minimum current value IZK = 0.25 mA (from the 1N4740A Zener diode datasheet). Maximum current can be calculated from the power specification ratings, PD(.max) = 1 Watt as follows: IZM = PD(.max) / VZ = 1W/10V=100 mA (1.3.10) For the minimum Zener current, the voltage across the 220 Ω resistor is: VR = IZK x R = 0.25 mA x 220 Ω = 55 mV (1.3.11) Since VIN = VR + VZ, then VIN(.min) = VR + VZ = 55 mV + 10 V = 10.055 V (1.3.12) For the maximum Zener current, the voltage across the 220 Ω resistor is: VR = IZM x R = 100 mA x 220 Ω = 22 V (1.3.13) Therefore VIN(.max) = VR + VZ = 22 V + 10 V = 32 V (1.3.14) This shows that the Zener diode 1N4740A can ideally regulate voltage from 10.055 V to 32 V, and maintain an approximate 10 V output. Example of Zener Regulation with a variable load (load regulation) Figure 1.3.4 shows a Zener voltage regulator with a variable load resistor across its terminal: Figure 1.3.4 Zener diode as regulator for load regulation. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 48  ECE - 1051 : Basic Electronics To understand the Zener Regulation with a variable load (load regulation), consider the following example. Design a zener regulator circuit to meet the following specifications: Load voltage=8V, input voltage=30 V, Load current=0-50 mA, Izmin= 5 mA, P = 1W Vin(min) = Vin(max)= 30 V. ILmin= 0A, ILmax= 50 mA PZ 1 V 8 I Z max    125mA , RL min  o   160 Vo 8 I L max 50mA To find current limiting series resistance, Vin min  Vo 30  8 Rmax    400 I L max  I Z min 0.05  0.005 Vin max  Vo 30  8 Rmin    176 I L min  I Z max 0  0.125 Example Problem 1: 1. An 1N756 zener diode is used as a 12 V regulator in the circuit shown below: What is the smallest load resistor that can be used before losing regulation? Assume the ideal model for the zener diode? Solutions: The no load zener current INL = (VIN – VZ) / R = (24 V-12 V) / 470 Ω = 25.5 mA This is the maximum load current in regulation, therefore the minimum value of load resistance RL(min)= VZ / INL = 12 V/ 25.5 mA = 470 Ω. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 49  ECE - 1051 : Basic Electronics Note that if RL is less than 470 Ω it will draw more of the total current away from the zener diode and IZ will be reduced below IZK. This will cause the zener diode to come out of break down and hence output will not be regulated. 2. A certain zener diode has a V z = 7.5 V and an Rz = 5 Ω at a certain current. Draw the equivalent circuit. Solutions: The equivalent circuit is shown below. 3. When the reverse current in a particular zener diode increases from 20 mA to 30 mA, the zener voltage changes from 5.6 V to 5.65 V. What is the resistance of this device? Solutions: VZ 5.65  5.6 RZ    5 I Z 30mA  20mA 4. Determine the minimum input voltage required for regulation to be established in the figure shown below. Assume an ideal zener diode with minimum zener current = 1.5 mA and Vz = 14 V. Solutions: VIN (min)  VZ  I ZK R  14  (1.5mA)(560)  14.8V Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 50  ECE - 1051 : Basic Electronics 1.3.2 IC Voltage Regulators IC voltage regulators are versatile, relatively inexpensive and are available with features such as programmable output, current / voltage boosting IC voltage regulators are available as 1. Fixed voltage regulator 2. Adjustable voltage regulators Fixed voltage regulators: 78XX series are three terminal positive voltage regulators. In 78XX, XX indicates the output voltage. They are available as 7805, 7806, 7808, 7815, 7818, and 7819. 79XX series are negative fixed voltage regulators which are complements to the 78XX series devices. MC7805 is a 3-terminal positive voltage regulator. It is designed for a wide range of applications. An example is shown in Figure 1.3.5. Figure 1.3.5: IC voltage regulators The internal limiting and thermal shutdown features of this regulator makes it essentially immune to overload. When used as a replacement for a Zener diode-resistor combination, an effective improvement in output impedance can be obtained together with lower-bias current. For output current up to 1A, no external components are required. The input capacitor is used to cancel the inductive effects due to long distributive leads and the output capacitor to improve the transient response. IC regulator like LM117, LM317, LM338 are adjustable voltage regulators. The LM117 series of adjustable 3-terminal positive voltage regulators is capable of supplying in excess of 1.5A over a 1.2V to 37V output range. They are exceptionally easy to use and require only two external resistors to set the output voltage. Further, both line and load regulations are better than standard fixed regulators. Normally, no capacitors are needed unless the device is situated more than 6 inches from the input filter capacitors in which case an input bypass is needed. An optional output capacitor can be added to improve transient response. Summary 1. Zener diodes are used as voltage references, regulators, and limiters. 2. Zener diodes are available in many voltage ratings ranging from 1.8 V to 200 V. Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 51  ECE - 1051 : Basic Electronics Exercise problems 1. Define Rectification, filtering and regulation. 2. List the applications of Varactor diode. 3. (a) Consider the circuit shown below. The Zener Diode regulates at 50V over a range of diode current from 5 mA to 40 mA. Supply voltage V = 200V. Calculate