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Diodes 1. Diode Class of non-linear circuits –having non-linear v-i Characteristics Uses –Generation of : DC voltage from the ac power supply Different.

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Presentation on theme: "Diodes 1. Diode Class of non-linear circuits –having non-linear v-i Characteristics Uses –Generation of : DC voltage from the ac power supply Different."— Presentation transcript:

1 Diodes 1

2 Diode Class of non-linear circuits –having non-linear v-i Characteristics Uses –Generation of : DC voltage from the ac power supply Different wave (square wave, pulse) form generation –Protection Circuits –Digital logic & memory circuits

3 Creating a Diode A diode allows current to flow in one direction but not the other. When you put N-type and P-type silicon together gives a diode its unique properties.

4 .

5 Diode Equivalent circuit in the reverse direction Equivalent circuit in the forward direction.

6 Reverse Bias -ve voltage is applied to Anode Current through diode = 0 ( cut off operation ) Diode act as open circuit Forward Bias +ve voltage applied to Anode Current flows through diode voltage Drop is zero (Turned on) Diode is short circuit Operation

7 The two modes of operation of ideal diodes Forward biased Forward Current 10 mA Reverse biased Reverse Voltage 10 V

8 Ex 3.2

9 Rectifier circuit Input waveform Equivalent circuit when v i  0 Equivalent circuit when v i ≤ 0 Waveform across diode Output waveform.

10 Exercise 3-3

11

12

13 Battery Charger

14 OR gate v y = v A +v B +v C v y = v A.v B.v C Diode logic gates AND gate (in a positive-logic system)

15 Figure 3.6 Circuits for Example 3.2. Diodes are ideal, Find the value of I and V

16 Example 3.2. Assumption Both Diodes are conducting

17 Node A Node B Assumption Both Diodes are conducting Not Possible Thus assumption of both diode conducting is wrong

18 Example 3.2(b). Assumption # 2 Diodes 1 is not conducting Diodes 2 is conducting Assumption is correct

19 Figure E3.4 Diodes are ideal, Find the value of I and V

20 Figure E3.4 Diodes are ideal, Find the value of I and V I= 2mA V= 0V I= 0A V= 5V I= 0A V= -5V I= 2mA V= 0V

21 Figure E3.4 Diodes are ideal, Find the value of I and V I= 3mA V= 3V I= 4mA V= 1V

22 Figure P3.2 Diodes are ideal, Find the value of I and V

23 Figure P3.2 Diodes are ideal, Find the value of I and V Diode is conducting I = 0.6 mA V = -3V Diode is cut-off I = 0 mA V = 3V Diode is conducting I = 0.6 mA V = 3V Diode is cut-off I = 0 mA V = -3V

24 D1 Cut-Off & D2 Conducting I = 3mA Problem 3-3 D1 Cut-Off & D2 Conducting I = 1mA, V=1 V Diodes are ideal, Find the value of I and V

25 Figure P3.4 In ideal diodes circuits, v 1 i s a 1-kHz, 10V peak sine wave. Sketch the waveform of v o

26 In ideal diodes circuits, v 1 i s a 1-kHz, 10V peak sine wave. Sketch the waveform of v o V p+ = 10V V p- = 0V f = 1 K-Hz V p+ = 0V V p- = - 10V f = 1 K-Hz V o = 0V

27 Figure P3.4 In ideal diodes circuits, v 1 s a 1-kHz, 10V peak sine wave. Sketch the waveform of v o V p+ = 10V V p- = 0V f = 1 K-Hz V p+ = 10V V p- = -10V f = 1 K-Hz V p+ = 10V V p- = 0V f = 1 K-Hz

28 Figure P3.4 In ideal diodes circuits, v 1 s a 1-kHz, 10V peak sine wave. Sketch the waveform of v o

29 Figure P3.4 In ideal diodes circuits, v 1 s a 1-kHz, 10V peak sine wave. Sketch the waveform of v o V p+ = 0V V p- = -10V f = 1 K-Hz V 0 = 0VV p+ = 10V V p- = -5V f = 1 K-Hz

30 Figure P3.4 In ideal diodes circuits, v 1 s a 1-kHz, 10V peak sine wave. Sketch the waveform of v o V p+ = 10V V p- = -5V f = 1 K-Hz

31 Problem 3-4(k) For Vi > 0 V D1 is cutoff D2 is conducting v o =1V For Vi < 0 V is conducting D2 is cutoff v o =v i +1V - 9 V

32 Problem 3-4(k)

33 Figure P3.6 X = A. B X = A + B

34 Problem 3-4 (c) v o =zero

35 Problem 3-4(f) +ve Half Cycle with 10 V peak at 1 KHz Vi is a 1 kH z 10-V peak sine wave.

36 Problem 3-4(h) v o =zero

37 Problem 3.5 v i is 10 V peak sine wave and I = 100 mA current source. B is battery of 4.5 V. Sketch and label the i B 4.5 v 100 mA

38 Solution P3-5 vovo 4.5 v 100 mA

39 Problem mA 4.5 v

40 Problem mA

41 REVERSE POLARITY PROTECTOR

42 The diode in this circuit protects a radio or a recorder etc... In the event that the battery or power source is connected the wrong way round, the diode does not allow current to flow. REVERSE POLARITY PROTECTOR

43 D1& D2 Conducting I 1 =1mA I 3 =0.5 mA I 2 =0.5 mA V= 0 V D1=off, D2=On I 1 = I 3 =0.66 mA V = -1.7 V Problem 3-9 I1I1 I3I3 2 I1I1 I3I3 2

44 Problem 3-10 D conducting I=0.225 mA V=4.5V D is not conducting I=0A V=-2V

45 Problem 3-16 VREDGREEN 3VOnOffD 1 conducts 0 VOffOff -3 VOffOnD 2 conducts

46 Quiz No 3 DE 28 EE -A Sketch v O if v i is 8 sin  Find out the conduction angle for the diode & fraction of the cycle the diode is conducting

47 Solution Quiz No 3 8V I1I1 I2I2 v i /2 I

48 Sketch v O if v i is 10 sin  Find out the conduction angle for the diode & fraction of the cycle the diode is conducts D 1 never conducts Vi<5V D2 is cut-off, Vo=5V Vi>5V D2 is conducts +12 V 5 D1D1 D2D2

49 Quiz No 3 DE 27 CE -B D 1 never conducts Vi<5V D2 is cut-off, Vo=Vi Vi>5V D2 is conducts Sketch v O if v i is 10 sin  Find out the conduction angle for the diode & fraction of the cycle the diode is conducts

50 Problem Assume the diodes are ideal, sketch v o if the input is 10sin  (9) Find out the conduction angles for Diode D 1 & D 2 (4) and the fraction of the cycle these diodes conduct. (2)

51

52 -2V

53 The earliest commercial semiconductor devices mostly used Germanium. This element has 32 electrons per atom and melts at 985 °C. It has now largely fallen into disuse because it is much rarer and more expensive than Silicon and has no real advantages for most purposes. Semiconductor materials Germanium

54 Current flow in Semiconductors An electric current can flow through a semiconductor as a result of the movement of holes and/or free electrons. There are two important processes that account for current flow in semiconductors. These processes are called drift and diffusion.

55 Drift Applying an electric field across a semiconductor will cause holes and free electrons to drift through the crystal The total current is equal to the sum of hole current (to the right) and electron current (to the left).

56 Diffusion A drop of ink in a glass of water diffuses through the water until it is evenly distributed. The same process, called diffusion, occurs with semiconductors. If some extra free electrons are introduced into a p-type semiconductor, the free electrons will redistribute themselves so that the concentration is more uniform. The free electrons will tend to move to the right. This net motion of charge carriers constitutes a diffusion current. The free electrons move away from the region of highest concentration. The higher the localized concentration, the greater will be the rate at which electrons move away. The same process applies to holes in an n-type semiconductor. Note that when a few minority carriers are diffusing through a sample, they will encounter a large number of majority carriers. Some recombination will occur. A number of both types of carrier will be lost.

57 The circles represent the inner core of silicon atoms, with +4 indicating its positive charge of +4q, which is neutralized by the charge of the four valence electrons. Observe how the covalent bonds are formed by sharing of the valence electrons. At 0 K, all bonds are intact and no free electrons are available for current conduction. Two-dimensional representation of the silicon crystal.

58 14 Electrons

59 Silicon Crystal All silicon atoms bond perfectly to four neighbors Leaving no free electrons to conduct electric current This makes a silicon crystal an insulator rather than a conductor.

60 Silicon Crystal Carbon, silicon and germanium -- each has four electrons in its outer orbital. The four valence electrons form perfect covalent bonds with four neighboring atoms, creating a lattice.atoms In silicon, the crystalline form is a silvery, metallic-looking substance. All of the outer electrons in a silicon crystal are involved in perfect covalent bonds, so they can't move around. In spite of four valence electrons, a pure silicon crystal is nearly an insulator -- very little electricity will flow through it.

61 Silicon and Germanium

62 Silicon Lattice

63 At room temperature, some of the covalent bonds are broken by thermal ionization. Each broken bond gives rise to a free electron and a hole, both of which become available for current conduction.

64 Intrinsic Semiconductor Electrons and holes

65 Semiconductor Current

66 The Valence Band The valence band is the band made up of the occupied molecular orbitals and is lower in energy than the so-called conduction band. It is generally completely full in semi-conductors. When heated, electrons from this band jump out of the band across the band gap and into the conduction band, making the material conductive.

67 The Fermi Level The Fermi Level is defined as the highest occupied molecular orbital in the valence band at 0 K, so that there are many states available to accept electrons, if the case were a metal. It should be noted that this is not the case in insulators and semiconductors since the valence and conduction bands are separated. Therefore the Fermi- Level is located in the band gap. The probability of the occupation of an energy level is based on the Fermi function

68 Conduction Band The conduction band is the band of orbitals that are high in energy and are generally empty. In reference to conductivity in semiconductors, it is the band that accepts the electrons from the valence band.

69 Doping Silicon You can change the behavior of silicon and turn it into a conductor by doping it. In doping, you mix a small amount of an impurity into the silicon crystal. There are two types of impurities: –.N-type – P-type -. A minute amount of either N-type or P-type doping turns a silicon crystal from a good insulator into a viable conductor -- hence the name "semiconductor." N-type and P-type silicon are not that amazing by themselves; but when you put them together, you get some very interesting behavior at the junction.

70 . The Doping of Semiconductors

71 Valence Electrons

72 N - Type Silicon Doping In N-type doping, phosphorus or arsenic is added to the silicon in small quantities.phosphorusarsenic Phosphorus and arsenic each have five outer electrons, The fifth electron has nothing to bond to, So it's free to move around. It takes only a very small quantity of the impurity to create enough free electrons to allow an electric current to flow through the silicon. N-type silicon is a good conductor. Electrons have a negative charge, hence the name N-type

73 A silicon crystal doped by a pentavalent element. Each dopant atom donates a free electron and is thus called a donor. The doped semiconductor becomes n type.

74 N Type

75 N-Type Semiconductor

76 P – Type Silicon Doping In P-type doping, boron or gallium is the dopant.borongallium Boron and gallium each have only three outer electrons. When mixed into the silicon lattice, they form "holes" in the lattice where a silicon electron has nothing to bond to. The absence of an electron creates the effect of a positive charge, hence the name P-type. Holes can conduct current. A hole happily accepts an electron from a neighbor, moving the hole over a space. P-type silicon is a good conductor.

77 A silicon crystal doped with a trivalent impurity. Each dopant atom gives rise to a hole The semiconductor becomes p type.

78 P Type

79 P-Type Semiconductor

80 n and p Type Semiconductors

81 Bands for Doped Semiconductors

82 P-N Junction

83 In practice, both the p and n region are part of the same silicon crystal pn junction is formed within a single silicon crystal by creating region of different doping (p & n regions) Atoms are held in the position by bonds – called convalent bond formed by its four reliance electrons. At low temperature, no free electrons are available

84 At room temperature, some of the bonds are broken by thermal ionization & some electrons are freed. Parent atom is left with positive charge after electron leaves. This attracts electrons form a neighboring atoms & other atoms is positively charge. Process repeats & Holes are available to conduct electric current Electrons fill the hole & this process is known as recombination.

85 Two Mechanisms Diffusion –Associated with random motion of electrons & holes due to thermal agitation –with uniform concentration of free electrons & holes no net flow of charge is resulted (no current) –if concentration of say free electrons (or holes) is higher in one part of silicon then the other (doping), electrons will diffuse from the region of higher concentration to the region of lower concentration

86 Two Mechanisms –This diffusion process given rise to a net flow of charge & diffusion current Drift –Carrier drift occurs when an electric field is applied across a piece of silicon –Electronic & Holes are accelerated by the electric field & acquire a velocity component (superimposed on the velocity of then thermal motion) called Drift velocity –Causing Drift current flow

87 Doped Semiconductor Material in which one kind (Electrons & Holes) predominate – majority of charged carrier are electron for n type & majority of charged carrier are holes for p types –Phosphorus impurity donated electrons (donor) –Boron impurity donated holes (Acceptor)

88 p-n Junction P Junction –Concentration of holes is high –Majority charge carrier are hole N Junction –Concentration of electron is high –Majority charge carrier are electron

89 Diffusion Current I D Hole diffuse across the junction from the p side to the n side & similarly electron Two current components add together to form the diffusion current with direction from p to n side

90 Depletion Region Recombination of Hole & electrons take close to the junction thus depletion region is formed on both side of the junction Region acts as a barrier that has to be overcome for holes to diffuse into the n region & same is true for electron thus causing Diffusion Current I D to flow from p side to n side

91 Drift Current I s Diffusion current due to majority carrier diffusion A component due to minority carrier drift exists across the junction

92 p-n Junction Due to electric field applied across the p-n junction –Electrons are moved by drift from p to n junction –Holes are moved by drift from n to p junction –Add together to form the drift current I s –Direction of I s is from n to p junction –Drift current is carried by thermally generated minority p carriers –Drift current value depends upon temperature –Independent of the value of the depletion layer voltage v o Under operative circuit conditions no external current flows I D = I S

93 p-n Junction open circuit Junction Built in Voltage V o will depends upon –Temperature –Doping Concentration Depletion region width –Same on both side of p & n junction if doping levels are equal –If doping levels are not equal the depletion layer will extend deeper into the more lightly doped material

94 (a)The pn junction with no applied voltage (open-circuited terminals). (b) The potential distribution along an axis perpendicular to the junction.

95 Depletion Region

96 Forward Biased Conduction

97 The polarity of applied voltage which can't produce any current is called Reverse Bias. The polarity of applied voltage which causes charge to flow through the diode is called Forward Bias.

98

99 The pn junction excited by a constant-current source I in the reverse direction. To avoid breakdown, I is kept smaller than I S. Note that the depletion layer widens and the barrier voltage increases by V R volts, which appears between the terminals as a reverse voltage.

100 The charge stored on either side of the depletion layer as a function of the reverse voltage V R.

101 The pn junction excited by a reverse-current source I, where I > I S. The junction breaks down, and a voltage V Z, with the polarity indicated, develops across the junction.

102 The pn junction excited by a constant-current source supplying a current I in the forward direction. The depletion layer narrows and the barrier voltage decreases by V volts, which appears as an external voltage in the forward direction.

103 Minority-carrier distribution in a forward-biased pn junction. It is assumed that the p region is more heavily doped than the n region; N N D.

104 Terminal Characteristics of a Junction Diode

105 The diode i– v relationship with some scales expanded and others compressed in order to reveal details. The diode i–v relationship

106 Terminal Characteristics of a Junction Diode Forward Biased Region v > 0 Reversed Biased Region v < 0 Breakdown Region v < - V ZK

107 Forward Biased Region I s Saturation current – Scale Current –I s is constant at a given temperature –I s is directly proportional to Cross-Sectional region of the diode, I s doubles if cross-sectional area is double –I s is A for small size diode –Doubles in value for every 10 O C rise in temperature

108 Thermal Voltage V T –V T = kT/q K = Boltzmann’s constant = 1.38 X Joules/Kelvin T = Absolute Temperature in Kelvin (273 +Temp in C o ) q = Magnitude of charge = 1.6 X Coulombs –V 20 o C is 25.2m V, ~ 25 mV n is 1 or 2 depending on the material and the physical structure of the diode –n = 1 for Germanium Diode & n=2 for Silicon Forward Biased Region

109 Switching of a Diode The switching speed of a diode depends upon: – its construction and fabrication. In general the smaller the chip the faster it switches, other things being equal. The chip geometry, doping levels, and the temperature at nativity determine switching speeds. The reverse recovery time, t rr, is the time it takes a diode to switch from on to off.

110 b Relationship of the current i to the voltage v holds good over many decades of current (seven decades, a factor of 10 7 Forward Biased Region i >> I s

111 Forward Biased Region

112 for v drop changes by for n = 1 for n = 2 Forward Biased Region

113 At a constant current, the voltage drop decreases by approximately 2 mV for every 1  C increase in temperature. Illustrating the temperature dependence of the diode forward characteristic

114 If V=1V at 20 o C, Find V at 40 0 C and 0 0 C At 20 o C Reverse current I s = 1V/1M Ω= 1μ A Since the reverse leakage current doubles for every 10 0 C increase, At 40 0 C I = 4*1 = 4 μ A V = 4 μ A * 1MΩ = 4.0 V At 0 C I = ¼ μ A V = 0.25 V IsIs Figure E3.9

115 Forward biased Diode Characteristics Example 3.3 A silicon diode displays a forward voltage of 0.7 V at a current of 1mA. Find I s at n=1 & 2

116 Ex 3.7 Silicon Diode with n=1 has V D i=1mA. Find voltage drop at i=0.1mA & 10mA

117 Solution P3-18 (a)At what forward voltage does a diode for which n=2 conduct a current equal to 1000Is ? (b)In term if Is what current flows in the same diode when its forward voltage is 0.7 V

118 Problem 3-23 The circuit shown utilizes three identical diodes having n=1 and Is= A. Find the value of the current I required to obtain an output voltage Vo=2 V. Assume n=1 If a current of 1mA is drawn away from the output terminal by a load, what if the change in the output voltage. Assume n=1

119 Solution 3-23 The circuit shown utilizes three identical diodes having n=1 and I s = A. Find the value of the current I required to obtain an output voltage V o =2 V. If a current of 1mA is drawn away from the output terminal by a load, what if the change in the output voltage.

120 Problem 3-25 In the circuit shown, both diode have n=1, but D 1 has 10 times the junction area of D 2. What value of V results?

121 Solution 3-25(a) In the circuit shown, both diode have n=1, but D 1 has 10 times the junction area of D 2. What value of V results?

122 solution 2-25 (b) To obtain a value of 50 mV, what current I 2 id needed.

123 Problem 3-26 For the circuit shown, both diodes are identical, conducting 10mA at 0.7 V and 100 mA at 0.8 V. Find ‘n’ Find the value of R for which V = 80 m V.

124 Solution 3-26 (a) Find η Find R if Vo=80mV

125 Figure P3.28

126 Problem 3.36 Assuming identical diodes for which V D I D =1mA. Find R if V 0 = 3 V

127 Modeling the Diode Forward Characteristics

128 A simple circuit used to illustrate the analysis of circuits in which the diode is forward conducting.

129 Graphical analysis of the circuit using the exponential diode model.

130 Iterative Analysis using the Exponential Model Determined the diode current I D and Diode voltage V D with V DD =5V and R =1000 ohms. Diode has a current of a V D of.7 V, and that its voltage drop changes by 0.1 V for every decade change in current.

131 Solution

132 The Piecewise-Linear Model

133 Approximating the diode forward characteristic with two straight lines: the piecewise-linear model.

134 The Piecewise-Linear Model Exponential curve is approx into two straight lines Line No 1 with zero slope & Line 2 with a slope of 1/r d The voltage change of less than 50 mV is observed in case the current change from 0.1 mA to 10 mA.

135 Piecewise-linear model of the diode forward characteristic and its equivalent circuit representation.

136 Piecewise-linear model

137 The Constant – Voltage Drop Model

138 Constant – Voltage Drop Model Forward conducting diode exhibits a constant voltage drop V D The voltage change of less than 50 mV is observed in case the current change from 0.1 mA to 10 mA. Model is used when –Detailed information about diode characteristics in not available

139 Constant-voltage-drop model

140 The constant-voltage-drop model of the diode forward characteristics and its equivalent-circuit representation.

141 Ex 3.12 Design a cct to provide an output of 2.4 V with available diodes have a current of a V D of 0.7 V, and that its voltage drop changes by 0.1 V for every decade change in current.

142 Voltage drop changes by 0.1 V for every decade change in current Solution Ex 3.12

143 Solution

144 The Small – Signal Model A small ac signal is superimposed on the DC components. First determined dc Operating Point Then small signal operation around the operating point –Small portion of the curve is approximated as almost linear segment of the diode characteristics.

145 The Small – Signal Model

146 Figure 3.17 Development of the diode small-signal model. Note that the numerical values shown are for a diode with n = 2.

147 The Small – Signal Model

148

149 Modeling the Diode Forward Characteristic

150 Table 3.1 (Continued)

151

152

153 + - VDVD IDID + - vdvd Exp 3-6

154 Solution Input variation of 10% resulted in output voltage variation of mV(0.8%) Voltage regulation

155 Exercise 3-16 Design a circuit shown so that Vo=3v when I L =0 A and Vo changes by 40 mV per 1mA of diode current. (a) Find the value of R (b)The junction area of each diode relative to a diode with ).7 V drop at 1mA current. Assume n=1

156 Excercise 3-16 Why 4 diodes and not 5? Diodes will not conduct at 0.6 V

157 Diode Forward Drop in Voltage Regulation Small signal model is used. Voltage remains constant in spite of : –Changes in load current –Changes in the dc power supply voltages One diode provides constant voltage of 0.7 V and for greater voltages diodes can be connected in series.

158 Example 3-7 A string of three diodes is used to provide a constant voltage of about 2.1 V. We want to calculate the percentage change in this regulated voltage caused by (a) a + 10 % change to the power supply voltage (b) Connection of a 1 K ohms load resistance, Assume n=2

159 Solution Exp 3-7

160 P 3-53 In a particular cct application, ten “20 mA diodes” ( a 20 mA diode is a diode that provides a 0.7 V drop when the current thru it is 20 mA) connected in parallel operate at a total current of 0.1 A. For the diodes closely matched, with n=1, what current flows in each. What is the corresponding small signal resistance of each diode and of the combination?

161 If each of the 20 mA diode has a series resistance of 0.2 ohm associated with the wire bonds to the junction. What is the equivalent resistance of the 10 parallel connected diodes? What connection resistance would single diode need in order to be totally equivalent?

162 The diode i–v relationship

163 Reversed Biased Diode

164 Leakage current: In the reverse direction there is a small leakage current up until the reverse breakdown voltage is reached. This leakage is undesirable, obviously the lower the better. Diodes are intended to operate below their breakdown voltage.

165 The Reversed Biased Region Current in reserved biased diode circuit is due to leakage current & increases with increase in reverse voltage Leakage current is proportional to the junction area & temperature but doubles for every 10 o C rise in temperature

166 Breakdown Region Once reverse voltage exceeds a threshold value of diode V ZK, this voltage is called breakdown voltage. V ZK Z – Zener, K – Knee At breakdown knee reverse current increases rapidly with associated small increase in voltage drop Diode breakdown is not destructive if power dissipated by diode is limited by external circuitry. Vertical line for current gives property of voltage regulation

167 The diode i– v characteristic with the breakdown region shown in some detail.

168 Zener Diode

169 Operation in the Reverse Breakdown Region Very steep i-v curve at breakdown with almost constant voltage drop region Used the designing voltage regulator Diode manufactured to operate specifically in the Breakdown region called Zener or Breakdown Diode

170 IZIZ - V Z + Zener Diode : Symbol

171 Model: Zener Manufacturer specify Zener Voltage V z at a specified Zener test current I z, the Max. power that the device can safely dissipate v at max 70mA r z Dynamic resistance of the Zener and is the inverse of the slope of the almost linear i-v curve at operating point Q Lower r z, the more constant Zener Voltage The most common range of zener voltage is 3.3 volts to 75 volts,

172 Model for the zener diode.

173 Model: Zener

174 Designing of the Zener shunt regulator + - VoVo Zener regulator Supply voltage includes a large ripple component Vo is an output of the zener regulator that is as constant as possible in spite of the ripples in the supply voltage V S and the variations in the load current Voltage regulator performance can be measured Line Regulation & Load Regulation Line Regulation = ΔV o /ΔV s Load Regulation = ΔV o /ΔI L

175 Expression of performance : Zener regulator I ILIL + - VoVo Only the first term on right hand side is desirable one Second and third terms depend upon Supply Voltage V s and Load current I L Line Regulation = Load Regulation =

176 Expression of performance : Zener regulator I ILIL + - VoVo An important consideration for the design is To ensure that current through the zener diode never becomes too low i.e less than I ZK or I zmin Minimum zener current I zmin occurs when Supply Voltage V s is at its minimum V Smin Load current I L is at its maximum I Lmax Above design can be made be selecting proper value of resistor R

177 The circuit with the zener diode replaced with its equivalent circuit model. Example 3.8

178 Exp 3-8

179 Solution No Load Change in voltage => A.C components only Line Regulation = rZrZ

180 With Load Solution

181

182 Example 3-8

183 a) Find No Load Line Regulation Depending upon the manufacturer provide Data First calculate V zo if V z =6.8 V & I z =5mA, r Z =20 ohm

184 Line Regulation Now connecting the Zener diode in the Cct as shown Calculate actual I z and resulting Vo Thus establishing operating Point Now carry out Small Signal Analysis Suppress DC source and calculate resultant change in Vo Use voltage divider rule

185 b) Find v O if load resistance R L connected & draws 1mA and load regulation

186 Check exact Calculations 1mA drawn by load would decrease by same amount so Load Regulation

187 c) for

188 1) Check Zener at Breakdown region

189

190

191 Zener is not operating

192 e) Min value of for which the diode still operates in the breakdown region at Breakdown Region

193

194 Temperature Effect Temperature coefficient of Zener is Known as “Temco” & expressed in mv/c o Temco depends upon: –Zener Voltage –Operating Current Zener Diode whose Vz < 5V temco is negative vz = 5V tempco is zero vz > 5V temco is positive

195 Temco compensation by use using positive temperature coefficient diode of app 2mv/c 0 in series in forward biased configuration temco component Zener diode have been replaced with voltage regulator I.C => more efficient & flexible Temperature Effect

196 Problem D3.68 Design a 7.5-V zener regulator circuit using a 7.5-V zener specified at 12mA. The zener has an incremental resistance r z = 30 Ω and a knee current of 0.5mA. The regulator operates from a 10-V supply and has a 1.2-kΩ load. (a)What is the value of R you have chosen? (b)What is the regulator output voltage when the supply is 10% high? Is 10% low? (c)What is the output voltage when both the supply is 10% high and the load is removed? (d)What is the smallest possible load resistor that can be used while the zener operates at a current no lower than the knee current while the supply is 10% low?

197 Solution 3-68

198 Design a 7.5-V zener regulator circuit using a 7.5-V zener specified at 12mA. The zener has an incremental resistance r z = 30 Ω and a knee current of 0.5mA. The regulator operates from a 10-V supply and has a 1.2-kΩ load. (a)What is the value of R you have chosen?

199 (b)What is the regulator output voltage when the supply is 10% high? Is 10% low? (c)What is the output voltage when both the supply is 10% high and the load is removed?

200 (c)What is the smallest possible load resistor that can be used while the zener operates at a current no lower than the knee current while the supply is 10% low? I ZK =0.5mA, V ZO =7.14 V 1 3 2

201

202 Rectifier Circuit Power Supply Power supply must supply dc voltage to be constant in spite of –variation is ac line voltage –Variation in current drawn by load, that is variable load resistance

203 Rectifier Circuits Transformers – Step Down/Up –Electrical isolation => Minimize risk of Electrical shock Rectifier –converts sinusoidal input into uni-polar output – pulsating dc with nonzero average components Unsuitable for equipments Needs filtering and smoothing

204 Filter –Smoothes out pulsating dc but still some time-dependent components-(ripple) remain in the output Voltage Regulation –Reduces ripples –Stabilizes magnitude of dc output against variation in load current –Regulation by Zener Diode or Voltage regulator I.C Rectifier Circuits

205 Half Wave Rectifier Half-wave rectifier. Equivalent circuit of the half-wave rectifier with the diode replaced with its battery-plus-resistance model.

206 Transfer characteristic of the rectifier circuit Input and output waveforms, assuming that r D >> R. Half Wave Rectifier

207 Current handling capacity Peak Inverse Voltage (PIV) –To withstand breakdown due to PIV –PIV = V s –Select diode having PIV 50% greater then expected Rectification of very small voltage < 0.7v precision rectifier are used Selection Diode

208 Full Wave Rectifier

209 Transfer characteristic assuming a constant- voltage-drop model for the diodes

210 Input and output waveforms.

211 Full Wave Rectifier Provides unipolar output during both half cycle +ve & -ve Uses Centre-tapped Transformer to provide two equal output +ve half cycle D1 Conducts D2 Off -ve half cycle D1 Off D2 Conducts Current flows in same direction through ‘R’ during both half cycles

212 Diode in Reverse biased state - V s + V o PIV = 2V s - V DO Twice as in case of half wave rectifier Full Wave Rectifier

213 Full –Wave Rectifier

214 For positive Half Cycle v i >0 D 1 conducts, D 2 is non-conducting V=0 v o1

215 Full –Wave Rectifier (Negative Half Cycle) For Negative Half Cycle v i < 0 D 1 is Off, D 2 conducts V = v i /R 1 v 01 v o1 /R V o1 /2R At inverting terminal of A 1 node v 01

216 Full –Wave Rectifier 0 0 v 01 i V o1 /2R

217 Bridge Rectifier

218 The bridge rectifier: (a) circuit; (b) input and output waveforms.

219 Bridge Rectifier

220

221 Don't require centre tape Transformer Uses Similar configuration to that of Wheatstone Bridge Requires four diode During positive half cycle current for right to left flows through D1, R & D2, D3&D4 cut off. During –ve cycle current flows from right to left through D3, R& D4, D1&D2 cut off Voltage Vo = V s -2V D

222 Peak Inverse Voltage –PIV => consider loop D3, R & D2 –V D3 (res) = V o + V D2 –V o = V s – 2V D –PIV = V s – 2V D + V D = V s – V D –Half of PIV for Full wave Rectifier Bridge Rectifier D4D4 D1D1 D2D2 D3D3

223 Advantages –No centre tapped transformer –Secondary winding contains half an many turns are required for the secondary winding of centre tapped transformer for full wave rectifier –Less cost – most popular circuit –Peak Inverse voltage half as that of full wave rectifier Bridge Rectifier

224 Quiz no 2 DE26 EE) Find the value of R for which V = 100 m V. For the circuit shown, both diodes are identical, conducting Group #1 1mA at 0.6 V and 10 mA at 0.7 V. Group #2 10mA at 0.65 V and 100 mA at 0.75 V. Group #3 0.1mA at 0.7 V and 100 mA at 0.8 V. V

225 Quiz N0 2 DE26 CE In the circuit shown, both diode have n=1, What value of V results if Gp # 1 D 1 has 5 times the junction area of D 2 Gp # 2 D 1 has 15 times the junction area of D 2 Gp # 3 D 1 has 100 times the junction area of D 2. V D2 D1

226 Peak Rectifier Pulsating dc output from rectifier in unsuitable as a dc power supply for electronic circuit Simplest way to reduce the variation is to place a capacitor across the load – smoothing, filter or reservoir capacitor Capacitor charges during positive half cycle to V S – (ideal diode) & maintains it during negative half cycle as diode is reversed biased

227

228 Figure 3.28 (a) A simple circuit used to illustrate the effect of a filter capacitor. (b) Input and output waveforms assuming an ideal diode.

229 Peak detector with Load

230 Capacitor charges through diode to peak during first positive quarter cycle and then discharges through R during the entire cycle until the time at which V s exceeds the capacitor voltage Diode conducts & charges capacitor to V peak Select value of C so as discharge time constant is much greater than the discharge interval CR >> T

231 Figure 3.29 Voltage and current waveforms in the peak rectifier circuit with CR<

232 Charge / Discharge Cycle

233 Peak detector with Load

234

235 Figure 3.30 Waveforms in the full-wave peak rectifier.

236 Peak Rectifier: Observations Diode conduct for an interval Δt near peak of input, and recharges capacitor to V p (makes up the lost) Conduction of diode begins at time t 1, at which V i equals the exponentially decaying output V o Conduction stops shortly after the peak input voltage V p and exact value of t 2 can be calculated by putting i D =0 in equation During the diode-off-interval capacitor discharges through R & the decays exponentially with a time constant → CR V o = V p - V r (V r => peak to peak ripple voltage)

237 When V r is small –V o = V peak –i L is almost constant –DC components of i L Accurate value of output dc voltage Average Value Peak Rectifier : Output Voltage

238 Charge / Discharge Cycle

239 During Discharge cycle At the end of discharge cycle Since CR >> T Peak Rectifier : Ripple Voltage

240 Ripple Voltage Peak Rectifier : Ripple Voltage

241 Peak Rectifier : Conduction Interval When Vr<

242 Deduction

243 Average Diode Current –During Conduction

244

245 As waveform of is almost right angle r triangle Deduction

246 Observations Diode current flows for short interval and must replenish the charge lost by the capacitor. Discharge interval is long & discharge is through high resistance Maximum diode current

247 Example N0 3-9 Consider a peak rectifier fed by a 60 Hz sinusoidal having a peak value of Vp = 100 V. Let the load resistance R =10 k Ohms. (a) Find the value of the capacitance C that will result in peak to peak ripple of 2 V (b) Calculate the fraction of the cycle during which the diode is conduction (c) Calculate the average and peak value of the diode current.

248 Example 3.9 Find value of C for V r =2V (peak to peak) Find fraction of cycles that diode conducts => Diode conducts of cycle

249 Solution Exp 3-9 Find &

250 Full wave peak Detector In full wave rectifier, the capacitor discharge for almost T/2 time interval. that mean ripple frequency is twice the input, so

251 Full Wave Peak Detector Deduction –Having same V P, f, R & V r; the size of capacitor required is half that of the capacitor in the half wave rectifier –Current through each diode of full wave peak rectifier is half that flows through the diode in half wave rectifier

252 Capacitor Charge Voltage (v p ) Ideal Diode –Capacitor will charge to v p Real diode in Half and full wave rectifier –Capacitor will charge to v p- v D Real diode in Bridge wave rectifier –Capacitor will charge to v p- 2v D All equations may be amended accordingly.

253 Applications Peak Rectifier – Peak detector is used for –Detecting the peak of the an input signal for signal processing systems –Demodulator for amplitude modulated (AM) signals

254 .

255 Precision Half Wave Rectifier Super Diode Normal Diodes V D = 0.7v are used for rectifier of input of much larger amplitude then V D For smaller signals detection, demodulation or rectification Operational Amplifiers (Op Amp) are used

256

257 D3.86 It is required to use a peak rectifier to design a dc power supply that provides an average dc output voltage of 15 V on which a maximum 1-V ripple is allowed. The rectifier feeds a load of 150Ω. The rectifier is fed from the line voltage (120 V rms, 60Hz) through a transformer. The diodes available have 0.7-V drop when conducting. If the designer opts for the half-wave circuit: –Specify the rms voltage that must appear across the transformer secondary –Find the required value of the filter capacitor. –Find the maximum reverse voltage that will appear across the diode, and specify the PIV rating of the diode. –Calculate the average current through the diode during conduction –Calculate the peak diode current.

258

259

260

261

262 Problem D3.87 It is required to use a peak rectifier to design a dc power supply that provides an average dc output voltage of 15 V on which a maximum 1-V ripple is allowed. The rectifier feeds a load of 150Ω. The rectifier is fed from the line voltage (120 V rms, 60Hz) through a center-tapped transformer. The diodes available have 0.7-V drop when conducting. If the designer opts for the Full-wave circuit: –Specify the rms voltage that must appear across the transformer secondary –Find the required value of the filter capacitor. –Find the maximum reverse voltage that will appear across the diode, and specify the PIV rating of the diode. –Calculate the average current through the diode during conduction –Calculate the peak diode current.

263

264

265

266 Wave form Generation Limiting Clamping Limiter Circuit –V o is limited between two levels – upper (L + ) and lower (L - ) thresholds

267 Figure 3.33 Applying a sine wave to a limiter can result in clipping off its two peaks.

268 Figure 3.34 Soft limiting.

269 Wave form Generation Limiting / Clamping Double Limiter –Clips off both negative & positive peaks Single Limiter Clips off only one side of the input peak Application –Limits the inputs to operation Amplifier to a limit lower than the breakdown voltage of transistors of input stage of operational Amplifier –Half / Full Rectifier for Battery Charger –Threshold and limiting

270 Figure 3.35 A variety of basic limiting circuits.

271 Threshold and limiting : Zener Diodes Threshold and limiting can be achieved by using strings of diode and/or by connecting a dc voltage in series with the diode(s). or make use of two zener diodes in series. These are Commercially Available and are called Double-Anode Zener

272 Figure E3.27

273 Solution Ex 3-27

274 D C Restorer The output waveform will have its lower peak “Clamped” to O V therefore known as “Clamped Capacitor” Output waveform will have a finite average value & is entirely different and unrelated to the average value of the input waveform

275 Application T X R DC Restorers

276 DC Restorer : Applications Pulse signal being transmitted through a capacitive coupled circuit or ac coupled circuit loses whatever dc components its originally had. Feeding the resultant pulse waveform to a Clamping circuit provides it with a well determined dc components Process is known as DC Restoration & circuit as DC Restorer

277 Restoring dc components is useful as dc components of a pulse is an effective measure of its Duty Cycle Duty Cycle of a pulse waveform is used in pulse Modulation & carries Information Demodulation could e achieved by feeding the received Signal to DC restorer and then using a simple RC low pass filter to separate the average of the output waveform from the superimposed pulse. DC Restorer : Applications

278 Figure 3.36 The clamped capacitor or dc restorer with a square-wave input and no load.

279 Figure 3.37 The clamped capacitor with a load resistance R.

280 Connecting load to DC Restorer

281

282

283

284 Deductions In steady state: the charge lost by the capacitor during the interval t 0 to t 1 is recovered during the interval t 1 to t 2 Average dc, diode current & output waveform can be calculated

285 Figure 3.38 Voltage doubler: (a) circuit; (b) waveform of the voltage across D 1.

286 Figure P3.97

287 Figure P3.98

288 Figure P3.102

289 Figure P3.103

290 Figure P3.105

291 vivi VCVC vovo

292 The Voltage Doubler

293 Special Diode Type Schottky-Barrier Diode (SBD) Shottky-Barrier Diode is formed by bringing metal into contact with a moderately doped ‘n’ type semiconductor material Resulting in flow of the conducting current in one direction from metal anode to the semiconductor cathode and acts as an open circuit in the other direction

294 Gets two important properties –SBD switches on-off faster due to current conducts due to majority carrier b (electrons) –Forward voltage drop is lower then P-n junction diode Schottky-Barrier Diode (SBD)

295 Varactor Variable Capacitor –Depletion layer acts as junction capacitance –Depletion layer varies Capacitance –Used for voltage controlled Tuning Circuit

296 Varactor When a reverse voltage is applied to a p-n junction, the depletion region, is essentially devoid of carriers and behaves as the dielectric of a capacitor. The depletion region increases as reverse voltage across it increases; and since capacitance varies inversely as dielectric thickness, the junction capacitance will decrease as the voltage across the p-n junction increases. By varying the reverse voltage across a p-n junction the junction capacitance can be varied.

297 Semiconductor diodes The tunnel diode, the current through the device decreases as the voltage is increased within a certain range; this property, known as negative resistance, makes it useful as an amplifier. Gunn diodes are negative-resistance diodes that are the basis of some microwave oscillators. Light-sensitive or photosensitive diodes can be used to measure illumination; the voltage drop across them depends on the amount of light that strikes them.

298 SCR (Thyristor) The Silicon Controlled Rectifier (SCR) is simply a conventional rectifier controlled by a gate signal. A gate signal controls the rectifier conduction. The rectifier circuit (anode-cathode) has a low forward resistance and a high reverse resistance. It is controlled from an off state (high resistance) to the on state (low resistance) by a signal applied to the third terminal, the gate. Most SCR applications are in power switching, phase control, chopper, and inverter circuits.

299

300 Photodiode If reversed biased PN junction is exposed to incident light – the photons impacting the junction cause covalent – bond to break thus give rise to current known as a photocurrent & is proportional to the intensity of incident light. Converts Light energy into a electrical signals

301 Photodiode Photodiode are manufactured using Gallium Arsenide (GaAs) Photodiodes are important element of optoelectronics or photonics circuit (Combination of Electronics & optics) used for signal processing, storage & transmission

302 Photodiode : Applications Fiber optics Transmission of telephonic & TV signals Opto-storage are CD ROM computer disks Wide bandwidth & low signal attenuation. Solar Cell – light energy into Electrical energy

303 Light Emitting Diode (LED) Inverse of Photodiode Converts a forward biased current into light GaAs used for manufacturing LEDs Used as electronics displays Coherent light into a narrow bandwidth laser diodes Fiber Optics & CD ROM

304 LED

305 Double heterostructure laser.

306 Optoisolator LED & Photodiode Electrical to light Light to electrical Provides complete electrical isolation between electrical circuits Reduces the effects of electrical interference on signal being fixed within a system Reduces risk of shock Can be implemented over long distance fiber optics communication links

307 Laser Pointer.

308 Laser Microphone

309 End

310 Problem Sketch and label the transfer Characteristics of the circuit shown over a + 10 V range of the input signal. All diodes are V D =0.7 1 mA with n=1. What are the slopes of the characteristic at the extreme + 10 V levels?

311 +1 V -2 V ViVi V0V0

312 Problem 3-103

313

314 Ist Sessional Q No 1 (12 Marks) In the circuit shown, input voltage is a 1kHz, 10 V peak to peak sine wave. The diode is an ideal diode. (a)Sketch the waveform resulting at output terminal vO. (b)What are its positive and negative peak values?

315 Q No 2 (15 Marks) A circuit utilizes three identical diodes connected in series having n=1 and I S = A. (a)Find the value of current required to obtain an output voltage of 2 V across the three diodes combined. (b)If a current of 1 mA is drawn away from the output terminal by a load (i)What is the change in output voltage? (ii)What is the value of the load? Ist Sessional

316 Q No 3 (13 Marks) For the circuit shown, sketch the output for the sine wave input of 10 volts peak. Label the positive and negative peak values assuming that CR >>T. Ist Sessional

317 Q No 4 (10 Marks) 9.25 V zener diode exhibits its nominal voltage at a test current of 28 mA. At this current the incremental resistance is specified as 7 ohms. –(a)Find V ZO of the zener model. –(b)Find the zener voltage at a current of 10 mA. Ist Sessional

318 Q No 5 (20 Marks)Consider a bridge rectifier circuit with a filter capacitor C placed across the load resistor R for the case in which the transformer secondary delivers a sinusoid of 12 V (rms) having the 60 Hz frequency and assuming VD = 0.8 V and a load resistance of 100 ohms. –Find the value of C that results in a ripple voltage no larger than 1 V peak to peak. –Find the diode conduction angle. –Find the load current. –What is the average load current? Ist Sessional

319 Q No 6 (10 Marks) In a circuit shown, the output voltage is 2.4 V. Assuming that the diodes are identical and are having 0.7 V drop at 1mA. –(a) Find the current following through the resistor R. –(b) What the value of resistor R. Ist Sessional

320 Figure 3.31 The “superdiode” precision half-wave rectifier and its almost-ideal transfer characteristic. Note that when v I > 0 and the diode conducts, the op amp supplies the load current, and the source is conveniently buffered, an added advantage. Not shown are the op-amp power supplies.

321 Figure P3.82

322 Figure P3.91

323 Figure P3.92

324 Figure P3.93

325 Figure P3.105

326

327 Quiz DE28 EE -B (10 Marks) 9.25 V zener diode exhibits its nominal voltage at a test current of 28 mA. At this current the incremental resistance is specified as 7 ohms. –(a)Find V ZO of the zener model. –(b)Find the zener voltage at a current of 10 mA.

328 Quiz DE 28 EE -A A zener diode whose nominal voltage is 10 V at 10 mA has an incremental resistance of 50 Ω. (a)What is the value of V ZO of the zener model? (b)What voltage do you expect if the diode current is doubled?


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