2Diode Class of non-linear circuits Uses having non-linear v-i CharacteristicsUsesGeneration of :DC voltage from the ac power supplyDifferent wave (square wave, pulse) form generationProtection CircuitsDigital logic & memory circuits
3Creating a DiodeA 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.A diode is the simplest possible semiconductor device. A diode allows current to flow in one direction but not the other. You may have seen turnstiles at a stadium or a subway station that let people go through in only one direction. A diode is a one-way turnstile for electrons.When you put N-type and P-type silicon together as shown in this diagram, you get a very interesting phenomenon that gives a diode its unique properties.
5Diode Equivalent circuit in the reverse direction Equivalent circuit in the forward direction.
6Reverse Bias Forward Bias Operation -ve voltage is applied to Anode Current through diode = 0 (cut off operation)Diode act as open circuitForward Bias+ve voltage applied to AnodeCurrent flows through diodevoltage Drop is zero (Turned on)Diode is short circuit
7The two modes of operation of ideal diodes sedr42021_0302a.jpgReverse biasedReverse Voltage 10 VForward biasedForward Current 10 mA
9Rectifier circuit Equivalent circuit when vi 0 Waveform across diode Input waveformEquivalent circuit when vi 0sedr42021_0303a.jpgOutput waveform.Equivalent circuit when vi ≤ 0Waveform across diode
14Diode logic gates vy = vA.vB.vC vy = vA+vB+vC OR gate AND gate (in a positive-logic system)sedr42021_0305a.jpgvy = vA.vB.vCvy = vA+vB+vC
15Diodes are ideal , Find the value of I and V sedr42021_0306a.jpgDiodes are ideal , Find the value of I and VFigure 3.6 Circuits for Example 3.2.
16Both Diodes are conducting Example 3.2.sedr42021_0306a.jpgAssumptionBoth Diodes are conducting
17Both Diodes are conducting AssumptionBoth Diodes are conductingNode ANode BNot PossibleThus assumption of both diodeconducting is wrong
18Example 3.2(b). Assumption is correct Assumption # 2 Diodes 1 is not conductingDiodes 2 is conductingsedr42021_0306a.jpgAssumption is correct
19Diodes are ideal , Find the value of I and V sedr42021_e0304a.jpgFigure E3.4Diodes are ideal , Find the value of I and V
20Diodes are ideal , Find the value of I and V Figure E3.4Diodes are ideal , Find the value of I and VI= 2mAV= 0VI= 0AV= 5VI= 0AV= -5VI= 2mAV= 0Vsedr42021_e0304a.jpg
21Figure E3.4 Diodes are ideal , Find the value of I and V sedr42021_e0304a.jpgI= 3mAV= 3VI= 4mAV= 1V
22Diodes are ideal , Find the value of I and V sedr42021_p03002a.jpgDiodes are ideal , Find the value of I and VFigure P3.2
23Figure P3.2 Diodes are ideal , Find the value of I and V Diode is conductingI = 0.6 mAV = -3Vsedr42021_p03002a.jpgDiode is cut-offI = 0 mAV = 3VDiode is conductingI = 0.6 mAV = 3VDiode is cut-offI = 0 mAV = -3V
24Problem 3-3 D1 Cut-Off & D2 Conducting I = 3mA Diodes are ideal , Find the value of I and Vsedr42021_p03003a.jpgD1 Cut-Off & D2 ConductingI = 3mAD1 Cut-Off & D2 ConductingI = 1mA , V=1 V
25In ideal diodes circuits, v1 is a 1-kHz, 10V peak sine wave. sedr42021_p03004a.jpgFigure P3.4In ideal diodes circuits, v1 is a 1-kHz, 10V peak sine wave.Sketch the waveform of vo
26In ideal diodes circuits, v1 is a 1-kHz, 10V peak sine wave. Sketch the waveform of voVp+ = 10VVp- = 0Vf = 1 K-HzVp+ = 0VVp- = - 10Vf = 1 K-HzVo = 0Vsedr42021_p03004a.jpg
27Sketch the waveform of vo Figure P3.4 In ideal diodes circuits, v1 s a 1-kHz, 10V peak sine wave.Sketch the waveform of voVp+ = 10VVp- = -10Vf = 1 K-HzVp+ = 10VVp- = 0Vf = 1 K-HzVp+ = 10VVp- = 0Vf = 1 K-Hzsedr42021_p03004a.jpg
28Sketch the waveform of vo Figure P3.4 In ideal diodes circuits, v1 s a 1-kHz, 10V peak sine wave.Sketch the waveform of vosedr42021_p03004g.jpg
29In ideal diodes circuits, v1 s a 1-kHz, 10V peak sine wave. Figure P3.4In ideal diodes circuits, v1 s a 1-kHz, 10V peak sine wave.Sketch the waveform of voVp+ = 0VVp- = -10Vf = 1 K-HzV0 = 0VVp+ = 10VVp- = -5Vf = 1 K-Hzsedr42021_p03004g.jpg
30Sketch the waveform of vo Figure P3.4 In ideal diodes circuits, v1 s a 1-kHz, 10V peak sine wave.Sketch the waveform of voVp+ = 10VVp- = -5Vf = 1 K-Hzsedr42021_p03004g.jpg
31Problem 3-4(k) For Vi >0 V D1 is cutoff D2 is conducting vo=1V For Vi < 0 V is conducting D2 is cutoff vo=vi+1V- 9 V
42REVERSE POLARITY PROTECTOR 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.
48Vi<5V D2 is cut-off, Vo=5V Vi>5V D2 is conducts Sketch vO if vi is 10 sin Find out the conduction angle for the diode &fraction of the cycle the diode is conductsD1 never conductsVi<5V D2 is cut-off, Vo=5VVi>5V D2 is conducts+12 V5D1D2
49Quiz No 3 DE 27 CE -B D1 never conducts Vi<5V D2 is cut-off, Vo=Vi Sketch vO if vi is 10 sin Find out the conduction angle for the diode &fraction of the cycle the diode is conductsD1 never conductsVi<5V D2 is cut-off, Vo=ViVi>5V D2 is conducts
50ProblemAssume the diodes are ideal, sketch vo if the input is 10sin (9)Find out the conduction angles for Diode D1 & D2 (4) and the fraction of the cycle these diodes conduct. (2)
53Semiconductor materials GermaniumThe 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.
54Current 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.
55DriftApplying an electric field across a semiconductor will cause holes and free electrons to drift through the crystalThe total current is equal to the sum of hole current (to the right) and electron current (to the left).
56DiffusionA 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.
57Two-dimensional representation of the silicon crystal. sedr42021_0340.jpgThe 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.
58Two-dimensional representation of the silicon crystal. sedr42021_0340.jpg14 Electrons
59Silicon Crystal All silicon atoms bond perfectly to four neighbors Silicon CrystalAll silicon atoms bond perfectly to four neighborsLeaving no free electrons to conduct electric currentThis makes a silicon crystal an insulator rather than a conductor.
60Silicon CrystalCarbon, 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.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.Metals tend to be good conductors of electricity because they usually have "free electrons" that can move easily between atoms, and electricity involves the flow of electrons. While silicon crystals look metallic, they are not, in fact, metals. All of the outer electrons in a silicon crystal are involved in perfect covalent bonds, so they can't move around. A pure silicon crystal is nearly an insulator -- very little electricity will flow through it.Carbon, silicon and germanium (germanium, like silicon, is also a semiconductor) have a unique property in their electron structure -- each has four electrons in its outer orbital. This allows them to form nice crystals. The four electrons form perfect covalent bonds with four neighboring atoms, creating a lattice. In carbon, we know the crystalline form as diamond. In silicon, the crystalline form is a silvery, metallic-looking substance
63sedr42021_0341.jpgAt 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.
66The Valence BandThe 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.
67The Fermi LevelThe 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
68Conduction BandThe 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.
69Doping SiliconYou 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-typeP-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.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 - In N-type doping, phosphorus or arsenic is added to the silicon in small quantities. Phosphorus and arsenic each have five outer electrons, so they're out of place when they get into the silicon lattice. 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.P-type - In P-type doping, boron or gallium is the dopant. 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.A minute amount of either N-type or P-type doping turns a silicon crystal from a good insulator into a viable (but not great) 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
70The Doping of Semiconductors .The Doping of SemiconductorsThe addition of a small percentage of foreign atoms in the regular crystal lattice of silicon or germanium produces dramatic changes in their electrical properties, producing n-type and p-type semiconductors.Pentavalent impuritiesImpurity atomw with 5 valence electrons produce n-type semiconductors by contributing extra electrons.Trivalent impuritiesImpurity atoms with 3 valence electrons produce p-type semiconductors by producing a "hole" or electron deficiency.
72N - Type Silicon DopingIn N-type doping, phosphorus or arsenic is added to the silicon in small quantities.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
73A silicon crystal doped by a pentavalent element. sedr42021_0343.jpgA 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.
75N-Type Semiconductor N-Type Semiconductor The addition of pentavalent impurities such as antimony, arsenic or phosphorous contributes free electrons, greatly increasing the conductivity of the intrinsic semiconductor. Phosphorous may be added by diffusion of phosphine gas (PH3).
76P – Type Silicon DopingIn P-type doping, boron or gallium is the dopant.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.
77A silicon crystal doped with a trivalent impurity. sedr42021_0344.jpgA silicon crystal doped with a trivalent impurity.Each dopant atom gives rise to a holeThe semiconductor becomes p type.
79P-Type Semiconductor P-Type Semiconductor P-Type SemiconductorThe addition of trivalent impurities such as boron, aluminum or gallium to an intrinsic semiconductor creates deficiencies of valence electrons,called "holes". It is typical to use B2H6 diborane gas to diffuse boron into the silicon material.
81Bands for Doped Semiconductors The application of band theory to n-type and p-type semiconductors shows that extra levels have been added by the impurities. In n-type material there are electron energy levels near the top of the band gap so that they can be easily excited into the conduction band. In p-type material, extra holes in the band gap allow excitation of valence band electrons, leaving mobile holes in the valence band
83P-N JunctionIn practice, both the p and n region are part of the same silicon crystalpn 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
84At 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 currentElectrons fill the hole & this process is known as recombination.
85Two Mechanisms Diffusion Associated with random motion of electrons & holes due to thermal agitationwith 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
86Two MechanismsThis diffusion process given rise to a net flow of charge & diffusion currentDriftCarrier drift occurs when an electric field is applied across a piece of siliconElectronic & Holes are accelerated by the electric field & acquire a velocity component (superimposed on the velocity of then thermal motion) called Drift velocityCausing Drift current flow
87Doped SemiconductorMaterial in which one kind (Electrons & Holes) predominate – majority of charged carrier are electron for n type & majority of charged carrier are holes for p typesPhosphorus impurity donated electrons (donor)Boron impurity donated holes (Acceptor)
88p-n Junction P Junction N Junction Concentration of holes is high Majority charge carrier are holeN JunctionConcentration of electron is highMajority charge carrier are electron
89Diffusion Current IDHole diffuse across the junction from the p side to the n side & similarly electronTwo current components add together to form the diffusion current with direction from p to n side
90Depletion RegionRecombination of Hole & electrons take close to the junction thus depletion region is formed on both side of the junctionRegion 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 ID to flow from p side to n side
91Drift Current Is Diffusion current due to majority carrier diffusion A component due to minority carrier drift exists across the junction
92p-n Junction Due to electric field applied across the p-n junction Electrons are moved by drift from p to n junctionHoles are moved by drift from n to p junctionAdd together to form the drift current IsDirection of Isis from n to p junctionDrift current is carried by thermally generated minority p carriersDrift current value depends upon temperatureIndependent of the value of the depletion layer voltage voUnder operative circuit conditions no external current flowsID = IS
93p-n Junction open circuit Junction Built in Voltage Vo will depends uponTemperatureDoping ConcentrationDepletion region widthSame on both side of p & n junction if doping levels are equalIf doping levels are not equal the depletion layer will extend deeper into the more lightly doped material
94The pn junction with no applied voltage (open-circuited terminals). sedr42021_0345a.jpgThe pn junction with no applied voltage (open-circuited terminals).(b) The potential distribution along an axis perpendicular to the junction.
99To avoid breakdown, I is kept smaller than IS. The pn junction excited by a constant-current source I in the reverse direction.To avoid breakdown, I is kept smaller than IS.Note that the depletion layer widens and the barrier voltage increases by VR volts, which appears between the terminals as a reverse voltage.sedr42021_0346.jpg
100sedr42021_0347.jpgThe charge stored on either side of the depletion layer as a function of the reverse voltage VR.
101The pn junction excited by a reverse-current source I, where I > IS. The junction breaks down, and a voltage VZ , with the polarity indicated, develops across the junction.sedr42021_0348.jpg
102The 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.sedr42021_0349.jpg
103Minority-carrier distribution in a forward-biased pn junction. sedr42021_0350.jpgMinority-carrier distribution in a forward-biased pn junction.It is assumed that the p region is more heavily doped than the n region; NA @ ND.
105The diode i–v relationship sedr42021_0308.jpgThe diode i–v relationship with some scales expanded and others compressed in order to reveal details.
106Terminal Characteristics of a Junction Diode Forward Biased Region v > 0Reversed Biased Region v < 0Breakdown Region v < -VZK
107Forward Biased Region Is Saturation current – Scale Current Is is constant at a given temperatureIs is directly proportional to Cross-Sectional region of the diode, Is doubles if cross-sectional area is doubleIs is A for small size diodeDoubles in value for every 10OC rise in temperature
108Forward Biased Region Thermal Voltage VT VT = kT/qK = Boltzmann’s constant = 1.38 X Joules/KelvinT = Absolute Temperature in Kelvin (273 +Temp in Co)q = Magnitude of charge = 1.6 X Coulombs20oC is 25.2mV, ~ 25 mVn is 1 or 2 depending on the material and the physical structure of the dioden = 1 for Germanium Diode & n=2 for Silicon
109Switching 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, trr, is the time it takes a diode to switch from on to off.
110i >> Is Forward Biased Region b Relationship of the current i to the voltage v holds good over many decades of current (seven decades, a factor of 107
112Forward Biased Regionfor v drop changes byfor n = 1for n = 2
113Illustrating the temperature dependence of the diode forward characteristic sedr42021_0309.jpgAt a constant current, the voltage drop decreases by approximately 2 mV for every 1C increase in temperature.
114At 20o C Reverse current Is = 1V/1M Ω= 1μ A Figure E3.9If V=1V at 20o C, Find V at 400C and 00CIssedr42021_e0309.jpgAt 20o C Reverse current Is = 1V/1M Ω= 1μ ASince the reverse leakage current doubles for every 100 C increase,At 400 C I = 4*1 = 4 μ A V = 4 μ A * 1MΩ = 4.0 VAt 0 C I = ¼ μ A V = 0.25 V
115Forward biased Diode Characteristics Example 3.3 A silicon diode displays a forward voltage of 0.7 V at a current of 1mA. Find Is at n=1 & 2
116Ex 3.7Silicon Diode with n=1 has i=1mA. Find voltage drop at i=0.1mA & 10mA
117Solution 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
118Problem 3-23The 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=1If 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
119Solution 3-23The 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.If a current of 1mA is drawn away from the output terminalby a load, what if the change in the output voltage.
120Problem 3-25In the circuit shown, both diode have n=1, but D1 has 10 times the junction area of D2. What value of V results?
121In the circuit shown, both diode have n=1, but D1 has 10 times the junction area of D2. What value of V results?Solution 3-25(a)
122To obtain a value of 50 mV, what current I2 id needed. solution 2-25 (b)To obtain a value of 50 mV, what current I2 id needed.
123Problem 3-26For 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.
128A simple circuit used to illustrate the analysis of circuits in which the diode is forward conducting.sedr42021_0310.jpg
129Graphical analysis of the circuit using the exponential diode model. sedr42021_0311.jpg
130Iterative Analysis using the Exponential Model Determined the diode current ID and Diode voltage VD with VDD =5V and R =1000 ohms. Diode has a current of a VD of .7 V, and that its voltage drop changes by 0.1 V for every decade change in current.
133Approximating the diode forward characteristic with two straight lines: the piecewise-linear model. sedr42021_0312.jpg
134The Piecewise-Linear Model Exponential curve is approx into two straight linesLine No 1 with zero slope & Line 2 with a slope of 1/rdThe voltage change of less than 50 mV is observed in case the current change from 0.1 mA to 10 mA.
135Piecewise-linear model of the diode forward characteristic and its equivalent circuit representation.sedr42021_0313a.jpg
138Constant – Voltage Drop Model Forward conducting diode exhibits a constant voltage drop VDThe voltage change of less than 50 mV is observed in case the current change from 0.1 mA to 10 mA.Model is used whenDetailed information about diode characteristics in not available
144The Small – Signal Model A small ac signal is superimposed on the DC components.First determined dc Operating PointThen small signal operation around the operating pointSmall portion of the curve is approximated as almost linear segment of the diode characteristics.
154Solution Input variation of 10% resulted in output voltage variation of mV(0.8%) Voltage regulation
155Exercise 3-16Design a circuit shown so that Vo=3v when IL =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
156Excercise 3-16Why 4 diodes and not 5? Diodes will not conduct at 0.6 V
157Diode Forward Drop in Voltage Regulation Small signal model is used.Voltage remains constant in spite of :Changes in load currentChanges in the dc power supply voltagesOne diode provides constant voltage of 0.7 V and for greater voltages diodes can be connected in series.
158Example 3-7A 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
160P 3-53In 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?
161If each of the 20 mA diode has a series resistance of 0 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?
164Leakage 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.
165The Reversed Biased Region Current in reserved biased diode circuit is due to leakage current & increases with increase in reverse voltageLeakage current is proportional to the junction area & temperature but doubles for every 10oC rise in temperature
166Breakdown RegionOnce reverse voltage exceeds a threshold value of diode VZK, this voltage is called breakdown voltage.VZK Z – Zener, K – KneeAt breakdown knee reverse current increases rapidly with associated small increase in voltage dropDiode breakdown is not destructive if power dissipated by diode is limited by external circuitry.Vertical line for current gives property of voltage regulation
167The diode i–v characteristic with the breakdown region shown in some detail. sedr42021_0321.jpg
169Zener Diode Operation in the Reverse Breakdown Region Very steep i-v curve at breakdown with almost constant voltage drop regionUsed the designing voltage regulatorDiode manufactured to operate specifically in the Breakdown region called Zener or Breakdown Diode
171Model: ZenerManufacturer specify Zener Voltage Vz at a specified Zener test current Iz, the Max. power that the device can safely dissipate v at max 70mArz Dynamic resistance of the Zener and is the inverse of the slope of the almost linear i-v curve at operating point QLower rz, the more constant Zener VoltageThe most common range of zener voltage is 3.3 volts to 75 volts,
174Designing of the Zener shunt regulator +Supply voltage includesa large ripple componentVo-Zener regulatorVo is an output of the zener regulatorthat is as constant as possible in spite ofthe ripples in the supply voltage VSand the variations in the load currentVoltage regulator performance can be measuredLine Regulation & Load RegulationLine Regulation = ΔVo/ΔVsLoad Regulation = ΔVo/ΔIL
175Expression of performance : Zener regulator +IVo-ILOnly the first term on right hand side is desirable oneSecond and third terms depend upon Supply Voltage Vs and Load current ILLine Regulation =Load Regulation =
176Expression of performance : Zener regulator IL+-VoAn important consideration for the design isTo ensure that current through the zener diode never becomes too low i.e less than IZK or IzminMinimum zener current Izmin occurs whenSupply Voltage Vs is at its minimum VSminLoad current IL is at its maximum ILmaxAbove design can be made be selecting proper value of resistor R
177Example 3.8The circuit with the zener diode replaced with its equivalent circuit model.sedr42021_0323a.jpg
183a) Find No Load Line Regulation Depending upon the manufacturer provide DataFirst calculate Vzo if Vz =6.8 V & Iz=5mA, rZ=20 ohm
184Line Regulation Now connecting the Zener diode in the Cct as shown Calculate actual Iz and resulting VoThus establishing operating PointNow carry out Small Signal AnalysisSuppress DC source and calculate resultant change in VoUse voltage divider ruleLine Regulation
185b) Find vO if load resistance RL connected & draws 1mA and load regulation
1861mA drawn by load would decrease by same amount so Load RegulationCheckexact Calculations
194Temperature EffectTemperature coefficient of Zener is Known as “Temco” & expressed in mv/coTemco depends upon:Zener VoltageOperating CurrentZener Diode whoseVz < 5V temco is negativevz = 5V tempco is zerovz > 5V temco is positive
195Temperature EffectTemco compensation by use using positive temperature coefficient diode of app 2mv/c0 in series in forward biased configuration temco componentZener diode have been replaced with voltage regulator I.C => more efficient & flexible
196Problem D3.68Design a 7.5-V zener regulator circuit using a 7.5-V zener specified at 12mA. The zener has an incremental resistance rz = 30 Ω and a knee current of 0.5mA. The regulator operates from a 10-V supply and has a 1.2-kΩ load.What is the value of R you have chosen?What is the regulator output voltage when the supply is 10% high? Is 10% low?What is the output voltage when both the supply is 10% high and the load is removed?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?
198Design a 7. 5-V zener regulator circuit using a 7 Design a 7.5-V zener regulator circuit using a 7.5-V zener specified at 12mA. The zener has an incremental resistance rz = 30 Ω and a knee current of 0.5mA. The regulator operates from a 10-V supply and has a 1.2-kΩ load.What is the value of R you have chosen?
199What is the regulator output voltage when the supply is 10% high What is the regulator output voltage when the supply is 10% high? Is 10% low?What is the output voltage when both the supply is 10% high and the load is removed?
200What 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? IZK=0.5mA, VZO=7.14 V321
202Rectifier Circuit Power Supply Power supply must supply dc voltage to be constant in spite ofvariation is ac line voltageVariation in current drawn by load, that is variable load resistance
203Rectifier Circuits Transformers Rectifier Step Down/Up Electrical isolation => Minimize risk of Electrical shockRectifierconverts sinusoidal input into uni-polar output – pulsating dc with nonzero average componentsUnsuitable for equipmentsNeeds filtering and smoothing
204Rectifier Circuits Filter Voltage Regulation Smoothes out pulsating dc but still some time-dependent components-(ripple) remain in the outputVoltage RegulationReduces ripplesStabilizes magnitude of dc output against variation in load currentRegulation by Zener Diode or Voltage regulator I.C
205Half Wave Rectifier Half-wave rectifier. Equivalent circuit of the half-wave rectifier with the diodereplaced with its battery-plus-resistance model.
206Half Wave Rectifier Transfer characteristic of the rectifier circuit sedr42021_0325a.jpgInput and output waveforms, assuming that rD >> R.
207Selection Diode Current handling capacity Peak Inverse Voltage (PIV) To withstand breakdown due to PIVPIV = VsSelect diode having PIV 50% greater then expectedRectification of very small voltage < 0.7v precision rectifier are used
209Transfer characteristic assuming a constant-voltage-drop model for the diodes sedr42021_0326a.jpg
210Input and output waveforms. sedr42021_0326a.jpg
211Full Wave RectifierProvides unipolar output during both half cycle +ve & -veUses Centre-tapped Transformer to provide two equal output+ve half cycle D1 ConductsD2 Off-ve half cycle D1 OffD2 ConductsCurrent flows in same direction through ‘R’ during both half cycles
212Full Wave Rectifier Diode in Reverse biased state Anode @ - Vs + VoPIV = 2Vs - VDOTwice as in case of half wave rectifier
221Bridge Rectifier Don't require centre tape Transformer Uses Similar configuration to that of Wheatstone BridgeRequires four diodeDuring 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 offVoltage Vo = Vs -2VD
222Bridge Rectifier Peak Inverse Voltage PIV => consider loop D3, R & D2VD3(res) = Vo + VD2Vo = Vs – 2VDPIV = Vs – 2VD + VD = Vs – VDHalf of PIV for Full wave RectifierD4D2D3
223Bridge Rectifier 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 rectifierLess cost – most popular circuitPeak Inverse voltage half as that of full wave rectifier
224Quiz no 2 DE26 EE) V Find the value of R for which V = 100 m V. For the circuit shown, both diodes are identical, conductingGroup # mA at 0.6 V and 10 mA at 0.7 V.Group # mA at 0.65 V and 100 mA at 0.75 V.Group # mA at 0.7 V and 100 mA at 0.8 V.V
225Quiz N0 2 DE26 CEIn the circuit shown, both diode have n=1, What value of V results ifGp # 1 D1 has 5 times the junction area of D2Gp # 2 D1 has 15 times the junction area of D2Gp # 3 D1 has 100 times the junction area of D2.D1D2V
226Peak RectifierPulsating dc output from rectifier in unsuitable as a dc power supply for electronic circuitSimplest way to reduce the variation is to place a capacitor across the load – smoothing, filter or reservoir capacitorCapacitor charges during positive half cycle to VS – (ideal diode) & maintains it during negative half cycle as diode is reversed biased
230Peak detector with Load 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 Vs exceeds the capacitor voltageDiode conducts & charges capacitor to VpeakSelect value of C so as discharge time constant is much greater than the discharge intervalCR >> T
231Figure 3.29 Voltage and current waveforms in the peak rectifier circuit with CR<<T. sedr42021_0329a.jpg
235Figure 3.30 Waveforms in the full-wave peak rectifier. sedr42021_0330.jpg
236Peak Rectifier: Observations Diode conduct for an interval Δt near peak of input, and recharges capacitor to Vp (makes up the lost)Conduction of diode begins at time t1, at which Vi equals the exponentially decaying output VoConduction stops shortly after the peak input voltage Vp and exact value of t2 can be calculated by putting iD=0 in equationDuring the diode-off-interval capacitor discharges through R & the decays exponentially with a time constant → CRVo = Vp - Vr (Vr => peak to peak ripple voltage)
237Peak Rectifier : Output Voltage When Vr is smallVo = VpeakiL is almost constantDC components of iLAccurate value of output dc voltage Average Value
245DeductionAs waveform of is almost right angle r triangle
246ObservationsDiode current flows for short interval and must replenish the charge lost by the capacitor. Discharge interval is long & discharge is through high resistanceMaximum diode current
247Example N0 3-9Consider 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.
248Example 3.9 Find value of C for Vr=2V (peak to peak) Find fraction of cycles that diode conducts=> Diode conducts of cycle
250Full 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
251Full Wave Peak Detector DeductionHaving same VP, f, R & Vr; the size of capacitor required is half that of the capacitor in the half wave rectifierCurrent through each diode of full wave peak rectifier is half that flows through the diode in half wave rectifier
252Capacitor Charge Voltage (vp) Ideal DiodeCapacitor will charge to vpReal diode in Half and full wave rectifierCapacitor will charge to vp-vDReal diode in Bridge wave rectifierCapacitor will charge to vp-2vDAll equations may be amended accordingly.
253Applications Peak Rectifier – Peak detector is used for Detecting the peak of the an input signal for signal processing systemsDemodulator for amplitude modulated (AM) signals
255Precision Half Wave Rectifier Super Diode Normal Diodes VD= 0.7v are used for rectifier of input of much larger amplitude then VDFor smaller signals detection, demodulation or rectification Operational Amplifiers (Op Amp) are used
257D3.86It 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 secondaryFind 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 conductionCalculate the peak diode current.
262Problem D3.87It 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 secondaryFind 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 conductionCalculate the peak diode current.
269Wave form Generation Limiting / Clamping Double LimiterClips off both negative & positive peaksSingle LimiterClips off only one side of the input peakApplicationLimits the inputs to operation Amplifier to a limit lower than the breakdown voltage of transistors of input stage of operational AmplifierHalf / Full Rectifier for Battery ChargerThreshold and limiting
270Figure 3.35 A variety of basic limiting circuits. sedr42021_0335.jpg
271Threshold 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
274D C RestorerThe 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
276DC 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 componentsProcess is known as DC Restoration & circuit as DC Restorer
277DC Restorer : Applications Restoring dc components is useful as dc components of a pulse is an effective measure of its Duty CycleDuty Cycle of a pulse waveform is used in pulse Modulation & carries InformationDemodulation 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.
278Figure 3.36 The clamped capacitor or dc restorer with a square-wave input and no load. sedr42021_0336.jpg
279Figure 3.37 The clamped capacitor with a load resistance R. sedr42021_0337.jpg
293Special Diode Type Schottky-Barrier Diode (SBD) Shottky-Barrier Diode is formed by bringing metal into contact with a moderately doped ‘n’ type semiconductor materialResulting 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
294Schottky-Barrier Diode (SBD) Gets two important propertiesSBD switches on-off faster due to current conducts due to majority carrier b (electrons)Forward voltage drop is lower then P-n junction diode
295Varactor Variable Capacitor Depletion layer acts as junction capacitanceDepletion layer varies CapacitanceUsed for voltage controlled Tuning Circuit
296VaractorWhen 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 .
297Semiconductor diodesThe 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.
298SCR (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.
300PhotodiodeIf 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
301Photodiode 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
302Photodiode : Applications Fiber optics Transmission of telephonic & TV signalsOpto-storage are CD ROM computer disksWide bandwidth & low signal attenuation.Solar Cell – light energy into Electrical energy
303Light Emitting Diode (LED) Inverse of PhotodiodeConverts a forward biased current into lightGaAs used for manufacturing LEDsUsed as electronics displaysCoherent light into a narrow bandwidth laser diodesFiber Optics & CD ROM
306Optoisolator LED & Photodiode Electrical to light Light to electrical Provides complete electrical isolation between electrical circuitsReduces the effects of electrical interference on signal being fixed within a systemReduces risk of shockCan be implemented over long distance fiber optics communication links
310Problem 3-103Sketch and label the transfer Characteristics of the circuit shown over a + 10 V range of the input signal. All diodes are VD =0.7 1 mA with n=1.What are the slopes of the characteristic at the extreme + 10 V levels?
314Ist SessionalQ 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?
315Ist SessionalQ No 2 (15 Marks) A circuit utilizes three identical diodes connected in series having n=1 and IS= 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?
316Ist SessionalQ 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.
317Ist SessionalQ No 4 (10 Marks) 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 VZO of the zener model.(b) Find the zener voltage at a current of 10 mA.
318Ist SessionalQ 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?
319Ist SessionalQ 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.
320sedr42021_0331a.jpgFigure The “superdiode” precision half-wave rectifier and its almost-ideal transfer characteristic. Note that when vI > 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.
327Quiz DE28 EE -B(10 Marks) 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 VZO of the zener model.(b) Find the zener voltage at a current of 10 mA.
328Quiz DE 28 EE -AA zener diode whose nominal voltage is 10 V at 10 mA has an incremental resistance of 50 Ω.What is the value of VZO of the zener model?What voltage do you expect if the diode current is doubled?