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:
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
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.
Diode Equivalent circuit in the reverse direction Equivalent circuit in the forward direction.
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
The two modes of operation of ideal diodes Forward biased Forward Current 10 mA Reverse biased Reverse Voltage 10 V
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)
Figure 3.6 Circuits for Example 3.2. Diodes are ideal, Find the value of I and V
Example 3.2. Assumption Both Diodes are conducting
Node A Node B Assumption Both Diodes are conducting Not Possible Thus assumption of both diode conducting is wrong
Example 3.2(b). Assumption # 2 Diodes 1 is not conducting Diodes 2 is conducting Assumption is correct
Figure E3.4 Diodes are ideal, Find the value of I and V
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
Figure E3.4 Diodes are ideal, Find the value of I and V I= 3mA V= 3V I= 4mA V= 1V
Figure P3.2 Diodes are ideal, Find the value of I and V
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
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
Figure P3.4 In ideal diodes circuits, v 1 i s a 1-kHz, 10V peak sine wave. Sketch the waveform of v o
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
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
Figure P3.4 In ideal diodes circuits, v 1 s a 1-kHz, 10V peak sine wave. Sketch the waveform of v o
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
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
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
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
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
Problem 3-10 D conducting I=0.225 mA V=4.5V D is not conducting I=0A V=-2V
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
Solution Quiz No 3 8V I1I1 I2I2 v i /2 I 10-10-07
22-10-07 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
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
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)
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
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.
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).
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.
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.
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.
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.
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.
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
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.
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.
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
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.
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.
A silicon crystal doped with a trivalent impurity. Each dopant atom gives rise to a hole The semiconductor becomes p type.
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
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.
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
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
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)
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
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
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
Drift Current I s Diffusion current due to majority carrier diffusion A component due to minority carrier drift exists across the junction
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
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
(a)The pn junction with no applied voltage (open-circuited terminals). (b) The potential distribution along an axis perpendicular to the junction.
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.
The charge stored on either side of the depletion layer as a function of the reverse voltage V R.
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.
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.
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 A @ N D.
The diode i– v relationship with some scales expanded and others compressed in order to reveal details. The diode i–v relationship
Terminal Characteristics of a Junction Diode Forward Biased Region v > 0 Reversed Biased Region v < 0 Breakdown Region v < - V ZK
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 10 -15 A for small size diode –Doubles in value for every 10 O C rise in temperature
Thermal Voltage V T –V T = kT/q K = Boltzmann’s constant = 1.38 X 10 -23 Joules/Kelvin T = Absolute Temperature in Kelvin (273 +Temp in C o ) q = Magnitude of charge = 1.6 X 10 -19 Coulombs –V T @ 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
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.
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
for v drop changes by for n = 1 for n = 2 Forward Biased Region
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
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
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
Ex 3.7 Silicon Diode with n=1 has V D =0.7V @ i=1mA. Find voltage drop at i=0.1mA & 10mA
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
Problem 3-23 The circuit shown utilizes three identical diodes having n=1 and Is= 10 -14 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
Solution 3-23 The circuit shown utilizes three identical diodes having n=1 and I s = 10 -14 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.
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?
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?
solution 2-25 (b) To obtain a value of 50 mV, what current I 2 id needed.
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.
A simple circuit used to illustrate the analysis of circuits in which the diode is forward conducting.
Graphical analysis of the circuit using the exponential diode model.
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 1mA @ a V D of.7 V, and that its voltage drop changes by 0.1 V for every decade change in current.
Approximating the diode forward characteristic with two straight lines: the piecewise-linear model.
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.
Piecewise-linear model of the diode forward characteristic and its equivalent circuit representation.
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
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.
Solution Input variation of 10% resulted in output voltage variation of 0.7+5.4mV(0.8%) Voltage regulation
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
Excercise 3-16 Why 4 diodes and not 5? Diodes will not conduct at 0.6 V
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.
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
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?
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?
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.
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
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
The diode i– v characteristic with the breakdown region shown in some detail.
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
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 0.5 W @ 6.8 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,
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
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 =
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
The circuit with the zener diode replaced with its equivalent circuit model. Example 3.8
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
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
b) Find v O if load resistance R L connected & draws 1mA and load regulation
Check exact Calculations 1mA drawn by load would decrease by same amount so Load Regulation
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
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
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?
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?
(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
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
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
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
Half Wave Rectifier Half-wave rectifier. Equivalent circuit of the half-wave rectifier with the diode replaced with its battery-plus-resistance model.
Transfer characteristic of the rectifier circuit Input and output waveforms, assuming that r D >> R. Half Wave Rectifier
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
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
Diode in Reverse biased state Anode @ - V s Cathode @ + V o PIV = 2V s - V DO Twice as in case of half wave rectifier Full Wave Rectifier
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
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
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
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
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
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
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
Figure 3.29 Voltage and current waveforms in the peak rectifier circuit with CR<
"name": "Figure 3.29 Voltage and current waveforms in the peak rectifier circuit with CR<
Figure 3.30 Waveforms in the full-wave peak rectifier.
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)
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
As waveform of is almost right angle r triangle Deduction
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
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.
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
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
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
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.
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
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
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.
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.
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
Figure 3.35 A variety of basic limiting circuits.
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
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
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
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
Figure 3.36 The clamped capacitor or dc restorer with a square-wave input and no load.
Figure 3.37 The clamped capacitor with a load resistance R.
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
Figure 3.38 Voltage doubler: (a) circuit; (b) waveform of the voltage across D 1.
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
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)
Varactor Variable Capacitor –Depletion layer acts as junction capacitance –Depletion layer varies Capacitance –Used for voltage controlled Tuning Circuit
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.
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.
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.
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
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
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
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
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
Problem 3-103 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 V @ 1 mA with n=1. What are the slopes of the characteristic at the extreme + 10 V levels?
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?
Q No 2 (15 Marks) A circuit utilizes three identical diodes connected in series having n=1 and I S = 10 -14 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
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
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
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
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
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.
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.
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?