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# Class Note for ECE 3336 at UH

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G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Section 91 Electrical Conduction in Semiconductor Devices Section 92 The pn Junction and the Semiconductor Diode Problem 91 Solution Known quantities The Ionized Acceptor Density for a doped Silicon N N 1017 N 0 In Find a If this material is an N or P type extrinsic semiconductor b Which are the majority and which the minority charge carriers c The density of majority and minority carriers Analysis a Each acceptor dopant atom introduces an additional positive charge carrier and a negative atomic ion The ion is NOT a charge carrier The density of positive carriers holes increases because of the doping so the material is extrinsic P type Silicon b The majority carriers are the positive carriers or valence band holes the minority carriers are the negative carriers or conduction band free electrons n2 c CNE ppa 0 np0 N 0 CPE 39 n quot pa p 2 39 r 2 r 2 0 N 0 13W ppalNa nw0 Ppa Using the quadratic equation n PW Ngt Ngt2 4 ngt g 1owgtg mwgt2492251030 51016 52210 1r02210 3 m where the negative answer is physically impossible Now use the CPE to obtain the minority carrier density 2 32 npa quoti 2 2510 17 22021015 i3 pp 102210 m Note that because of the doping the hole density is now about 100 times the electron density The thermally produced carriers present in the intrinsic Silicon before doping has a small effect on the carrier densities in the extrinsic Silicon At higher doping levels the effect becomes negligible As temperature increases the densities of the thermally produced carriers increase and their effect on the final carrier densities increase At very high temperatures about 175 C for Silicon the thermally produced carriers primarily determine the final carrier densities and the doping has a negligible effect ie the semiconductor behaves as an intrinsic material This is why semiconductors cannot operate in high temperature environments 92 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Problem 92 Solution Known quantities The intrinsic Silicon is doped 1 N N 1017 73 N N 51018 1 In E Find a If the silicon is an N or P type extrinsic semiconductor b Which are the majority and which the minority charge carriers c The density of majority and minority carriers Analysis a N Type b Majority Conduction band free electrons 2 Negative carriers Minority Valence band holes 2 Positive carriers c nwzzwio18 1 m3 1 pm 4r5910 E Problem 93 Solution Find Describe the microscopic structure of semiconductor materials What are the three most commonly used semiconductor materials Analysis Semiconductor materials are crystalline with the atoms arranged in a repeated three dimensional array The distance between atoms in the array is the quotlattice constantquot Each atom of a semiconductor has four valence electrons These electrons participate in covalent bonds with the valance electrons of other atoms For certain materials with the properties above quantumwave mechanics predicts that the valance electrons may have a total energy kinetic plus Coulombic potential energyl within certain quotallowedquot bands The two most important bands are the valence band containing the valence electrons in covalent bonds and the conduction band containing conduction or free electrons which have obtained enough energy to escape from its covalent bond Separating these two allowed bands is the quotenergy gapquot extending over those energies which the electrons are quotforbiddenquot to have For semiconductor materials the energy gap is on the order of l electronVolt eV Silicon is the most common semiconductor material and is used in a variety of applications and devices Germanium is used in some optical devices and other special purpose devices Gallium Arsenide is a compound IIIV semiconductor material One atom has three and the other has five valence electrons giving an average of four per atom It is used in microwave optical and very high speed digital devices Problem 94 Solution Find Describe the thermal production of charge carriers in a semiconductor and how this process limits the operation of a semiconductor device 93 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Analysis At a temperature of absolute zero ALL valence electrons in a semiconductor are contained in a covalent bond and there are NO charge carriers The internal or thermal energy of a solid material is caused by the vibration of the atoms and electrons about their equilibrium position As the temperature of the material increases its thermal vibrational energy increases Some electrons will gain sufficient energy to escape the covalent bond in the valence band and quotjumpquot past the energy gap into the conduction band As a consequence TWO charge carriers are generated The conduction or free electron in the conduction band is a negative charge carrier The vacancy in the valence band covalent bond or quotholequot is a positive charge carrier A conduction band electron may also give up energy and recombine with a valence band hole The generation and recombination rates both increase with temperature At any particular temperature they are equal and produce equal equilibrium densities of electrons and holes The equilibrium carrier densities increase with temperature For Silicon at T 300 K approximately room temperature nw pi0 151010 4510 136 cm In A number of carriers is a dimensionless quantity and may be omitted from the units Almost all semiconductors devices are quotdopedquot to achieve DIFFERENT densities of positive and negative carriers A quotPtypequot semiconductor has a higher density of positive carriers and an quotNtypequot semiconductor has a higher density of negative carriers However at high temperatures the density of thermally produced carriers becomes very large and significantly reduces or nullifies the effects of the doping ie the positive and negative carrier densities become nearly equal For this reason semiconductor devices cannot be used in high temperature applications The limit in temperature depends on the semiconductor material Problem 95 Solution Find Describe the properties of donor and acceptor dopant atoms and how they affect the densities of charge carriers in a semiconductor material Analysis An quotintrinsicquot semiconductor material is undoped When dopant atoms are added the material becomes an quotextrinsicquot semiconductor Doping results in the replacement of an intrinsic atom with a dopant atom As few as one out of every million intrinsic atoms may be replaced A quotdonorquot dopant atom has 5 valence electrons Only 4 are required to complete the bonding structure in the semiconductor material The 5th electron requires very little energy to escape to the conduction band and become a negative charge carrier This leaves behind a donor atom with one missing electron or a negative atomic ion The ion is immobile and cannot move through the material therefore IT IS NOT A CHARGE CARRIER Each donor contributes an additional negative carrier to the material The increased density of negative carriers results in an increased recombination rate which reduces the density of positive carriers The PRODUCT of the two densities remains constant Materials doped with donor atoms are N type extrinsic semiconductors The majority carriers are conduction band electrons the minority carriers are valence band holes An quotacceptorquot dopant atom has 3 valence electrons however 4 are required to complete the bonding structure in the semiconductor material The quotmissingquot 4th electron causes a vacancy or hole in the bonding structure Another valence electron may move to and occupy this hole thus eliminating it and generating another hole and a negative atomic ion The ion is immobile and cannot move through the material therefore IT IS NOT A CHARGE CARRIER Each acceptor contributes an additional positive carrier to the material The increased density of positive carriers results in an increased recombination rate which reduces the density of negative carriers The PRODUCT of the two densities remains constant Materials doped with acceptor atoms are P type 94 G Runonu Puncuples and Applucatuons or Electsucal Enguneeuung Problem solutuons chapteu 9 exuunsuc senuconductous The muouuty cauuueus aue valance band holes the nunouuty cauuueus aue conductuon band electsons Frublem 96 Solution Find Physucally descuube the behavuou 0f the chauge cauuueus and uonuzed dopunt atoms un the vucunuty Of a senuconductou FN Jmcuun that causes the potentual eneugyl bones that tends to puevent chauge cauuueus mm cuossung the uunctuon Ammm Semuconductou atoms aue not shown un the two ugues The cuucles uepuesent uonuzed dopunt atoms and the uncuucled plus and munus sugns uepuesentchauge cauuueus V p The dotted lune un the rust glle uepuesents how rau the m depletuonspace chauge uegon extends unto the F and N ueguons Neau the umctuon the negatuve cauuueus un the N nuteuual uecounbune wuth the posutuve cauuueus un the F nuteuual Thus rouuns a small uegon on eutheu sude or the Juncuun whuch us depleted orchauge cauuueus howeveu the uonuzed dopant atoms aue ununobule and uennun Thelafale un the N nuteuual the uegon has a net posutuve chauge and un the F nuteuual the ueguon has a net negatuve chauge Thus ueguon us called eutheu a depleuun ueguon depleted 0f cauuueusl ox a space chauge ueguon due to the dopant uonsl lra conductuon bund flee electxon a unauouuty cauuueul un the N mateuual tuues to cuoss unto the F nnteuual ut encounteus and us uepelled by the net negatuve chauge due to the uonuzed acceptou alums un the depleted paut uf leF nnteual If a valence band hole a muouuty cauuueul un the F nnteual tuues to cuoss unto the the N nnteual ut encomteus and us uepelled by the net P A posutuve charge due to the uonuzed donou atoms un the depleted part Of the N nnteual The uepulsuon 0f the cauuueus us chauacteuuzed as a Canlumblc potentual eneugyl bahHer Wuth no voltage applued acuoss the Jmcuun ohms law leqmles the cuuuent to be zeso Actually veuy veuy snnll equal but opposute cuuents do aw acuoss the Juncuun but the net cuuent us zeso The bauuueu can be decueased by applyung a fulvmd bms voltage acuoss the Jmcuun Thus allows moue cauuueus to cuoss the Juncuun and when thus voltage us geateu than a cemm value a 7 v for Slhwn a suguucantcuuuent nulluampsl flows The bauuueu can be uncueased by applyung a reverse blas voltage acuoss the Jmcuun Thus uncueases the bauuueu and reweu unuouuty cauuueus have smacuent eneugy to cuoss the Juncuun ue the cuuuent essentually ceases Howeveu these us a VERY VERY snnll reverse samuatuon cmlant m the remtoamps uangel due to the nunouuty cauuueus Sunce the nunouuty cauuueus aue themnlly puoduced thus cuuuent us dependent on mperatme 95 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Section 93 Circuit Models for the Semiconductor Diode Focus on Methodology Determining the conduction state of an ideal diode 1 Assume a diode conduction state on or off Substitute the ideal diode circuit model into the circuit short circuit if on open circuit if off Solve for diode current and voltage using linear circuit analysis techniques 4 If the solution is consistent with the assumption then the initial assumption was correct if not the diode conduction state is opposite to that initially assumed For example if the diode has been assumed off but the diode voltage computed after replacing the diode with an open circuit is a forward bias then it must be true that the actual state of the diode is on E Focus on Methodology Determining the operating point of a diode 1 Reduce the circuit to a Thevenin or Norton equivalent circuit with the diode as the load 2 Write the load line equation 915 3 Solve numerically two simultaneous equations in two unknowns the loadline equation and the diode equation for the diode current and voltage or 4 Solve graphically by finding the intersection of the diode curve eg from a a data sheet with the loadline The intersection of the two curves is the diode operating point Problem 97 Solution Known quantities The circuit of Figure P97 Find A plot of VL versus v3 Analysis For VS lt 0 the diode is reverse biased and VL 0 For VS 2 0 the diode is forward biased and V V RLHRI L s A R5 RL HR1 VL R R Where Ol tan 1 RS RL HR1 96 GRszzumPnnmp1 s and Appmauans ufElccmcal Engmeamg Pmblcm wluuuns Chap39u 9 Problem 53 Solulio Known qunlilies The cucun of 1ng P97 usmg an offset dmdc mead Find A pm of V versus VSV Amlysis The cucun can be 11313ch as shawn m m fulluwmg gures RS V7 0 VS 9 L Fur v lt V7 the dmdcxs reverse biased and vL 039 anmr Vv ehavc vL0ltvltVy1 R For v 2 V7 Le v 2 V7 1 ijthc made 3 award bsascd and Luis A u 0 V V 9V7 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Problem 99 Solution Known quantities For the circuit of Figure P97 VS 2 6 V and the resistances R1 R5 2 RL 2 le Find Determine iD and VD graphically Assumptions Forward voltage vs Use the diode characteristic of the 1N461A mm forward current Analysis lt Replace the diode with an open circuit and V DOC 3 V 10 E 1 6 E Replace the diode With a short circuit and 1D 2 mA 3 SC 2 1500 E 1 0 n These are the end points of the load line g The load line is superimposed on the diode characteristic in the figure H39 L m L From the intersection of the load line and the diode characteristic 0 U l we see that iD 2515 mA and VD 5 075V 392 0 4 0390 0398 1390 1392 VF Forward voltage volts Problem 910 Solution Known quantities The current I 1 mA that make the diode to be above the knee of its i V characteristic Find a The value of R to establish a 5 mA current in the circuit b With the value of R established in the preceding part what is the minimum value to which the voltage E could be reduced and still maintain diode current above the knee Assumptions V 07 v Analysis a R 5 0 28609 5 10 b 1 Emiquot 0 7 110 3 860 gt Emquot 08607 2156 v 98 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Problem 911 Solution Known quantities The circuit of Figure P911 driven by a sinusoidal source of 50 V rms V7 07 V R 220 Q Find a The maximum forward current b The peak inverse voltage across the diode Analysis a Fm mfO7318mA b V 50 70r7v revw Problem 912 Solution Known quantities The configurations shown if Figure P912 Find Which diode are forward biased and which are reverse biased Analysis a reversebiased b forwardbiased c reversebiased d forwardbiased e forwardbiased Problem 913 Solution Known quantities The configuration shown if Figure P913 Find The range of Vin for which D1 is forwardbiased Analysis The diode D1 is clearly forwardbiased for any Vin gt O Problem 914 Solution Known quantities The configurations of Figure P914 Find Determine which diodes are forwardbiased and which are reversebiased Determine the output voltage 99 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Assumptions The drop across each forward biased diode is 07 V Analysis a D2 and D4 are forward biased D1 and D3 are reverse biased vow 5 07 43 V b D1 and D2 are reverse biased D3 is forward biased vow 10 07 93 V c D1 is reverse biased D2 is forward biased vow 5 07 43 V Problem 915 Solution v Known quantities The circuit of Figure P915 vs t 10 sin 20007rt Find The output waveform and the voltage transfer characteristic gt39 Assumptions The diode is ideal Analysis 0 0 5 1 15 2 tms For VS lt8 V v04 V For v52 8 V v0 1152 The voltage transfer characteristic is A Va 8 V O 8 16 VS Problem 916 Solution Known quantities The circuit of Figure P915 vs t 10sin20007rt 6 Find 5 The output waveform and the voltage transfer characteristic Assumptions 4 I The diode has an offset V 06 V E l 3 3 Analysis 3 2 For VS lt68 V v04Vy 34 V For 152 68 V v0 1252 The voltage transfer characteristic is 1 A 0 V0 0 05 1 15 2 25 tsec 3 6 8 x 10 V O 68 136 910 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Vs Problem 917 Solution Known quantities Same as Problem 915 but with VS I 15 sin 20007L39t the battery equal to l V and the resistors of 1kg Find The output waveform and the voltage transfer characteristic Assumptions The diode has an offset V 06 V and I D 200 9 Analysis VS 1 V7 1000 r0 vs 2 1 1 1 7 1000 FD 1000 For VS lt 21 V3 08 V V0 For 1232 08 V 120 1232 The voltage transfer characteristic is A V0 08 i E 0 08 l 6 VS gt0 t sec X 103 911 G RIZZUHL Ptlholples aha Apphoatlohs otEleomoal Ehglheetlhg Problem solutlohs Chaptet 9 Problem 918 Solution Knnwn quantities VD Theclrcult otFlgutepa18 thedlude IS tahtloated from Slllcun and ID 10 5V7 71 At T300 K10 25010 A V Ezzs mV v5 42V110caowtmv q w377R7kQ S nd Detetmlhe uslhg supetposltloh the DC ot Qrpumt ouneht though the dlude a Uslhg the DC ottset model totthe dlude b By Itemther sulvlngthe clrcult ohataotetlstlo aha devloe ohataotetlstlo Analysis a Supptess the AC campunent ufthe some voltage Construct the DC equlvaleht mot uslhg the threshold ot offset voltage model The Q polht IS at the mtersectlun ufthe devloe made ohataotetlstlo and the clrcult ohataotenstlo glveh by the KVL below Hete the devlce ohataotenstlo IS apptomated by the threshold voltage model glvlhg the apptomate Q polht at the uppet tlght Assume the dlude 15 oh Then VD VM 07 V KVL b In the totwam blased tegloh Wlth slghmoaht CUnduCtlUn 912 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Vi Device ID IOeVT VD VT Inf Circuit KVL VS IDRVD 0 A simultaneous solution the lower left Q point of the device and circuit characteristics is required To do this iteratively initially assume a value for the diode voltage say 07 V for a Silicon device Then 1 Using the initial or new diode voltage and the circuit characteristic determine a new diode current 2 Using this new diode current and the device characteristic determine a new diode voltage ITERATE or REPEAT until convergence is obtained Voltage New Current New Voltage 700 mV 05 ma 3772 mV 3772 mV 05461 ma 3795 mV 3795 mV 05458 ma 3795 mV ltltlt Convergence Convergence is obtained after only 3 iterations and IDQ 546 uA VDQ 3795 mV This is a much more accurate solution than that of Part a The two solutions differ significantly because the reverse saturation current given is atypically large Problem 919 Solution Known quantities VD The circuit of Figure P918 the diode is fabricated from Silicon and ID 10 eVT l At kT T300K 1O 2ro310 15 A VT 7z26 mV v5 53 V7casatmV 1 rad 03777 R4i6 kg S Find Determine using superposition and the offset voltage model for the diode the DC or Qpoint current through the diode Analysis Suppress the AC component of the source voltage Construct the DC equivalent circuit using the threshold or offset voltage model The Q point is at the intersection of the device diode characteristic and the circuit characterich given by the KVL below Here the device characterich is approximated by the threshold voltage model giving the approximate Q point at the upper right The DC source voltage will tend to make the diode conduct Assume the diode is on Then Vs VD VD VDW 017 V KVL VS 1DQR VD 01DQ TW10 mA The current is positive so the assumption above that the diode is on is valid 913 G Rszzum Pnncxples and Apphcauuns er Exeemeax Engmeenng Problem winner s Chapter 9 Frublem 920 Solution Known quanti es The cucml 0f Fxgme P9 13 me made 5 abnca39ed from thcun and I D E326 mV v I rad a 377 i R 7 kg The DC upemung pawns 1mZ s T 300 K 10 25010 AnT Find The eqmvalant mmlhxgml AC resistance er me mode at room mpemmre at me Q paint gwan Ammm We y D 41 z 914 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Problem 921 Solution Known quantities Ln The circuit of Figure P918 the diode is fabricated from Silicon and ID 10 eVT l At T 300 K 10 2031045 A VT 1LT z 26 mv VS 53 V 70005at mV 1 rad y y a 377 7 R 46 kQThe DC operatingpointis IDQ 10 mA VDQ 07 V S Find The equivalent smallsignal AC resistance of the diode at room temperature at the Q point given Analysis DdeD 1 1V J 4269 dIDQ dID 118 D9 D9 dV 0 D Q Q Problem 922 Solution Known quantities A diode with the iv characterich in Figure 98 connected to a 5 V source and a load resistance of 200 9 Find a The load current and voltage b The power dissipated by the diode c The load current and voltage if the load is changed to 100 Q and 500 2 Analysis a The operating point can be determined by using the loadline analysis The load line is Dlode Irv curve 0 05 r r l 5 VD 5 VD 7 7 7 0 0457 D RL 200 200 0 04 7 The load voltage is 0 035 7 003 gt vL 5 vD 25 074426V D ode current A o 0 o o m m m The load current is obtained by the figure as intersection of the two characteristics and is equal to 0021 A o S m 001 0 005 b The power dissipated by the diode is 0 0 2 04 0 6 0 8 1 Dlode voltage V 915 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Dlode Irv curve PD vDiD 074 0021 15 mV 005 i i 0 045 c For RLlOO 9 we have 0 04 Eio o n i i i i i i v i i i VD E 0757 V 00357 f 0037 l iDzL 00424A 1 100 a o 025 vL5 0757424V E 002 1 RL500 Q l 0 015 1 Similarly for RL 500 2 we have 001 l VD 20717V 0005 J 3 0 i 0 0 2 04 0 8 0 8 I39D 1L 004283 A Dlodevoltage v 100 VL 5 0717 4283 V Problem 923 Solution Known quantities A diode with the iv characterich in Figure 932 connected in series to a 2 V source and a load resistance of 200 2 Find a The load current and voltage b The power dissipated by the diode c The load current and voltage if the load is changed to 100 Q and 300 2 Analysis a The operating point can be determined by using the loadline analysis The load line is 2 v 2 v 1D D 7 7D001 0005 VD RL 200 200 By drawing this line on the top of Figure 932 the following operating point is obtained iD iL z6mA VD 073V VL 2 vD 2 073127 V b PD vDiD 073 0006 438 mV c For RLleO 2 we have VD x 0825 V 2 0825 1L E 7 100 VL z 2 0825 1175 V iD 1175 mA 916 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 d For RL3OO 9 we have szmv 2 07 39 2743 A ID IL In sz2 0713v Problem 924 Solution Known quantities Ln The circuit of Figure P918 the diode is fabricated from Silicon and ID 10 8 VT l At kT T 300 K 10 25010 12 A VT 7 26 mV v5 42 V 110005at mV 4 a 377 E R 7 kQThe DC operating pointis IDQ 0548 mA VDQ 0365 V S rd 47r459 Find Determine using superposition the AC voltage across the diode and the AC current through it Analysis Suppress the DC component of the source voltage Replace the diode With its AC equivalent resistance then r V V d d SRrd 4745 V 110z0 740i6z0 Vv t 740i6caswt v d 700047i45 H do 1d V1M15i6110 uA id 15i6lcasatuA rd 4745 The total solution is then id 0548103 15i6lcasat10 6 A vd t 0365 740613046001076 v 917 G Rszzum Pnncxples and Apphcauuns 0f 5194mm Engmeenng Pmblem mluuuns mpg 9 13 3V 1 9 j M 4 Frublem 925 Solution Known quanti es TheeucmlufogmeP as The dlude 15 bxnamed from Sthcan and R 22 kg 39152 3 V Find The mmmum value 0f vn mm have winch the mode W111 canductwnh a sxgm canlcmmnt Ammm A dmdefabncawd mm sxhcun wll canduathth a mgm cantcunanufnhasa rumm bmeqml w or larger um mm 17 V KVL 7v IDR VD sz At pmm 0f canducuun 1 0v vM 07 V vn 007337 V 395 DR VD ng 913 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Section 94 Rectifier Circuits Problem 926 Solution Known quantities The circuit of Figure P926 The input voltage is sinusoidal with an amplitude of 5 V Find The average value of the output voltage Assumptions Vy 07 V Analysis The capacitor will charge to 5 V 07 V 43 V and therefore the input sine wave will be shifted up 43 V to produce the output As a result after the cycle the capacitor builds up its stored charge during the third quarter cycle the average value of the output will be 43 V Problem 927 Solution Known quantities The rectifier circuit of Figure P927 Vt A Sin2727100t V The conduction must begin during each positive halfcycle at an angle no greater than 5 Find The minimum peak valueA that the AC source must produce Assumptions V07v Analysis Am sin5 07 Am 0 7 0 7 803V sin 5 00872 Problem 928 Solution Known quantities A halfwave rectifier is to provide an average voltage of 50 V at its output Find a Draw a schematic diagram of the circuit b Sketch the output voltage waveshape c Determine the peak value of the input voltage d Sketch the input voltage waveshape e The rms voltage at the input Analysis a 919 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 N M V 9 R2 v b v t 0 VW 0318mek 50 gt vmk 157 2 V d vll t 6 15 111i2 V V mm J5 Problem 929 Solution Known quantities A rectifier circuit similar to that of Figure 925 Load Resistance 100 2 AC source voltage 30 V rms Find The peak and average current in the load Assumptions Ideal diode Analysis The peak voltage is V VRW 305 V j W if 0424 A The average current in the load is 920 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 1 M0135 A RAV n Problem 930 Solution Known quantities A rectifier circuit similar to that of Figure 925 Load Resistance 220 2 AC source voltage 25 V rms Find The peak and average current in the load Assumptions Ideal diode Analysis The peak voltage is 7 0353 A V VRW 255 V 1RW W 215805 The average current in the load is IR 1 W01125A RAV n Problem 931 Solution Known quantities The fullwave power supply of Figure P931 The diodes are 1N461 with a rated peak reverse voltage equal to 25 V and are fabricated from Silicon n 005883 C 80 HF V1 g 170003377t V Find a The actual peak reverse voltage across each diode b The reasons for which these diodes are or are not suitable for the specification given Analysis a At wt 2 0 D is on At wt 2 727 D is off and the reverse voltage across it is maximum V an 170005883 10 V KVL v t VD1VLt 0 At at 0 VM VD70n Vm 0 j Vm Vw VDM 10 0i7 93 V At at 72 VwVD1 Vm 0 VD1 Vw Vm 10 9r3 193 v b The actual peak reverse voltage 193 V is less than the rated peak reverse voltage 25 V by a barely adequate margin of safety Therefore the diodes are suitable for the specifications given 921 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Problem 932 Solution Known quantities The fullwave power supply of Figure P93l n 005883 C 80 HF Vbquotg 1700056770 V The diodes are 1N4727 switching diodes fabricated from Silicon and with the following rated W performances PM 500 mW at T 25 C derated 3 n T 25 125 C and 4 LEW T 125175 C VW 30 v Find a The actual peak reverse voltage across each diode b The reasons for which these diodes are or are not suitable for the specification given Analysis a At cot 0 D1 is on At cot 727 D1 is off and the reverse voltage across it is maximum V an 1700117 v KVL 39 v tvD1vLt 0 At at 0 V VM Vm 0 Vm an 4M 17 oi7 163 v At at 72 VVD1 Vm 0 VD1 Vw Vm 17 163 33r3 v b The actual peak reverse voltage 333 V is greater than the rated peak reverse voltage 30 V Therefore the diodes are not suitable for the specifications given Problem 933 Solution Known quantities The fullwave DC power supply of Figure P93l the load voltage of Figure P933 L 60 mA VL 5 v V 5 VW 170005atVa377 E S Find a The turns ratio b The capacitor C Analysis VL Vr 5 01254r875 v Limin a Vm VLV 50i1255i125 V V VJ1 VJ2 VJO casat KVL v tvD1vLt0 Atat0 Vw V VL 0 VJO VDim Vm 0i75i1255i825 V Dian n V 00343 V i0 922 G Rlzzunl Pnnclples and Applueeueus er Elemlcal Eugmeeuug Fmblam spluueus clmples 9 will WV f in I an m blKVL vfztvmvlt 0 At r 5 v wxat2 The expande dlselmge emue capaclwr can be expressed Vtvzmvz0ivzme 7 W 02 6 Vme W M I t vLt2vLm Vme W c 2 1le p1 V 1 L Vm Nave An appruxlmm but censemuve wlue an an be ebmuel w usmg me appwxmmuun 1527 Than Ceray2000uF DVL ln nus value ls censemuve because u gves a smaller upple vulmg um um spemed Freblem 534 Solution Known quanli es ThefullrwaveDC puwex supply ufFlgure m 31 IL 600 mA v 50 V v 3 rad up 170wswt V a 377 i u Find a Thelmns mm d The capaclwr C Analysis a vm VLV 50252 VVLM V vn N mum KVL rv rvm vlt0 Max0 7v w w 0 vm W vm 0752527 V 1m 923 G Rszzum Pnncxples and Apphcauuns er Elemml Engmeenng Fmblam mluuuns Chapter 9 3 x If 2n wl bJKVL V2tvmvzt0 At t v cusat2rVDW 7v 729 ms The exppnemm dmhexge quhe eepmex can be expressed vzrvztegtvzogtrvzegtgte owmroy wimpquot vmeT c 7 1g V m V lnT 1093pF Frublem 935 Solution Known quanti es ThefullrwaveDC power supply ufogure m 311 5 mA V 10 V v 20 vi 170wsmt V a 377 E s Fin 39 a Thele mm b Thecapacxwr C Ammm a vm vy v 10111 V VLW 7 10719 V veewmmtax KVL V tvmvlt0 rVerVVme UVXOVgtWVM07167V 0688 924 Gt Rluum Pnnclplc and Appllmuun of Elemlea Engnccnng Problem soluttonx chapter 9 I t a 3 2 1r n Zn E mKVL39 v t vmvLtO At 2 Va coswr2gtrvom rvmv The cxpuncrmal dlxchargc ofthe oapaotor m be cxprcxxcd math twmw wvm 70W vLt2VMn Vmef C VLwn V47 Note An approxtmate but causewach value of c m be obtatned by uxmgthc appruxlmzuun 1534uF m2 n1hen C Macisuh V DVL n w Vm Thl value 1 cumcrvauvc because lt gvc a ma cr rtpple vultzg than that peofted Problem 936 Salufian Known quantities The fll rwavc recufter of gure P936 wtth a 12 V m supply Find 3 Sketch the lnput source vultzg v t and the uutput vultagc vL t and state whroh dludc are on and whlch are offtfthe dludc have an offset voltage ofO6 V and the frequency of the uurcc l 60 Hz b Sketch the uutpu voltage tf RL 1000 a and aoapaotor placed acmxx RL to provtde umc ftltenng ha a value of 8 uFV c A part bl wtth the capaottance equal to 100 uFV Analysis 3 The lnput source vultzg r huwn below together wrth the rccn cd load voltage 12 V m 1697 V 1ka 93925 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 V v rmm t I D1134 D213 D1134 V S OH on on b Thetime constant 7 CR is CR 10003105 8 ms 1 The period of the input sinusoid is T 16i7 ms 1 f Since the capacitor initial voltage is VC01697 1 2 1577 V and the final value is VC m 0 V VC 1 is given by VCt15i77 effT Therefore at t T we have VCT15i77ETTT 1i96 V The output waveform is shown below c The time constantis CR 2 100010010762 100 ms Note that CR gtgt T VC01697 1i2 1577 V and the final value is VC 00 0 V VC t is given by VC t 15 77871T and therefore VCT15i77ETTT 1334 V The output waveform is shown below 926 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Problem 937 Solution Known quantities The fullwave bridge power supply of Figure P937 the diodes are 1N659 with a rated peak reverse voltage 50 V n 02941 RL 25 kg C 700 H F v 170005at V a 377 E Find a The actual peak reverse voltage across the diodes b Explain why these diodes are or are not suitable for the specifications given Analysis a Vlm nvlm 50003at V V 50 V At at 0 the source voltage has the polarity shown therefore DI and D3 are conducting and D2 and D4 are off At cot 727 D1 and D3 are off and the voltage across them is the peak reverse voltage KVL 39 vxt VD1 VLt vm 0 At at 0 vL0Vm VD1VD3 an 486 v At at 7r VD17T VD3 727 V COS7T VL 727 V V w m VD1 VD3 Vm 50 486 493 v b The diodes are not suitable because the rated and actual peak reverse voltages are about the same Problem 938 Solution Known quantities The fullwave bridge power supply of Figure P937 the diodes are T151 with a rated peak reverse voltage 10 V and are fabricated from Silicon n 00423 VL 51 V V 02 V L 25 mA VW 156 casat v a 377 E S Find a The actual peak reverse voltage across the diodes 927 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 b Explain why these diodes are or are not suitable for the specifications given Analysis 1 a V an 66 V vt 66casax V Vm VL EV 52 V At at 0 the source voltage has the polarity shown therefore DI and D3 are conducting and D2 and D4 are off KVL 39 V tVD4 VL 0 VD2 0 At at 0 vm 0 VM 0 VL 0 V 0050 V V mm 1 VD2 VD4 VD E 66 5i2 5i9 V b The diodes are suitable because the actual PRV 59 V is significantly less than the rated PRV 10 V Problem 939 Solution Known quantities The fullwave bridge power supply of Figure P937 the diodes are fabricated from Silicon 752366 deg VL 10 V V 1V 1L 650mA vm 170003atVa377 E S Find The value of the average and the peak current through each diode Analysis Diodes DI and D3 will conduct half of the load current and Diodes D2 and D4 will conduct the other half Therefore 1 11L 325 mA 2 Dime The waveforms of the diode currents are complex but can be roughly approximated as triangular recall area of triangle 2 bhZ 1 2n 1 PI I IL ID13 ID24 m ij ID13 600 ID24 2nDpk W 272 0 2 2 1D i ML 128 A p 2 2 Problem 940 Solution Known quantities The fullwave bridge power supply of Figure P937 the diodes are fabricated from Silicon VL 53 V rad V 06 V L 85 mA mm 156casat V 0377 7 S Find a The turns ratio n 928 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 b The capacitor C Analysis a First determine the maximum and minimum voltage across the load resistance and capacitor VL V 5r3 035 V imin Vm VLV 5i30i35i6 V VL The amplitude of the supply voltage can now be determined KVL 39 Vm casat vD1 VD3 VL t 0 Att 0 VM VD VDM VL 0 VJO ZVDW Vm 0i70i7 5i6 7 V ran V ni0io4487 V127 b KVL VltvD2 vD4 VLt0 At t tza Via 005at2 VDran VDran Verm39 1 2V V at2 cas DHquot L mm 2725 rad V The exponential discharge of the capacitor can be expressed w 7 7 Lay vLt vLcgtltgtvL0vLcgtltgt E 0V 0e R Ve W h VLt2Verin V e mVLCa c m Latl V COVL lnW V771 1023uF Problem 941 Solution Known quantities The fullwave bridge power supply of Figure P937 the diodes are fabricated from Silicon VL 10 V rad V 24 v1L 250 mA VW 156casat V a 377 7 S Find a The turns ratio n b The capacitor C Analysis a First determine the maximum and minimum voltage across the load resistance and capacitor VLVY 10 1i28i8 v Limin Vm VL V 101T211T2 v V The amplitude of the supply voltage can now be determined KVL 39 Vm casat vD1 VD3 VL t 0 Att 0 VM VD VDM VL 0 VJO ZVDW Vm 0i70i7 11i212i6 V am 929 G Rizzoni Principles and Applications of Electrical Engineering 0 008077 n i0 b KVL vxtvm vD4 VLt0 At t t2 Vw casat2 VDM VDM WWquot at cas 1 W 1505 rad The exponential discharge of the capacitor can be expressed 7 7 Lay vLt vL ltgt vL0 who 0 Vm 0e R Vme W ILW Z VLt2 mm Vme W C i 6913 H F COVL Vm Problem solutions Chapter 9 930 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Section 95 DC Power Supplies Zener Diodes and Voltage Regulations Problem 942 Solution Known quantities The piecewise characteristic that passes through the points 10 V 5 HA 0 0 05 V 5 mA and 1 V 50 mA Find Determine the piecewise linear model and using that model solve for i and V Analysis Assume that the diode is forwardbiased and operating in the region between 05V 5mA and 1V 50mA If this is true then the diode can be modeled by the resistance AV 1 0 5 0 5 VD if 73 773 1111 9 in series with a battery having value AID 50 510 4510 V 05 510 31111 0444 V 2 0444 1 l 1 11 This solution is within the range initially assumed justifying the assumption Then 139 14 mA and v04440014111106 v Problem 943 Solution Known quantities The output voltage at 56 V Find Determine the minimum value of RL for which the output voltage remains atjust 56 V Analysis R 1856 124RL 10080 RL 8129Q Rme mm mm Problem 944 Solution Known quantities The output voltage 25 V The input voltage that varies from 35 to 40 V The maximum load current is 75 mA The maximum current 250 mA for the Zener diode used Find Determine the minimum and the maximum value for the series resistance 931 G Rlzzunl Pnnmples and Aple natluns of EIeomuaI Engrneenng Problem sulutruns Chapter 9 Analysis 7 40725 7 REM M 25075325mAgtRS 17521629 W 325x103 is O7575mAgtRS ijdzem W M 75gtlt10 Problem 945 Solution Knnwn quantities The H oharaotellstlo Ufa seml Conductor the mmlmum current at the quotkneequot ufthe curve 5 mA and the manmum current 90 mA Find Detemnne the Zenet tesl smrme and Zenet vulmge ufthe dlude Analysis The Zenet voltage ls evaluated at the mlddIe of the rated leg on of uperatlun le mldway between the knee of the omve and manmum rated current VZSV The Zenel lesl smnte ls detelmmed from the slope ufthe lrv ohataoten sue over the rated leg on at Uperatlun 1 AvD VD rvm 2 7 259 Slope AtD AiD 3104 AvD Note The manmum Zener current ls dltently related to the manmum power the Zener dlude can dlsslpate wlthuut breaking out ln smoke and ames Ln the spenlflnatlun sheet elthet at both may be speol ed 932 lt1 szzunL Pnnaplcs and Apphmuons ufElccmml Engmeamg Problem salmon Cth39cr 9 Problem 946 Solution Known quaint i5 The chcr diode med m 1N5231E m npplc componcnl of m mum volmgg obtained from a DC powersupplx v V prhcrc 4 20 V V 250 mVgt R ggog IL 65mAgt VL 5111 m39L PW 05Wgt 1M 10 mA mm De cmnnc an mmm mud Currant an mode can handle when txcecdmg as power 1mnnun Amlys Only an and power is rcquucdfmm an mformzuon gym Ugc an EC model of an Zanch New um power 13 man m m cqmvalcnl Ana resistance and m m cqmvalcnl mum Now an daemon of m cumnz and m polanty of m mum w 4 771501227 R2 7778 mA where m ncgmvc answer 13 mmd because an Zane current by dcfmmun ows m m daemon shown Problem 947 Solution Known qnnnmiu T39thcncrdmdeusciLhc was V 12 V Rz 11Sngt PRM 04 W Judaldmknccofthc my 1m 025 mAgt rm 7000 mm De cmnnc the mmm rated Currant the diode can handle when cxcecdmg as power um39anun Amlysis Only the ma power is rcquucdfmm the mformzuon gmn Ugc the DC model of an 2an New um power 13 man m m cqmvalcnl Ana resistance and m m cqmvalcnl mum Now an daemon of m cumnz and m polanty of m mum P 119 Izvzm P PW Then I 1H PW Igmxz Mgz n2 2 74PM 23155541 R2 R where m ncgmvc amch 13 mm bmux m Ana cumnz by Mnman ows m m dlrccuon shown 26mA 27m 933 G anzenn Rnncnpies and Appincaciens er Eieccsncai Engnneenng Rmbienn seinciens cinapces 9 Frublem 948 Solution Known quanti es v 5 V10 R 1552 IHM I 70 t 20 mA Find uecemnne cine mxmnnn and minimum wine er R w nanncann cine zenen dnede cnnenc wncinnn iLs speci ed inman Ammm Censcincc DC eqmvalant cncnnc KCL V V R SmAJM 30 mAVS 23V 0 5 1 1L KVL 7v 7113 w 0 Than 1 v 1sz v 7v iv 7v 7113 1 I I I A imximum Zenes cmenc is caused by Minimum wine afR Maximum sumce vdmge Minimum load cmianl Mnnnmnm Zenes vdmge V m 12W Han him A mimmmn Zenen cmencns caused by Maxnnnnnn wine 0 Minimum some vamge Maxnnnnnn iead cnnenc Maxnnnnnn zenex yeicage V Vz m m R 3687 l M 12m hm Nece cinac cine minimum wine 0f R EXCEEDS cine maxnnnnnn wine 0 Tinns means cinac cinese ns ne wine 0 far winncin aii cine speci cacienswnii be mecmdesaii candmuns A wine chan be cindsen bnc candmuns my uccm where cine zenen cnnenc exceeds iLs mmmn wine or raiis beiew ncs mimmmn wine Tinnspmbienn can be suiyed by 1 Chummg anecines Zenen dnede wncin dnrresenc minimum and mxmnnn cmiaan 2 Reiaxnng cine speci cacaens en wmce voltage and load cnnenc Nece cinac cine neiaciensinnps becween minimum and maxnnnnnn wines ALWAYS sTARTs WITH THE QUESTION OFWHAT WILLCAUSE A MINIMUM oRMAxiMuM zENER CURRENT in mine Zenen cnnenc exceeds ncs mximum med wine nc wnii bnnn np nrnc raiis beiew iLs minimum wine cine dnede wnii ieaye cine Zenen iegun and cease ce regime cine yeicage Tinese nepnesenc woRsT CASE cendncnuns a pmcedme rneqnnenciy nsed in design Nece cinac a zenen diode accs inine an eieccxncai surge ink for cnnenc hawevei cine analugy is nec exacc A liquid smge canknegniaces ne nanncannscenscanc pnessme and md aw mes ic dues cinns by cempexaniy scenng excess uid winen aw mes nncnease 0i snppiynngexcia md winen aw mes decnease Tine Zenen surge mnk pnmniy negniaces iead vamge as cine iead nesnscance and cineserene cine iead currency or wmce vdmge and cineseme cine cmenc snppined by cine some cinanges ic cempexaniy Stores 934 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 excess current When load current decreases or source voltage increases and quotsuppliesquot extra current When load current increases or source voltage decreases Problem 949 Solution Known quantities VZ 12 Vi10 Rz 9o1 1L 31i2105 mA Find Determine the maximum and minimum value of R to maintain the Zener diode current Within its specified limits 3025 mA 12 80 mA VS 25 105 v Z imin Analysis Construct DC equivalent circuit V V KCL Iz IL 0 KVL VZ IZRZ VL 0 Then Vs VL Vs Vz Isz I Z I L I Z I L A maximum Zener current is caused by Minimum value of R Maximum source voltage Minimum load current Minimum Zener voltage R I V5mx VZmin IZmdez 1664Q m PM I A minimum Zener current is caused by Maximum value of R Minimum source voltage Maximum load current Maximum Zener voltage R I V5mmx V2mx IZminRz 182 6 Q m I Em VL VZ IZRZ R Limin Zirm39n Problem 950 Solution Known quantities The diode used 1N474OA Vz 10 Vi5 R2 7 Q 1 VS 14Jr2 V R19089 PRM 1W Find Determine the maximum and minimum value of the load current to maintain the Zener diode current Within its specified limits 10 mA 12 91 mA Zirm39n Analysis Construct DC equivalent circuit VL V KCL Tsuz 1L 0 KVL VZ IZRZ VL 0 Then 935 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 V V V V I R VL VZ ZRZ L z Iz A minimum and maximum Zener current is caused by respectively Minimum source voltage Maximum source voltage Maximum load current Minimum load current Maximum Zener voltage Minimum Zener voltage Substituting With these extreme values and solving for the load current V V I R mm w 127mm 2 6222 mA R V V I R 1Wquot SW Z W Z 12m 2051 mA The minimum load current EXCEEDS the maximum load current Problem 951 Solution Known quantities The diode used 1N963 Vz 12 Vi10 Rz 115 Q I VS25i2VR47OQP 0i4W Rama Find Determine the maximum and minimum value of the load current to maintain the Zener diode current Within its specified limits 25 mA 12 326 mA Z imin Analysis Construct DC equivalent circuit V V KCL IZ IL 0 KVL VZ IZRZ VL 0 R Then V V V V I R VLVZIZRZIL7S L IZ S Z Z Z IZ R R A minimum and maximum Zener current is caused by respectively Minimum source voltage Maximum source voltage Maximum load current Minimum load current Maximum Zener voltage Minimum Zener voltage Substituting With these extreme values and solving for the load current V V I R mm W 127mm 21829 mA V V I R mm 5 Z 2 Z IZW 1i07 mA Problem 952 Solution Known quantities The diode used 1N474OA Vz 10 Vi5 R2 7 9 1M 10 mA PM 91 mA IL35i10mARSOQi5 PRmi 1W 936 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Find Determine the maximum and minimum value of the source voltage to maintain the Zener diode current within its specified limits Analysis Construct DC equivalent circuit V V KCL TIZIL0J VL VZ IZRZVL0 Then VL 2 V2 IZRZ VSVLRIZILVZIZRZRIZIL Minimum and Maximum Zener currents are caused by respectively Minimum source voltage Maximum source voltage Maximum load current Minimum load current Maximum Zener voltage Minimum Zener voltage Maximum R Minimum R Substituting with these extreme values and solving for the source voltage Vsrmin Vzrmax lintle Rm 127mm Lima 1519 m VSW Vzm IZWRZ Rm 1H 1Lm18r95 mA Problem 953 Solution Known quantities The diode used 1N474OA The source voltage is obtained from a DC power supply that has a DC and a ripple component VS VS Vr where VS 16 V V 2 V 127mm 10 mA 127W 91 mA IL 35 mA RZ 79 RSOQ VL 10 V VZ 10 V Find Determine the ripple voltage across the load Analysis Construct the AC equivalent circuit Suppress the DC component of all voltages and currents in the circuit Suppress also the DC source in the Zener diode model 2gs7gR I 3quot R L Z R 6i83Q L L Note VS Vr R V VL7W1574 mV RR 5 1 Problem 954 Solution Known quantities The diode used lN523lB The source voltage is obtained from a DC power supply that has a DC and a ripple component VS VS Vr where VS 20 V Vr 250 mV 127mm 10 mA PM 05 W IL 65 mA RZ 17Q R2209 VL 5l V VZ 5l V 937 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Find Determine the ripple voltage across the load Analysis Construct the AC equivalent circuit Note that in this case the load resistance is much smaller than the Zener resistance and Will cause a small reduction in the ripple voltage across the load V R R RL 7L78i46Q qu 13i979 I L R2 R L Note VS V qu V VL 7 14r93 mV RR5q Problem 955 Solution Known quantities The diode used 1N970 The source voltage is obtained from a DC power supply that has a DC and a ripple component VS VS VrWhere VS 30 V V 3 V 127mm 15 A 127W 15 A L 8 A RZ 33Q RIQ VL 24v V2 24 v Find Determine the ripple voltage across the load Analysis Construct the AC equivalent circuit Note that in this case the load resistance is much smaller than the Zener resistance and Will cause a small reduction in the ripple voltage across the load V R R RL iL3Q qu 27SOQ IL RZRL Note VS Vr R WVL 22v 938 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Section 96 Signal Processing Applications Section 97 Photodiodes Problem 956 Solution Known quantities The range for the input voltage 0 S V S 10 V for the circuit of Figure P956 Find Determine the iV characteristic of the circuit Analysis With the variables defined in the circuit below we can compute the following currents V V V 1 2 forv24V I3 fort26V 100 100 For OSVS4V I00lv For 431236 V I 002v 004 For 631310 V I 003v O1 The resulting iv characteristic is shown below i ll mA 1 1009 7 100 l w l 4V V 10052 6039 I 1 1009 6vl 40 I l l 13 T I 20 39 L J Device 0 2 4 6 8 10 V Problem 957 Solution Known quantities The input voltage waveform and the circuit of Figure P957 Find Determine the output voltage Analysis For Vin lt 507 V VOW Vin 06 98 06 The input and output voltage waveforms are sketched below When viquot 2 507 V vm 507 vi 507 6085x10 3 m 939 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 v V 100 5131 507 Problem 958 Solution Known quantities The circuit of Figure P958 where the switch is close at time t t1 Find The currents IS 13 SW for the following conditions a t t1 b t t c what will happen to the battery after the switch closes Repeat all considering the circuit of Figure P958b if the diode has an offset voltage of 06 V Analysis For Figure P958 a a At t ti before the switch S1 closes we have V V ISW0 IS 13w0831A S B b At tt1we have 5 13 A 18 096 A 15W IS IB 13i96 A c The battery voltage will drop quickly because of the small resistance in the circuit For Figure P958 b a At t twe have 940 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 V V V B W025 A RSRB b At tt1we have 15 ISW 13 A 130 c The battery will not be drained because of the large reverse resistance of the diode Problem 959 Solution Known quantities The circuit in Figure P959 The diode has Vy 07 V The input voltage is sinusoidal with an amplitude of 5 V Find Determine the average value of the output voltage Analysis The capacitor will charge to 5 V 07 V 43 V and therefore the input sine wave will be shifted up 43 V to produce the output As a result after the cycle the capacitor builds up its stored charge during the third quarter cycle the average value of the output will be 43 V Problem 960 Solution Known quantities The LED circuit in Figure 914 Diode operating point VLED 217 V LEDZO mA V325 V Find Determine the LED power consumption and the power required by the voltage source Analysis The power consumption is PLED VLEDILED 34 mW The power required at the source is P5 VSILED 100 mW Problem 961 Solution Known quantities The LED circuit in Figure 914 Diode operating point VLED 15 V ImD3O mA V325 V Find Determine the LED power consumption and the power required by the voltage source Analysis The power consumption is PLED VLEDILED 45 mW The power required at the source is P5 VSILED 150 mW 941 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 9 Chapter 9 Instructor Notes After a brief introduction to semiconductors materials in Section 91 Section 92 introduces the pn junction and the semiconductor diode equation and iv curves These sections could be bypassed in favor of a more intuitive presentation provided in Section 93 with an explanation of the basic circuit behavior of the diode beginning with largesignal models This explanation is supplemented by the sidebar Make the Connection Hydraulic Check Valves pp 466467 which continues the electrichydraulic analogy introduced in earlier chapters and provides an intuitive explanation of the operation of the semiconductor diodesl The box Focus on Methodology Determining the Conduction State of a Diode p 467 helps the student understand how diode state of conduction can be determined The largesignal circuit model material is probably sufficient for the purposes of most introductory courses 5nstructors who are interested in introducing the subject of smallsignal models in preparation for a the study of transistor smallsignal amplifiers may find the rest of the section covering smallsignal diode models pp 474483 of interest The concept of operating point is introduced with a review of the loadline equation and the definition of the quiescent operating point of a device The box Focus on Methodology Determining the Operating Point ofa Diode p 475 summarizes this material Section 38 can be recalled or introduced for the first time at this point The solution methods used in Examples 95 96 97 emphasize the use of simple circuit models for the diode together with the use of Thevenin equivalent circuits this method is quite general and will reinforce the importance and understanding of the concept of equivalent circuits in analyzing more advanced circuits The use of Device Data Sheets is continued in this chapter with a summary data sheet on a general purpose diode The box Focus on Methodology Using device data sheets pp 476477 is designed to familiarize the student with the basic contents and function of device data sheets The instructor may wish to expand on this introductory presentation by asking students to locate data sheets in the accompanying CDROM or on the web and to identify specific devices and their parameters for use in homework problems Section 94 discusses various diode rectifier circuits Section 95 introduces DC power supplies Zener diode circuits and voltage regulation Section 96 analyzes various signal processing circuits and introduces two application examples in the boxes Focus on Measurements Peak Detector Circuit for Capacitive Displacement Transducer pp 499501 which is tied to two earlier boxeson the capacitive displacement transducer pp 147148 and pp 175177 and Focus on Measurements Diode Thermometer pp 502503 The latter example can be tied to a laboratory experiment2 Finally Section 97 introduces photodiodes and solar cells and includes the box Focus on Measurements OptoIsolators p 506 The homework problems present a graded variety of problems mostly related to the 17 examples and application examples presented in the text Learning Objectives 1 Understand the basic principles underlying the physics of semiconductor devices in general and of the pn junction in particular Become familiar with the diode equation and iv characteristic Sections 1 2 2 Use various circuit models of the semiconductor diode in simple circuits These are divided into two classes large signal model useful to study rectifier circuits and small signal models useful in signal processing applications Section 2 3 Study practical fullwave rectifier circuits and learn to analyze and determine the practical specifications of a rectifier using largesignal diode models Section 3 4 Understand the basic operation of Zener diodes as voltage references and use simple circuit models to analyze elementary voltage regulators Section 4 5 Use the diode models presented in Section 2 to analyze the operation of various practical diode circuits in signal processing applications Section 5 6 Understand the basic principle of operation of photodiodes including solar cells photosensors and lightemitting diodes Section 6 1 With many thanks to Bill Ribbens who first suggested this idea to me some 20 years ago 2 G Rizzoni A Practical Introduction to Electronic Instrumentation 3 d Edition KendallHunt 1998 91

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