1 DOCUMENT RESUME SE 004 647 ED 021 740 24 By-Balabanian Norman; Kirwin Gerald J. DIODE AND DIODE CIRCUITS APROGRAMMED TEXT. Syracuse Univ. N.Y. Dept. of ElectricalEngineering. Spons Agency-Office of Education(DH(W), Washington D.C. Bureau ofResearch Report No- 4 Bureau No- BR-5-0796 Pub Date 64 Contract- OEC- 4-10-102 Note-177p. EDRS Price MF-S0.75 HC-S7.16 *ELECTRONICS *ENGINEERING EDUCATIM*INSTRUCTIONAL Desc_riptors-*COLLEGE SCIENCE. ELECTRICITY, UNDERGRADUATE STLOY MATRIALS PHYSICAL SCIENCES*PROGRAMED INSTRUCTIOK TEXTBOOKS Identifiers-United States Office ofEducation University of Syracuse This programed text on diodeand diode circuits wasdeveloped under contract Office of Education as Number4 in a series of materialsfor with the Uniteci States regular use in anelectrical engineering sequence.It is intended as a supplement to a text and otherinstructional material. (DI-1) U.S. DEPARTMENT OF HEALTH, EDUCATION & OFFICE OF EDUCATION WELFARE s--,67 cit Mo.' 4 THIS DOCUMENT HAS BEEN REPRODUCED EXACTLY AS RECEIVED FROM THE , POSITIONPERSONSTATED OR ORDO ORGANIZATION POLICY.NOT NECESSARILY ORIGINATING REPRESENT IT. OffiCIAL POINTS Of VIEW OR OPINIONS OfFICE Of EDUCATION ?Rog DIODE' AND DIODE CIRCUITS A Programmed Text CD Norman Balabanian and Gerald J. Kirwin by 1-4 El ectri cal Engineering Department Syracuse Uni versi ty COPYRIGHTED"PERMISSION TO REPRODUCE BEENTHIS 6RANTED ciCD BYTO ERIC AND ORGANIZATIONS MATERVA OPERATING L4.1CD Copyright, 1964 UNDERTHEEDUCATION.THE ERIC AGREEMENTS COPYRIGHT SYSTEM FURTHER REQUIRES WITH OWNER." REPRODUCTION OUTSIDE THE U.S. OFFICEPERMISSION OF Of 11011 A_EMEKNEMti-2121Et DIODE AND DIODE CIRCUITS by Un Norman Balabanian andElectrical Gerald J. EngineeringKirwin Department Syracuse University U.S.Contract Office No. of4-10-102 Education OE Copyright, 1964 L.3 Aorer,Aormerr LIiIi L I1=1 L=1.. L'"7-1_ E -1L _I t 1 L F-1 r ft1
. fl Section I DIODES 3. symbolicelement formas depicted in Fig. byla. the An ideal resistor is an Thestraight current line is proportionalcurrent-voltagehypothetical curvenetwori element to the voltage in this which is shown in in Fig. lb. Question: The(b)(a) resistance, Ft, is The slopereciprocal of the of the current-voltagedetermined as: curve. slope of the current-voltage curve. / ., 2 Answer: (b) ofThe (Note:the reciprocal current-voltage of the slopecurve. the abscissa is voltage And the.. ordinate is ,- current.) . Many electrical conductors, such as copper and. aluminum) exhibit a behavior 3 proportionalitysuchthe thattemperature the voltage and currentbetweenof the voltageconductor and currentbe held constant.are proportional. apply, it is But in order usually necessary that the strict that Question: Arentit in of a Ideep amperes coil of copper ice water wire is held atin athe copper bath. A voltage of wire. constant temperature Then the voltage V volts results by immersingis increased in a cur- to (b)(a)2V volts. the newold currentpower Determine the to the new. to the old; following ratios: 11.0010.110.4.1 Answer: (a) 2 .1 (b) as1/4 (Rememberthe square that power varies of the current.) Fig. 2 (b) LJ There are nany important and useful devices in electrical engineering in5 whichthroughcan thebe thedescribedcurrent-voltage origin. as curve being a deviceOne suchrequiring device only is calleda small a does not assume the form of a diode. In simple voltageterms, ato diode produce cur- straight line voltagerentdirection. in is one necessary direction, to producethe the same magnitude (In many instances, the voltage accompanying so-called forward direction. of current in the reverse However,appreciable a much reverse greater cur- renttypical would diode be so in high Fig. as2a. to cause The above behavior is picturedThe schematic graphically symbol by of the Fig. current-voltage 2b curve complete breakdown and destruction will be used to repre- of the device.) for a actualsent andiode actual points diode. in the forward Note that the orientation of the arrowheaddirection. at the center of the symbol for an Question: powerCompare when the i power= -I amperes.dissipated in (Assume I to be some positive number.)the above diode when i = I, with the Answer: Thein powerthe reverse dissipated direction, i = with the current willforwardsame be magnitudemuch direction. greater of current is in This is athan re- when the the issult necessary of the much to sustain greater voltage that the current ,11.0010. neededindirection. the reversefor the current direction over what is in the forward Fig. 3 -1 We have seen that a diode tends to provide an easy path for current through 7 - An direction.itselfideal diodein the is forward an hypothetical direction while tending This mode of operation leads us to the network element having the following three to prevent current in the raverse concept (2 an ideal diode. properties: (b)(a) thereits voltageis no reverse is zero current; when its current is forward (positive); ferentIn order symbol to distinguish shown in Fig. an ideal (c) its current is zero when its voltage 3 will be used. diode from an actual diodel.the slightly is reverse (negative). dif- terms. See(a) if you can write the three properties of an ideal diode in mathematical i (c)(b) 8 Answer: .( c(b)(a) ) vi i > .= 0 0for for i v>. < 0 0 Since we shall be making continual reference to the plot of current versus 9 voltagedesignate"i-v curve"when the discussing forplot this of voltageplot.diodes, againstwe shall current. adopt Likewise the expression the convenient abbreviation, M v-i curve" will serve to Question: Fromthe thei-v idealcurve diodefor an definition ideal diode. in the previous frame, try to construct 1r=1 rm =1 1=3 =1 1=3 concept known as the forward. 11 tionand reverseof the i-vresistance characteristic When considering diodes of a diode. there is a very usefulof Fig. 4. This concept may be This plot is i:Itroduced.indistinguishable by considera- fram for positive thatofvalues vof to aof iresistor. vis only. constant Thus, when v is positive,Let us sorestrictfor oura fixed value of attention to this i-v curve0. also is i. Mbreover, the ratio Question: Whati-vcurve must curve is happen to for anto approach the same form as ideal diode? if the first quadrant the vertical segment position of this i-v of the 12 1: Answer: 0 must approach 900. .. -I oil...w,. 13 of Fig. 4, recall Continuing our that the slope of discussion of the first this line and the quadrant segment resistance are related. of the i-v curve Question: What8 approacheslimiting value 900? of résistaace is associated with this line as, Answer: 0 ohms.
Fig. 4 (repeated) 15 an ideal diode has zero WeIn returnview of to what Fig. you have just done, 4 and focusresistance our attention to forward on the current. third quadrant it should seem natural to say that segment ofis segmentthis seen i-v to of curvebecome the idealas horizontal, e isdiode made i-v totaking curve. on approach zero. the same position as the horizontal In this instance the segment Question: resistanceIn view of theof theabove ideal interpretation, diode when the voltage is negative?what can you say about the Answer:16 It will be infinite. 17 andfor non-zero. which we will find the forward and. reverse resistancesLater in this to bediscussion finite, positivewe shall be talking about other than ideal diodes conducting,diode is equivalent the switch to is a onswitch and thethat voltage is either is zero.onAnother or off. description of an ideal diode is.often usefUlly employed. When the diode 14 An ideal Complete the following sentence: When the diode is non-conducting, 18 Answer: voltagethe(or switch the is currentnegativeis off andis (or zero.)the zero). diode .211 Fig. 5 ORM.ii 19 equivalent to a short When the diode current circuit. is positive, the diode voltage is zero, making it L I Question: voltage.byConsider its equivalent the circuit and in calculate the diodeFig. current 5. Redraw it with the and the diode diode replaced HI AmarammanowiewaseliftlissWINSIIIIIIIMINIgeormor T 10 volts T 10 volts VW\ 2g2 1lilt i = 10 2 = 5 amps Fig. 6 I I 21 theis equivalent&Lode by its to equivalentan open When the diode voltage circuit. andie negative, calculate the the Redraw the circuit in Fig. diode currentdiode is currentzero and the 6 having replaced and the diodediode voltage. 7.17 1,4,1Air VLept . 22 25 2 22 111111M ci a .111111111INS 10 volts 11111111=1111 10 volts v i= = -100 amps volts . Fig. 7 23 rl-1 diode is in its open or closed In some networks it is possible to tell rather simplycondition. This will often be the case whether the ideal when [] athere is only one source,network whose of resistors and ideal diodes.Consider'the arrangement of elements as voltage or current is constant, depicted in Fig. 7. connected to What is the 1 only possible direction for actual current in the 10 volt source? Answer: from(Upward c to throughd through the the source.) source. 25 d. 0 volts a Fig. 7 ( repeated) 25 impossible for With reference to the current to go the 25 ohm resistor from a to b through in the network this resistor? of Fig. 7, why is it El =1 26 1 Answer: Sinceacrossin parallel, the the diode 25 current ohmand resistorresistor from a wouldareto b 25 2 allowedpositivecause the by quantity. diodethe diode. voltage to be a This is not 10 volts + v b Fig. 1 (repeated) Then the diode will have a reverse Suppose that current in the 25 ohm voltage and hence zero current. resistor is in the direction from b to a. Question: Rememberingvoltage source that isthe upward,'assuming only possible direction the current in the 25 ohm, for current in the Li resistorWhatopen conclusioncircuit), to be from howare b can weto faceda with? the Kcl be satisfied at the junction (which insures that the diode is an a? 28 El Answer: theUnder Kcl the cannot conditions be satisfied postulated, at 25thatjunction obit current resistor a. cannot in anyexist direction. in the We can only conclude
1.,I.MYMM,OPONiONMINN.. 29 Fl sideration it must be equivalent to Since the 25 ohm resistor can carry no current in the Circuit under con- by their respective equivalents) Redraw the circuit with both the diode and the 25 and determine the diode ohm resistor replacedcurrent and voltage. Answer:30 an open circuit. IWO .0E101/111M 10 v 5 SI v . = 1010 - 5i = 05 volts = 2 amps. (Answer all of the questions below, before checking your answers.) 31 dissipation of power. Another feature of the ideal diode which deserves special attention is the 2)1) TheInequals powerthe casethe absorbed produceof the by idealof any electricaldiode, current element in having and the forward direction two terminals Lj 3) Similarly,poweris always must accompaniedbewhen the diode by zerovoltage voltage, is negative, which for this case. tells us that the the current is 4) idealWrite diodea general dissipates statement powers. telling the conditions under which the Answer:32 1)2) voltagezero and current 4)3) sipatesIdealzero diode power. never dis-
.. 33 diode is.easily_remembered in A brief and. handy rule that accurately states the following form: the behavior of the ideal "Forwardvoltage current means means zero cutrent." voltage while p. 34 Answer:. meansvoltageForward zero and.current current. reverse means voltage zero Fig. 8 INN MONINIIII Sometimes a voltage source is connected directly in series with an ideal 35 behaviordiode as indicatedconsidered This combination can in Fig. in terms of the be viewed froma. voltage and current the terminals a and associated with these b, and the network interminals. this arrangement Remember that there The diode is often because with no external is only one current said to be back voltageshared the by diodethe twobiased by the Vo voltageelements isvoltlioattery andnegative. that thethatthree voltage, were fundamental written vab, propertiesevalsfor the the ideal sumof this ofdiode two the battery and alone on page terminal element in diode voltages. analogy with those Write the (b)(a)(c) 7. 11/111111111111=1011011/111111111111111111MMI1111.1.... Answer:36 (a) i must always be positive (or zero). (b)(c) iv is zero whenab v = E o whenever i is positive. ab is less thanV o . Kvlthe(This diodethe is diode truevoltage becausevoltage, is negative. i v= =0 v when7ab -V By o ). the circuit and construct the Suppose that an ideal diode is back biased with a 10i-v plot of this arrangement. volt battery. Draw 3 8 10 voI 1 t s 1 I MM. i v 1 v 39 network Of Fig. 9. We continue to examine the network response of the ideal diode using the Here a diode is back biased by the 10v source. The current, fora i,function uswill also be of to:determinea thefunction voltage, of the thevl, diode voltageis tbe voltage, requiredv v, result.as a function of vl. 1 and a graphical plot of the current as It will prove instructive You(b)(a) can get things started:* calculating: athe literal 6urrent, expression i, when forv the diode voltage, vl in terms of i and v1. 1 = 0 10 volts Answer:. (a) (b) v i= = 0 when v vl - 2i - 10 1 = 0 . v _ I I s __L------"'IMMII v 1 I Fig. 9 (repeated) 1 i To determine whether the diode is conducting or not, we need to examine its off),voltage.let's we use can this dispose expression. of half our job by assuming the.diode to be non-conducting. You have just calculated the diode voltage in quite general terms so Since the diode has only two possible states (on or nessi = of0. our assumption of no diode current? Examine the expression for diodeUnder voltage, this condition)imposing the what assumed range restrictionof values of vi will insure the correct- Answer : v = v 1 - 10 < 0 2 SI Fi g . 9 ( repeated) 43 conducting. We have determined that when v It follows that when v 1 exceedsis less 10than volts 10 voltsthe diode the diodewill isconduct. non- Draw the equivaJent circuit that . applies to this situation for vl > 10 alland valuescalculate of vl.i as a function v IIMINNIMO 1 Finally, plot i as a function of v 1 for 5 10 15 20 i =. 02 for = v 1 2-10 2 vlfor v < 10 1 > 10 Fig. 10 45 as a function of v As a finale to this little problem, let us now consider 1. As derived earlier, v = v 1 - 10 - 21 is the generalthe diode voltage orknowequation v that forthe.diode diode voltage.Voltage is zero. 1 - 10. This result is indicated graphically in Fig. 10. However, when the diode conducts, (i Thus the diode voltage is either zero 0), we samplecareful rule statement for ideal describing diodes (pagethe behavior 33) of this Usingcircuit Fig. and 10 showing and the that plot the of i versus vl, as developed earlier, write a not violated. 1S flammo.' Answer:46 When v 1 > 10 the diode conducts Whenrequi:eingzero. v, < 10that the its diode voltage voltage be This is the case in Fig. is10. negativeuhichi versus is requiring the vi. case zero.currentin the plot of some practical utility in which again there is aOurThe next voltage example, vl varies shown with in Fig.time. 11, will be concerned with a circuit of Our interest is in determining the vari- back biased diode. variationofation v of the of outputv voltage v2 with time.2 on v 1 ; then when the variation of v (The output tarminals, marked a and b in the figure, are to 1 withLet ustime first is givendetermine we can the find dependence the voltageremain openVo. circuited.) Write an expression for v 2* Our first objective is to find a plot v2 against vl.2 in terms of the diode voltage v and the battery -1WMM Ma go pa .1 41, Li Answer:48 = + v 0 v2 Vo ElEl Fig. 11 (repeated) r-Ei 01k In view of the permissible values of diode voltage, answer the following: 1)2)3) v WhatisIn either whatis maximum state does the maximumvalue of v 2 ? or value of v 2 place the diode? NA* Answer:50 (1) ... either negative or (2)(3) Vo conducting zero. Fig. 11 repeated) 51 valueWhen the of diodev. But is clearly,on, Thus, when the diOde is Off the output the output voltage some value of v remainsI at Vo no voltagewill cause v2 the will'he less than Vo. matterdiode what to changethe state,in the to diagram and When the diode is switch off. find an expression off replace it by an for v2. appropriate equivalent circuit 52
(since thereno = current) is
°"4 Thus, the output voltage v2 is equal to the input voltage vl when the diode 53 is off. To summarize: v 2 = V o (a constant) when the diode is on. Furthermore, this is The only problem remaining is to determine for what value v-2 vi (a variable)the maximum when the value diode for is v2. off. of v the diode switches from off to on. 1) andState complete.the the value following:of v You should do the following: 1 at which the diode switches, 1 2)3) V2v2 = Vvi when v o when v 1 Answer:54 1) Diode switches when v 1 = V o . 2)3) v2v2 = =v V o1 whenwhen v v1 1 isis less more than than V . V o. . _ appropriate4; It is noN: possible to plot v2 against v Make this plot and label it
LJ
against v we see that the output voltage v is limited 57 calledto those a limiter.values of v From the plot of v In the present1 case only the lesspositive than V2 o 1 . The circuit undei discussion is accordingly values of the output 2 ,11, voltage are limited. ItSuppose is desired vl has to the plot time the variation time variation of v2for Vo . 5 volts.shown in Fig..12. From the v 2 - v I plot we see that v 2 will duplicate v 1 whenever v 1 is less than 5 volts. When v 1 is greater . than 51 v 2 will remain at 5. Hence the plot of v 2 will take theshow form the of plot Fig. of 13. vl dotted. Plot v 2 against time for the same v 1 , assuming V o = -5. On the same diagram 1. 411 / s _ \ / / / / \ v.. / / OT+ V / \ / / gc 59 In Fig. 14 is shown a slightly modified. version of the previous circuit. Li that of the former circuit, Fig. Describe in your own words how the behavior11. of this circuit compares with Answer: The behavior is very similar. valuestheHowever, negative that this are excursions circuitno more willnegative of v2limit to ofeffectthan v2. -V upon the positive values o volts. There will be no Fig. 12 (repeated) 61 of combining the two most recent circuits, that we have negativestudied, valuesinto an There is a way 0: a voltage.arrangemént that circuit SQ that simultaneously limits the positive amplitude is limited both the positive to and 8 ifvolts the andsource the Try to design a is that of Fig. negative amplitude 3,2. to -3 volts. Sketch the limited voltage A 1 El 2 1 a - ElD Fig. 15 ii 00 Increasing the complexity somewhat, let us moveon to the network of Fig. 15. 63 withThe current,the source i, voltagewill vary is withthe desiredthe source You might at first be tempted to throw up relationship.your hands at the seeming voltage, vl, and. a plot of this difficulty current ofleads this to'problem, the complete but do solution.not despair. Our method of attacic*will be comprised of three steps. You will see that a series of simple At first we shall steps weassumethis shall isthe examinewhen diode the thisto diode be open is actually open. expression, noting when the diode voltage and calculate the diode voltage Finally, the diode is assumed to for this case. is negative, for be Then closedcase, andvl =i 0. As a usefUl check on some of our 1 versus vl determined(It may provefor this helpful case. to redraw the circuit into later results, determine i1 in the simpler form.) limiting 64 4 s2 Fig. 15 (repeated) i 1 - 4 +5 1 - 7 amps. a RemeMber that the ideal diode is a two state device, equivalent t9 either 65 a casesshort Inspectionmust be considered.circuit or of the network an open cirCuit. indicates that TO solve this with the diode open problem Completely circuiteit, both: theand simplest calculate il.* Redraw the situation occurs. circuit into equivalent fört for the case of the diode open
67 opento be(orclios0010 open (this is the simpler The basic question, 48 Always, We can begin tol'ansWer this Vestion case) and determining the diode is updwr what ;944itions by assuming the diode will the diode be voltage under .. thisclearlydrawing assumption, labeled?)the approreiate circuit T444 You shou34 do now. diagram with the voltage across (zyX eug4est that.you begin by each e)ement L 1 I717 E.7_ ETA :=1
.1110=111=.11
CO k0 69 perhaps it is well to Although it may appear summarize what we have that we have not accomplishedlearned at this &great point: deal yet, We know that, when (a)(b) i the diode voltage is v = 1 = 1 amp. the diode is open: 4 - v 1 volts. value for v It follows that when MOeover, when vl .< 1 separating conduction 4 4.the- v diode conducts and 1 is negative, the from non-conduction, diode is open. its voltage is zero. is therefore v The'critical 1 = 4 volts.voltage as a function There is now sufficient of vl. informationThat to constructis your present a e' task. plot of the diode r .-2 -4 -6 Fig. 16 0n[ 71 for v 1 To complete< the4. In this case the diode solution of this problem we is a short circuit and the must now consider the network simplifies situation ,Among totherestrictions that variety of Fig. of andways solve fbr the The desired current, ill 16. of determining i1 we iscurrent one of ithe currents in a Fbr this purpose, you are choose to impose KV1 and two loop netwqrk. asked to: Kcl (a)(b) ExpressWrite thei3 inKV1 terms equations of i 1°that result from summingand i2. voltages around theallequations two other meshes. termsso that on the Simplify by eliminating and i2 :terms appear on:theright side. 2 and rearranging these left side and 171 1.-=ETA LA L....J = 1
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H
H I
1 ErN
It n 0r-lal N'Hal I I H H v-1 .r4 CU K1
cd *sm...... 7 3 informationequations for to i plotin terms i against of vl. vl for all The final step in the solution of When this is accomplished there will this problem is the valuessolving of of these be enough verifyresult the with correctness your previous of your result answer that by i setting vl Solve= the previous equations for the 1 desired i 12as a function7 of vl. when v 1 0, =and 0. comparing the .Also Answer: i _ 15 - 2vi 7 for v < 4 1 Whenthe vprevious result. 1 . 0, i = 1.2. 7 which eaecks .fta. 75 All of the required information is at hand for the construction 15-2v1 of a plot vof i 1 1 4. against v Construct the desired 1 . Thus we know that i plot. 1 = 1 for v 1 > 4 and i 1 7 1..."1- L-.-i I7-1 =I L___1 I-1
t
V.
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a,
AINIMINIONNIMMOMPOR Sumraary 77 Fill1. in all of The ideal diode these statements is a switch-like before looking at element having answers. exactly modes of 2. Asbehavior. a switch it is either fully is zero, and or completely when completely off the current 4.3. WhenisThere fully is onnever any the current in the direction nor any 5. Thein theforward forward and and direction.reverse resistances of an ideal diode are respectively, 79 Whetherstates thedepends ideal upon diode a)conditions is actually in If the circuit causes the diode the circuit to which init oneis connected.or the other voltageof the twoto be negative, the current pOssible Thusl b) willIf thebe zerocircuit and causesthe diode the isdiode equivalent current and the diode is equivalent to to be positive,to the 7. A thefruitfUl following approach three to steps:a) the analysis Assume the diode to be open and of a network containing calculatea diode the desired quantity. involves b)c) CalculateDetermine the the desired conditions quantity under when which the the diode isabove closed. assumption is valid by MI` Ahswers:1.a0 two modes of behavior. 4.3.2. voltageonin or the completely isreverse zero directionoff. nor any voltage in the forward direction. current is zero. 6.5. b)a)zero and infinite. ... the voltagecurrent will be zero and the diode is equivalent to aan short open circuit. 7. b) examining the diode voltage tendsthisunder voltage tothe be above positive is negative,open thecircuit diode the Conditions. diodewill becomeis indeed a short an open circuit. circuit. If the voltage When Section II DIODES AND DIODE CIRCUITS introducedis a first-order as a model In the preceding section a approximation of of a physical hypothetical device,the behavior of an device. This model has a actualthe ideal diode. diode, was behavior which It ac- countsreversecurrent. for the current impossible.In fact, the major effects of ideal diode makes easy forward forward current current and difficult flow too easy and reverse reverse The ideal diode resistance. calls for forward resistance and LiI i Answer:82 zero forward resistance - infinite reverse resistance -
I! Figure+ 17v - o 83 coincidentlies close withto them. Pt typical i-v curve the positive Furthermore) it of a physical diodecurrent and negative voltage is not a straight is shown in Fig. 17. axes, line but is but it is not The curve Answer:84 curved (nonlinear) / A ,.1 DPP Figure 18 1 1 INNINI11 The notion of forward and, reverse resistance was introduced as the 85 respectively.inverseeven when slope we of the confine ourselves toSince the slope of i-v curves, evaluated theeither i-v curvethe forward of Fig. or reverse in the first and, third 17 is not constant, region, then quadrants, a tunateuniquecurve becauseis composed many forward. or reverse of straight-line simple and useful resistance does not segnents. techniques are available For this reason, exist. This is most unfor- whenstraight the i-v lines can linessometimes marked be usedA and to A device, having B in Fig. approximate straightthe i-v curvesegments as for 18. its i-v curve as indicated by the dashed indicated by the linesresistancesistor A and B in Fig. State the values ofRr as properties 18, has a behavior in each of the linesthe A forwardand resistance B. region like that of a re- Rf and the reverse Answer:86 Forward resistance is the line 11, the reciprocal Reverse resistanceofslope the slopeof theis ofthe reciprocal line B. of the r111.10. -414.1... , ..] 87 NommerasIM1111.11 two straight The dashed-line lines, one for each characteristic of Fig. portion or piece 18 is not a single, but of the entire voltage rather range.approximation Of the The straight This approximation lines are drawn in i-v curve. is, therefore, such a way that called the piecewise depends they come "closest" linear orto current bethe applicable. actual over curve. which the above What it means to be closest A upon the range approximation is of voltage 88Answer: piecewise linear approximation M. 11. Figure 19 E Consider, for example, the forward portion of a diode i-v curve drawn 89 range,1 involt, Fig. 0 andto19. 1we volt. are interested in a Suppose the voltages we expect across The line iabeled. A would be suitable. piecewise linear approximation over this the diode do not exceed On the other markedhand, ifB,A wouldtheyields region lead a much toof largeinterestbetter errors extended near over approximation over the range from 0 to 2 the 2-volt point. the range oA ;to volts,The second the line line, volts. isline called. segments, a piecewise one for linearthe forward. model A hypothetical element having an region,diode, and.or simply one for a modelthe reverse diode,i-v region) forcurve in the form of two straight- brevity.properties. Draw a typical i-v curve for a model diode and. label its pertinent Answer:90 LI IlLslope equals Rf 1.1 f -- forward resistance Slope equals Rr1 R r -- reverse resistance Figure 20 114 91 thecurrent, scale axis.of the Figure 20 shows the Similarly, the negativenegative current axis is exact i-v curve of a voltage axis has a different from that of the diode, drawn to scale. different scale from positiveVote that thethe positive first voltage A straight-line qUadrant behavior of the exact axis.segment, shown in dotted diode curve. form, is chosen to The slope of this line approximate resistanceis 2.2 of 0.45 ohms. Using a clear plasticamps./volts. Therefore, the piecewise straightedge, construct your linear diode has a forward best choice of straight lineyouristic. to choice approximate Determine the reverse of straight line. the entire third quadrant resistance of the model segment of the i-v diode corresponding to character- 92 Answer: matingvaryThe answer withline. to this questions will the choice of approxi- One choice yields a -4 15%Yourreverse of answer this resistance one.should beof within230 10- lc. ohms. -- Figure 21 - So far, we have the i-v characteristic of a piecewise linear model diode. 93 Theducesobjective. development to a resistance, of an It should 'be clear that equivalent circuitequal to for the the forward what is required is a resistance of the model diode, model diode is our next network which re- whenresistanceequivalent the voltage of circuit the is diode.positive. must reduce Likewise,to another when resistance the voltage -1.s equal to the reverse negative, the thediode definitions shown in Fig. of 21. To start, consider and vi.) the parallelConstruct combination the i1-v1 curve of a resistor and an ofideal this network. (Note / 1 Answer:911- I. - D0 lope evals El1 v 0 EWNW.. \ 111. I 95 iiI of forward Let us and_ reversedescribe the i 1-v1 resistances. behavior of the network of Fig. 21 in terms L_ b)a) TheThe forward reverse resistanceresistance equals equals ohms. Li Irr"774-- 96 Answer: a)b) TheTheequals forward reverse zero resistance ohms.resistance equals R 14111,011 1 ohms. . Figure 22 rammi.1111.1.-Nilma.... 44 To achieve a non-zero forward resistance, let us modify the network by 97 whenaddingHowever, the a diodesecond when the is closed resistancethe in diode is open, R1 network reducesseries, to asa indicated and R2 become resistanceseries connected. inof Fig.R 22. We see that2 ohms. Hence, this network is Complete the a) The ideal diode equivalent to a sortfollowing: is closed when v of two-state 1 is resistor. b)c)d) TheThe ideal reverseforward diode resistance ofOf the is open when vl'is networknetwork equals equals Answer:98 a) v The> O.ideal diode is closed when b) vThe < ideal0. diode is open when c) RThe forward2 resistance equalsohms. RThe reverse1 resistance+R equals 2 ohms. II 99 nowequivalent be clear. resistance, R2. The behavior of the network For one polarity of voltage, v However, for vl <: 0 the network is of Fig. 22 as a "two-state resistor" 1 > 0, the network reduces to an equivalent to should theindicating larger resistance, the appropriate R1iR2 Construct the i-v characteristic curve slopes. ohms. for the network of Fig. 22, 100Answer 7, slope equals R01 slope equals R 1 +R21 101 [1 propershould now convince you that Comparison of this il-vichoice of B1 and :R2. these two curvescurve canwith be that made of the model diode, page 90 identical by the -7 Fig.R 22 is fequivalent and the reverse resistanceDetermine expressions for RI and R2 to the model diode. R r of the model diode so that in terms of -i,Le forward resistance the network of 102 i Answer : R2R1 = ,-,Rr Rf -Rf 1 Suppose a certain diode has its i-v curve approximated by a model diode 103 whoseresistor.Draw forward an equivalent and reverse network resistances forare the2 ohms diode and and 100,000 specify ohms, the respectively. resistance of each Answer:104 99,9982 7r-i Remark: v2 theance ratioare typical of reverse practical to forward values. resistance has beenThese values of forward and reverse resist- If anything, v I OMNI. vo Li thereunderstated. is a large difference between the forward and We learn from this example that, when MO mosttoreverse equalalways resistances,Rr the without case inappreciable the practical resistor error.situations. R1 may be taken This is al- Figure 23 105 assumelet us thatreturn an toactual the limiterdiode is circuit used. of Fig. 9. To illustrate the use of the model diode in making circuit calculations, The arrangement is indicated in Fig. 23. This time, however, we shall Oncevarying again, input our voltageinterest vl. will center on the variationThe actualof v2 indiode response behavior to will be represented by a model diode and a circuitthecalculations network ofmade Fig. using 23) thereplacing equivalent the actualcircuit diode of the by modelits model diode. diode equivalent Redraw Answer1o6 :
o OM 107 will be in one of As in the original two states: limiter, the ideal open or closed. diode in this equivalent network We must determine 2)1) thethe relationship relationship three things: between vl and v2 when the diode is open;closed; and First, consider the 3) thestate. critical value diode to be open. of vl, the value Draw the equivalent at which the diode circuit and changes ternscalculate of v both the 1 V 0 , and the three ideal diode voltage, v, resistors. and the output voltage, v21 in Li Answer108 V - R+R 1.+R -L 2 ' 1 0 t"- v2 - R 1 +R2 2 vl R-FR R1 .-1.R 2 VO 109 Remember that these expressions apply when the ideal diode is open, which meansat which that thethe diodediode changesvoltage state,v is equals . Hence, the critical value of v 1' Answer:110 v is'negative v 1 equals V .11111 0 111 withoccurs the whendiode v closed and write an expression relatingIt v2remains to examine the network when the ideal 1 >. V 0 Looking again at Fig. 23, draw an equivalent network diode is closed, which to 1/1. I , / , Answer112 : 2 ...... v2 - R+R R2 2 vl + R+RR -V 02 t 113 have now obtained the following results: To summarize, we v 2 R R1+R2 R 1 2 v 1 R R1+R2 V 0 when v : V 1 0 v 2 R+R R2 2 v I R+R 2 V when v0 1 > V 0to the forward and reverse resistancesresistance ofR the Remembering that Ri arid f' and the reverse diode, rewrite R2 are related resistance, R these expressions r in terms of the forward ! . R+R 0 V f R+R R + 1 v f : f R R+R V 0R+R V > = 1 v2 IT when and, ' 0 V -r R + 1 r : r0 RV < 1 v When - ;: Answer 114 115 Let us rearrange v2 1+-- 1 R v 1 these equations+ 1 REr 0' V when v somewhat,as follows:1 -< 0V V 2 t :: : "' 1 R V 1 + 1 R , 117 O. 1 when v > V 1 0 . a straight-line Note the grert segmentRf when v2 similarity between 1-+ R' is plotted against these expressions, v1. Observe how the respec-each of which defines tive slopes and a)Suppose As compared we use as intercepts differ. with the slope typical values: for v1 < Rr = 100,000a, Vo (diode open), the slope Rf = 2, and R = 2,000Q.for As interceptcompared for v1 .> Vo (diode closed) is about with the intercept Vo (diode closed) is for v1 < Voabout times smaller.(diode open), the times larger. 116 .111.... Answer: b)a) aboutabout 1,000 50 times times larger smaller 04MM , r 117 Labelplot of the v slopes and With these typical 2 against v 1 interceptsusing and the critical literal valuevalues for orders of expressions which are repeated magnitude in mind, construct a of 1/1. below. [I v 1+--R- v 1 + R- Vo, when 1 R va. Vo v2 1+-- 1 R v 1 + 1 RRf 0 V 5 when v > V 1 0 Answer slope equals R+R R f f zero intercept Li R+R RV 0f ( intercept equals v slope equalsR +R R r R+R r 0 Figure 25 119 This is the output-input curve of the limiter when a model diode is used, beinstead diodecompared ofinstead an with ideal of a the output-input curvemodel diodediode, is used. to approximate of a limiterThis curvethe is actual diode. circuit in which anreproduced in Fig. This curve is to ideal 25. ofthe Fig.values 25 for R Compare the ideal f and. rlimiter curve of that will change Fig. 25 with that the curve of Fig. of Fig. 24, and 24 into the curve give Rf r J 1 1 L_I I 1 = i 1 L_I L1 L:1 e I
1
..
1 121 ideal as possible. It is, of course, desirable that the limiter operate as close to the whenclosed. the diode is open Consider each of the two From Fig. 24 the values are: andehe (extrapolated) 0Y1 regionsiftFig. 24A focus attention on crndintercept when the diode is the slope b)a) extrapolatedslope when diode intercept is open: when diode is closed: fa ad, 120 ANOIr. Lr_ j Answer slope: R+Iir Rr intercept: R+Rf R '0 ft
11. , 123 - olated intercept TheIt desired follows that whenvalues the for diode the this condition slopeis closed when are,the requires that respectively,diode unity is open and both: the extrap-and Vo. a) - R+Rr Rr equal . and. b) R+R R equal . 121+ Answer b)a) unity.unity,
1 1111.1110.1100111.1111111610.101MILOY. ~ ~ , _,zsaraitgaa",' 125 limiter, it is necessary To make the non-ideal that both expressionslimiter behave as closely as Rr/(Rr+R) and RAR+Rf) be as possible to the ideal nearlyare theequal restrictions to unity as Considering that Rr and placed upon the resistor possible. Ri are fixed by the R by the nature of the diode, what above conditions? Answer:126 R +R r 3.. 1 requires R << R r R+R±. 1 requires R >> Rf oo. LP 127 aresecond incompatible, requires Rbut to becausebe a large of thenumber. wide range The first of these conditions requires R to be a small number, while the Taken to the limit) these conditions between R and Rr, a suitable value2are <: 2 of Rand R can100,000 be found. ohms, respectively, tht: 100,000 ohms. For example, if the forward and reverse resistances above conditions require that f ofcondition the expressions and solve approachesunity for the resultant to the R samein terms degree.A ofgood R compromise is obtained by requirlmg that R be chosen so that each f andImpose R this comprondse r -177. I L _I L = t___.] L .7. -1. 129 and.the reversevalue of R In order to get a resistances of that will satisfy feeling for the diode to be the compromise orders of magnitude, 2 and 100,000 condition. assume theohms. Calculate forward
.imiaMYWMMOVOIEIMMIUMIAMIIONOWN. 130 Answer: R = 2 x 10 !:: 447 ohms. 100,0002 ohms andohms. is This is certainly- nowhere near as large as much greater than 131 Calculations The linearized made with model diode this model has been will not introduced as a yield the exact answers; sort of compromise, however,Our we gainmaycomputational in expect theaccepting this work to secure results to model is a be answers.close considerable estiAates ofOur loss reduction inis, of course, exact behazior. the amount a reduction of in notthe callaccuracy Because you for any response of the answers.have been to thisworking so very frame. hard) we will give you a rest and 133 Complete all of the following before Summary for Section 11checking your answers: 1) realisticIn devisedorder to thanwhich represent actual accounts forthe the diode behavior small voltage model, another with a modelin the scheme has been that is more direction 2) andweThis thedesignate better small model is calledmore briefly as in the reverse the the model direction.diode. diode, which 3) Thei-vThe main curve model reason is diode hasfor usingan equivalent linear segments circuit consisting to approximate of two the diode 4)5) andThe tionsone equivalent in the same circuit for the general way as model diode is before. Each of the used in circuit possiblecalcula- bethestates determined. circuit of the variables causingmodel diode must the model bediode considered. to The critical values of must 134 Answer: idealcurrentforward diode 9 2)3) piecewisethe great linear ease ofmodel calculation affordeA. by such straight-line segments. r- 4)5) two ...resistors and one totwo change possible state ...... ideal diode. 1, DIODES AND DIODE CIRCUITS 135 Sectionexact solutionIII of a We continue our diode by the idealized network containingdiscussion a with a considerationand oversimplifiedof somediode. Up until now, we models, e.g., the techniques for the have repre- models,idealsented. diode onlyan actual someand of the piecewise-linear the features of the the ideal diode, diode. actual diode are In the case takesof each into account accurately repre- onlyof these the twoswitch- likesented.further character refinement of the and The simplest model, actual diode.includes the additional These models are very The piecewise-linear effects of small forward useful and are often diode constitutes a used in voltage andapwoximate.sentationthe small analysis reverse of anof actualdiode current. circuit60diode and thus answers However, neither one they yiAad are provides the exact repre- necessarily only ofcircuit analysis consisting are Because the actual inapplicable. of constant resistors diode i-v curve is Determination of the exact and ideal diodes. not linear, there solution of this The previousis nomethods equivalent T.] 136non-linearto this end problem that our requires a new attackefforts will be directed in this and it is section.elementto of a network compute theWhen current, you complete containing resistors,this section, you voltage, or power in any will be able sources, mation.andgraphical a single and diode. they Also included will yield answersThe techniques are will be a consideration without approxi- essentially of ofmowerelements this dissipation dissipation in the network. within the diode andby proper choice of the other the control (a) Figure 26 (b) Lin 137 non-linearelement, let diode us whose To get some idea consider the simple of the problemi-v curve is given arrangementpresented of-Fig. by the presence in Fig. 26(b). 26(a), containing a of a non-linear li valuesexplicitly of Vo inand terms Ro. Assume the diode or only ihe Cani-v youcurve write is givena known quantities? single equation that in full detail, It so, write the along with the gives the current, equation now. numerical i Answer:138 Sincethe diodethere i-vis nocurve simple in way terms of an equation, of expressing thatquantities.it isexpresses not possible i to explicitly in terms of known write a single equation is the graphical representation Refer again to the circuit of Fig. of one relation between v 26(a). The i-v curve of Fig. and i, the diode 13926(h) 010 current and. voltage. These quantities are also under the influence of the voltage source,involving the 'diode NOM. Vo,current and the and. resistor, voltage that Ro. reflects the influenceCan ofyou write en equation Vo and Ro? If so, do it. rw, 0111 140Answer: (This equationforms mayand takestill several be correct.) different
4111.1% 141 rentunknowns, to satisfy namely v Thus, the circuit the equation, v . and i, and therefore connected to the diode V0-11100 is not enough, However, this equation requires the diode by itself, to voltage and cur-contains two determine v tionsand i can be Clearly a second uniquely. obtained. independent relationWhat is this second is requiredrelation? before the unique solu- Answer: The i-v curve of the diode, Fig. 26. 911.1.0. pa.1
IOWA/ ci rlationships simultaneously. Hence, v and i) the diode current and One is the i-v curve of the diode. voltage, are required to satisfy two. The second 1.43 diodeis howthe i-v toequation, curvefind thewere v solution= known,V0-iR0, perhapssatisfying which an the external circuit requires. bothalgebraic relations. procedure could be used. If the equation of the The problem How- V plicatedever,convert this forthe equation thisi-v curvepurpose. is of the diode to unknown and, in any event) would probably Therefore, the only alternative, since we algebraic form, is to convert the relation be too com- cannot plotv = Vwith slope and intercept values. 0Plot the curve-iR of i vs. v resulting 0 to graphical form. from the above equation. Labei the Answer : *maw,
=1.1.,.11 source and the resistor. Thus, it is clear that this line is FUrthermorp, the line does not have any completely specified by the voltage direct practicalasdependence the load applications line.upon the diode. the resistor, Ro, The name "load line" derives from the For ease of reference, we shall refer represents the part of the circuit fact that in many to this line to1,000 which <: theR < useful2,000 ohms.power is fed, i.e., Suppose that Ro is permitted to range over Similarly, suppose the source voltage is the circuit load. the values defined. by confined in accordanceingfirst these 9uadrant withconditions. theof theexpression i-v curve 10 which<:Vo
Irsmarmamillml Answer:154 T10 volts 162 MA0,...... 0 a 1 ANN/5/62 o0 ba ± T7 Ll volts 0 b d. Figure 31 (b) 155 in Fig. TheFirst, network let ofut Fig. 31(b). calculate the values 31(a) contains a single of i1 and. v, diode whose i-v curve the diode current andis given voltage.isdiode convenient to a Thevenin to This can be done do this in two steps. equivalent (i.e., a by reducing the.networkConsiderseries that isconnection theconnected terminals c of Ro and and d; ignor- Vo). to the It Theveniningequivalent the entireequivalent. network. branch containing the Calculate the numerical diode, reduce the values of Vo, and Ro, remainder to a simple in the j 156 Answer : 152 30 volts a. Figure 32 157 network, which is now The branch containing represented "by the the diode may now Thevenin equivalent. be returned to the The result restis of the F indicatedequivalent in connectionFig. 32. of a View the network from the terminals resistor and a voltage source. a-1) and redUce it to a.simple series- 158Answer : r r easily by drawing 159 the loadvalues line of on The current i i 1 theand_ i-v v curve1 and -the voltage v for the diode, ean now be determinedFig. 31(b). EO this and state (repeated) 31(a) Figure 201'2 ,...... 12 ' 4.1.....I. v I ve...... ------2------0.3 = ..L. in Answer: 160 i Refer again to the network of Fig. 31(a). Suppose that the current i2 miningofis i required. the effect of a 2 does depend upon the diode It is clear that, while non-linear element upon the and its i-v curve. i2 is not the diode current somewhere elseThis in is a case of deter- current, the value ofthe inetwork. 2 Remember thatusing i whatever approach you 1 and v are now known quantities. like. Calculate the value Answer:162 Two approaches are presented belowiayield thethe valueform of two equations which . 1.0 amperes b)a) 1202012 = 60(i1+i2) + 2012 = 10 + + 10i1 . the result is: When these equations are solved, i 2 = 1.275 amp. 6 Figure 33 (b) volts r" 4 fl 163 can be constructed Somtimes the i directly1 on the i-v curve-v 1 curve for the network of Fig. of the diode in the 33(a) is required. following manner. It loadSupposecurrent line Ro isof equals slopeindicated -5 .7, ohms. in2 Fig. 33(h) by the intersecting the voltage axisTo find i corresponding to vi = 1 point a. Since this current occursat v = 1. volt, draw the The resultant whenuntilTo determinev it intersects the 1 equals 1) it can be current for v the v = 4 point on plotted directly above1 = 4) the load line is moved the voltage axis. the point v = 1, point The intersection b isparallel to itself at. procedurethen translated will yieldto a Construct the i 1 positionthe-v entire directly locus aboveof1 v = curve for R 0 . 10 ohms. 4) bt. Repetition of this Answer161i- : the real diode,165 namely,of power the in an up until now dissipation of electrical device a very important and power, has results in a not beenpractical temperatureconsidered. feature of rise. The dissipation Because such certain limits,deviceseach device.it operate is necessaryWhen the properly only actual powerto provide a when the dissipationmaximum does temperature does loower-dissipation not exceed not exceed rating forthis maximum, theby device the diode. can Return to the be expected network of Fig. to operate 31(a) and calculate in the proper way. the power dipsipated
111.01111110111MONOM Answer166 p = A 1 v P = i(0.3)(12)1 v = (0.35)(12) = watts 4.2 0 Figure 34 E] Ti 167 andi-v current, curve of the Since instantaneous power limiting this power is diode over which dissipation egals the operation may takeequivalent to restricting place. product of the The naturethe of rangethis of the voltage a restrictionresulting fromis conveniently the power Using the diode plot restriction,of i.e.,Fig. 34,P constructdisplayed by plotting ma directly= oni v.this plotthe relation between the i and. v maymaximumcase. be power dissipation usedWhat withouttype of dangercurve iscurve. this? Labelto the the diode. Assume that Pmax = region of the diode curve 0.400 watts for this that Answer:: El useful part of curve volts 169 ftefera) again to Suppose that a the diode time varying whose i-v plot voltage source is is on page connected168. in series withthatin an thethe 8 diode?ohmvoltage resistor and source may this diode.have without What ip thecausing excessive maximum value power dissipationin volts valuethe ofdiode? series b) If the source resistance necessary voltage is limited to 2.5 to prevent excessivevolts, dissipation what is the minimum in 170 Answer: a)b) v max = 3.75 v Rmin , Figure 35 =1101. 171 In Fig. 35 are shown the i-v plot and the maximum-pawer hyperbola for a dis- givensipation diode. in the a) Suppose that the For the load diode. voltage Voline is drawnheld there is clearly anconstant and the excessive power series resistance Whatincreased percentage until Instead. of' increasing theincrease diode in R is dissipation is reduced R, the diode powerrequired? to the maximummay be reduced by lowering permissible. Vo.exactly the maximum b)What per cent reduction in Vo is permissible power? necessary so that the diode dissipates 1 Ale. .. . . U : 3 .111,, 7.I. . . I.. . r 1 i i._...I . . ...I. L. . .I _ _ r a ) - p
.,
,
, 173 FillSummary: in all blanks before checking answers. 1. tainingIn this sectionresistors we and one learned how to calculate non-linear diode, using a quantities in networks con- graphical construction. directly on the 2.3. TheThediode method coordinates i-v involvescurve. of the point superimposing the of intersection between the curve 4. reducedandIn general, to a seriesthe rest of line equal the diode equivalent connection of a the network, exclusive current and voltage. of thedode, must and a be 1. . loammmearmoomerawannevearea. Suzmnary174 (continued) 6.5. TheTheexcessive power maximum-powr dissipation temperature hyperbola inrise. a diode is a mustplot beof the equations: . to prevent damage from .4110.- 1111.11m1 ,1 Answers: 175 2)1)3) loaddiode line curve and load line ... tz4 4)6)5) mustreliAstorvi be= plimited and a max voltage source ET Cr