电子电路分析与设计 DonaldA.Neamen著课后答案(英文版).pdf

Electronic Circuit Analysis and Design,2nd editionSolutions ManualChapter 1Exercise SolutionsELI一打卬(-静)GaAs:%=(2.1 x 1014)(300)3/2p(4r/H U(86 x 1Q-)(3OO)EL4(a)n0=5xlOl6cmt pa naa 三叫八=(L6xl0-l9)(1350)(5x10l(i)=(10n)(l0l7f(1.5 x 10 叫二(0.026)In%=0而9；Vb.n=Nd-A.=5 x 10”*majority carrier:=5。W cm7 minority carrierb.匕，=(0.026)In)-Vo=0.650 0.2n 幺=0.450 VE1.134=Jq(4)+Kd，fc=(4-Vp)/4 mAmdBy trul and error:Id=0.864 mA aad Vp=0.535 VEl.14Vp 0.5 2.25 mA0 6 0.105 mA0 7 4.93 mAEl.111=4 P(w)1110-=(】)(箴)t(0.026)In(1010)2 VpVD m 0.399 VTest Your UnderstandingChapter 1:Exercise SolutionsEl.15Power dissipatioa in diode.=1。%1.05 mW=/q(0.7)=Ad=1.5 mAEl.17/c 0.8 石94=b=-.=30.&mSVt 0.026-El.18VT 0.026 r 0.026%=石5。=丁=.Id=0.32 mA%=4”乙 4 G SEl.197q N Is exp(胃)Vr=VV In(受)IO*pn junedom Vo=(0.026)In(Q-it)=0.539-Schottky Dicdei VD-(0.026)In-d=0-99 VEL16EL20For the pn junedon diodeV匕0(77)=(0 0-6,ln(7TTiVD=0.6871 VSchoaky diode voltage will be smaller=0.5371-0.265=0.4221 VElectronic Circuit Analysis and Design,2tHi editionSolutions ManualEL21Power=I 匕10=1(3.6)0 I=1.79 mAR10 5.6-R-10-5.6 1.79n R=2,46 k6EL22Vz=zo+IzzSo 匕o=Kz-IzrzVzo=5.20-(10-3)(20)=5.20-0.02=5.18ThenVz=5.1B+(10 x 10巧(20)=Yz=5.38 VElectronic Circuit Analysis and Design,2ml editionSolutions ManualProblem Solutions5.23 x IQT372 exp(a)%=(i)Silicon,T=275K=(5.23304)(275广，一响如叱)(叫%=1.90 xL09 cm-3ni(ii)T=325K=(5.23#0。(325亡以3叫 3叫/=8.71xl0,c cm3%(b)GaAs(i)T=275K=(2.111寸)(275产的厂乂275)&=134x10s(ii)T=325K=(210 xl 巧(325)m/W*W3)2=1.63x107 cma.时卬（去）-i.i1.91b.Chapter 12(86 x 10-)(T)6.40 x 103TBy trial tad atror.T a 367*K=10f cm-3】八5,23、1产/(2(疝二片行1.91 x 10-7=T3 expBy trixl and error,T1268*Ka.Nd=3 x IO15 cm 7=8 一 typeM=乂=3 x W cm-3(15 x IO10)35 x 1CU-=po=4.5 x 10 加一3b.Nd=5 x 10”cm-3=一type，o=Na=5 x I。cm-3n.=(2.10 x 10k)(300)3/2 exp(“二)、-,八 2(86 x 1O-)(300)=(2.10 x 10h)(300)3/2(1.65 x 10-12)=UO x 10 cm-3n?(18x106)2 po=.=囚=6.48 x 10 cmc M 5 x-1.4&.N.=1016 cm 7=p type=I。cmt”。=9(L5 蓝量=所=2.25 x 104 cmb.GermaniumNa,=io cnj7 n p-typePq h Nq=1/6 Cir.7*叫叫 F(标瑞丽:=(1.66 x 1014)(300)3/1(2.79 x iL)=2.4 x 1013 cm-3J%以-5 76 乂屋加7L5(a)n9=5xlO15 cnT3川(15xlOlc)1 4 wp.=-=-=P,=45x10 cnf3八 n0 5x10”匕(b)n0 凡=n-type(c)三 Nd=5x10”cm3L6&Add DonorsM=7 x 10”加核b.W4ntpc=IC6 cm_3=So n?=(IO。)(7 x 1015)=7 x 1021=B，r，exP(tf)7 x】产=(5.23 x H竹exp(而滞历)By trial and error.T=324 KElectronic Circuit Analysis and Design,211d editionSolutions Manual/=(22)(15)(10-)=7=3.3/nA1.7A%=(0.026)In(叫/=J A=aEAA*(L8 x 10)ab.(0.026)1x1(10为(10)(1.5 x 10)a1.131.8C.VU=(0.026)In(10力：10”1J=a E=tr=g=空(1.8 x IM 产。=7.08(ohm-cm)一1.9=%【nNM 叫2=(0.026)InN.(06)(a)For n-type,T=(L6xl 0-19)(8500)For 10、4 N/10,9 cmy=L364 b4L36xI0(Q_cm 厂For N=QIS For N&=1018cmcmLIO(b)J=oE=y(0.1)=0.136S J 1.36x10 A/cm2A p=l.Qi*lO.um1.14*（。3）隘）%=巧E 刁.iz，n m八 1.(10u)(10u)=%=0.57S Vr200tr(T0.017332828.41.5 x 1010)22500.021673952.8|（心（1。叫I3000.0265196Jb.Vbt=(0.026)In=%=O.757 V3500.030336547.9(1.5 x 10lc)214000.034678000.0%,=(0.026)In=Vt=0.937 V4500.03909545.9c.(1.5 x 10叩5000.04333IL 180.31.12心（叫叫（再段湎14,=vyin%,=%lnChapter 1:Problem SolutionsT匕，20012561.4052506.02 x 1031.3893001.80 x 101.3703501.09 x 1041.3494002.44 x 10fl1.3274502.80 x IO101.3025002.00 x 10n1.2771.15c，=c，G 勃Tk。；铲卜皿a.Va=1 VG=(念)I，=0.634 也b.Vk=5 V/x -1/2=(i4 onj 1.16C=(0.02)0+*)=0.00743 pFFor%=15 V,Ci=(0.021+亲)=0.0118 pF、0.00743 4-0.0118 C)(avg)=-=0.00962 pF%Q)=vc(final)+(vc(mH/aZ)-vc(jina/)el/r wheret=RC=RCj(avg)=(47x103)(0.00962x10与 orr=452xl0H0 jThen%(r)=L5=O+(5-O)日-=e，=八=rlnf15 1 U5 J“=5.44x10M0 s(b)For V,=0V,Cj=CjC=0.02 pFFor%=35V,J=(O.O2(1+高=0.00863 pFr、0.02+0.00863Cj(avg)=-=0.0143 pFM=RCj(avg)=6,72x10一”$vc(r)=final+(yc(initial)-vc(finalr,r3S=5+(0-5-，1/r=5(1-c*so that t2=8.09x10-1 sL17%=(0.026)In n1T 1。叼0 737 V&.VG,=1 VCj=(0.25)(1+57)=64 pFf _ 1 _】*2t7lC-21r万：2 二：五64 x 10一”)fo=8.38 MHzElectronic Circuit Analysis and Design、2“editionSolutions Manualb.vR=10 vG(+搐广=0.0663 pF0 2斤，C.2，10-3；(0J663 x 1。-“)衣13.2 MHz1J8VD=V；rln(0 10)=Vb=0.0599 V1.20(a)/=(7)150 x10二 l(T%Sr-1)三 lOTeThenM黑巧=(。26M窄射Or%=0.430 V(b).,z.f 150 x10)Or%=0549 V1.211.19g=2190Id=/sP(焉)IL w(寂而)北=142 2支:八/。,8b.Id=(1.42 x 10)exp 2(0026)1 口=6.82 0AI-。-)6P(05?)4/三L公皿AZ=(1L)p(56)=1=QJQS AI=(1)P(点劫=/=4厢 A/=(1。-力 exp=!=22.5 yA1=(L(器)=，=/=(io-13)exp(5)*I=钝途 mAC.10-3=Is=P(嬴)/s=2.03 x 107s 人=(2.03 x】。-)exp(0 02&)Id=46.8 1nA1.22l5 doubles for every SC increase in temperature.。=10一12 a atT=300KFor/$=01。72 人=T=295KFor!s=50 xl0-12 A.(2)*=50 n 烈=5.64Where n equals number of 5C increases.ThenT=(564)(5)=282 KSo2954 TM 328,2 KChapter 1:Problem Solutio咆1.231.26寿焉-5=1555 器翳=2J 2.147X1/Vr100*C=373 K=Vr=0.03220%-55,=216K Vp=0.01865Zo(lOO)Zd(-55)=(2.147 x 10)6(藤)exP(岛/q(100)Id(-55)=(Z147 x 10。)(】.237 x 10)=(97374 x 10)-2.83 x 1031.24Vo=Vrln(10)n=59.9 m、亍 60 mYb 皿=VTln(100)=AVd h 9.7 mV 二】M 叱1.25%”=(笥2 h 0.450.4S=IdRtr+%,%=%由(%)By(rial and error:Iq=2,6 A,=0.402 VR=lo八九/小p(匍t1.5=(10 X 10e)(30 x 10T)*P I*)-1”。P(胃A+外By trial and error.Yu=0,Q46 V,1.5 0.046 r c i mAThen Id=-=Iq=Q-45 b.Reverse-Bias7=/s=30 nAVR=(30 x 10-t)10 x 10e)=0.30 VVo=-1.3+0.3=Vd=L2 V1.27Is=2 x 10-13A%=0.60 V/i=/seXP 偌)=(2 x I。*(S)=2.105 mA7a=0.60 mA7i=h+Ir=2.705 mA%m 修)=(0 026ln(嚷彘3=0.6065必=2%,+%=%=1.81 VElectronic Circuit Analysis and Design.2nd editionSolutions Manual(b)Diode is conductingor10-0.610iD=0.94 mAV。=3-5=(0.94)(10)-5n%=44V(c)Diode is reverse biased5 叫+/J10)-5Is=5 x io A/=Q.5C mA c2/0.3 x 10-3=(0026)tnV5xio-J益 h 0.479 V=(0.5 x 1。7)(4.7 x 103)+0.479+V4 工 2.17 V(b)P=(05)(0.479)orP=024mW0 VD=-10V 1.31Minimum diode current for Vw(min)Z0(min)=2 mA.%=0.7 Vr 0.7,5-0.7 4.3=-.Z.=-二-风 叫We haveso小 4.3 0.7.(1)=+2&R1.29(a)Assume diode is conducting.Then,%叫=0,7VSo that G=黑=233 必 r 1.2-0.7 s%=10,=50Then lD=IRl-=50-233Or一=26(b)Let&=50 Hl Diode is cutoff.Vp=(12)=0.45 V0 30+50 1 7Since Vfl peakb.For Idq=0.1 mA。=(奠=260Q 0.1id=O.OSIqq=5 A peak-copeak匕=U=(260)(5)“V=va 1,30 mV peak-o-peakSchottky:I=Is exp(京)匕=%In=Q26)In(喂需二 0.1796 VThen匕 of pn junction=0.1796+0.30=0.4796,I 0.5 x 10-32-E=/。47%exp exp/=4.87 x 10 A”(小茄)”Electronic Circuit Analysis and Design,2*editionSolutions Manual1.39pn junction ZD=05 mAThen%=(0.026)ln()=0521VSchottky diode匕=(0.026)1,)=0.281VThenVo(p)-%(S)_ 0521-0281 二 05 05R=480nI=10n exp(管)=5 X 10-/g=mAPz=(11)(5=Pz=74.8 mWb.12-6.80.2-=/z=26 mA%=变1 X 100=136%Pz=(26)(6.8)=176.8 mW176.8-74.874.8x IOC=136%1.43Izrz=(0.1)(20|=2mVV2Q=6.8-0.002=6.798 V10-6.798 0.5+0.02n!z=6.158 mA%=z=zo+Zzrz=6.798+(0.006153)(20)%=6.921 Vb.Z=Jz+lx10-%-6198%0.50 0,020 丁 110 6.798.fl,1.11而+而。宣+,+1359.9=VJ(53)%=6.791 VVo=6.791-6.921 V；=-0 13 VElectronic Circuil Analysis and Design.2nd editionSolutions ManualChapter 2Exercise SolutionsE2.1E2.3t/,=l20sin(2z60l),V；=07.R=2.5 kQFull-wave reedierTurns raao I：2=f=sVw=120-0.T=119.3 VVr=L19.3-100=19.3 V.=皿2fRVr-2(60)(26 X 10)(19.31C-2.06 x KT=20.6 x l6=C=20.6 gFE2.4小=50 sin(2?r60t),V；=0.7t R=10 kflFull-wive rectifierVe(50-14)2fRVr _ 2(60)(10 x 10D(2)C=2.025 x 10*5=20.25 x 108=C=3E2.5Using Eq.(2-10)vi 24 sin at=12.5 3tl-sin-*(J=31.7*By symmetry.皿=180 31.7=148.3x 100%=32.4%=73 sin(2r60t)vr=J!zlfRC-p/75O K 二 二,_ _ _fCVr(60)(50 x 10-6*4)R=6.25 kAU 34=属=混=287%=X 100%=9.14%10 Vps$14 V,Vi=5.620 Al s(min)-O.SVz-_Vps(min)-VzHMmi.)Vpnuu)-0.9匕0.1Vs(m.K)_ 14-5.6(280)-10-5.6(56)=10-(0.9)(5.6)-(0.1)(14)_ 2352-246.4=XS6Iz(xnax)=591.5 mAPower(min)=Zz(mwc)-Vz=(0.5915)(5.6)Power 工 3,31 W_ V$(mxx)-匕_14-5.6凡 /z(m)+(min)=03915+0.0568.4=0.6475R,w 13nE2.7For 7(min)and ZL(max),then(Minimum Zener current is zero.)For%(max)and/min),thenVtlmin)=nominal)+/z(nun)rz=5.6+.10-5.6，z(nnn)=-0.280=0.0565 AV(min)=5.6+(0.0585)(1.5)=5.689w d 6.487-5.688%Reg=-0.143=14.3%0(max)=I，：。9-0=230 mAThe characteristic of the minimum Zener current being one-tenth of the maximum value is violated.The proper circuit operation is questionable.E2.llE2.8Zz(nun)=噎蜉)/-/Jmax)so一八 10-9 r/30=-Jt(max)0.0153 u Or/max)=35.4E2.9%Regulation=Vt(tn&x)-VL(tnin)V(nominal)Vt(nominaj)=5.6%VZ(mg=VI(nominal)+Zz(maix)rz=5.6-h(0 5915)(1.5)=6.487Test Your UnderstandingChaDter 2:Exercise Solutions匕=0.7 vFor%=1.8 VEZ15。-2/12E2.12For 匕=0,v0(max)=-2KNow,Avo=87.so thatvQ(min)=一10/E2.13As“s goes negative.D turns on tnd 为=+5 VAs vs gg positive.D turns ofifE2.16Output,a square wave oscillNng between+5 andE2.17a.%=0.6 V for tU 匕Summary 0 3.3.D2 airns off and when”了 9.4.“0=10Electronic Circuit Analysis and Design,2n，editionSolutions ManualE2.18A.If=AI-(0.8)(16 x 10-)湍猾羽(。IPz=12.8 mAb.We have“=12.8)(1)=12.8 volts.The diode mult be reverse bi&sed so that Vps 12.8 volts.E2.19The equivalent circuit u5-11-0.2 r/+R=15 mAr,+A=0.207 kQ207nA=207-15=R=192nElectronic Circuit Analysis and Design,2nd edition Solutions ManualProblem SolutionsChapter 2(a)v5(max)=40 V(b)PIV=卜式max=40 V1(c)(d)50%2.4%=s-2匕=vs(max)=+2Ka.For c(inajc)=25 V=y$max)=25+2(0.7)=26.4 VNi 160 Ni 仁“瓦二笳i*瓦二b.For 也(max)=100 V=vs(m，x)=101.4 VFrom part(a)PZV=2vs(max)-=2(26.4)-0.7orPJV=52.1Vor,from part(b)P7V=2(101.4)-0.7 or PW=202.1VElectronic Circuit Analysis and Design,2nd editionSolutions ManualA,(ni)=24 V=(mo)=24+2(0.7)1(ma)=2$.4 V25.4“s(rnis)=-2=-=F(rms,二】7.96 VVm _ 外 I C=a2fRC 2fVrR3D.m”工 5.Q8 A2.6(a)vXmax)=24+0.7=24JVvJ(nrw)=vrms)=175 V Q=危n=丽=(60)(150)(05)orC=5333/(c)For the haIRwave rectifier用 I嗨)=+4*离j orJ=10。4%。)=24 sin 必Nowjjggb?。城We have fbr 4 4 4/24sinr-12.7“一R-To find 玉 and.，24 sin x,=12.7士=0558”/x2=a-0.558=2.584 radThen西-I成牌用半 吟借卜皿喏-同掾k orFraction of time diode is conducting=殳*xl00%=空竺T吧1xlOO%2万 Inor Fraction=32.2%Power ratingD。.z k2.R K24sinx-12,71“F=悭u旬天产=盛力(24)3-2。2m+”7)%=短皎4)任-X12.7X24X-COSX)*+(12.7)2 J.For R=L190 th5Pf=17.9 WChapter 2:Problem Solutions2.9R=150Q0.1%(max)=%(max)+匕=15+0.7241For%0,(6=0)orvJ(max)=15.7 VThen%()=*=ll.1V108y _ 乙=二乙二 尸 2fRC=2fRV,=2(601150)(0.4)orC=2083FPZV=2vr(max)-Vy=2(I5.7)-0.7 orPZV=30.7 V2.10For 叫 0=%(ms)=3.04 VVi=6.3 V.凡=120,Vz=4.8r 6.3 4.8,m 人1.Ii=I?=125 mA/l=Z/-Zz=125 Iz25 120 mA=404Rl S 192Rb.Pz=Izz=(100)(48)=Pz=480 mWPl=/lVq=(120)(4.8)=Pl=576 mW2.13%=0Voltage across Rc+Ri=%.20-10,c,h=-222=。L=430 0A今 11=26.3 mA/z=7/-Zt Iz=18.7 mAg)=皿Electronic Circuit Analysis and Design.2M editionSolutions Manualb.Pz(ni3)=400 mW a Zz(m&x)=黑=40 mA n/4(znixi)=-/z(max)=45-40a 1Mmin)=5 mA=找n&n 2 kCFor 凡 h ITSfi，:57,1 mA/l=26.3 mA/z=30.8 mAZz(majc)=40 mA=/.(min)=5T.1-40=17.1 mARl=Kl=585。2.14a.From Eq.(223)r.、500(20-1可-5015-10|”(max)=15 _(o.9)(io)_(q)(20)_ 5000-2so=41z(e3=1.1875 A7z(min)=0.11375 AFront Eq.(2-21(b)艮=1187：5+50,&=8 08Ob.Pz=(1.1875)(10)=Pz=1L9 WPl=/t(maxJVc=(0.5)(10)=Pl=S W2.15(a)As approximation,assume/2(max)and Zz(min)are the same as in problem 2-14.Vo(max)=Vo(nom)+1z(m=)rz=20+(0453)=20.906Vo(min)=Vb(nom)+/z(xnin)rz=20+(0.0453)(2)=20.0906 cr 20.906-20.0906,rtziorb.%Reg=-x 100%=%Reg=4，8%2.16%Reg=Vjmax,-Vjmin)VL(nom)X100%_ 匕Som)+/z(max.(匕(Mm)+/z(min)q)VL(nomj=，z(m秋)式血嗽3)=0056So7z(max)-/z(min)=0.1 ANowZt(max)=-=0.012 A,Z,(min)=0.006 A500 ia/1000NowR：匕Kmin).十/z(min)ZL(max)28=、八=/z(min)=0.020A Umm)+0.012 Zk，Then/式 max)=0.1+0.02=0.12 Aandr=%的驮)-%r Z2(max)+Zt(min)or28。=德离=Vw(max)=41.3 V2.17Using Figure 2.17a.=20/2S%=15 s s(min);It 二 Iz(min)+/L(max)=5+20=25 mA-Vz 15-10凡-1一L-今 凡=2000h 25-b.For V2s(m&x)=n ff(max)=75 mAFar=0 n/z(max)75 mAVzo=Vj-lzrz=10-(0.025)(5)=9.875 VVi(mAX)=9.875+(0.075)(5)=10,25%(rain)=9.875+(0.005)(5)=9.90%=0.35 VAUc.%Reg /-x 100%=%Reg=3.5%Vo(nom)-8-2.18From Equation(2.21(a)R 二匕s(min)一%=24-16 i Zz(min)+/t(tnax)40+400or&=182QAlso y VV=C=2/RC 2JRVrR 或&+r,=182+2=20.2。ThenC=；-JI一r=C=9901 卢 2(60X1)(202).-ChaDter 2:Problem Solutions119匕=%+Izrz 匕(nom)M 8 V8=Vi。+(0)(0.5)=分7.95 Vr%(mq)一%(nom)12-8/i-=“n 1.33Hi 3For II=Q.2 A n Jz=1.133 AFer=1 A n/z=0 333 A匕(max)=Vzo+7z(max)rz=7.95+(1.133)(0.5)=8.5165V(ndn)=Vzo-+Iz(min)rz=7.95+(0.333)(0.5)=34165匕=0.4 V%R4=*%=5.0%Vq(nom)8 *-V VK=r 2JRC 2加匕=4+9=3+05=35。Then一幅忐)(0.8)=七吧:2.20(a)For-10 4 0.both diodes are conducting=%=0For 0 4 v,4 3,Zener not in breakdown,so1=0，%=0For vf32.21(a)Vi=ixl5=5V=fofi/f e丁 4-+7=”(1+亍+=vo(2q)Vf+8.55=%(2.5)=%=泮/+342vi=5.7 n 外=5.7 町015彳d1工9.42At v,=10 V,%=35V,(j=0.35 mA(b)For v；3,&=.At%=10V.i,Q,35mAElectronic Circuit Analysis and DesiRn.2nd editionSolutions Manual2.22(a)For D off,%=(耿20)-10=333 VThen for v,M 3.33+0.7=4.03 V=%=333 VFor v,4.03.=vr-0.7;For v,=10,v0=93产。s%7_?.rs/1-:1 t 5 2(b)For,44.03V,iD=0For”403,+1222 二%一(_1。),10 20Which yields iD-0.605For vf=10,iD=0395 mA&，6 IO223。弋-100+10 _.%=i(10)+10.T=12.5 Vb.U J ixa加 y2.24*5-y-h 1KI一r-刃.法“KV；=0.6 V=15 sin 3t5 lz。C 一T.4工：二父/二鼻一-n.il2.25a.匕二0L.U LJVy=0.6二4,LJ LTChapter 2:Problem SolutionsFor%=。；V；10 V2.26One possible example is shown.2.30L will tend to block the transient signals Dz will limit the volUgcto+14 V and-0.7 V.Power ratings depends on number of pulses per second and duration of pulse.For circuit in Figure P2.27(b)(i)For 匕=+3/2.27Electro口ic Circuit Analysis and Design.2nd editionSolutions Manual231For Figure P2.27(a)b.必=%=5 V Di and Di on.Ds off 10=7(9.5)+0.6+(05)+5 I=0,5I mA2.3210 0.69.S+0.5=Adi=0.94 mA1/2=0b.=Is(9.5)=%=8.93 VS-0.6 Aoi=0.44 mA&=0=/。，3.5)。匕=4.18 V“9.5)ID2 iSime u&)10=+0.6+1(9.5)=T=0.964 tnA=%=9.16 v1-20.482 mAIo=Idi=o/m=Q2=0.226 mA%=1。-I(9.S)=10-(0.451)(9.5)=%=3.72 Vc.%=5 V,%=0 D off,Dt,D on1 Di=0I Di=Im !=7.6 0.389=，dj=7.01 inAd,4=5 V,%=2 V D off,Di,Zh on%=4.4 VIds=1d?/=3.6 0,589 0 la=3.01 jpA2.352.33L J=/a=0 K=jj)b.10 二“9.5)+0.6+Z(O,S)nnId、=0.94 mA=口%=10-1(9.5)=%=L07 VC.10 /(9.5)+0.6+Z(0.5)+S0=Im=0.44 mA 4=0%=10-Z(9.S)=%=S.82 VdL 1。=/(9.S)+0.6+j(0.5)=/_=C,964 mA4di=Jg=3 心，=pi=0.482 mA%=10-7(9.5)=*-0“2 V2.34工 V Vi st 0=D),Dz,Dj.on =4,4 Y/i=/m 工 OS，=/di=Ta 7.6 mAa=16+%2-1=2(7.6)-0.589=Ins=14.6 mA(a)D on,D2 offt Da onSo ID2=0NowK=T).6P,10-0.6-(46)1。=2+6Iol=125mAU)_0.6_(125)(2)n 9=6”-0.6-(-5).&=2=22 mA/s=/iu-/m=22-L*=3。95(b)Dy on,D,on,D、offSo/s=010-0.6-4.4 5 二or10t=0.833 mAr 4.4-(-5)9.4 z=&+a记=0.94 ndIK=IK2-%=0.94-0833=，g=0.107 初 匕=4 禺-5=(0.94X5)-5=匕=T).3 9 Chapter 2:Problem Solutions(c)All diodes are on 匕=44夕，匕=4.60“=05/w4=山一乎n 用=100Ijft=05+05=1 mA=4二 卜。6)，R3=L5 mA=/6 二)=&=2,9302.36For Vj small,both diodes off“岛卜=。吟 When vf-vo=0.6,D turns on.So we have vt-0.0909vf=0,6=,=0.66,v0=0.06 For Dy on.()；-%+=言 which yields2,0.6%=When%=0.6,D2 turns on.Then a z 2v;-0.6.-0.6=-=v.=3.9 V12,Now fbr vz 3.9匕-0.6_%+为.%=殳+%-065 5 05 05Which yields 2v+54%=-；For vz=10=vo=U5 F2.38Ai=5 m R2=10 kQ D and D?on*%510=1.86-1.0Im=0.86 mAb.A】=10 kfl,R?=5 k。，Di aS,D?on151.2872.37%=IRz 10 a%=-3.57 V2.3915-(%+0.7)%+0.7,%10=120-+20Vi=6.975 V n Ad=6349 mA20-For vi 0,when D)turns off _ 10 _ 0.7 cI=se 0.465 mA20=7(10 kQ)=4.65 VElectronic Circuit Analysis and Design.2nd editionSolutions Manual2.402.43a.%,=%-=：vb.%1=0.6 V,%2=1.2 Vc.治=0.6 V,侬=L2 VLogic“(T$ud degrades u it goes through addidorul logic cafes.&.%=15 V,=10 V Diode off匕工 7.S V,匕=5 V o Vp=-2.5 V/d=0b.%=10 V,%=IS V Diode oaV3=0.6 V2.44倍/D%)OR传/匕)2.45.10 1.5 002 QA A A etcI=,=12 mA=0.012R+10 g 2=,=mnR=681 了口2.46叼=10-1,7-8(0.7$)n%=2.3 Y2.41”r Q,D off,Qq an.10-2.S&I=-=0.5 mA Immo=10 (O.S)(5)n”=7.5 V for 0 4 肛 W J.3For vi=30 V,Di off.为=10 VDetenmne vr when 匕=10T*7-2.57=3-14-10=Z(io)+2.5=I=0.75 ehA“1=(0.7S)(2S)+2.3=21.25%二 i巧i=1 V,/=0.3 mA l+(0.8)(2)k=2.6 V2.48Ipm=e。人0.6 x IO-3=(1)(1.6 x 10-,!)(1017)人=3k5 x IQ-?cm，242t.%1%H 0b.%i=4.4 V,%=3.8 VG.-1=4,4 V,%=3.8 VLope*1*level decr&des as it goes through Additional logicgates.Electronic Cir