14.3.3解:(a)D的正向压降可忽略,+-uoui+-DR(a)ui负半周,D截止,;0ou;iouuui正半周,D导通,uo的波形如下,30ui/Vωt-3030uo/Vωt14.3.3解:(b)D的正向压降可忽略,+-uoui+-DR(b)ui负半周,D截止,;iouu;0ouui正半周,D导通,uo的波形如下,30ui/Vωt-30-30uo/Vωt14.3.4解:D的正向压降可忽略,(2)VA=+3V,VB=0V,DA截止、DB导通,VY=0V。mAIIImARVVIRDDYRBA54.121,08.39.31212(1)VA=VB=0V,可见DA、DB加正向偏压,DA、DB导通,所以VY=0V。mAImAImAIBADDR08.3,0,08.3(3)VA=VB=+3V,DA、DB导通,VY=3V。mAIIImAIRDDRBA15.121,31.29.331214.4.1解:ZMZRRZZRZRIImAIIImARUImARUEI2,1.91.11020,1.119.01020212211如果IZ超过IZM,增加R1的值,来减小IZ14.5.2解:(1):正常工作(2)工作不正常mWPmWUIPmAIIVUUCMCECCCMCCEOBRCE1003020,15)((3)工作不正常mAIICMC20mWPmWUIPCMCECC10012014.5.4(a)解:mAkVImARUIIBCCCCSB12.050624.015012,,晶体管处于放大区BBII(b)mAkVImARUIIBCCCCSB26.047122.05.14012,,晶体管处于饱和区BBII(c)发射结反偏,晶体管处于截止区15.2.1解:mAIIARUIBCBCCB25024012,VRIUUCCCCCE63212(3)在静态时,UC1=UBE=0.6V,左负右正;UC2=UCE=6V,左右正负;+UCCRBRCT++–UBEUCE–ICIBIE(1)画出直流通路如右图(2)画直流负载线,M(12,0),N(0,4)交点Q如图所示,读出VUmAICEC62,UCE/VIC/mAO412IB=50μAQNM2615.2.5解:(a)发射结正偏,集电结反偏,满足直流条件;信号能顺利输入,能顺利输出,满足交流条件;所以该电路能放大交流信号。-UCC++(d)集电结正偏,且信号不能顺利输入,所以该电路不能放大交流信号。+UCC++15.3.1解:微变等效电路见右(1)输出端开路的电压放大倍数为,iIbIcIoUbIβiUrbeRBRCRLEBC+-+-1508.0340beCurRA(2)RL=6kΩ的电压放大倍数为,1008.06//340//beLCurRRA15.3.4解:VURrRUoLLoL2.121.53.31.50015.4.2解:(1)计算静态工作点,(2)微变等效电路见右,VURRRVCCBBBB58.510103324212AIImARUVIICBEBEBEC506632.332.35.16.058.5VRRIUUECCCCCE06.8)5.13.3(32.324)(iUiIbIcIoUbIβSErbeRBRCRLEBC+-+-+-RS15.4.2解:(3)计算rbe,(4)有载电压放大倍数为,kIrEbe72.032.3266720026)1(2007.18372.01.5//3.366beLurRA(5)空载电压放大倍数为,5.30272.03.366beCurRA(6)kRrkrrRRrcobebeBBi3.372.0////2115.6.1解:(1)电压放大倍数为,(2)输入、输出电阻为,98.01)501(11)501()1()1(EbeEuRrRA50////215050100016])1(//[//2121BBSSSbeoEbeBBiRRRRRrrkRrRRr其中15.7.3解:(1)静态时,ARRUUIEBBEEEB10)1(2AIIAIIBEBCm51.0)1(,m5.0VIRVVIRUVVIRUVBBBEEEEECCCCC1.08.02,45.3(2)mVuuuumVuuuuiiididiiicic22,5221212121(3)121)1(2icEbeBCococuRrRRuuk8.25126)501(200rbe式中mVuuococ39.22115.7.4解:(4)(5)mVuuumVuuuodocoodoco4.37,2.42222111mVurRRumVurRRuidbeBCodidbeBCod8.39,8.392211(6)odoooodododocococumVuuumVuuuuuu6.796.79,021212115.9.3解:25.11155.12121DmiouRguuA16.2.1解:50105001RRAFufkRRRF8.9500//10//12mVAuuufio500)50(1016.2.5解:VuRRRuRRRuii5.342132144333434FoiiRuuRuuRuu2211Vuuoo5.5,15.325.3225.3116.2.8解:第一级为电压跟随器Vuuio21第二级为反相比例运算电路VuRRuRRuoFiFo4)2(12112116.2.12证:iDDiuRRuRuRuii122121,,41423iiiii413RuRuRuuDioD整理得:iouRRRRRRu143232/16.2.20解:(1)t=0~10mstVdtdtuCRuiFo1005101105011631(2)t=10~20msKtdtCRuFo100)5(112,1,10KVumstoVtuo)2100(u0的波形如右图,17.1.1解:(b)电路如右图A1R4ufud+R3R2R1u0ui∞∞A2+--+R4为A2的局部反馈,交直流反馈,同理可判断为负反馈R1、R2为级间反馈,交直流反馈。瞬时极性标注各点极性如图所示。反馈电压uf使净输入电压ud减小,所以为负反馈。ud=ui-ufCu0ufud+R3R2R1ui∞++17.1.1解:(c)电路如右图R1、R2为反馈电阻,电容C将交流旁路到地,所以是直流反馈。瞬时极性标注各点极性如图所示。反馈电压uf使净输入电压ud减小,所以为负反馈。ud=ui-uf17.2.1解:(a)电路如下图R4、C为反馈元件,电容C将直流信号隔断,所以是交流流反馈。瞬时极性标注各点极性如图所示,id↓=ii-if↑,所以为负反馈。反馈信号与输入信号接同一输入端,所以为并联反馈。A2R3idC+R4∞A1+R2R1u0ui∞-+iiif因为if=-u0/R4,所以为电压反馈。17.2.6解:(1)7501.030013001AFAAf(2)%501.030011%2011AFAAAAff17.2.7解:20049.0100011000100FAAAuuufVAuuufi1.02020VFuuf098.0049.020Vuuufid002.0098.01.017.3.3解:(+)(+)(+)(-)-UCCCRRC满足相位平衡条件。000360,0,360fafa即可能振荡。17.3.5解(a)L1(-)RC(-)(+)(+)RECRb1Rb2+UccL2满足相位平衡条件。000360,180,180fafa产生反馈电压。即可能振荡。2L17.3.5解:(b)满足相位平衡条件。000360,180,180fafa产生反馈电压。即可能振荡。2CC1(-)RC(-)(+)(+)RELRb1Rb2+UccC218.1.3解:(1)ARUIVUULooo27.0,279.0(2)VUUDRM7.4621.118.2.3解:变压器二次侧电压为,VUU252.102二极管的最高反向电压为,VUUDRM4.3522mAIIoD752流过二极管的电流为,可选二极管为2CZ52B,最大整流电流100mA,反向工作峰值电压50V。取,25TCRL20015.03000IURLFRCL250200/05.0/05.0选250μF,耐压50V的电解电容。半波整流电容滤波,二极管的最高反向电压是桥式的二倍,为,222U18.3.1解:(1)S1断开,S2闭合,无电容滤波,VUU1822.289.09.01mARUUIR10102.1618301(2)S1,S2均闭合,有电容滤波,R=0,U1=24V直接加在6V稳压管上,可能损坏。反接DZ,U0=0.7V,达不到稳压6V。mARUIL31026300mAIIIRZ70VUU2422.282.12.11mARUUIR15102.1624301mAI30mAIIIRZ12018.3.6解:运放为电压跟随器,输出端电位和同相输入端相同,又,UUVUUXX5)7805(23的管脚电压所以,VURRRRRUXXPP96.6121min0VURRRRUXXP73.17121max020.2.2解:根据与非门的状态表,输出波形如下,ABCY20.2.3解:根据逻辑图,Y的表达式如下,CBACY输出波形如下,信号输出信号输出BBBYCAAAYC,1,0,1,1ABCY20.5.1解:(5)根据Y的表达式,逻辑图如下,A≥1≥1&BYC20.5.2解:(2)将Y的表达式转换为与非表达式,逻辑图如下,CBACBAYA&BYC=1=120.5.3解:(2)将Y的表达式转换为与非表达式,逻辑图如下,CBACBAYA&BYC=1=1=120.5.5解:(2)BBAABBACCABCABBAABCY)((5)11DDABCABCDCBAABCY20.5.6证明:(3)右边左边BABABABABAABBAABBAAB))((20.5.7解:(1)画出卡诺图,CBABCAABY所以,化简为Y=B20.6.3解:(1)由逻辑图写出逻辑表达式,CBAY(2)由逻辑表达式列写状态表ABCY00000011010101101001101011001111(3)由状态表分析逻辑功能。当有奇数个1时,输出为1,所以该电路为判奇电路。20.6.11解:(1)由逻辑图写出逻辑表达式,SDCBAY1开锁信号,(2)由逻辑表达式可见,当ABCD取值1001时,且S=1,Y1=1,Y2=0,开锁且不报警;其它取值,Y1=0,Y2=1,不能开锁,且报警。所以该电路的密码为1001。SDCBAY2报警信号,20.6.14解:(1)列状态表,(2)写逻辑表达式,(3)画逻辑图ABCD总分YABCD总分Y00000010001000015010016000104010105000119110111010100201100300