電子技術(shù)答案

UD11D22b)>1截止。U21222U。即<。11.6（1）U=U=0VABABAB,10sinVutiuoDD解：UUU，，U6V，1kU10VRRLZZ⑵VUIZLLZZ。①a)a)c)BII=CBIBII=CBBII=CBUbIDDDUGSUUUGS-DSNIDSSP為N思考與習(xí)題）圖2-52電路與波形圖2-53題2.3圖VR和RCbQVRVVRCb電容C斷開時(shí)電壓放大倍數(shù)、輸入電阻輸出電阻。e圖2-55題2.5電路R30UBQVU5.31.8(mA)R3Ce1.8IUcece26(mV)300()(3001036.7K1.8beIEQ////1ioc=-150URA0.974OL)RuUrbeiEr[r)Re]//RRibeb1b2ocr0.45kQ。（2）beVRbReIV0.7I0.17(mA)eUe（2）kbero)R（3）R//R=0.48kRLr)RLeLbeLIUm解：g(RR)gRmdLmLrRRR519kiodrR//[r)R]Au=β（Rc//r)/rbe1；111i22)RARR//R2Lr)Ru2LL2LAu=AuAu12rr//R//Rirbe2rR//1oe22b）rR//[r)R]Au=β（Rc//r)/rbe1；111i22RA2c2r)Ru22e2Au=AuAu12rr//R//RirRoVRLuiuUViimU(210)228212.5W解：(1)Pom2RoL2VV(V2)P=PTPCComπom410WRLVU220210PV22.5WCComRLPoP0.56VU225W(2)PCC2RoL2VVV2()omP==6.85W(V=V)CCom4TπRL2V2P31.85WCCπRVmπL78.5%4uRUiLPVV/2)29W解：PCC2RomLV24VV和Vuuu)/20.05Vidi1i2解：uuu)/20.55Vici1i2Au。o+－＋∞2.1mV＋+－+uO－2mV－uuu)/20.05mVidi1i2解：uAu100.05mV0.5Vo4udidf1f2圖3-23題3.4電路Sf到1F到到FF到F到FF31F31F31F2F310fA0Aff0ff00V1d））圖3-26題3.8電路2A∞AΩ2BDC.0F1Ω解：⑴圖中A接接C92121VV思考與習(xí)題R，10k，VR2IO－＋∞uAO＋uIRR)RRuuf3RoI12350-1Vuu10oI220kuRR12O(uoI21O圖4-33題3.5電路uuuoI和uu12FuORuFI1RU1MOuI1－＋∞2T3TutR2OA＋uuI2I2UMR2TOta）b）()uuuoI1I2I1I212u時(shí)u＝tiouo圖4-35題4.7電路及輸入波形1解：－5udtu=-2.5VO1RCI011155u－O2RCORC5u02.5v-2.5vu與uIO解：圖4-36題4.8電路RRRUU222RRZTRZ21212121uO+UZUuU0I-UZ圖uUIuR3IRuRuRUVDD解：當(dāng)u超過正常值時(shí)，uU，輸出為負(fù)使VT導(dǎo)通，指示燈亮。二極管o1uIVD和電阻R在起限流保護(hù)作用。3圖解：，，Z2：345T1msCR3V1u和O1O∞C∞C=0.003μFR=16.7KΩ1URZ2思考與習(xí)題U為25V，輸出直流電流IOO為U2U=2R=U/IΩLOOT≥(3～5)2RL≥(240～400)μF11II200mA100mA22UU220.8V29.5VRM2URC2LUoUoUoUUUoooUUo2UoUUUoooUR由oLUUUZLoI=UUUURImaxLIminLIIIIZmaxLminZminLmaxU2UP2~3ILRUUIIIIUR與Ro21R2RU)5o1RR1與的比值為。21IRRRQ1PP1.25解：U1.25(I)RRoQp1RP思考題與習(xí)題A)8282A)2282828)圖圖6-51題電路圖13468高電平輸出的有Y27Y5圖6-52題電路圖324高電平輸出的有Y17YABCACBCYABBABYABCAYAABCBC解：YABCACBCABCBCBCABCYABBABBABYABCACCYAABCBCACACYYABCABADCBDY=Y=mm解：YBCBC(2)YBCD(3)YACDABC(4)YABCDCDY=Y=mmY=Y=dmACABABAC0解：YADBDCD(2)YCD或YCD(3)Y或YBCD(4)YB思考與習(xí)題圖題73．解：Y表7-18真值表AB00110011CY0110101000011111010101解：YABCYAC解：YABBCABBCAB00110011CY1011001100011111010101該電路實(shí)現(xiàn)B的異或邏輯功能。=AB+AC=AB+C)解：當(dāng)S10S為時(shí)F=01010ABBCAB00110011CY0010111000011111010101YABCAAABBB2,。1010AB當(dāng)≥A10000000011111111B10011001100110011B00101010101010101Y1000110011101111000111100001111YABCDDABCDABCDABCDDDABCDABCDABCDCDABBCABAD利用3線74LS138譯碼器和門電路實(shí)現(xiàn)FABCABCABCABCFABCABCABCABCmmmmYYYYFmmm=5047045704578選1Y=C。YABCmmmm12478選1AAADDDDDDDD21012470356CAB00110011CY0001011100011111010101(1)用與非門YABBCYmmmm35678選1AAADDDDDDDD21035670124YmmmmmmmmYYYY356735673567QJ1Q0KQ0Q1QRDDSSQDDDDSSQDDQCPDQ1DQ1CPDDCP1QJCPKJKJQ1Q1KQ0QCP00圖DQ、Q和端Z12Z113J221312331313121212312351122111212121212Y21時(shí)至?xí)r至由定時(shí)器構(gòu)成的多諧振蕩器如圖8-63VR、12u和uCOT=0.7(R+2R)C=4.13s12,圖VCCUU。O解：UU=8VΔ=4Vu波形如圖847所示.O圖思考與習(xí)題V10VT形R，REFf為解：（18D）（）,U=-397×10/2=-3.88VOU=-255×10/2=-2.49VO（F8）=（248）,U=-248×10/2=-2.27VOT形RfU解：輸出電壓的取值范圍:～0（200）=512）,5=-512×U/2,U=-10V8位解：UR/(2R),UR