--------------------------------------------------品文檔放大電路中的反饋習(xí)題解：(1)BB(2)D(3)C(4)C品文檔(5)ABBAB解：(1)A(2)B(3)C(4)D(5)B(6)A果。果(3)√(4)×品文檔圖P6.4----------------------------------------------------------------------------------------------------品文檔圖P6.5--------------------------------------------------品文檔374fofoR+R12fofoR+R12foR+R13U=1RU=1RoU1Ro4I.=R2oRR2fffoR+R+R249ffif1if=24978LRR29A=A=oii1f11(f)ifif1if1if1(f)2ifs2--------------------------------------------------品文檔ufOIIufOII11O122O2OV。V品文檔BB2u=u↑→u=u↓(u=u↓)→i=i↓→u↓(u↓)→i↓→u↓B1B2C1C2B4B5E4E5R5B3C3R1u=u↑←C1C2(1)A＝？fff11=0.0051A相對(duì)變化率為A的相對(duì)變化率的，故A的相對(duì)變化率約為f1＋AFf品文檔uufuAF必=100fAFuFZ2OO123.U12~123.U，Z1RRR即123.6V~123.6V121--------------------------------------------------品文檔為f＝104Hzf＝f＝105HzL2L3L2L3L1L2L3LL3(a)(b)(c)(d)所示。--------------------------------------------------品文檔IL (2)若u＝0~5V時(shí)，Ii＝0~10mA，LF--------------------------------------------------1品文檔1f2ffR+R+R1f61ff=R＝18.5kΩf品文檔圖P6.202----------------------------------------------------------------------------------------------------品文檔圖P6.21圖P6.22品文檔uR解圖P6.22oioioiRRRRRofffoRRRffif1H2πRCf品文檔ofifof圖P6.23fA=OA=O=O=A(r∥R)iuodidfIiidfi1F=F=uROfodidfRodidfRffA(r∥R)A=odidffArR).1odidfRf解圖P6.23ioifofiidfoof----------------------------------------------------------------------------------------------------品文檔R必idf=fifA.(r∥R).1AododidfRfofArRAodridodidfRfidfofodfifooARA=odfiffodofoodfifofR=r+R∥Riidfoofoofu=u.idfridIridIuF=FR品文檔rAA=idffrRrRfr∥(R+R)R必ofofrRArRidfofi