【正文】 ?? ? ? ? ?1 1 1/ / 0 . 9 5i b b eR R r k? ? ?? ?2 2 2 21 2 3 . 7 5i b e eR r R k?? ? ? ? ?2222 2 2( 1 )cube eRArR??? ? ? ???22122 2 2603( 1 )cu u ube eRA A ArR??? ? ? ???2 2ocR R k? ? ?祝大家考研成功 習(xí)題 2 2 4 P 1 0 9?1= 4 0?2= 2 0UB E Q 1= 0 . 7 V|UB E Q 2|= 0 . 7 VVCC= 1 2 VVEE= 1 2 VRc1= 3 . 9 k ?Rc2= 3 . 3 k ?Re2=2 k ?Rb= 2 0 0 k ?解：① uS=0 時， uC2= uO=022 / 3. 64CQ E E cI V R m A??2 2 2/ 0 . 1 8B Q C QI I m A???2 2 2 3 .8 2E Q B Q CQI I I m A? ? ?2 2 2 4 . 3 6E Q C C E Q eU V I R V? ? ?1 2 2 2 6CQ B Q E Q B E QU U U U V? ? ? ?Rc1uS uO+ VCCRe2Rc2VT1VT2 VEERbRi1 11( ) / 2. 14CR CC CQ cI V U R m A? ? ?1122. 32CCQ R B QI I I m A? ? ?1 1 1/ 0 .0 5 8B Q CQI I m A???1 1 1 2. 38E Q CQ B QI I I m A? ? ?11 1 2 1 1 . 3E Q E E B QU V V U V? ? ? ? ? ?，Rc1uS uO+ VCCRe2Rc2VT1VT2 VEERb?1= 4 0?2= 2 0UB E Q 1= 0 . 7 V|UB E Q 2|= 0 . 7 VVCC= 1 2 VVEE= 1 2 VRc1= 3 . 9 k ?Rc2= 3 . 3 k ?Re2=2 k ?Rb= 2 0 0 k ?祝大家考研成功 ?1= 4 0?2= 2 0Rc1= 3 . 9 k ?Rc2= 3 . 3 k ?Re2=2 k ?Rb= 2 0 0 k ?IEQ 1= 2 . 3 8 m AIEQ 2= 3 . 8 2 m A②111263 0 0 ( 1 ) 7 4 8beEQrI?? ? ? ? ?222263 0 0 ( 1 ) 4 4 3beEQrI?? ? ? ? ?1 1 27 4 8 3 . 3i i b e o cR R r R R k? ? ? ? ? ? ?，2 2 2 2( 1 ) 4 2 . 4 4 3i b e eR r R k?? ? ? ? ?RiRi2Rc1uS uO+ VCCRe2Rc2VT1VT2 VEERb?1= 4 0?2= 2 0Rc1= 3 . 9 k ?Rc2= 3 . 3 k ?Ri 1= 7 4 8 ?Ri 2= 4 2 . 4 4 3 k ?rbe 1= 7 4 8 ?1 1 1 211( // )191c c iui beu R RAur?? ? ? ? ?222121 .5 6ocuciuRAuR?? ? ? ? ?29721 ??? uuu AAAuiuc 1RiRi2Rc1uS uO+ VCCRe2Rc2VT1VT2 VEERb祝大家考研成功 Rb1C11C12RE 1Rb 21+ VCCRC 2C22Rb 22CE 2RE 2RLRSsUoU?VT1 VT2? = 1 0 0rbe 1= 5 . 3 k ?rbe 2= 6 . 2 k ?VCC= 1 2 VRS= 20 k ?Rb 1= 1 . 5 M ?RE 1= 7 . 5 k ?Rb 21= 9 1 k ?Rb 22= 3 0 k ?RC2= 12 k ?RE 2= 5 . 1 k ?iU?RSRb 1rbe 111 bI??RE 1Rb 21Rb 22rbe 222 bI??RC 2RLoU?解：微變等效電路：習(xí)題 2 2 5 P 1 1 0Ri 2Ri 11 1 1 1 2// [ ( 1 ) ( // ) ]252i i b b e E iR R R r R Rk?? ? ? ?????? kRR Co 122iU?RSRb 1rbe 111 bI??RE 1Rb 21Rb 22rbe 222 bI??RC 2RLoU?2 2 1 2 2 2/ / / / 4 . 8 6i b b b eR R R r k? ? ?①? = 1 0 0rbe 1= 5 . 3 k ?rbe 2= 6 . 2 k ?VCC= 1 2 VRS= 20 k ?Rb 1= 1 . 5 M ?RE 1= 7 . 5 k ?Rb 21= 9 1 k ?Rb 22= 3 0 k ?RC2= 12 k ?RE 2= 5 . 1 k ?祝大家考研成功 ? = 1 0 0rbe 1= 5 . 3 k ?rbe 2= 6 . 2 k ?VCC= 1 2 VRS= 2 0 k ?RE 1= 7 . 5 k ?RC2= 12 k ?RE 2= 5 . 1 k ?Ri= 2 5 2 k ?Ri 2= 4 . 8 6 k ?)//)(1()//)(1(211211?????iEbeiEuRRrRRA??22219 ubeRAr?? ? ? ?12 1 8 9 . 6u u uA A A? ? ? ?Ri 2Ri 1iU?RSRb 1rbe 111 bI??RE 1Rb 21Rb 22rbe 222 bI??RC 2RLoU?,0LSRR? ? ?② 時20SRk??時17 ius uiSRAARR? ? ??Ri 2Ri 1iU?RSRb 1rbe 111 bI??RE 1Rb 21Rb 22rbe 222 bI??RC 2RLoU?? = 1 0 0rbe 1= 5 . 3 k ?rbe 2= 6 . 2 k ?VCC= 1 2 VRS= 2 0 k ?RE 1= 7 . 5 k ?RC2= 12 k ?RE 2= 5 . 1 k ?Ri= 2 5 2 k ?Ri 2= 4 . 8 6 k ?③ 當(dāng) RS= 2 0 k ? 時，如果去掉射極輸出器，則2 4 . 8 6iiR R k? ? ?us uiSRAARR? ? ??2 1 9 3 . 5uuAA ? ? ?祝大家考研成功 習(xí)題 2 26 P 1 1 0? = 1 0 0rbe 1= 6 . 2 k ?rbe 2= 1 . 6 k ?VCC= 1 2 VRb 11= 9 1 k ?Rb 12= 3 0 k ?Rc 1= 12 k ?Re 1= 5 . 1 k ?Rb 2= 1 8 0 k ?Re 2= 3 . 6 k ?RL= 3 . 6 k ?iU?Rb 11rbe 111 bI??Rb 2Rc 1rbe 222 bI??Re 2RLoU?Rb 12微變等效電路：Rb 11C11C12Re 1Rb 2+ VCCC22Ce 1Re 2RLiU? oU?VT1VT2Rb 12Rc 1Ri 1Ri 22 2 2 2/ / [ ( 1 ) ( / / ) ] 9 0 . 8i b b e e LR R r R R k?? ? ? ? ?2 1 22( // )// 1 2 21b e c boer R RRR??? ? ??1 1 2 1[ ( / / ) ] / 1 7 6u c i b eA R R r?? ? ? ?iU?Rb 11rbe 111 bI??Rb 2Rc 1rbe 222 bI??Re 2RLoU?Rb 121 1 1 1 2 1/ / / / 4 . 8 6i i b b b eR R R R r k? ? ? ?①? = 1 0 0rbe 1= 6 . 2 k ?rbe 2= 1 . 6 k ?Rb 11= 9 1 k ?Rb 12= 3 0 k ?Rc 1= 12 k ?Re 1= 5 . 1 k ?Rb 2= 1 8 0 k ?Re 2= 3 . 6