3计算BW (3)计算R20→Cb开路, Rb Cb两端的等效阻抗 Rc 1/2RL odl u b'e i(Tb+Rb)∈=Rn;i bb+ rbtib el n2=Rbi+(+R·gm)R rbb b' 20 12=(+gnk)+R b'e RL Tambre 1.14 ROcHe t r20 Cb 2r (4) 1.14 R rbb 20 CreRs+[Rs(1+8mR)+RI.CK 2丌 +u2 L b b'e 8mlbiel 休息1休息2返回Cb‘c gmub‘e rb‘e rb‘e Cb‘e Cb‘c gmub‘e uid1 (4)  ( )    2 1 C R' R 1 g R R C 1.14 2 1 R C R C 1.14 f b e S S m L L b c 1 0 b e 2 0 b c H   +  + +   =  + =     3 计算Ｂw + u0d1 uid1 _ 1/ 2RL （3） 计算Ｒ20 →Ｃb'e 开路， Ｃb'c 两端的等效阻抗 ( ) ( ) ( ) ( ) b m L L 2 20 2 b b m L b bb b b e bb b b e b e R 1 g R R i u R u R i i R i g R R i r R r i r R r u = =  +  +  =   + +    =    + +   +  =    R20 + ub‘e _ i i + u2 - 休息1 休息2 返回