解:第二级是分压式偏置电路 +24v R Bll IMS 82kQ10k2 BI T2 RE510Q2 R el 27kQ2 43kQ 75kQ2 市CB 0.96 B2 mA=19. 2 uA B2 50 CE2 CC I(2(R(2+R2+RB2) =24-0.96(10+051+75)V=6.7v第二级是分压式偏置电路 mA 19 .2 μA 50 0 .96 2 C2 B 2 = = =  I I 24 0 .96(10 0 .51 7 .5)V 6 .71V ( ) CE2 CC C2 C2 E2 E2 = − + + = U = U − I R + R + R 解: RB1 C1 C2 RE1 + + + – RC2 C3 CE + + +24V + – UO  Ui  RB1  RB2  T1 T2 RE2  RE1  1M 27k 82k 43k 7.5k 510 10k