(2)计算r;和r0 bI b2 c2 b1 be2 B1 C2 R B E1 E Vi2 26 26 rm2=200+(1+B) 200+51 Q=1.58k2 0.96 2=R1∥l22+(1+1)z}=14k2 27×14 k2=922k2 27+14b 2 I c 2 I rbe2 RC2 U o  rbe1 U i  RB1 RB1  RB2  b1 I b2 βI b1 βIc1 I RE1 U o 1  +_ +_ +_ RE 2  (2) 计算 r i 和 r 0 Ω 1.58 kΩ 0 .96 26 200 51 26 200 (1 ) E be2 = + + = + = I r  // // (1 )  14 k ri2 = RB 1  RB 2  rbe2 + +  RE2 = Ω k 9 .22 k 27 14 27 14 // L1 E1 i 2 Ω = Ω + R = R r = i 2 r