VCEOIVcc-1corRoI-(col +lBoure =lcc-lcoI (ra +rei) (12-0.93×7.8)V=47V V2=lB+pa2=(726+0.7)V=796V Bo≈lm2=(c-V2)Ra2=(12-796)/39mA=404/39mA=104mA C=lo2Ra2=(104×4.3)V=447V EQ2=e2-B2=(447-7.96v=-345V (2)求电压放大倍数 先计算三极管的输入电阻 bel=rb+(1+B) 26(mV) =3009+101 Q2=3.1 kQ +(1+B) 26(mV) 300g+101×—9=2.8kg 电压增益 A,=B(RA)=-100×5128 式中R2=2 42=-(2∥R)_100×43 =-1536 2.8 A,=A1A2=-58.3×(-153.6)=8955 如果求从vs算起的电压增益,需计算输入电阻 R1=r1∥Rb1∥R2=3.l∥5l∥20=255k2 RI 2.55 R+R1-1+255 ×(-58.3)=-41.9 A=A,A,2=-41.9×(-1536)=6436 2用开路电压法求电压增益 第一级的开路电压增益(12 0.93 7.8) V 4.7 V = ( ) ( ) CEQ1 cc CQ1 c1 CQ1 BQ1 e1 cc CQ1 c1 e1 = −  = V V − I R − I + I R  V − I R + R VE2 =VB2 +VBE2 = (7.26 + 0.7) V = 7.96V (1.04 4.3) V 4.47 V ( )/ [(12 7.96)/ 3.9]mA 4.04/ 3.9 mA 1.04 mA C2 CQ2 c2 EQ2 CQ2 CC E2 e2 = =  =  = − = − = = V I R I I V V R VCEQ2 = VC2 −VE2 = (4.47 − 7.96) V = −3.45 V （2）求电压放大倍数 先计算三极管的输入电阻 + + =  +   =  + + =  +   =  2.8 k 1.04 26 300 101 (mA) 26(mV) = (1 ) 3.1k 0.93 26 300 101 (mA) 26(mV) = (1 ) E2 be2 b b E1 be1 b b I r r I r r   电压增益 i 2 be2 be1 c1 i 2 1 58.3 3.1 ( // ) 100 (5.1// 2.8) = R r r R R Av = = −  − = − 式中  153.6 2.8 ( // ) 100 4.3 = be2 c2 L 2 = −  − = − r R R Av  Av = Av1Av2 = −58.3(−153.6) = 8955 如果求从 VS算起的电压增益，需计算输入电阻 Ri1 = rbe1 // Rb1 // Rb2 = 3.1// 51// 20 = 2.55k ( 58.3) 41.9 1 2.55 2.55 1 S i1 i1 s1  − = − + = + v = Av R R R A Avs = Avs1Av2 = −41.9(−153.6) = 6436 2 用开路电压法求电压增益 第一级的开路电压增益