(12 =0-(12-1×3)V=-9V rs=2000+(1+B l=200+(1+80)=23kn IEsR.3+I VBEsI 1×3+0.7 A=0.37mA 10×10 =[12-0.37×10-(-0.7)]V=9V 26 =120014+50)0.37a=378k2 TE-2 2=0.74 mA (Ra+Ra)IE=(12-0.7)V R。=12-0.710×0.74ka=5.2k (2)总的电压增益 (Ra‖Ra) AvD2=2(1+Rb) 2=rk3+(1+B3)Ras =(2.3+81×3)ka=245.3k0 R。‖Ra=9.6k ×9.6 Aw2=2(3.78+1)≈50 R An=-r3+(1+B2)R 80×12 2.3+(1+80)3 Av=Av·Av2=-50×3.9=-195 (3)当v,=5mV时,输出电压 Uo=Ay·v1=-195×5×103=-0.98V (4)当输出端接一负载电阻R1=12k时 B3(Ra‖R1) Avr= (1+B3) 0×6 3+(1+80)3 97.5 629电路如图题6.2.9所示,设所有BJT的B=20,rk=2.5kn,r=200k0,FET的 4mS,其他参数如图所示。求:(1)两级放大电路的电压增益A=Av:Av2;(2)R和P