下面电路图的Q点怎么算

2026-05-20 23:42:39
推荐回答(1个)
回答1:

此题直流等效图中C2开路,
IB(R1+R2)+0.7+(1+β)IBxRc=Vcc
IB=(Vcc-0.7)/(R1+R2+(1+β)Rc)
Ic=βIB
Vce=12-(1+β)IBxRc