如圖所示為雙極性電晶體 Q 接成共射極組態,若逆向飽和電流為ICBO 且電流增益為 β,試問電晶體Q的截止條件為何?

AiB = 0,iC = 0,iE = -ICBO
BiE = 0,iC = ICBO,iB = -ICBO正確答案
CiB = 0,iC =βICBO,iE = -(1+β)ICBO
DiB = -ICBO,iC = ICBO,iE = (1+β)ICBO
答案與詳解

截止條件下射極與地斷開,iE=0;集極-基極接面反偏,仍有漏電流iC=ICBO流入集極;由KCL:iB+iC=iE=0,故iB=-ICBO(電流從基極流出),三者完全符合KCL,為正確答案。
