38.如【圖38】所示,若電晶體



【Sampo Lin】評論
求解
【yum49123003】評論
Vo/Vi=Ri/(50+Ri)*(1/(re+1))---兩次分壓=50.73/(50+50.73)*(1/(0.02+1))=0.49, 其中Ri=100//(2+101*1)
【楊景程】評論
因信號源有電阻Rs造成分壓,故Zi = Rs // Zi'先求出Zi' = Vb / ib = (ib*rπ + ie*Re) / ib 因 ie = (1+β)ib Re = (Re//RL) = 1K歐姆故 Zi' = (ib*rπ + (1+β)ib *Re) / ib = rπ + (1+β) *Re = 2k + (1+101)*1k = 103K歐姆----------------------------------------Zi = Rb // Zi' = 100K // 103K 約等於50K歐姆----------------------------------------AV = (Vo/Vi) * (Zi / RS + Zi)Vi = ib*rπ +ie*Re = ib*rπ + (1+β)ib *ReVo = ie*Re =(1+β)ib *ReVo / Vi = (1+β) *Re / rπ + (1+β) *Re ...