38.如【圖38】所示，若電晶體 = 2kΩ，β= 100，則約為多少？

(A) 50
(B) 1
(C) 0.9
(D) 0.5

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 

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Zi = Rb // Zi'

    = 100K // 103K

    約等於50K歐姆

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AV = (Vo/Vi) * (Zi / RS + Zi)

Vi  =  ib*rπ +ie*Re

     =  ib*rπ + (1+β)ib *Re

Vo = ie*Re

     =(1+β)ib *Re

Vo / Vi = (1+β) *Re / rπ + (1+β) *Re

           = (101) * 1K / 2K + 101K

           = 0.9806

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最後在乘上分壓即為答案

(Vo / Vi)*(Zi / RS + Zi)

=  0.9806 * (50K / 50K +50K) 

=  0.9806 * 0.5歐姆

=  0.4903