【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】62.有一個OPA的迴轉率SR為2V/μs，當輸入信號在10μs內變動0.5V，則在不失真的情況下，該放大器之Av =？(A) 10 (B) 20 (C) 30 (D) 40
【評論內容】
迴轉率(SR)的定義為放大器在一定時間內的最大電壓輸出 SR = dV(t)/dt
--------------------------------------------換個說法SR即表示輸出電壓(Vout)
而d(v)/dt 則為輸入電壓(Vin)
--------------------------------------
所以依照電壓增益Av = Vout / Vin的公式
Vout = SR = 2V / us
= 2MV；
-----------------------------------------
Vin = dV(t) / dt
= 0.5V / 10us
= 0.05MV；
-----------------------------------------
Av = 2MV / 0.05MV
= 40
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...
【評論主題】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
【評論內容】
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 *Re
Vo = ie*Re
=(1+β)ib *Re
Vo / Vi = (1+β) *Re / rπ + (1+β) *Re
...