102年普考電子學 第二題 - 普考
By George
at 2013-07-09T22:51
at 2013-07-09T22:51
Table of Contents
※ 引述《Iampenguin (我是企鵝)》之銘言:
: 題目來源: 102年普考電子學 第二題
: http://ppt.cc/vz4Q
: 這題老師給的答案Rin=r拍+(1+貝塔)*(ro並RL)
: Ro = (Rsig+r拍)/1+貝塔 並 ro
: 我寫的跟老師一樣
: 但第4小題的Ai跟老師的不同
: 老師是給
: Ai=(1+貝塔)*(ro / ro+RL)*(Rsig / Rsig+Rin)
: 我的解法如下:
: http://ppt.cc/gcWl
: 我的想法是BJT跟他右邊的等效電路流過RL的電流io應相同
: 想問一下我這樣的解題想法問題出在哪裡
: 謝謝~
老師這題的答案有做修正了
Ro=ro Ai=1+B
可是我自己二邊的電路做了驗證
還是覺的怪怪的
不知道問題出在哪
好像比較多人的答案跟老師一樣是Ro=ro Ai=1+B
有人的答案跟我寫的是一樣的嗎@@
老師的答案做驗證
http://ppt.cc/~qYT
我的答案做驗證
http://ppt.cc/aCXg
--
: 題目來源: 102年普考電子學 第二題
: http://ppt.cc/vz4Q
: 這題老師給的答案Rin=r拍+(1+貝塔)*(ro並RL)
: Ro = (Rsig+r拍)/1+貝塔 並 ro
: 我寫的跟老師一樣
: 但第4小題的Ai跟老師的不同
: 老師是給
: Ai=(1+貝塔)*(ro / ro+RL)*(Rsig / Rsig+Rin)
: 我的解法如下:
: http://ppt.cc/gcWl
: 我的想法是BJT跟他右邊的等效電路流過RL的電流io應相同
: 想問一下我這樣的解題想法問題出在哪裡
: 謝謝~
老師這題的答案有做修正了
Ro=ro Ai=1+B
可是我自己二邊的電路做了驗證
還是覺的怪怪的
不知道問題出在哪
好像比較多人的答案跟老師一樣是Ro=ro Ai=1+B
有人的答案跟我寫的是一樣的嗎@@
老師的答案做驗證
http://ppt.cc/~qYT
我的答案做驗證
http://ppt.cc/aCXg
--
Tags:
普考
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