96專利商標特考電路學第三題 - 特考

By Tom
at 2019-07-04T02:02
at 2019-07-04T02:02
Table of Contents
大大好:
在做這題時 https://imgur.com/BlUYRpL,求出答案和書本解答不一樣,所以想
來請教高手。
個人解法:
先考慮二極體不導通的狀態,求得Vc=100-90e(-100t),並求出二極體開始導通的時間4.0
5ms;
接著考慮二極體導通的狀況,求電容兩端戴維寧等效電路,端點開路得Vth=70V,端點短
路之電流=Vth/Rth,得Rth=70歐姆,因此:
0.04>t>0 Vc=100-90e(-100t)
t>0.04 Vc=70-30e[-7/1000(t-0.04)]
但書上答案是 0.04>t>0 Vc=100-90e(-100t)
t>0.04 Vc=70-30e[-200(t-0.04)]
--
在做這題時 https://imgur.com/BlUYRpL,求出答案和書本解答不一樣,所以想
來請教高手。
個人解法:
先考慮二極體不導通的狀態,求得Vc=100-90e(-100t),並求出二極體開始導通的時間4.0
5ms;
接著考慮二極體導通的狀況,求電容兩端戴維寧等效電路,端點開路得Vth=70V,端點短
路之電流=Vth/Rth,得Rth=70歐姆,因此:
0.04>t>0 Vc=100-90e(-100t)
t>0.04 Vc=70-30e[-7/1000(t-0.04)]
但書上答案是 0.04>t>0 Vc=100-90e(-100t)
t>0.04 Vc=70-30e[-200(t-0.04)]
--
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