國營101年電子學、電路學(兩題) - 考試
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By Harry
at 2013-05-29T00:00
at 2013-05-29T00:00
Table of Contents
56 漣波因素r = Vr(rms) / Vdc *100%
其中:Vr(rms) = √(Vrms^2 - Vdc^2) = √【(Vm/√2)^2 - (0.636Vm)^2】
= 0.309Vm
Vrms = Vm / √2 ; Vdc = 0.636Vm
則r = (0.309Vm / 0.636Vm) *100% ≒48%
35 題意知Po(max) = 2W
PD(av) = ton / T * VCE*Ic(sat) = 1μ/(1/200K) *0.2*100m
= 4mW
則η(max)= Po(max)/ Po(max) + PD(av)
= 2 / (2+0.004) *100% = 99.8%
以上。
※ 引述《cenacenacena (科科)》之銘言:
: 56.試求全波整流電路之輸出波形(未濾波前)的連波百分比為何?
: (A)55% (B)48% (C)32% (D)63%
: 答案為(B)
: 35.有一個協調C類放大器,電源20V且最大輸出功率為2W,若推動此放大器的輸入訊號
: 頻率為200KHz,導通的時間為每週期1us,且導通時Ic(sat) = 100mA,VCE = 0.2V
: ,則此放大器的最大效率為何?
: (A)98.2% (B)99.5% (C)99.8% (D)93.5%
: 答案為(C)
: 請教大大一下如何算
: 照著公式去算得到答案皆不在選項= =
: 囧
--
其中:Vr(rms) = √(Vrms^2 - Vdc^2) = √【(Vm/√2)^2 - (0.636Vm)^2】
= 0.309Vm
Vrms = Vm / √2 ; Vdc = 0.636Vm
則r = (0.309Vm / 0.636Vm) *100% ≒48%
35 題意知Po(max) = 2W
PD(av) = ton / T * VCE*Ic(sat) = 1μ/(1/200K) *0.2*100m
= 4mW
則η(max)= Po(max)/ Po(max) + PD(av)
= 2 / (2+0.004) *100% = 99.8%
以上。
※ 引述《cenacenacena (科科)》之銘言:
: 56.試求全波整流電路之輸出波形(未濾波前)的連波百分比為何?
: (A)55% (B)48% (C)32% (D)63%
: 答案為(B)
: 35.有一個協調C類放大器,電源20V且最大輸出功率為2W,若推動此放大器的輸入訊號
: 頻率為200KHz,導通的時間為每週期1us,且導通時Ic(sat) = 100mA,VCE = 0.2V
: ,則此放大器的最大效率為何?
: (A)98.2% (B)99.5% (C)99.8% (D)93.5%
: 答案為(C)
: 請教大大一下如何算
: 照著公式去算得到答案皆不在選項= =
: 囧
--
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By Catherine
at 2013-05-29T15:25
at 2013-05-29T15:25
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By Zanna
at 2013-06-03T04:07
at 2013-06-03T04:07
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