作者coolboychiu ()
看板Physics
標題[問題] 全功率帶寬
時間Sun Oct 9 23:20:25 2011
剛剛在看電子學，有一句話看不懂，跟同學後還是不懂，
想說來物理版問問看，以下就是我們看不懂的地方：
full-power bandwidth:
It's the frequency at which an output sinusoid with amplitude equal to
the rated output voltage of the op amp begins to show distortion due to
slew-rate limiting.
我已經卡快一個多小時了，請各位幫幫忙，感恩~
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我把全文打上來好了，有點長，麻煩各位幫個忙，感恩
昨天熬夜都在想這個，想不出來好難過
Op-amp slew-rate limiting can cause nonlinear distortion in sinusoidal
waveforms. Consider the unity-gain follower with a sine-wave input given by
v_i = V_i*sin(wt)
The rate of change of this waveform is given by
dv_i
---- = wV_i*cos(wt)
dt
with a maximum value of wV_i. This maximum occurs at the zero crossings of
the input sinudoid. Now if wV_i exceeds the slew rate of the op amp, the
output will be distorted. Observe that the output cannot keep up with the
large rate of change of the sinusoid at its zero crossings, and the op amp
slews.
The op-amp data sheets usually specify a frequency f_M called the full-power
bandwidth. (接下來就是說明什麼是fpbd，就是上面那段) If we denote the rated
output voltage V_o,max, then f_M is related to SR as follows:
w_M*V_o,max = Slew rate (SR)
Thus,
f_M = SR/(2*pi*V_o,max)
It should be obvious that output sinusoids of amplitudes smaller than V_o,max
will show slew-rate distortion at frequencies higher than w_M. In fact, at a
frequency w higher than w_M, the maximum amplitude of the undistorted output
sinusoid is given by
V_o = V_o,max*(w_M/w)
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◆ From: 140.112.248.193
1F：推 herbert570:在該頻率的時候，輸出弦波會因為slew-rate的關係沒辦法 10/10 00:42
2F：→ herbert570:跟上輸入弦波，而產生distortion 10/10 00:42
3F：→ herbert570: ↑會"開始"因為 10/10 00:42
4F：推 Leadgen:頻寬不夠，所以速度跟不上? 10/10 12:29
※ 編輯: coolboychiu 來自: 140.112.248.193 (10/10 13:58)
5F：→ threedices:op amp慢半拍 跟不上輸入的訊號.. 10/10 13:43