看板 Physics 關於我們 聯絡資訊
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~ ----------------------------------------------------------------------- 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) -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.112.248.193
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
herbert570:跟上輸入弦波,而產生distortion 10/10 00:42
herbert570: ↑會"開始"因為 10/10 00:42
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)
threedices:op amp慢半拍 跟不上輸入的訊號.. 10/10 13:43