請ACT30 FAE回復(fù)這個問題:
做調(diào)試時,發(fā)現(xiàn)圖中200R的電阻對空載文波影響特別大,但是改成300R以后問題解決.請問能該嗎,對整機(jī)的可靠性是否有影響?這個RC回路是什么作用?
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500) {this.resized=true; this.width=500; this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標(biāo)滾輪可自由縮放';this.style.cursor='hand'}" onclick="if(!this.resized) {return true;} else {window.open('http://u.dianyuan.com/bbs/u/52/96751177818752.jpg');}" onmousewheel="return imgzoom(this);">@szlhb2008
[圖片]500){this.resized=true;this.width=500;this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標(biāo)滾輪可自由縮放';this.style.cursor='hand'}"onclick="if(!this.resized){returntrue;}else{window.open('http://u.dianyuan.com/bbs/u/52/96751177818752.jpg');}"onmousewheel="returnimgzoom(this);">
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1. Single Capacitor Compensation
500) {this.resized=true; this.width=500; this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標(biāo)滾輪可自由縮放';this.style.cursor='hand'}" onclick="if(!this.resized) {return true;} else {window.open('http://u.dianyuan.com/bbs/u/53/95251178858944.bmp');}" onmousewheel="return imgzoom(this);">
Vfb = Iq * (1/(S*C1)), where S = 2*π*f
Vfb/Iq = 1/(S*C1)
Vfb/Iq = (1/S) * (1/C1)
Pole at frequency = 0 Hz
Gain = (1/C1)
Therefore, small C1 will get a higher gain and faster transient response but easier to oscillate.
2. Capacitor-Resistor Compensation
500) {this.resized=true; this.width=500; this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標(biāo)滾輪可自由縮放';this.style.cursor='hand'}" onclick="if(!this.resized) {return true;} else {window.open('http://u.dianyuan.com/bbs/u/53/95251178858989.bmp');}" onmousewheel="return imgzoom(this);">
Vfb = Iq * (1/(S*C1) + R1), where S = 2*π*f
Vfb/Iq = 1/(S*C1) + R1
Vfb/Iq = (1+S*R1*C1) / (1/S*C1)
Vfb/Iq = (1/S) * (1/C1) * (1+S*R1*C1)
Pole at frequency = 0 Hz
Gain = (1/C1)
Zero at frequency = 1/(2*π*R1*C1)
Small C1 will get a higher gain and therefore faster transient response but easier to oscillate.
Larger R1 will also get a higher gain and phase lead and therefore faster transient response but easier to oscillate.
500) {this.resized=true; this.width=500; this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標(biāo)滾輪可自由縮放';this.style.cursor='hand'}" onclick="if(!this.resized) {return true;} else {window.open('http://u.dianyuan.com/bbs/u/53/95251178858944.bmp');}" onmousewheel="return imgzoom(this);">
Vfb = Iq * (1/(S*C1)), where S = 2*π*f
Vfb/Iq = 1/(S*C1)
Vfb/Iq = (1/S) * (1/C1)
Pole at frequency = 0 Hz
Gain = (1/C1)
Therefore, small C1 will get a higher gain and faster transient response but easier to oscillate.
2. Capacitor-Resistor Compensation
500) {this.resized=true; this.width=500; this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標(biāo)滾輪可自由縮放';this.style.cursor='hand'}" onclick="if(!this.resized) {return true;} else {window.open('http://u.dianyuan.com/bbs/u/53/95251178858989.bmp');}" onmousewheel="return imgzoom(this);">
Vfb = Iq * (1/(S*C1) + R1), where S = 2*π*f
Vfb/Iq = 1/(S*C1) + R1
Vfb/Iq = (1+S*R1*C1) / (1/S*C1)
Vfb/Iq = (1/S) * (1/C1) * (1+S*R1*C1)
Pole at frequency = 0 Hz
Gain = (1/C1)
Zero at frequency = 1/(2*π*R1*C1)
Small C1 will get a higher gain and therefore faster transient response but easier to oscillate.
Larger R1 will also get a higher gain and phase lead and therefore faster transient response but easier to oscillate.
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@jerome
1.SingleCapacitorCompensation[圖片]500){this.resized=true;this.width=500;this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標(biāo)滾輪可自由縮放';this.style.cursor='hand'}"onclick="if(!this.resized){returntrue;}else{window.open('http://u.dianyuan.com/bbs/u/53/95251178858944.bmp');}"onmousewheel="returnimgzoom(this);">Vfb=Iq*(1/(S*C1)),whereS=2*π*fVfb/Iq=1/(S*C1)Vfb/Iq=(1/S)*(1/C1)Poleatfrequency=0HzGain=(1/C1)Therefore,smallC1willgetahighergainandfastertransientresponsebuteasiertooscillate.2.Capacitor-ResistorCompensation[圖片]500){this.resized=true;this.width=500;this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標(biāo)滾輪可自由縮放';this.style.cursor='hand'}"onclick="if(!this.resized){returntrue;}else{window.open('http://u.dianyuan.com/bbs/u/53/95251178858989.bmp');}"onmousewheel="returnimgzoom(this);">Vfb=Iq*(1/(S*C1)+R1),whereS=2*π*fVfb/Iq=1/(S*C1)+R1Vfb/Iq=(1+S*R1*C1)/(1/S*C1)Vfb/Iq=(1/S)*(1/C1)*(1+S*R1*C1)Poleatfrequency=0HzGain=(1/C1)Zeroatfrequency=1/(2*π*R1*C1)SmallC1willgetahighergainandthereforefastertransientresponsebuteasiertooscillate.LargerR1willalsogetahighergainandphaseleadandthereforefastertransientresponsebuteasiertooscillate.
高人啊,講的太好了!
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@jerome
1.SingleCapacitorCompensation[圖片]500){this.resized=true;this.width=500;this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標(biāo)滾輪可自由縮放';this.style.cursor='hand'}"onclick="if(!this.resized){returntrue;}else{window.open('http://u.dianyuan.com/bbs/u/53/95251178858944.bmp');}"onmousewheel="returnimgzoom(this);">Vfb=Iq*(1/(S*C1)),whereS=2*π*fVfb/Iq=1/(S*C1)Vfb/Iq=(1/S)*(1/C1)Poleatfrequency=0HzGain=(1/C1)Therefore,smallC1willgetahighergainandfastertransientresponsebuteasiertooscillate.2.Capacitor-ResistorCompensation[圖片]500){this.resized=true;this.width=500;this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標(biāo)滾輪可自由縮放';this.style.cursor='hand'}"onclick="if(!this.resized){returntrue;}else{window.open('http://u.dianyuan.com/bbs/u/53/95251178858989.bmp');}"onmousewheel="returnimgzoom(this);">Vfb=Iq*(1/(S*C1)+R1),whereS=2*π*fVfb/Iq=1/(S*C1)+R1Vfb/Iq=(1+S*R1*C1)/(1/S*C1)Vfb/Iq=(1/S)*(1/C1)*(1+S*R1*C1)Poleatfrequency=0HzGain=(1/C1)Zeroatfrequency=1/(2*π*R1*C1)SmallC1willgetahighergainandthereforefastertransientresponsebuteasiertooscillate.LargerR1willalsogetahighergainandphaseleadandthereforefastertransientresponsebuteasiertooscillate.
又學(xué)到了一點,呵呵----
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