| 
 | 
 
 
发表于 2012-9-14 23:30:58
|
显示全部楼层
 
 
 
回复 4# feynmancgz  
 
诚心请教版主, 
   I'm in school so that I cannot type in Chinese. I'm sorry. 
 
   Intuitively, feedback signal will influence the poles and  
   feedforward  will influence the zeros. But the most  
   important ideal is to identify the feedback signals and  
   feedforward signals in a system.   1)关于如何“Intuitively”去找零点,理论上就是让输出为零的频率; 
        a)RC串联,有时会引入零点为0频率或无穷大频率; 
 
        b)差分结构的射极耦合端,RC并联呈现无穷大电阻会给共模增益引入零点,CMRR一个极点; 
        c)两个节点间两条支路电流抵消产生零点,即版主说的前馈结构; 
        d)消灭前馈通路可以去掉零点,比如Miller补偿的调零电阻结构和前馈隔断结构;如何较好的理解反馈影响极点,电路结构变了,极点会变(呃,是没错),或是? 
        ....... 
        是否有更直观的指导原则,或是这些实例的积累就是直观理解?Intuitively 
   2)极点就是因为电容高频时的低阻抗让增益减小,与传输函数分母为零导致无穷大有些差别,呵呵..... 
      或是其它直观的解释? 
       极点就归结为节点对地的电容和等效电阻。 
 
   For Gmf in 9.33, the signal will directly feedforward  
   from input to output, so it's just feedforward signal. 
   It will only effect the zeros. 
   1)没问题,令Vo为零,三路电流之和为零可以得到零点。没有其它零点了么?对于9.33确实如此。 
   或是大原则就是这样,如9.34,5路电流之和为零的方程包含了两个零点。 
   2)两级放大器,一级一级找零点,如常规二级放大器中的电流镜零点和米勒电容零点; 
    ?会不会有遗漏。正确的传输函数不会有遗漏,想更直观物理上的零点。 
   For Gmf1 in9.34, because Cm2 close the loop, the signal  
   from Gmf1 will also influence the poles. As there is another 
   loop in the Cm2 loop, its effect on the poles becomes  
   complicated as you can see from the fomulas. In addition, 
   since Vo is much stronger than V1 and V2, so the feedback 
   signal from Cm2*V0 is dominant, so the dominant pole is  
   still determinated by the outer loop, i.e. Cm2 loop.  
 
   There is another way to understand this. 
   In 9.33, the signal from Gmf is Gmf*V1, it definitely will 
   increase the DC gain, it seems that it will feedback to influence 
   the poles, but it will not. Why? 
   because when V1 becomes stronger, the signal Gmf*V1 becomes 
   stronger, but the signal V2=Gm1*R1*V1 will also becomes stronger 
   which means feedforward effect of V2 becomes much stronger than 
   the so called "feedback effect" of V1. But when I close the loop 
   through Cm2 as indicated in 9.34, the feedback signal Gm1*V1 will  
   feedback through Cm2 and amplified by Gm1 which makes this  
   signal becomes strong enough to effect the poles. 
   But I think this explanation is not absolutely correct, it's just a kind  
   of intuitive way to understand the effects of signals. You may be  
   confused about this explanation, it's OK, you can just ignore it. 
   版主咋得到这个解释的?共享一下,多年修道所得,或是有南山捷径? 
 
   Another thing I want to say is that Miller effect is a Two-port feedback 
   effect, I call the feedforward effect of Miller capacitor "parasitic effect", 
   because we have some techiniques to cut the feedforward path.no one  
   else name it like this, so you can also ignore this. What I want to say 
   actually is that in Miller effect, what will influence the poles is the signals 
   inside the two terminals of Miller capacitor. |   
 
 
 
 |