关于Folded-cascode中，input-cascod管子作用的直观分析！

aqishisi

呵呵，如果我理解的不透彻，估计唐长文老师也中枪了。。。楼主我真心劝你仿一下，加深一下对电路的理解，好好思考下，不是你想得那么简单的，到时希望你别说仿真器也错了。。。

vdslafe

同意楼上的，求求楼主您防一下吧。把结果贴出来。

feynmancgz

回复 53# JoyShockley


   This post is so hot. I come to say something about my opinion on this.
   I think actually, you and aqishisi are talking about totally different things.
   The impedance at the folded node is high when the frequency is low ? Yes, of course.
   Then the effect of Miller effect of the Cgd of the input transistors is huge, of course!
   Does this miller has influence on the amplifier itself ? Yes, kind of. but it has huge effect on the previous circuit which drives this amplifier. This is what you want to say, right ?
   Ok, Cgd is very small, its effect is really huge ? Yes, maybe, sometimes.
   BUT, BUT,
   The pole at the folded node is located at low frequency because of its high resistance calue? NO, I don't think so.
   The pole depends both on the capacitance and resistance.
   The capacitance at the output of this amplifier is normally large, say 5pf, the resistance at this node is also very large, so this pole is located at very very low frequency.
   BUT how about the pole at the folded node ? The capacitance ? so small. The resistance ? about at the same level as ro, much smaller than gm*ro*ro at the output node. This pole is located at much high frequency compared to the pole at the output node.
   When you calculate the impedance at some node, you always take all the capacitances and resistances into account. ok, ignore most of them properly.
   So, when the impedance at the output node starts to decrease ? It's the time when the first pole starts to have effect, from then on, the impedance at the output will decrease with 20dB/dec. So, at the same time the impedance at the folded node also decrease very fast, also 20dB/dec at the beginning, then the speed of the decreasing will becomes slow. so when the pole related to the folded node wants to have effect on the circuit, it actually becomes larger. That's the reason everybody says the resistance looking into the source of M8 and M9 is only 1/gm.
   The pole for each node will not change when the frequency goes high ? No, It changes. The only pole that will not change is the first pole, it is always there. That's the reason people call it "the dominant pole". It's really dominant, it has huge effect on the other poles. Does other poles have effect on the dominant pole ? Yes, they do. but their effects can be ignored.
   Sum up, why the resistance looking into the source of M8 and M9 is only 1/gm ? The only reason is the we have capacitors at the output node, and it's huge.    The resistance looking into the source of M8 and M9 is about ro ? Yes, it' true. you are right ! But it's only valid when the frequency is lower than the first pole, then it decreases.

   Another thing I should tell you is that why the professors from Berkeley says we want to avoid miller effect ? Because it decreases the frequency of the pole at the output of the previous circuit. This pole may be the dominant pole for the previous circuit, maybe even becomes the dominant pole for the whole chain.

feynmancgz

回复 53# JoyShockley


   Then I want to say something about Miller effect.
   The analysis of original Miller effect is very simple. But when we put into circuits, especially when other capacitors comes in, the analysis of Miller effect is very complex. That's why no textbook talks about this, all of them still use the very original simple model to analyze it. In my memory, at least there are three different models to analyze Miller effect at different frequencies. It's very complex, I won't say more about this.

firevortex

回复 46# kboost


   楼主关于那个方向的阻抗计算并没有错，你可以把59页M1漏端再串联一个电阻做推导，看看关系，就像主楼给出的那张图一样。但是我并不认为那个结点会因为加入cascode管而改变零极点分布，感觉楼主看的方向错了，应该从M6漏端看，而不是M1漏端看。

firevortex

回复 63# feynmancgz


   顶这个

lonerinuestc

我觉得楼主是自己意淫了一种情况：cascode处极点频率比输出极点频率还要低。理论上确实是可能的，但是实际中压根不会出现。

    根据楼主的思路，我们还能意淫出另外一种情况：两级运放的负载电容超级大的时候，这时候连频率补偿都不用了，不是吗？这在理论上确实是可能的，但实际中几乎不会给你这样发挥的机会。

     其实我们刚学电路的时候也会和楼主一样有这些奇怪的想法，只是体现的方式没有这么偏激罢了。

JoyShockley

回复 61# aqishisi


   你的疑惑是为什么在传输函数和零极点分析时，会看不到这个高阻，对吧。
   那个时候计算这个点的极点的时候的确是gm/Cp（Cp是该点的对地寄生电容)，我完全承认的结论；

  这个时候有两种理解：

1. 看这个极点的时候，已经在较高频率了，我们简化分析，认为输出极点的电容短路（因为高频和大电容），得到1/gm； 同样，和我以前说的没有差别，你们想看到1/gm只能是高频下的一种简化。而绝对不是你说：   

“既然是恒流源，其电流就是恒定的，小信号电流是不可能从这上面走的，这个电流的流向和频率没关系，和电容的大小也没关系。如果楼主不信，可以仿一下就知道了。其实，你有点异想天开了。”

JoyShockley

回复 67# lonerinuestc


   请不要随意指责别人偏激，这样我认为你也很偏激。

JoyShockley

回复 63# feynmancgz


   感谢版主的回复，你说的很对，那一点是高阻，但是在做零极点分析的时候，将输出电容短路，哪一点近似1/gm； 所以，前提依然是高频，大电容的条件