关于格雷书上密勒补偿权威性的质疑。

feynmancgz

回复 1# jxjxhwx


   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.

   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.
   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.

   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.

feynmancgz

回复 5# jxjxhwx


   the widely used way to model Miller capacitor is not accurate
   check the file attached, this paper is very interesting

feynmancgz

回复 7# jxjxhwx


   Gray is right.
   It's hard to say that's a bug
   It's just about different approximations
   In most cases, very accurate approximations is not useful
   as simple as possible, but not simpler. hehe~

pusher_yxg

遇到高手了.....