Hi,

 I am not exactly sure what the answer to question 1), but if Rz very large, it will looks open to me for the miller capacitor
2) the high frequency pole should be P3=-1/RzCi, Ci is the capacitor of first stage.

Judy

hi Tosei
           thanks for your response,but third pole will come because of 3 degree's of freedom...

hi Judy,
          thanks for your response,i know the location,i am able to find first two poles intuitively but i am not able to find 3rd pole.so with out lot of math can you please tell me..

Thanks,
Rajasekhar.

hi Aaron,
            your answer is absolutely correct but in case of no nulling resistor, but if nulling resistor is there then you can distribute whatever voltage you want according to kvl remaining voltage will come across Rz.
i hope u understand..

Thanks,
rajasekhar.

raja.cedt wrote on May 12^th, 2009, 8:11pm:


Hi Rajasekhar,

Thanks for your clarifications. However I still disagree about the third pole: having a third degree of freedom (which is true) does not necessarily mean you have a a three pole system. Actually such extra degree of freedom is the one that allows to cancel the RHP by means of the nulling resistor. In other words the third degree of freedom is a capacitor in series  with the signal, i.e. is places a zero, while the other two nodes are shunting the signal, thus generate poles.
As an example of a system with two degree of freedoms - but one pole - just consider the passive filter attached. The transfer function shows one pole and one zero, and clearly two degrees of freedom.

Going back to the nulling resistor case, for me the effect of the nulling resistor is two-fold:
1) Affects the location of the first and second poles
2) Brings the zero from the RHP to the LHP (this is the third degree of freedom that does not exist without Rz)

I might be missing something but I do not see how the third poles is generated here.

Regards
Tosei

Hi Rajasekhar,

What you say is true and I might have created some confusion with my last post.
I think the question is: is the nulling resistor creating a third degree of freedom in the circuit, or still keeps the third complex impedance (the Miller cap) state as a function of the other two.
As is suggested in Aaron's reference, withouth nulling resistor the Miller cap voltage-current state is fully determined by the voltage/current conditions on the other two reactive components, therefore there are only two linearly independent derivatives (capacitor currents) in your system. From algebra point of view: your system is described by three differential equations, one of which is linearly dependent --> second order system

It looks to me that adding the nulling resistor does not change this situation: still the other two caps conditions in the system set the voltage accross Rz and Cm, and therefore the current is also set and cannot be freely changed. Thus it looks to me like the case without Rz.
In other words the differential equation describing the state of the Miller cap with or without Rz is linearly dependent and can be fully described with the diff eqs for the other two caps.

Please check: "Analysis of Multi-stage amplifiers-Frequency Compensation" Ka Nang Leung and Philip K. T. Mok, IEEE TRANSACTIONS ON CIRCUITS AND SYSTEMS—I: FUNDAMENTAL THEORY AND APPLICATIONS, VOL. 48, NO. 9, SEPTEMBER 2001

This paper describes different compensation approaches, among which you obviously will find the two-stage miller amplifier. The full transfer function with nulling resistor is there and the system is a second order one.

Regards
Tosei

Hi Rajasekhar,

Have you arrived to any conclusion on this?
I cannot find in my Gray and Meyer book the reference you are indicating...could you tell me which page it is located at?

Thanks
Tosei

hi Tosei,
           if possible can you read pole splitting post  here and give me your answer ,because i am very keen to know that answer for that and in my profession i did many times miller compensation.

Thanks,
Rajasekhar.