The Designer's Guide Community Forum

The Designer's Guide Community Forum
https://designers-guide.org/forum/YaBB.pl
Design >> Analog Design >> Miller compensation
https://designers-guide.org/forum/YaBB.pl?num=1242094210

Message started by raja.cedt on May 11^th, 2009, 7:10pm

Title: Miller compensation
Post by raja.cedt on May 11^th, 2009, 7:10pm

hi,
i think this is very basics doubt, can any one please tell me what is the impact of zero nulling resistor in two stage miller compensated op amp means
1. How it effects the already existing poles i know that if Rz=very large no pole splitting
2.And it introduce third pole, how to estimate that?

Thanks,
Rajasekhar.

Title: Re: Miller compensation
Post by circuit_cook on May 12^th, 2009, 3:54pm

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

Title: Re: Miller compensation
Post by thechopper on May 12^th, 2009, 6:14pm

I agree with Judy on #1. IF Rz tends to ∞ Miller cap will be disconnected.
As for #2 it does not introduce a third pole. The ideal value for Rz is 1/gm2, where gm2 is the 2nd stage transconductance. For such value the RHP is cancelled out. IF larger values for Rz are used (usually the case to account for process tolerances) then the zero in the overall transfer function will be located in the LHP, which should not significantly affect stability.
Rz should not be made too large though because of 1. It is recommended to make it Rz < Rout1/10, where Rout1 is the output impedance of the first stage in order to still have an effective pole-splitting.

Cheers
Tosei

Title: Re: Miller compensation
Post by raja.cedt on May 12^th, 2009, 7:40pm

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.

Title: Re: Miller compensation
Post by aaron_do on May 12^th, 2009, 8:04pm

Hi Rajasekhar,

Quote:

third pole will come because of 3 degree's of freedom...

According to T. H. Lee, "Because the capacitors form a loop, KVL tells us that the voltage across any two automatically determines that across the third. Hence, there are actually only two degrees of freedom and the transfer function is therefore indeed of the second
order" Here he is talking about the output capacitance of the first stage, the miller capacitance and the output capacitance of the second stage.

Here is a link to the document (pg. 2 paragraph 2)

http://www.ece.arizona.edu/~ece304/Pdf/Lee_PoleSplit.pdf

cheers,
Aaron

Title: Re: Miller compensation
Post by raja.cedt on May 12^th, 2009, 8:11pm

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.

Title: Re: Miller compensation
Post by aaron_do on May 12^th, 2009, 8:28pm

Thanks. I didn't think of that :)

Aaron

Title: Re: Miller compensation
Post by thechopper on May 13^th, 2009, 6:39pm

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

hi but if nulling resistor is there then you can distribute whatever voltage you want according to kvl remaining voltage will come across Rz.

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

Title: Re: Miller compensation
Post by raja.cedt on May 13^th, 2009, 8:43pm

hi Tosei,
this example is fine,but for two stage op amp i derived the transfer functions, in that three poles has come.But with your answer i got one basic doubt, i thought no of poles is equal to no of degree of freedom is it true?

Thanks,
rajasekhar.

Title: Re: Miller compensation
Post by thechopper on May 14^th, 2009, 6:33pm

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

Title: Re: Miller compensation
Post by raja.cedt on May 14^th, 2009, 7:48pm

hi Tosei,
thanks for your reply, i will go through the paper and let you know.But if you have gray and mayer book just read in that he has given third pole.

Thanks,
Rajasekhar.

Title: Re: Miller compensation
Post by thechopper on May 29^th, 2009, 7:51pm

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

Title: Re: Miller compensation
Post by raja.cedt on May 29^th, 2009, 10:39pm

hi Tosei,
thanks for your interest.In Indian addition gary mayer you can find page no 650 or you can refer 'considerations for fats settling operational amplifiers by DAVID J ALLSTOT'

'https://ccnet.stanford.edu/cgi-bin/course.cgi?cc=ee315a&action=handout_download&handout_id=ID124201514713595' refer handout 15 in this site.

Thanks,
Rajasekhar.

Title: Re: Miller compensation
Post by raja.cedt on May 30^th, 2009, 6:57am

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.