Hi guys,

I have been working on 3-stage LDO design.

I found that in the papers, the transfer functions have been simplified. But when I was trying to derive the transfer function, the denominator ends up in a very complicated form.

For example, most paper shows a very simple TF with three poles (1+s/Wp1)*(1+s/Wp2)*(1+s/Wp3).

But when I was trying to derive the transfer function with Matlab symbolic calculation, the denominator always becomes something like 1+ A*s + B*s^2 + c*s^3+D*s^4+E*s^5.

My question is, is there a better way to derive the TG?

Or, how to simplify the 5th-order poly into a closed-form like those in the paper.

Can anybody help?

Thanks a lot.!

raja.cedt wrote on Dec 14^th, 2012, 12:23am:


Hi Raj, as you recommend to "extend it for any order" - may I ask you if YOU have done something similar already?
I have some doubts regarding such an "extension".

Answer to wandola: I suggest to use a symbolic calculator to determine the pole distribution and to group the poles according to the classical scheme (s-sp).

Thanks to Raj and Buddy.

The schematic is shown below.

I am actually trying to derive the LDO TF from the paper titled as "Pole-Zero Analysis of Low-Dropout (LDO) Regulators: A Tutorial Overview" presented in 2012 Int' Conf on VLSI design.

In the paper, it is a three stage LDO with two feedback paths. I assume at least u will have 5 poles. And this is verified with MaTLAB.

To Buddy, I did use the MATLAB symbolic calculation. And the denominator is a 5-th order polynomial. But in the paper, the final TF is approximated to a 3-pole system. I didn't know how to simplify the TF.

If u have matlab on hand, u can try my script to see the TF.

%------- My script.
% ---- The voltage V1 is Vp in my sript, V2 is Vq in my script.
syms Z1 Z2 Z3
>> eq6='R1/(1+s*C1*R1)=Z1'
>> eq7='R2/(1+s*Cgs*R2)=Z2'
>> eq8='Ro/(1+s*Co*Ro)=Z3'
>> syms ic1 ic2 Vx Vy Vo Vp Vq
>> eq9='s*Cc1*(Vo-Vx)=ic1'
>> eq10='s*Cc2*(Vq-Vy)/(1+s*Cc2*Rc2)=ic2'
>> eq1='(gm1*Vs-gmBU*Vx+gmCG*Vy)*Z1=Vp'
>> eq2='(-gm2*Vp-ic2)*Z2=Vq'
>> eq3='(-gmp*Vq-ic1)*Z3=Vo'
>> eq4='ic1/gmBU=Vx'
>> eq5='ic2/gmCG=Vy'
>> [ic1 ic2 Vo Vp Vq Vx Vy Z1 Z2 Z3]=solve(eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,ic1, ic2,Vo,Vp,Vq,Vx,Vy,Z1,Z2,Z3)
Note: [ic1 ic2 Vo Vp Vq Vx Vy Z1 Z2 Z3]<-- the variables in the bracket must be in alphabetic order. Otherwise the results will be in wrong order and u don't which one is which.

Vo =
-(R1*R2*Vs*gm1*gm2*(gmCG + Cc2*s + Cc2*Rc2*gmCG*s)*(gmBU + Cc1*s + Cc1*Co*Ro*s^2 + Cc1*Ro*gmBU*s + Co*Ro*gmBU*s))/(gmBU*gmCG + Cc2*gmBU*s + Cc1*gmCG*s + Cc1*Cc2*s^2 + C1*R1*gmBU*gmCG*s + Cc2*R2*gmBU*gmCG*s + Cgs*R2*gmBU*gmCG*s + Cc2*Rc2*gmBU*gmCG*s + Cc1*Ro*gmBU*gmCG*s + Co*Ro*gmBU*gmCG*s + C1*Cc1*Cc2*R1*s^3 + Cc1*Cc2*Cgs*R2*s^3 + Cc1*Cc2*Co*Ro*s^3 + C1*Cc2*R1*gmBU*s^2 + C1*Cc1*R1*gmCG*s^2 + Cc1*Cc2*R2*gmCG*s^2 + Cc2*Cgs*R2*gmBU*s^2 + Cc1*Cgs*R2*gmCG*s^2 + Cc1*Cc2*Rc2*gmCG*s^2 + Cc1*Cc2*Ro*gmBU*s^2 + Cc2*Co*Ro*gmBU*s^2 + Cc1*Co*Ro*gmCG*s^2 + C1*Cc1*Cc2*R1*R2*gmCG*s^3 + C1*Cc2*Cgs*R1*R2*gmBU*s^3 + C1*Cc1*Cgs*R1*R2*gmCG*s^3 + C1*Cc1*Cc2*R1*Rc2*gmCG*s^3 + C1*Cc1*Cc2*R1*Ro*gmBU*s^3 + C1*Cc2*Co*R1*Ro*gmBU*s^3 + C1*Cc1*Co*R1*Ro*gmCG*s^3 + Cc1*Cc2*Cgs*R2*Rc2*gmCG*s^3 + Cc1*Cc2*Cgs*R2*Ro*gmBU*s^3 + Cc1*Cc2*Co*R2*Ro*gmCG*s^3 + Cc2*Cgs*Co*R2*Ro*gmBU*s^3 + Cc1*Cgs*Co*R2*Ro*gmCG*s^3 + Cc1*Cc2*Co*Rc2*Ro*gmCG*s^3 + C1*Cc2*R1*R2*gmBU*gmCG*s^2 + C1*Cgs*R1*R2*gmBU*gmCG*s^2 + Cc1*Cc2*R1*R2*gm2*gmCG*s^2 + C1*Cc2*R1*Rc2*gmBU*gmCG*s^2 + C1*Cc1*R1*Ro*gmBU*gmCG*s^2 + C1*Co*R1*Ro*gmBU*gmCG*s^2 + Cc2*Cgs*R2*Rc2*gmBU*gmCG*s^2 + Cc1*Cc2*R2*Ro*gmBU*gmCG*s^2 + Cc1*Cgs*R2*Ro*gmBU*gmCG*s^2 + Cc2*Co*R2*Ro*gmBU*gmCG*s^2 + Cgs*Co*R2*Ro*gmBU*gmCG*s^2 + Cc1*Cc2*Rc2*Ro*gmBU*gmCG*s^2 + Cc2*Co*Rc2*Ro*gmBU*gmCG*s^2 + C1*Cc1*Cc2*Cgs*R1*R2*s^4 + C1*Cc1*Cc2*Co*R1*Ro*s^4 + Cc1*Cc2*Cgs*Co*R2*Ro*s^4 + Cc2*R1*R2*gm2*gmBU*gmCG*s + C1*Cc1*Cc2*Cgs*Co*R1*R2*Ro*s^5 + C1*Cc1*Cc2*Cgs*R1*R2*Rc2*gmCG*s^4 + C1*Cc1*Cc2*Cgs*R1*R2*Ro*gmBU*s^4 + C1*Cc1*Cc2*Co*R1*R2*Ro*gmCG*s^4 + C1*Cc2*Cgs*Co*R1*R2*Ro*gmBU*s^4 + C1*Cc1*Cgs*Co*R1*R2*Ro*gmCG*s^4 + C1*Cc1*Cc2*Co*R1*Rc2*Ro*gmCG*s^4 + Cc1*Cc2*Cgs*Co*R2*Rc2*Ro*gmCG*s^4 + Cc1*R1*R2*Ro*gm2*gmBU*gmCG*gmp*s + C1*Cc2*Cgs*R1*R2*Rc2*gmBU*gmCG*s^3 + C1*Cc1*Cc2*R1*R2*Ro*gmBU*gmCG*s^3 + C1*Cc1*Cgs*R1*R2*Ro*gmBU*gmCG*s^3 + C1*Cc2*Co*R1*R2*Ro*gmBU*gmCG*s^3 + C1*Cgs*Co*R1*R2*Ro*gmBU*gmCG*s^3 + C1*Cc1*Cc2*R1*Rc2*Ro*gmBU*gmCG*s^3 + Cc1*Cc2*Co*R1*R2*Ro*gm2*gmCG*s^3 + C1*Cc2*Co*R1*Rc2*Ro*gmBU*gmCG*s^3 + Cc1*Cc2*Cgs*R2*Rc2*Ro*gmBU*gmCG*s^3 + Cc2*Cgs*Co*R2*Rc2*Ro*gmBU*gmCG*s^3 + Cc1*Cc2*R1*R2*Ro*gm2*gmBU*gmCG*s^2 + Cc1*Cc2*R1*R2*Ro*gm2*gmBU*gmp*s^2 + Cc2*Co*R1*R2*Ro*gm2*gmBU*gmCG*s^2 + C1*Cc1*Cc2*Cgs*Co*R1*R2*Rc2*Ro*gmCG*s^5 + C1*Cc1*Cc2*Cgs*R1*R2*Rc2*Ro*gmBU*gmCG*s^4 + C1*Cc2*Cgs*Co*R1*R2*Rc2*Ro*gmBU*gmCG*s^4 + Cc1*Cc2*R1*R2*Rc2*Ro*gm2*gmBU*gmCG*gmp*s^2)

%----------------
From above expression, I could simplify the denominator with the usual assumptions: Cc1, Cc2 >> C1, Cc1, Cc2 << Co, CgsPass, CgdPass << Vo, gm(x)*R(x)>>1.

However, i still got a 5-th order poly.

As suggested by Raj using the approximation, what happen if two poles are close and in the same order??

Thank you guys!

Let's open this for discussion.

I suggest to draw the root locus as in my opinion it gives more insight.