wandola
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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.
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