在 voltage mode 的 DC-DC 中，由于功率级中的电感和电容会在环路中引入一对极点，从而会导致环路相位的裕度的问题。为保持环路的稳定，通常的方法是在环路中加上 Type-III 补偿。

具体的 type-III 补偿的计算或者说补偿电路中的零极点的设置，很多论文中都有不同的方法，这里主要参考的是这篇文章中具体通过 ESR 零点位置来选择两种不同零点位置设置的计算方法。

具体的计算 voltage-mode buck 变换器 Type III 补偿的 matlab 脚本如下：

% calculate the type III compensation network for voltage mode buck converter
%
%                    -------------------Cc2-----|   
%                    |            |             |  
%                    |   | \      |--Rc1--Cc1---|                                                  
%   ---------Rf1---------|   \                  |
%       |            |   |     \ _______________|____
%       |--Rf3--Cf3--|   |     /
%                        |   /
%                        | /
%  Rf1>>Rf3, Cc1>>Cc2
%

clc;
clear;

% setting the main parameters of the buck converter
vin=5;                  % 3.3v or 5v input voltage
vo=1.2;                 % 1.2v output voltage
d=vo/vin;               % the duty cycle in ccm
fs=3e5;                 % the switching frequency (Hz)
imax=0.5;               % the max load current (A) 
L=22e-6;                % the inductance in the LC filter
C=47e-6;                % the capacitance in the LC filter
ESR=0.07;               % the esr of C
vm=1.25;                % the peak-peak value of the sawtooth wave
delt_I=(vin-vo)/L*d/fs; % the peak-to-peak current ripple (A)
R=vo/(0.5*delt_I);      % the equivalent load resistance at the CCM/DCM boundary
%R=1e6;

% calculate the loop-gain without the compensation
s=tf('s');
H1=(vin/vm)*(1+s*ESR*C)/(1+s*(L/R+ESR*C)+s^2*L*C);
figure(1);
margin(H1);
grid on;

% calculate the pole/zero of the power-stage
fc=0.1*fs;              % the cross over frequency
fesr=1/2/pi/ESR/C;      % the ESR zero
fpo=1/2/pi/(sqrt(L*C)); % the LC corner frequency

% choose method 1 or 2 based on the fesr 
if fesr < 0.5*fs          
    method=1;           % calculate the pole/zero based on method 1
    fz1=0.75*fpo;
    fz2=1*fpo;
    fp2=1*fesr;
    fp3=0.5*fs;
else    
    method=2;            % calculate the pole/zero based on method 2
    theta=60;            % the boosted phase
    %k=(1+sind(theta))/(1-sind(theta));
    k=(tand(theta/2+45))^2;
    fz2=fc/sqrt(k);
    fp2=fc*sqrt(k);
    fz1=0.5*min(fz2,fpo);
    fp3=0.5*fs;
end

% calculate the R/C of the compensation network
Rc1=1.0e5;              % choose 100k as a start point
Cf3=(2*pi*fc*L*C/Rc1)*(vm/vin);
Cc1=1/(2*pi*fz1*Rc1);
Cc2=1/(2*pi*fp3*Rc1);
Rf3=1/(2*pi*fp2*Cf3);
Rf1=1/(2*pi*fz2*Cf3)-Rf3;

%% calculate the loop-gain with the compensation
Gc=(1+s*Rc1*Cc1)*(1+s*(Rf1+Rf3)*Cf3)/Rf1/(Cc1+Cc2)/s/(1+s*Rc1*Cc1*Cc2/(Cc1+Cc2))/(1+s*Rf3*Cf3);
H2=H1*Gc;
[Gm,Pm,Wg,Wp] = margin(H2);
figure(2);
margin(H2);
grid on;

fprintf(1,'Type III compensation, method %g. \n', method);
fprintf(1,'Phase Margin= %g deg @ %gk Hz\n', Pm,Wp/2/pi/1e3);
fprintf(1,'calculated R/C of the compensation network: \n ');
fprintf(1,'Rf1=%gk ohm\n Rf3=%gk ohm\n Cf3=%gpF\n Rc1=%gk ohm\n Cc1=%gpF\n Cc2=%gpF\n', Rf1*1e-3, Rf3*1e-3,Cf3*1e12, Rc1*1e-3, Cc1*1e12, Cc2*1e12);

可以看到补偿前后的环路特性