matlab code 研究一下吧

mayluli1981 · 發表於 2008-11-6 22:18:46

Vg=28 Vo=15 Ro=3
H=5/15 分壓
Vref=5
% VRM
s = tf('s');
% Input voltage Vin
Vin=28
% Output voltage Vo
Vo=15
% LOAD R
RL=3
% PWM operating frequency Fsw
Fsw=100e3
% PWM operating period Ts
Ts=1/Fsw
% duty cycle D
D=Vo/Vin
% Inductance L
L=50e-6
% Output capacitance
Co=500e-6
%power stage Gvd
%Gvd=(Vin*((s*Co*RL*rc)+RL))/((s^2*L*Co*(RL+rc))+(s*(L+(Co*rc*rL)+Co*RL*(rc+rL)))+RL+rL)
Gvd=(Vin*RL)/((s^2*L*Co*RL)+s*L+RL)
figure(1)
margin(Gvd)
F0=1/(2*pi*sqrt(L*Co))
%Fesr=1/(2*pi*Co*rc)
Fc=Fsw/20
% 分壓比
H=1/3
% PWM Vm
Vm=4
% uncompensation loop gain Tu0
Tu0=H*Vin/Vm
% 補償zero Fz
Fz=Fc*sqrt((1-sin (52*(pi/180)))/(1+sin (52*(pi/180))))
% 補償pole Fp
Fp=Fc*sqrt((1+sin (52*(pi/180)))/(1-sin (52*(pi/180))))
Wz=2*pi*Fz
R1=2e3
C1=1/(R1*Wz)
Wp=2*pi*Fp
R3=R1/(Wp*R1*C1-1)
% compensation gain Gc0
Gc0=(((Fc/F0)^2)/Tu0)*sqrt(Fz/Fp)
R2=Gc0*(R1+R3)
% OPA gain:A;OPA open loop gain:A0 ; dominant pole

; compensator close feedback gain:Gcl
A0=1
p=10e3*2*pi
A=A0*p/(s+p)
Gc=(A0*p*R2+s*A0*p*R1*R2*C1)/(s^2*(R1*R2*C1+C1*R1*R3)+s*(A0*p*R1*R3*C1+p*R1*R3*C1+p*R1*R2*C1+R1+R2+R3)+(A0*p*R1+A0*p*R3+p*(R1+R3+R2)))
figure(2)
bode(Gc,A)
% total loop gain T
T=Tu0*Gc0*(1+s/Wz)*RL/((1+s/Wp)*((s^2*L*Co*RL)+s*L+RL))
% total colse loop gain Gcl
Gcl=(Gc*(1/Vm)*Gvd)/(1+Gc*(1/Vm)*Gvd*H)
figure(3)
bode(Gvd,Gcl,Gc,A,T)