直流稳压电源控制设计PPT

2021-07-03 本文已影响0人 b6b299e7d230

参数计算

$f=10kHz$
$T_s=\frac{1}{f}=0.0001s$
$K_s=\frac{\Delta{U_d}}{\Delta{U_c}}=\frac{50}{5}=10$
$W_s=\frac{10}{0.0001s+1}$
$\alpha=\frac{U_n^*}{U_d}=\frac{5}{50}=0.1$
$\beta=\frac{U_i^*}{I_{dm}}=\frac{5}{2.5}=2$
$\frac{1}{T_{oi}}=(\frac{1}{5}\sim\frac{1}{10})\frac{1}{T_{pwm}}=\frac{1}{10}\frac{1}{T_{pwm}}=0.001s$
$W_{ACR}(s)\frac{10}{(0.0001s+1)}\frac{2}{(0.001s+1)}=\frac{K_1}{S(TS+1)}$
$T=T_s=0.0001s$
$\tau=T_{oi}=0.001$
$1.K_IT=0.5$
$K_I=5000$
$\frac{1}{3T_{pwm}}=3333<5000$
$2.K_IT=0.25$
$K_I=2500$
$K=0.125$
$W_{ACR}(s)=0.125+\frac{125}{S}$
$W_{CI}=\frac{1}{\beta}$ X $\frac{\frac{K_I}{S(T_sS+1)}}{1+\frac{K_I}{S(T_s+1)}}\approx\frac{1250}{s+2500}$
$a=0$
$b=4\times10^{-8}$
$c=4\times10^{-4}$
$W_C=2500\leq min(\sqrt\frac{1}{b},\sqrt\frac{c}{a})=0.5\times10^{-4}$
$W_{CI}\times\frac{0.1}{T_{oi}S+1}\times W_{AUR}(S)=\frac{K_V}{S(T_s+1)}$ = $\frac{2500}{S+5000}\times\frac{0.1}{0.001S+1}\times W_{AUR}(S)$
$W_{AUR}(S)=\frac{K(\tau s+1)}{\tau s}$
$\tau=T_{oi}=0.001$
$K_V=\frac{0.25}{0.0004}=625$
$K=\frac{K_V}{0.1\times2\times1000}=3.125$
$W_{AUR}=3.125+\frac{312.5}{S}$

直流稳压电源控制设计PPT

参数计算

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