Chapter 5 The Discontinuous Conduction ModeChapter 5 The Dis

Chapter 5 The Discontinuous Conduction Mode

这一章讲The Discontinuous Conduction Mode. 考虑电感ripple很大, 负载电流很小, 电感电流为零的场景. DCM会导致conversion ratio M随着负载变化, 输出阻抗变大.

对于buck converter 电感纹波 (1/2 Ipeak-peak)

\Delta i_{L}=\frac{V_{g}-V}{2L}DT_{s}

随着R增大, I减小, 当I= $\Delta iL$ 时系统进入DCM.

I > $\Delta iL$ for CCM

I < $\Delta iL$ for DCM

可以用K或者R进行DCM和CCM判断评价

K < Kcrit(D) for DCM, K > Kcrit(D) for CCM

K=2L/(R*Ts). K is a measure of the tendency of a converter to operate in the discontinuous conduction mode.

R < Rcrit(D) for CCM, R > Rcrit(D) for DCM

R为负载电阻

Buck, boost, buck-boost K和Rcrit如下图所示

K > Kcrit(D) or R < Rcrit(D) for CCM

K < Kcrit(D) or R > Rcrit(D) for DCM

5.2 Analysis of the Conversion Ratio M(D, K)

在计算DCM的电压转换比我们依然采用chapter 2的电感伏秒平衡和电容电荷平衡. 另外假设输出电压保持稳定,忽略ripple, 但是不能忽略电感电流的纹波!

考虑Buck Converter DCM场景

列出电感电压波形

列出电感电流和电容电流波形

\left\langle v_{L}(t) \right\rangle=D_{1}(V_{g}-V)+D_{2}(-V)+D_{3}(0)=0

\left\langle i_{L}(t) \right\rangle=\frac{V}{R}

\left\langle i_{L}(t) \right\rangle=\frac{1}{2}i_{pk}(D_{1}+D_{2})=(V_{g}-V)(\frac{D_{1}T_{s}}{2L})(D_{1}+D_{2})=\frac{V}{R}

可得Conversion Ratio M(D1,K)=V/Vg

M=\frac{V}{V_{g}}=\frac{2}{1+\sqrt{1+\frac{4K}{D^{2}_{1}}}}

where

K=\frac{2L}{RT_{s}}

Valid for K<Kcrit

因此

DCM会导致输出电压V变大(相比于CCM).

Boost converter在DCM和buck类似

I > $\Delta$ iL for CCM I < $\Delta$ iL for DCM

K > Kcrit(D) for CCM K < Kcrit(D) for DCM

K=\frac{2L}{RT_{s}}

Kcrit(D)=D*(1-D)^2

利用电感伏秒和电容电荷平衡,可得

\frac{V}{V_{g}}=M(D_{1},K)=\frac{1+\sqrt{1+\frac{4D^{2}_{1}}{K}}}{2}

Buck, boost, buck-boost的K和conversion ratio M总结如下

进入DCM后, 输出电压由负载决定, 因此导致converter输出impedance变大.