Question

A buck-boost converter has following specifications:

$V_d=12V,V_0=15V,I_0=250mA,L=150\mu H,C=470\mu Fand{f}_s=20kHz$

The value of $\Delta V_0$ (peak-to-peak) for the converter will be _______ mV (up to 2 decimal place).

Answer 1

For buck-boost diode converter:

Given, $V_d=12V,V_0=15V,I_0=0.25A,L=150\mu H,C=470\mu F,f_s=20kHz$

Assuming continuous conduction,

$\frac{V_0}{V_d}=\frac{D}{1-D}$

$\frac{15}{12}=\frac{D}{1-D}\Rightarrow \frac{1}{D}-1=\frac{12}{15}$

$\frac{1}{D}=\frac{12+15}{15}\Rightarrow D=\frac{15}{27}=0.556$

$I_{OB}=\frac{T_s{V}_0(1-D{)}^2}{2L}$

$=\frac{15\times (1-0.556{)}^2}{2\times 150\times {10}^{-6}\times 20\times {10}^3}=0.493A$

$R=\frac{V_0}{I_0}=60\Omega$

$As{I}_0<I_{OB},converterisoperatingindiscontinuousmode.$

For discontinuous mode of operation,

$I_{OB,max}=\frac{T_s{V}_0}{2L}=\frac{15}{2\times 150\times {10}^{-6}\times 20\times {10}^3}=2.5A$

$D=\frac{V_0}{V_d}\sqrt{\frac{I_0}{I_{OB,max}}}=\frac{15}{12}\sqrt{\frac{0.25}{2.5}}=0.395$

$\Delta V_0=\frac{1}{C}[(1-D\frac{V_d}{V_0})\frac{I_0}{f_s}+\frac{L{I}_0}{2R}]$

$=\frac{1}{470\times {10}^{-6}}[(1-\frac{0.395\times 12}{15})\times \frac{0.25}{20\times {10}^3}+\frac{150\times {10}^{-6}\times 0.25}{2\times 60}]$

$=18.86mV$