Question

A Boost converter as shown in figure has a very large output capacitor filter. Output voltage waveform has been drawn below. Maximum average output current in ampere at the edge of continuous conduction is

• A. $\frac{T_s{V}_0}{16L}$
• B. $\frac{2T_s{V}_0}{27L}$
• C. $\frac{T_s{V}_0}{8L}$
• D. $\frac{T_s{V}_0}{9L}$

Answer 1

The correct option is B $\frac{2T_s{V}_0}{27L}$

At the boundary condition
Average inductor current is,
$I_{LB}=\frac{i_{Lpeak}}{2}=\frac{1}{2}\times \frac{V_s}{L}{t}_{on}$
$V_0$ is constant as C is very large.


In boost, $V_0=\frac{V_s}{(1-D)}$
$\therefore I_{LB}=\frac{1}{2}\times \frac{V_0(1-D)t_{on}}{L}$
$I_{LB}=\frac{T_s{V}_{0}D(1-D)}{2L}$ $(D=\frac{t_{on}}{T_s})$
$V_0{I}_0=V_s{I}_s$
$I_{OB}=(1-D)I_s$
$I_{OB}=(1-D)I_{LB}$
$I_{OB}=\frac{T_s{V}_0}{2L}(D^3-2D^2+D)$
For maximum output current
$\frac{d{I}_{OB}}{dD}=0$
$\therefore 3D^2-4D+1=0$
$\Rightarrow D=1,\frac{1}{3}$
$\therefore D=\frac{1}{3}$
$I_{OBmax}=\frac{T_s{V}_0}{L}\frac{1}{3}{(1-\frac{1}{3})}^2$
$I_{OBmax}=\frac{2T_s{V}_0}{27L}$