Question

A step up converter shown in figure has maximum power output of 120 W and switching frequency is 50 kHz. For the converter operating in discontinuous current conduction mode and assuming capacitance to be very large, Maximum value of inductor is

• A. $27\mu H$
• B. $54\mu H$
• C. $3\mu H$
• D. $9\mu H$

Answer 1

The correct option is D $9\mu H$

For step up converter, $V_0=\frac{V_s}{(1-\alpha )}$
$\alpha =(1-\frac{V_s}{V_0})$
Range of $\alpha$ for $V_B=(12-36)V$
$\alpha =(0.25-0.75$)
Average output current$I_{OB}$ is minimum for $\alpha =0.75$
$\therefore$ Maximum value of $L$ for discontinuous condition is when
$\alpha =0.75$
$\Delta I_L=\frac{V_0\alpha (1-\alpha )}{fL}$
$\Delta I=\frac{V_0\alpha (1-\alpha {)}^2}{fL}$

At boundary condition,
$\frac{\Delta I}{2}=I_{OB}$
$\Delta I=2I_{OB}$
$P_{max}=120W$
$I_{OB}\times V_0=120$
$I_{OB}=\frac{120}{48}$
$I_{OB}=2.5A$
$\therefore 2\times 2.5=\frac{48\times 0.75\times (1-0.75{)}^2}{50\times {10}^3\times L}$
$L=9\mu H$