A boost converter has following parameters:
$V_{s} = 20 V, D = 0.6, L = 100 μ H$
$R = 50 Ω, C = 100 μ F, f = 15 k H z$

If the converter is operating in discontinuous mode of operation, then time interval for which inductor current remains zero in each switching time period of converter will be ___________ $μ s e c$ .

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Solution

For discontinuous mode of operation of boost converter,

As average inductor voltage is zero and average current in diode is same as the load current, we can write,
$V_{s} D T + (V_{s} - V_{0}) D_{1} T = 0$

$W h e r e D_{1} i s t h e t i m e i n t e r v a l w h e n i n d u c t o r c u r r e n t d e c r e a s e s$

$V_{0} = V_{s} (\frac{D + D_{1}}{D_{1}}) . . (i)$

Average current in diode,
$I_{D} = \frac{1}{T} (\frac{1}{2} \times I_{m a x} D_{1} T) = \frac{1}{2} I_{m a x} D_{1} (i i)$

We also know,

$V_{s} = L \frac{Δ I_{L}}{Δ t}$
$o r I_{m a x} = \frac{V_{s} D T}{L} (i i i)$

Substituting equation (iii) in equation (ii),
$I_{D} = \frac{1}{2} (\frac{V_{s} D T}{L}) D_{1}$

As diode current is equal to load current,
$\frac{1}{2} (\frac{V_{s} D T}{L}) D_{1} = \frac{V_{0}}{R}$
$o r, D_{1} = (\frac{V_{0}}{V_{s}}) (\frac{2 L}{R D T})$

Substituting equation (iv) value in equation (i),

$V_{0} = V_{s} (\frac{D}{D_{1}} + 1)$
$V_{0} = V_{s} ⎡ ⎣ \frac{D}{\frac{V_{0}}{V_{s}} \frac{2 L}{R D T}} + 1 ⎤ ⎦$

$o r, {(\frac{V_{0}}{V_{s}})}^{2} - \frac{V_{0}}{V_{s}} - \frac{D^{2} R T}{2 L} = 0$

$∴ \frac{V_{0}}{V_{s}} = \frac{1}{2} (1 + \sqrt{1 + \frac{2 D^{2} R T}{L}})$

$∴ V_{0} = \frac{20}{2} [1 + \sqrt{1 + \frac{2 (0.6)^{2} (50)}{100 \times 10^{- 6} \times 15000}}] = 60 V$

now using equation (i)

$V_{0} = V_{s} \frac{(D + D_{1})}{D_{1}}$

$60 = 20 (\frac{0.6 + D_{1}}{D_{1}})$

$o r, 3 = \frac{0.6}{D_{1}} + 1$

$\frac{0.6}{D_{1}} = 2$
$o r D_{1} = 0.3$

$∴ c u r r e n t i n i n d u c t o r d e c r e a s e s f o r 0.3 T w h e r e T i s t i m e p e r i o d a n d i n c r e a s e f o r 0.6 T .$
$∴ t i m e i n t e r v a l f o r z e r o i n d u c t o r c u r r e n t,$

$= T - 0.6 T - 0.3 T$
$= 0.1 T = 0.1 \cdot \frac{1}{15000} = 6.67 μ s e c$

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