A buck converter feeding a variable resistive load is shown in the figure. The switching frequency of the switch S is \(100\;kHz\) and the duty ratio is 0.6. The output voltage \({V_0}\) is \(36V\). Assume that all the components are ideal, and that the output voltage is ripple-free. The value of \(R\) (in Ohm) that will make the inductor current (\({i_L}\)) just continuous is _________.

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GATE EE 2015 Official Paper: Shift 2

CT 1: Ratio and Proportion

2672

10 Questions
16 Marks
30 Mins

__ Concept__:

current waveform for a buck converter

**Boundary of continuous & discontinuous conduction mode**

**For continuous conduction,**

I_{L(max)} = I_{L} + \(\frac{Δ I_L}{2} \) ---(1)

IL(min) = IL + \(\frac{Δ I_L}{2} \) ---(2)

__ Calculation__:

given information,

f = 100 kHz

D = 0.6

V_{0} = 36 V

V_{s} = 60 V

At the boundary of continuous conduction

I_{L(min) }= 0

From eq(2)

\(⇒ 0 = I_L - \frac{Δ I_L}{2}\)

\(\rm ⇒ I_L = \frac{Δ I_L}{2}\)

The inductor ripple current is given as,

\(\rm Δ I_L = \frac{V_0(V_S - V_0)}{fLV_s}\)

\(\rm ⇒ Δ I_L = \frac{36(60-36)}{100 \times 10^3 \times 5 \times 10^{-3} \times 60}\)

⇒ ΔI_{L} = 28.8 mA

∵ \( I_L = \frac{\Delta I_L}{2}\)

\(\rm \Rightarrow I_L = \frac{28.8}{2} = 14.4 mA\)

\(\Rightarrow \frac{V_0}{R} = 14.4 mA \left( \because I_L = \frac{V_0}{R} \right)\)

\(\Rightarrow R = \frac{36}{14.4 \times 10^{-3}}\)

∴ R = 2500 Ω