GATE Electrical Engineering (EE) 2019 Solved Paper

Question : 42 of 65

Marks: +1, -0

In a DC-DC boost converter, the duty ratio is controlled to regulate the output voltage at

48 {V}

. The input DC voltage is

24 {V}

. The output power is

120 {W}

. The switching frequency is

50 {kHz}

. Assume ideal components and a very large output filter capacitor. The converter operates at the boundary between continuous and discontinuous conduction modes. The value of the boost inductor (in

µ {H}

) is

Your Answer:

Validate

Solution:

DC-DC boost converter

{V}_{0} \;\; =48 {V}

{V}_{ {s}} \;\; =24 {V}

{V}_{0} \;\; =\;{ {V}_{ {s}}}/{1- {D}} ⇒ 48=\;{24}/{1- {D}}

{D}=0.5

{T}=\;{1}/{ {f}_{ {s}}}=\;{1}/{50 × 10^{3}}=20 µ {s}

output power

( {P})=120

watts

{f}=50 {kHz}

Inductor operates at the boundary of continuous and discontinuous conduction In boost converter,
average value of supply current

( {I}_{ {SA}})=

Average value of inductor current

( {I}_{ {LA}})

{I}_{ {SA}}= {I}_{ {LA}}=\;{120}/{24}=5 {Amps}

{I}_{ {LA}}=\;{0+ {I}_{ {Lmx}}}/{2} ⇒ {I}_{ {Lmx}}=2 × {I}_{ {LA}}=2 × 5=10 {Amps}

current ripple

((∆ {I}_{ {L}})= {I}_{ {Lmx}}-0= {I}_{ {Lmx}}=10

∴ \;{ {V}_{ {s}}}/{ {L}} × {DT}=10 ⇒ \;{24}/{ {L}} × 0.5 × 20=10

{L} \;\; =\;{24 × 0.5 × 20}/{10} µ {H}

=24 µ {H}

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