GATE Electrical Engineering (EE) 2020 Solved Paper

Question : 51 of 65

Marks: +1, -0

In the dc-dc converter circuit shown, switch

{Q}

is switched at a frequency of

10 {kHz}

with a duty ratio of

0.6

. All components of the circuit are ideal, and the initial current in the inductor is zero. Energy stored in the inductor in

{mJ}

(rounded off to 2 decimal places) at the end of 10 complete switching cycles is _____________

Solution:

{f}=10 {KHz}

{T}=\;{1}/{ {f}}=\;{1}/{10 × 10^{3}}=100 µ {sec}

{D}=0.6

{T}_{ {ON}}= {DT}=0.6 × 100=6 µ {sec}

{T}_{ {OFF}}=(1- {D}) {T}=40 µ {sec}

Consider Ist cycle:
During

T_{\;\text "ON "\;}

V_{L}=V_{S}=50

volt

V_{L}=L \;{d i}/{d t}

∫↖{I_{\max} }↙{{I}_{\min }} {di}=\;{V_{L}}/{L} ∫ {dt}

=∫↙{0}↖{I_{\max }} {di}=\;{50}/{10 × 10^{-3}} ∫↙{0}^{T_{O N}} {dt}

I_{\max }=5 × 10^{3} × T_{\;\text "ON "\;}

=5 × 10^{3} × 60 × 10^{-6}

=300 × 10^{-3}=300 {mA}

During

T_{\;\text "off "\;}(V_{L}=-50

volt )

∫↖{I_{\max} }↙{{I}_{\min }}d i=\;{-50}/{10 × 10^{-3}} ∫↙{T O N}↖{T} d t

(I_{\min }-300)=-5 × 10^{3} × 0.4 × 10 × 10^{-6}

=-200 {mA}

I_{\min }=100 {mA}

Current in Inductor after Ist cycle

=100 {mA}

Current in Inductor after

10^{\;\text "th "\;}

cycle

=100 × 10=1000 {mA}=1 {A}

,
Energy stored in Inductor

==\;{1}/{2} {LI}^{2}

=\;{1}/{2} × 10 {mH} ×(1)^{2}

=5 {m}]

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