GATE Electrical Engineering (EE) 2018 Solved Paper

Question : 41 of 65

Marks: +1, -0

A dc to dc converter shown in the figure is charging a battery bank, B2 whose voltage is constant at

150 {V}

. B 1 is another batter bank whose voltage is constant at

50 {V}

. The value of the inductor. Is

5 {mH}

and the ideal switch, is operated with a switching frequency of

5 {kHz}

with a duty ratio of

0.4

. Once the circuit has attained steady state and assuming the diode to be ideal, the power transferred from to (in Watt) is ________ (upto 2 decimal places).

Your Answer:

Validate

Solution:

For continuous conduction:

{V}_{0}=\;{ {V}_{ {s}}}/{1- {D}}=\;{50}/{1-0.4}=\;{50}/{0.6}=83.33 {V}

Since

V_{0}<150

, it is discontinuous mode

When S ON

{V}_{ {s}}= {L} \;{d I_{L}}/{d t} ⇒ \;{ {V}_{5}}/{ {L}}=\;{d I_{L}}/{d t}

{I}_{ {P}}=\;{ {V}_{ {s}}}/{ {L}} × {DT}

..............(i)
When S OFF

{V}_{ {s}}- {V}_{0}= {L} \;{d l_{L}}/{d t} ⇒ \;{ {V}_{ {s}}- {V}_{0}}/{ {L}}=\;{d I_{L}}/{d t}

I_{ {P}}=\;{ {V}_{5}- {V}_{0}}/{ {L}} ×( {D}-β) {T}

....(ii)
From (i) and (ii)

\;{ {V}_{ {s}}}/{ {L}} × {DT}=\;{ {V}_{ {s}}- {V}_{0}}/{ {L}} ×( {D}-β) {T}

Solving, we get

{V}_{0}=\;{β}/{β- {D}}

150=\;{β}/{β-0.4} × 50

β=0.6

The power is transferred to

150 {V}

during DT to

β {T}

From (i)

{I}_{ {P}}=\;{50}/{5 × 10^{-3}} × 0.4 × \;{1}/{5 × 10^{3}}=0.8 {A}

Power transferred to

150 {V}

source

{P}=\;{1}/{ {T}} × \;{1}/{2} × {I}_{ {P}} × {V}_{0} ×(β- {D}) {T}=\;{1}/{2} × 0.8 × 150 ×(0.6-0.4)=12 {W}

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