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Q12GATE 2018NAT

A dc to dc converter shown in the figure is charging a battery bank, B2 whose voltage is constant at 150 V. B1 is another battery bank whose voltage is constant at 50 V. The value of the inductor, L is 5 mH and the ideal switch, S 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 D to be ideal, the power transferred from B1 to B2 (in Watt) is ___________ (up to 2 decimal places). .

🚀 Solve in Practice Mode

📖 Explanation

During $T_{on}$ the circuit behaves as,

\begin{aligned} V_s&=L\frac{di}{dt} \\ di&=\frac{V_s}{L}dt \end{aligned}

Integrating on both sides,

\begin{aligned} I_P&=\frac{V_s}{L}T_{on}\\ T_{on}&=\alpha T=\frac{\alpha }{f}=80 \times 10^{-6}s\\ I_P&=\frac{50}{5 \times 10^{-3}}\times (80 \times 10^{-6})\\ &=0.8A \end{aligned}

During $T_{off}$ , it is $T_{on}\leq t\leq \beta T$

\begin{aligned} \text{KVL, } V_L&=V_s-V_0 \\ (V_L)_{avg}&=0 \\ V_sT_{on}+&(V_s-V_0)(\beta T-T_{on}) =0 \\ V_s\beta T &=V_0(\beta T-T_{on}) \\ \frac{V_0}{V_s}&=\frac{\beta }{\beta -\alpha } \\ \beta &=0.6\\ &\text{Fromt the graph of } i_L, \\ I_{L(avg)}&=\frac{\frac{1}{2} \times b \times h}{T}\\ &=\frac{\frac{1}{2} \times \beta T \times I_P}{T}\\ &=\frac{1}{2} \times 0.6 \times 0.8=0.24A\\ &\text{Power transferred to } B_2 \text{ is,}\\ P&=V_s \times (I_L)_{avg}\\ \text{ where } I_s&=I_L\\ P&=50 \times 0.24\\ &=12W \end{aligned}

Explanation Figure 2

Explanation Figure 3