GATE 2018 EE Q.50: solution and answer

GATE 2018 · EEQ.50 · 2 marks · NAT

The figure shows two buck converters connected in parallel. The common input dc voltage for the converters has a value of 100 V. The converters have inductors of identical value. The load resistance is 1 Ω. The capacitor voltage has negligible ripple. Both converters operate in the continuous conduction mode. The switching frequency is 1 kHz, and the switch control signals are as shown. The circuit operates in the steady state. Assuming that the converters share the load equally, the average value of i_S1, the current of switch S1 (in Ampere), is _____ (up to 2 decimal places).

GATE 2018 EE question 50: A 100 V dc source (+ at top) feeds two buck stages sharing the same input rails. — Figure from the GATE 2018 EE paper. It is part of the question.

Answer and solution

ANSWER

The average value of $i_{S 1}$ is $12.5 A$ .

Official answer: 11.50 to 13.50Source: Official answer key, GATE 2018 EE (IIT Guwahati)

Switch S1 is on from $0$ to $0.5 ms$ in a period of $T = \frac{1}{1 kHz} = 1 ms$ , so the duty ratio is $D = \frac{0.5}{1} = 0.5$ . S2 is on in the other half of the period, so it has the same $D = 0.5$ .
Both converters run in continuous conduction from the same $100 V$ input with the same $D$ . With negligible capacitor ripple, $V_{o} = D V_{s} = 0.5 \times 100 = 50 V$ .
Load current: $I_{o} = \frac{V _{o}}{R} = \frac{50}{1} = 50 A$ .
The converters share the load equally, so each average inductor current is $I_{L} = \frac{50}{2} = 25 A$ .
In a buck converter the switch carries the inductor current only while it is on, a fraction $D$ of each period. The diode carries the inductor current for the rest of the period. The ripple does not change the average, because the average of the switch current over a period is $D$ times the average inductor current.
Average switch current: $I_{S 1, avg} = D I_{L} = 0.5 \times 25 = 12.5 A$ .
Check by power balance: each converter delivers $50 \times 25 = 1250 W$ , so the current drawn from the $100 V$ source is $\frac{1250}{100} = 12.5 A$ . This current flows only through S1, so it is the average of $i_{S 1}$ .

KEY IDEA

Each buck converter has $D = 0.5$ , so $V_{o} = 50 V$ and the load takes $50 A$ . Each converter supplies $25 A$ through its inductor, and the switch conducts that current for the fraction $D$ of the period, giving $D I_{L}$ .

WATCH OUT FOR

Giving $25 A$ , the average inductor current, as the average switch current. The switch conducts only during the on-time, so its average is $D I_{L}$ . Another slip is taking the full $50 A$ load current as the share of one converter.

From the GATE 2018 EE paper. Every question of this paper · Download the full paper · All 9 DC-DC converters questions, with notes

GATE 2018 EE Q.50: solution and answer

Answer and solution

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