GATE 2018 Electrical Engineering (EE) Power Electronics.

Section 01

1-Mark Questions

QQuestion 1 1 Mark

A single phase fully controlled rectifier is supplying a load with an anti-parallel diode as shown in the figure. All switches and diodes are ideal. Which one of the following is true for instantaneous load voltage and current?

AOptions

\(v_o \geq 0\) \& \(i_o \geq 0\)
\(v_o \geq 0\) \& \(i_o \leq 0\)
\(v_o \leq 0\) \& \(i_o \geq 0\)
\(v_o \leq 0\) \& \(i_o \leq 0\)

SSolution

The diode sits across the load with its anode on the negative rail and its cathode on the positive rail.

Whenever the load voltage tries to go negative this diode becomes forward biased and clamps it, so

\[v_o \geq 0\]

When the source reverses and the thyristors cease conducting, the inductive load current freewheels through the same diode. Every element in the load path - the four thyristors and the diode - conducts in one direction only, so

\[i_o \geq 0\]

Correct answer: A

QQuestion 2 1 Mark

Four power semiconductor devices are shown in the figure along with their relevant terminals. The device(s) that can carry dc current continuously in the direction shown when gated appropriately is (are)

AOptions

Triac only
Triac and MOSFET
Triac and GTO
Thyristor and Triac

SSolution

The arrow marks the direction of the dc current \(I\) in each device: it enters at the cathode of the thyristor and of the GTO, at \(MT_2\) of the triac, and at the source of the MOSFET.

Thyristor: conducts only from anode to cathode; the direction shown is its blocking direction, and no gate signal changes that.

GTO: also a latching, unidirectional device from anode to cathode, so the direction shown is blocked.

Triac: a bidirectional device; with the appropriate gate polarity it latches for current from \(MT_2\) to \(MT_1\).

MOSFET: the channel is symmetric, so with \(V_{GS}\) applied it carries current from source to drain as readily as from drain to source.

Only the triac and the MOSFET can carry the current in the direction marked.

Correct answer: B

Section 02

2-Mark Questions

QQuestion 3 2 Mark

A phase controlled single phase rectifier, supplied by an AC source, feeds power to an R-L-E load as shown in the figure. The rectifier output voltage has an average value given by \(V_o = \frac{V_m}{2\pi}(3 + \cos\alpha)\), where \(V_m = 80\pi\) volts and \(\alpha\) is the firing angle. If the power delivered to the lossless battery is 1600 W, \(\alpha\) in degree is\_\_\_\_\_\_\_\_ (up to 2 decimal places).

SSolution

Given:

\(V_m = 80\pi\) V
Average output voltage: \(V_o = \frac{80\pi}{2\pi}(3 + \cos\alpha) = 40(3 + \cos\alpha)\)
Battery voltage: \(E = 80\) V
Resistance: \(R = 2\) \(\Omega\)
Power to battery: \(P_{battery} = 1600\) W

Power to battery: \(P_{battery} = EI\)

\[I = \frac{1600}{80} = 20 \text{ A}\]

Voltage equation:

\[V_o = E + IR\]

\[40(3 + \cos\alpha) = 80 + 20 \times 2\]

\[40(3 + \cos\alpha) = 120\]

\[3 + \cos\alpha = 3\]

\[\cos\alpha = 0\]

\[\alpha = 90°\]

Answer: 90.00 degrees

QQuestion 4 2 Mark

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 \(\Omega\). 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).

SSolution

For buck converter:

\[V_o = D \times V_{in}\]

From timing diagram: - \(S_1\) duty cycle: \(D_1 = 0.5\) - \(S_2\) duty cycle: \(D_2 = 0.5\) - They operate in complementary fashion (interleaved)

Output voltage:

\[V_o = 0.5 \times 100 = 50 \text{ V}\]

Total load current:

\[I_o = \frac{V_o}{R} = \frac{50}{1} = 50 \text{ A}\]

Since converters share equally:

\[I_{o1} = I_{o2} = 25 \text{ A}\]

Average switch current (switch S1):

\[I_{S1,avg} = D_1 \times I_{o1} = 0.5 \times 25 = 12.5 \text{ A}\]

Answer: 12.50 A (range: 11.50 to 13.50)

QQuestion 5 2 Mark

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).

SSolution

With \(S\) closed the inductor is connected across B1 and its current ramps up; with \(S\) open the current flows through \(D\) into B2 and, since \(V_{B2} > V_{B1}\), it ramps down. Both terminal voltages are fixed, so volt-second balance cannot be satisfied by a continuous current: the inductor current is discontinuous.

Peak inductor current (\(T = 1/f = 200\) \(\mu\)s, \(DT = 80\) \(\mu\)s):

\[I_{pk} = \frac{V_{B1}DT}{L} = \frac{50 \times 80 \times 10^{-6}}{5 \times 10^{-3}} = 0.8 \text{ A}\]

Fall time (the inductor sees \(V_{B1} - V_{B2} = -100\) V while \(D\) conducts):

\[t_f = \frac{LI_{pk}}{V_{B2} - V_{B1}} = \frac{5 \times 10^{-3} \times 0.8}{100} = 40 \text{ } \mu\text{s}\]

This is shorter than the off interval of 120 \(\mu\)s, confirming discontinuous conduction.

Average current drawn from B1:

\[I_{B1} = \frac{1}{T}\cdot\frac{1}{2}I_{pk}(DT + t_f) = \frac{0.5 \times 0.8 \times 120 \times 10^{-6}}{200 \times 10^{-6}} = 0.24 \text{ A}\]

\[P = V_{B1}I_{B1} = 50 \times 0.24 = 12 \text{ W}\]

Checking at the receiving end, \(I_{B2} = \frac{1}{T}\cdot\frac{1}{2}I_{pk}t_f = 0.08\) A and \(P = 150 \times 0.08 = 12\) W.

Answer: 12.00 W