GATE 2007 Electrical Engineering (EE) Power Electronics.

Section 01

1-Mark Questions

QQuestion 1 1 Mark

Three-terminal linear voltage regulator connected to 10 \(\Omega\) load. If \(V_{in}\) = 10 V, power dissipated in transistor is

AOptions

0.6 W
2.4 W
4.2 W
5.4 W

SSolution

Linear voltage regulator:

Typical configuration provides regulated output voltage (e.g., 5V or other standard values).

Assumptions:

Common regulator: 5V output

\[V_{out} = 5 \text{ V}\]

\[V_{in} = 10 \text{ V}\]

\[R_L = 10 \text{ } \Omega\]

Load current:

\[I_L = \frac{V_{out}}{R_L} = \frac{5}{10} = 0.5 \text{ A}\]

Transistor voltage drop:

\[V_{transistor} = V_{in} - V_{out} = 10 - 5 = 5 \text{ V}\]

Power dissipation in transistor:

\[P_{transistor} = V_{transistor} \times I_L = 5 \times 0.5 = 2.5 \text{ W}\]

Closest answer: 2.4 W

Correct answer: B

QQuestion 2 1 Mark

Single-phase fully controlled thyristor bridge AC-DC converter operating at firing angle 25° and overlap angle 10° with constant DC output current 20 A. Fundamental power factor (displacement factor) at input AC mains is

AOptions

0.78
0.827
0.866
0.9

SSolution

Displacement factor with overlap:

Effective displacement angle:

\[\phi_1 = \alpha + \frac{\mu}{2}\]

where:

\(\alpha = 25°\) (firing angle)
\(\mu = 10°\) (overlap angle)

\[\phi_1 = 25° + \frac{10°}{2} = 25° + 5° = 30°\]

Displacement factor:

\[\text{DF} = \cos\phi_1 = \cos 30° = 0.866\]

Note: Displacement factor considers only fundamental component phase shift, not harmonics.

Correct answer: C

QQuestion 3 1 Mark

Three-phase fully-controlled thyristor bridge inverter feeds 50 kW power at 420 V DC to three-phase 415 V (line), 50 Hz AC mains. DC link current constant. RMS current of thyristor is

AOptions

119.05 A
79.37 A
68.73 A
39.68 A

SSolution

Given:

Power: \(P = 50\) kW
DC voltage: \(V_{dc} = 420\) V
Operating as inverter

DC current:

\[I_{dc} = \frac{P}{V_{dc}} = \frac{50000}{420} = 119.05 \text{ A}\]

Thyristor current:

In three-phase bridge, each thyristor conducts for 120°.

For constant DC current:

\[I_{thyristor,rms} = I_{dc}\sqrt{\frac{120°}{360°}} = I_{dc}\sqrt{\frac{1}{3}}\]

\[I_{thyristor,rms} = \frac{119.05}{\sqrt{3}} = 68.73 \text{ A}\]

Correct answer: C

QQuestion 4 1 Mark

Single phase full-wave half-controlled bridge converter feeds inductive load. Two SCRs connected to common DC bus. Converter must have freewheeling diode

AOptions

because converter inherently does not provide freewheeling
because converter does not provide freewheeling for high triggering angles
or else freewheeling action will cause shorting of AC supply
or else if gate pulse to one SCR is missed, it will cause high load current in other SCR

SSolution

Half-controlled bridge:

The two SCRs share the common DC bus (common cathode); the two diodes form the lower half of the bridge.

Freewheeling is inherent:

Let \(T_1\) and \(D_2\) conduct during the positive half cycle. When the supply reverses at \(\omega t = \pi\), \(D_1\) becomes forward biased and takes over from \(D_2\), and the load current circulates through \(T_1\) and \(D_1\). Both devices sit on the same supply terminal, so the load is short-circuited by the converter itself, \(v_o = 0\), and the AC supply carries no current. This path exists at every triggering angle, which rules out options (a), (b) and (c).

Why the diode is still required:

Because that freewheeling path keeps \(T_1\) in conduction throughout the negative half cycle, \(T_1\) never recovers its blocking state. If the gate pulse to \(T_2\) is missed, \(T_1\) simply picks up the following positive half cycle as well: the converter half-waves and one SCR carries the entire load current, cycle after cycle. A freewheeling diode across the load takes over the circulating current so that the conducting SCR turns off at the end of each half cycle.

Correct answer: D

QQuestion 5 1 Mark

"Six MOSFETs connected in bridge configuration (no other power device) MUST be operated as Voltage Source Inverter". This statement is

AOptions

True, because MOSFETs are voltage driven
True, because MOSFETs have inherently anti-parallel diodes
False, because it can be operated both as CSI or VSI
False, because MOSFETs can be operated as constant current sources

SSolution

MOSFET characteristics:

Body diode:

MOSFETs have inherent anti-parallel body diode:

Provides bidirectional current capability
Prevents reverse voltage blocking
Essential for VSI operation (freewheeling)

VSI requirements:

Switches must:

Block voltage in one direction
Conduct current bidirectionally
MOSFETs + body diode satisfy this

CSI requirements:

Switches must:

Block voltage bidirectionally
Conduct current in one direction
MOSFETs cannot block reverse voltage (body diode conducts)

Conclusion:

Due to body diode, MOSFETs in bridge MUST operate as VSI.

Cannot operate as CSI because body diode prevents reverse voltage blocking.

Correct answer: B

Section 02

2-Mark Questions

QQuestion 6 2 Mark

In a transformer, zero voltage regulation at full load is

AOptions

not possible
possible at unity power factor load
possible at leading power factor load
possible at lagging power factor load

SSolution

Voltage regulation:

\[\text{VR} = \frac{V_{NL} - V_{FL}}{V_{FL}} \times 100%\]

For zero regulation: \(V_{NL} = V_{FL}\)

Voltage drop equation:

\[\Delta V = I(R\cos\phi \pm X\sin\phi)\]

where + for lagging, - for leading.

For zero regulation:

\[I(R\cos\phi - X\sin\phi) = 0 \quad \Rightarrow \quad \tan\phi = \frac{R}{X}\]

The minus sign belongs to the leading case, so a solution exists only when the load current leads the terminal voltage. At unity or lagging power factor the drop \(I(R\cos\phi + X\sin\phi)\) is strictly positive and the regulation can never vanish.

Correct answer: C — zero full-load regulation is possible at a leading power factor, the one satisfying \(\tan\phi = R/X\).

QQuestion 7 2 Mark

Input signal \(V_{in}\) is 1 kHz square wave alternating between +7V and -7V with 50% duty cycle. Circuit delivers power to load \(R_L = 10\) \(\Omega\). Both transistors have same high current gain. Circuit efficiency for given input is \begin{center}

AOptions

SSolution

Class-B push-pull (complementary emitter follower):

For \(V_{in} = +7\) V the NPN conducts as an emitter follower; for \(V_{in} = -7\) V the PNP conducts. Each loses one base-emitter drop, so the output is a square wave of amplitude

\[V_o = \pm(7 - 0.7) = \pm 6.3 \text{ V}\]

Output power: for a square wave the rms value equals the peak value, \(V_{o,rms} = 6.3\) V:

\[P_{out} = \frac{V_{o,rms}^2}{R_L} = \frac{6.3^2}{10} = 3.969 \text{ W}\]

Supply power: the load current magnitude is \(|I_o| = 6.3/10 = 0.63\) A at all times, drawn from \(+V_{CC}\) for half the period and from \(-V_{CC}\) for the other half, so each rail supplies an average of \(0.315\) A:

\[P_{dc} = 10(0.315) + 10(0.315) = 6.3 \text{ W}\]

Efficiency:

\[\eta = \frac{P_{out}}{P_{dc}} = \frac{3.969}{6.3} = 0.63 = 63\%\]

Correct answer: C

QQuestion 8 2 Mark

Single-phase voltage source inverter controlled in single pulse-width modulated mode with pulse width 150° in each half cycle. THD is defined as \(\text{THD} = \frac{\sqrt{V_{rms}^2 - V_1^2}}{V_1} \times 100\). THD of output AC voltage waveform is

AOptions

65.65%
48.42%
31.83%
30.49%

SSolution

Pulse width = 150° in each half cycle

Fourier analysis:

For square pulse of width \(2\delta\) centered at 0°:

\[V_n = \frac{4V_{dc}}{n\pi}\sin(n\delta)\]

With \(\delta = 75°\) (half of 150°):

Fundamental:

\[V_1 = \frac{4V_{dc}}{\pi}\sin(75°) = \frac{4V_{dc}}{\pi}(0.9659) = 1.229V_{dc}\]

RMS voltage:

For pulse of width 150° in each half:

\[V_{rms} = V_{dc}\sqrt{\frac{150°}{180°}} = V_{dc}\sqrt{0.833} = 0.913V_{dc}\]

For a square wave with pulse width \(2\delta\):

\[V_{rms} = V_{dc}\sqrt{\frac{2\delta}{\pi}}\]

With \(\delta = 75° = 1.309\) rad:

\[V_{rms} = V_{dc}\sqrt{\frac{2 \times 1.309}{\pi}} = V_{dc}\sqrt{0.833} = 0.913V_{dc}\]

\[V_1 = \frac{4V_{dc}}{\pi}\sin(75°) = 1.2298V_{dc} \text{ (peak)}, \qquad V_{1,rms} = \frac{1.2298V_{dc}}{\sqrt{2}} = 0.8696V_{dc}\]

THD:

\[\text{THD} = \frac{\sqrt{V_{rms}^2 - V_{1,rms}^2}}{V_{1,rms}} = \frac{\sqrt{0.8333 - 0.7562}}{0.8696}\]

\[= \frac{0.2777}{0.8696} = 0.3192 = 31.9\%\]

which is the listed value of 31.83%.

Correct answer: C

QQuestion 9 2 Mark

Current commutated DC-DC chopper where \(Th_M\) is main SCR and \(Th_{Aux}\) is auxiliary SCR. Load current constant at 10 A. \(Th_M\) is ON. \(Th_{Aux}\) triggered at t=0. \(Th_M\) is turned OFF between

AOptions

0 \(\mu\)s < t \(\leq\) 25 \(\mu\)s
25 \(\mu\)s < t \(\leq\) 50 \(\mu\)s
50 \(\mu\)s < t \(\leq\) 75 \(\mu\)s
75 \(\mu\)s < t \(\leq\) 100 \(\mu\)s

SSolution

Current commutation chopper:

From the circuit: \(C = 10\) \(\mu\)F, \(L = 25.28\) \(\mu\)H, \(V_s = 230\) V, load current \(I_o = 10\) A.

Stage 1 — capacitor reversal. The capacitor starts charged to \(V_s\). Firing \(Th_{Aux}\) at \(t = 0\) rings \(C\) with \(L\) through \(Th_{Aux}\) alone; \(Th_M\) still carries the full load current. The ring lasts one half period, at the end of which the capacitor voltage has reversed and \(Th_{Aux}\) turns off:

\[t_1 = \pi\sqrt{LC} = \pi\sqrt{(25.28 \times 10^{-6})(10 \times 10^{-6})} = 49.95 \text{ } \mu\text{s}\]

Stage 2 — current transfer. The reversed capacitor now drives the resonant current back through the diode and through \(Th_M\), against the load current. Its peak value is

\[I_p = V_s\sqrt{\frac{C}{L}} = 230\sqrt{\frac{10}{25.28}} = 144.7 \text{ A}\]

The main thyristor current is \(I_o - I_p\sin\omega t\), which reaches zero when

\[\omega t_2 = \sin^{-1}\frac{10}{144.7} = 0.0692 \text{ rad}, \qquad t_2 = 0.0692\sqrt{LC} = 1.10 \text{ } \mu\text{s}\]

Turn-off instant:

\[t = t_1 + t_2 = 49.95 + 1.10 = 51.05 \text{ } \mu\text{s}\]

Correct answer: C