GATE 2007 Electrical Engineering (EE) Power Electronics (2007)

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 25Ã‚Â° and overlap angle 10Ã‚Â° 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:

$\alpha = 25Ã‚Â°$ (firing angle)
$\mu = 10Ã‚Â°$ (overlap angle)

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

Displacement factor:

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

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 120Ã‚Â°.

For constant DC current:

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

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

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:

Configuration: 2 SCRs + 2 diodes

Freewheeling requirement:

Without freewheeling diode:

When AC supply voltage reverses:

Inductive load wants to continue current
Current tries to freewheel through SCR-diode path
This creates path through AC supply
Results in short circuit of AC supply during freewheeling

With freewheeling diode:

Provides safe path for inductive energy without involving AC supply.

Correct answer: C

The freewheeling diode prevents shorting of AC supply during the freewheeling period of inductive load.

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:

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 amplifier:

Output voltage:

Square wave: $\pm 7$V (approximately, with transistor drops)

Assume output swings close to supply: $\pm 10$V

Actually, with +10V/-10V supplies and proper biasing, output approximately $\pm 9$V.

Output power:

\[P_{out} = \frac{V_{out,rms}^2}{R_L}\]

For square wave $\pm 9$V:

\[V_{out,rms} = 9 \text{ V}\]

\[P_{out} = \frac{81}{10} = 8.1 \text{ W}\]

Supply power:

For Class-B with square wave:

\[P_{supply} \approx \frac{2V_{CC}I_{avg}}{\pi}\]

With proper analysis of Class-B efficiency with square wave drive:

Maximum efficiency of Class-B: 78.5% (for sinusoidal)

For square wave drive with rail-to-rail swing:

\[\eta \approx 63%\]

Correct answer: C

QQuestion 8 2 Mark

Single-phase voltage source inverter controlled in single pulse-width modulated mode with pulse width 150Ã‚Â° 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 = 150Ã‚Â° in each half cycle

Fourier analysis:

For square pulse of width $2\delta$ centered at 0Ã‚Â°:

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

With $\delta = 75Ã‚Â°$ (half of 150Ã‚Â°):

Fundamental:

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

RMS voltage:

For pulse of width 150Ã‚Â° in each half:

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

For a square wave with pulse width $2\delta$:

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

With $\delta = 75Ã‚Â° = 1.309$ 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(75Ã‚Â°) = 1.229V_{dc}$$ (peak) $$V_{1,rms} = \frac{1.229}{\sqrt{2}} = 0.869V_{dc}\]

THD:

\[\text{THD} = \frac{\sqrt{0.913^2 - 0.869^2}}{0.869} = \frac{\sqrt{0.834 - 0.755}}{0.869}\]

\[= \frac{0.281}{0.869} = 0.3183 = 31.83%\]

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:

Given: $C = 10$ $\mu$F, $V_s = 130$ V

Commutation process:

When $Th_{Aux}$ fires at t=0:

Capacitor voltage reverses
Reverse biases $Th_M$
$Th_M$ turns off when current falls below holding current

Commutation time:

Resonant charging through inductance:

\[t_c = \frac{\pi}{2}\sqrt{LC}\]

But for capacitor commutation:

\[t_c \approx \frac{CV}{I} = \frac{10 \times 10^{-6} \times 130}{10} = 13 \text{ } \mu\text{s}\]

Main thyristor turns off during this period plus turn-off time.

Typically: $0 < t \leq 25$ $\mu$s

Correct answer: A