GATE 2020 Electrical Engineering (EE) Power Electronics (2020)

Section 01

0-Mark Questions

QQuestion 1 0 Mark

A double pulse measurement for an inductively loaded circuit controlled by an IGBT switch is carried out to evaluate the reverse recovery characteristics of the diode $D$. The point on the diode current vs time plot (indicate choice by entering 1, 2, 3 or 4) at which the IGBT experiences the highest current stress is \_\_\_\_\_\_\_\_.

SSolution

During diode reverse recovery, the IGBT must conduct:

The load current
The reverse recovery current of the diode

The highest current stress occurs at point 3, which represents the peak reverse recovery current. At this instant, the IGBT conducts both the load current and the maximum reverse recovery current simultaneously.

Answer: \boxed{3}

QQuestion 2 0 Mark

A common-source amplifier with a drain resistance, $R_D = 4.7$ k$\Omega$, is powered using a 10 V power supply. Assuming that the transconductance, $g_m$, is 520 $\mu$A/V, the voltage gain of the amplifier is closest to:

AOptions

1.22
$-1.22$
2.44 \CorrectChoice $-2.44$

SSolution

For a common-source amplifier, the voltage gain is:

\[A_v = -g_m R_D\]

Given: $g_m = 520$ $\mu$A/V $= 520 \times 10^{-6}$ A/V, $R_D = 4.7$ k$\Omega = 4700$ $\Omega$

\[A_v = -520 \times 10^{-6} \times 4700 = \boxed{-2.44}\]

The negative sign indicates 180Ã‚Â° phase shift between input and output.

Correct answer: D

QQuestion 3 0 Mark

A single-phase, full-bridge diode rectifier fed from a 230 V, 50 Hz sinusoidal source supplies a series combination of finite resistance $R$ and a very large inductance $L$. The two most dominant frequency components in the source current are:

AOptions

150 Hz, 250 Hz \CorrectChoice 50 Hz, 150 Hz
50 Hz, 100 Hz
50 Hz, 0 Hz

SSolution

For a single-phase full-bridge diode rectifier with highly inductive load:

The output current is nearly dc (due to large $L$). The input current is a square wave at supply frequency.

Fourier analysis of the square wave input current shows:

Fundamental component at 50 Hz (supply frequency)
Odd harmonics: 3rd (150 Hz), 5th (250 Hz), etc.

The two most dominant components are:

50 Hz (fundamental)
150 Hz (3rd harmonic)

Correct answer: B

QQuestion 4 0 Mark

A single-phase inverter is fed from a 100 V dc source and is controlled using a quasi-square wave modulation scheme. The angle $\sigma$ is adjusted to entirely eliminate the 3rd harmonic component from the output voltage. Under this condition, the magnitude of the 5th harmonic component as a percentage of the magnitude of the fundamental component is \_\_\_\_\_\_\_\_ (rounded off to 2 decimal places).

SSolution

For quasi-square wave modulation:

\[V_n = \frac{4V_s}{n\pi} \cos(n\sigma)\]

To eliminate 3rd harmonic: $V_3 = 0$

\[\cos(3\sigma) = 0 \Rightarrow 3\sigma = \frac{\pi}{2} \Rightarrow \sigma = \frac{\pi}{6}\]

Ratio of 5th to fundamental:

\[\frac{V_5}{V_1} = \frac{\cos(5\sigma)}{5\cos(\sigma)} = \frac{\cos(5\pi/6)}{5\cos(\pi/6)}\]

\[= \frac{-\sqrt{3}/2}{5 \times \sqrt{3}/2} = \frac{-1}{5}\]

Magnitude: $\left|\frac{V_5}{V_1}\right| = \frac{1}{5} = 0.2 = \boxed{20.00%}$$

QQuestion 5 0 Mark

Thyristor $T_1$ is triggered at angle $\alpha$ (in degrees), and $T_2$ at angle $180Ã‚Â° + \alpha$, in each cycle of the sinusoidal input voltage. Assume both thyristors to be ideal. To control the load power over the range 0 to 2 kW, the minimum range of variation in $\alpha$ is:

AOptions

$0Ã‚Â°$ to $60Ã‚Â°$ \CorrectChoice $60Ã‚Â°$ to $180Ã‚Â°$
$0Ã‚Â°$ to $120Ã‚Â°$
$60Ã‚Â°$ to $120Ã‚Â°$

SSolution

Given: Load impedance $= 10\angle -60Ã‚Â°$ $\Omega$, Supply $= 200$ V, 50 Hz

Load: $Z = 10\angle -60Ã‚Â° = 5 + j8.66$ $\Omega$ (R-L load)

For ac voltage controller with R-L load, maximum power occurs when:

\[\alpha = \phi = 60Ã‚Â°\]

For minimum power (zero): $\alpha = 180Ã‚Â°$

Therefore, to control power from 0 to maximum (2 kW), the firing angle must vary from:

\[\alpha = \boxed{60Ã‚Â° \text{ to } 180Ã‚Â°}\]

Correct answer: B

QQuestion 6 0 Mark

A single-phase, full-bridge, fully controlled thyristor rectifier feeds a load comprising a 10 $\Omega$ resistance in series with a very large inductance. The rectifier is fed from an ideal 230 V, 50 Hz sinusoidal source through cables which have negligible internal resistance and a total inductance of 2.28 mH. If the thyristors are triggered at an angle $\alpha = 45Ã‚Â°$, the commutation overlap angle in degree (rounded off to 2 decimal places) is \_\_\_\_\_\_\_\_.

SSolution

Given: $V_s = 230$ V, $f = 50$ Hz, $\alpha = 45Ã‚Â°$, $R = 10$ $\Omega$, $L_s = 2.28$ mH

Average output voltage without overlap:

\[V_{d0} = \frac{2V_m}{\pi}\cos\alpha = \frac{2 \times 230\sqrt{2}}{\pi}\cos 45Ã‚Â° = 146.42 \text{ V}\]

Load current: $I_0 = \frac{V_{d0} - \Delta V_d}{R}$

Voltage drop due to commutation:

\[\Delta V_d = \frac{2\omega L_s I_0}{\pi} = 4fL_s I_0\]

Therefore: $I_0 \times 10 = 146.42 - 4 \times 50 \times 2.28 \times 10^{-3} \times I_0$

\[I_0(10 + 0.456) = 146.42\]

\[I_0 = 14.0036 \text{ A}\]

Using overlap equation:

\[\cos\alpha - \cos(\alpha + \mu) = \frac{2\omega L_s I_0}{V_m}\]

\[\cos 45Ã‚Â° - \cos(45Ã‚Â° + \mu) = \frac{4 \times 50 \times 2.28 \times 10^{-3} \times 14.0036}{230\sqrt{2}}\]

\[0.7071 - \cos(45Ã‚Â° + \mu) = 0.06166\]

\[\cos(45Ã‚Â° + \mu) = 0.6454\]

\[45Ã‚Â° + \mu = 49.80Ã‚Â°\]

\[\mu = \boxed{4.80Ã‚Â°}\]

QQuestion 7 0 Mark

A non-ideal Si-based pn junction diode is tested by sweeping the bias applied across its terminals from $-5$ V to $+5$ V. The effective thermal voltage, $V_T$, for the diode is measured to be $(29 \pm 2)$ mV. The resolution of the voltage source in the measurement range is 1 mV. The percentage uncertainty (rounded off to 2 decimal places) in the measured current at a bias voltage of 0.02 V is \_\_\_\_\_\_\_\_.

SSolution

Diode equation: $I = I_0\left[\exp\left(\frac{V_D}{\eta V_T}\right) - 1\right]$

Given: $V_T = (29 \pm 2)$ mV, $V_D = 20$ mV, $W_{V_D} = 1$ mV, $W_{V_T} = 2$ mV

Taking natural log: $\ln(I) = \ln(I_0) + \frac{V_D}{\eta V_T}$

For $\eta = 1$, differentiating:

\[\frac{\partial I}{I} = -\frac{V_D}{V_T^2}\partial V_T + \frac{1}{V_T}\partial V_D\]

Uncertainty propagation:

\[W_{res} = I\sqrt{\left(\frac{V_D}{V_T^2}W_{V_T}\right)^2 + \left(\frac{1}{V_T}W_{V_D}\right)^2}\]

\[\frac{W_{res}}{I} = \sqrt{\left(\frac{0.02}{(0.029)^2} \times 0.002\right)^2 + \left(\frac{1}{0.029} \times 0.001\right)^2}\]

\[= \sqrt{(0.04756)^2 + (0.03448)^2} = \sqrt{0.002262 + 0.001189}\]

\[= \sqrt{0.003451} = 0.0587\]

Percentage uncertainty: $\boxed{5.87%}$$

QQuestion 8 0 Mark

Consider the diode circuit shown. The diode obeys $I_D = I_S\left[\exp\left(\frac{V_D}{nV_T}\right) - 1\right]$, where $n > 1$, $V_T > 0$. The circuit is biased so that $V > 0$ and $I < 0$. To transfer maximum power from the current source ($I_1$) to a resistive load (not shown) at the output, what values of $R_1$ and $R_2$ would you choose?

AOptions

Small $R_1$ and small $R_2$.
Large $R_1$ and large $R_2$. \CorrectChoice Small $R_1$ and large $R_2$.
Large $R_1$ and small $R_2$.

SSolution

For maximum power transfer from current source $I_1$:

Analysis:

The voltage across the diode: $V_D = \frac{R_2}{R_1 + R_2} \times V$
For maximum power to the load, we need maximum voltage across the diode
Small $R_1$ minimizes voltage drop across $R_1$
Large $R_2$ maximizes the voltage divider ratio

With small $R_1$ and large $R_2$: $V_D \approx V$ (maximum possible)

This ensures maximum voltage and current capability for the load connected across the diode terminals.

Correct answer: C

QQuestion 9 0 Mark

A dc-dc converter circuit with switch $Q$ switched at a frequency of 10 kHz with a duty ratio of 0.6. All components are ideal, and the initial current in the inductor is zero. Energy stored in the inductor in mJ (rounded off to 2 decimal places) at the end of 10 complete switching cycles is \_\_\_\_\_\_\_\_.

SSolution

Given: Buck-boost converter, $L = 10$ mH, $V_{in} = 50$ V, $V_{out} = 50$ V, $D = 0.6$, $f = 10$ kHz

Time period: $T = \frac{1}{10 \times 10^3} = 100$ $\mu$s

During ON time ($D \cdot T = 60$ $\mu$s): Inductor charges

\[\frac{di}{dt} = \frac{V_{in}}{L} = \frac{50}{10 \times 10^{-3}} = 5000 \text{ A/s}\]

During OFF time ($D' \cdot T = 40$ $\mu$s): Inductor discharges

\[\frac{di}{dt} = \frac{-V_{out}}{L} = \frac{-50}{10 \times 10^{-3}} = -5000 \text{ A/s}\]

Current rise per cycle: $\Delta i_{rise} = 5000 \times 60 \times 10^{-6} = 0.3$ A

Current fall per cycle: $\Delta i_{fall} = 5000 \times 40 \times 10^{-6} = 0.2$ A

Net rise per cycle: $0.3 - 0.2 = 0.1$ A

After 10 cycles: $i_L = 10 \times 0.1 = 1$ A

Energy stored:

\[E = \frac{1}{2}Li_L^2 = \frac{1}{2} \times 10 \times 10^{-3} \times 1^2 = 5 \times 10^{-3} \text{ J} = \boxed{5.00 \text{ mJ}}\]

QQuestion 10 0 Mark

A non-ideal diode is biased with a voltage of $-0.03$ V, and a diode current of $I_1$ is measured. The thermal voltage is 26 mV and the ideality factor for the diode is 15/13. The voltage, in V, at which the measured current increases to $1.5I_1$ is closest to:

AOptions

$-4.50$ \CorrectChoice $-0.09$
$-0.02$
$-1.50$

SSolution

Diode equation with ideality factor: $I = I_0\left[\exp\left(\frac{V_D}{\eta V_T}\right) - 1\right]$

Given: $\eta = \frac{15}{13}$, $V_T = 26$ mV, so $\eta V_T = \frac{15}{13} \times 26 = 30$ mV

At $V_{D1} = -0.03$ V $= -30$ mV:

\[I_1 = I_0[e^{-30/30} - 1] = I_0[e^{-1} - 1] = I_0[0.368 - 1] = -0.632 I_0\]

When current is $1.5I_1 = 1.5 \times (-0.632 I_0) = -0.948 I_0$:

\[-0.948 I_0 = I_0[e^{V_{D2}/30} - 1]\]

\[e^{V_{D2}/30} = 1 - 0.948 = 0.052\]

\[\frac{V_{D2}}{30} = \ln(0.052) = -2.956\]

\[V_{D2} = -2.956 \times 30 = -88.68 \text{ mV} \approx \boxed{-0.09 \text{ V}}\]

Correct answer: B