Question 1 |

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 = 45^{\circ},
the commutation overlap angle in degree (rounded off to 2 decimal places) is_____.

2.4 | |

4.8 | |

6.4 | |

8.2 |

Question 1 Explanation:

\begin{aligned}&1-\phi\text{ SCR bridge rectifier} \\ \alpha &=45^{\circ},\; \; R=10\, \Omega\\ &\text{supply 230 V, 50 Hz}\\ L_{s}&=2.28\, mH \\ \mu &=? \\ \Delta V_{d}&=\frac{V_{m}}{\pi }[\cos \alpha -\cos (\alpha +\mu )]\\ &= 4fL_{s}I_{0} \\ V_{0}&=\frac{2V_{m}}{\pi }\cos \alpha -4fL_{s}I_{0} \\ I_{0}R&=\frac{2V_{m}}{\pi}\cos \alpha -4fL_{s}I_{0} \\ &\text{Find }I_{0} \\ I_{0}\times 10&=\frac{2\times 230\sqrt{2}}{\pi } \cos 45 \\&-4\times 50\times 2.28\times 10^{-3}I_{0} \\ I_{0}(10+0.456)&=146.42 \\ I_{0}&=\frac{146.49}{10.456}=14.0036\: A \\ \Delta V_{d0 }&=\frac{230\sqrt{2}}{\pi }[\cos 45-\cos (45+\mu)] \\ &=4\times 50\times 2.28\times 10^{-3}\times 14 \\ &=6.384 \\ \cos 45^{\circ}-\cos (45^{\circ}+\mu )&=0.061659 \\ 45+\mu &=49.80 \\ \therefore \, \, \mu =4.80^{\circ}\end{aligned}

Question 2 |

In the dc-dc converter circuit shown, switch Q is switched at a frequency of 10 kHz
with a duty ratio of 0.6. All components of the circuit 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 ________.

10 | |

5 | |

15 | |

20 |

Question 2 Explanation:

Buck boost converter,

D=0.6\rightarrow \text{store energy}

D=\frac{T_{ON}}{T}=0.6

T_{ON}=0.6\: T\rightarrow \text{store energy}

T_{OFF}=0.4\: T\rightarrow \text{releasing energy}

For one cycle: Rise in current for 0.2T

For 10 cycles: Find rise in current (0.2T) \times 10 = 2T

i=\frac{50}{L}t

i=\frac{50}{L}(2T)=\frac{50\times 2}{LP}=\frac{100}{10\cdot 10^{-3}\times 10\cdot 10^{3}}=1\, A

\therefore \, \text{Energy stored}=\frac{1}{2}Li^{2}=\frac{1}{2}\times (10\cdot 10^{-3})(1)^{2}=5\: mJ

D=0.6\rightarrow \text{store energy}

D=\frac{T_{ON}}{T}=0.6

T_{ON}=0.6\: T\rightarrow \text{store energy}

T_{OFF}=0.4\: T\rightarrow \text{releasing energy}

For one cycle: Rise in current for 0.2T

For 10 cycles: Find rise in current (0.2T) \times 10 = 2T

i=\frac{50}{L}t

i=\frac{50}{L}(2T)=\frac{50\times 2}{LP}=\frac{100}{10\cdot 10^{-3}\times 10\cdot 10^{3}}=1\, A

\therefore \, \text{Energy stored}=\frac{1}{2}Li^{2}=\frac{1}{2}\times (10\cdot 10^{-3})(1)^{2}=5\: mJ

Question 3 |

A resistor and a capacitor are connected in series to a 10 V dc supply through a switch.
The switch is closed at t=0, and the capacitor voltage is found to cross 0 V at t=0.4 \tau,
where \tau is the circuit time constant. The absolute value of percentage change required
in the initial capacitor voltage if the zero crossing has to happen at t=0.2 \tau is _______
(rounded off to 2 decimal places).

24.24 | |

78.83 | |

12.45 | |

54.99 |

Question 3 Explanation:

If initial charge polarities on the capacitor is opposite to the supply voltage then only
the capacitor voltage crosses the zero line.

\begin{aligned} V_{c}(t) \; \Rightarrow \;& \text{Final value} \\ &+ (\text{Initial value - Final value}) e^{-t/\tau }\\ 0&=10+(-V_{0}-10)e^{-0.4} \\ 10&=(V_{0}+10)e^{-0.4} \\ V_{0}&=4.918 V \\ \text{Now, } t&=0.2\tau \\ 0&=10+(-{V_{0}}'-10)e^{-0.2} \\ {V_{0}}'&=2.214 \\ \%\text{change in voltage} &= \frac{4.918-2.214}{4.918}\times 100 \%\\ &=54.99\% \end{aligned}

\begin{aligned} V_{c}(t) \; \Rightarrow \;& \text{Final value} \\ &+ (\text{Initial value - Final value}) e^{-t/\tau }\\ 0&=10+(-V_{0}-10)e^{-0.4} \\ 10&=(V_{0}+10)e^{-0.4} \\ V_{0}&=4.918 V \\ \text{Now, } t&=0.2\tau \\ 0&=10+(-{V_{0}}'-10)e^{-0.2} \\ {V_{0}}'&=2.214 \\ \%\text{change in voltage} &= \frac{4.918-2.214}{4.918}\times 100 \%\\ &=54.99\% \end{aligned}

Question 4 |

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:

-0.02 | |

-0.09 | |

-1.5 | |

-4.5 |

Question 4 Explanation:

\begin{aligned}
I_{1}&=i_{0}\left [ e^{\frac{0.03}{15/13\times 26mV}}-1 \right ] \\ V_{D}&=\text{ -ve '1' can not be neglected}\\&\text{in diode current equation} \\ I_{1}&=I_{0}[e^{-30mV/30mV}-1] \\ &=I_{0}[e^{-1}-1] \\ &=-0.64\: I_{0} \\ 1.5I_{1}&=I_{0}[e^{V_{D}/30mV}-1] \\ -1.5\times 0.64 I_{0}&=I_{0}[e^{V_{D}/30mV}-1]\\-0.96&=e^{V_{D2}/30mV}-1\\1-0.96&=e^{V_{D2}/30mV}\\0.04&=e^{V_{D2}/30mV}\\ 30 mV \ln (0.04)&=V_{D}\\ V_{D}&=-0.09 V
\end{aligned}

Question 5 |

Consider the diode circuit shown below. The diode, D, obeys the current-voltage
characteristic I_D=I_S\left ( exp\left ( \frac{V_D}{nV_T} \right )-1 \right ), where n \gt 1, V_T \gt 0, V_D is the voltage across
the diode and I_D is the current through it. The circuit is biased so that voltage, V \gt 0 and current, I \lt 0. If you had to design this circuit to transfer maximum power from
the current source (I_1) to a resistive load (not shown) at the output, what values R_1 \; and \; R_2 would you choose?

Large R_1 and large R_2. | |

Small R_1 and small R_2. | |

Large R_1 and small R_2. | |

Small R_1 and large R_2. |

Question 5 Explanation:

R_{1}-low, R_{2}- high

V_{D}=V\times \frac{R_{2}}{R_{1}+R_{2}}

If R_{2} is large V_{D} becomes high

If R_{1} is less V_{D}=V

So for maximum power, R_{1} is small and R_{2} is large.

V_{D}=V\times \frac{R_{2}}{R_{1}+R_{2}}

If R_{2} is large V_{D} becomes high

If R_{1} is less V_{D}=V

So for maximum power, R_{1} is small and R_{2} is large.

Question 6 |

A single-phase inverter is fed from a 100 V dc source and is controlled using a quasisquare wave modulation scheme to produce an output waveform, v(t). as shown. The
angle \sigma is adjusted to entirely eliminate the 3^{rd} harmonic component from the output
voltage. Under this condition, for v(t), the magnitude of the 5^{th} harmonic component as
a percentage of the magnitude of the fundamental component is _______(rounded off
to 2 decimal places).

15 | |

10 | |

20 | |

25 |

Question 6 Explanation:

Using result,

\begin{aligned} V_{n}&=\frac{4V_{s}}{n\pi }\cos n\sigma\ \text{For, } V_{3}=0\\ \cos 3\sigma &=0 \\ 3\sigma &=\frac{\pi }{2} \\ \sigma &=\frac{\pi }{6}\\ \text{Now, } \frac{V_{5}}{V_{1}}&=\frac{\cos 5\sigma }{5\cos\sigma }\\&=\frac{\cos 5\pi /6}{5\cos \pi /6}=-\frac{1}{5} \\ \% \left | \frac{V_{5}}{V_{1}} \right |&=\frac{1}{5}\times 100=20\% \end{aligned}

\begin{aligned} V_{n}&=\frac{4V_{s}}{n\pi }\cos n\sigma\ \text{For, } V_{3}=0\\ \cos 3\sigma &=0 \\ 3\sigma &=\frac{\pi }{2} \\ \sigma &=\frac{\pi }{6}\\ \text{Now, } \frac{V_{5}}{V_{1}}&=\frac{\cos 5\sigma }{5\cos\sigma }\\&=\frac{\cos 5\pi /6}{5\cos \pi /6}=-\frac{1}{5} \\ \% \left | \frac{V_{5}}{V_{1}} \right |&=\frac{1}{5}\times 100=20\% \end{aligned}

Question 7 |

A double pulse measurement for an inductively loaded circuit controlled by the IGBT
switch is carried out to evaluate the reverse recovery characteristics of the diode. D,
represented approximately as a piecewise linear plot of current vs time at diode turn-off.
L_{par} is a parasitic inductance due to the wiring of the circuit, and is in series with the
diode. The point on the plot (indicate your choice by entering 1. 2, 3 or 4) at which the
IGBT experiences the highest current stress is ______.

1 | |

2 | |

3 | |

4 |

Question 7 Explanation:

Using KCL, I_s=I_L-I_D

For inductively loaded circuits, load can be assumed to be constant.

\therefore \; I_s is maximum when, I_D is minimum, i.e. at point 3.

Therefore, IGBT experiences highest current stress at point 3.

Question 8 |

Thyristor T_1 is triggered at an angle \alpha (in degree), and T_2 at angle 180^{\circ} + \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:

0^{\circ} \; to \; 60^{\circ} | |

0^{\circ} \; to \; 120^{\circ} | |

60^{\circ} \; to \; 120^{\circ} | |

60^{\circ} \; to \; 180^{\circ} |

Question 8 Explanation:

As per GATE official answer key, MTA (Marks to All)

As load is capacitive, for any \alpha :

\begin{aligned} V_{or}&=\frac{V_m}{\sqrt{2\pi}}\left [ \pi - \alpha +\frac{1}{2} \sin 2\alpha \right ]^{1/2}\\ P_0&=\left ( \frac{V_{or}}{z} \right )^2 \times R\\ &=\frac{V_{or}^2}{100} \times 5=\frac{V_{or}^2}{20}\\ \text{For }\alpha &=0^{\circ}\\ V_{or}&=\frac{V_m}{\sqrt{2}}=200\\ P_0&=\frac{200}{20}=2KW \end{aligned}

For \alpha =120^{\circ}, as current distinguishes, so P=0.

So, \alpha should be 0^{\circ} \text{ to } 120^{\circ}.

As load is capacitive, for any \alpha :

\begin{aligned} V_{or}&=\frac{V_m}{\sqrt{2\pi}}\left [ \pi - \alpha +\frac{1}{2} \sin 2\alpha \right ]^{1/2}\\ P_0&=\left ( \frac{V_{or}}{z} \right )^2 \times R\\ &=\frac{V_{or}^2}{100} \times 5=\frac{V_{or}^2}{20}\\ \text{For }\alpha &=0^{\circ}\\ V_{or}&=\frac{V_m}{\sqrt{2}}=200\\ P_0&=\frac{200}{20}=2KW \end{aligned}

For \alpha =120^{\circ}, as current distinguishes, so P=0.

So, \alpha should be 0^{\circ} \text{ to } 120^{\circ}.

Question 9 |

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:

50 Hz, 0 Hz | |

50 Hz, 100 Hz | |

50 Hz, 150 Hz | |

50 Hz, 250 Hz |

Question 9 Explanation:

For full bridge rectifier, High inductive load

\begin{aligned} I_{source}&=\sum_{n=1,3,5}^{\infty }\frac{4I_0}{n \pi} \sin nd \sin \frac{n \pi}{2} \sin (n\omega t+\phi _n) \\ 2d &=\pi , \;\; d=\frac{\pi}{2}\\ I_{source}&= \frac{4I_0}{n \pi} \sum_{n=1,3,5}^{\infty } \sin ^2 n\left ( \frac{\pi}{2} \right ) \sin (n\omega t+\phi _n)\\ I_{source}&\neq 0 \end{aligned} For n=1,3,5,...

Since the most dominant frequency component will be f, 3f, f = 50

So dominant frequency = 50 Hz, 150 Hz.

\begin{aligned} I_{source}&=\sum_{n=1,3,5}^{\infty }\frac{4I_0}{n \pi} \sin nd \sin \frac{n \pi}{2} \sin (n\omega t+\phi _n) \\ 2d &=\pi , \;\; d=\frac{\pi}{2}\\ I_{source}&= \frac{4I_0}{n \pi} \sum_{n=1,3,5}^{\infty } \sin ^2 n\left ( \frac{\pi}{2} \right ) \sin (n\omega t+\phi _n)\\ I_{source}&\neq 0 \end{aligned} For n=1,3,5,...

Since the most dominant frequency component will be f, 3f, f = 50

So dominant frequency = 50 Hz, 150 Hz.

Question 10 |

A single-phase fully-controlled thyristor converter is used to obtain an average voltage of 180 V with 10 A constant current to feed a DC load. It is fed from single-phase AC supply of 230 V, 50 Hz. Neglect the source impedance. The power factor (round off to two decimal places) of AC mains is________

0.25 | |

0.55 | |

0.78 | |

0.95 |

Question 10 Explanation:

V_{sr}I_{sr} \cos \phi =V_0I_0

For single-phase fully controlled converter,

I_0=I_{sr}=10A

\cos \phi =\frac{V_0}{V_{sr}}=\frac{180}{230}=0.78

For single-phase fully controlled converter,

I_0=I_{sr}=10A

\cos \phi =\frac{V_0}{V_{sr}}=\frac{180}{230}=0.78

There are 10 questions to complete.