Question 1 |

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 1 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 2 |

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 2 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 3 |

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 3 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 4 |

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)

Triac only | |

Triac and MOSFET | |

Triac and GTO | |

Thyristor and Triac |

Question 5 |

For the power semiconductor devices IGBT, MOSFET, Diode and Thyristor, which one of the
following statements is TRUE?

All of the four are majority carrier devices. | |

All the four are minority carrier devices | |

IGBT and MOSFET are majority carrier devices, whereas Diode and Thyristor are minority
carrier devices. | |

MOSFET is majority carrier device, whereas IGBT, Diode, Thyristor are minority carrier
devices. |

Question 6 |

The voltage (v_{s}) across and the current (i_{s}) through a semiconductor switch during a turn-ON transition are shown in figure. The energy dissipated during the turn-ON transition, in mJ, is _______.

25 | |

50 | |

75 | |

100 |

Question 6 Explanation:

\begin{aligned} \text{Energy}&=\int_{0}^{T_1}V\cdot i dt+\int_{0}^{T_2}V\cdot i dt\\ &=V\left [ \frac{1}{2}IT_1 \right ]+I\left [ \frac{1}{2}VT_2 \right ]\\ &=600\left [ \frac{150}{2}\times 1 \times 10^{-6} \right ]\\ &+100\left [ \frac{1}{2} \times 600 \times 1 \times 10^{-6} \right ]\\ \text{ Energy}&=75mJ \end{aligned}

Question 7 |

A steady dc current of 100 A is flowing through a power module (S,D) as shown in Figure (a). The V-I characteristics of the IGBT (S) and the diode (D) are shown in Figures (b) and (c), respectively.
The conduction power loss in the power module (S,D), in watts, is ________.

100 | |

135 | |

240 | |

170 |

Question 7 Explanation:

No current flows through IGBT. So current flows only in diode. Equivalent circuit of diode is as shown below,

\begin{aligned} V&=IR+0.7\\ &=(100 \times 0.01)+0.7\\ V&=1.7V \end{aligned}

Power loss during conduction,

\begin{aligned} P&=V \times I\\ &= 1.7 \times 100=170W \end{aligned}

\begin{aligned} V&=IR+0.7\\ &=(100 \times 0.01)+0.7\\ V&=1.7V \end{aligned}

Power loss during conduction,

\begin{aligned} P&=V \times I\\ &= 1.7 \times 100=170W \end{aligned}

Question 8 |

The circuit shown is meant to supply a resistive load R_{L} from two separate DC voltage sources. The switches S1 and S2 are controlled so that only one of them is ON at any instant. S1 is turned on for 0.2 ms and S2 is turned on for 0.3 ms in a 0.5 ms switching cycle time period. Assuming continuous conduction of the inductor current and negligible ripple on the capacitor voltage, the output voltage V_{0} (in Volt) across R_{L} is ___________.

3 | |

7 | |

9 | |

11 |

Question 8 Explanation:

S_1 is turn ON for 0.2 ms

S_2 is turn ON for 0.3 ms.

Switching cycle time period is 0.5 ms.

the output voltage V_0 across R_L is

\begin{aligned} V_0&= \frac{1}{0.5 \times 10^{-3}}\left [ \int_{0}^{0.2}10dt +\int_{0.2}^{0.5}5 dt \right ]\\ &= \frac{1}{0.5 \times 10^{-3}} [10(0.2)+5(0.5-0.2)]\\ &=\frac{1}{0.5 \times 10^{-3}} [2 \times 10^{-3}+1.5 \times 10^{-3}] \\ &= \frac{3.5 \times 10^{-3}}{0.5 \times 10^{-3}}=7V \end{aligned}

Question 9 |

Figure shows four electronic switches (i), (ii), (iii) and (iv). Which of the switches
can block voltages of either polarity (applied between terminals 'a' and 'b') when
the active device is in the OFF state ?

(i), (ii) and (iii) | |

(ii), (iii) and (iv) | |

(ii) and (iii) | |

(i) and (iv) |

Question 9 Explanation:

As given that the active device is in OFF state it means the device can block the voltage when 'a' is positive with respect to 'b' and when 'b' is positive with respect to 'a' then the diode is reverse baised and voltage is blocked.

When 'a' is positive with respect to 'b', as given the device is OFF so it will block the voltage.

When 'b' is positive with respect to 'a' it will block the voltage.

Where as the other two devices will conduct when 'b' is positive with respect to 'a'.

Question 10 |

The typical ratio of latching current to holding current in a 20 A thyristor is

5 | |

2 | |

1 | |

0.5 |

Question 10 Explanation:

For medium power thyristor of rating 6 A to 60 A the ratio of the latching current to holding current is 1.5 to 2.

There are 10 questions to complete.