Question 1 |

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 1 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 2 |

For the circuit shown in the figure below, assume that diodes D_1,D_2 and D_3 are ideal.

The DC components of voltages v_1 \; and \; v_2, respectively are

The DC components of voltages v_1 \; and \; v_2, respectively are

0 V and 1 V | |

-0.5 V and 0.5 V | |

1 V and 0.5 V | |

1 V and 1 V |

Question 2 Explanation:

\begin{aligned} V_{2\; avg}&=\frac{V_m}{\pi}=\frac{\pi/2}{\pi}=\frac{1}{2}=0.5V \\ V_{1\; avg} &=\frac{1}{2 \pi}[ \int_{0}^{\pi}\frac{\pi}{2}\sin 100 \pi t\cdot d(\omega t)\\ &+\int_{\pi}^{2\pi} \pi \sin 100 \pi t\cdot d(\omega t) ]\\ &= -0.5V \end{aligned}

Question 3 |

Consider a HVDC link which uses thyristor based line-commutated converters as shown in the figure. For a power flow of 750 MW from System 1 to System 2, the voltages at the two ends, and the current, are given by: V_{1}=500 kV, V_{2}=485 kV and V_{3}=1.5 kA. If the direction of power flow is to be reversed (that is, from System 2 to System 1) without changing the electrical connections, then which one of the following combinations is feasible?

V_{1}=-500 kV,V_{2}=-485 kV and latex]I=1.5kA[/latex] | |

V_{1}=-485 kV,V_{2}=500 kV and latex]I=1.5kA[/latex] | |

V_{1}=500 kV,V_{2}=-485 kV and latex]I=-1.5kA[/latex] | |

V_{1}=-500 kV,V_{2}=-485 kV and latex]I=-1.5kA[/latex] |

Question 3 Explanation:

To maintain the direction of power flow from system 2 to system 1, the voltage V_1=-485kV and voltage V_2=500kV and I=1.5kA.

Since, current cannot flow in reverse direction. Option (B) is correct answer.

Since, current cannot flow in reverse direction. Option (B) is correct answer.

Question 4 |

The SCR in the circuit shown has a latching current of 40 mA. A gate pulse of
50 \mus is applied to the SCR. The maximum value of R in \Omega to ensure successful
firing of the SCR is ______.

4050 | |

5560 | |

6060 | |

8015 |

Question 4 Explanation:

Let us assume the SCR is conducting,

\begin{aligned} &I_{ss}=\frac{100}{500}=0.2A\\ &[\because \text{inductor will be dhort circuited in DC}]\\ &i(t)=I_{ss}(1-e^{-t/\tau })\\ &\tau =\frac{L}{R}=\frac{200 \times 10^{-3}}{500}\\ &\;\;=4 \times 10^{-4}\; sec\\ &\text{Given }t=50 \times 10^{-6}\; sec\\ &\therefore \; i(t)=0.2\left ( 1-e^{\frac{50 \times 10^{-6}}{4 \times 10^{-4}}} \right )=23.5A \end{aligned}

V=I \times R

R=\frac{V}{I}=\frac{100}{16.5 \times 10^{-3}} =6060 \Omega

\begin{aligned} &I_{ss}=\frac{100}{500}=0.2A\\ &[\because \text{inductor will be dhort circuited in DC}]\\ &i(t)=I_{ss}(1-e^{-t/\tau })\\ &\tau =\frac{L}{R}=\frac{200 \times 10^{-3}}{500}\\ &\;\;=4 \times 10^{-4}\; sec\\ &\text{Given }t=50 \times 10^{-6}\; sec\\ &\therefore \; i(t)=0.2\left ( 1-e^{\frac{50 \times 10^{-6}}{4 \times 10^{-4}}} \right )=23.5A \end{aligned}

V=I \times R

R=\frac{V}{I}=\frac{100}{16.5 \times 10^{-3}} =6060 \Omega

Question 5 |

A single-phase SCR based ac regulator is feeding power to a load consisting of 5 \Omega
resistance and 16 mH inductance. The input supply is 230 V, 50 Hz ac. The maximum firing angle at which the voltage across the device becomes zero all throughout and the rms value of current through SCR, under this operating condition, are

30^{\circ} and 46 A | |

30^{\circ} and 23 A | |

45^{\circ} and 23 A | |

45^{\circ} and 32 A |

Question 5 Explanation:

The maximum firing angle at which the voltage across the device becomes '\phi ' = load angle.

\begin{aligned} \phi &= \tan ^{-1} \left ( \frac{\omega L}{R} \right )\\ &= \tan^{-1}\left ( \frac{2 \pi \times 50 \times 16 \times 10^{-3}}{5} \right ) \\ \phi &=45.15\simeq 45^{\circ} \end{aligned}

Rms value of current through SCR is

\begin{aligned} I_{T_{rms}} &=\sqrt{\left [ \frac{1}{2 \pi}\int_{\alpha }^{\pi+\alpha }\left ( \frac{V_m}{z}\sin (\omega t-\phi ) \right )^2 d(\omega t) \right ]}\\ &=\sqrt{\frac{V_m^2}{2 \pi z^2} \int_{\alpha }^{\pi+\alpha } \left [ \frac{1-\cos 2(\omega t-\phi )}{2} \right ] d(\omega t)}\\ &=\sqrt{\frac{V_m^2}{2 \times 2 \pi z^2} \left [\pi- \left.\begin{matrix} \frac{\sin 2(\omega t-\phi )}{2} \end{matrix}\right|_\alpha ^{(\pi+\alpha )} \right ] }\\ &=\frac{V_m}{2z \sqrt{\pi} } \sqrt{\left [ \pi+\frac{-\sin 2(\pi+\alpha -\alpha )+\sin 2(\alpha -\alpha )}{2} \right ] }\\ &=\frac{V_m}{2z \sqrt{\pi} }\sqrt{\pi}=\frac{V_m}{2z}\\ I_{T_{rms}}&=\frac{230\sqrt{2}}{2 \times \sqrt{5^2(2 \pi \times 50 \times 16 \times 10^{-3})^2}}\\ &=22.93\simeq 23A\\ I_{T_{rms}}&=23A \end{aligned}

\begin{aligned} \phi &= \tan ^{-1} \left ( \frac{\omega L}{R} \right )\\ &= \tan^{-1}\left ( \frac{2 \pi \times 50 \times 16 \times 10^{-3}}{5} \right ) \\ \phi &=45.15\simeq 45^{\circ} \end{aligned}

Rms value of current through SCR is

\begin{aligned} I_{T_{rms}} &=\sqrt{\left [ \frac{1}{2 \pi}\int_{\alpha }^{\pi+\alpha }\left ( \frac{V_m}{z}\sin (\omega t-\phi ) \right )^2 d(\omega t) \right ]}\\ &=\sqrt{\frac{V_m^2}{2 \pi z^2} \int_{\alpha }^{\pi+\alpha } \left [ \frac{1-\cos 2(\omega t-\phi )}{2} \right ] d(\omega t)}\\ &=\sqrt{\frac{V_m^2}{2 \times 2 \pi z^2} \left [\pi- \left.\begin{matrix} \frac{\sin 2(\omega t-\phi )}{2} \end{matrix}\right|_\alpha ^{(\pi+\alpha )} \right ] }\\ &=\frac{V_m}{2z \sqrt{\pi} } \sqrt{\left [ \pi+\frac{-\sin 2(\pi+\alpha -\alpha )+\sin 2(\alpha -\alpha )}{2} \right ] }\\ &=\frac{V_m}{2z \sqrt{\pi} }\sqrt{\pi}=\frac{V_m}{2z}\\ I_{T_{rms}}&=\frac{230\sqrt{2}}{2 \times \sqrt{5^2(2 \pi \times 50 \times 16 \times 10^{-3})^2}}\\ &=22.93\simeq 23A\\ I_{T_{rms}}&=23A \end{aligned}

There are 5 questions to complete.