Q21GATE 2011MCQ

A three phase current source inverter used for the speed control of an induction motor is to be realized using MOSFET switches as shown below. Switches $S_1 \; to \; S_6$ are identical switches. The proper configuration for realizing switches $S_1 \; to \; S_6$ is

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Q23GATE 2008MCQ

A 3-phase voltage source inverter is operated in 180 $^{\circ}$ conduction mode. Which one of the following statements is true ?

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📖 Explanation

In 3-phase VSI working in $180^{\circ}$ mode Pole voltage is a squre wave. Therefore 3rd harmonic is present in pole voltage. In line voltage waveform even and tripplen harmonics are not present.

Q24GATE 2008MCQ

A single phase source inverter is feeding a purely inductive load as shown in the figure The inverter is operated at 50 Hz in 180 $^{\circ}$ square wave mode. Assume that the load current does not have any dc component. The peak value of the inductor current $i_0$ will be

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📖 Explanation

Load current increases linearly from $-I_P \; to \;I_P$ during $0 \lt \omega t\leq \pi$ , where $I_P=$ peak value of $i_0$

\begin{aligned} \Rightarrow \; t_1&\frac{\pi}{\omega }=\frac{\pi}{2 \pi 50}=\frac{1}{100}sec \\ \text{At } t&=0, i_0=-I_P \text{ and }t=t_1, i_0=I_P \\ V_0 &=L\frac{di}{dt} \\ \text{During } & 0\lt \omega t\leq \pi \\ V_0&=V_s =200V\\ \Rightarrow \; 200 &=0.1\left ( \frac{I_P-(-I_P)}{t_1-0} \right ) \\ \Rightarrow \; 2I_P&= \frac{200}{0.1}\times t_1\\ I_P &=\frac{2000}{2}\times \frac{1}{100}=10A \end{aligned}

Q25GATE 2007MCQ

A single-phase voltages source inverter is controlled in a single pulse-width modulated mode with a pulse width of 150 $^{\circ}$ in each half cycle. Total harmonic distortion is defined as $THD=\frac{\sqrt{V^{2}_{rms}-V_{1}^{2}}}{V_{1}} \times 100$ , where $V_1$ is the rms value of the fundamental component of the output voltage. The THD of output ac voltage waveform is

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📖 Explanation

Pulse width $=2d=150^{\circ}$

\begin{aligned} \Rightarrow \; d&=75^{\circ} \\ V_{rms}&=\sqrt{\frac{1}{\pi}\int_{\pi/2-d}^{\pi/2+d}V_s^2 d (\omega t)} \\ &= V_s\sqrt{\frac{2d}{\pi}}=V_s\sqrt{\frac{5}{6}}=0.913V_s \end{aligned}

Output voltage in fourier series, $V_0=\sum_{n=1,3,5}^{\infty }\frac{4V_s}{n \pi} \sin \frac{n \pi}{2} \sin nd \sin n\omega t$ RMS value of fundamental component of output voltage, $V_1=\frac{1}{\sqrt{2}}\left[ \frac{4V_s}{\pi} \sin 75^{\circ} \right]\$ =0.87V_s$ Total harmonic Distortion,

\begin{aligned} THD &=\sqrt{\frac{V_{rms}^{2-}V_1^2}{V_1^2} } \times 100 \\ &= \sqrt{\frac{(0.913V_s)^{2-}(0.87V_s)^2}{(0.87V_s)^2}} \times 100\\ &= 31.83 \% \end{aligned}

Q26GATE 2007MCQ

"Six MOSFETs connected in a bridge configuration (having no other power device) must be operated as a Voltage Source Inverter (VSI)". This statement is

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Q27GATE 2007MCQ

A voltage source inverter is used to control the speed of three-phase, 50 Hz, squirrel cage induction motor. Its slip for rated torque is 4%. The flux is maintained at rated value. If the stator resistance and rotational losses are neglected, then the frequency of the impressed voltage to obtain twice the rated torque at starting should be

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Q28GATE 2006MCQ

A single-phase inverter is operated in PWM mode generating a single-pulse of width 2d in the centre of each half cycle as shown in figure. It is found that the output voltage is free from 5th harmonic for pulse width 144 $^{\circ}$ . What will be percentage of 3rd harmonic present in the output voltage ( $V_{o3}/V_{o1max}$ ) ?

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📖 Explanation

$\left ( \frac{V_{03}}{V_{01}} \right )_{max}=\frac{\frac{4V_s}{3 \pi}\sin 3d}{\frac{4V_s}{ \pi}}=19.6\%$

Q29GATE 2005MCQ

The output voltage waveform of a three-phase square-wave inverter contains

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📖 Explanation

The output voltage in both $180^{\circ}$ and $120^{\circ}$ conduction mode can be described by the fourier series as follow: In $180^{\circ}$ conduction mode, line output voltage $V_{ab}=\sum_{n=1,3,5}^{\infty }\frac{4V_s}{n \pi}\cos \frac{n \pi }{6} \sin n \left ( \omega t +\frac{\pi}{6}\right )\;\;...(i)$ line to neutral output voltage $V_{a0}=\sum_{n=6k\pm 1}^{\infty }\frac{2V_s}{n \pi}\sin n \omega t \;\;...(ii)$ where k=0,1,2,... In $120^{\circ}$ conduction mode line to neutral output voltage $V_{a0}=\sum_{n=1,3,5}^{\infty }\frac{2V_s}{n \pi}\cos \frac{n \pi }{6} \sin n \left ( \omega t +\frac{\pi}{6}\right )\;\;...(iii)$ Line output voltage $V_{ab}=\sum_{n=6k\pm 1}^{\infty }\frac{3V_s}{n \pi} \sin n \left ( \omega t +\frac{\pi}{3}\right )\;\;...(iv)$ where k=0,1,2... It is clear from equation (i) to (iv), output voltage wave form contains only odd-harmonic.

Q30GATE 2003MCQ

With reference to the output wave form given in the figure, the output of the converter will be free from 5th harmonic when

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📖 Explanation

To eliminate fifth harmonic,

\begin{aligned} &5d=0, \pi, 2 \pi\Rightarrow d\Rightarrow 0, \frac{\pi}{5},\frac{2 \pi}{5} \\ \therefore & \text{ Pulse width }=2d=\alpha =0,\frac{2\pi}{5},\frac{4 \pi}{5}\\ &= (0^{\circ},72^{\circ},144^{\circ}) \end{aligned}

Q31GATE 2003MCQ

An inverter has a periodic output voltage with the output wave form as shown in figure When the conduction angle $\alpha =120^{\circ}$ , the rms fundamental component of the output voltage is

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📖 Explanation

Output voltage, $V_0=\sum_{n=1,3,5...}^{\infty }\frac{4V_s}{n \pi} \sin nd \;\sin \frac{n \pi}{2} \;\sin n\omega t$ R.M.S. value of fundamental component, n=1 $=\frac{4V_s}{\sqrt{2}\pi} \sin d \times 1$ $\alpha =120, \;\therefore \;\; 2d=120^{\circ}, \;\therefore \;\; d=60^{\circ}$ $=\frac{4V_s}{\sqrt{2}\pi} \sin 60^{\circ}=0.78 V_s=0.78V$

Q32GATE 2002MCQ

Figure shows an inverter circuit with a dc source voltage $V_{s}$ . The semiconductor switches of the inverter are operated in such a manner that the pole voltages $V_{10}$ and $V_{20}$ are as shown in figure. What is the rms value of the pole-to-pole voltage $V_{12}$ ?

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📖 Explanation

\begin{aligned} V_{12}&=V_{10}-V_{20} \\ V_{rms}&=V_s\sqrt{\frac{\phi }{\pi}} \end{aligned}

Q33GATE 2001MCQ

A single-phase full-bridge voltage source inverter feeds a purely inductive load, a shown in figure, where $T_{1}$ , $T_{2}$ , $T_{3}$ , $T_{4}$ are power transistors and $D_{1}$ , $D_{2}$ , $D_{3}$ , $D_{4}$ are feedback diodes. The inverter is operated in square-wave mode with a frequency of 50 Hz. If the average load current is zero, what is the time duration of conduction of each feedback diode in a cycle?

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📖 Explanation

Conduction time $=\frac{T}{4}=\frac{20}{4}=5 \; ms$