Inverters

Question 1
The output voltage of a single-phase full bridge voltage source inverter is controlled by unipolar PWM with one pulse per half cycle. For the fundamental rms component of output voltageto be 75% of DC voltage, the required pulse width in degrees (round off up to one decimal place) is _________
A
82.6
B
98.6
C
112.8
D
168.2
GATE EE 2019   Power Electronics
Question 1 Explanation: 
\begin{aligned} V_{01,rms}&=\frac{2\sqrt{2}}{\pi}\\ 0.9 \sin d \; V_s&=0.75V_s\\ d&=56.44^{\circ}\\ 2d&=112.88^{\circ}\\ \text{Pulse width }&=112.88^{\circ} \end{aligned}
Question 2
The figure below shows a half-bridge voltage source inverter supplying an RL-load with R=40\Omega and L=(\frac{0.3}{\pi })H. The desired fundamental frequency of the load voltage is 50 Hz. The switch control signals of the converter are generated using sinusoidal pulse width modulation with modulation index. M = 0.6. At 50 Hz, the RL-load draws an active power of 1.44 kW. The value of DC source voltage V_{DC} in volts is
A
300\sqrt{2}
B
500
C
500\sqrt{2}
D
1000\sqrt{2}
GATE EE 2017-SET-2   Power Electronics
Question 2 Explanation: 
\begin{aligned} V_{01 peak}&=M\cdot \frac{V_s}{2} \;\;\;M\rightarrow \text{Modulation index}\\ V_{01peak}&=\frac{M\cdot \frac{V_s}{2}}{\sqrt{2}}=M\cdot \frac{V_s}{2\sqrt{2}}\\ V_{01peak}&=\frac{0.6V_s}{2\sqrt{2}} \end{aligned}

\begin{aligned} V_{01}&=\frac{0.3}{\sqrt{2}}V_s\\ |Z_1|&=\sqrt{R^2+(\omega L)^2}\\ &=\sqrt{40^2+(2 \pi f L)^2}\\ &=\sqrt{40^2+\left (2 \pi \times 50 \times \frac{0.3}{\pi} \right )^2}\\ &=\sqrt{40^2+30^2}=50\\ \phi _1&=\tan ^{-1}\frac{\omega L}{R}=\tan ^{-1}\frac{30}{40}=36.896^{\circ}\\ \text{Active power} &=V_{01}I_{01} \cos \phi \\ &=V_{01}\frac{V_{01}}{|Z_1|} \cos \phi \\ 1.44 \times 10^3&=\left (\frac{0.3}{\sqrt{2}}V_s \right )^2\cdot \frac{1}{50}(\cos 36.896^{\circ} )\\ \therefore \;\;V_s^2&=\frac{1.44 \times 10^3 \times 100}{0.3^2 (0.8)}\\ &=20 \times 10^5\\ V_s&=\sqrt{2} \cdot 10^3=1000\sqrt{2}\\ V_{DC}&=\frac{V_s}{2}=500\sqrt{2}V \end{aligned}
Question 3
A three-phase voltage source inverter with ideal devices operating in 180^{\circ} conduction mode is feeding a balanced star-connected resistive load. The DC voltage input is V_{dc}. The peak of the fundamental component of the phase voltage is
A
\frac{V_{dc}}{\pi}
B
\frac{2V_{dc}}{\pi}
C
\frac{3V_{dc}}{\pi}
D
\frac{4V_{dc}}{\pi}
GATE EE 2017-SET-2   Power Electronics
Question 3 Explanation: 
3-\phi \;\text{ VSI }180^{\circ} \text{ mode:}

\begin{aligned} V_{Rn}&=\frac{6 V_{dc}/3}{n \pi} \sin n\omega t\\ &= \frac{2 \times V_{dc}}{n \pi} \sin n\omega t \\ V_{Rn} &= \frac{2 V_{dc}}{ \pi} \sin \omega t \end{aligned}
Question 4
In the converter circuit shown below, the switches are controlled such that the load voltage v_o(t) is a 400 Hz square wave. The RMS value of the fundamental component of v_o(t) in volts is _______.
A
220
B
440
C
198
D
110
GATE EE 2017-SET-1   Power Electronics
Question 4 Explanation: 


V_{01}=\frac{4V_s}{\sqrt{2} \pi}=\frac{4 \times 220}{\sqrt{2} \pi}=198.06V
Question 5
A 3-phase voltage source inverter is supplied from a 600V DC source as shown in the figure below. For a star connected resistive load of 20 \Omega per phase, the load power for 120^{\circ} device conduction, in kW is __________.
A
18
B
9
C
245
D
90
GATE EE 2017-SET-1   Power Electronics
Question 5 Explanation: 


\begin{aligned} V_{rms}&=\sqrt{\frac{1}{\pi}(300)^2\frac{2 \pi}{3}}\\ &=300\sqrt{\frac{2}{3}}=244.94V\\ 3\frac{V^2}{R}&=\frac{3 \times (244.94)^2}{20}=9kW \end{aligned}
Question 6
A three-phase Voltage Source Inverter (VSI) as shown in the figure is feeding a delta connected resistive load of 30 \Omega/phase. If it is fed from a 600 V battery, with 180^{\circ} conduction of solid-state devices, the power consumed by the load, in kW, is __________.
A
10
B
20
C
24
D
30
GATE EE 2016-SET-2   Power Electronics
Question 6 Explanation: 
\begin{aligned} V_L&=V_{ph}=\sqrt{\frac{2}{3}}V_s\\ V_{ph}&=\sqrt{\frac{2}{3}} \times 600\\ P&=3\frac{V_{ph}^2}{R}=\frac{3 \times \frac{2}{3} \times 600^2}{30}\\ &=24kW \end{aligned}
Question 7
A single-phase full-bridge voltage source inverter (VSI) is fed from a 300 V battery. A pulse of 120^{\circ} duration is used to trigger the appropriate devices in each half-cycle. The rms value of the fundamental component of the output voltage, in volts, is
A
234
B
245
C
300
D
331
GATE EE 2016-SET-1   Power Electronics
Question 7 Explanation: 
\begin{aligned} V_{01(rms)}&=\frac{2\sqrt{2}}{\pi}V_s \cdot \sin d\\ \text{Pulse width} & \text{ where } 2d=120^{\circ}\\ d=60^{\circ}\\ V_{01(rms)}&=\frac{2\sqrt{2}}{\pi}V_s \sin 60^{\circ}\\ &=\frac{2\sqrt{2}}{\pi}V_s \frac{\sqrt{3}}{2}=234V \end{aligned}
Question 8
The switches T1 and T2 in Figure (a) are switched in a complementary fashion with sinusoidal pulse width modulation technique. The modulating voltage v_{m}(t) = 0.8 \sin (200 \pi t) V and the triangular carrier voltage (v_{c}) are as shown in Figure (b). The carrier frequency is 5 kHz. The peak value of the 100 Hz component of the load current (i_{L}), in ampere, is ________ .
A
10
B
100
C
150
D
20
GATE EE 2016-SET-1   Power Electronics
Question 8 Explanation: 
\begin{aligned} m_a&=0.8\\ (V_{01})_{peak}&=m_a\frac{V_d}{2}\;\;[m_a\leq 1]\\ &=0.8 \times 250=200V\\ (I_{01})_{peak}&=\frac{(V_{01})_{peak}}{Z_1}\\ &=\frac{200}{\sqrt{R^2}+(\omega L)^2}\\ &=\frac{200}{\sqrt{12^2+16^2}}=10A \end{aligned}
Question 9
The single-phase full-bridge voltage source inverter (VSI), shown in figure, has an output frequency of 50 Hz. It uses unipolar pulse width modulation with switching frequency of 50 kHz and modulation index of 0.7. For V_{in} =100 V DC, L=9.55 mH, C=63.66 \muF, and R=5 \Omega, the amplitude of the fundamental component in the output voltage V_o (in Volt) under steady-state is ________.
A
12.25
B
40.25
C
62.75
D
82.75
GATE EE 2015-SET-1   Power Electronics
Question 9 Explanation: 
The rms value of the fundamental component of output voltage,
\begin{aligned} V_{R1}&=\frac{V_s}{\sqrt{2}}\times m_i \\ &= \frac{100}{\sqrt{2}} \times 0.7=49.49V \end{aligned}
Amplitude of the value is peak value (i.e.)
V_{R_{peak}}=\sqrt{2} \times 49.49=70V

V_{R_{peak}} is the amplitude of fundamental voltage at the VSI. V_{0_{(peak)}} is the amplitude of fundamental voltage at the resistor.
\begin{aligned} Z&=\frac{R(-jX_C)}{R-X_Cj} \\ &= \frac{5(-j50)}{5-j50}\\ &= 4.97\angle -5.71^{\circ}\Omega \\ \text{where, } X_C&= \frac{1}{\omega C}\\ &=\frac{1}{2 \pi \times 50 \times 63.66 \times 10^{-6}} \\ &= 50\Omega \\ X_L &=\omega L \\ &=2 \pi \times 50 \times 9.55 \times 10^{-3} \\ &= 3 \Omega\\ V_{0(peak)}&=V_{R_{peak}} \times \frac{4.97\angle -5.71^{\circ}}{(4.97\angle -5.71^{\circ})+j3}\\ &=62.75\angle -32.5^{\circ}\;V \end{aligned}
Therefore, the amplitude of the fundamental component in the output voltage,
V_0=62.75V
Question 10
A single-phase voltage source inverter shown in figure is feeding power to a load. The triggering pulses of the devices are also shown in the figure.

If the load current is sinusoidal and is zero at 0,\pi ,2\pi.., the node voltage V_{AO} has the waveform
A
A
B
B
C
C
D
D
GATE EE 2014-SET-3   Power Electronics
Question 10 Explanation: 
\theta \leq \omega t \leq \pi-\theta \;\;(S_{11}S_4\rightarrow ON)

\begin{aligned} V_A &=V_{DC}, \; V_0=\frac{V_{DC}}{2} \\ V_{AO}&=\frac{V_{DC}}{2}\;\; (\pi-\theta \lt \omega t \lt z) \end{aligned}

\begin{aligned} V_A &=0\\ V_{AO}&=V\cdot 0-\frac{V_{DC}}{2}\\ &=-\frac{V_{DC}}{2} \end{aligned}
\pi +\theta \leq \omega t\leq 2 \pi

V_A=0,\;\; V_0=\frac{V_d}{2}
There are 10 questions to complete.
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