Inverters

Question 1
A 3-phase grid-connected voltage source converter with DC link voltage of 1000 V is switched using sinusoidal Pulse Width Modulation (PWM) technique. If the grid phase current is 10 A and the 3-phase complex power supplied by the converter is given by (-4000 - j3000) VA, then the modulation index used in sinusoidal PWM is ___________. (round off to two decimal places)
A
0.23
B
0.47
C
0.64
D
0.87
GATE EE 2022   Power Electronics
Question 1 Explanation: 
Apparent power, S = (-4000 - j3000) VA
or S=5000\angle -143.13^{\circ}VA
S=\sqrt{3}V_LI_L
V_L=\frac{5000}{\sqrt{3} \times 10}=288.675V
We have,
Peak, V_{01}=\sqrt{3}M_A\frac{V_s}{2}
Put the values
\sqrt{2}\times 288.675=\sqrt{3}M_A \times \frac{1000}{2}
\Rightarrow \; M_A=0.471
Question 2
Consider an ideal full-bridge single-phase DC-AC inverter with a DC bus voltage magnitude of 1000 V. The inverter output voltage v(t) shown below, is obtained when diagonal switches of the inverter are switched with 50 % duty cycle. The inverter feeds a load with a sinusoidal current given by, i(t)=10 \sin (\omega t-\frac{\pi}{3})A, where \omega =\frac{2\pi}{T}. The active power, in watts, delivered to the load is _________. (round off to nearest integer)

A
2154
B
3254
C
3181
D
4578
GATE EE 2022   Power Electronics
Question 2 Explanation: 
For 1-\phi invertor, RMS value of fundamental component
V_{01}=\frac{2\sqrt{2}V_s}{\pi}
Now, Power output
=V_{01}I_{or} \cos \frac{\pi}{3}=\frac{2\sqrt{2}}{\pi} \times 1000 \times \frac{10}{\sqrt{2}} \times \frac{1}{2}=3183.098W
Question 3
A single-phase full-bridge inverter fed by a \text{325 V DC} produces a symmetric quasi-square waveform across 'ab' as shown. To achieve a modulation index of 0.8, the angle \theta expressed in degrees should be _______________. (Round off to 2 decimal places.)
(Modulation index is defined as the ratio of the peak of the fundamental component of V_{ab} to the applied \text{DC} value.)

A
51.1
B
28.06
C
58.3
D
69.24
GATE EE 2021   Power Electronics
Question 3 Explanation: 

\begin{aligned} \widehat{V_{01}} &=m_{a} V_{S}=0.8 \times 325=260 \mathrm{~V} \\ \widehat{V_{01}} &=\frac{4 V_{S}}{\pi} \sin d=260 \\ \frac{4(325)}{\pi} \sin d &=260 \\ \sin d &=\frac{260 \times \pi}{4 \times 325}=0.628 \\ d &=38.9 \\ \therefore \qquad \qquad \theta &=\frac{\pi}{2}-d=90^{\circ}-38.9=51.1 \end{aligned}
Question 4
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 4 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 5
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 5 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 6
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 6 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 7
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 7 Explanation: 


V_{01}=\frac{4V_s}{\sqrt{2} \pi}=\frac{4 \times 220}{\sqrt{2} \pi}=198.06V
Question 8
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 8 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 9
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 9 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 10
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 10 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}
There are 10 questions to complete.