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)
0.23 | |
0.47 | |
0.64 | |
0.87 |
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
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)
2154 | |
3254 | |
3181 | |
4578 |
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
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.)
(Modulation index is defined as the ratio of the peak of the fundamental component of V_{ab} to the applied \text{DC} value.)
51.1 | |
28.06 | |
58.3 | |
69.24 |
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 _________
82.6 | |
98.6 | |
112.8 | |
168.2 |
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
300\sqrt{2} | |
500 | |
500\sqrt{2} | |
1000\sqrt{2} |
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}
\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}
There are 5 questions to complete.