Q4GATE 2024NAT

A single-phase full bridge voltage source inverter (VSI) feeds a purely inductive load. The inverter output voltage is a square wave in $180^{\circ}$ conduction mode. The fundamental frequency of the output voltage is $50 \mathrm{~Hz}$ . If the DC input voltage of the inverter is $100 \mathrm{~V}$ and the value of the load inductance is $20 \mathrm{mH}$ , the peak-to-peak load current in amperes is ______ (rounded off to the nearest integer).

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} & 1-\phi \mathrm{VSI}, f_{1}=f_{0}=50 \mathrm{~Hz} \\ & V_{S}=100 \mathrm{~V} \end{aligned}

$V_{0}=V_{S} \quad\left[0 \leq t \leq \frac{T}{2}\right]$

\begin{array}{rlrl} L \cdot \frac{d i}{d t} & =V_{S} \\ \frac{d i}{d t} & =\frac{V_{S}}{L} \\ \int d i_{0} & =\frac{V_{S}}{L} \int d t \\ i_{0} & =\frac{V_{S}}{L} t+K & \\ -I_{m} & =\frac{V_{S}}{L}(0)+K & \\ i_{0} & =\frac{V_{S}}{L} \cdot t-I_{m} & \text { At } t=0, i_{0}=-I_{m} \end{array}

\text { At } t=\frac{T}{2}, \quad \begin{aligned} i_{0} & =I_{M} \\ i_{0} & =\frac{V_{S}}{L} \cdot t-I_{m} \\ I_{M} & =\frac{V_{S}}{L} \cdot \frac{I}{2}-I_{m} \\ 2 I_{m} & =\frac{V_{S}}{2 f L} \\ I_{m} & =\frac{V_{S}}{4 f L} \end{aligned}

$\therefore$ Peak to peak load current, $2 I_{m}=\frac{V_{S}}{2 f L}=\frac{100}{2.50\left(20 \cdot 10^{-3}\right)}=50 \mathrm{~A}$

Q10GATE 2017MCQ

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

🚀 Solve in Practice Mode

📖 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}

Q16GATE 2015NAT

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 $\mu$ F, and R=5 $\Omega$ , the amplitude of the fundamental component in the output voltage $V_o$ (in Volt) under steady-state is ________.

🚀 Solve in Practice Mode

📖 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, $V_0=62.75V$

Q18GATE 2014NAT

The figure shows one period of the output voltage of an inverter. $\alpha$ should be chosen such that $60^{\circ} \lt \alpha \lt 90^{\circ}$ . If rms value of the fundamental component is 50V, then $\alpha$ in degree is _____.

🚀 Solve in Practice Mode

📖 Explanation

The output voltage $V_0$ can be reprresnted by fourier series as under:

\begin{aligned} V_0&=\sum_{n=1,3,5,...}^{\infty }a_n \cos n\omega t+b_n \sin n\omega t\\ \text{where, } a_n&=\frac{2}{T}\int_{0}^{T}f(t)\cos n\omega t \;d(\omega t) \\ b_n&=\frac{2}{T}\int_{0}^{T}f(t)\sin n\omega t \;d(\omega t) \\ a_n&=\frac{1}{\pi} \$[\int_{0}^{\alpha }100 \cos \omega t \;d(\omega t) +\int_{\alpha }^{180^{\circ}-\alpha }(-100) \cos \omega t \;d(\omega t) \\ &+ \int_{180^{\circ}-\alpha}^{180^{\circ} }100 \cos \omega t \;d(\omega t) +\int_{180^{\circ} }^{180^{\circ}+\alpha }(-100) \cos \omega t \;d(\omega t)\\ &+ \int_{180^{\circ}+\alpha}^{360^{\circ} -\alpha }100 \cos \omega t \;d(\omega t) +\int_{360^{\circ}-\alpha }^{360^{\circ} }(-100) \cos \omega t \;d(\omega t)]\$\\ a_n &=\frac{100}{\pi}\$[\sin \alpha - \sin (180^{\circ} -\alpha ) + \sin \alpha \\ &+\sin (180^{\circ}) -\sin (180^{\circ} -\alpha )\\ &- \sin (180^{\circ} +\alpha ) + \sin 180^{\circ}\\ &+ \sin (360^{\circ} -\alpha )- \sin (180^{\circ} +\alpha ) \\ &- \sin (360^{\circ} ) + \sin (360^{\circ} -\alpha )]\$\\ a_n &=0 \\ b_n&=\frac{1}{\pi} \$[\int_{0}^{\alpha }100 \sin \omega t \;d(\omega t) +\int_{\alpha }^{180^{\circ}-\alpha }(-100) \sin \omega t \;d(\omega t) \\ &+ \int_{180^{\circ}-\alpha}^{180^{\circ} }100 \sin \omega t \;d(\omega t) +\int_{180^{\circ} }^{180^{\circ}+\alpha }(-100) \sin \omega t \;d(\omega t)\\ &+ \int_{180^{\circ}+\alpha}^{360^{\circ} -\alpha }100 \sin \omega t \;d(\omega t) +\int_{360^{\circ}-\alpha }^{360^{\circ} }(-100) \sin \omega t \;d(\omega t)]\$\\ b_n &=\frac{100}{\pi}\$[-\cos \alpha +1+ \cos (180^{\circ} -\alpha ) - \cos \alpha \\ &-\cos (180^{\circ}) +\cos (180^{\circ} -\alpha )\\ &+ \cos (180^{\circ} +\alpha ) - \cos 180^{\circ}\\ &- \cos (360^{\circ} -\alpha )+ \cos (180^{\circ} +\alpha ) \\ &+ \cos (360^{\circ} ) - \cos (360^{\circ} -\alpha )]\$\\ b_n&=\frac{100}{\pi}[4-8 \cos \alpha ] =50\sqrt{2}\\ \cos \alpha &=0.22231=77.15^{\circ} \end{aligned}

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