Question 1 |
The single phase rectifier consisting of three thyristors T_{1}, T_{2}, T_{3} and a diode D_{1} feed power to a 10 \mathrm{~A} constant current load. T_{1} and T_{3} are fired at \alpha=60^{\circ} and T_{2} is fired at \alpha=240^{\circ}. The reference for \alpha is the positive zero crossing of \mathrm{V}_{\text {in }}. The average voltage \mathrm{V}_{o} across the load in volts is _____ (Round off to 2 decimal places).
25.35 | |
36.47 | |
39.79 | |
48.64 |
Question 1 Explanation:
\therefore Average output voltage,
\begin{gathered} \mathrm{V}_{o}=\frac{1}{2 \pi}\left[\int_{\alpha}^{\pi} \mathrm{V}_{\mathrm{m}} \sin \omega t d \omega t+\int_{\pi+\alpha}^{2 \pi+\alpha}-\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t} d \omega t\right] \\ =\frac{V_{m}}{2 \pi}[1+3 \cos \alpha] \end{gathered}
Put the values,
\begin{aligned} V_{o} & =\frac{100}{2 \pi}\left[1+3 \cos 60^{\circ}\right] \\ & =39.79 \mathrm{~V} \end{aligned}
Question 2 |
The circuit shown in the figure has reached steady state with thyristor 'T' in OFF condition. Assume that the latching and holding currents of the thyristor are zero.
The thyristor is turned ON at t=0 sec. The duration in microseconds for which the thyristor would conduct, before it turns off, is ___ (Round off to 2 decimal places).
The thyristor is turned ON at t=0 sec. The duration in microseconds for which the thyristor would conduct, before it turns off, is ___ (Round off to 2 decimal places).
7.33 | |
5.36 | |
4.25 | |
8.23 |
Question 2 Explanation:
Case (i) t\lt 0 sec :
Steady state circuit :
Case (ii) \mathrm{t} \gt 0 sec:
Thyristor is ON
Redraw the circuit :
\mathrm{I}_{4 \Omega}=\frac{100}{4}=25 \mathrm{~A}
and
\begin{aligned} \mathrm{i}_{\mathrm{C}} & =\mathrm{I}_{\mathrm{P}} \sin \omega_{0} \mathrm{t} \\ & =\mathrm{V}_{\mathrm{S}} \sqrt{\frac{C}{L}} \sin \omega_{0} \mathrm{t} \\ & =100 \sqrt{\frac{1}{4}} \sin \omega_{0} \mathrm{t}=50 \sin \omega_{0} \mathrm{t} \end{aligned}
\therefore Current through thyristor :
\mathrm{I}_{\mathrm{T}}=25+50 \sin \omega_{0} \mathrm{t}
When \mathrm{I}_{\mathrm{T}}=0 \Rightarrow Thyristor is turn off.
0=25+50 \sin \omega_{0} t
\Rightarrow \quad \omega_{0} \mathrm{t}=210^{\circ} \times \frac{\pi}{180^{\circ}}
\Rightarrow \quad \mathrm{t}=\sqrt{\mathrm{LC}} \times \frac{7 \pi}{6}
=\sqrt{10^{-6} \times 4 \times 10^{-6}} \times \frac{7 \pi}{6}
=7.33 \mu \mathrm{sec}
Steady state circuit :
Case (ii) \mathrm{t} \gt 0 sec:
Thyristor is ON
Redraw the circuit :
\mathrm{I}_{4 \Omega}=\frac{100}{4}=25 \mathrm{~A}
and
\begin{aligned} \mathrm{i}_{\mathrm{C}} & =\mathrm{I}_{\mathrm{P}} \sin \omega_{0} \mathrm{t} \\ & =\mathrm{V}_{\mathrm{S}} \sqrt{\frac{C}{L}} \sin \omega_{0} \mathrm{t} \\ & =100 \sqrt{\frac{1}{4}} \sin \omega_{0} \mathrm{t}=50 \sin \omega_{0} \mathrm{t} \end{aligned}
\therefore Current through thyristor :
\mathrm{I}_{\mathrm{T}}=25+50 \sin \omega_{0} \mathrm{t}
When \mathrm{I}_{\mathrm{T}}=0 \Rightarrow Thyristor is turn off.
0=25+50 \sin \omega_{0} t
\Rightarrow \quad \omega_{0} \mathrm{t}=210^{\circ} \times \frac{\pi}{180^{\circ}}
\Rightarrow \quad \mathrm{t}=\sqrt{\mathrm{LC}} \times \frac{7 \pi}{6}
=\sqrt{10^{-6} \times 4 \times 10^{-6}} \times \frac{7 \pi}{6}
=7.33 \mu \mathrm{sec}
Question 3 |
A semiconductor switch needs to block voltage V of only one polarity (V \gt 0) during OFF state as shown in figure (i) and carry current in both directions during ON state as shown in figure (ii). Which of the following switch combination(s) will realize the same?
A | |
B | |
C | |
D |
Question 3 Explanation:
I-V characteristic of given switch :
this characteristic shows by options (A) and (D) switch.
this characteristic shows by options (A) and (D) switch.
Question 4 |
The chopper circuit shown in figure (i) feeds power to a 5 A DC constant current source. The switching frequency of the chopper is 100 \mathrm{kHz}. All the components can be assumed to be ideal. The gate signals of switches S_{1} and S_{2} are shown in figure (ii). Average voltage across the 5 \mathrm{~A} current source is
10 \mathrm{~V} | |
6 \mathrm{~V} | |
12 \mathrm{~V} | |
20 \mathrm{~V} |
Question 4 Explanation:
Output Voltage Waveform :
\therefore Average voltage, \mathrm{V}_{0}=\frac{1}{10}[20 \times 3]=6 \mathrm{~V}
\therefore Average voltage, \mathrm{V}_{0}=\frac{1}{10}[20 \times 3]=6 \mathrm{~V}
Question 5 |
The steady state current flowing through the inductor of a DC-DC buck boost
converter is given in the figure below. If the peak-to-peak ripple in the output
voltage of the converter is 1 V, then the value of the output capacitor, in \mu F , is
_______________. (round off to nearest integer)
124 | |
148 | |
165 | |
182 |
Question 5 Explanation:
We have for buck boost converter,
\begin{aligned} \Delta V_c&= \frac{I_o}{fC}\\ where, \; \alpha &=\frac{T_{ON}}{T}\frac{20}{20+30}=0.4 \\ I_{L(Avg)}&=\frac{16+12}{2}=14A \end{aligned}
For buck-boost converter,
\begin{aligned} \frac{I_{o(Avg)}}{I_{L(Avg)}}&=\frac{1}{1-\alpha }\\ \Rightarrow I_L&=I_o(1-\alpha )\\ &=14(1-0.4)=8.4A\\ Given: \; \Delta V_c&=1V\\ Therefore, \; 1&=\frac{0.4 \times 8.4 \times 50 \times 1^{-6}}{C}\\ C&=168\mu F \end{aligned}
\begin{aligned} \Delta V_c&= \frac{I_o}{fC}\\ where, \; \alpha &=\frac{T_{ON}}{T}\frac{20}{20+30}=0.4 \\ I_{L(Avg)}&=\frac{16+12}{2}=14A \end{aligned}
For buck-boost converter,
\begin{aligned} \frac{I_{o(Avg)}}{I_{L(Avg)}}&=\frac{1}{1-\alpha }\\ \Rightarrow I_L&=I_o(1-\alpha )\\ &=14(1-0.4)=8.4A\\ Given: \; \Delta V_c&=1V\\ Therefore, \; 1&=\frac{0.4 \times 8.4 \times 50 \times 1^{-6}}{C}\\ C&=168\mu F \end{aligned}
There are 5 questions to complete.