Phase Controlled Rectifiers

Question 1
A single-phase, full-bridge, fully controlled thyristor rectifier feeds a load comprising a 10\Omega resistance in series with a very large inductance. The rectifier is fed from an ideal 230 V, 50 Hz sinusoidal source through cables which have negligible internal resistance and a total inductance of 2.28 mH. If the thyristors are triggered at an angle \alpha = 45^{\circ}, the commutation overlap angle in degree (rounded off to 2 decimal places) is_____.
A
2.4
B
4.8
C
6.4
D
8.2
GATE EE 2020   Power Electronics
Question 1 Explanation: 
\begin{aligned}&1-\phi\text{ SCR bridge rectifier} \\ \alpha &=45^{\circ},\; \; R=10\, \Omega\\ &\text{supply 230 V, 50 Hz}\\ L_{s}&=2.28\, mH \\ \mu &=? \\ \Delta V_{d}&=\frac{V_{m}}{\pi }[\cos \alpha -\cos (\alpha +\mu )]\\ &= 4fL_{s}I_{0} \\ V_{0}&=\frac{2V_{m}}{\pi }\cos \alpha -4fL_{s}I_{0} \\ I_{0}R&=\frac{2V_{m}}{\pi}\cos \alpha -4fL_{s}I_{0} \\ &\text{Find }I_{0} \\ I_{0}\times 10&=\frac{2\times 230\sqrt{2}}{\pi } \cos 45 \\&-4\times 50\times 2.28\times 10^{-3}I_{0} \\ I_{0}(10+0.456)&=146.42 \\ I_{0}&=\frac{146.49}{10.456}=14.0036\: A \\ \Delta V_{d0 }&=\frac{230\sqrt{2}}{\pi }[\cos 45-\cos (45+\mu)] \\ &=4\times 50\times 2.28\times 10^{-3}\times 14 \\ &=6.384 \\ \cos 45^{\circ}-\cos (45^{\circ}+\mu )&=0.061659 \\ 45+\mu &=49.80 \\ \therefore \, \, \mu =4.80^{\circ}\end{aligned}
Question 2
Consider the diode circuit shown below. The diode, D, obeys the current-voltage characteristic I_D=I_S\left ( exp\left ( \frac{V_D}{nV_T} \right )-1 \right ), where n \gt 1, V_T \gt 0, V_D is the voltage across the diode and I_D is the current through it. The circuit is biased so that voltage, V \gt 0 and current, I \lt 0. If you had to design this circuit to transfer maximum power from the current source (I_1) to a resistive load (not shown) at the output, what values R_1 \; and \; R_2 would you choose?
A
Large R_1 and large R_2.
B
Small R_1 and small R_2.
C
Large R_1 and small R_2.
D
Small R_1 and large R_2.
GATE EE 2020   Power Electronics
Question 2 Explanation: 
R_{1}-low, R_{2}- high
V_{D}=V\times \frac{R_{2}}{R_{1}+R_{2}}
If R_{2} is large V_{D} becomes high
If R_{1} is less V_{D}=V
So for maximum power, R_{1} is small and R_{2} is large.
Question 3
Thyristor T_1 is triggered at an angle \alpha (in degree), and T_2 at angle 180^{\circ} + \alpha, in each cycle of the sinusoidal input voltage. Assume both thyristors to be ideal. To control the load power over the range 0 to 2 kW, the minimum range of variation in \alpha is:
A
0^{\circ} \; to \; 60^{\circ}
B
0^{\circ} \; to \; 120^{\circ}
C
60^{\circ} \; to \; 120^{\circ}
D
60^{\circ} \; to \; 180^{\circ}
GATE EE 2020   Power Electronics
Question 3 Explanation: 
As per GATE official answer key, MTA (Marks to All)


As load is capacitive, for any \alpha :

\begin{aligned} V_{or}&=\frac{V_m}{\sqrt{2\pi}}\left [ \pi - \alpha +\frac{1}{2} \sin 2\alpha \right ]^{1/2}\\ P_0&=\left ( \frac{V_{or}}{z} \right )^2 \times R\\ &=\frac{V_{or}^2}{100} \times 5=\frac{V_{or}^2}{20}\\ \text{For }\alpha &=0^{\circ}\\ V_{or}&=\frac{V_m}{\sqrt{2}}=200\\ P_0&=\frac{200}{20}=2KW \end{aligned}
For \alpha =120^{\circ}, as current distinguishes, so P=0.
So, \alpha should be 0^{\circ} \text{ to } 120^{\circ}.
Question 4
A single-phase, full-bridge diode rectifier fed from a 230 V, 50 Hz sinusoidal source supplies a series combination of finite resistance. R, and a very large inductance, L, The two most dominant frequency components in the source current are:
A
50 Hz, 0 Hz
B
50 Hz, 100 Hz
C
50 Hz, 150 Hz
D
50 Hz, 250 Hz
GATE EE 2020   Power Electronics
Question 4 Explanation: 
For full bridge rectifier, High inductive load

\begin{aligned} I_{source}&=\sum_{n=1,3,5}^{\infty }\frac{4I_0}{n \pi} \sin nd \sin \frac{n \pi}{2} \sin (n\omega t+\phi _n) \\ 2d &=\pi , \;\; d=\frac{\pi}{2}\\ I_{source}&= \frac{4I_0}{n \pi} \sum_{n=1,3,5}^{\infty } \sin ^2 n\left ( \frac{\pi}{2} \right ) \sin (n\omega t+\phi _n)\\ I_{source}&\neq 0 \end{aligned} For n=1,3,5,...
Since the most dominant frequency component will be f, 3f, f = 50
So dominant frequency = 50 Hz, 150 Hz.
Question 5
A single-phase fully-controlled thyristor converter is used to obtain an average voltage of 180 V with 10 A constant current to feed a DC load. It is fed from single-phase AC supply of 230 V, 50 Hz. Neglect the source impedance. The power factor (round off to two decimal places) of AC mains is________
A
0.25
B
0.55
C
0.78
D
0.95
GATE EE 2019   Power Electronics
Question 5 Explanation: 
V_{sr}I_{sr} \cos \phi =V_0I_0
For single-phase fully controlled converter,
I_0=I_{sr}=10A
\cos \phi =\frac{V_0}{V_{sr}}=\frac{180}{230}=0.78
Question 6
A fully-controlled three-phase bridge converter is working from a 415V, 50 Hz AC supply. It is supplying constant current of 100 A at 400 V to a DC load. Assume large inductive smoothing and neglect overlap. The rms value of the AC line currentin amperes (round off to two decimal places) is ________.
A
81.65
B
41.25
C
66.45
D
69.85
GATE EE 2019   Power Electronics
Question 6 Explanation: 
AC line current rms =(I_s)_{rms}=I_0\sqrt{\frac{2}{3}}=100 \times \sqrt{\frac{2}{3}}=81.65A
Question 7
A six-pulse thyristor bridge rectifier is connected to a balanced three-phase, 50Hz AC source. Assuming that the DC output current of the rectifier is constant, the lowest harmonic component in the AC input current is
A
100Hz
B
150Hz
C
250Hz
D
300Hz
GATE EE 2019   Power Electronics
Question 7 Explanation: 
For 6 pulse converter harmonics present in AC current are 6k\pm 1
Lowest order harmonic =5
Lowest harmonic frequency = 5 x 50 =250Hz
Question 8
A phase controlled single phase rectifier, supplied by an AC source, feeds power to an R-L-E load as shown in the figure. The rectifier output voltage has an average value given by
V_{0}=\frac{V_{m}}{2 \pi}(3+ \cos \alpha), where V_{m}=80\pi volts and \alpha is the firing angle. If the power delivered to the lossless battery is 1600 W, \alpha in degree is________ (up to 2 decimal places).
A
45
B
90
C
60
D
30
GATE EE 2018   Power Electronics
Question 8 Explanation: 
\begin{aligned} V_0&=\frac{V_m}{2 \pi}(3+\cos \alpha )\\ E_bI_0&=1600W\\ I_0&=\frac{1600}{80}=20A\\ V_0&=E_b+I_0R_a\\ \frac{V_m}{2 \pi}(3+\cos \alpha )&=80+(20 \times 2)\\ \frac{80 \pi}{2 \pi}(3+\cos \alpha )&=80+40\\ \alpha &=90^{\circ} \end{aligned}
Question 9
The waveform of the current drawn by a semi-converter from a sinusoidal AC voltage source is shown in the figure. If I_0=20 A, the rms value of fundamental component of the current is ___________A (up to 2 decimal places).
A
17.39
B
18.92
C
24.67
D
14.86
GATE EE 2018   Power Electronics
Question 9 Explanation: 
\begin{aligned} i_{s1}&=\frac{4I_0}{\pi}\cos \frac{\alpha }{2}\\ I_{s1(rms)}&=\frac{2\sqrt{2}}{\pi}I_0\cdot \cos \frac{\alpha }{2}\\ &=\frac{2\sqrt{2}}{\pi} \times 20 \times \cos \left ( \frac{30^{\circ}}{2} \right )\\ &=17.39A \end{aligned}
Question 10
A single phase fully controlled rectifier is supplying a load with an anti-parallel diode as shown in the figure. All switches and diodes are ideal. Which one of the following is true for instantaneous load voltage and current?
A
v_{0}\geq 0 \text{ and } i_{0} \lt 0
B
v_{0} \lt 0 \text{ and } i_{0}\lt  0
C
v_{0}\geq 0 \text{ and } i_{0}\geq 0
D
v_{0} \lt 0 \text{ and } i_{0}\geq 0
GATE EE 2018   Power Electronics
There are 10 questions to complete.