Phase Controlled Rectifiers

Question 1
For the ideal AC-DC rectifier circuit shown in the figure below, the load current magnitude is I_{dc} = 15 A and is ripple free. The thyristors are fired with a delay angle of 45^{\circ}. The amplitude of the fundamental component of the source current, in amperes, is __________. (round off to two decimal places)

A
12.45
B
25.32
C
14.25
D
17.64
GATE EE 2022   Power Electronics
Question 1 Explanation: 
Given rectifier circuit is a 1-\phi semiconverter Waveform of source current:

\therefore \; i_{sn}=\sum_{n=1,3,5...}^{\infty }\frac{4I_o}{n\pi} \cos \left ( \frac{n\alpha }{2} \right ) \sin (n\omega t+\phi _n)
Now, fundamental component,
I_{s1}=\frac{4I_o}{\pi} \cos \left ( \frac{\alpha }{2} \right )=\frac{4 \times 15}{\pi} \cos \left ( \frac{45^{\circ} }{2} \right )=17.64A
Question 2
The voltage at the input of an AC-DC rectifier is given by v(t)=230\sqrt{2}\sin \omega t where \omega = 2 \pi \times 50 rad/s. The input current drawn by the rectifier is given by
i(t)=10\sin \left (\omega t -\frac{\pi}{3} \right )+4\sin \left (3\omega t -\frac{\pi}{6} \right )+3\sin \left (5\omega t -\frac{\pi}{3} \right )
The input power factor, (rounded off to two decimal places), is, _________ lag.
A
0.44
B
0.22
C
0.66
D
0.88
GATE EE 2022   Power Electronics
Question 2 Explanation: 
We have,
\begin{aligned} PF&=g.FDF \\ g&=\frac{I_{{s1}}}{I_{sr}} \\ &= \frac{10/\sqrt{2}}{\sqrt{\left ( \frac{10}{\sqrt{2}} \right )^2+\left ( \frac{4}{\sqrt{2}} \right )^2+\left ( \frac{3}{\sqrt{2}} \right )^2}}\\ &=0.894 \\ FDF&= \cos \frac{\pi}{3}=0.5 \\ PF&=0.894 \times 0.5=0.447 \end{aligned}
Question 3
A single-phase full-bridge diode rectifier feeds a resistive load of 50 \Omega from a 200 V, 50 Hz single phase AC supply. If the diodes are ideal, then the active power, in watts, drawn by the load is _____________. (round off to nearest integer).
A
200
B
1600
C
600
D
800
GATE EE 2022   Power Electronics
Question 3 Explanation: 
For 1-\phi full bridge diode rectifier:
V_{or}=\frac{V_m}{\sqrt{2}} =\frac{\sqrt{2} \times 200}{\sqrt{2}}=200V
Active power drawn by the load
P_{o}=\frac{V_{or}^2}{R} =\frac{200^2}{50}=800W
Question 4
The waveform shown in solid line is obtained by clipping a full-wave rectified sinusoid (shown dashed). The ratio of the \text{RMS} value of the full-wave rectified waveform to the \text{RMS} value of the clipped waveform is _______________. (Round off to 2 decimal places.)

A
1.21
B
2.24
C
1.82
D
0.65
GATE EE 2021   Power Electronics
Question 4 Explanation: 


We know,
RMS value of full wave rectified sine =\frac{V_{m}}{\sqrt{2}}=0.707 V_{m}
For clipped wave form
As 0 \rightarrow \frac{\pi}{4} and \frac{3 \pi}{4} \rightarrow \pi wave form are identical
\begin{aligned} &\text { rms value of clipped wave } \\ &\left.=\sqrt{\frac{1}{\pi}\left[2 \int_{0}^{\pi / 4} V_{m}^{2} \sin ^{2} \omega t+\int_{\pi / 4}^{3 \pi / 4}\left(0.707 V_{m}\right)^{2} d \omega t\right.}\right] \\ &= \sqrt{\frac{1}{\pi}\left(\int_{0}^{\pi / 4} \frac{2 V_{m}^{2}}{2}[1-\cos 2 \omega t] d \omega t+\frac{1}{2} V_{m}^{2} \times \frac{\pi}{2}\right)^{1 / 2}}\\ &=\left[\frac{1}{\pi}\left[V_{m}^{2}\left(\frac{\pi}{4}-\frac{1}{2} \sin \frac{\pi}{2}\right)+\frac{V_{m}^{2}}{4} \pi\right]\right]^{1 / 2} \\ &=\left[\frac{V_{m}^{2}}{\pi}\left[\frac{\pi}{4}-\frac{1}{2}\right]+\frac{V_{m}^{2}}{4}\right]^{1 / 2}=0.5838 V_{m} \\ \text { Ratio } &=\frac{0.707}{0.5838}=1.21 \end{aligned}
Question 5
In the circuit shown, the input V_{i} is a sinusoidal \text{AC} voltage having an \text{RMS} value of 230\;V\pm 20\%. The worst-case peak-inverse voltage seen across any diode is ____________V. (Round off to 2 decimal places.)

A
390.32
B
124.25
C
258.36
D
471.56
GATE EE 2021   Power Electronics
Question 5 Explanation: 
\begin{aligned} \left(V_{D}\right)_{\text {peak }} \text { for worst case } &=(230+20 \%) \times \sqrt{2} \\ &=\left(230+230 \times \frac{20}{100}\right) \times \sqrt{2}=390.32 \mathrm{~V} \end{aligned}
Question 6
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 6 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 7
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 7 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 8
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 8 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 9
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 9 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 10
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 10 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
There are 10 questions to complete.