# 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.