Single Phase Rectifiers Questions and Answers

Q: Assuming ideal diode and no ripple, what is the dc load current in the circuit shown below ?

The dc load current is about 6.8 mA.With an ideal diode and no ripple, the capacitor charges to the peak of the secondary voltage and holds it, so the load sees a steady dc voltage equal to that peak.\[ \begin{aligned} V_{2} &= \frac{120}{5} = 24\ \text{V rms} \\ V_{m} &= 24\sqrt{2} \approx 33.9\ \text{V} \\ I_{dc} &= \frac{33.9}{5000} \approx 6.8\ \text{mA} \end{aligned} \]4.8 mA is the rms secondary voltage divided by the load, \(24/5000\), which forgets that the capacitor charges to the peak. 0.9 A and 1.28 A are hundreds of times too large for a 5 kΩ load.

The average DC output of a single-phase, half-wave controlled rectifier will be maximum if the firing angle is ______.

A.\(0^\circ\)
B.\(30^\circ\)
C.\(60^\circ\)
D.\(90^\circ\)

Exam Year(s)

DSSSB AE - 2023,

Answer: A

For a single phase half wave controlled rectifier with a resistive load the average output voltage is

\[ V_{dc} = \frac{V_m}{2\pi}(1 + \cos\alpha) \]

This is greatest when \(\cos\alpha\) is greatest, that is at \(\alpha = 0^\circ\), where \(V_{dc} = V_m/\pi\), the value of an uncontrolled rectifier. As the firing angle increases, \(V_{dc}\) falls, to \(V_m/(2\pi)\) at 90 degrees and to zero at 180 degrees. The answer is \(0^\circ\), option A.

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For a single-phase semi-converter, the input voltage is \(V_m \sin\omega t\). If the firing angle is \(90^\circ\), find the average DC output voltage.

A.\(\dfrac{V_m}{2\pi}\)
B.\(\dfrac{V_m}{\pi}\)
C.\(\dfrac{2V_m}{\pi}\)
D.\(\dfrac{V_m}{\sqrt{2}\,\pi}\)

Exam Year(s)

DSSSB AE - 2023,

Answer: B

For a single phase semi-converter (half controlled bridge) with a firing angle \(\alpha\), the average output voltage is

\[ V_{dc} = \frac{V_m}{\pi}(1 + \cos\alpha) \]

With \(\alpha = 90^\circ\), \(\cos\alpha = 0\), so

\[ V_{dc} = \frac{V_m}{\pi} \]

The value \(2V_m/\pi\) is the output at \(\alpha = 0\), where the bridge acts as an uncontrolled rectifier, and \(V_m/2\pi\) is the half wave controlled value at the same angle. The answer is option B.

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For a three-phase bridge diode rectifier, the output current is \(I_{dc}\). The RMS diode current is ______.

A.\(0.409\,I_{dc}\)
B.\(0.579\,I_{dc}\)
C.\(0.785\,I_{dc}\)
D.\(4.09\,I_{dc}\)

Exam Year(s)

DSSSB AE - 2023,

Answer: B

In a three phase bridge rectifier with a constant (highly inductive) load current \(I_{dc}\), each diode carries the current for one third of the cycle, that is for \(120^\circ\).

\[ \begin{aligned} I_{D,rms} \\ &= \sqrt{\frac{1}{2\pi}\int_0^{2\pi/3} I_{dc}^2\, d\theta} \\ &= I_{dc}\sqrt{\frac{1}{3}} = 0.577\,I_{dc} \end{aligned} \]

The nearest value offered is \(0.579\,I_{dc}\), option B (the exact figure is 0.5774). The average diode current is \(I_{dc}/3\), and 0.409 and 0.785 are the factors for other rectifier connections.

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In the given semi-converter, the maximum value of AC input voltage is 300 V. The SCR is fixed at \(30^\circ\) angle. Compute the output average DC voltage.

Single-phase semi-converter bridge: AC mains Vs, thyristors T1 and T2 in the upper arms and diodes D2 and D1 in the lower arms, with a resistor R carrying the output voltage Vo

A.17.828 V
B.1.7828 V
C.178.28 V
D.1782.8 V

Exam Year(s)

DSSSB AE - 2023,

Answer: C

For a single phase semi-converter, the average output voltage is

\[ V_{dc} = \frac{V_m}{\pi}(1 + \cos\alpha) \]

\[ \begin{aligned} V_{dc} = \frac{300}{\pi}(1 + \cos 30^\circ) \\ &= 95.49 \times 1.866 = 178.2\ \text{V} \end{aligned} \]

The nearest option is 178.28 V. The other values are 10 times larger or smaller or 100 times smaller. The answer is option C.

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Identify the circuit shown below:

A circuit with four diodes connected as a bridge, the alternating supply on the left and the positive and negative output terminals on the right

A.Half wave rectifier
B.Centre-tapped full wave rectifier
C.Uncontrolled Bridge rectifier
D.Controlled bridge rectifier

Exam Year(s)

DSSSB - 2023,

Answer: C

The circuit shown is an uncontrolled bridge rectifier. It has four diodes in a bridge, two conducting in each half cycle of the supply, so the load gets a direct voltage from both half cycles of the alternating supply.

A half wave rectifier has one diode, and a centre-tapped full wave rectifier needs two diodes and a transformer with a tapped secondary. A controlled bridge would use thyristors in place of diodes. The picture has only diodes, so the third option is correct.

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The ripple frequency of the bridge rectifier is 120 Hz. What will be the frequency of the supply?

A.30 Hz
B.60 Hz
C.90 Hz
D.120 Hz

Exam Year(s)

DSSSB - 2023,

Answer: B

The frequency of the supply is 60 Hz. In a single phase bridge rectifier, the output ripple frequency is twice the supply frequency, because both half cycles are rectified.

\[ f_{ripple} = 2f \]

\[ f = \frac{120}{2} = 60\ \text{Hz} \]

A supply of 30 Hz would give a ripple of 60 Hz, 90 Hz would give 180 Hz and 120 Hz would give 240 Hz, none of which is 120 Hz. So the second option is correct.

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How many SCRs are required in single-phase half-wave controlled rectifier?

Exam Year(s)

DSSSB - 2023,

Answer: B

One SCR is needed in a single phase half wave controlled rectifier. The thyristor conducts in the positive half cycle, after it is triggered at the firing angle, and blocks in the negative half cycle, so the one device controls the output.

Zero would be an uncontrolled circuit with a diode. Two SCRs would be used in a full wave controlled rectifier with a centre tap or in a half controlled bridge, and four would be used in a full controlled bridge. So the second option is the answer.

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Assuming ideal diode and no ripple, what is the dc load current in the circuit shown below ?

A 120 V 60 Hz source on a 5 to 1 transformer; the secondary feeds diode D, then a 100 microfarad capacitor in parallel with a 5 kilohm load, returning to ground

A.4.8 mA
B.0.9 A
C.1.28 A
D.6.8 mA

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: D

The dc load current is about 6.8 mA.

With an ideal diode and no ripple, the capacitor charges to the peak of the secondary voltage and holds it, so the load sees a steady dc voltage equal to that peak.

\[ \begin{aligned} V_{2} &= \frac{120}{5} = 24\ \text{V rms} \\ V_{m} &= 24\sqrt{2} \approx 33.9\ \text{V} \\ I_{dc} &= \frac{33.9}{5000} \approx 6.8\ \text{mA} \end{aligned} \]

4.8 mA is the rms secondary voltage divided by the load, \(24/5000\), which forgets that the capacitor charges to the peak. 0.9 A and 1.28 A are hundreds of times too large for a 5 kΩ load.

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