Phase Controlled Rectifiers MCQ Quiz - Objective Question with Answer for Phase Controlled Rectifiers - Download Free PDF

Concept

The average output voltage for a full-bridge rectifier is given by:

\(V_{o(avg)}={2V_m \over \pi}cosα\)

where, V_m = Maximum value of input voltage

cos α = Firing angle

Explanation

Given, α = 120° 

\(V_{o(avg)}={2V_m \over \pi}cos(120)\)

\(V_{o(avg)}={-V_m \over \pi}\)

Since the result is negative, the average load voltage is negative.

Nature of Load Current:

• The load is highly inductive, meaning the current is continuous (i.e., it does not become zero).
• Even though the average voltage is negative, the current remains positive due to the energy stored in the inductor.

The correct answer is option 2.

Explanation:

Phase Sequence in Three-Phase Systems

Definition: The phase sequence (or phase rotation) in a three-phase system refers to the order in which the three phases (commonly labeled as A, B, and C) reach their respective maximum positive values. This sequence can be either ABC or ACB, and it is crucial for the correct operation of three-phase equipment, especially motors and other rotating machinery.

Working Principle: In a three-phase system, three sinusoidal voltages of equal magnitude and frequency are generated, with each voltage phase-shifted by 120 degrees from the others. The standard phase sequence ensures that the voltages reach their peak values in a specific order (e.g., A first, then B, then C). If the phase sequence is changed (e.g., from ABC to ACB), the direction of rotation of the magnetic field in motors will reverse, which can cause the motor to run in the opposite direction.

Correct Option Analysis:

The correct option is:

Option 3: Phase current changes by angle but not by magnitude.

This option correctly describes the effect of changing the phase sequence on the phase currents in a three-phase system. When the phase sequence is altered, the phase currents will shift in phase angle by 120 degrees, but their magnitudes will remain unchanged. This is because the phase sequence change does not affect the amplitude of the sinusoidal currents, only their relative timing.

Detailed Explanation:

In a three-phase system, the voltage and current waveforms are typically represented as:

Original Phase Sequence (ABC):

• Phase A: V_A(t) = V_msin(ωt)
• Phase B: V_B(t) = V_msin(ωt - 120°)
• Phase C: V_C(t) = V_msin(ωt - 240°)

Here, V_m is the peak voltage, ω is the angular frequency, and t is time.

When the phase sequence changes from ABC to ACB, the voltage waveforms become:

• Phase A: V_A(t) = V_msin(ωt)
• Phase C: V_C(t) = V_msin(ωt - 120°)
• Phase B: V_B(t) = V_msin(ωt - 240°)

Consequently, the phase currents will also shift in phase angle by 120 degrees, but their magnitudes will remain the same. This shift in phase angle is crucial for devices that rely on the direction of the rotating magnetic field, such as induction motors, as it will cause them to rotate in the opposite direction.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Magnitude of phase power is changed.

This option is incorrect because the magnitude of the phase power in a balanced three-phase system does not depend on the phase sequence. Phase power is primarily determined by the voltage, current, and power factor. Changing the phase sequence only affects the direction of rotation of the magnetic field, not the magnitude of the power.

Option 2: Magnitude of phase current is changed.

This option is also incorrect because changing the phase sequence does not affect the magnitude of the phase currents. The phase currents will have the same amplitude but will be phase-shifted by 120 degrees.

Option 4: Total power consumed will change.

This option is incorrect because the total power consumed in a balanced three-phase system is the sum of the power consumed in each phase. Since the power in each phase remains unchanged regardless of the phase sequence, the total power consumed will also remain unchanged.

Conclusion:

Understanding the impact of phase sequence on three-phase systems is crucial for the correct operation of equipment, especially motors. Changing the phase sequence results in a phase shift of the currents by 120 degrees, but their magnitudes remain unchanged. This phase shift can reverse the direction of rotation of motors, which is essential information for ensuring the proper functioning of three-phase machinery.

Explanation

• The firing angle (α) is the point at which the SCR is triggered (turned on).
• The extinction angle (β) is the point at which the SCR turns off (due to current dropping below the holding current or due to forced commutation).
• The conduction angle(γ) represents the duration for which the SCR remains conducting.

Since conduction starts at α and ends at β, the conduction angle is given by:

Conduction angle = β - α

Single phase full wave rectifier

Case 1: From α < ωt < π + α

T1 and T2 conduct. 

So, Vo = V_s

Case 2: From π + α < ωt < 2π + α 

T₃ and T₄ conduct.

So, Vo = -Vs

\(V_{o(avg)}={1\over \pi}\int_{\alpha}^{\pi+\alpha}V_m \space sin\omega t \space d\omega t\)

\(V_{o(avg)}={V_m\over \pi}(\mathrm{cos\space \omega t})_{\pi +\alpha}^{\alpha}\)

\(V_{o(avg)}={V_m\over \pi}(\mathrm{cos \alpha-cos(\pi+\alpha)})\)

\(V_{o(avg)}=\frac{2 v_{m}}{\pi} \cos \alpha\)

1ϕ semi converter

Case 1: From 0 < ωt < α, T₂ and D2 conduct. So, Vo = 0

Case 2: From α < ωt < π, T1 and D2 conduct. So, Vo = Vs

Case 3: From π < ωt < π+α, T₁ and D1 conduct . So, Vo = 0

Case 4: From π+α < ωt < 2π, T₂ and D₁ conduct. So, Vo = -Vs

The output waveform is given below:

Explanation

• The free-wheeling diode D₃ turns ON when SCR conducts beyond π. An SCR can be conducted beyond π only if the load is inductive.
• The load is resistive. So, the free-wheeling diode D3 does not turn ON. So, the current through diode D3 is 0A.

Concept:

Ripple frequency at the output = m × supply frequency

fo = m × fs

Where m = types of the pulse converter

Calculation:

A three-phase full-wave AC to DC converter is a 6-pulse converter

Number of pulses (m) = 6

fo = 6 × supply voltage frequency

∴ f₀ = 6 x 50

f₀ = 300 Hz

Concept:

Center tapped full wave rectifier:

• The Center tapped full-wave rectifier is a device used to convert the AC input voltage into DC voltage at the output terminals.
• It employs a transformer with the secondary winding tapped at the center point. And it uses only two diodes, which are connected to the opposite ends of a center-tapped transformer as shown in the figure below.
• The center tap is usually considered as the ground point or the zero voltage reference point.

     ​

Analysis:

The DC output voltage or average output voltage can be calculated as follows,

\({{\rm{V}}_0} = {{\rm{V}}_{{\rm{dc}}}} = \frac{1}{π }\mathop \smallint \limits_0^π {{\rm{V}}_{\rm{m}}}sin\omega t\;d\omega t\)

\( = \;\frac{{{V_m}}}{π }\left. {\left( { - \cos \omega t} \right)} \right|\begin{array}{*{20}{c}} π \\ 0 \end{array}\)

\( = \frac{{{V_m}}}{π }\left( { - cosπ - \left( { - cos 0^\circ } \right)} \right)\)

\( = \frac{{{V_m}}}{π }\left( { - \left( { - 1} \right) + 1} \right)\)

V₀ = 2V_m / π 

Now we can calculate the average or mean current of load by dividing the average load voltage by load resistance R_L. Therefore mean load current is given by

I₀ = V₀ / R_L

If the internal resistance of the diode is given in that case mean load current I₀ = V₀ / (R_L + r)

Where r = internal resistance of the diode.

Calculation:

Given that 

Rms value of supply voltage V = 50 V

The internal resistance of diode r = 20 Ω 

The load resistance R_L = 980 Ω 

Maximum voltage on the secondary side V_m = √2 V = √2 × 50 = 70.7 V

Average or DC output voltage V₀ = \(\frac{(2 × 70.7) }{π} = 45\; V\)

Average or mean load current is

I0 = V0 / (RL + r) = \(\frac{45 }{(980 + 20)} \) = 45 mA​

Full wave rectifier

Case 1: During +ve half

D_o ON and D₁ OFF

V_o = V_s

Case 2: During -ve half

Do OFF and D1 ON

Vo = -Vs

The output waveform is:

The rectification efficiency is the ratio of the DC output power to the AC input power. 

\(V_{o(avg)}={2V_m\over \pi}\) and \(I_{o(avg)}={2V_m\over \pi R}\)

\(V_{o(rms)}={V_m\over \sqrt{2}}\) and \(I_{o(rms)}={V_m\over \sqrt{2}R}\)

% η = \({V_{o(avg)}\times I_{o(avg)}\over V_{o(rms) \times I_{o(rms)}}}\)

% η = \({{2V_m\over \pi}\times{2V_m\over \pi R}\over {V_m\over \sqrt{2}}\times {V_m\over \sqrt{2}R}}\)

% η = 81.2%

Mistake Points The rectification efficiency of half wave rectifier is 40.6%

Concept:

In a three-phase semi-converter,

The conduction period of each thyristor = π – α

The conduction period of freewheeling diode = β – 60°

Where α is the firing angle

β is the extinction angle

Calculation:

Given that, firing angle (α) = 120°

Extinction angle (β) = 110°

The conduction period of each thyristor = π – α = 180 – 120 = 60°

The conduction period of freewheeling diode = β – 60° = 110 – 60 = 50°

Concept:

Considered V_m is the maximum value of AC input voltage of converter and V₀ is the average output voltage converter and α is delay angle.

For single-phase semi converter or delayed full-wave rectifier,

\({V_0} = \frac{{{V_m}}}{π }\left( {1 + cosα } \right) \)

Calculation:

Given that,

\({V_0} = \frac{{{V_m}}}{3} \)

Hence, the equation becomes,

\(\frac{{{V_m}}}{3} = \frac{{{V_m}}}{π }\left( {1 + cosα } \right) \)

or,

\(\frac{π }{3} = \left( {1 + cosα } \right) \)

cos α = 0.047
α = cos^-1(0.047)

Concept:

• A dual converter is an electronic converter or circuit which comprises of two converters.
• One performs as a rectifier and the other performs as the inverter.
• Two full converters are arranged in an anti-parallel pattern and linked to the same DC load.
• These converters can provide four-quadrant operations.

Modes of operation of Dual converter:

Non-circulating current mode:

• There is no circulating current between the converters in this mode as only one converter will perform at a time
• During converter 1 operation, firing angle (α₁) will be between 0 to 90°
• During converter 2 operation, firing angle (α₂) will be between 0 and 90°

Circulating current Mode:

• In this mode, there is a circulating current as the two converters will be in the ON condition at the same time.
• The firing angles are adjusted such that
• Firing angle of converter 1 (α₁) + firing angle of converter 2 (α₂) = 180°
• Converter 1 as rectifier when 0° < α₁ < 90° and converter 2 as inverter when 90° < α₂ < 180°
• Converter 1 as inverter when 90° < α₁ < 180° and converter 2 as rectifier when 0° < α₂ < 90°
• Four quadrant operation of dual converter is shown below

Application of dual converter:

• Direction and speed control of DC motors
• Applicable wherever the reversible DC required
• Industrial variable speed DC drives.

Calculation:

Given that, firing angle of one bridge is 30° (i.e. α₁ = 30°)

We know that firing angles can never be greater than 180°.

i.e. α₁ + α₂ = 180°

∴ 30° + α₂ = 180°

α₂ = 180° - 30°

α₂ = 150°

The average output voltage of different 1-ϕ rectifier circuits is:

A. Uncontrolled half wave - \(\frac{{{V_{peak}}}}{\pi }\)

B. Controlled half wave - \(\frac{{{V_{peak}}}}{{2\pi }}\left( {1 + \cos \alpha } \right)\)

C. Controlled full wave - \(\frac{{2{V_{peak}}}}{\pi }\cos \alpha \)

D. Semi controlled - \(\frac{{{V_{peak}}}}{\pi }\left( {1 + \cos \alpha } \right)\)

Concept:

The average output voltage of a full-wave controlled rectifier with R load is given by:

\({V_0} = \frac{{{V_m}}}{\pi }\left( {1 + \cos \alpha } \right)\)

Where V_m is the maximum value of supply voltage

α is the firing angle or delay angle

Average load current:

\({I_0} = \frac{{{V_m}}}{{\pi R}}\left( {1 + \cos \alpha } \right) = \frac{{{I_m}}}{\pi }\left( {1 + \cos \alpha } \right)\)

Calculation:

Given that, the average value of the average load current is equal to half of its maximum value.

\(\frac{{{I_m}}}{2} = \frac{{{I_m}}}{\pi }\left( {1 + \cos \theta } \right)\)

Peak inverse voltage:

PIV is the maximum voltage appearing across the p-n junction diode when it is non-conducting (reverse polarized)

Freewheeling diode:

• A freewheeling diode placed across the inductive load will provide a path for the release of energy stored in the inductor while the load voltage drops to zero.
• The freewheeling diode prevents the load voltage from becoming negative. Whenever load voltage tends to go negative, FD comes into play.
• As a result, the load current is transferred from the main thyristor to FD, allowing the thyristor to regain its forward blocking capability.
• The advantages are the input power factor is improved, load current waveform is improved and better load performance.