Effect of Source Impedance on Converters MCQ Quiz in मराठी - Objective Question with Answer for Effect of Source Impedance on Converters - मोफत PDF डाउनलोड करा

Concept:

The average output voltage is given by, \({V_1} = \frac{{2{V_m}}}{\pi }\cos \alpha \)

The average output current, \({I_1} = \frac{{{V_0}}}{R}\)

If SCR T₃ is damaged and gets open circuited, the circuit acts as half wave rectifier & only 2 SCR's conducts

The average output voltage = \({{\rm{V}}_{\rm{o}}} = \frac{1}{{2{\rm{\pi }}}}\mathop \smallint \limits_{\rm{\alpha }}^{{\rm{\pi }} + {\rm{\alpha }}} {{\rm{V}}_{\rm{m}}}{\rm{sin\omega t\;d}}\left( {{\rm{\omega t}}} \right)\)

\(V_{2} = {V_{m} \over \pi}\times cos\alpha\)

The average output current, \({I_2} = \frac{{{V_2}}}{R}\)

\(\frac{{{I_1}}}{{{I_2}}} = \frac{2}{1}\)

Calculation:

Given that, I₁ = 4 A,

\({I_2} = \frac{{{4}}}{2} = 2\;A\)

Concept:

The average output voltage of a three-phase fully controlled bridge converter by considering the effect of source inductance is

\({V_0} = \frac{{3\;{V_{mL}}}}{\pi }\cos \alpha - \frac{{3\;\omega {L_s}}}{\pi }{I_0}\)

Calculation:

Given that,

Load current (I₀) = 8 A

Firing angle (α) = 45°

Supply line voltage (V_L) = 400 V

Output voltage (V₀) = 320 V

Frequency (f) = 50 Hz

\( \Rightarrow 320 = \frac{{3 \times \sqrt 2 \times 400}}{\pi } \times \cos 45^\circ - \frac{{3 \times 2\pi \times 50 \times {L_s}}}{\pi } \times 8\)

The average output voltage of a single-phase full converter with source inductance is

\({V_o} = \frac{{2{V_m}}}{\pi }\cos \alpha - \frac{{2\omega {L_S}}}{\pi }{I_o}\)

Given that, load power, kW

P_o = 4 kW

⇒ 4 V_oI_o = 4000

\( \Rightarrow \left[ {\frac{{2 \times 230 \times \sqrt 2 }}{\pi }\cos 30 - \frac{{2 \times 2\pi \times 50 \times 1.4 \times {{10}^{ - 3}}}}{\pi }{I_o}} \right]{I_o} = 4000\)

⇒ 179.33 I_o – 0.28 I²_o = 4000

⇒ 0.28 I²_o – 179.33 I_o + 4000 = 0

⇒ I_o = 23.14 A

\( - \cos \left( {\alpha + \mu } \right) + \cos \alpha = \frac{{2\omega {L_S}}}{{{V_m}}}{I_o}\)

\( \Rightarrow \cos 30^\circ - \cos \left( {30 + \mu } \right) = \frac{{2 \times 100\pi \times 1.4 \times {{10}^{ - 3}}}}{{230 \times \sqrt 2 }} \times 23.14\)

⇒ μ = 6.54°

Active power (p_o) \( = {V_{S1}}{I_{S1}}\cos \left( {\alpha + \frac{\mu }{2}} \right)\)

\( \Rightarrow 4000 = 230 \times {I_{S1}} \times \cos \left( {30 + \frac{{6.54}}{2}} \right)\)

Source inductance (L_s) in the converters:

• Source inductance has a significant impact on the converter performance because its presence alters the output voltage of the converter.
• The output voltage reduces as the load current reduces.
• In addition, the input current and output voltage waveforms change significantly.

Effect of L_s:

• It reduces the average DC output voltage of the converter.
• It limits the maximum range of the firing angle α.
• The triggering angle extends to α' = α + μ 

           μ = overlapped angle due to L_s​. 

• It increases the distortion factor as well as the power factor(slightly increases)

Effect of L_s on 3-∅ controlled converter:

The 3-∅ controlled converter with Ls is shown below.

The effect of L_s is shown below waveforms.

The amount of reduced output voltage is showing in the shaded part of the above waveform that happened because of L_s.

α = 25°

μ = 10°

\(\begin{array}{l} {I_o} = \frac{{{v_m}}}{{\omega {L_S}}}\left[ {\cos \alpha - \cos \left( {\alpha + \mu } \right)} \right]\\ 20 = \frac{{230 \times \sqrt 2 }}{{2\pi \times 50{L_S}}}\left[ {\cos 25^\circ - \cos \left( {25^\circ + 10^\circ } \right)} \right] \end{array}\)

L_S = 0.0045 H

\(\begin{array}{l} {V_o} = \frac{{2{V_m}\cos \alpha }}{\pi } - \frac{{\omega {L_S}{I_o}}}{\pi }\\ = \frac{{2 \times 230\sqrt 2 \cos 25^\circ }}{\pi } - \frac{{2 \times 3.14 \times 50 \times 4.5 \times {{10}^{ - 3}} \times 20}}{{3.14}} \end{array}\)

= 178.74 V

power factor

\(\begin{array}{l} = \frac{{{V_o}{I_o}}}{{{V_S}{I_S}}}\\ = \frac{{178.25 \times 20}}{{230 \times 20}} = 0.78 \end{array}\)

Explanation:

Full Bridge Converter

Definition: A full bridge converter is a type of switched-mode power supply (SMPS) topology widely used in high-power applications. It is designed to efficiently convert electrical energy from one form to another, typically stepping up or stepping down DC voltage. The full bridge converter uses four switching devices (such as MOSFETs or IGBTs) arranged in a bridge configuration to transfer energy from the input to the output via a transformer. The primary advantage of this topology is its ability to handle higher power levels with improved efficiency.

Working Principle:

In a full bridge converter, the input DC voltage is applied across the primary winding of a transformer using four switches arranged in a full-bridge configuration. By alternately turning on and off pairs of switches, the converter generates an AC waveform at the transformer primary. This AC waveform is then stepped up or down by the transformer and rectified on the secondary side to produce the desired DC output voltage.

The switching sequence ensures that the transformer core operates efficiently, avoiding saturation and minimizing losses. The use of a transformer also provides isolation between the input and output, which is critical for safety and noise reduction in many applications.

Advantages:

• Capable of handling high power levels, making it suitable for industrial and commercial applications.
• Provides electrical isolation between input and output through the use of a transformer.
• High efficiency due to reduced switching losses and optimized transformer operation.
• Flexibility in stepping up or stepping down voltage levels using the transformer turns ratio.
• Supports bidirectional power flow when combined with appropriate control strategies.

Disadvantages:

• Increased complexity due to the requirement of four switches and associated control circuitry.
• Higher cost compared to simpler topologies like flyback or push-pull converters.
• Requires precise control of switching to avoid issues like shoot-through or transformer saturation.

Applications:

• High-power industrial power supplies.
• Electric vehicle chargers.
• Uninterruptible power supplies (UPS).
• Telecommunication power systems.
• Renewable energy systems, such as solar inverters and wind turbine controllers.

Correct Option Analysis:

The correct option is:

Option 1: Full bridge converter

The full bridge converter is the preferred choice for high-power applications due to its ability to handle large amounts of power efficiently and reliably. The use of four switches in a bridge configuration allows for effective energy transfer and voltage transformation through a transformer. Additionally, the inherent electrical isolation provided by the transformer enhances safety and noise immunity, making it ideal for demanding industrial and commercial environments. Its versatility and performance advantages make it the go-to topology for high-power SMPS designs.

Additional Information

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

Option 2: Push-pull converter

The push-pull converter is another SMPS topology that uses two switches to alternately drive the primary winding of a center-tapped transformer. While it is suitable for medium power applications, it is not ideal for high-power scenarios due to limitations such as transformer core saturation and increased losses in the center-tapped winding. Additionally, the push-pull topology is less efficient than the full bridge converter at higher power levels, making it less suitable for demanding applications.

Option 3: Flyback converter

The flyback converter is a simple and cost-effective SMPS topology that uses a single switch and a transformer to convert voltage levels. It is primarily used in low to medium power applications due to its simplicity and low component count. However, it is not well-suited for high-power applications because of limitations like higher switching losses, reduced efficiency, and increased stress on components. The flyback topology is more commonly found in applications such as mobile chargers, LED drivers, and small power supplies.

Option 4: Half bridge converter

The half bridge converter is a simplified version of the full bridge converter, using only two switches to drive the transformer. While it is more efficient and less complex than the push-pull topology, it is still not as capable as the full bridge converter for handling very high power levels. The half bridge converter is typically used in medium power applications where cost and complexity need to be balanced with performance requirements.

Conclusion:

The full bridge converter stands out as the most suitable topology for high-power applications due to its superior efficiency, ability to handle large power levels, and inherent electrical isolation. While other topologies like push-pull, flyback, and half bridge converters have their own advantages in specific scenarios, they are not as effective as the full bridge converter for demanding high-power designs. Engineers often choose the full bridge topology for applications requiring reliable and efficient power conversion at elevated power levels.

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.

Concept:

The average output voltage of a 3ϕ full wave rectifier is:

\(V_{o(avg)}={3\sqrt{3}V_{mp}\over \pi}cosα\)

\(V_{o(avg)}=I_{o(avg)}R+E\)

Calculation:

\(V_{o(avg)}=(15\times 10)+250\)

\(V_{o(avg)}=400\space V\)

\(400={3\sqrt{3}({415\sqrt{2}\over \sqrt{3}})\over \pi}cosα\)

α = 44.49° 

Concept:

The Fourier series for the source current is:

\(I_{s(n)}=\sum_{1,3,5...}^{\infty}{4I_o\over n\pi }sin \space ({n\pi \over 3})sin(n\omega t-nα)\)

IPF = DF × CDF

Displacement factor (DF) = cosα 

Current distortion factor (CDF) = \({I_{s1(rms)}\over I_{s(rms)}}\)

CDF = \({4I_{o}/\sqrt{2}\pi\over I_o/\sqrt{2\over 3}}=0.955\)

IPF = cos (45°) × 0.955

IPF = 0.675

Concept:

The average output voltage of a single-phase full converter by considering the effect of source-inductance is

\({V_0} = \frac{{2{V_m}}}{\pi }\cos \alpha - \frac{{2\omega {L_s}}}{\pi }{I_0}\;\)       ---(1)

And

\(\cos \alpha - \cos \left( {\alpha + \mu } \right) = \frac{{2\omega {L_s}}}{\pi }{I_0}\)        ---(2)

Calculation:

Given that, load current (I₀) = 5 A

Firing angle (α) = 45°

Supply voltage (V_m) = 330 V

Load voltage (V₀) = 140 V

From equation (1)

\( \Rightarrow 140 = \frac{{2 \times 330}}{\pi }\cos 45^\circ - \frac{{2 \times 2\pi \times 50 \times {L_s}}}{\pi } \times 5\)

⇒ L_s = 8.55 × 10^-3

From equation (2)

\( \Rightarrow \cos \left( {45} \right) - \cos \left( {45 + \mu } \right) = \frac{{8.55 \times {{10}^{ - 3}} \times 2 \times 2\pi \times 50}}{{330}} \times 5\)