Three Phase Thyristor Converter Circuits MCQ Quiz - Objective Question with Answer for Three Phase Thyristor Converter Circuits - Download Free PDF

Explanation:

Calculation of DC Voltage Generated by the Inverter

To calculate the DC voltage generated by the inverter, let us analyze the given data step-by-step:

• DC source voltage, V_source = 16 kV = 16000 V
• Internal resistance of the DC source, R_internal = 1 Ω
• DC current supplied by the source, I_DC = 900 A
• Inverter output voltage, V_inverter = 12 kV = 12000 V

Step 1: Voltage Drop Across the Internal Resistance

The internal resistance of the DC source causes a voltage drop, which can be calculated using Ohm's Law:

Voltage Drop (V_drop) = I_DC × R_internal

Substitute the given values:

V_drop = 900 A × 1 Ω = 900 V

Step 2: DC Voltage at the Terminals of the Source

The DC voltage at the terminals of the source is the source voltage minus the voltage drop across the internal resistance:

V_terminals = V_source - V_drop

Substitute the values:

V_terminals = 16000 V - 900 V = 15100 V

Step 3: DC Voltage Generated by the Inverter

The inverter generates a DC voltage that corresponds to the terminal voltage of the source because it is directly connected to the source.

Thus, the DC voltage generated by the inverter is:

V_inverter = V_terminals = 15100 V

Final Answer: The DC voltage generated by the inverter is 15100 V.

Analysis of Other Options:

Let us now analyze why the other options are incorrect:

Option 1: 27000 V

This option is incorrect because the DC source voltage is only 16000 V, and the internal resistance causes a voltage drop. The DC voltage generated by the inverter cannot exceed the source voltage. The value of 27000 V is unrealistic and does not match the calculations.

Option 2: 15100 V

This is the correct answer as derived above. The terminal voltage of the DC source, after accounting for the voltage drop across the internal resistance, is 15100 V, which is the voltage generated by the inverter.

Option 3: 16000 V

This option assumes that there is no voltage drop across the internal resistance of the DC source. However, the problem explicitly states that the internal resistance is 1 Ω and the current is 900 A, leading to a voltage drop of 900 V. Thus, the terminal voltage is less than 16000 V, making this option incorrect.

Option 4: 16400 V

This option is incorrect as it exceeds the source voltage of 16000 V. Additionally, it does not account for the voltage drop caused by the internal resistance of the source. Such a value is not physically possible in this scenario.

Conclusion:

The correct answer is Option 2: 15100 V. This value accurately accounts for the voltage drop across the internal resistance of the DC source and represents the DC voltage generated by the inverter.

Additional Information

Important Considerations:

• The internal resistance of the DC source plays a significant role in determining the terminal voltage. For high currents, even small resistances can cause substantial voltage drops.
• In practical applications, minimizing the internal resistance of the source is crucial to maximize the efficiency of the system.
• Six-pulse inverters are commonly used in industrial applications due to their simplicity and cost-effectiveness. However, they may introduce harmonics into the system, which need to be mitigated using filters.

Concept:

• A 3ϕ semi converter works in a continuous mode of operation for α ≤ 60°.
• In a continuous mode of operation, the free-wheeling diode does not conduct.
• For α > 60°, the mode of operation becomes discontinuous in nature.
• In a discontinuous mode of operation, the free-wheeling diode conducts for α - 60°.

​Calculation:

Given, α = 90° 

Conduction period of free-wheeling diode = 90° - 60° 

Conduction period of free-wheeling diode = 30° 

Multiquadrant Operation of Electric Drives:

A quadrant diagram is drawn by plotting the speed-torque characteristics of the load/motor for all four quadrant operations.

There are basically two modes of operation:

(i) Motoring Mode: The electrical energy is converted into mechanical energy and supports the motion.

(ii) Braking mode: The mechanical energy is converted into electrical energy and the motor works as a generator and opposes the motion. The fig. below shows the torque and speed coordinates for both forward and reverse motions.

We know that motor can provide motoring and braking operation for both forward and reverse directions.

The power is given as Power = Speed × Torque.

Now, if the power developed is positive then the operation is motoring. If the power developed is negative, the operation is braking.

Quadrant I - Forward Motoring:

In this region, the power and torque are both positive, so the power developed is positive and the machine works as a motor supplying mechanical energy.

Quadrant II - Forward Braking:

In this region, the speed is positive, but torque is negative, so power developed is negative and the machine works under braking opposing the motion.

Quadrant III - Reverse Motoring:

In this region, the speed is negative and the torque is also negative, so power developed is positive and so the machine works under reverse motoring mode.

Quadrant IV - Reverse Braking:

In this region, the torque is positive and the speed is negative, so the power developed is negative, so the machine works as a braking mode in the reverse direction.

Three-Phase Full Converters:

If all diode is replaced by thyristor, a three-phase full converter bridge formed as shown below,

Explanation:

To operate the motor in reverse motoring the current should be negative. So for that the firing angle α should be greater than π/2.

Additional Information

Four-Quadrant Operation of the Hoist:

As the load is resistive the peak instantaneous output voltage is 

Vm = √2 Vrms

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:

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°

In a six-pulse thyristor bridge rectifier, the harmonics present are = 6 k ± 1

So, the harmonics are = 5, 7, 11, 13, ...

lowest harmonic component = 5^th harmonic supply frequency = 50 Hz

5^th harmonic frequency = 5f = 250 Hz

3-phase semi-converter:

Circuit diagram:

Where, 

V_a V_b V_c are the input voltages

i_a i_b i_c are the input currents

T₁ T₃ T₅ are the thyristors

D₄ D₆ D₂ are the diodes

V₀ i₀ are output voltage and current respectively

R is the load resistance

• This is obtained by a series connection of a 3 pulse controlled converter and a 3 pulse uncontrolled one.
• The three arms of the former consist of thyristors and the three arms of the latter comprise diodes.
• The thyristors commutate at the phase angle at which they are fired. The diodes commu­tate at the natural firing instant α = 0.
• The thyristors conduct for 120° and are fired at intervals of 120°.
• A thyristor and diode conduct at any given time the diode is forward biased at the natural firing instant.
• A thyristor conducts, even if it is reverse biased until the next thyristor in the sequence is fired.
• Thus there is a natural free wheeling of the load current via the incoming diode and outgoing thyristor.
• The load voltage is zero during the freewheeling period.
• Freewheeling due to the conduction of the diode does not allow negative excursions of load voltage.
• This reduces the ripple content in the output voltage.
• The ripple frequency of the output voltage at α = 0 is 6f.
• For α
• A 3-phase semi-converter works as a six-pulse converter when the firing angle α is between 0° to 60°

Concept:

The output voltage of a three-phase full converter is given by

V_mp is maximum supply phase voltage

α is firing angle

Calculation:

Given that, rms supply line voltage (V_rms) = 440 V

Maximum supply phase voltage 

Firing angle (α) = 45°

Concept: Given circuit is a three-phase half wave rectifier circuit. In this circuit any diode will conduct when voltage across them will be maximum and remaining will be off.

Let diode D₁ belongs to R-phase, D₂ belongs to Y phase and D₃ belongs to B-phase. 

Hence, diode D₁ will conduct when voltage in R-phase is maximum.

Let us see the waveform of three phase voltages.

V_RN = V_m sin wt

VYN = Vm sin (wt - 120°)

VBN = Vm sin (wt + 120°)

It is clearly visible from the waveform that R-phase voltage is maximum from 30° to 150° 

∴ Diode D₁ would conduct from 30 to 150 degrees.

Therefore, correct option is (3).

Concept:

For a fully controlled three phase bridge rectifier,

Calculation:

Given that, output load current (I₀) = 100 A

In a fully controlled three-phase bridge converter,

RMS value of the A.C. line current

A

RMS current

Fundamental current

Concept:

• A 3ϕ semi converter works in a continuous mode of operation for α ≤ 60°
• In a continuous mode of operation, the free-wheeling diode does not conduct.
• For α > 60°, the mode of operation becomes discontinuous in nature.
• In a discontinuous mode of operation, the free-wheeling diode conducts for α - 60°.

​Calculation:

Given, α = 90° 

Conduction period of free-wheeling diode = 90° - 60° 

Conduction period of free-wheeling diode = 30° 

Three-Phase Full Converters:

If all diode is replaced by thyristor, a three-phase full converter bridge formed as shown below,

• A three-phase input supply is connected to the terminal A, B, C, and RLE load is connected across the output terminal.
• It worked as three-phase AC to DC converter for firing angle delay 0
• The positive group of SCRs is fired at an interval of 120° and a similarly negative group of SCRs fired at an interval of 120°, but SCR of both groups are fired at an internal of 60° or commutation occurs at every 60°.
• At any time tow SCRs, one from the positive group and one from the negative group must be conducted together for the source to energize the load.
   

Waveform:

Voltage and Current Equation:

The average output voltage (V₀) is given by,

Where V_m is the maximum value of line voltage.

Voltage and current waveform for α = 30° and for constant current can be drawn as,

From waveform RMS value of Thyristor current (I_T = i_T1) will be,

Where I₀ is constant DC load current.

Application:

Given,

I₀ = 150 A

From the above concept,

The rms value of the current flowing through each thyristor of converter is