Test: Phase Controlled Rectifiers - Electrical Engineering (EE) MCQ

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 <α ≤ 90.
• 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.

Concept:

Phase-controlled converters:

• It is a device which is used to convert A.C sinusoidal into D.C.
• In this the A.C input is fed from transformer stepdown side and then input is converted into D.C output.
• Here we are talking about single phase fully controlled converter.

Single phase full converter with resistive load:

• An A.C voltage source of V_s = V_msin(wt) is connected.
• Four controlled thyristors and a resistive load is connected.
• Let Vo be the output voltage.

Working:

• Let thyristor T₁ and T₄ are triggered at a firing angle of 'α'.
• During the first cycle from '0 to π' the current starts flowing at firing 'αand flows through V_s→ T₁​→ R_L​→ T₄​→ V_s.
• During the second cycle from 'π→2π' the thyristor T₂ and T₃ are triggered at (π + α)
• The current flows through V_s→ T​₂→ R_L​→ T₃​→ V_s.

Current flowing through a resistive load will be 

I = V_o/R

We can see from the above graph that the current will be zero from '0 to α' and the current will be non-zero from α to π.

So, the current will be non-zero and zero from 'π - α' and 'α' respectively.

Concept:

The average output voltage of 3 phase 6 pulse converter is given by 

here V_ml is the maximum value of line-to-line voltage 

The voltage drop in the resistance is given as

2RI0, here 2 resistance will come in to picture in every 60º conduction of two phases

so this voltage drop is subtracted from the final output of this converter.

Solution:

The average output voltage is equal to

R = 4.19 Ω 

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°

Concept:

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

Where V_m is the maximum value of supply voltage

α is the firing angle or delay angle

Average load current:

Calculation:

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

Concept:

Considered Vm 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,

Calculation:

Given that,

V₀ = V_m/3

Hence, the equation becomes,

or,
π/3 = (1 + cosα)
cos α = 0.047
α = cos^-1(0.047)

Total harmonic distortion:

• It is a common measurement of the level of harmonic distortion present in power systems.
• THD can be related to either current harmonics or voltage harmonics.
• It is defined as the ratio of r.m.s value of all the harmonic components to the r.m.s value of the fundamental component.
• Mathematically, it can be represented as

Where,
V_or = Total rms value

V­₀₁ = Fundamental rms value

Distortion factor (DF): It is the ratio of the fundamental component to r.m.s value of the waveform.

Distortion factor (DF) and total harmonic distortion (THD) are related as

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.

Concept:

Fourier series of source current in a full bridge rectifier is given by

For n = 1, 3, 5, ......

The most dominant frequency components in the above expression are f, 3f

Application:

A single-phase, full-bridge diode rectifier fed from a 230 V, 50 Hz sinusoidal source

Dominant frequencies = f, 3f

= 50 and 150 Hz

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 = 5th harmonic supply frequency = 50 Hz

5th harmonic frequency = 5f = 250 Hz