Open access peer-reviewed chapter

# Five-Phase Line Commutated Rectifiers

Written By

Mahmoud I. Masoud

Submitted: 18 June 2018 Reviewed: 10 September 2018 Published: 11 April 2019

DOI: 10.5772/intechopen.81436

From the Edited Volume

## Electric Power Conversion

Edited by Marian Găiceanu

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## Abstract

Multiphase systems including multiphase generators or motors, especially five-phase, offer improved performance compared to three-phase counterpart. Five-phase generators could generate power in applications such as, but not limited to, wind power generation, electric vehicles, aerospace, and oil and gas. The five-phase generator output requires converter system such as AC-DC converters. This chapter introduces the basic construction and performance analysis for uncontrolled/controlled five-phase line commutated rectifier guided by numerical examples. This rectifier is suitable for wind energy applications to be the intermediate stage between five-phase generator and DC load or inverter stage. The filtration for AC side and effects of source inductances are detailed in other references. Here, this chapter gives the reader a quick idea about the analysis and performance of multiphase line commutated rectifiers and specifically five phase.

### Keywords

• five phase
• multiphase generators
• wind turbine
• controlled and uncontrolled rectifiers
• AC-DC converters
• line commutated rectifiers

## 1. Introduction

This chapter introduces uncontrolled/controlled line-commutated five-phase rectifier as we target high-power wind energy applications. The study includes the analysis of both AC side and DC side. Moreover, the gating signal generation is introduced. The study includes ideal source to represent the PM generator. Effect of source inductance and filtration of AC side are introduced in Refs. [3, 5].

## 2. Five-phase source

The phase voltage of the five-phase source, which is the output of the five-phase generator, can be written as

v jn = V ph _ max sin ω t 2 πk 5 E1

where j represents phases a, b, c, d, and e, respectively, and k equals 0, 1, 2, 3, and 4, respectively. The five-phase source has two line voltages (adjacent and nonadjacent) [3, 5]. The line voltages can be calculated using phasor diagram shown in Figure 1.

The line voltage for adjacent phases has amplitude 1.1756 Vph_max and leads to the phase voltage by 54°; see Figure 1a. The line voltage for nonadjacent phases has amplitude 1.902 Vph_max and leads to the phase voltage by 18°; see Figure 1b.

Example 1: For a 200 V (peak value-phase voltage) five-phase source, calculate and plot the phase and line voltage waveforms.

Solution

Phase voltages are

v a = 200 sin ω t , v b = 200 sin ω t 2 π 5 , v c = 200 sin ω t 4 π 5 , v d = 200 sin ω t + 4 π 5 , v e = 200 sin ω t + 2 π 5 V

Figure 2 shows the line voltages for adjacent and non-adjacent phase voltages.

The line voltage for adjacent phases is

v ab = 200 1.1756 sin ω t + 3 π 10 = 235.12 sin ω t + 3 π 10 V

The line voltage for nonadjacent phases is

v ac = 200 1.902 sin ω t + π 10 = 380.4 sin ω t + π 10 V

## 3. Uncontrolled five-phase rectifier

The five-phase uncontrolled rectifier is shown in Figure 3. It consists of 10 diodes where the load is fed from five-phase half wave connection, five switches D1D5, and return path being via another half wave connection, five switches D6D10, to one of the five supply lines.

The load voltage or output DC voltage can be calculated by the addition of the two five-phase half-wave voltages, relative to the supply neutral point n, appearing at the positive and negative sides of the load, respectively. The voltage is 10 pulses in nature having its maximum instantaneous value of that of line voltages for nonadjacent phases and leads the phase voltage by π/10. The load voltage follows in turn 10 sinusoidal voltages during one cycle, those being vac, vad, vbd, vbe, vce, vca, vda, vdb, veb, and vec. The average voltage can be calculated with the aid of Figure 4. The repeated period accommodates 36° and the limits are −18° to +18°.

The average or mean load voltage can be calculated as

V mean = 1 π / 5 π 10 π 10 1.902 V ph _ max cos ωt dωt E2
V mean = 1.87 V ph _ max E3

The waveform of supply voltage (phase voltage), positive voltage with respect to neutral (VP), negative voltage with respect to neutral (VN), load voltage, diode 1 voltage, diode 1 current, and supply current are shown in Figure 5 parts ‘a’ to ‘e’. The peak reverse voltage for one diode is the line voltage of nonadjacent phases and as indicated in Figure 5c.

The ripple voltage for DC side is

% Voltage ripple = Maximum value Minimum value V mean 100 E4

The DC-side harmonics are generated at 10kf, where f is the fundamental frequency of the AC source or generated power frequency and k equals (1, 2, 3, etc.). The lowest order harmonic represents 2.1% of the average DC value as shown in Figure 6.

The supply current is a quasi-square wave in its nature as shown in Figure 4e. The load current is constant as the load is considered highly inductive. Switch current operates for only 2π/5 from one cycle; accordingly, the switch average current can be calculated as in Eq. (5) which represents 20% of the DC-link current.

I av _ switch = 1 2 π 3 π 10 7 π 10 I dc dωt = 0.2 I dc E5

This means five-phase rectifier could have switches with lower ratings than three-phase counterpart which feed the same load as the average value of switch current in three-phase system is 33.3% [8, 15]. Part ‘e’ in Figure 5 shows that supply current is a quasi-square wave where input current comprises 2π/5 alternating polarity blocks, with each phase displaced to others by 2π/5. The input phase current can be expressed as [5, 8, 11, 15].

i s t = n = 1 , 3 , 5 , 4 I dc sin 5 sin nωt E6

Substituting n = 5 (or any odd multiple of 5) into Eq. (6) results in is5(t) = 0. The fundamental input current (n = 1) is

i s 1 t = 0.7484 I dc sin ω t E7

and rms value of fundamental current is

I s 1 = 0.529 I dc E8

The rms value of supply current is calculated as

I s = 1 π 3 π 10 7 π 10 I dc 2 t = 2 5 I dc = 0.6324 I dc E9

The harmonic contents of the supply current normalized to the fundamental component (Isn/Is1) are shown in Figure 7. The harmonic factor (HFn) is defined as

HF n % = I sn I s 1 100 E10

Example 2: A 240 V, 50 Hz, five-phase uncontrolled rectifier is connected to highly inductive load and takes 25 A. Calculate:

2. The value of lowest order harmonic in DC side

3. Maximum and minimum DC side voltage and voltage ripple

4. The root mean square value of supply current

5. The root mean square value of the supply current fundamental

6. The third, fifth, and seventh harmonic factors for supply current

Solution

Figure 8 shows the load voltage and supply voltage for the uncontrolled rectifier when the supply voltage has phase and rms value of 240 V.

1. The average voltage is calculated using Eq. (3):

V mean = 1.87 V ph _ max = 1.87 240 2 = 634.7 V
1. The value of lowest harmonic of DC-side voltage is 2.1% of average DC value and occurs at 10 times the supply voltage frequency, i.e. 500 Hz:

Lowest order harmonic = 0.021 634.7 = 13.32 V
1. The maximum voltage in DC side is the peak value of line voltage for nonadjacent phases:

V max = 1.902 V ph _ max = 1.902 240 2 = 645.5 V

The minimum voltage in DC side is the value of line voltage for nonadjacent phases at angle 18°:

V max = 1.902 V ph _ max cos 18 = 613.96 V

The percentage of ripple voltage can be calculated using Eq. (4):

% Voltage ripple = Maximum value Minimum value V mean 100 = 645.6 613.96 634.7 100 = 4.98 %
1. The rms of supply current is calculated using Eq. (9):

I s = 0.6324 I dc = 15.81 A
1. The root mean square value of supply current fundamental is calculated using Eq. (8):

I s 1 = 0.529 I dc = 0.529 25 = 13.225 A
1. The third harmonic of supply current represents 53.9% of the fundamental = 7.13 A:

The fifth harmonic of supply current = 0.

The seventh harmonic of supply current represents 23.1% of the fundamental = 3.05 A.

Example 3: A highly inductive DC load requires 100 A at 475 V. Give design details for the diode requirement using uncontrolled five-phase full-wave rectifier.

Solution

The required load voltage is 475 V; hence the supply voltage can be calculated using Eq. (3):

V mean = 1.87 V ph _ max = 475 = 1.87 V ph _ max V ph _ max = 254 V

The required rms value of phase voltage is 179.6 ≅ 180 V.

The line voltage of nonadjacent phases is

V L = 1.902 V ph _ max = 1.902 179.6 2 = 483 V

Hence, the peak reverse voltage is 483 V. Hence, each diode should withstand maximum voltage of 483 V and 100 A.

## 4. Fully controlled five-phase rectifier

The five-phase fully controlled rectifier is similar to the uncontrolled rectifier, but it has 10 thyristors instead of diodes as shown in Figure 9. The challenge with five-phase controlled rectifier is gating signal generation where the development and implementation of gating signals algorithm use nonadjacent line voltages. Delaying the commutation of thyristors for a certain firing delay angle ‘α’ controls the mean load voltage.

The mean load voltage can be evaluated with the aid of Figures 10c and 12 where it shows one pulse of the output load voltage:

V mean = 1 2 π 10 π 10 + α π 10 + α 1.902 V ph _ max cos ω t t E11
V mean = 1.87 V ph _ max cos α E12

The angle between the fundamental input current and fundamental input phase voltage, ϕ, is called the displacement angle. The displacement factor is defined as [3, 5, 16].

DPF = cos ϕ 1 = cos α E13

The supply power factor is defined as the ratio of the supply power delivered P to apparent supply power S [16]:

pf = P S E14

In the case of non-sinusoidal current and voltage, the active input power to the converter delivered by the five-phase source (e.g. output of wind turbine generator) is

P in = 5 2 π 0 2 π v s 1 t i s 1 t t = 5 V s 1 I s 1 cos ϕ 1 E15

where Vs1 and Is1 are the rms values of the fundamental components of the input phase voltage and current, respectively. The apparent power is given by

S = 5 V s rms I s rms E16

where Vs-rms and Is-rms are the rms values of the input phase voltage and current, respectively. The displacement factor (DPF) of the fundamental current was obtained in Eq. (13). By substituting (13), (15), and (16) into (14), the power factor can be expressed as

pf = V s 1 I s 1 cos ϕ 1 V s rms I s rms E17

if the source is pure sinusoidal supply Vs1 = Vs-rms.

The output power (load power) can be calculated using

P out = 1 T 0 T v t i t dt = V mean I mean E18

The efficiency is

η = P out P in E19

The total harmonic distortion, THD, is given by [3, 5, 15]

THD = I s 2 I s 1 2 I s 1 E20

For a five-phase rectifier, the THD is 65.5%.

Example 4: A five-phase fully controlled rectifier is fed from 220 V (phase value), 50 Hz AC source. The DC load is highly inductive and requires 100 A at 475 V. Give design details for the thyristor requirement and the required firing angle. What is maximum DC load voltage that the converter can deliver to the load?

Solution

The required load voltage is 475 V and the supply voltage is 220 V. The firing angle can be calculated using Eq. (12):

V mean = 1.87 V ph _ max cos α = 475 = 1.87 200 2 cos α cos α = 0.8998 α = 26.1

The peak line voltage of nonadjacent phases is

V L = 1.902 V ph _ max = 1.902 220 2 = 591.8 V

Hence, the peak reverse voltage is ≅592 V. Hence, each thyristor should withstand maximum voltage of 592 V and 100 A.

The maximum possible DC load voltage is when α is adjusted to zero:

V mean = 1.87 V ph _ max = 1.87 220 2 = 581.8 V

Example 5: A five-phase fully controlled rectifier is fed from 230 V (phase value), 15 Hz AC source. The load is 45 Ω resistance and 1 H inductance (highly inductive load). Calculate for firing angle 36°, the converter power factor. Simulate your converter.

Solution

V mean = 1.87 V ph _ max cos α = 1.87 230 2 cos 36 = 492 V

I mean = V mean R = 492 25 = 19.68 Ω

From Eq. (8), the rms value for fundamental supply current is 10.41 A.

The angle of fundamental current is the firing angle.

From Eq. (9), the rms value of supply current is 12.44 A.

The power factor is calculated using Eq. (17):

pf = V s 1 I s 1 cos ϕ 1 V s rms I s rms = 230 10.41 cos 36 230 12.44 = 0.677

The load voltage and supply current are shown in Figure 13. It should be noted that as the firing angle is 36°, the load voltage shown in Figure 13a starts for phase ‘a’ at 90° which equals 54°, the intersection point between two adjacent phases, and 36° which is the firing angle. The supply current for phase ‘a’ is shown in Figure 13b. Switch 1 accommodates 72° and starts at 90°. Switch 6 is shifted 180° for the same phase.

## 5. Gating pulse generation

A pulse generator synchronized on the source voltage provides the trigger pulses for the 10 thyristors that construct the 10-pulse rectifier. The pulses are generated by α degrees after increasing zero crossings of the commutation voltages which is indicated in Figures 10a and 12. The input to the gating signal generator is 10 signals represent five-line voltages for nonadjacent phases and its inverted signals, and multiplexed, then determines the zero-crossing point and counts the delay angle α and hence applies the pulse for the specified switch through gate drive circuits as shown in Figure 14. A double-pulse technique is used to ensure that the specified gate signal fires its own thyristor [3, 5].

A sample of gating signal generation is shown in Figure 15 where an example for gating signal generator output with delay angle 36° with double-pulse technique is indicated. The phase voltages of five-phase permanent magnet generator are measured as shown in part a where the rotational speed tends to have a frequency around 18 Hz and rms value of 56 V. The delay angle is calculated from the zero-crossing point determined by the pulse generator as indicated in Figure 15b where gate signals 1, 2, and 6 confirm that the phase shift between gates 1 and 2 is 72° and between gates 1 and 6 is 180°. The delay angle is calculated after 54° from the start of the phase voltage [3, 5]. If the delay angle is increased to 36°, the results show that the relation between g1, g2, and g6 will be the same but shifted to 90° from the starting of phase voltage as shown in Figure 15c.

Example 6: For the converter given in ‘Example 5’, plot the gating signals

Solution

Figure 16 shows the gating signals for firing angle 36°.

## 6. Conclusion

This chapter introduces the five-phase, 10-pulse, line commutated rectifier. The relation between phase voltage and line voltage for five-phase source is introduced where line voltage equals 1.1756Vph_max and leads the phase voltage by 54° for adjacent phases, while it equals 1.902Vph_max and leads the phase voltage by 18° for nonadjacent phases. The converter performance and analysis for both AC side and DC side are detailed for both uncontrolled and controlled rectifiers. The average voltage of the DC side is 1.87 Vph_max with ten pulses per one cycle, while it is 1.654 Vph_max with six pulses for three-phase counterpart. Moreover, the DC-side voltage ripple is around 5%, while it is around 14% for full-wave three-phase converter. The switch (diode or thyristor) carries only 20% of the load current during one cycle, while it carries 33.3% in three-phase counterpart, and no-phase shifting transformer is required like 12-pulse converter. Also, the implementation of gating signal generation is introduced. The gating signal generation adopts double-pulse techniques and ensures the real-time synchronization between the five-phase source and gating signals. The phase shift pulse of adjacent phases is 72°, while the phase shift between pulses applied for switches on the same phase adopts 180°.

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Written By

Mahmoud I. Masoud

Submitted: 18 June 2018 Reviewed: 10 September 2018 Published: 11 April 2019