8. Three-phase AC regulators

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AC to AC CONVERTERS

(AC Voltage Regulator)

Part – 3, Unit - II

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8. Three-phase AC regulators

• The power rating of a single phase load circuit is allowed normally up to 1kW or kVA only and hardly up to 2 or 3 kVA.

• A single phase load causes unbalanced line current due to different current of each phase.

• A single phase AC regulator adds further distortion in the line current as well as unbalancing in the three-phase line current.

• Although unbalancing is allowed at certain level (about 15%) but, in general, it is avoided.

• Therefore for high power applications (above few kW), three phase systems is employed.

AC-AC Converter POWER ELECTRONICS – II, EEC 3210 Delivered by Prof. M. S. Jamil Asghar, EED, AMU, Aligarh, India 2

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8. Three-phase AC regulators

Circuit configurations: Figure 15 shows different three-phase ac regulator circuits. The

configurations are for both Y and Δ connected loads. However, the circuit configurations shown in Fig. 15(b) & 15(d) are commonly used.

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8. Three-phase AC regulators

Fig 16: Different other circuit configurations Y- and delta- connected loads.

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The circuit configurations shown in figure 15(a) and 15(c) are basically three independent single-phase ac regulators connected in Y and Δ modes,

respectively. Each switch (thyristor pair) controls independently (irrespective of the voltage of) other phases, the output voltage across each load.

The analysis of single-phase ac regulator for single phase load is valid for these cases.

The equation of output voltage, current for etc., of single-phase circuits are valid but the input voltage to regulator may be phase voltage (Fig.15b) or line voltage (Fig. 15c).

The other diode based configurations (Fig. 16), are not commonly used as they suffer from different drawbacks (asymmetric current in positive and negative half cycles, dc offset current which reaches to distribution transformer causing saturation of its core etc. )

It is important to note that when all switches (of each phase) are ON, the load

voltage of each phase would be just like in case of normal three-phase system

(as if there is no any other switch).

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The phase voltages and its associated line voltages are given here (Fig. 17). Each phase voltage (vectors) are associated with it and these line voltages are leading and lagging at 30° with the phase voltage vector.

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 

 ² ³ ^sin ²  ^sin( ³⁰ ⁾

) 30

sin(

2 3

o A C

A

o A B

t V

v

t V

v

t V

v

ForPhaseA











 

 ² ³ ^sin( ²  ^sin( ¹²⁰ ⁾ ¹²⁰ ³⁰ ⁾

) 30 120

sin(

2 3

o o

B C

o B

o o

B A

t V

v

t V

v

t V

v

ForPhaseB

















 

 ² ³ ^sin ²  ^sin( ⁽ ¹²⁰ ¹²⁰ ⁾ ³⁰ ⁾

) 30 120

sin(

2 3

o o

CA

o C

o o

CB

t V

v

t t

V v

t V

v

ForPhaseC

















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Three-phase, three-wire , Y-connection

When switch of TWO phases are ON, the load voltage of conducting phases would be half of its line voltage for Y-connected loads. For Δ modes, when two switches are ON, full line voltage appears at load.

The operating condition of AC regulator with 3-phase, 3-wire system (i)Discontinuous conduction mode (90°≤ α ≤ 150°)

(ii) Continuous conduction mode-I (60° ≤ α ≤ 90°) (iii) Continuous conduction mode-II (0° ≤ α ≤ 60°) (iv) Continuous conduction mode-II (0° ≤ α ≤ 60°)

EEC 3210 POWER ELECTRONICS - II AC to AC CONVERTERS (PART – 2, UNIT – II) 578

AC-AC Converter POWER ELECTRONICS – II, EEC 3210 Delivered by Prof. M. S. Jamil Asghar, EED, AMU, Aligarh, India

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(i) Discontinuous conduction mode (90º   150º)

Only two phases conduct at a time and the control range of  is from 90º to 150 only. In this mode either two thyristors of different phases conduct together or none conducts at all (Fig. 18).

The load voltage or load current becomes discontinuous and half of the line voltage appears across each load (R load). The condition is similar to a single-phase ac regulator system. Thus the same equations and analysis are valid. As V_AB is leading with respect to V_A, the commutation of thyristor takes place when V_AB becomes zero at t =18030 = /6 (Fig. 18).

The load voltage v_OA,v_OB and v_OC are either zero or the half of line voltage, v_L. The output is discontinuous with four voltage pulses in each cycle. The load or output voltage at appear at each phase load, are

v_OA= v_AB/2, v_AC/2 or zero

v_OB =v_BC/2, v_BA/2 or zero, and v_OC = v_CA/2, v_CB/2 or zero

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   

2 / 1 6

/ 5

2

0

sin / 6

2 2 3 2

4 1

 





 





 



 

 

 

^^ ^

  

 ^t ^d ^t

V

_A

V

 

2 / 1

6 /

2

0

sin

2 2 3

2 1

 





 





 



 

 









 

 ^t ^d ^t

V

_A

V

2 / 1 2 / 1

2 4 cos

2 3 4 sin 1 6

5 2

3 1

6 ) (

2 2 sin ) 1

( 6 2

3 1



















   





 



 



 



    





 



 

 

V V

The load or output voltage at appear at each phase load, contains two identical

voltage pulses of v_AB/2 and v_AC/2 in positive half-cycle as well as two identical pulses in the negative-half cycle (total number of pulses are four). Therefore, with the help of voltage equation of v_AB/2, the rms value of load voltage of phase A is

To simplify the integration, the reference is changed by /6, i.e. add /6. Thus

The analysis will be similar to that of a single-phase ac regulator. Therefore,

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2 / 1

0

sin 2

2 2 1

2 sin } 1

1 {

 

 



 



 



   

    

V 

V

2 / 1

0 sin 2

2 ) 1 ( 6

2 2 sin

)} 1 ( 6

1 {



 



 



 



     



     



L

A V

V

In case of RL loads (Fig. 19), the commutation does not take place at 150 or 5/6, but at  which is higher than 5/6. The voltage equation of single phase ac regulator with RL load will hold good. i.e.

Thus,

a A

B

C

T1

T2 T5

T6 T3

T4 R

c b

n

VOC

VOB

VOA

VAB

VBC

VCA

L

R L

Fig.19 Three-phase ac regulator with Y-connected RL load.

Fig. 20 deleted.

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Continuous conduction mode-I (60º   90º)

In this mode, there is no idle period and conduction always takes place. Thus two thyristors of different phases always conduct (Fig. 21).



t MODE II



v_0A

V_m V_Lm/2

vA

V_AB/2

v_0A v_0A

v0A

v_0A v0A

V_AC/2

 = 90^o

Fig. 21 Three-phase, 3-wire, Y-connected R load.

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MODE -I (=135 deg.)

 t



0

^o

60

^o

120

^o

180

^o

240

^o

300

^o

360

^o

 = 135

^o

Fig.22 Output voltageand current for 3-phase ac regulator (Y-connected RL load) V

_AB

/2

V

_AC

/2 V

_A

V

_0A

V

_0A

i

_0A

v

_0A

V

_mA

V

_Lm

/2

i

_0A

=i

_0AB

i

_0A

i

_0A

=i

_AC

135

^o

+15

^o

+15

^o

A Phase= RED B Phase= Yellow C Phase= Blue

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 

2 4 cos

2 3 4 sin

1 3

2 3 1

6 ) (

2 2 sin

) 1 ( 6

2 ) 2 (

3 1

2 sin 2 3 2 1

) 6 (

2 sin 2 3 2

4 1

2 / 1

2 / 1 2

/ 1 2

/ 6 /

2

2 / 1 3

/

2 0

 





 





 



 

  



 

 



 



 



     



 





 





 



 

 

 





 





 



 

 



 



 









 



 





 













V V

t d

V t

t d

V t v

_A

The load voltage is always half of the line voltage, v_L/2 i.e. either v_AB/2 or v_AC/2 for the load voltage, v_OAwhich is given by

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Continuous conduction mode-II (060º)

In this case, in some part of cycle, all phases hence switches of all phases of regulator conduct and in this portion, the load voltage (V_OA) is equal to the phase voltage, V (Fig.

22). In some part of cycle, only two phases and hence of thyristors of two phases conduct which depend upon the circuit conditions. Therefore, for example, the load voltage, v_OA will be either the phase voltage, v_A or the half of line voltage (v_AB/2 or v_AC/2). The rms value of the load voltage can be found accordingly.

Example: In a three-phase, three-phase, three wire system, the input line voltage is 400V (rms value) and load resistance, R=10. Find the output voltage of each phase load when the switching angle is 120 and 135.

Solution: Use the relevant equations and find the desired values. It is a case of discontinuous conduction. There would be two voltage pulses in each half cycle.

Example: In a three-phase, three-phase, three wire system, the input line voltage is 400V (rms value) and load reactance, X =10 at 50 Hz. Find the output voltage of each phase load when the switching angle is 135.

Solution: It is a case of discontinuous conduction of purely inductive load. The line voltage (V_AB) is leading by 30, therefore it becomes zero at t=150 (after conduction this period is 15 only).

During this period energy is stored in reactor (both current and voltage positive). When voltage becomes zero (at t=150) current continues to flow and reactive power becomes negative (current positive and voltage negative). Therefore, the current continues to flow for the same period (15). Thus  can be found accordingly.

The same situation appears in single-phase system with L load. The conduction period is identical with reference to voltage waveform, when the voltage passes through zero crossing instant and therefore=2.

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Delta-connected loads:

Here all systems are three-phase, three wire system where switches are either connected in series with the load in each delta branch or switches are in the line terminals. For the sake of simplicity only the case of R loads are discussed.

Load in series with switch: When the switch (SCR pair or Triac) of each phase is in series with the load in each branch, the conduction of each thyristor is independent of the other. Therefore each load behaves as a single-phase load. Similarly, load voltage of one phase is independent of other phases.

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However, the line current is sum of instantaneous currents of two phases, i.e.

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i_A= i_AB+ i_AC= i_AB i_CA i_B = i_BC + i_BA= i_BC  i_AB i_C = i_CA + i_CB= i_CA  i_BC

where i_A, i_B, i_C and i_AB, i_BC, i_CA are line current and load current or phase currents, respectively.

Here the line voltage (v_AB=v_L) is taken as a reference. The rms value of the output voltage and current of phase A, are

where V_AB, V_BC and V_CA are the line voltage.

Figure 20 shows the load voltage phase and line currents for =120. Since the line current is sum of the phase current (i_A= i_AB+ i_AC).

Example: For a three-phase, three-wire delta system, RL loads are connected in series with reverse-parallel SCR switch (Fig. 24). Draw the output voltage and line current.

Solution: Each branch is independent of other phases. Thus it is like three independent, single-phase Ac regulator with R load. The line voltage and phase voltage are same.

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Delta connection-Switches in line wires: In this case, all the switches are connected in line wire (at A, B, C terminals) and all loads form a  configuration without any switch. Two thyristors of different phases must be conducting for the flow of current. When current is flowing in one branch of delta (resistor), it also flows in two resistors of other phases with half magnitude due to double series impedance (2R). Thus other phases also contribute their half load voltage at AB terminals. The output voltage for each load (branch) can be found as

Also, the line currents, I_AI_B=(I_ABI_CA)(I_BCI_AB) which gives,

I_AB= (I_AI_B)/3 I_BC= (I_BI_C)/3 I_CA= (I_CI_A)/3

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Example: Three load resistors of R=10 each are connected in series with reverse- parallel SCR switch. The line voltage is 400V. Find the output voltage and current of each branch.

Solution: Each branch is independent of other phases. Thus it is like three independent, single-phase Ac regulator with R load. The equation of single AC regulator for R load is valid. Also find the current accordingly.

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%%%%% THIRD PART ENDS HERE %%%%%

||

*************** END OF UNIT – II ***************

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Thank you.

8. Three-phase AC regulators

AC to AC CONVERTERS