AC Voltage Controllers Guide

POWER ELECTRONICS
POWER ELECTRONICS

AC VOLTAGE
CONTROLLERS

Dr. Adel Gastli
Email: adel@gastli.net
http://adel.gastli.net

INTRODUCTION
• Purpose: control the output rms voltage
using SCR- or Triac-type switch.
• Name: AC Voltage Controller, AC to
controlled AC converters or AC regulators
• Types: there are single-phase and three-
phase types of ac voltage controllers

Dr. Adel Gastli AC VOLTAGE CONTROLLERS 2

CHAPTER’S CONTENT
CHAPTER’S
1. 1-PHASE AC VOLTAGE CONTROLLERS
2. 3-PHASE AC VOLTAGE CONTROLLERS
3. CYCLOCONVERTERS
4. APPLICATIONS & SUMMARY


Section 1

1-PHASE AC VOLTAGE
CONTROLLERS
T2
iL iL

≡
T1 Triac
vL ZL vL ZL

vs = Vsm sin(ωt ) vs = Vsm sin(ωt )


SECTION’S CONTENTS
SECTION’S
1. ON-OFF CONTROL
2. PHASE CONTROL
3. SECTION SUMMARY


ON-OFF CONTROL
Integral half cycle control. Usually used for resistive load.
iL vs
Vsm

vL R
0
t
n
vs = Vsm sin(ωt )
N
T : period

N : number of half cycle during period T
n : number of half cycles during switch on
Similar to
V0 rms = Vsrms
n Vsm
=
n
= Vs k chopper
Duty principle.
N 2 N cycle

PHASE CONTROL: BI-DIRECTIONAL
1. Resistive Load
iL vs
Vsm

vL R 0 α π +α 2π + α ωt
vs = Vsm sin(ωt )

2
1 π Vsm π
V0 rms = ∫ v dθ = ∫α sin 2 θ ⋅ dθ
2

π 0
L
π
1⎛ sin (2α ) ⎞
V0 rms = Vs ⎜ π -α + ⎟
π⎝ 2 ⎠


SIMULINK SIMULATION

ia
(sp_ac_reg.mdl)
a
i
k +
g -

Triac iA

g

g
+
+ v v0 R=10Ω
v vs -
120V -
50Hz vA

Press to
Plot Results

(sp_ac_regm.m)


1.5

Gate signal
1

0.5

0
0 200 400 600 800 1000
200
v
Voltage, (V)

s
v0
0
α = 45o
-200
0 200 400 600 800 1000
20
Current, (A)

0

-20
0 200 400 600 800 1000
Angle, (Deg)


2. Inductive Load Three cases to be distinguished:
i0 i. α > θ
L ii. α < θ
v0 R
iii. α = θ
i) α > θ : Discontinuous current
vs = Vsm sin(ωt )
Current equation is obtained similarly to
Chapter 10 (single-phase controlled
⎛ ωL ⎞
θ = tan −1 ⎜ ⎟ rectifier).
⎝ R ⎠
⎡ ⎜ ⎟ ⎜ −t ⎟ ⎤
⎛ R ⎞⎛ α ⎞

⎢sin (ωt − θ ) − sin (α − θ )e ⎝ L ⎠⎝ ω ⎠ ⎥
2Vs
i1 =
vs Z ⎢ ⎥
Vsm v0 ⎣ ⎦
i0 β is obtained by taking i1(β)=0.
β
0 φα π +α 2π + α ωt 1⎛ sin 2α sin 2β ⎞
V0 rms = Vs ⎜ β −α + − ⎟
π⎝ 2 2 ⎠


ii) α < θ : Not Practical because conduction angle cannot
exceed π
β > α +π Conduction in one alternance.

To be avoided

If triac gate pulse is large enough then we will obtain
continuous conduction.

iii) α = θ : Continuous conduction
Vsm
β = α +π V0 rms = = Vsrms
2


SIMULINK SIMULATION

ia
(sp_ac_reg.mdl)
a
i
k +
g -
Triac iA

g

g
+ R=10Ω
+ v v0 L=10mH
v vs -
-
120V
50Hz

Press to
Plot Results

(sp_ac_regm.m)

120V
Dr. Adel Gastli 50Hz R=10Ω AC VOLTAGE CONTROLLERS 12

α = 45o > θ = 17.44o
1.5

Gate signal
1

0.5

0
0 200 400 600 800 1000
200
vs
Voltage, (V)

v0
0
α = 45o
-200
0 200 400 600 800 1000
20
Current, (A)

0

-20
0 200 400 600 800 1000
Angle, (Deg)


α = 10o < θ = 17.44o Short gate pulse
1.5
Gate signal

1

0.5

0
0 200 400 600 800 1000
200
Voltage, (V)

vs
v0
0

-200
0 200 400 600 800 1000
20
Current, (A)

0

-20
0 200 400 600 800 1000
Angle, (Deg)


α = 10o < θ = 17.44o Long gate pulse

1.5

Gate signal
1

0.5

0
0 200 400 600 800 1000
200
vs
Voltage, (V)

v0
0

-200
0 200 400 600 800 1000
20
Current, (A)

0

-20
0 200 400 600 800 1000
Angle, (Deg)


α = θ = 17.44o Continuous conduction

1.5
Gate signal

1

0.5

0
0 200 400 600 800 1000
200
vs
Voltage, (V)

v0
0

-200
0 200 400 600 800 1000
20
Current, (A)

0

-20
0 200 400 600 800 1000
Angle, (Deg)


SECTION SUMMARY

Single-phase AC voltage rms can be controlled
by on-off control or phase delay control.
By controlling the phase delay it is possible to
control the AC output voltage rms value
between 0 and the source rms voltage value.
It is important to know beforehand the load
angle in order to be able to control the output
voltage properly.


Section 2

3-PHASE AC VOLTAGE
CONTROLLERS
i0 A
vA
i0 B ZL
vB
ZL ZL
i0C
vC


SECTION’S
1. TOPOLOGIES
2. OPERATION
3. SIMULINK SIMULATION


TOPOLOGIES
i0 A 1
vA i0 A 1
T1 vA
i0 B 2 ZL T1
vB N i0 B 2 ZL
T2 ZL ZL vB
i0C 3 T2 ZL ZL
vC i0C 3
T3 vC
N T3

Equivalent to 3 single-phases Is studied in this chapter
1
vA
i0 A 1 T1 ZL
vA
T1
i0 B 2 ZL
T3
vB ZL ZL
T2 ZL ZL
i0C 3 vB 2 3
vC
T3 T2
vC

OPERATION
i0 A 1
vA ⎧0 ≤ α < 60o ⇒ alternate between 2 and 3 switches
T1 ⎪ o
⎪60 ≤ α < 90 ⇒ only 2 switches conduct at a time
o
i0 B 2 ZL
vB N ⎨ o
⎪90 ≤ α < 150 ⇒ 0 and 2 switches conduct at a time
o
T2 ZL ZL
i0C 3 ⎪α > 150o ⇒ there is no conduction v = 0V
vC ⎩ out
T3

line on line off v12 v1N
A,B, C None VAB VA
A,B C VAB VAB/2
A,C B VAC/2 VAC/2
B,C A -VBC/2 0
None A,B,C 0 0
A Impossible
B Impossible
C Impossible

SIMULINK SIMULATION
ia
(sp_ac_reg.mdl) Triac
+
i
Converter -
+ iA 10Ω
Ain Aout
-
v vab
A A
i
ib
Bin Bout +
- B B
iB C C
Cin Cout

i
ic 3- Phase
+ Y-connected Load
-
+
-
v vAB iC +
v van
-
+
-
v vBC

+
-
v vCA

Press to
+ Plot Results
-
v vAN

120V
vA vB vC
50Hz
(sp_ac_regm.m)


α = 30o Purely resistive load
0.5vab va 0.5vac
va
Voltage, (V)
200
va
0

-200

0 50 100 150 200 250 300 350
α π π 2π 2π π
1.5 +α +α
3 3 3 3
Gate signal

1 Pulse width
0.5
50%

0
0 50 100 150 200 250 300 350
20
Current, (A)

0

-20
0 50 100 150 200 250 300 350
Angle, (Deg)


0.5vab 0.5vac
Voltage, (V)

200

0

-200

0 50α 1002π 150 π
2 200 250 300 350
1.5 +α
3 3
Gate signal

1

0.5

0
0 50 100 150 200 250 300 350
20
Current, (A)

0

-20
0 50 100 150 200 250 300 350
Angle, (Deg)


0.5vab 0.5vac

Voltage, (V)
200

0

-200

0 100 200 300 400 500
α 5π π + α 7π
1.5
6 3 6
Gate signal

1

0.5

0
0 100 200 300 400 500
20
Current, (A)

0

-20
0 100 200 300 400 500
Angle, (Deg)


Voltage, (V)

200

0

-200

0 100 200 300 400 500
1.5
Gate signal

1

0.5

0
0 100 200 300 400 500
20
Current, (A)

0

-20
0 100 200 300 400 500
Angle, (Deg)


MATHEMATICAL ANALYSIS: (Resistive Load)
T1
iL ia

A a
T4
vAN van
R=
T3 1o
B ib R=
N 1o
b
vBN
T6
R=
vCN
1o
T5
C ic

T2
⎛ π⎞
v AN = 2Vs sin (θ ) v AB = 6Vs sin ⎜θ + ⎟
⎝ 6⎠
⎛ 2π ⎞ π⎞
vBN = 2Vs sin ⎜θ − ⎟ ⎛
⎝ 3 ⎠ vBC = 6Vs sin ⎜θ − ⎟
⎝ 2⎠
⎛ 2π ⎞ π⎞
vBN = 2Vs sin ⎜θ + ⎟ ⎛
⎝ 3 ⎠ vCA = 6Vs sin ⎜θ − ⎟
⎝ 6⎠

0.5vab va 0.5vac
0 ≤ α < 60 o
va
Voltage, (V)

200
va
0

-200

0 50 100
150 200 250 300 350
α π π 2π 2π π
+α +α
3 3 3 3

2π
van d (ωt )
1
V0 = ∫
2

2π 0

1 ⎡ π /3 2 π / 3+α v
2
2π / 3 2π / 3+α v
2
π ⎤
va d (ωt ) + ∫ d (ω )t + ∫ va d (ωt ) + ∫ d (ωt ) + ∫ va d (ωt )⎥
π ⎢ ∫α
= ab 2 ac 2
π /3 π / 3+α 2π / 3 2π / 3+α
⎣ 4 4 ⎦

1 ⎡ π / 3 sin 2 ωt π / 3+α sin (ωt + π / 6 ) 2π / 3 sin ωt 2π / 3+α sin (ωt − π / 6 ) sin 2 ωt ⎤
2 2 2
π
V0 = 6Vs ⎢ ∫α d (ωt ) + ∫ d (ωt ) + ∫ d (ωt ) + ∫ d (ωt ) + ∫ d (ωt )⎥
π⎣ 3 π /3 4 π / 3+α 3 2π / 3 4 2π / 3+α 3 ⎦


1 ⎡ π / 3 sin 2 ωt π / 3+α sin (ωt + π / 6 ) 2π / 3 sin ωt 2π / 3+α sin (ωt − π / 6 ) sin 2 ωt ⎤
2 2 2
π
V0 = 6Vs ⎢ ∫α d (ωt ) + ∫ d (ωt ) + ∫ d (ωt ) + ∫ d (ωt ) + ∫ d (ωt )⎥
π⎣ 3 π /3 4 π / 3+α 3 2π / 3 4 2π / 3+α 3 ⎦

1 ⎛ π α sin 2α ⎞ rms Output
V0 = 6Vs ⎜ − + ⎟ phase voltage
π ⎝6 4 8 ⎠

V0
I0 = P0 = 3I 0V0
R
rms output Output power
Phase Current


0.5vab 0.5vac
60o ≤ α < 90o
Voltage, (V)

200

0

-200

0 50α π
100 150 π
2 200 250 300 350
+α +α
3 3
2π
van d (ωt )
1
V0 = ∫
2

2π 0

1 ⎡ π / 3+α vab
2
2π / 3+α v
2
⎤
+ d (ωt )∫ d (ωt )⎥
π ⎢ ∫α
= ac
π / 3+α
⎣ 4 4 ⎦

1 ⎡ π / 3+α sin 2 (ωt + π / 6 ) 2π / 3+α sin (ωt − π / 6 )
2
⎤
d (ωt ) + ∫ d (ωt )⎥
π ⎢ ∫α
V0 = 6Vs
π / 3+α
⎣ 4 4 ⎦
1 ⎡ π / 3+α +π / 6 sin 2 ωt 2π / 3+α −π / 6 sin ωt
2
⎤
d (ωt ) + ∫ d (ωt )⎥
π ⎢ ∫α +π / 6
V0 = 6Vs
π / 3+α −π / 6
⎣ 4 4 ⎦


1 ⎡ π / 2+α sin 2 ωt π / 2 +α sin ωt
2
⎤
d (ωt ) + ∫ d (ωt )⎥
π ⎢ ∫α +π / 6 4
V0 = 6Vs
π / 6 +α
⎣ 4 ⎦

1 ⎛ π 3 sin 2α 3 cos 2α ⎞ rms Output
V0 = 6Vs ⎜ + + ⎟ phase voltage
π⎝⎜ 12 16 16 ⎟
⎠

V0
I0 = P0 = 3I 0V0
R
Phase Current


0.5vab 0.5vac
90 ≤ α < 150
o o
Voltage, (V)

200

0

-200

0 100 200 300 400 500
α 5π π + α 7π
6 3 6
2π
van d (ωt )
1
V0 = ∫
2

2π 0

1 ⎡ 5π / 6 vab
2
7π / 6 v
2
⎤
d (ωt ) + ∫ d (ωt )⎥
π ⎢ ∫α
= ac
π / 3+α 4
⎣ 4 ⎦
1 ⎡ 5π / 6 sin 2 (ωt + π / 6) 7π / 6 sin (ωt − π / 6 )
2
⎤
d (ωt ) + ∫
π ⎢ ∫α
V0 = 6Vs d (ωt )⎥
π / 3+α
⎣ 4 4 ⎦

1 ⎡ 5π / 6+π / 6 sin 2 ωt 7π / 6 −π / 6 sin ωt
2
⎤
π ⎢ ∫α +π / 6
V0 = 6Vs d (ωt ) + ∫ d (ωt )⎥
π / 3+α −π / 6
⎣ 4 4 ⎦

1⎡ π sin 2 ωt π sin 2 ωt ⎤
V0 = 6Vs ⎢ ∫π / 6+α d (ωt ) + ∫ d (ωt )⎥
π⎣ 4 π / 6 +α 4 ⎦

1 ⎛ 5π α sin 2α 3 cos 2α ⎞ rms Output
V0 = 6Vs ⎜ ⎟ phase voltage
π⎝⎜ 24 − 4 + 16 + 16 ⎟
⎠

V0
I0 = P0 = 3I 0V0
R
Phase Current


Section 3

CYCLOCONVERTERS


SECTION’S
1. SINGLE-PHASE
2. THREE-PHASE


SINGLE-PHASE CYCLOCONVERTER
P-converter N-converter
+ −
T1 T3 T2' T4' Variable voltage
variable frequency
Load

vs v01 v02
converter output
T4 T2 T3' T1' voltage.
− +

vs fs
f out = (n = number of positive half cycles)
Vsm n
N-converter π
1 2
on V0 = ∫ v s dθ
π α
0 α
t
1⎛ sin 2α ⎞
P-converter = Vs ⎜π − α + ⎟
on π⎝ π ⎠


THREE-PHASE CYCLOCONVERTER

Three-phase/single-phase cycloconverter
P-converter N-converter
+ −
T1 T3 T5 T2' T6' T4'
A

Load
B v01 v02 C
B
C A
T4 T6 T2 T5' T3' T1'
− +


Three-phase/three-phase cycloconverter

Three-phase supply
P N P N P N

Phase a Phase b Phase c
load load load

Neutral

18 thyristors are used to obtain a 3 full-wave three-phase
converter.


Section 4

APPLICATIONS & SUMMARY


APPLICATIONS

Light dimmer.

Soft starting of ac motors
(compressors, pumps, etc.)

Variable speed drives for
appliances an tools.
vp
Transformer static tap changers.
v0


SUMMARY
Ac voltage controllers can use on-off control or
phase-angle control.
The on-off control is more suitable for systems
having a high time constant.
Due to the switching characteristics of thyristors,
an inductive load makes the solutions of
equations describing the performance of
controllers very complex and simulation is more
convenient.
The input power factor of controllers, which vary
with the delay angle, is generally poor, especially
at low voltage output range.


AC Voltage Controllers Guide

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