LECTURE #14

AC Voltage Controllers

The main purpose of AC voltage controller or AC regulator is to regulate the AC output

voltage apply to the load. Some of the applications of such AC voltage control are as
follows:
a) Control of heating in industrial heating loads.
b) Control of illumination in a lighting control system
c) Control of the RMS magnitude of the voltage in a transformer tap changing
system
d) Speed control of industrial AC drive.

14.1 Circuit topology

Triac based circuit.

Inverse parallel connection

IGBT based circuit with PWM control

Two types of control are normally used:

(i) Phase Control

(ii) Integral Cycle Control (On-Off Control)

14.2 Phase control

(1) Single phase bidirectional controllers with resistive loads

π+α

During the positive half-cycle of input voltage the power flow is controlled by varying the
delay angle of thyristor T1; and thyristor T2 controls the power flow during the negative
half-cycle of input voltage.
The load rms value

ˆ sin ωt , and the delay angles of thyristor T1 is α and T2 is

The input voltage, vs (t) = Vs
π+α

The rms output voltage,

1  
π 2π

 ∫ ( v s ) dθ + ∫ ( v s ) dθ 
2 2
Vo = (1)
2π  α π+α 

π π
2 1 2
Vo = ∫
2π α
( vs ) dθ = ∫ Vˆ s sin θ dθ
2

πα
( )

V̂s 1 sin 2α  (2)

Vo = π−α + 
2 π 2 

NB:
• Varying the delay angle α from 0 to π can vary the RMS output voltage from
Vs to 0.

As the output voltage is periodical but non-sinusoidal, it consists of the fundamental as

well as higher harmonic components. The DC component is zero because of the
symmetry of the positive and negative waveform. Employing the Fourier analysis the
load voltage can be written as:

∞
vload (t ) = ∑ a
n =1
n cos nω t + bn sin nωt 

where
 2 ⋅π
1 ⌠ 2
T
vs ( t) ⋅cos ( n⋅ωt) d( ωt)
T ∫0
an ⋅ ⇔ a1 = vs cos ( nωt )dt
π ⌡0
2 ⋅π
1 ⌠ T
⋅ vs ( t) ⋅sin( n⋅ωt) d( ωt) 2
T ∫0
bn ⇔ b1 = vs sin ( nωt )dt
π ⌡0

ˆ sin ωt
and vs (t) = V For intervals ( α, π ) and ( π + α, 2π )
s

In the case of an induction motor load, the fundamental component VLoad _1 is of

interest

T
2
T ∫0
a1 = vs cos (1ωt )dt

2  ˆ ˆ sin ωt cos (1ωt )dωt 

π 2π
a1 =  ∫ Vs sin ωt cos (1ωt )dωt + ∫ V s
2π  α π+α 
ˆ π
4V 
 ∫ sin ωt cos ωtdωt 
s
a1 =
2π  α 
ˆ π 1
2V 
a1 = s  ∫ ( sin(ωt − ωt) + sin(ωt + ωt) )dωt 
π α 2 

V̂s  
π
a1 =  ∫ ( sin 0 + sin 2ωt )dωt 
π α 

V̂s 1 π
a1 =
π  2 ( − cos 2ωt ) α 

V̂s  cos 2α − 1 
a1 =
π  2 

T
2
and b1 = ∫ vs sin (1ωt )dt
T0

V̂s  sin 2α 
b1 =  π − α + 2 
π
 vload_1 can now be written as

vload_1( t) a1 ⋅cos ( ωt) + b1 ⋅sin( ωt)

vload_1( t) (
Vload_1_mag⋅sin ωt + φ load_1 )
where  a1 
Vload_1_mag
2
a1 + b1
2 φ load_1 atan 
 b1 

⋅ a1 + b1 
1 2 2
Vload_1_rms
2

The power factor, assuming input power = output power

Real_Power
Power_Factor
Apparent_Input_Power

2
Where Vload_rms
Real_Power
R
 Vˆ   Vload _ rms 
Apparent_Input =  s   
 2   R 

Vload _ rms Vload _ rms

Power_Factor = =
Therefore  Vˆs  Vrms _ input
 
 2 

Example: (Rashid page 508)

A single-phase full-wave AC voltage controller has a resistive load R = 10 Ω and the
input voltage is Vs = 120 V, 50 Hz. The delay angles of thyristors T1 and T2 are equal: α1
= α2 = π/2. Determine (a) the RMS output voltage Vo (b) the input PF (c) the average
current of thyristors IA and (d) the RMS current of thyristors IR
(2) Single phase bidirectional controllers with resistive and inductive loads
Phase control [Rashid –pg509][Lander-pg253]

α ⇒ T1 β π + α ⇒ T2

Analysis

REFER: (Rashid page 509)

Assume THY1 is fired during the positive half cycle and carries the load current. Due to
the inductance in the circuit, the current of THY1 will not fall to zero at ωt = π , when the
input voltage starts to be negative. THY1 continues to conduct until its current iTHY1 falls

to zero at ωt = β (extinction angle).

With inductive loads, the continuation of conduction beyond the voltage zero means that
there is no control below a certain firing angle delay. Firing control is only possible after
cessation of current (i.e current goes to zero).

When the thyristor THY1 conducts, the voltage equation becomes

di1
V̂s sin ωt = Ri1 + L
dt

The above equation has a solution in the form of:

V̂s
i1 = sin(ωt − φ) + A1e− (R / L)t
Z

where load impedance Z = R 2 + (ωL) 2 and load angle φ = tan −1 (ωL / R)

when ωt = α, i1 = 0 then

V̂s R(α)
A1 = − sin(α − φ)e L ω
Z
therefore
V̂s 
sin(ωt − φ) − sin(α − φ)e L ω 
R ( α −t )
i1 =
Z  

The current i1 falls to zero at angle ωt = β (extinction angle). Thus

R ( α −t )
L ω
sin(β − φ) − sin(α − φ)e =0

β can be found using iterative method (make initial guess of β, find β such as the left-
hand-side equation equals=0)
The rms output voltage

1  ˆ 
β π+β
2 2
Vo =
2π  α
(
 ∫ Vs sin θ dθ + ∫ V ) (
ˆ sin θ dθ 
s  ) (3)
π+α 

β
2 2
Vo = ∫
2π α
Vs (
ˆ sin θ dθ )

V̂s 1 sin 2α sin 2β  (4)

Vo = β − α + − 
2 π 2 2 

NOTES:
1) For inductive load, continuous gate pulse is required (i.e. duration of pulse= π −α )
for the converter to work properly.
; θ = θ = tan −1 
α θ ≤α ≤π ωL 
2) Control range of the delay angle :  = load angle
 R 
3) If α ≤ θ and continuous gate pulse, load current would not change with α , T1 turn-
on at ωt = θ and T2 turn-on at ωt = π + θ . Hence, continuous and sinusoidal
current will be obtained and the output voltage follows the input.

α
 ωL 
θ = tan −1   = load angle
 R 
14.3 INTEGRAL CYCLE

The thyristor switch connects the AC supply to load for a time tn. The on-time N usually
consists of an integral number of cycles. The thyristors are turned on at the zero-
crossing of AC input voltage.
Analysis
(i) Harmonic and sub harmonic properties
a0 = 0
N
2π
T
1
an =
π ∫ 0
Vˆs sin(ω t ) cos( nω t ) d (ωt )

VˆsT N
an = 2 2
(1 − cos(2nπ ))
π (T − n ) T

N
2π
T
1
bn =
π ∫
0
Vˆs sin(ωt )sin(nωt )d (ωt )

VˆsT N
bn = 2 2
( − sin(2nπ ))
π (T − n ) T

The magnitude of nth harmonic is found to be

2 2
c n = a n + bn
 2Vˆs T N
cn = 2 2
(sin( nπ ))
π (T − n ) T

The rms load voltage is given as

N 2π ˆ 2
Vrms _ load = ∫
2πT 0
(
Vs sin ωt dωt )
Vˆs N
Vrms _ load =
2 T

The average power is given as

Vˆs 2 N
Power =
2R T

Vˆs N
Vload _ rms 2 T N
Power_Factor = = =
The power factor is  Vˆs  Vˆ T
s
 
 2  2

Comparison between integral-cycle and phase angle controlled

PHASE ANGLE INTEGRAL CYCLE

1. Could be used to control light 1. Unsuitable for light control i.e. light
dimmer due to present of sub-
harmonic components
2. Could be used for motor control. 2. Unsuitable for motor control. Motor
will cogging
3. Could be used to control heater. 3. Suitable for loads with a large
thermal time period such as electric
furnace
4. Generate radio frequency 4. No radio frequency interference
 interference
5. Effect of voltage dip (ripple voltage) 6. Significant effect due to voltage dips
is less. Voltage dip is due to initial because of sub-harmonic
large starting current component.

14.4 Cycloconverter

Applicable for low speed and very large horsepower synchronous and induction motors.
The firing angles in each phase are cyclically controlled to yield a low-frequency
sinusoidal output. The output voltage is derived directly from the line-frequency input
without an intermediate dc link.