d

1. AC-AC Converters
I. Voltage Regulators
II. Cycloconverters.
1

2. Introduction
• A power electronic ac/ac converter, in generic form,
accepts electric power from one system and converts it
for delivery to another ac system with waveforms of
different amplitude, frequency, and phase.
• The ac/ac converters employed to vary the rms voltage
across the load at constant frequency are known as ac
voltage controllers or ac regulators.
• The ac/ac power converters in which ac power at one
frequency is directly converted to ac power at another
frequency are known as cycloconverters,
2

3. Basic Power circuit of single phase
AC/AC Converter.
3

4. Single-Phase AC/AC Voltage
Controller with R-load
Waveforms for single-phase ac full-wave
voltage controller with R-load.
4

5. Cnd…
• For a full-wave, symmetrical phase control, the SCRs T1
and T2 in the above Fig. are gated at α and π +β,
respectively, from the zero crossing of the input voltage
and
• By varying α, the power flow to the load is controlled
through voltage control in alternate half-cycles.
• As long as one SCR is carrying current, the other SCR
remains reverse-biased by the voltage drop across the
conducting SCR.
• The principle of operation in each half-cycle is similar to
that of the controlled half-wave rectifier (L3) and one can
use the same approach for analysis of the circuit.
5

6. Analysis with R-Load
R
V
I rms
rms
NB: Vo can be varied from Vs to 0 by varying α from 0 to π.
R
V
I
V
V
V
rms
o
rms
s
m
rms
o
,
,
2
)
2
sin(
1
2
)
2
sin(
1
2















6

7. Example-1
A Triac controller provides variable power
to a 200Ω resistive load from a 240V,
50Hz supply.
Determine the values of rms load current,
power and converter power factor for firing
angle delays of
(a) α = 30°
(b) α = 150°
7

8. Cnd..
2
,
sin
)
cos
1
(
2
)
(
)
sin(
2
1
.
2
)
2
(sin
1
2
/
2
)
2
(sin
1
2
)
/
(
/
*
,
,
,
,
,
,
,
2
,
,
,
,
,
0
rms
o
rms
SCR
m
m
avg
SCR
m
rms
s
rms
o
rms
o
rms
s
rms
o
rms
s
rms
s
rms
o
rms
I
I
is
SCR
each
in
current
rms
the
current
line
the
of
half
one
carries
SCR
each
ce
R
V
t
d
R
t
V
I
is
current
SCR
average
the
symmetry
of
because
zero
is
current
source
average
the
Vm
V
V
V
R
V
V
R
V
I
V
p
s
p
pf
load
the
of
Factor
power
The
R
V
I
is
source
the
and
load
the
in
current
rms
















 














8

9. Example-2
• The single-phase ac voltage controller below
has a 120—v rms 60Hz source. The load
resistance is determine
a) The delay angle required to deliver 500w to the
load,
b) the rms source current
c) the rms and average current in the SCRs
d) the power factor
9

10. Operation with RL Load.
• Due to the inductance, the
current carried by the SCR T1
may not fall to zero at ωt = π
• when the input voltage goes
negative and may continue
until ωt = β, the extinction
angle, As shown.
• The conduction angle of the SCR
depends on the firing delay angle α
and the load impedance angle
• θ = β - α
10

11.  
.
)
sin(
)
sin(
/
0
0
)
sin(
)
sin(
)
sin(
)
sin(
)
sin(
)
sin(
0
/
.
.
0
tan
)
/
(
tan
]
)
(
[
)
sin(
sin
Re
(
/
)
(
0
/
/
1
2
/
1
2
2
)
/
(
0
1
equation
above
the
from
y
numericall
obtained
be
can
angle
Extinction
e
t
or
t
at
again
i
diagram
wave
above
the
from
seen
is
It
otherwise
t
for
e
t
Z
V
e
t
Z
V
i
as
i
gives
equation
above
the
in
value
this
of
oin
substituti
e
Z
V
A
Ae
Z
V
Therfore
t
at
e
i
t
at
i
which
to
according
condition
initial
the
from
obtained
be
can
A
t
cons
R
L
L
R
Z
Ae
t
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V
i
form
the
of
is
equation
this
of
solution
the
t
dt
di
L
Ri
t
v
V
give
conducts
T
when
load
inductive
sitive
series
for
KVL
The
L
R
o
t
m
t
L
R
m
o
L
R
m
L
R
m
o
t
L
R
m
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o
m
s

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
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
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



 





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


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
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
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

11

12. Example-3
For the single-phase voltage controller, the source is
120v rms at 60Hz, and the load is a serious
combination with R=20Ω and L=50mH. The delay
angle α is 90°.
Determine
a) An expression for load current for the first half-period.
b) the rms load current
c) the rms SCR current,
d) The average SCR current
e) The power delivered to the load
f) The power factor
12

13. TRIAC BURST-FIRING
CONTROLLER
• Burst-firing is also known as 'on-off control', 'integral-
cycle control' and 'cycle syncopation'.
• The method allows a number of complete supply voltage
cycles (or half-cycles) through to the load and then
blanks out other cycles.
• The power switch is turned on at mains voltage zero.
• The gate pulse burst firing circuit in its simplest form
consists of a zero voltage detector, a pulse width
modulator and logic circuitry.
• The basic Triac burst-firing controller circuit is shown
below.
13

14. The basic Triac burst-firing
controller circuit
14

15. Analysis
• Analysis of the basic
controller performance is
simple, since switching is
at voltage zero and rms
values are readily
obtained, without the
need for numerical
methods.
• Root-mean-square load
voltage and current:
)
(
2
)
2
5
.
0
(
2
]
2
/
)
2
(sin
[
5
.
0
2
)
2
cos(
1
(
5
.
0
2
sin
)
2
/
1
(
2
2
0
2
0
2
2
0
2
2
2
0
2
m
n
n
s
m
m
n
n
m
n
m
n
m
rms
t
t
t
V
m
n
V
n
m
V
t
t
m
V
t
d
t
m
V
t
td
V
m
V



























15

16. Example-4
• A burst-firing a.c. to a.c. power controller
is connected between a 250V, 50Hz
supply and a 100Ω resistive load.
• Determine the values of load voltage,
current and power when
(a) tn = 40 ms and tf = 120 ms,
(b) tn = 120 ms and tf = 40 ms.
• Sketch load voltage waveforms for each
case.
16

17. Solution for Example-4
17

18. Cycloconverters (Introduction)
• Converts input power at one frequency to
output power at a different frequency with
one-stage conversion.
• It is thus a one stage frequency changer.
• Basically they are two types
- step-down, fo < fs
- step-up, fo > fs .
Where fo - output frequency
fs - input frequency.
18
18

19. Application
• Speed control of high-power ac drives
• Induction heating.
• Static VAR generation.
• For converting variable-speed alternator
voltage to constant frequency output
voltage for use as power supply in aircraft
or shipboards.
19
19

20. 20
Principle of operation
Principles of CC ( Frequency changer)
Fixed
frequency
supply
P Group
N Group
Load
V
i
20

21. 21
1) mid-point cycloconverters
21

22. cnd
22
22

23. cnd
• As it can be seen from above slides(#21),
terminal ‘a’ is positive with respect to terminal ‘b’
• Thus in this positive half cycle, both SCRs P1 &
N2 are forward biased ωt = 0 to ωt=π.
• So by sequential switching of the positive and
negative devices, a waveforms like that of step-
up can be generated (slide#22) generated.
• In this particular example fo = 6fs is generated.
NB! i) Thyristors require forced commutation for
their switching off.
ii)1-phase seldom used in practice. -23
23

24. 2) Bridge type Cycloconverters
24
Input and output waveforms of a 50 to 162 3-Hz cyclo-converter with RL load.

25. Operation with R-Load
• The above figure shows the input and output voltage
waveforms with a pure R-load for a 50 to 16 2/3 Hz
cycloconverter.
• The P- and N-converters operate for all alternate To/2
periods.
• The output frequency (1/To) can be varied by varying To
and the voltage magnitude by varying the firing angle α
of the SCRs.
• As shown in the figure, three cycles of the ac input wave
are combined to produce one cycle of the output
frequency to reduce the supply frequency to one-third
across the load.
25

26. Bridge type Cyclo-converters with
Integral voltage control
26
NB:If αp is the delay angle for the positive cycle then αn is equal to π- αp.