3. Macam Konverter AC-AC
Direct AC-AC Voltage Converters
Cycloconverters
Indirect AC-AC Converters:
- DC Link
- AC Link
Matrix Converters
4. Direct AC-AC Voltage Converters
Komponen utama:
- TRIAC
T1
- Antiparallel connected is io
two thyristors
+
Teknik kendali:
- Integral cycle vs T2 R vo
- Phase control
5. Integral cycle control
Almost no transient or
EMI problems
Can only be used for
applications with large vs
“inertia”.
The control ratio is vo
limited.
6. Phase-Controlled AC-AC Voltage
Converters : Resistive Load
Input voltage is
sinusoidal. is
T1
io
+
Output voltage and vs T2 R vo
current waveforms
are the same. vs
is
Input and output
current waveforms
α
are the same. vo io
Power factor is less
than unity.
7. Phase-Controlled AC-AC Voltage
Converters : Resistive Load
Tegangan output :
⎡ π − α + 1 sin 2α ⎤
1/ 2
2Vs2 π
Vo = ∫α sin θdθ = Vs ⎢
2 2
π π ⎥
⎣ ⎦
Vo
I s = Io =
R
VA sumber :
VAs = Vs I s
Daya Beban :
P = Vo I o = Vo2 / R
Faktor Daya :
⎡ π − α + 1 sin 2α ⎤
1/ 2
PF = P / VAs = ⎢ 2
⎥
⎣ π ⎦
8. Phase-Controlled AC-AC Voltage
Converters : Inductive Load
Output voltage and
T1
current waveforms +
is io
are different. vs T2 R
vo
L
Output and input
current waveforms vs
is
are the same.
Source power factor α vo
io
is less than the load
power factor.
9. Three-Phase AC-AC Voltage Converters :
Resistive Loads
va vb vc
va
T1 ia
iA vA
N + A n
R
T2
vb T3
iB vB
+ B
R vab
T4
vc
T5
iC vC
+ C
R
T6
14. DC Link AC-AC Converters
AC source AC load
Voltage - Type Voltage - Type
Rectifier Inverter
AC source AC load
Current - Type Current - Type
Rectifier Inverter
15. AC Link AC-AC Converter
AC source AC load
Voltage - Type Voltage - Type
Rectifier Inverter
AC source AC load
Current - Type Current - Type
Rectifier Inverter
17. Matrix Converters
S1
is io euo S1
S2 S2 u
u
+
S3
es vo Load
evo S4
S3 v S5 v
0 n
S6
S4
ewo
S7
w S8 w
S9
18. PWM Technique for Single-Phase
Matrix Converter
⎛ 2T ⎞e e r
v uo = ⎜ ON 1 − 1⎟ s = s v u
⎜ T ⎟ 2
⎝ s ⎠ 2
⎛ 2T ⎞e e r
+ S2 v vo = ⎜ ON 2 − 1⎟ s = s v v
⎜ T ⎟ 2
S1 ⎝ s ⎠ 2
vo
es u v
e
2
(r r
) e r
v uv = s v u − v v = s v uv
2
Let e s = V s sin ω s t and v uv = Vo sin ω o t
Load
Then
S3 S4 2Vo sin ω o t
v uv =
r
V s sin ω s t
with ω o < ω s