06/11/2015
by ZEN-15 1
Presented by:Moh. Zaenal Efendi (ZEN)
[email protected] JOSS
Industrial Electrical EngineeringElectronics Engineering Polytechnic of Surabaya
POWER ELECTRONICS
1
PART-1
INTRODUCTION TO POWER ELECTRONICS SINGLE PHASE UNCONTROLLED RECTIFIER THREE PHASE UNCONTROLLED RECTIFIER SINGLE PHASE CONTROLLED RECTIFIER THREE PHASE CONTROLLED RECTIFIER SINGLE PHASE AC-AC VOLTAGE
CONTROLLER
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Generally: Electronics: Solid State Electronics Devices and
their Driving Circuits. Power: Static and Dynamic Requirements for
Generation, Conversion and Transmission of Power.
Control: The Steady State and Dynamic Stability of the Closed Loop system.
POWER ELECTRONICS may be defined as the application of Solid State Electronics for the Control and conversion of Power.
3
Definition of Power Electronics
DEFINITION:To convert, or to process and control the flow of electric power by supplying voltages and currents in a form that is optimally suited for user loads.
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Detailed Block Diagram of Power Electronics System
Filter&
RectifyPE Circuit Load
Filter&
Rectify
ControlCircuit
Mechanical VariableFeedback
Electrical VariableFeedback
Input
Form ofelectricalenergy
ElectricalMechanical
Pre-stage Power proc. stage Post stage Output
Form of elec. ormechan. energy
Switc
hD
rives
Process feedbacksignals and decide
on control
Could generateundesirablewaveforms
Mostly ac linevoltage (singleor three phase)
Mostlyunregulate
d dcvoltage
Interface betweencontrol and power
circuits 5
Applications Static applications involves non-rotating or moving mechanical
components.Examples:
DC Power supply Un-interruptible power supply, Power generation
and transmission (HVDC), Electroplating, Welding, Heating, Cooling, Electronic ballast
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Applications
Drive applicationscontains moving or rotating components such
as motors.Examples:
Electric trains, Electric vehicles, Air-conditioning System, Pumps, Compressor,
Conveyer Belt (Factory automation).
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Other Applications
Heating, coolingElectroplating, Welding
Photovoltaic Systems.eV (fuel cell, Solar)
Wind-electric systems.8
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Charge Controller 12V Battery
12VDC/220VAC
Power Inverter
12V/100W
Solar Panel
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Power Electronics Converters
AC to DC: RECTIFIER
DC to DC: CHOPPER
DC to AC: INVERTER
AC to AC: CYCLOCONVERTER
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Types Of Power Conversion
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14Superconducting Magnet Energy Storage System
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FUEL CELL DRIVER
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Power Electronics
AC TO DC CONVERSION (RECTIFIER)
Re-edited by : ZEN (ELIN-PENS)20
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Overview Single-phase, half wave
rectifier Uncontrolled R load R-L load R-C load Free wheeling diode Controlled
Single-phase, full wave rectifier R load R-L load, Continuous and
discontinuous current mode
controlled
Three-phase rectifier uncontrolled controlled
21
Rectifiers
DEFINITION: Converting AC (from mains or other AC source) to DC power by using power diodes or by controlling the firing angles of thyristors/controllable switches.
Basic block diagram
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Rectifiers
Input can be single or multi-phase (e.g. 3-phase).
Output can be made fixed or variable
Applications: DC welder, DC motor drive, Battery charger,
DC power supply, HVDC23
Root-Mean-Squares (RMS)
2
0
.21 td
2(.)
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Root Mean Squares of f
2)( fStep 1:
2
0
2)(21 tdfStep 2:
2
0
2)(21 tdfStep 3:
25
Concept of RMS
tv
v2 Average of v2
Square root of the average of v2 Average
of v=026
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Performance parameters
27
)()()( dcodcodco IVP
)()()( rmsormsoaco IVP
)(
)(
aco
dco
PP
)(
)(
dco
rmso
VV
FF
11
;
2
2
)(
)(
2)(
2)(
)(
FFVV
RF
VVVVVRF
dco
rmso
dcormsoacdco
acRipple Factor :
Form Factor :
Effiency:
DC Output Power:
AC Output Power:
Performance parameters
28
Power Factor :
Crest Factor:
Transformer Ratio:
Transformer Utilization Factor (TUF):
sekunders
primers
VV
a(
)(
)()(
2)(
)()(
2)(
)()(
)(
//
rmsormss
dco
rmsormss
dco
rmssrmss
dco
VVV
RVVRV
IVP
TUF
)(
)(
rmss
peaks
II
CF
)(
)(
)()(
2)(
)()(
2)(
///
rmss
rmso
rmsormss
rmso
rmssrmss
rmso
VV
RVVRV
IVRV
SPpf
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Single-Phase Uncontrolled Half-Wave Rectifier
Resistor Load
Considering the diode is ideal, the voltage at R-load during forward biased is the positive cycle of voltage source, while for negative biased, the voltage is zero.
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Resistor Load
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Output Voltage
mm
o
mDCaveo
ss
VVV
tdtVVVV
voltageoutputDCtVtvSourceVoltageGiven
318.0
)sin(21
,""),sin()(,
0
VV
22V
V
1)(1)((2V
V
0)(-Cos-Cos2V
V
tCos2V
V
dt0dttSinV21V
v(t)dtT1VV
m
m
m
m
m
m
o(dc)
o(dc)
o(dc)
o(dc)
0o(dc)
2
0o(dc)
T
0o(dc)avg
31
RMS OUTPUT VOLTAGE
2
4
0sin2102sin
21
4
2sin21
4
)(2cos121
2
)(sin(21
)(
2
)(
)(
0
2
)(
0
2
)(
0
2)(
mrmso
mrmso
mrmso
mrmso
mrmso
mrmso
VV
VV
VV
ttVV
tdVV
tdtVV
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Half-wave with R-C load In some applications in which a constant output
is desirable, a series inductor is replaced by a parallel capacitor.
The purpose of capacitor is to reduce the variation in the output voltage, making it more like dc.
The resistance may represent an external load, while the capacitor is a filterof rectifier circuit.
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Half-wave with R-C load Assume the capacitor is
uncharged, and as source positively increased, diode is forward biased
Capacitor is charged to Vm as input voltage reaches its positive peak at t = /2.
As diode is on, the output voltage is the same as source voltage, and capacitor charges.
As source decreases after t = /2, the capacitor discharges into load resistor. As diode is reversed biased, the load is isolated from source, and the output voltage (capacitive charge) decaying exponentially with time constant RC.
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Half-wave with R-C load
The effectiveness of capacitor filter is determined by the variation in output voltage, or expressed as maximum and minimum output voltage, which is peak-to-peak ripple voltage.
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Half-wave with R-C load (Ripple Voltage) The ripple can be approximated as:
fRCV
RCVV mmo
2
The ripple factor
The average output voltage:
2)(
omdco
VVV
)(;22:; dcodcoac
dc
ac VVVVwhereVVRF
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Assuming ideal diodes
offdiodeseVonetV
tvRCt
m
m
, sin
onpair diode ,|sin|)(
)/()(0
the angle where the diodes become reverse biased, which is the same as for the half-wave rectifierand is
)RC(Tan)RC(Tan 11 2t
)2sin( sin )/()( mRC
m VeV0sin)(sin )/()2( RCe
= solved numerically for
Peak-to-peak variation (ripple)
)sin1(|)2sin(| mmmo VVVV
In practical circuits where RC
2 , 2
minimal output voltage occurs at 2t
)/(2)/()222(
0 )2(RC
m
RC
m eVeVv
fRCV
RCV
RCV
eVeVVV
mm
m
RCm
RCmmo
2
211
1 )/()/(
f
xxxe
TheoryTaylor
x
2
...321
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Battery Charger To supply a dc source
from an ac source
The diode will remain off as long as the voltage of ac source is less than dc voltage.
Diode starts to conduct at t=. Given by,
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For Vs>E, diode D conducts. The angle whwn the diode starts conducting can be found from condition:
Vm sin = E
and = arcsin(E/Vm)
Diode will be tirned off when Vs
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The average charging current:
EEVR
I
tdR
EtVI
mdc
mdc
2cos22
1
)(sin21
Eficiency:
RIPIEPPP
PrmsRdcdc
Rdc
dc
2;;
The rms battery current:
2
22
2
22
2
cos42sin2
222
1
)(sin21
rmsrms
mmm
rms
mrms
II
EVVEVR
I
tdR
EtVI
41
FULL WAVE RECTIFIERThe objective of full wave rectifier is to
produce a voltage or current which is purely dc or has some specified dc component.
While the purpose of full wave rectifier is basically the same as that of half-wave rectifier, full wave rectifier have some fundamental advantages.
The output of the full wave rectifier has inherently less ripple than half wave rectifier.
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FULL WAVE RECTIFIER Can be asBridge rectifierCenter-tapped transformer
Center-tapped rectifier requires center-tap transformer. Bridge does not.
Center tap requires only two diodes, compared to four for bridge. Hence, per half-cycle only one diode volt-drop is experienced. Conduction losses is half of bridge.
Bridge Rectifier
Center-tapped Transformer Rectifier
43
44
RV
RV
I
VtdtVV
mdcodco
mmdco
2
2)()(sin1
)()(
0)(
The dc component of the output voltage is the average value, and load current is simply the resistor voltage divided by resistance.
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Bridge waveforms
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Center-tapped waveforms
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Full-wave with R-C load (Ripple Voltage)
fRCV
RCVV mmo 2
Ripple can be approximated as:
2)(
omdco
VVV
the average output voltage:
)(;22:; dcodcoac
dc
ac VVVVwhereVVRF
The ripple factor:
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Fullwave Rectifier Battery Charger
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Key waveforms
49
The average charging current:
EEVR
I
EEVR
I
tdR
EtVI
mdc
mdc
mdc
2cos21
2cos22
2
)(sin22
Eficiency:
RIPIEPPP
PrmsRdcdc
Rdc
dc
2;;
The rms battery current:
2
22
2
22
22
2
22
2
cos42sin2
22
1
cos42sin2
222
2
)(sin22
rmsrms
mmm
rms
mmm
rms
mrms
II
EVVEVR
I
EVVEVR
I
tdR
EtVI
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THREE PHASE UNCONTROLLED RECTIFIER
Edited by : M. Zaenal Efendi
51
THREE PHASE UNCONTROLLED RECTIFIER
Three phase rectifiers are commonly used in industry to produce a dc voltage and current for large loads high power levels output power in excess of 4 kW for a 220/380 AC system. They have several advantages to offer compared with the single-phase diode rectifiers.
These include: Lower output voltage ripple No triple harmonics in a three-wire configuration As already mentioned higher power handling capability
for a given supply-side voltage and current.
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Review on three phase voltage and current
Star connection
phLline
phNLLLline
IIIVVVV
33
53
Review on three phase voltage and current
Delta connection
phLline
phNLLLline
III
VVVV
3
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Delta-Star Transformer
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Delta-Delta Transformer
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POWER IN THREE PHASE SYSTEM
ACTIVE/TRUE POWER in LOAD
RV
P
RVP
CosIVP
CosIVP
ph
NL
LLL
LNL
2
2
3
3
3
3
57
POWER IN THREE PHASE SYSTEM
REACTIVE POWER in LOAD
XV
Q
XVQ
SinIVQ
SinIVQ
ph
NL
LLL
LNL
2
2
3
33
3
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POWER IN THREE PHASE SYSTEM
APPARENT POWER in LOAD
ZV
S
ZVS
IVS
IVS
ph
NL
LLL
LNL
2
2
3
33
3
59
3 Phase Half-wave Rectifier
This rectifier may be considered as 3 single phase halfwave rectifiers and can be considered as a half-wave type. The k-th diode will conduct during the period when the voltage of k-th phase is higher than other phases. The conduction period of each diode is 2/3.
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Input and output voltage of 3 phase Half-wave Rectifier
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The average output voltage of 3 phase halfwave rectifier
3
3sin
3sin2
3/2
3sin
3sin
3/2
sin3/2
cos3/2
)()(
)()(
)()(
3
3
)()(
3
3
)()(
NLmdco
NLmdco
NLmdco
NLmdco
NLmdco
VV
VV
VV
tV
V
tdtV
V
63
))(()(
)(max)()(
)max(,)(
)(
)(
)()(
17.1
8274.02
33
3/23
3/3
sin
NLrmssdco
NLsdco
NLsdco
NLm
dco
NLmdco
VVVV
VV
VV
VV
The average output voltage of 3 phase halfwave rectifier
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))(()(
)(max)()(
)()(
3/
0
22)(
189.1841.0
32sin
21
323
)(cos3/2
2
NLrmssrmso
NLsrmso
NLmrmso
mrmso
VVVV
VV
tdtVV
The rms output voltage of 3 phase halfwave rectifier
65
The rms input current of 3 phase halfwave rectifier
RV
RV
RVI
I
NLrmss
NLrmss
rmsormsormss
))((
))((
)()()(
687.03
189.133
The rms output current of 3 phase halfwave rectifier
RV
I rmsormso)(
)(
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Performance parameters
67
)()()( dcodcodco IVP
)()()( rmsormsoaco IVP
)(
)(
aco
dco
PP
)(
)(
dco
rmso
VV
FF
Ripple Factor :
Form Factor :
Effiency:
DC Output Power:
AC Output Power:
11
;
2
2
)(
)(
2)(
2)(
)(
FFVV
RF
VVVVVRF
dco
rmso
dcormsoacdco
ac
Power Factor
)()(,
2)(
)(
3/
rmssNLrmss
rmso
aco
IVRV
PF
SP
PF
where:
2)max(,
)(,NLs
NLrmss
VV
RV
RV
RVI
I NLrmssNLrmssrmsormsormss)(,)(,)()(
)(
687.03
189.133
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Transformer Utilization Factor
;3
/
)()(,
2)(
)(
rmssNLrmss
dco
dco
IVRV
VAP
TUF
where:
2)max(,
)(,NLs
NLrmss
VV
RV
RV
RVI
I NLrmssNLrmssrmsormsormss)(,)(,)()(
)(
687.03
189.133
69
Example
The 3 phase halfwave rectifier is operated from 380 V 50 Hz supply at secondary side and the load resistance is R=20 . If the source inductance is negligible, Determine (a) Rectification efficiency,(b) Form factor(c) Ripple factor(d) Power Factor(e) TUF
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VVVVV NLsNLrmsS 1.3112*220,22063.2193380
)(max,,)(,
max,max,
)( 827.0233
ss
dco VV
V R
VR
VI ssdc
max,max, 827.0233
max,)( 8407.0 srmso VV RV
I srmsomax,
)(
8407.0
%767.96)()(
)()(
)(
)( rmsormso
dcodco
aco
dco
IVIV
PP
%657.101)(
)( dco
rmso
VV
FF
%28.1811 22)(
2)(
)(
2)(
2)(
FFVV
VVV
VVRF
dco
rmso
dco
dcormso
dc
ac
Solution :
71
3-phase Full-wave rectifiers3-phase bridge rectifier is very common in high-power applications. This is a full-wave rectifier. It can operate with or without a transformer and gives 6-pulse ripple on the output voltage.
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Two series diode are always conducting (one feeds into the circuit, the other forms the return path). The diodes are numbered in order of conduction and each diode conducts for 120o. The conduction sequence for diodes is D1D2; D2D3; D3D4; D4D5; D5D6 and D6D1. The pair of diodes which are connected between that pair of supply lines having the highest amount of line-line voltage will conduct.
phlineline VV 3
There are 6 combinations of VL-L. Considering one period of the source to be 360o, a transition of the highest line-line voltage must take place 360o/6=60o.Because of six transition that occur for each period of the source voltage, the circuit is called A SIX-PULSE RECTIFIER
73
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Input and output voltage of 3 phase Full-wave Rectifier
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Considers only one of the six segments. Obtain its average over 60 degres
LLm
LLm
LLm
LLmdco
V
V
tV
tdtVV
,
,
3/23/
,
3/2
3/,)(
955.0
3
cos3
sin3/
1
3 phase full-wave average voltage
77
LLm
LLm
NLm
NLm
NLm
wavehalfdcdco
V
V
orV
V
tdtV
VV
,
,
,
,
6/
0,
)()(
955.0
3654.1
33
cos6/2
2
2
3 phase full-wave average voltage
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LLmLLm
NLmNLm
NLmrmso
VV
VV
tdtVV
,,
,,
6/
0
22,)(
95575.04
3923
3
6554.14
3923
cos6/2
2
RV
I rmsormso)(
)(
3 phase full-wave rms output voltage
)(
)(
2)(
)(
2)(
)( ;
aco
dco
rmsoaco
dcodco
PP
R
VP
R
VP
3 phase full-wave rms output currentEfficiency of 3 phase full-wave rectifier
79
Power Factor
rmssLLrmss
rmso
sNLrmss
rmso
aco
IVRV
orIV
RVS
PPF
,)(,
2)(
)(,
2)(
)(
3/
;3
/
RV
I
orRV
Iwhere
II
II
tdtII
LLmm
NLmm
mrmss
mrmss
mrmss
)(
)(
)(
)(
26/
0
2)(
;3
:
;7804.0
22sin
21
62
)(cos28
where:
2
2
)()(,
)()(,
LLmLLrmss
NLmNLrmss
VV
and
VV
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Transformer Utilization Factor
rmssLLrmss
dco
sNLrmss
dco
dco
IVRV
orIV
RVVA
PTUF
,)(,
2)(
)(,
2)(
)(
3/
;3
/
RV
I
orRV
Iwhere
II
II
tdtII
LLmm
NLmm
mrmss
mrmss
mrmss
)(
)(
)(
)(
26/
0
2)(
;3
:
;7804.0
22sin
21
62
)(cos28
where:
2
2
)()(,
)()(,
LLmLLrmss
NLmNLrmss
VV
and
VV
81
Example
The 3 phase full-wave rectifier is operated from 380 V; 50 Hz supply and the load resistance is R=20 . Determine (a) The efficiency, (b) Form factor (c) Ripple factor(d) Power factor(e) TUF
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VVV LLmdco 14.513955.0 ,)(
AR
VI dcodco 66.25
)()(
VVV LLmrmso 544.51395575.0 ,)(
AR
VI rmsormso 68.25
)()(
%83.99)()(
)()(
)(
)( rmsormso
dcodco
aco
dco
IVIV
PP
Solution :
%08.100)(
)( dco
rmso
VV
FF
%411 22)(
2)(
)(
2)(
2)(
FFVV
VVV
VVRF
dco
rmso
dco
dcormso
dc
ac
83
Homework
1. A three phase halfwave rectifier is supplied by 380V-rms, 50Hz. The primary and secondary of input transformer are connected delta-wye and has turns ratio of n=3:1. The load is a 100 ohm. Determine:a. The average load current; b. The rms load current; c. The rms source current; d. The power factor; e. Efficiency; f. Form factor; g. Ripple factor; h. TUF
2. A three phase fullwave rectifier is supplied by 380V-rms, 50Hz. The primary and secondary of input transformer are connected delta-delta and has turns ratio of n=3:1. The load is a 100 ohm. Determine :a. The average load current; b. The rms load current; c. The rms source current; d. The power factor; e. Efficiency; f. Form factor; g. Ripple factor; h. TUF
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Power Electronics
SINGLE PHASE CONTROLLED AC TO DC RECTIFIER
Presented by : M. Zaenal Efendi
85
Type of input: Fixed voltage, fixed frequency ac power supply.
Type of output: Variable dc output voltage
Type of commutation: Natural / AC line commutation
LineCommutated
Converter
+
-
DC OutputV0(dc)
ACInput
Voltage
86
Block Diagram
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Applications of Phase Controlled Rectifiers
DC motor control in steel mills, paper and textile mills employing dc motor drives.
AC fed traction system using dc traction motor.
Electro-chemical and electro-metallurgical processes.
Magnet power supplies.
Portable hand tool drives.
87
Classification of Phase Controlled Rectifiers Single Phase Controlled Rectifiers. Three Phase Controlled Rectifiers.
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Single Phase Half-Wave ThyristorConverter with a Resistive Load
89
The Controlled Half-wave Rectifier
Previously discussed are classified as uncontrolled rectifiers.
Once the source and load parameters are established, the dc level of the output and power transferred to the load are fixed quantities.
A way to control the output is to use SCR instead of diode. Two condition must be met before SCR can conduct: The SCR must be forward biased (VSCR>0) Current must be applied to the gate of SCR
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Controlled, Half-wave R load
A gate signal is applied at t = , where is the delay/firing angle.
91
Average load Voltage (D.C. Voltage)
DC Voltage
max
max
Maximum average (dc) o/pvoltage is obtained when 0 and the maximum dc output voltage
1 cos0 ; cos 0 12
mdmdc
mdmdc
VV V
VV V
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Control Characteristic
VO(dc)
Trigger angle in degrees0 60 120 180
Vdm
0.2 Vdm
0.6Vdm
93
22
0
12
2 2
The RMS output voltage is given by
1 .2
Output voltage sin ; for to
1 sin .2
OO RMS
O m
mO RMS
V v d t
v V t t
V V t d t
Root Mean Square Values
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Root Mean Square Values
RMS Current
For a resistive load VRMS = IRMS R
Note: for = 0 VRMS = 0.5 Vm
2)2sin(1
2
)()]sin([21,
resistor,by absorbedpower Average
0
2,
22
m
mrmso
rms
V
tdtVVwhere
RVRIP rms
95
Performance Parameters
22
2)cos1(
4
RVP mdc
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Example 1 Design a circuit to produce an average voltage of
40V across 100 load resistor from a 120Vrms 50 Hz ac source. Determine the power absorbed by the resistor, efficiency, FF, RF and the power factor.
To obtain Vo=25% of Vdm, determine firing angle, FF, RF, pF and TUF
97
Example 1 (Cont)
Solution
rad
VV
o
so
07.12.61
]cos1[2
212040
]cos1[2
In such that to achieved 40V average voltage, the delay angle must be
If an uncontrolled diode is used, the average voltage would be
That means, some reducing average resistor to the design must be made. A series resistor or inductor could be added to an uncontrolled rectifier, while controlled rectifier has advantage of not altering the load or introducing the losses
V
VV mrmso
6.752
)07.1(2sin07.112
21202
)2sin(12,
WR
VP rms 1.57100
6.75 22
63.0
1006.75)120(
1.57
pf
VVV so 54)120(2
98
06/11/2015
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99
SINGLE PHASE: FULL WAVE BRIDGE CONTROLLED RECTIFIERThere are 2 types of FW Bridge Controlled Rectifiers:
1. Fully Controlled Bridge Converter (Full Converter)
2. Half Controlled Bridge Converter (Semi-Converter)
Single Phase Full Wave Controlled Rectifier With R Load
100
06/11/2015
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101
DC Output Voltage
0
1 .
1 sin .
cos
cos cos ; cos 1
1 cos
dc OO dct
dc mO dc
mdcO dc
mdcO dc
mdcO dc
V V v d t
V V V t d t
VV V t
VV V
VV V
102
06/11/2015
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RMS Output Voltage
2sin212sin
21
2
2sin21
2
)(2cos121
)(sin(1
2
)(
2
)(
2
)(
2)(
mrmso
mrmso
mrmso
mrmso
VV
ttVV
tdVV
tdtVV
103
22sin1
42sin
221
42sin2sin
221
2sin212sin
21
21
)()(
)(
)(
rmssmrmso
mrmso
mrmso
VVV
VV
VV
Power Electronics
THREE PHASE CONTROLLED AC TO DC RECTIFIER
Presented by : M. Zaenal Efendi104
06/11/2015
by ZEN-15 53
105
3 Phase Controlled Rectifiers Operate from 3 phase ac supply voltage.
They provide higher dc output voltage.
Higher dc output power.
Higher output voltage ripple frequency.
Filtering requirements are simplified for smoothing out load voltage and load current.
Extensively used in high power variable speed industrial dc drives.
Three single phase half-wave converters can be connected together to form a three phase half-wave converter.
105
Vector Diagram of 3 Phase Supply Voltages
106
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3-pulse (Half Wave) Controlled Rectifier With Resistive Load
107
0
0
300
300
600
600
900
900
1200
1200
1500
1500
1800
1800
2100
2100
2400
2400
2700
2700
3000
3000
3300
3300
3600
3600
3900
3900
4200
4200
Vs
V0
Van
=0
=150
Vbn Vcn
t
Van Vbn Vcn
t
Waveformes
=00
=150
For R- Loads the current waveform is similar
to the voltage waveforms.
The FWD has no effect for small values of .
For >30 the FWD sets the load voltage at 0
leading to higher DC output voltage.
108
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0
0
300
300
600
600
900
900
1200
1200
1500
1500
1800
1800
2100
2100
2400
2400
2700
2700
3000
3000
3300
3300
3600
3600
3900
3900
4200
4200
V0
=300
Van Vbn Vcn
t
V0
=600
Van Vbn Vcn
t
=300
=600
Waveforms for different
109
01
0 01
02
0 02
0
306
30 180 ;
sin5 1506
150 300 ;
sin 120
O an m
O bn m
T is triggered at t
T conducts from to
v v V t
T is triggered at t
T conducts from to
v v V t
Average Voltage (DC Voltage) >30
110
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0
0
0
0
0
0
180
30
0 0
180
30
180
30
3 .2
sin ; for 30 to 180
3 sin .2
3 sin .2
dc O
O an m
dc m
mdc
V v d t
v v V t t
V V t d t
VV t d t
Average Voltage (DC Voltage)
111
112
)30cos(12
3
)30cos(180cos2
32
3
)()(
)()(
180
30
)()( cos
oNLmdco
ooNLmdco
NLmdco
VV
VV
VV t
o
o
DC Output Voltage of 3-pulse (Half Wave) Controlled Rectifier With Resistive Load
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113
RMS Output Voltage of 3-pulse (HWR) Controlled Rectifier With Resistive Load
260sin81
42453
)()(sin23
)()(
6
22)()(
oNLmrmso
NLmrmso
VV
tdtVV
114
The RMS input current of Half Wave Controlled Rectifier
RVI
I rmsormsormss
33
)()()(
So, we can calculate the power factor dan TUF of Half Wave Controlled Rectifier
)()(,
2)(
)(
3/
rmssNLrmss
rmso
aco
IVRV
PF
SP
PF
;
3/
)()(,
2)(
)(
rmssNLrmss
dco
dco
IVRV
VAP
TUF
06/11/2015
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115
Three Phase Full Converter(Full Wave Controlled Rectifier)
3 Phase Fully Controlled Full Wave Bridge Converter.
Known as a 6-pulse converter.
Used in industrial applications up to 120kW output power.
Two quadrant operation is possible.
115
Circuit Layout
116
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Circuit of Three Phase Full Converter
117
Waveforms
= 0118
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Waveforms
= 30 119
120
The thyristors are triggered at an interval of / 3.
The frequency of output ripple voltage is 6fS. T1 is triggered at t = (/6 + ), T6 is already conducting
when T1 is turned ON.
During the interval (/6 + ) to (/2 + ), T1 and T6 conduct together & the output load voltage is equal to vab = (van vbn)
Operation of the circuit
120
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T2 is triggered at t = (/2 + ), T6 turns off naturally as it is reverse biased as soon as T2 is triggered.
During the interval (/2 + ) to (5/6 + ), T1 and T2conduct together & the output load voltage vO = vac = (van vcn)
Thyristors are numbered in the order in which they are triggered.
The thyristor triggering sequence is 12, 23, 34, 45, 56, 61, 12, 23, 34,
Operation of the circuit
121
Sequence of Gate Pulses of Thyristor
122
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Table of the thyristor pair in conduction at any instant.
123
The corresponding line-to-line supply voltages are
3 sin6
3 sin2
3 sin2
RY ab an bn m
YB bc bn cn m
BR ca cn an m
v v v v V t
v v v v V t
v v v v V t
Three phase line voltages
124
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125
2
6
6 . ; 2
3 sin6
dc OO dc
O ab m
V V v d t
v v V t
The output load voltage consists of 6 voltage pulses over a period of 2 radians, Hence the average output voltage is calculated as
DC output voltage
125
The maximum dc voltage is given by
126
)cos(3
)cos(33
)()6
sin(33
)()(
)()(
2
6)()(
LLmdco
NLmdco
NLmdco
VV
or
VV
tdtVV
)()(
max)(
333 LLmNLmdco
VVV
DC Output Voltage of FULL WAVE Controlled Rectifier With Resistive Load
06/11/2015
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127
RMS Output Voltage of Full Wave Controlled Rectifier With Resistive Load
2cos4
3321
2cos4
33213
)()6
(sin33
)()(
)()(
2
6
22)()(
LLmrmso
NLmrmso
NLmrmso
VV
or
VV
tdtVV
127
128
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129
The RMS input current of Full Wave Controlled Rectifier
)()( 64
rmsormss II
So, we can calculate the power factor dan TUF of Full Wave Controlled Rectifier
rmssLLrmss
dco
sNLrmss
dco
dco
IVRV
orIV
RVVA
PTUF
,)(,
2)(
)(,
2)(
)(
3/
;3
/
rmssLLrmss
rmso
sNLrmss
rmso
aco
IVRV
orIV
RVS
PPF
,)(,
2)(
)(,
2)(
)(
3/
;3
/
130
AC-AC VOLTAGE CONTROLLER
RE-EDITED BY ZENELIN
Power Electronics
06/11/2015
by ZEN-15 66
131131
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
AC Voltage Controller (RMS Voltage Controllers)
ACVoltage
Controller
V0(RMS)
fS
Variable AC RMS O/P Voltage
ACInput
Voltagefs
Vs
fs
132132
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
Applications Lighting / Illumination control in ac power circuits. Induction heating. Industrial heating & Domestic heating. Transformer tap changing (on load transformer
tap changing). Speed control of induction motors (single phase
and poly phase ac induction motor control). AC magnet controls.
06/11/2015
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133133
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
Single Phase Full Wave AC Voltage Controller
Principle of AC Phase Control
134134
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
Single Phase Full Wave AC Voltage Controller using Triac
06/11/2015
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135135
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
136
The principle of operation for a single-phase ac voltage controller using phase control is similar to the controlled half-wave rectifier.An analysis identical to that done for the controlled half-wave rectifier can be done on a half cycle for the voltage controller. Then, by symmetry, the result can be extrapolated to describe the operation for the entire period. Some basic observations about cycloconverters are as follows:
The SCRs cannot conduct simultaneously.The load voltage is the same as the source voltage when either SCR is on. The load voltage is zero when both SCRs are off.
06/11/2015
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137137
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
Input supply voltage
sin 2 sinOutput voltage across the load resistor
sin ;for to & to 2
S m S
O L m
v V t V t
v v V tt t
Equations
138138
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
1
2
Thyristor is triggered at is triggered at
Output load currentsin sin ;
for to & to 2
O mO m
L L
T tT t
v V ti I tR R
t t
06/11/2015
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139139
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
To Derive an Expression forthe RMS Value of Output Voltage
2
2 2 2
0
The RMS value of output voltage
12
For a FW ac voltage controller, we can see that two half cycles of o/p voltage w.f. are symmetrical & o/p pulse time period is radians. We
LO RMS L RMSV V v d t
can also calculatethe RMS o/p voltage by using the expression
140140
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
2 2 2
02
2 2
0
1 sin
1 ;2
sin ;For to & to 2
mL RMS
LL RMS
L O m
V V t d t
V v d t
v v V tt t
06/11/2015
by ZEN-15 71
141141
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
2
22 2
22 2 2 2
22
Hence
1 sin sin2
1 sin . sin .2
1 cos 2 1 cos 22 2 2
L RMS
m m
m m
m
V
V t d t V t d t
V t d t V t d t
V t td t d t
142142
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
2 22
2 22
2
2
cos2 . cos2 .2 2
sin2 sin24 2 2
1 1sin2 sin2 sin4 sin24 2 2
1 12 0 sin2 0 sin24 2 2
m
m
m
m
V d t t d t d t t d t
V t tt t
V
V
06/11/2015
by ZEN-15 72
143143
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
2
2
2
sin 2sin 224 2 2
sin 2 2sin 224 2 2
sin 2 12 sin 2 .cos2 cos2 .sin 24 2 2
sin 2 0 & cos2 1;
m
m
m
V
V
V
144144
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
22
2
22
Therefore,sin 2 sin 22
4 2 2
2 sin 24
2 2 sin 24
mL RMS
m
mL RMS
VV
V
VV
06/11/2015
by ZEN-15 73
145145
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
Taking the square root, we get
2 2 sin 22
2 2 sin 22 2
1 2 2 sin 222
mL RMS
mL RMS
mL RMS
VV
VV
VV
146146
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
1 sin 222 22
1 sin 222
1 sin 22
1 sin 22
mL RMS
mL RMS
L RMS i RMS
SL RMS
VV
VV
V V
V V
06/11/2015
by ZEN-15 74
147147
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
M aximum RM S voltage will be applied to the load when 0, in that case the full sine wave appears across the load. RM S load voltage will be the same as
the RM S supply voltage .2
W hen is increased th
mV
e RM S loadvoltage decreases.
148148
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
0
0
0
1 sin 2 0022
1 022
2
mL RMS
mL RMS
mSL RMS i RMS
VV
VV
VV V V
06/11/2015
by ZEN-15 75
149149
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
Control Characteristic of Single Phase FW AC Voltage Controller With Resistive Load
1 sin 2 ;
2
Where RMS value of 2
input supply voltage
SO RMS
mS
V V
VV
150150
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
06/11/2015
by ZEN-15 76
151151
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
VO(RMS)
Trigger angle in degrees
0 60 120 180
VS
0.2 VS
0.6VS
152Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
Need for Isolation
Pulse Transformer
G1
K1G2
K2
GateTriggerPulse
Generator
152
06/11/2015
by ZEN-15 77
153Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
Problems in AC Voltage Controllers
154154
Prof. M. Madhusudhan Rao, E&C Dept., MSRIT
A single phase full wave controller has an input voltage of 120 V (RMS) and a load resistance of 6 ohm. The firing angle of thyristor is 90o. Find RMS output voltage Power output Input power factor If the output power decreases to 20%
of the maximum power, determine : firing angle and Vo
06/11/2015
by ZEN-15 78
155
0
12
12
90 , 120 V , 62
RM S Value of O utput V oltage
1 sin 22
1 sin 1801202 2
84.85 V olts
S
O S
O
O
V R
V V
V
V
156
2
2
RMS Output Current84.85 14.14 A
6Load Power
14.14 6 1200 wattsI/P Current is same as Load Current
14.14 AmpsInput Supply Volt-Amp
120 14.14 1696.8
OO
O O
O
S O
S S
VIRP I R
P
I I
V I VA
06/11/2015
by ZEN-15 79
Question
220 V and 2.2 kW heater is fed by a singlephase AC chopper which is connected 220 VAC grid.
a) Calculate the resistance of the heaterb) For =90, find load voltage and powerc) To obtain the output volatge 15 % of
maximum voltage, determine the delayangle.
06/11/2015
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159
Find the RMS and average current flowing through the heater shown in figure. The delay angle of both the SCRs is 450.
SCR2
SCR1 io+
1 kW, 220Vheater
1-220V
ac
References1. Muh. H. Rashid, Power Electronics, Circuit, Devices and
Applications, Prentice Hall, 19932. Daniel W. Hart, Introduction to Power Electronics, Prentice
Hall, 19973. William Shepherd, Power Converter Circuits, Taylor&Francis,
20044. Mohd Shawal Jadin, Hand-out of Power Electronics5. Yasser. G. Hegazy, Hand-out of Power Electronics6. M. Madhusudhan Rao, , Hand-out of Power Electronics7. Univ of Central Florida, Hand-out of Power Electronics
160