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EE432 POWER ELECTRONICS
Dr. Ali M. Eltamaly, King Saud University
Dr. Ali M. EltamalyKing Saud University
Dr. Ali M. Eltamaly, King Saud University
Chapter 1
Introduction
1.1 .Definition of Power Electronics
Power electronics refers to control and conversion of electrical power by power semiconductor devices wherein these devices operate as switches
Electronic power converter•Rectifier converting an AC voltage to a DC voltage, •Inverter converting a DC voltage to an AC voltage, •Chopper or a switch-mode power supply that converts a DC voltage to another DC voltage, and •Cycloconverter and cycloinverter converting an AC voltage to another AC voltage.
Dr. Ali M. Eltamaly, King Saud University
1.2 Rectificationuncontrolled and controlled rectifiers
DC-To-AC Conversion
•Emergency lighting systems, •AC variable speed drives, •Uninterrupted power supplies, and, •Frequency converters.
DC-to-DC Conversion
•Step-down switch-mode power supply, •Step-up chopper, •Fly-back converter, and ,•Resonant converter.
typical applications•DC drive, •Battery charger, and,•DC power supply.
Dr. Ali M. Eltamaly, King Saud University
1.5 AC-TO-AC Conversion
cycloconverter or a Matrix converter converts
Adjustable Speed Drives (ASD)
Dr. Ali M. Eltamaly, King Saud University
Diode Circuits or Uncontrolled Rectifier
Rectification: The process of converting the alternating voltages
and currents to direct currents
Dr. Ali M. Eltamaly, King Saud University
The main disadvantages of half wave rectifier are :
• High ripple factor,• Low rectification efficiency,• Low transformer utilization factor,
and,• DC saturation of transformer
secondary winding.
Dr. Ali M. Eltamaly, King Saud University
Performance Parameters
22dcrmsac VVV
acdc PP / rectification effeciency
dcrms VVFF /
11 22
222
FFVV
VVV
VV
RFdc
rms
dc
dcrms
dc
ac
form factor
ripple factor
Dr. Ali M. Eltamaly, King Saud University
121
2
21
21
2
S
S
S
SSi
II
IIITHD
121
2
21
21
2
S
S
S
SSv
VV
VVV
THD
FaactorntDisplacemeFactorDistortionII
IVIV
IVPPF
S
S
SS
SS
SS
*
coscos1
111
Dr. Ali M. Eltamaly, King Saud University
Single-phase half-wave diode rectifier with resistive load.
Dr. Ali M. Eltamaly, King Saud University
0
sin21 m
mdcV
tdtVV
2sin
21
0
22 mmrms
VtdtVV
RV
RVI mdc
dc
RV
RV
I mrmsrms 2
the load and diode currents RV
II mDS 2
Dr. Ali M. Eltamaly, King Saud University
Example 1: The rectifier shown in Fig.2.1 has a pure resistive load of R Determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) Peak inverse voltage (PIV) of diode D1.
mm
mdcVV
tdtVV ))0cos(cos(2
)sin(21
0 RV
RV
I mdcdc
2)sin(
21
0
2 mmrms
VtVV
R
VI m
rms 2
%53.40
2*
2
*
**
RVV
RVV
IVIV
PP
mm
mm
rmsrms
dcdc
ac
dc
57.12
2
m
m
dc
rmsV
V
VVFF
211.1157.11 22 FFVV
RFdc
ac
)d (It is clear from Fig2.2 that the PIV is mV
Dr. Ali M. Eltamaly, King Saud University
Half Wave Diode Rectifier With R-L Load
Fig.2.3 Half Wave Diode Rectifier With R-L Load
Dr. Ali M. Eltamaly, King Saud University
ttViRdtdiL m 0),(sin*
Divide the above equation by L we get:
ttL
ViLR
dtdi m 0),(sin*
Adtt
LVeeti mdt
LRdt
LR
sin*)(
Adtt
LVeeti mt
LRt
LR
sin*)(
t
LR
m AetLtRLwR
Vti
cossin)( 222
wLjRZ
Dr. Ali M. Eltamaly, King Saud University
2222 LwRZ
cosZR sinZL RL tan
R
wLZ
t
LR
m AettZ
Vti
cossinsincos)(
t
LR
m AetZ
Vti
sin)(
tansinsin)(t
mt
LR
m AetZ
VAetZ
Vti
00 i sinZ
VA m
tansinsin)(t
m etZ
Vti
Dr. Ali M. Eltamaly, King Saud University
tansinsin)(t
m etZ
Vti
0sinsin)( tan
eZ
Vi m
)cos1(*2
sin*2 0
mmdc
Vtdt
VV
)2sin(1(5.0*2
)sin(*21
0
2
VmdwttVV mrms
Dr. Ali M. Eltamaly, King Saud University
)cos1(*2
sin*2 0
mmdc
Vtdt
VV
)2sin(1(5.0*2
)sin(*21
0
2
VmdwttVV mrms
Dr. Ali M. Eltamaly, King Saud University
Half wave diode rectifier with free wheeling diode
Dr. Ali M. Eltamaly, King Saud University
tansinsin)(t
m etZ
Vti t0
20* tforiRtdidL
tan)(
t
eBti Bi )(
BeZ
Vi m
)sin(sin)( tan
tantansinsin)(
tm eeZ
Vti 2 t
Dr. Ali M. Eltamaly, King Saud University
Example 2 A diode circuit shown in Fig.2.3 with R=10 , L=20mH, and VS=220 2 sin314t.
(a) Determine the expression for the current though the load in the period 20 t and determine the conduction angle .
(b) If we connect free wheeling diode through the load as shown in Fig.2.5 Determine the expression for the current though the load in the period of 30 t .
Solution: (a) For the period of t0 , the expression of the load current can be obtained from (2.24) as following:
.561.010
10*20*314tantan3
11 radRL
and 628343.0tan
8084.11)10*20*314(10)( 23222 LRZ
t
tm
et
etZ
Vti
5915.1
tan
*532.0561.0sin8084.11
2220
sinsin)(
tetti 5915.1*0171.14561.0sin3479.26)(
Dr. Ali M. Eltamaly, King Saud University
The value of can be obtained from the above equation by substituting for 0)( i . Then, 5915.1*0171.14561.0sin3479.260 e
By using the numerical analysis we can get the value of . The simplest method is by using the simple iteration technique by assuming
5915.1*0171.14561.0sin3479.26 e and substitute different values for in the region 2 till we get the minimum value of then the corresponding value of is the required value. The narrow intervals mean an accurate values of . The following table shows the relation between and :
1.1 6.49518
1.12 4.872781.14 3.231861.16 1.578851.18 -0.0798081.2 -1.73761
It is clear from the above table that 18.1 rad. The current in 2 wt will be zero due to the diode will block the negative current to flow.
Dr. Ali M. Eltamaly, King Saud University
(b) In case of free-wheeling diode as shown in Fig.2.5, we have to divide the operation of this circuit into three parts. The first one when t0 (D1 “ON”, D2 “OFF”), the second case when 2 t (D1 “OFF” and D2 “ON”) and the last one when
32 t (D1 “ON”, D2 “OFF”). In the first part ( t0 ) the expression for the load current can be obtained as
In case (a). Then: wtetwti 5915.1*0171.14561.0sin3479.26)( for t0
the current at t is starting value for the current in the next part. Then Aei 1124.14*0171.14561.0sin3479.26)( 5915.1
In the second part 2 t , the expression for the load current can be obtained from (2.30) as following:
tan)(
t
eBti where AiB 1124.14)(
Then teti 5915.11124.14)( for ( 2 t ) The current at 2t is starting value for the current in the next part. Then
Ai 095103.0)2( In the last part ( 32 t ) the expression for the load current can be obtained
from (2.36):
tan2
sin22sin)(
tmm eZ
VitZ
Vti
25915.1532.0*3479.26095103.08442.6sin3479.26)( tetti
25915.11131.148442.6sin3479.26)( tetti for ( 32 t )
Dr. Ali M. Eltamaly, King Saud University
Single-Phase Full-Wave Diode Rectifier Center-Tap Diode Rectifier
Dr. Ali M. Eltamaly, King Saud University
m
mdcVtdtVV 2sin1
0
RVI m
dc 2
2
sin1
0
2 mmrms
VtdtVV
R
VI mrms 2
PIV of each diode = mV2
RVII m
DS 2
Example 3. The rectifier in Fig.2.8 has a purely resistive load of R Determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) TUF (e) Peak inverse voltage (PIV) of diode D1 and(f) Crest factor of transformer secondary current.
Dr. Ali M. Eltamaly, King Saud University
%05.81
2*
2
2*2
**
RVV
RVV
IVIV
PP
mm
mm
rmsrms
dcdc
ac
dc
11.1222
2
m
m
dc
rmsV
V
VVFF
483.0111.11 22 FFVV
RFdc
ac
The PIV is mV2
Dr. Ali M. Eltamaly, King Saud University
Center-Tap Diode Rectifier With R-L Load
Dr. Ali M. Eltamaly, King Saud University
Dr. Ali M. Eltamaly, King Saud University
)sin(* tViRtdidL m
tansinsin)(t
m etZ
Vti t0
tansin)(
t
m AetZ
Vti
i()=i(2)=i(3)=……..=Io
tansinsin)( eZ
ViI mo
tansin)(
AeZ
VIi mo
sinZ
VIA mo
Dr. Ali M. Eltamaly, King Saud University
tansinsin)(
tm
om e
ZVIt
ZVti
tantansinsin)(
t
o
tm eIetZ
Vti
tan2
tan2
sin2sin)(
t
o
tm eIewtZ
Vti 32 t
Dr. Ali M. Eltamaly, King Saud University
Single-Phase Full Bridge Diode Rectifier With Resistive Load
Dr. Ali M. Eltamaly, King Saud University
Example 4 single-phase diode bridge rectfier has a purely resistive load of R=15 ohms and, VS=300 sin 314 t and unity transformer ratio. Determine (a) The efficiency, (b) Form factor, (c) Ripple factor, (d) The peak inverse voltage, (PIV) of each diode, , and, (e) Input power factor.
VV
tdtVV mmdc 956.190
2sin1
0
A
RV
I mdc 7324.12
2
VVtdtVV mmrms 132.212
2sin1
2/1
0
2
%06.81rmsrms
dcdc
ac
dc
IVIV
PP
11.1dc
rms
VV
FF
482.011 22
222
FFVV
VVV
VVRF
dc
rms
dc
dcrms
dc
ac The PIV=300V
Input power factor = 1cosRe
SS
SS
IVIV
PowerApperantPoweral
Dr. Ali M. Eltamaly, King Saud University
.............,5,3,14
cos0cos2
cos2
sin*20
0
nforn
In
nI
tnn
ItdtnIb
oo
oon
)..........9sin917sin
715sin
513sin
31(sin*4)( tttttIti o
%46151
131
111
91
71
51
31))((
2222222
tITHD s
24
1o
SI
I
%34.4814
21
24
1))((2
2
2
1
o
o
S
Ss I
III
tITHD
Dr. Ali M. Eltamaly, King Saud University
Example 5 solve Example 4 if the load is 30 A pure DCFrom example 4 Vdc= 190.986 V, Vrms=212.132 V AIdc 30 and rmsI = 30 A
%90rmsrms
dcdc
ac
dcIVIV
PP 11.1
dc
rmsVVFF
482.011 22
222
FF
VV
VVV
VVRF
dc
rms
dc
dcrms
dc
ac
The PIV=Vm=300V AII o
S 01.272
30*42
41
Input Power factor= PowerApperant
PoweralRe
LagI
IIV
IV
S
S
SS
SS 9.01*30
01.27cos*cos* 11
Full Bridge Single-phase Diode Rectifier with DC Load Current
.............,5,3,14
cos0cos2
cos2
sin*20
0
nforn
In
nI
tnn
ItdtnIb
oo
oon
)..........9sin917sin
715sin
513sin
31(sin*4)( tttttIti o
%46151
131
111
91
71
51
31))((
2222222
tITHD s
24
1o
SI
I
%34.4814
21
24
1))((2
2
2
1
o
o
S
Ss I
III
tITHD
Example 5 solve Example 4 if the load is 30 A pure DCFrom example 4 Vdc= 190.986 V, Vrms=212.132 V AIdc 30 and rmsI = 30 A
%90rmsrms
dcdc
ac
dcIVIV
PP 11.1
dc
rmsVVFF
482.011 22
222
FF
VV
VVV
VVRF
dc
rms
dc
dcrms
dc
ac
The PIV=Vm=300V AII o
S 01.272
30*42
41
Input Power factor= PowerApperant
PoweralRe
LagI
IIV
IV
S
S
SS
SS 9.01*30
01.27cos*cos* 11
Effect Of LS On Current Commutation Of Single-Phase Diode Bridge Rectifier .
Dr. Ali M. Eltamaly, King Saud University
0dtdiLV S
sS
Multiply the above equation by td then, SsS diLtdV
o
o
I
ISs
u
m diLtdtV
sin
Then; osm ILuV 2coscos Then; osm ILuV 2cos1
Then; m
osV
ILu 21cos
Then;
m
osV
ILu 21cos 1
m
osV
ILut
21cos1 1
Dr. Ali M. Eltamaly, King Saud University
dtdiLv S
srd
oS
I
ISS
u
rd ILdiLtdvo
o
2
the total reduction per period is:
oS
u
rd ILtdv
42
oSoS
rd ILfILV 42
4
u
rd tdv
is the reduction area in one commutation period t
osm
rdceinducsourcewithoutdcactualdc IfLVVVV 42tan
Dr. Ali M. Eltamaly, King Saud University
the rms value of the supply current
2u
oI
oI 2u
uIo2
uIo2
sI
2u
2u
22 u
2
2u
2u
2u
2u
22
u
usI
2
Dr. Ali M. Eltamaly, King Saud University
]2
[2 2/
0
2
2/
22
u
uo
os tdItdt
uI
I
322
228342 23
2
2 uIuuu
II oo
s
2u
oI
oI 2u
uIo2
uIo2
sI
2u
2u
22
u
2
2u
2u
2u
2u
22 u
usI
2
Dr. Ali M. Eltamaly, King Saud University
2u
oI
oI2u
uIo2
uIo2
sI
2u
2u
22 u
2
2u
2u
2u
2u
22 u
usI
sJ 2u
2u
2u
2u
sI 0 0 0 0
sI uIo2
uIo2
uIo2
uIo2
sJ 2u
2u
2u
2u
sI 0 0 0 0
sI uIo2
uIo2
uIo2
uIo2
It is an odd function, then 0 no aa
m
sss
m
sssn tnJ
ntnJ
nb
11sin1cos1
2sin
2sin
2sin
2sin2*11 unununun
uI
nnb o
n
2sin*8
2nu
unIb o
n
2sin*8
1u
uIb o
2sin*
28
1u
uII o
S
Dr. Ali M. Eltamaly, King Saud University
32
sin2
32
2cos
2sin4
2cos
322
2sin*
28
2cos*
21
uu
uuu
uu
u
uI
uu
Iu
IIpf
o
o
S
S
Dr. Ali M. Eltamaly, King Saud University
2sin*
28
1u
uII o
S
322 2 uII o
s
Dr. Ali M. Eltamaly, King Saud University
m
osV
ILu 21cos 1 oSoS
rd ILfIL
V 42
4
osm
rdceinducsourcewithoutdcactualdc IfLVVVV 42tan
322 2 uII o
s
2sin*
28
1u
uI
I oS
32
sin2uu
upf
Example 6 Single phase diode bridge rectifier connected to 11 kV, 50 Hz, source inductance Ls=5mH supply to feed 200 A pure DC load, find: (i) Average DC output voltage, (ii) Power factor. And (iii) Determine the THD of the utility line current.
VVm 155562*11000
VV actualdc 9703200*005.0*50*415556*2
m
osV
ILu 21cos 1 oSoS
rd ILfIL
V 42
4
osm
rdceinducsourcewithoutdcactualdc IfLVVVV 42tan
322 2 uII o
s
2sin*
28
1u
uI
I oS
32
sin2uu
upf
Example 6 Single phase diode bridge rectifier connected to 11 kV, 50 Hz, source inductance Ls=5mH supply to feed 200 A pure DC load, find: (i) Average DC output voltage, (ii) Power factor. And (iii) Determine the THD of the utility line current.
VVm 155562*11000
VV actualdc 9703200*005.0*50*415556*2
.285.015556
200*005.0*50**2*21cos21cos 11 radV
ILum
os
917.0
3285.
2285.0
285.0sin*2
32
sin*22
cos*1
uu
uuIIpf
S
S
AuII o
S 85.1933285.0
2200*2
322 22
Auu
II oS 46.179
2285.0sin*
285.0*2200*8
2sin*
28
1
%84.40146.17985.1931
22
1
S
Si I
ITHD
Three-Phase Half Wave Rectifier
mm
mdc VVtdtVV 827.0233sin
23 6/5
6/
RV
RVI mm
dc*827.0
**233
mmmrms VVtdtVV 8407.08
3*321sin
23 6/5
6/
2
RVI m
rms8407.0
RV
RVII mm
Sr 4854.03
08407
ThePIV of the diodes is mLL VV 32
Example 7 The rectifier in Fig.2.21 is operated from 460 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) Peak inverse voltage (PIV) of each diode.
VVVV mS 59.3752*58.265,58.2653
460
mm
dc VVV 827.0233
R
VRVI mm
dc0827
233
mrms VV 8407.0R
VI mrms
8407.0
%767.96rmsrms
dcdc
ac
dcIVIV
PP
%657.101dc
rmsVVFF
%28.1811 22
222
FF
VV
VVV
VVRF
dc
rms
dc
dcrms
dc
ac
The PIV= 3 Vm=650.54V
Three-Phase Half Wave Rectifier With DC Load Current and zero source induct
New
axis
321 3/
3/0
oo
ItdIa
harmonicstrepleanallfor
nfornI
nfornI
tnnIdwttnIa
o
o
oon
0
17,16,11,10,5,43*
,....14,13,8,7,2,13*
sincos*1 3/3/
3/
3/
...8cos
817cos
715cos
514cos
412cos
21cos3
3)( ttttttIItI OO
s
%24.1090924.119
*21
23
3/1))((
2
2
2
1
O
o
S
Ss I
III
tITHD
Example 8 Solve example 7 if the load current is 100 A pure DC
VVVV mm
dc 613.310827.0233
AIdc 100
VVV mrms 759.3158407.0
%37.98100*759.315100*613.310
rmsrms
dcdc
ac
dcIVIV
PP
%657.101dc
rmsVVFF
%28.1811 22
222
FF
VV
VVV
VVRF
dc
rms
dc
dcrms
dc
ac
The PIV= 3 Vm=650.54V
Dr. Ali M. Eltamaly, King Saud University
Three-Phase Half Wave Rectifier With Source Inductance
Dr. Ali M. Eltamaly, King Saud University
m
oVLIu
321cos 1
m
oVLIut
321cos1 1
oo
rd ILfLIV 32
3
om
Actualdc ILfVV 3233
Dr. Ali M. Eltamaly, King Saud University
Example 9 Three-phase half-wave diode rectifier connected to 66 kV, 50 Hz , 5mH supply to feed a DC load with 500 A DC, fined the average DC output voltage.
Solution: Vvm 538892*3
66000
(i) oceinduc
sourcewithout
dcActualdc ILfVV 3tan
VILfVV om
Actualdc 44190500*005.0*50*32
53889*3*33233
Three-Phase Full Wave Rectifier With Resistive Load
1 3 5
4 6 2
b
c
IL
V LIs
Ip
a
LLmLLm
mdc VVVVtdtVV 3505.1654.12333sin33 3/2
3/
RV
RV
RV
RVI LLLLmm
dc3505.123654.133
LLmmmrms VVVtdtVV 3516.16554.14
3*923sin33 3/2
3/
2
RVI m
rms6554.1
RV
RVI mm
r 9667.03
6554.1
RVI m
S 29667.0
Example 10 The rectifier shown in Fig.2.30 is operated from 460 V 50 Hz supply and the load resistance is R=20ohms. If the source inductance is negligible, determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) Peak inverse voltage (PIV) of each diode .
VVVV mm
dc 226.621654.133
A
RV
RVI mm
dc 0613.31654.133
VVVV mmrms 752.6216554.14
3*923
AR
VI mrms 0876.316554.1
%83.99rmsrms
dcdc
ac
dcIVIV
PP
%08.100dc
rmsVVFF
%411 22
222
FF
VV
VVV
VVRF
dc
rms
dc
dcrms
dc
ac
The PIV= 3 Vm=650.54V
Three-Phase Full Wave Rectifier With DC Load Current
....,.........15,14,12,10,9,8,6,4,3,2,0
.....),........3(132
),3(112
),3(7
2),3(
52
,32
cos2
sin*2
1311751
6/56/
6/5
6/
nforb
Ib
Ib
Ib
Ib
Ib
tnn
ItdtnIb
n
ooooo
oon
ttttv�tItI o
s
13sin13111sin
1117sin
715sin
51sin32)(
%31251
231
191
171
131
111
71
51))((
22222222
tITHD s
oS II32 oS II
3*2
1
%01.311/3*23/21))(( 2
2
1
S
Ss I
ItITHD
Power Factor =S
S
S
SII
II 11 )0cos(*
LL
oSV
ILu 21cos 1o
ord fLILIV 6
26
oLLrdceinducsourcewithoutdcactualdc fLIVVVV 635.1tan
632 2 uII o
S
2sin
621
uuI
I oS
63
sin*32
cos
632
2sin62
2cos
21
uu
uuuI
uu
Iu
IIpf
o
o
S
S
Example 11 Three phase diode bridge rectifier connected to tree phase 33kV, 50 Hz supply has 8 mH source inductance to feed 300A pure DC load current Find;Commutation time and commutation angle.DC output voltage.Power factor.Total harmonic distortion of line current.
oradu 61.14.2549.0
LL
oSV
ILu 21cos 1
dLLrdceinducsourcewithoutdcactualdc fLIVVVV 635.1tan
VVdcactual 43830300*008.*50*633000*35.1 9644.0
62549.0
3*2549.0
2549.0sin3
63
sin*3
uu
upf
AuII d
s 929.2396
2549.03
*300*263
2 22
Auu
II o
S 28.2332
2549.0sin*2*2549.0*
300*3432*2
sin2
341
9644.02
2549.0cos*929.23928.233
2cos*1
uIIpf
s
S
%05.24128.233
929.239122
1
S
Si I
ITHD
For three phase uncontrolled rectifier with pure DC current load without source inductance shown in Fig.1, draw the following: (a) The output voltage waveform in waveform (2) of Fig.2. (b) The currents waveforms of switches 1 to 6 in the waveforms (3) to (8) of Fig.2, (c) Secondary current of phase b in waveform (9) of Fig.2, (d) Primary current of phase B in waveform (10) of Fig.2, Then,
(e) Derive an equation of secondary current tib waveform by using Fourier transforms. (f) Derive an equation of Primary current tiB waveform by using Fourier transforms. (g) Find its input power factor at the supply (primary) terminals .
A
B
C
a
b
c
IA
IB
IC
Ia
Ib
Ic
IAB
IBC
ICA n
2:1D1 D3 D5
D4 D6 D2
30A
=I'a
=I'b
=I'C