1. No category

Dr. Ali M. Eltamaly
King Saud University
1
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 Rectification
uncontrolled 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
  Pdc / Pac
Vac 
2
Vrms
rectification effeciency
2
 Vdc
FF  Vrms /Vdc
form factor
ripple factor
Vac
RF 

Vdc
2
Vrms
2
 Vdc
Vdc

2
Vrms
2
Vdc
 1  FF 2  1
Dr. Ali M. Eltamaly, King Saud University
THDi 
THD v 
I S2  I S21

2
I S1
VS2  VS21
VS21

I S2
2
I S1
VS2
VS21
1
1
VS I S1 cos 1 I S1
P
PF 


cos 1
VS I S
VS I S
IS
 Distortion Factor * Displacement Faactor
Dr. Ali M. Eltamaly, King Saud University
Single-phase half-wave diode rectifier with resistive load.
Dr. Ali M. Eltamaly, King Saud University

Vdc
Vm
1

Vm sin t dt 

2 0


Vrms 
Vm
1
2
2
Vm sin t dt 

2 0
2
the load and diode currents
I dc
Vdc Vm


R  R
I rms
Vrms Vm


R
2R
IS  ID
Dr. Ali M. Eltamaly, King Saud University
Vm

2R
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.

Vdc
V
V
1

Vm sin(t ) dt  m ( cos  cos(0))  m
2
2


I dc
0

Vrms
V
1

(Vm sin t ) 2  m
2
2

I rms
Vdc Vm


R R
Vm

2R
0
Vm

Pdc
V *I
 dc dc
Pac Vrms * I rms
Vrms
FF 
Vdc
RF 
Vm
 R

 40.53%
Vm Vm
*
2 2R
*
Vm

2

  1.57
Vm 2

Vac
 FF 2  1  1.57 2  1  1.211
Vdc
(d) It is clear from Fig2.2 that the PIV is
Dr. Ali M. Eltamaly, King Saud University
Vm
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
di
L  R * i  Vm sin (t ), 0  t  
dt
Divide the above equation by L we get:
di R
V
 * i  m sin (t ), 0  t  
dt L
L
i (t )  e

R 
dt
L 

R
 t
i (t )  e L 

i (t ) 
e

R

dt V
L * m sin t dt  A
R
t
L
e
Vm
R w L
2

2 2
L


Vm
* sin t dt  A
L

R sin t  L cos t  
R
 t
Ae L
Dr. Ali M. Eltamaly, King Saud University
Z  R  j wL
R  Z cos
Z 2  R 2  w2 L2
L  Z sin 
tan  
V
i (t )  m cos  sin t  sin  cos t  
Z
Vm
i (t ) 
sin t    
Z
Z
L
wL
R

R
 t
Ae L
R
R
 t
Ae L

R
t
L
Vm
i (t ) 
sin t     Ae
Z
Vm
A

sin  
i0  0
Z
Vm

sin t     Ae
Z

t
tan 
t 


V 

i (t )  m  sin t     sin  e tan  
Z 



Dr. Ali M. Eltamaly, King Saud University
t 


Vm 
i (t ) 
sint     sin e tan 

Z 


 


Vm 
tan  




i( ) 
sin




sin

e

0

Z 



Vdc
Vm
Vm

*  sin t dt 
* (1  cos  )
2 0
2

Vrms
1
Vm
2

*  (Vm sin t ) dwt 
*   0.5(1  sin( 2  )
2 0
2 
Dr. Ali M. Eltamaly, King Saud University

Vdc
Vm
Vm

*  sin t dt 
* (1  cos  )
2 0
2

Vrms
1
Vm
2

*  (Vm sin t ) dwt 
*   0.5(1  sin( 2  )
2 0
2 
Dr. Ali M. Eltamaly, King Saud University
Half wave diode rectifier with free wheeling diode
Dr. Ali M. Eltamaly, King Saud University
t 


V 

i (t )  m  sin t     sin   e tan  
Z 



L
di
 R* i  0
dt
i (t )  B e
i ( ) 

t 
tan 
0  t  
for   t  2
i ( )  B
Vm
(sin      sin   e
Z


tan 
)B
  t 



Vm 
tan  
tan 




i (t ) 
sin




sin

e
e

Z 



  t  2
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 0  t  2
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 0  t  3 .
Solution: (a) For the period of 0  t   , the expression of the load current can be
obtained from (2.24) as following:
3
1 L
1 314 * 20 *10
  tan
 tan
 0.561 rad . and tan   0.628343
R
10
Z  R 2  (L) 2  10 2  (314 * 20 *10  3 ) 2  11.8084
t


V 
i (t )  m  sin t     sin   e tan 
Z 








220 2
sin t  0.561  0.532 * e 1.5915 t
11.8084

i (t )  26.3479 sint  0.561  14.0171* e 1.5915 t
Dr. Ali M. Eltamaly, King Saud University
The value of  can be obtained from the above equation by substituting for i (  )  0 . Then,
0  26.3479 sin   0.561  14.0171 * e 1.5915 
By using the numerical analysis we can get the value of . The simplest method is by
using
the
simple
iteration
technique
by
assuming
  26.3479 sin   0.561  14.0171 * e 1.5915  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 
1.12 
1.14 
1.16 
1.18 
1.2 

6.49518
4.87278
3.23186
1.57885
-0.079808
-1.73761
It is clear from the above table that   1.18  rad. The current in   wt  2 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 0  t   (D1 “ON”, D2 “OFF”), the
second case when   t  2 (D1 “OFF” and D2 “ON”) and the last one when
2  t  3 (D1 “ON”, D2 “OFF”).
 In the first part ( 0  t   ) the expression for the load current can be obtained as
In case (a). Then:
for 0  t  
i ( wt )  26.3479 sin t  0.561  14.0171* e 1.5915 wt
the current at t   is starting value for the current in the next part. Then
i ( )  26.3479 sin   0.561  14.0171 * e 1.5915   14.1124 A
 In the second part   t  2 , the expression for the load current can be obtained
from (2.30) as following:

t 
tan 
i (t )  B e
where B  i ( )  14.1124 A
Then i (t )  14.1124 e 1.5915t  
for (   t  2 )
The current at t  2 is starting value for the current in the next part. Then
i (2 )  0.095103 A
 In the last part ( 2  t  3 ) the expression for the load current can be obtained
from (2.36):
i (t ) 
Vm
V


sin t  2      i 2   m sin  e
Z
Z



t  2
tan 
 i (t )  26.3479 sin t  6.8442  0.095103  26.3479 * 0.532e 1.5915t  2 
 i (t )  26.3479 sin t  6.8442   14.1131e 1.5915t  2  for ( 2  t  3 )
Dr. Ali M. Eltamaly, King Saud University
Single-Phase Full-Wave Diode Rectifier
Center-Tap Diode Rectifier
Dr. Ali M. Eltamaly, King Saud University
Vdc 
1

Vrms 


Vm sin t dt 
0
1


 Vm sin t 
2
0

Vm
dt 
2
PIV of each diode =
IS  ID
2 Vm
I dc
2 Vm

 R
I rms
Vm

2 R
2Vm
Vm

2R
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
2 Vm
2 Vm
*

R

 81.05%
Vm
Vm
*
2
2R
Pdc
Vdc * I dc


Pac Vrms * I rms
Vrms
FF 
Vdc
Vm

2


 1.11
2 Vm 2 2

Vac
2
2
RF 
 FF  1  1.11  1  0.483
Vdc
The PIV is
2Vm
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
di
L
 R * i  Vm sin(t )
dt
t 


Vm 
tan  
i (t ) 
sin t     sin  e


Z 



Vm
i (t ) 
sin t       Ae
Z

0  t  
t 
tan 
i()=i(2)=i(3)=……..=Io
 


Vm 
tan  
I o  i ( ) 
sin      sin  e


Z 



Vm
i ( )  I o 
sin        Ae
Z
Vm
A  Io 
sin  
Z

 
tan 
Dr. Ali M. Eltamaly, King Saud University
i (t ) 
Vm
V


sin t        I o  m sin  e
Z
Z



t 
tan 
t  
t 



V
i (t )  m sin t       sin  e tan    I o e tan 

Z 


t  2 
t  2



V
i (t )  m sin wt  2     sin  e tan    I o e tan 

Z 


Dr. Ali M. Eltamaly, King Saud University
2  t  3
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.
Vdc 
1

Vm sin t dt 


2 Vm

0

Vrms
 190.956 V
I dc
2 Vm

 12.7324 A
 R
1/ 2
1

2
   Vm sin t  dt 
  0


Vm
 212.132 V
2
Pdc
Vdc I dc


 81.06 %
Pac Vrms I rms
Vrms
FF 
 1.11
Vdc
2
2
Vrms
 Vdc2
Vac
Vrms
2
RF 



1

FF
 1  0.482
2
Vdc
Vdc
Vdc
The PIV=300V
VS I S cos
Re al Power

1
Input power factor =
Apperant Power
VS I S
Dr. Ali M. Eltamaly, King Saud University
bn 
2

2 Io
 cos nt 0

n
I o * sin nt dt


0
2 Io
4 Io
cos 0  cos n  

for n  1, 3, 5, .......... ...
n
n
4 Io
1
1
1
1
i (t ) 
* (sin t  sin 3t  sin 5t  sin 7t  sin 9t  .......... )

3
5
7
9
2
2
2
2
2
2
2
1 1 1 1  1   1   1 
THD ( I s (t ))                       46%
 3   5   7   9   11   13   15 
4 Io
I S1 
2
2




2
 IS 
Io 

  1 
THD ( I s (t ))  
1 


4 Io
 I S1 


 2
 2

 4

Dr. Ali M. Eltamaly, King Saud University
2

  1  48.34%


Example 5 solve Example 4 if the load is 30 A pure DC
From example 4 Vdc= 190.986 V, Vrms=212.132 V I dc  30 A and I rms = 30 A
Pdc
Vdc I dc


 90 %
Pac Vrms I rms
Vrms
FF 
 1.11
Vdc
2
2
2
Vrms
 Vdc
Vac
Vrms
2
RF 



1

FF
 1  0.482
2
Vdc
Vdc
Vdc
The PIV=Vm=300V
4 Io
4 * 30
I S1 

 27.01A
2
2
Re al Power

Input Power factor=
Apperant Power
VS I S1 * cos
I S1 * cos
27.01


*1  0.9 Lag
VS I S
IS
30
Dr. Ali M. Eltamaly, King Saud University
Full Bridge Single-phase Diode Rectifier with DC Load Current
bn 
2

2 Io
 cos nt 0

n
I o * sin nt dt


0
2 Io
4 Io
cos 0  cos n  

for n  1, 3, 5, .......... ...
n
n
4 Io
1
1
1
1
i (t ) 
* (sin t  sin 3t  sin 5t  sin 7t  sin 9t  .......... )

3
5
7
9
2
2
2
2
2
2
2
1 1 1 1  1   1   1 
THD ( I s (t ))                       46%
 3   5   7   9   11   13   15 
4 Io
I S1 
2
2




2
 IS 
Io 

  1 
THD ( I s (t ))  
1 


4 Io
 I S1 


 2
 2

 4

2

  1  48.34%


Example 5 solve Example 4 if the load is 30 A pure DC
From example 4 Vdc= 190.986 V, Vrms=212.132 V I dc  30 A and I rms = 30 A
Pdc
Vdc I dc


 90 %
Pac Vrms I rms
Vrms
FF 
 1.11
Vdc
2
2
2
Vrms
 Vdc
Vac
Vrms
2
RF 



1

FF
 1  0.482
2
Vdc
Vdc
Vdc
The PIV=Vm=300V
4 Io
4 * 30
I S1 

 27.01A
2
2
Re al Power

Input Power factor=
Apperant Power
VS I S1 * cos
I S1 * cos
27.01


*1  0.9 Lag
VS I S
IS
30
Effect Of LS On Current Commutation Of Single-Phase Diode Bridge Rectifier.
diS
VS  Ls
0
dt
Multiply the above equation by dt then,
VS dt  Ls diS
 u
Io

Io
 Vm sin t dt  Ls  diS
Then; Vm cos   cos  u   2Ls I o
Then; Vm  1  cosu   2Ls I o
2Ls I o
Then; cosu   1 
Vm
 2Ls I o 

Then; u  cos 1 1 
Vm 

t 
u


 2Ls I o 

cos 11 

Vm 

1
Dr. Ali M. Eltamaly, King Saud University
vrd  Ls
diS
dt
 u
Io

Io
 vrd d t    LS diS  2 LS I o
 u
 vrd d t is the reduction area in one commutation period t

the total reduction per period is:
 u
2  vrd d t  4  LS I o

Vrd 
 4 LS I o
 4 f LS I o
2
Vdc actual  Vdc without sourceinduc tan ce  Vrd 
2Vm

 4 fLs I o
Dr. Ali M. Eltamaly, King Saud University
the rms value of the supply current
Is
u
Io


u
2
2
u

2

u
2
2
 Io

u
2
2 
Dr. Ali M.
I ¢ Eltamaly, King Saud University
s
u
2
Is
u
Io


u

2
2
u
2

u
2
2
u/2
u

2
 Io
I s¢
2
u

u
u 2
3
 2I o  4 u  u 
2 I o2   u 
Is  2  2
 2  

  3u 8 2 2 
  2 3 

u
2

2
 2I o 
I s  2 I[o 
t  dt   I o2 dt ]

u

0 
u/2
2
2 

u
2

u
u
Dr. Ali M.Eltamaly, King Saud
2 University

Is
u
Io

u
2
u

2
u

2
 Io
2
u

2
2 
u
2
I s¢

u
2



2I o
u

Is
0
2Io
u
I s¢
2Io
u
u

2
Js
u
2
u
2
u
2
2 
u
2
Dr. Ali M. Eltamaly, King Saud University
u
2
0


u
2
0
2I o
u

2I o
u

0
2Io
u
u
2
It is an odd function, then ao  an  0
Js

Is
0
2Io
u
I s¢
1
bn 
n
u
2
u
2
0


u
2
0
2I o
u

2I o
u

u
2
0
2Io
u
m

1m
¢
J
cos
n

t

J
sin
n

t
 s
s
s
 s
n
 s 1

s 1
1   1 2Io 
u
u  
 u
u


bn 
*
 sin n    sin n   sin n     sin n    

n  n u 
2
2  
 2
2


8I o
nu
bn  2 * sin
2
n u
8I o
u
b1 
* sin
u
2
8I o
u
I S1 
* sin
2 u
2
Dr. Ali M. Eltamaly, King Saud University
8I o
u
I S1 
* sin
2 u
2
2 I o2  u 
Is 

  2 3 
8I o
u
* sin
2 u
2
I S1
u 
u 
* cos   
cos  
IS
2
2
2 I o2  u 
 

  2 3
u u

4 sin   cos  
2 sin u 
2 2



 u
 u
u    
u    
 2 3
 2 3
pf 
Dr. Ali M. Eltamaly, King Saud University
 2Ls I o 

u  cos1 1 
Vm 

Vrd
 4 LS I o

 4 f LS I o
2
Vdc actual  Vdc without sourceinduc tan ce  Vrd 
2 I o2   u 
Is 
 

  2 3
8I o
u
I S1 
* sin
2
2 u
2Vm

pf 
 4 fLs I o
2 sinu 
 u 
u   
 2 3
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.
Vm  11000 * 2  15556V
Vdc actual 
2 *15556

 4 * 50 * 0.005 * 200  9703V
Dr. Ali M. Eltamaly, King Saud University
 2Ls I o 

u  cos1 1 
Vm 

Vrd
 4 LS I o

 4 f LS I o
2
Vdc actual  Vdc without sourceinduc tan ce  Vrd 
2 I o2   u 
Is 
 

  2 3
8I o
u
I S1 
* sin
2
2 u
2Vm

pf 
 4 fLs I o
2 sinu 
 u 
u   
 2 3
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.
Vm  11000 * 2  15556V
Vdc actual 
2 *15556

 4 * 50 * 0.005 * 200  9703V
1 
u  cos 1 

2Ls I o 
2 * 2 *  * 50 * 0.005 * 200 
  cos1 1 
  0.285 rad .
Vm 
15556


I S1
u
pf 
* cos  
IS
2
IS 
2 I o2   u 
 

  2 3
2 * sinu 
 u 
u   
 2 3

2 * sin0.285
  .285 
0.285   
3 
2
 0.917
2 * 2002   0.285 

 193.85 A



3 
2
8I o
u
8 * 200
 0.285 
I S1 
* sin 
* sin
  179.46 A
2
2 u
2  * 0.285
 2 
2
2
 IS 
 193.85 


THDi  
1  
  1  40.84%

 179.46 
 I S1 
Three-Phase Half Wave Rectifier
Vdc
3

2
Vrms
I rms
5 / 6

Vm sin t dt 
 /6
3

2
3 3 Vm
 0.827Vm
2
5 / 6
 Vm sin t 
2
 /6
0.8407 Vm

R
ThePIV of the diodes is
I dc
3 3 Vm 0.827 * Vm


2 * * R
R
1 3* 3
dt 

Vm  0.8407 Vm
2 8
08407 Vm
Vm
Ir  IS 
 0.4854
R
R 3
2 VLL  3 Vm
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.
460
VS 
 265.58 V , Vm  265.58 * 2  375.59 V
3
Vdc
3 3 Vm

 0.827 Vm
2
Vrms  0.8407 Vm
3 3 Vm 0827Vm
I dc 

2 R
R
0.8407 Vm
I rms 
R
Pdc
Vdc I dc


 96.767 %
Pac Vrms I rms
Vrms
FF 
 101.657 %
Vdc
2
2
2
Vrms
 Vdc
Vac
Vrms
2
RF 



1

FF
 1  18.28 %
2
Vdc
Vdc
Vdc
The PIV=
3
Vm=650.54V
Three-Phase Half Wave Rectifier With DC Load Current and zero source induct
New axis
1
a0 
2
an 
1

 /3
Io
 I o dt  3
 / 3
 /3
Io
sin nt //33

n
 I o * cos nt dwt
 / 3
Io

* 3 for n  1,2,7,8,13,14,....
n
Io

* 3 for n  4,5,10,11,16,17
n
 0 for all treplean harmonics
I s (t ) 
IO
3I O 
1
1
1
1
1


 cos t  cos 2t  cos 4t  cos 5t  cos 7t  cos 8t  ...
3
 
2
4
5
7
8

2
I 
THD ( I s (t ))   S 
 I S1 
2




I / 3
1   o
1 

 3I O 
  2 


2 * 2
 1  1.0924  109.24%
9
Example 8 Solve example 7 if the load current is 100 A pure DC
Vdc 
3 3 Vm
 0.827 Vm  310.613V
2
I dc  100A
Vrms  0.8407 Vm  315.759 V

Pdc
V I
310.613* 100
 dc dc 
 98.37 %
Pac Vrms I rms 315.759 *100
Vrms
FF 
 101.657 %
Vdc
2
2
2
Vrms
 Vdc
Vac
Vrms
2
RF 



1

FF
 1  18.28 %
2
Vdc
Vdc
Vdc
The PIV=
3
Vm=650.54V
Three-Phase Half Wave Rectifier With Source Inductance
Dr. Ali M. Eltamaly, King Saud University
2LI o 

u  cos 1 
3 Vm 

1 
2LI o 

t   cos 1 
 
3 Vm 

u
1
1 
3LI o
 3 f L Io
2
3 3 Vm

 3 f L Io
Actual
2
Vrd 
Vdc
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.
 66000 
Solution: vm  
 * 2  53889V
 3 
 3 f L Io
(i) Vdc Actual  Vdc without
source
induc tan ce
Vdc
Actual

3 3 Vm
3 * 3 * 53889
 3 f L Io 
 3 * 50 * 0.005 * 500  44190V
2
2
Dr. Ali M. Eltamaly, King Saud University
Three-Phase Full Wave Rectifier With Resistive Load
IL
Ip
Is
3
1
5
VL
a
b
c
4
6
2
Vdc 
I dc
3
2 / 3


3 Vm sin t dt 
 /3
3 3 Vm


3 2 VLL

 1.654Vm  1.3505VLL
3 3 Vm 1.654Vm 3 2 VLL 1.3505VLL




 R
R
R
R
Vrms 
3

I rms
2 / 3
 
3 Vm sin t
 /3

2
3 9* 3
dt 

Vm  1.6554 Vm  1.3516VLL
2
4
1.6554 Vm

R
1.6554 Vm
Vm
Ir 
 0.9667
R
R 3
I S  0.9667
2
Vm
R
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 .
Vdc 
3 3 Vm

 1.654Vm  621.226 V
3 3 Vm 1.654Vm
I dc 

 31.0613 A
 R
R
3 9* 3
Vrms 

Vm  1.6554 Vm  621.752 V
2
4
I rms
1.6554 Vm

 31.0876 A
R
Pdc
Vdc I dc


 99.83 %
Pac Vrms I rms
Vrms
FF 
 100.08 %
Vdc
2
2
2
Vrms
 Vdc
Vac
Vrms
2
RF 



1

FF
1  4 %
2
Vdc
Vdc
Vdc
The PIV=
3
Vm=650.54V
Three-Phase Full Wave Rectifier With DC Load Current
bn 
2

5 / 6
 I o * sin nt dt

 /6
2 Io
2 Io
2 Io
2 Io
( 3 ), b7 
( 3 ), b11 
( 3 ), b13 
( 3 ),.......

5
7
11
13
bn  0, for n  2,3,4,6,8,9,10,12,14,15,......... ....
b1 
2 Io
2 Io
 cos nt 5/ 6/ 6
n
I s (t ) 
3, b5 
2 3I o 
1
1
1
1

sin

t

sin
5
v•

t

sin
7

t

sin
11

t

sin
13

t


 
5
7
11
13

2
2
2
2
2
2
2
1 1  1   1   1   1   1   1 
THD ( I s (t ))                        
 5   7   11   13   17   19   23   25 
2
IS 
Io
3
2* 3
I S1 

2
 IS 
  1 
THD ( I s (t ))  
 I S1 
Io
2/3
2*3/
2
 1  31.01%
I S1
I S1
Power Factor =
* cos(0) 
IS
IS
2
 31%
1 
2 LS I o 
u  cos 1 

VLL 

Vrd
6 LI o

 6 fLI o
2
Vdc actual  Vdc without sourceinduc tan ce  Vrd  1.35VLL  6 fLIo
2 I o2   u 
IS 
 

  3 6
pf 
I S1
u
cos   
IS
2
I S1 
2 6 Io  u 
sin  
u
2
2 6 Io  u 
sin 
u
2
3 * sin u 
u
cos   
2
2 I o2   u 
 u 
u

 
 


  3 6
 3 6
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.
1 
2 LS I o 
u  cos 1 

VLL 

Vdc
actual
 Vdc
u  0.2549 rad.  14.61
without sourceinduc tan ce
o
 Vrd  1.35VLL  6 fLI d
Vdcactual1.35 * 33000  6 * 50 * .008 * 300  43830V
3 * sinu 
3 sin0.2549
pf 

 0.9644
 u 
u   
 3 6
  0.2549 
0.2549 *   
6 
3
2 I d2   u 
2 * 3002
Is 
 

  3 6

I S1 
  0.2549 
* 
  239.929 A
6 
3
4 3 Io
4 3 * 300
u
 0.2549 
sin  * 2 3 
* sin
  233.28 A
u 2
 * 0.2549 * 2
2
 2 
I S1
 u  233.28
 0.2549 
pf 
* cos  
* cos
  0.9644
Is
 2  239.929
 2 
2
2
 IS 
 239.929 


THDi  
1  
  1  24.05%

 233.28 
 I S1 
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 ib t  waveform by using Fourier transforms.
(f) Derive an equation of Primary current iB t  waveform by using Fourier transforms.
(g) Find its input power factor at the supply (primary) terminals .
IA
2:1
A
IAB
IC
C
IBC
Ib
n
b
=I'b
D5
30A
ICA
=I'a
B
D3
a
=I'C
IB
D1
Ia
c
D4
Ic
D6
D2