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8/10/2019 Single Phase Rectifiers
1/22
Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
2-1
2. Single Phase Rectifiers
The learning objectives of this chapter are:
To be able to classify the types of single phase rectifiers
To be able to describe the operation of single-phase rectifiers with different types ofload
To differentiate between the operation of single-phase uncontrolled and controlledrectifiers, following similar load conditions
To compare different types of rectifiers from the prospective of certain performanceparameters
To develop effective calculation methods for analyzing and design of single phase
rectifiers
To be able to simulate single phase rectifiers, perform lab measurements and comparethe results
To be able to formulate conclusions regarding the operation of single-phase rectifiersin normal and abnormal operating conditions
2.1.Single-phase uncontrolled rectifiers
A rectifier is a circuit that converts an AC signal into unidirectional one. In this paragraph
diode rectifiers are considered, and for simplicity ideal diodes are considered. When making
the assumption of ideal diodes, we mean diodes with zero forward voltage drop in
conduction, infinite internal resistance in blocking state (when reverse biased) and
instantaneous transition from on-state to off-state and inversely, depending of biasing
conditions.
2.1.1 Half-Wave Uncontrolled Rectifier
Applications: Low cost, low power supplies for electronics
For the single phase rectifiers, having connected certain loads at the output, the following
objectives are to be achieved:
a. Explain the operation
b.
Calculate the output (load) voltage average value VL,DC
c. Calculate the output (load) voltage effective (RMS) value VL,RMS
d. Calculate the output (load) current average value IL,DC
e.
Calculate the output (load) current effective (RMS) value IL,RMSIn addition to the above-mentioned, most of the characteristics defined in previous chapter
can be calculated or determined upon the necessity. These are:
Efficiency of the rectifier Form factor FF Ripple factor RF
Transformer utilization factor TUF
Peak inverse voltage of the diode PIV
Effective (RMS) value of the diode current
Crest factor of the input current CF Input power factor PF
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Once in conduction, the ideal diode D will let the power supply voltage to be
applied across the load:
dt
diiRitVvv LLLmSL +=== sin (2.2)
Equation (2.2) has the following solution:
( ) ( )
++
=t
L
R
et
RL
Vti
mL
sinsin222
, with
=
R
L arctan (2.3)
Note: Solving the equation (2.2.) is the key point for further calculations and for this purpose
will use Laplace Transform, when zero initial load current is considered:
( )
Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
2-3
( ) ( ) ( ) ( )
+
=+=
L
LLLL
TsL
sVsIsIsLRsV
1, with
R
LTL = (2.4)
( ) ( ){ }22
+==
s
VtvLsV mL (2.5)
Finally, the operational load current is expressed as:
( )( )
+
+=
L
mL
TsL
s
VsI
1
122
(2.6)
For finding the load current in time domain, we have to express the operational
current from (2.6) as a sum of simple fractions. Once determined, the coefficients A,Band C,
will give the possibility to find out the time expression of the current by applying Laplace
inversions:
( )
+
++
+
=
L
mL
Ts
C
js
B
js
A
L
VsI
1
(2.7)
Bringing to a common denominator (2.6) and (2.7) and imposing to have the same
numerator, a three-equation system is composed:
( ) ( )
( )
=+
=++
=++
1
01
0
2
CBAT
j
BA
T
BAj
CBA
L
L
(2.8)
Solving the above system, we found:
,
=
jT
j
B
L
12
1and
2
2
1
1
+
=
LT
C (2.9)
+
=
jT
j
A
L
12
1
From (2.7) and (2.9), by inverting the Laplace elementary fractions, we obtained:
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( )
+
+
+
+
= LT
t
e
T
tje
jT
j
tje
jT
jL
Vti
LLL
mL
2
2
1
1
12
1
12
1 (2.10)
Applying the definitions of the complex sine and cosine functions:
and ( )j
tje
tje
t2
sin
= , we obtain:( )
2cos
tje
tje
t
+=
( )( )
Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
2-4
+
+
=
sin
sinsin
1
1
2
2
LT
t
et
T
L
Vti
L
mL (2.11)
Finally, by performing the calculations, we obtain:
( ) ( )
++=
tL
R
etRL
V
ti m
L
sinsin222 , with
= R
L
arctan (2.3)
Due to the presence of the inductance, the conduction angle is higher than 1800and is
strongly dependent upon the resistance and inductance of the load.
To determine the value of the conduction angle of the diode, we have to solve the
following equation, where represents the conduction angle in radians:
( ) 0sinsin =
+
t
L
R
e
, (2.12)
which is equivalent to:
0cossin =
+
L
R
eL
R (2.12)
Equation (2.12) shows that the conduction angle is practically determined by the
values of load resistance and reactance. Solving it requires numerical methods, and is very
important when estimating the value of the load average voltage:
b. Calculate the output (load) voltage average value VL,DC
To calculate it, we apply the definition:
( ) ( ) (
cos12
sin2
1
0
.0, === m
mavgDCL
VttdVVV ) (2.13)
The above mentioned relationship can be expressed with respect to the angle , which
represents how far beyond =t , the current goes through the resistive inductive load:
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( ) ( ) (
cos12
sin2
1
0
.0, +=== +
mmavgDCL
VttdVVV ) (2.13)
Note: Meanwhile, it can be evaluated the contribution of the load inductance upon the overall
load voltage, using (2.3) and assuming the substitution += :
( ) ( )
( )
+==
sincos
222
tL
R
eL
Rt
RL
LV
dt
tdiLtv
mLL (2.14)
( ) ( )tdt
L
R
eL
Rt
RL
LVV
mDCXL
+=
+
sincos2
1
0222
, (2.15)
( )
+
+= ++
00
222, sinsin
2
tm
DCXLL
R
et
RL
LVV (2.15)
One step further in (2.15) will lead to:
( ) ( )
Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
2-5
( )
+
+++
=
sinsinsinsin
2 222,
L
R
e
RL
LVV
mDCXL , and after
subtracting the terms:
( ) 0sinsin2 222
, =
+
+=
L
R
e
RL
tVV
mDCXL , according to (2.12)
It is very important to demonstrate that the presence of the inductor modifies the value
of the DC output voltage, despite the fact that the average voltage across the inductive part of
the load is equal to zero.
c.
Calculate the output (load) voltage effective (RMS) value VL,RMS
The definition for the effective value is to be applied, in order to determine the value of the
effective (RMS) value of the load voltage:
( ) ( ) ( )
( ) ( )
( )tdV
tdttV
tdtVV mm
mRMSL
=
== 4
2sin
224
2sin
22sin
2
12
00
2
0
22,
(2.16)
d. Calculate the output (load) current average value IL,DC
( ) ( ) =
++
= tdt
L
R
et
RL
VI
mDCL
0222
, sinsin2
1
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( ) ( ) ( ) =
++
= tdt
L
R
etdt
RL
VI
mDCL
00222
, sinsin
2
( ) =
++=
0
0222
, sincos2
t
L
R
eR
L
tRL
V
I m
DCL
( )
++
++
+=
L
R
eR
L
RL
VI
mDCL 1sincos1
2222
, (2.17)
e. The output (load) current effective (RMS) value IL,RMS:
( ) ( )=
+
+
=
0
22
222, sinsin
21 tdtL
R
et
RL
VI mRMSL
( ) ( ) ( )
+
+
+
+
+
++
+
+=
sincos
1
sin2sin
2sin
2
24
2sin
2
2
12
22
222
L
RL
R
e
L
RR
LL
R
eR
L
RL
Vm
(2.18)
Note: In order to determine the effective (RMS) value of the load current , the
following preliminary calculation is necessary to be performed:
( ) ( ) ( ) =
+
+ tdt
L
R
ett
L
R
et
0
22 sinsin2sin
2
sin
( ) ( ) ( ) ( ) ( )
+
+
00
2
0
2 sinsin2sin
2
sin tdt
L
R
ettdt
L
R
etdt
Every single term is evaluated separately:
( ) ( ) ( ) ( )
4
2sin
24
22sin
2sin
00
2
++
=
=
+=
tttdt
( )( )
Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
2-6
+
+
=
=
+
22
0
2
0
2sin
2sin
2
2sin
2
2sin
2
R
LL
R
eR
LL
R
eR
Ltd
tL
R
e
t
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( ) ( )( )
+
+
=
sincos
1
sin2sinsin2
2
0L
RL
R
e
L
Rtd
tL
R
et
Finally, the overall integration appears as:
( ) ( ) ( ) ( )
( ) ( )
+
+
+
+
+
+++
=
+
+
sincos
1
sin2sin
2sin
2
2
4
2sin
2sinsin2sin
2
sin
2
22
0
22
L
RL
R
e
L
RR
LL
R
eR
L
tdt
L
R
ett
L
R
et
Example 2-1
For the circuit from Figure 2.1, lets consider: VVS 120= , = 25R , HL 2.0= , .Hzf 60=
Determine:
a. The output (load) voltage average value VL,DC
b. The output (load) voltage effective (RMS) value VL,RMS
c. The output (load) current average value IL,DC
d. The output (load) current effective (RMS) value IL,RMS
e.
The efficiency of the rectifier
f. The form factor FF
g.
The ripple factor RFh. The transformer utilization factor TUF
i. The peak inverse voltage of the diode PIV
j. The effective (RMS) value of the diode current
k. The crest factor of the input current CF
Solution: For a complete illustration, simulated results are presented in Figure 2.2 and
Figure 2.3.
a. First of all, we have to determine the conduction angle, which is beyond radians,
due to the presence of the inductor:
( )radR
fL
R
L25.166.71
25
2.0602arctan
2arctanarctan
0=
=
=
=
So, the total conduction angle is equal to ( )rad25.1+=
( ) ( ) (
cos12
sin2
1
0
.0, === m
mavgDCL
VttdVVV )=35.51V
b. The output (load) voltage effective (RMS) value VL,RMS:
( ) ( ) ( )
( ) ( )
( ) VtdV
tdttV
tdtVV mm
mRMSL 48.684
2sin
224
2sin
22sin
2
12
00
2
0
22, =
=
==
Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
2-7
8/10/2019 Single Phase Rectifiers
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c.
The output (load) current average value IL,DC:
( ) ( ) =
++
= tdt
L
R
et
RL
VI
mDCL
0222
, sinsin2
1
( )
++
++
+=
L
R
eR
L
RL
VI mDCL 1sincos1
2 222, =1.194A
d. The output (load) current effective (RMS) value IL,RMS:
( ) ( )=
+
+=
0
22
222, sinsin
21 td
tL
R
et
RL
VI mRMSL
( ) ( ) ( )
AL
RL
R
e
L
RR
LL
R
eR
L
RL
Vm 732.1
sincos
1
sin2sin
2sin
2
24
2sin
2
2
12
22
222=
+
+
+
+
+
++
+
+=
e.
The efficiency of the rectifier:
357.0732.148.68
194.151.35
,,
,,
,
,=
===
AV
AV
IV
IV
P
P
RMSLRMSL
DCLDCL
RMSL
DCL
f. The form factor:
( )928.1
561.35
48.68
.0
,
,
,====
V
V
V
V
V
VFF
avg
RMSL
DCL
RMSL
g.
The ripple factor is equal to:
( )648.11928.111 22
2
,
,
.0
,
,
,===
=== FF
V
V
V
V
V
VRF
DCL
RMSL
avg
ACL
DCL
ACL
h. The transformer utilization factor TUF:
204.0732.1120
194.151.35
,
,,,=
===
AV
VV
IV
IV
IV
PTUF
RMSLS
DCLDCL
SS
DCL
i. The peak inverse voltage of the diode PIV
VPIV 1702120
Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
2-8
8/10/2019 Single Phase Rectifiers
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j.
The effective (RMS) value of the diode current:
AII RMSLRMSDIODE 732.1,, ==
k. The crest factor of the input current CF
7.1732.1
95.2
,
,,max,=====
A
A
I
I
I
I
I
ICF
RMSL
peaks
S
peaks
S
s
In order to calculate the crest factor we have to determine the maximum value of the
input current (which in this case is basically the same like for the output current). This can be
solved by imposing:( )
( ) ( ) 0sincos0 =
=
tt
R
eL
Rt
td
tdiL
. In fact, this represents a
transient equation, which requires numerical methods for solving it. Graphical methods are
less accurate, but still convenient sometimes. So, by plotting the curve )( tiL we can find out
the maximum load current value as being AI peaks 95.2,
2.1.1.2. Half-wave uncontrolled rectifier with resistive load
a. Explain the Operation
When having purely resistive load, then 0=L , which means that: 0= . For such asituation will have the have-wave uncontrolled rectifier with purely resistive load, shown in
figure 2.2. The secondary winding of the transformer is considered as power supply.
The conduction duration equals a half of the cycle of the power supply voltage. During the
first (positive) half of the cycle of the power supply voltage, the diode D is in conduction.
Almost all of the power supply voltage appears across the load (because assuming that D is
ideal, we neglect the on state voltage drop across it). In the second (negative) half of the cycle
of the power supply voltage, the diode is reverse biased, so is blocked and the output voltage
is equal to zero.
b.
Calculate the output (load) voltage average value VL,DC
To calculate it, we apply the definition:
( ) ( ) ( ) ( ) mmm
mLDCL V
V
t
V
tdtVtdtvV 318.0cos2sin2
1
2
1
00
2
0, =====
(2.19)
c. Calculate the output (load) voltage effective (RMS) value VL,RMS
The definition for the effective value is to be applied, in order to determine the value
of the effective (RMS) value of the load:
( ) ( ) ( ) ( ) mmLRMSL VtdtVtdtvV 5.0sin2
1
2
1
0
22
2
0
2, ===
(2.20)
Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
2-9
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Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
2-11
Figure2.3Half-Wave Rectifier with resistive load
Example 2-2
For the circuit from Figure 2.1, lets consider: VVS 120= , = 100R , Hzf 60= .
Determine:
a. The output (load) voltage average value VL,DC
b.
The output (load) voltage effective (RMS) value VL,RMS
c. The output (load) current average value IL,DC
d.
The output (load) current effective (RMS) value IL,RMS
e. The efficiency of the rectifier
f. The form factor FF
g. The ripple factor RF
h.
The transformer utilization factor TUFi.
The peak inverse voltage of the diode PIV
j. The effective (RMS) value of the diode current
k. The crest factor of the input current CF
l. The input power factor PF
m. The harmonic content of the output (load) voltage
Solution: For a complete illustration, simulated results are presented in Figure 2.5 and
Figure 2.6.
a.
The output (load) voltage average value is calculated as:
( ) ( ) ( ) ( ) VVV
tV
tdtVtdtvV mmm
mLDCL 97.53318.0cos2
sin2
1
2
10
0
2
0
, ======
b.
The output (load) voltage effective (RMS) value is calculated as:
( ) ( ) ( ) ( ) VVtdtVtdtvV mmLRMSL 85.845.0sin2
1
2
1
0
22
2
0
2, ====
c. The output (load) current average value is calculated as:
VLvPvS
R
iL
D
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( ) ( ) ( ) ( ) AR
V
R
Vt
R
VtdtItdtiI
mmmmLDCL 54.0318.0cos
2sin
2
1
2
10
0
2
0
, ======
d. The output (load) current effective (RMS) value is calculated as:
( ) ( ) ( ) ( ) AR
VtdtV
Rtdtv
RI
mmLRMSL 8485.05.0sin
2
1
2
1
0
222
0
2, ====
e. The efficiency of the rectifier is equal to:
%45.401001
5.05.0
318.0318.0
100100[%],,
,,
,
,=
===
RVV
R
VV
IV
IV
P
P
mm
mm
RMSLRMSL
DCLDCL
ACL
DCL
f.
The form factor FF is:
57.1318.0
5.0
,
,===
m
m
DCL
RMSL
V
V
V
VFF
Figure2.4 Half-Wave Rectifier with resistive load: Voltage across the Load,
Voltage Across the Diode (anode to cathode) and Load (Diode)
Current(Simulations)
Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
2-12
8/10/2019 Single Phase Rectifiers
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g. The ripple factor RF is:
21.1157.11 22
,
,==
==
DCL
RMSL
DC
AC
V
V
V
VRF
h.
The transformer utilization factor TUF
286.05.0
2
318.0318.0
,,=
===
R
VV
R
VV
IV
IV
IV
PTUF
mm
mm
SS
DCLDCL
SS
DC
Note: The last value justifies the fact that we need a transformer with
5.31
TUF
apparent power than the value of the load DC power
i. The peak inverse voltage of the diode PIV, from the waveforms appears as:
mVPIV=
j. The effective (RMS) value of the diode current is:
AR
V
R
VII m
RMSL
RMSLRMSDIODE 8485.05.0,
,, ====
Figure2.5 Half-Wave Rectifier with inductive load: Voltage across the Load,
Voltage Across the Diode (anode to cathode) and Load (Diode)Current(Simulations)
Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
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Note: The last two calculations represent useful values when wanting to determine
the ratings of the diode used to build the rectifier with. In fact, when choosing a diode
from a manufacturers catalog, we have to impose the following conditions:
)log,, cataRMSDIODERMSDIODE
II andlogcata
PIVPIV
k. The crest factor of the input current is given by:
25.0
,max,====
R
V
R
V
I
I
I
ICF
m
m
S
peaks
S
s
l.
The input power factor can be calculated as following:
707.02
1
5.0
2
5.05.0
cos
,
,1 =====
R
VV
R
VV
IV
R
VV
IV
IVPF
mm
mm
SS
RMSL
RMSL
SS
SS
m. The harmonic content of the output voltage can be determined using Fourier
Series:
(2.23)( ) ( )
=
++=1
, sinsin
n
nnDCLL tnbtnaVtv
The average (DC) component was previously determined: VVV mDCL 97.53318.0, ==
( ) ( ) ( )
=
==== 5,4,3,2:,0
1:,2sinsin
1sin
1
0
2
0 nfor
nforVttdntVttdntva
m
mLn
(2.24)
( ) ( ) ( )
=
==== 8,6,4,2:,
1
2
7,5,3,1:,0
cossin1
cos1
20
2
0nfor
n
V
nfor
ttdntVttdntvb mmLn
(2.25)
Finally, from (2.23), (2.24) and (2.25), well assemble the output voltage as a
harmonic (Fourier) series:
( ) ...8cos63
26cos
35
24cos
15
22cos
3
2sin
2+= t
Vt
Vt
Vt
Vt
VVtv
mmmmmmL
(2.26)
For the data in Example 2.2, we can express (2.26) as:
( ) ...10cos09.18cos71.16cos09.34cos2.72cos36sin85.8497.53 += tttttttvL
Note: For determining the coefficients of the Fourier series, the following calculationsna
have been performed:
For , we have1=n ( ) ( ) ( )24
2sin
2sinsinsin
00
2
0
=
==
tttdtttdnt
Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
2-14
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For will apply the trigonometric identities:,...6,5,4,3,2=n
( )[ ] ( ) tnttnttnttn sinsincoscoscos1cos =+= +
(2.27)
( )[ ]
( ) tnttnttnttn , sinsincoscoscos1cos =+=+
They show to the following integral calculations:
( ) ( ) ( )[ ] ( ) ( ) ( )[ ]
( )
Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
2-15
( ) ( )[ ]
( )[ ]
( ) ( ) Ittdnt
nn
nttdnt
n
tnttdnttdtnttdntttdnt
==+
++
+=
=+
+=+=
00
0000
coscos1
0sin
1
1sincoscos
...01
1sincoscos1coscoscossinsin
Now, if considering: , then( )[ ] ( ) ( ) ( ) =+=
000
2sinsincoscos1cos Ittdntttdnttdtn
because: ( )[ ] ( ) ( )
( )( )( ) ( )
001
0sin
1
1sin01
1sin1cos
0==
=
= Inn
tn
n
tntdtn
Now was clarified that ( ) ,0sinsin1
0
== ttdntVa mn
when ....6,5,4,3,2=n
For determining the coefficients of the Fourier series, the following trigonometricnb
identities are available:
( )[ ] ( ) tnttnttnttn sincoscossinsin1sin =+= +
(2.28)( )[ ] ( ) tnttnttnttn sincoscossinsin1sin =+=+ +
For , we have and after calculating:1=n ( ) ( )[ ] ( ) ( )tdttntdtnttdnt
+=0
cossin
0
1sin
0
cossin
( )[ ] ( ) ( )[ ]
( )
( )
( )
( )
( )
=
+==+
++
+=
+
+=+
....10,7,6,4,2,1
2
9,7,5,3.,1,01
0cos
1
1cos01
1cos1sin
0
nn
nnn
n
n
tntdtn
(2.29)
So, forodd order numbers, ( )[ ] ( ) ( ) ( ) Kttdtnttdnttdtn ===+
000
cossincossin01sin
On the other hand:
( )[ ] ( ) ( )[ ]
( ) ( ) ( ) 02sincoscossin0
01
1cos1sin
000
==+==
= KKttdntttdntn
tntdtn
For even numbers, can ( )[ ] ( ) ( ) ( ) ( )tdtntttdntntdtn
+=+=+ 000sincoscossin1
2
1sin . This
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contributes to express one term under integration with respect to the another like in the next
identity:
( )( )
( )tdtntn
ttdnt
+=00
sincos1
2cossin (2.30)
Furthermore, when integrating the first trigonometric identity from (2.29), then obtain:
( )[ ] ( ) ( )[ ]
( ) ( )
Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
2-16
1
201
1cos1sin
0
=
= nn
tntdtn
(2.31)
From (2.30) and (2.31) applying the identity, we cab express the value of the following
integral as:
( ) ( ) ( ) ( )
( )200
1
2cossincossin2
1
2
1
2
ntdtnttdtnt
nn =
+=
(2.32.)
All of the relationships (2.27) through (2.32), explain the structure of the harmonicdecomposition of the load (output) voltage of the single phase half wave rectifier with
resistive load
2.1.1.3. Half-wave uncontrolled rectifier with resistive inductive load and freewheeling
diode
Figure2.6. Half-Wave Rectifier with resistive-inductive load and freewheeling diode
a. Explain the Operation
For the rectifier from figure 2.7, having the output connection across a resistive-
inductive load, an anti-parallel (freewheeling) diode is connected across the load. The
conduction interval of the diode D is diminished to the half of the cycle.
The freewheeling diode (FWD), actually, is preventing the appearance of the negative
voltage across the load.
When the power supply polarity is changing ( ) =t , the main diode will become
negatively polarized, and due to the magnetic energy accumulated in the inductor, the load
current will find a path through the loop composed by the free wheeling diode and the load.
During the second half of the cycle, when the power supply voltage becomes
negative, across the load will be applied the freewheeling diode voltage drop, which can beneglected. The overall output (load) voltage average value is increased.
FWD
D
VLvPvS
vR
vXL
R
L
iL
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The load current can stay continuous or can be discontinuous depending upon the load
constantR
L .Calculating the load current becomes an important issue in order to be able to
predict all the performance parameters. The power supply voltage is described by (2.1).
In the first half of the cycle, the ideal diode D will let the power supply voltage to
be applied across the load, and equation (2.2) describes the circuit.
In the second half of the cycle, the output (load) voltage dropped to zero:
dt
diiRiv LLLL +== 0 (2.33)
The operation of the circuit is cyclical, so the following conditions must be imposed
for determining the load current:
( )( )
+
+==+
L
R
L
R
mLL
e
e
RL
VIkI
1
11
sin12
222max, (2.34)
( )( )
+
+==+
L
R
L
R
L
R
mLL
e
ee
RL
VIkI
1
1sin
12222
max, (2.35)
It is to remind that
=
R
L arctan . In the above mentioned expressions, krepresents an integer
constant. For conditions like (2.34) and (2.35), the stabilized load current can be expressed
as:
( ) ( ) ( )
( ) ( ) ( )
++
++
=
+
+
+
+
+=
1212,1sin
122,
1
sinsin
222
222222
ktkt
L
R
eL
R
e
RL
Vti
ktke
e
e
RL
Vet
RL
Vti
mL
tL
R
L
R
L
R
mt
L
R
mL
(2.36)
We considered k a multiplication number, which show the fact that the rectifier operates
under steady state conditions, following the first few tens of cycles.
Note: Stabilized (steady state) current means that the rectifier has been operating for asignificant number of cycles.
Note: Solving the equations (2.33.) and (2.34) is the key point for further calculations and for
this purpose will use Laplace Transform, when the initial load current is considered:( )0LI
( ) ( ) ( ) ( ) ( ) ( ) ( )
+
+=+=
L
LLLLLL
TsL
LIsVsILIsIsLRsV
1
00 , with
=
tR
LTL
0
(2.37)
( ) ( ){ }22
+==
s
VtvLsV
mL (2.38)
Finally, the operational load current is expressed as:
Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
2-17
8/10/2019 Single Phase Rectifiers
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( )( )
( ) ( )
+
+
+
+=
+
++=
L
L
L
m
L
L
Lm
L
TsL
LI
TsL
s
V
I
TsL
LIs
V
sI1
0
10
1
02222
(2.39)
( ) ) ( )
( )
++
++=
L
LmL
TssL
IsLVsI
1
0
22
22
(2.39)
For the second half of the cycle 2 t .( )
Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
2-18
( ) ( ) ( ) ( )
+
=+=
L
LLLL
TsL
LIsILIsIsLR
10
(2.40)
For finding the load current in time domain, we have to express the operational
current from (2.39) as a sum of simple fractions. Once determined, the coefficientsA,Band
C, will give the possibility to find out the time expression of the current by applying Laplace
inversions:
( )
+
++
+
=
L
L
Ts
C
js
B
js
A
LsI
1
1
(2.41)
Bringing to a common denominator (2.39) and (2.41) and imposing to have the same
numerator, a three-equation system is composed:
( )
( ) ( )
( ) ( )
+=+
=++
=++
220
01
0
Lm
L
L
L
LIVCBAT
j
BAT
BAj
LICBA
(2.42)
Solving the above system, we found:
,
=
jT
j
VB
L
m
12
and ( )2
2
10
+
+=
L
mL
T
VLIC (2.43)
+
=
jT
j
VA
L
m
12
After introducing (2.43) into (2.41), the load current in Laplace form is changed into:
( ) ( )
+
+
+++
+
+
=
LL
mL
L
m
L
mL
Ts
T
VLI
js
Tjj
V
js
Tjj
V
LsI
1
1
10
1
12
1
12
1
2
2
(2.44)
From (2.44) by inverting the Laplace elementary fractions, we obtained:
( ) ( )
+
++
+
+
= LT
t
e
T
VLI
tje
jT
j
Vtje
jT
j
V
Lti
L
mL
L
m
L
mL
2
2 10
12
12
1 (2.45)
Applying the definitions of the complex sine and cosine functions,
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( )2
costj
etj
et
+= and ( )
j
tje
tje
t2
sin
= , after some intermediate calculations will
obtain:
( ) ( )
Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
2-19
LT
t
L
mL
L
m
L
mL e
T
VI
T
tL
VtLTVti
+++
+
=
2
2
2
2 10
11cossin (2.45)
Finally, by performing the calculations, we obtain:
( ) ( ) ( ) tL
R
Lm
L eIt
L
R
et
RL
Vti
+
++
= 0sinsin222
(2.46)
Previously, we described the load as having the power factor angle
=
R
L arctan
In a similar manner, when 2 t , after applying the Laplace transformation,
when describing the operational current we find:
( )( )
( )( ) ( )
tL
R
eIti
Ts
I
sLR
LIsi LL
L
LLL
=
+
=+
=1
(2.47)
When considering the rectifier operating for sufficient time, then we can express the
current for ( ) 12 += kt as:
( )( )
L
R
e
RL
V
k
L
R
e
L
R
e
L
R
e
RL
Vki
mm
kL
+=
+
+
+=+
1
1sin
12
12
1
1sinlim12
222222 (2.48)
When kt 2= , then:
( )
L
R
e
RL
L
R
eVL
R
e
k
L
R
e
L
R
e
L
R
e
RL
Vki
mm
kL
+
=
+
+
+=
1
1sin
12
12
1
1sinlim2
222222(2.49)
b. Calculate the output (load) voltage average value VL,DC
To calculate it, we apply the definition and obtain a similar value like for the half-wave, uncontrolled rectifier with purely resistive load (2.19). This is due to the presence of
the free-wheeling diode which closes the current path for the second half of the cycle when
the power supply voltage comes negative.
c. Calculate the output (load) voltage effective (RMS) value VL,RMS
The effective value is determined exactly like for the half-wave, uncontrolled rectifier
with purely resistive load (2.20).
d. Calculate the output (load) current average value IL,DC
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( ) ( ) ( ) ( ) ( ) ( )( ) ( )
++
+
+
+==
0
2
222
2
0
, 122sinsin2
1
2
1tdekItdekIet
RL
VtdtiI
tL
R
L
tL
R
L
tL
R
mLDCL
Considering the values determined in (2.48) and (2.49), we obtain:
+
+
+=
L
R
eL
R
eR
L
RL
VI mDCL
2
2sincos22 222
, (2.50)
e.
The output (load) current effective (RMS) value IL,RMS:
Figure2.7 Half-Wave Rectifier with Resistive inductive load and Free-wheeling
diode: Power Supply Voltage, Load Current, Load Voltage, Voltage
across the Inductive Reactance, Voltage Across the Diode (anode to
cathode)and Voltage across Free-Wheeling Diode(Simulations)
Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
2-20
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Figure2.8 Half-Wave Rectifier with Resistive Capacitive Load: Power Supply
Voltage, Load Voltage, Diode Current, Capacitor Current and
Resistor Current (Simulations)
Dr. Sorin Deleanu Dr. David Carpenter
Power Electronics. Theory and Applications
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Dr. Sorin Deleanu Dr. David Carpenter2-22
Bibliography:
[1] Agrawal, J. P, Power Electronic Systems. Theory and Design, Prentice Hall, Upper
Saddle River, New Jersey, Columbus, Ohio, 2001
[2] Rashid, M.H. Power Electronics Circuits, Devices and Applications, Pearson
Prentice Hall, Upper Saddle River, NJ 07458, 2003
[3] Deleanu, S. Contribuii Privind Acionarile Electrice de Curent Alternativ in
Traciune. Tezde Doctorat, Universiatea POLITEHNICA, Bucureti, ROMANIA,
Iulie 2001 (In Romanian).
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