CHAPTER THREE
DIODE RECTIFIERS
3.2
Single-Phase Half –Wave Rectifier:
Single phase half-wave rectifier is the simplest circuit, this circuit is not used in precise practical
applications due to high voltage ripples, and low efficiency. Therefore discussing this circuit aims
to compare further electrical circuits with this circuit.
Performance Parameters :
3.2.1.
Half Wave Rectifier:
The electrical circuit is shown in fig.3.1-a, where a resistive load is energized throughout this
rectifier, and the waveforms obtained from this circuit is illustrated on fig.3.1-b
fig.3.1-a: Electrical circuit
Fig.3.1-b: Circuit waveforms( source voltage, output voltage , and load current)
As well shown from the figure the diode will conduct only for the positive half wave of the
supply voltage, during the negative half wave the diode is in reverse biasing and there is no
output voltage. During the negative half wave the source voltage is applied across the diode,
therefore the diode must carry the peak value of the source voltage.
There are different types of rectifier circuits and the performances of a rectifier are normally
evaluated in terms of the following parameters:
1. The average value of the output ( load) voltage Vdc, and current Idc:
V DC =
π
1
T
∫ Vm sin( ω t ) d ω t =
0
I DC
=
Vm
π
V DC
R
2. The rms value of the output ( load) voltage Vrms, and current Irms :
Vrms =
π
1
T
∫ (Vm
0
sin ω t ) dω t =
2
Vm
2
Irms = Vrms
R
3. The load average and rms power: These are the power energized the load in form of
DC (average ) and AC (effective ) values. Usually the effective value is greater than the
average value due to output waveform shape.
P
P
= V
= V
DC
AC
. I DC
RMS
. I RMS
DC
4. The Rectifier Efficiency : This parameter characterized the ratio between the average
and effective power, and depends on the rectifier type and configuration:
P DC
P RMS
η=
5. The Transformer Utilization: This parameter characterized the ratio between the average
power and transformer secondary ( source) volt-ampere rating ( VA)rating , and characterized
the ratio between average output power and the appearance power energized the system (
transformer, rectifier, and load):
TUF =
P DC
( VA ) rating
Where (VA)rating= Vs.Is,
Vs= the secondary ( source) rms voltage,
Is=IRMS : the secondary rms current .
6. The Form Factor: This parameter characterized the ratio between the rms and average
voltage (the physical mean of this parameter is the difference between the root mean square of
the signal shape and the average value of this shape, therefore if the shape has pure dc value,
there is no difference and FF=1.
FF =
V RMS
V DC
7. The Ripple Factor: This parameter characterized the difference between the ac
component of the output voltage and dc component of the same voltage :
V AC
RF =
=
V DC
2
 V RMS 

 −1 =
 V DC 
FF 2 − 1
8. The Harmonic Factor: This is a measure of the distortion of a waveform, which
characterized the difference between the total rms ac current ( secondary current Is) and
fundamental component of ac source current, which can be defined by decomposing the
secondary current into Fourier series ( Harmonics Specter):
HF =
 I S − I S1 


IS 

2
2
=
 IS 

 −1
 IS1 
In the case of pure sinusoidal source current Is=Is1, therefore HF=0.
9. The Displacement angle: This parameter characterized the angle(Φ) between the
fundamental current Is1 and the source voltage :
DF = cos φ
10. The Power Factor: This parameter defines the input power factor:
PF =
VS.IS1
IS1
. cos φ = cos φ
VS.IS
IS
As well mentioned in the PF equation the input power factor depends on the load character
and on the source current shape.
11. The Crest factor: This parameter defines the measure of the peak input current (I S)peak
as compared with its rms value I S:
CF =
( IS ) peak
IS
Example 3.1:
Single phase rectifier has a purely resistive load of 10Ω , energized by voltage source of 220V
throughout two windings transformer with ratio 2:1.
Determine : 1- the average and rms voltage and current
2- the efficiency, TUF,
3- FF, RF, and the peak voltage across the diode (PIV).
4- the CF, and the input PF.
Solution:
1. The average and RMS voltage and current:
Vm
110 . 2
V DC =
=
= 49 . 51 V
π
π
;
V DC
49 . 51
I DC =
=
= 4 . 951 A
R
10
Vrms
Irms
Vm
2
Vrms
=
R
=
110 . 2
= 77 . 78 V
2
= 77 . 78 = 7 . 778 A
10
=
2. The efficiency and TUF:
P DC
V DC . I DC
49 . 51 * 4 . 951
245 . 124
η =
=
=
=
* 100 = 40 . 51 %
P AC
V RMS . I RMS
77 . 78 * 7 . 778
604 . 97
P DC
245 . 124
245 . 124
TUF =
=
=
* 100 = 28 . 6 %
( VA ) rating
Vs .Is
110 * 7 . 778
where Is = I RMS = 7 . 778 A .
3. The FF, RF, and PIV
FF =
V RMS
V DC
=
FF
RF
PIV
= Vm
=
2
77 . 78
= 1 . 5709
49 . 51
− 1 =
=
2 Vs
1 . 57
=
2
− 1 = 1 . 21 = 121 %
2 * 110
= 155 . 56 V .
. The CF, and input PF
( I S ) peak
Vm / R
2 * 110 / 10
=
=
= 2
IS
IS
7 . 778
Pac
V S .I S 1
IS1
0 . 5 2 Vm 2 / R
PF =
=
. cos φ =
.1 =
=
( VA ) rating
V S .I S
IS
0 .5 * 0 . 707 Vm 2 / R
CF =
=
0.5 2
= 0 . 707 lag .
0 . 5 * 0 . 707
Summary:
Taking into account the obtained rectifier parameters we conclude that this type rectifier
characterized with bad parameters presented by :
1. Low ( poor) transform utilization 28.6%, which means that the transformer must
be 1/0.286=3.49 times larger that when it is used to deliver power from a pure ac
voltage.
2. Low ( poor) rectification efficiency = 40.5%
3. Presence of current dc component in the secondary current causing additional
losses ( winding and core heating).
4. High ripples 121% greater than that when the source is pure dc
5. High ripple factor, which means that a filter with large capacitance is required
for smoothing the output voltage, therefore this yield high capacitor starting
current problem.
Therefore this type rectifier is rarely used due to the weakness in quality of it's power and
signal parameters.
3.2.2. The effect of freewheeling diode on the output voltage:
When a rectifier energized RL load, the conduction period of the diode D1 will extend beyond
180o until the current becomes zero at ωt=π +Φ .
Figure 3.2 illustrates the electrical circuit consist of RL load ( R=10Ω , L=20mH)
( fig.3.2a), and the obtained simulation performance of the waveforms ( Vs, Vo, Is) where it's
shown the diode will conduct for the time of π +Φ .
(a)
The diode conduction
for 10ms+Φ
Fig.3.2: Electrical circuit(a) and the circuit waveforms.
As well shown from the previous figures it's noted that the diode will conduct in the negative
half cycle for the time of Φ , therefore the average output voltage decreases due to load
inductance.
The average voltage taking into account Φ can be expressed as follows:
V DC =
π+ φ
∫ Vm
1
T
0
φ = tan
−1
sin( ω t ) d ω t =
Vm
(1 + cos φ ),
2π
ωL
.
R
As θ increases, the output average voltage decreases, which the main drawback of
existing inductance in the rectifier circuit.
Avoiding this drawback requires connecting a freewheeling diode Dm across the load as well
shown on fig.3.3(a,b) where the negative portion of the output voltage is eliminated, and the
result is keeping the average voltage at the rated value despite of the existing of L load.
Fig.3.3-a: Electrical circuit with RL load.
The diode conduction
for 10ms+Φ
The diode conduction
for 10ms only.
Fig.3.3-b: Circuit waveform
3.2.3. The effect of back voltage ( charger) the circuit performance:
When the rectifier energized Charging circuit with back voltage E, the conduction period of the
diode D1 will be less than half period (δ <π) depending on the back voltage value. In this case
the diode will conduct when Vs > E.
Figure 3.4 illustrates the electrical circuit energized battery charger with resistance play the role
of current limiter( fig.3.4a), and the obtained simulation performance of the waveforms ( Vs,
Vo, Is) where it's shown the diode will conduct for the time of δ .
The front angle α and back angle β depends on the peak value of the secondary voltage and back voltage. The
diode D1 will turned off when Vs< E.
Fig.3.4-a: Electrical charging circuit ( principle).
The diode
conduction angle δ
Back angle β
Front angle α
Fig.3.4-b: Circuit waveforms
The values of α , β and δ are defined as follows:
β = π− α ;
δ = β −α.
;
The charging current can be expressed :
Vs−E Vm.sinωt −E
io =
=
for α<ωt <β .
R
R
α = sin
−1
E
Vm
Example 3.2:
The battery voltage of fig.3.4a is E=24V and its capacity is 200Wh. The average charging
current be Idc=10A. The primary input voltage is Vp=240V, 50Hz, and the transformer has a
turn ratio of n=2:1.
Calculate: (a) the conducting angle δ of the diode,
(b) the current-limiting resistance R,
(c) the power rating PR of R,
(d) the charging time hO in hours,
(e) the rectifier efficiency η, and
(f) the PIV of the diode.
Solution:
E=12V, Vp=120V, Vs=Vp/n=240/2=120V, and Vm =
(a) the conducting angle δ of the diode:
α = sin
−1
2 Vs = 169 . 7 V ,
E
−1 24
= sin
= 8.13° ;
Vm
169.7
β = π − α = 180 − 8.13 = 171.87°
δ = 171.87 − 8.13 = 163.74°.
(b) the average charging current IDC :
I DC =
=
R =
=
=
1 β Vm . sin ω t − E
dω t,
2 π ∫α
R
1
( 2 Vm cos α + 2 E α − π E ).
2πR
1
( 2 Vm cos α + 2 E α − π E ).
2 π I DC
1
( 2 * 169 . 7 * COS ( 8 . 13 ) + 2
2 π . 10
4 . 2559 Ω .
*
24 * 0 . 1419
− π . 24 )
(c) the power rating PR of R,
PR=R.IRMS2
1 β  Vm . sin ω t − E 

 dω t
2 πR ∫α 
R

2
IRMS =
=
=
1   Vm

2 π R   2
1
2 π10
2


Vm 2
+ E 2  (π − 2α ) +
sin 2 α − 4 Vm .E. cos α 
2


  169 .7 2

169 .7 2
2 

+
24

(
π
−
2
* 0 . 141 ) +
sin 2 * 8 .13 − 4 * 169 .7 * 24 . cos 8 .13 

2

 2

= 6 .772 A
P R = 10 * 6 .772 2 = 458 .704 W
2
(d) the charging time hO in hours:
PDC=E. IDC= 24*10=240W
hPDC =200W; ho= hPDC /PDC= 200/240=0.8334 hrs.
(e) the rectifier efficiency η:
η=
PDC
240
=
* 100 = 34.35%
PDC + PR 240 + 458.70
(f) the PIV of the diode:
PIV=Vm+E =169.7+24193.7 V
Summary:
Taking into account the obtained rectifier parameters we conclude that :
1. The conducting time of the diode is less than π
2. Low ( poor) rectification efficiency 33.34% due to small conducting time
3. Great heating losses due to current limiting resistance
4. The diode must carry total voltage > Vm in reverse biasing.
3.3. Single-Phase Full –Wave Rectifier:
Single phase Full-wave rectifier is the popular circuit, applied in most industrial application due
to good rectifier parameters . There are two types rectifiers
- Full- wave center tap rectifier ( Midpoint)
- Full- Wave bridge rectifier (Gretz) circuit.
Both circuit characterized with identified rectifier parameters, except the secondary current,
efficiency and diode voltage.
Performance Parameters :
3.3.1. Full-Wave Center tap rectifier
The electrical circuit is shown on fig.3.5-a, where a resistive load is energized throughout this
rectifier, and the waveforms obtained from this circuit are illustrated on fig.3.5-b.
Fig.3.5-a: Electrical circuit
Each diode will conduct for half period ( D1- positive cycle, D2 negative cycle).
The output voltage of full wave rectified with less ripples. As well shown from the illustrated
waveforms on fig.3.5-b, the output rectified voltage has a unidirectional form among the full
period, also the voltage applied across the diode is twice the source value.
1. The average value of the output ( load) voltage Vdc, and current Idc:
V DC =
π
2
T
∫ Vm
0
sin( ω t ) d ω t =
2 . Vm
= 0 . 6366 Vm
π
I DC = V DC
R
Fig.3.5-b: Electrical circuit
Fig.3.5-b: Electrical circuit
2. The rms value of the output ( load) voltage Vrms, and current Irms:
Vrms =
π
2
T
∫ (Vm
0
sin ω t ) dω t =
2
Vm
= VS
2
Irms = Vrms
R
Example 3.3:
Single phase center tape rectifier with the 20Ω resistive load, 220V, 2:1 transformer ratio.
Determine : 1- the average and rms voltage and current
2- the efficiency, TUF,
3- FF, RF, and the peak voltage across the diode (PIV).
4- the CF, and the input PF.
Solution:
1. The average and RMS voltage and current:
2 Vm
110 . 2
= 2 *
= 99 . 03 V
π
π
;
V DC
99 . 03
=
=
= 9 . 903 A
R
10
V DC =
I DC
Vm
2
Vrms
=
R
110 . 2
= 110 V
2
= 110 = 11 A
10
=
Vrms
Irms
=
2. The efficiency and TUF:
P DC
V D C . ID C
99 . 03 * 9 . 903
980 . 694
=
=
=
* 100 = 81 %
P AC
V RMS . IRMS
110 * 11
1210
P DC
980 . 694
980 . 694
TUF =
=
=
* 100 = 57 . 3 %
( VA ) rating
2 . Vs . Is
2 * 110 * 7 . 778
where Is = I RMS / 2 = 7 . 778 A .
η =
3. The FF, RF, and PIV
FF =
V RMS
V DC
=
FF
RF
PIV
=
110
= 1 . 11
99 . 03
2
− 1 =
= 2 .Vm
= 2
2
1 . 11
*
2 Vs
− 1 = 0 . 483
= 2
*
2
*
= 48 . 3 %
110
= 311 . 126 V .
4. The diode average and rms current and PIV
I DAV
1 π
I DC
=
Im
sin
ω
t
d
ω
t
=
2 π ∫0
2
I DR =
1
2π
π
∫ (Im sin
ω t) dω t =
0
PIV = Vm =
2
I RMS
2
2 Vs .
5. The CF, and input PF
( IS) peak
Vm / R
Vm / R
=
=
=2
IS
IRMS / 2 Vm / 2 .R
Pac
V S.IS1
IS1
0 . 707 2 Vm 2 / R
PF =
=
. cos φ = . 1 =
=
( VA ) rating 2 .V S.IS
IS
0 .707 * Vm 2 / R
CF =
=
0 . 707 2
= 0 . 707 lag .
0 . 707
Summary:
Taking into account the obtained rectifier parameters we conclude that the rectifier
parameters has been improved comparing with previous case, but not all as follows:
1. Low transform utilization 57.3%, which means that the transformer must be
1/0.573=1.75 times larger that when it is used to deliver power from a pure ac
voltage.
2. good rectification efficiency = 81%
3. No dc component in the secondary current, therefore no additional losses in the
transformer core.
4. Acceptable ripples 111% greater than that when the source is pure dc
5. Acceptable ripple factor 48%.
6. The diode must carry twice voltage in the backward biasing, which the main
drawback of this type rectifier.
7.
Overcoming of this drawback and improving the transformer utilization, force us to use
Full-wave bridge rectifier, as well going to be described hereinafter.
3.3.2. Full-Wave Bridge rectifier
The electrical circuit is shown in fig.3.6-a, where a resistive load is energized throughout this
rectifier, and the waveforms obtained from this circuit are illustrated on fig.3.6-b
fig.3.6-a: Electrical circuit
fig.3.6-b: The main waveforms ( Vs,Vo, Is).
fig.3.6-c: The diode waveforms ( Vd, Id), and load current Io.
With purpose to evaluate the harmonic spectrum of the output voltage, and how can the
higher order harmonics can be predicted or fully eliminated, an example is going to be described
as follows:
Example 3.4:
Finding the Fourier Series of the Output Voltage for a Full-Wave Rectifier energizing R-L load.
Let Vs=110 V, R=10Ω, L=20mH.
Determine : 1- determine the mathematical expression of the output voltage
2- the percentage value of the highest order harmonics.
Solution:
The rectifier output voltage may be described by a Fourier Series as follows:
1- The Fourier series:
Vo ( t ) = V DC +
∞
∑
( A n . cos n ω t + B n sin n ω t ) ,
n = 2 , 4 ,..
where:
1 2π
2 π
2 Vm
Vo ( t ) d ( ω t ) =
Vm . sin ω t d ( ω t ) =
∫
∫
2π 0
2π 0
π
The Fourier coefficients can be found as follows:
1 2π
2 π
An =
Vo
.
cos
n
ω
t
d
(
ω
t
)
=
∫
∫ Vm . cos n ω t d ( ω t ) =
π 0
π 0
∞
4 Vm
−1
=
for n = 2 , 4 , 6 ,..
∑
π
n = 2 , 4 ( n − 1 )( n + 1 )
= 0 for n = 1 , 3 , 5 ,...
1 2π
2 π
Bn =
Vo
.
sin
n
ω
t
d
(
ω
t
)
=
Vm . sin n ω t d ( ω t ) = 0 .
π ∫0
π ∫0
V DC =
Substituting the values of An and Bn, the expression for the output voltage is:
2 Vm
4 Vm
4 Vm
4 Vm
Vo ( t ) =
−
cos 2 ω t −
cos 4 ω t −
cos 6 ω t
π
3π
15 π
35 π
1- The percentage values ….. There are even harmonic numbers only.
Vm=1.414*110=155.54 V
2nd order harmonic is :
4 Vm
4 * 155 . 54
=
= 66 . 013 V
3π
3π
Vo 2
Vo 2 % =
* 100
= 40 . 52 %
Vm
Vo 2 =
This means that this harmonic is characterized with great negative effect and leads to additional
large losses and waveform deformation.
The harmonic speed ( angular frequency) = 2*ω = 2( 2πf) = 2*2 π *50 =628.31 rad/s
Which means that the harmonic frequency is twice the rated i.e. f2=2 fs=100 Hz.
4nd order harmonic is :
4 Vm
4 * 155 . 54
=
= 13 . 202 V
15 π
15 π
Vo 2
Vo 4 % =
* 100
= 8 . 48 %
Vm
Vo 4 =
The negative effect of this harmonic is negligible comparing with previous one.
The harmonic speed ( angular frequency) = 2*ω = 4( 2πf) = 4*2 π *50 =1256.62 rad/s, which
means that the harmonic frequency is 4 times the rated i.e. f4=4 fs=200 Hz.
Now the issue is to eliminate ( reduce) the effect of the strongest harmonic which is in our case
the fourth one. This could be realized by connecting filtering circuit with RC filter as shown on
fig.3.7a:
Fig.3.7a: Single phase full wave rectifier with C- filter
Fig.3.7b: The obtained waveforms illustration the capacitor effect on the output voltage and load
current.
As well shown from fig3.7c, the magnitude of 2nd harmonic has been reduced to 24V, which
equals to 15.4% from the fundamental one.
Fig.3.7c: The Fourier analysis of the output voltage.
3.3.3. Full-Wave Bridge rectifier with RL & EMF:
Bridge rectifier energized RL load with back induced voltage E is illustrated on fig.3.8a, where
the load current flows when the supply voltage being greater than the back voltage E. The circuit
waveform are illustrated on fig.3.8b. The load current has discontinuous character due to the
value of back emf E.
Fig.3.8-a: Bridge rectifier energized R-L- E load
Fig.3.8-b: Circuit waveforms ( Vs, Vo, Io).
By increasing the back induced voltage, the conducting time of the diode decreases with reduced
magnitude as well shown on fig.3.8-c.
Fig .3.8-c: Circuit waveforms ( Vs, Vo, Io) for RL load with back emf.
At the same time keeping E at constant value, increasing the load inductance leads to further
deformation in the output voltage and current as well shown on fig3.8d.
Fig.3.8-d: Circuit waveforms ( Vs, Vo, Io) with high inductive load and back emf..
Summary:
In general Full wave rectifier is characterized with enhanced parameters as follows :
1. High transform utilization 81%, which means that the transformer must be
1/0.81=1.23 times larger that when it is used to deliver power from a pure ac
voltage.
2. good rectification efficiency = 81%
3. No dc component in the secondary current, therefore no additional losses in the
transformer core.
4. Acceptable ripples 111% greater than that when the source is pure dc
5. Acceptable ri pple factor 48%.
6. The diode must carry only the applied voltage in the backward biasing,
7. In the case of RL load and back emf, the current mode ( continuous or
discontinuous), therefore the diode conducting time depends on the load impedance
and phase shift.
8. Single phase rectifiers are used for load up to 15 kW. For larger power output,
three-phase and multiphase rectifiers are used.
Despite the good performances of the rectifier, but there is a need to improve the system
performances, reducing the trans former weight and capabilities, and energized loads with
larger power. This could be achieved by applying three phase rectifiers as well described
in the coming sections:
3.4. Three Phase Rectifiers
Three-phase rectifiers are classified into Half-wave, and Full-wave energized loads with various
impedances and back emf. Applying three-phase rectifiers aims to realize smooth rectified
voltage, increasing efficiency, utilization , and minimizing the parameters of the filter.
3.4.1: Three-Phase Half-Wave Rectifier
Figure 3.9-a illustrates three-phase half wave rectifier energized R load, where three diodes (
D1, D2, and D3) operates in series sequence, each one for a time of 120°. The operation
sequence is determined by the criteria : " the diode with maximum positive voltage applied
across it's terminals will conduct " , therefore each phase will pass the current for 120°, where
for the rest of time this phase will be off and doesn't participate in the rectification process.
Fig.3.9-a: Three-phase half-wave rectifier- Wye connected
Fig.3.9-b: Three-phase voltage, and output rectified voltage.
Fig.3.9-c: Phase current, and instantaneous rectified current.
Fig.3.9-d: Diode voltage
Fig.3.9-e: Phase current, and instantaneous rectified current at RL load.
1. The average, rms voltage and current:
π/q
2
q
π
V DC =
Vm . cos ω t d ( ω t ) = Vm
sin
∫
2π /q 0
π
q
q = 3
in the case of three − phase .
3 3
Vm = 0 . 8269 Vm
2π
V DC
I DC =
.
R
2 π/ q
(Vm . cos ωt )2 d (ωt ) = Vm q  π + 1 sin 2 π 
V RMS =
∫
2π / q 0
2π  q 2
q 
Three − phase ⇒ q = 3
∴ V RMS = 0 .84068 Vm
∴ V
DC
IRMS =
=
VRMS
.
R
2. The efficiency and TUF:
;
P DC
;
P AC
η=
P DC = V DC . I DC =
(0 .8269
Vm
)
2
= 0 . 6837
Vm
R
2
R
2
(
0 . 84068 Vm )
Vm 2
P AC = V RMS .I RMS =
= 0 . 70674
R
R
2
Vm
0 . 6837
P DC
0 . 6837
R
∴η =
=
* 100 % = 96 . 76 %
2 =
Vm
P AC
0 . 70674
0 . 70674
R
P DC
;
(VA )rating
(VA )rating = 3 .V S .I S
TUF =
1 π 1
2π 
 + sin
 = 0 .4854 Im
2π  q 2
q 
0
0 . 4854
Vm 2
∴ (VA )rating = 3 * 0 .707 Vm *
Vm = 1 . 0295
R
R
2
Vm
0 .6733
P DC
R = 0 .6733 * 100 % = 66 %
∴ TUF =
=
(VA )rating 1 .0295 Vm 2 1 .0295
R
π/q
2
2π
IS =
∫ Im
2
cos 2 ω t d ( ω t ) = Im
3. The FF, RF, PF and PIV
FF =
V RMS 0 .8406 Vm
=
* 100 % = 101 . 65 %
VDC 0 .8269 Vm
RF = FF 2 − 1 = 18 .24 %
0 .8406
PF = P AC =
= 0.6844
( VA ) rating 3 * 0 .707 * 0 .4854
4. The average and rms diode current and PIV
I DAV =
I DR =
PIV =
1
2π
π/3
1
2π
∫
Im . cos ω t d ( ω t ) = Im
0
π/3
∫ (Im . cos ω t )
2
1
π
sin
= 0 . 2757 Im
π
3
d ( ω t ) = I RMS /
3
0
3 .Vm =
6 Vs
Summary:
Taking into account the obtained rectifier parameters we conclude:
1. The output average voltage is 82% of the phase magnitude .
2. Satisfied trans former utilization 66%, which means that the transformer must be
1/0.66=1.55 times larger that when it is used to deliver power from a pure ac
voltage.
3. good rectification efficiency = 96.66%.
4. There is a dc component in the secondary current, therefore additional losses in the
transformer core. This reason explain the small value of TUF.
5. Good form factor 101%, and Acceptable ripples factor 18.24% greater than that
when the source is pure dc.
6. The diode must 33% of the total average dc current, 57% of the total rms current
, and must carry 1.73 Vm in the reverse biasing.
With purpose to enhanced the rectifier parameters ( reducing the ripples, increasing TUF,
increasing the rectifier capability, and elimination the dc component in the secondary current) a three-phase
full-wave rectifier is applied as follows: