CHAPTER THREE
DIODE RECTIFIERS
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 transformer 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:
3.4.2: Three-Phase Full-Wave Rectifier
Figure 3.10-a illustrates three-phase full- wave rectifier energized R load, where six diodes (
D1, D2, D3, D4, D5, & D6) operates in six groups, each group for a time of 60°, while each
diode operates for 120°. The operation sequence is determined by the criteria : " the diode with
maximum positive voltage applied across it's terminals will conduct, and the diode with
maximum negative voltage applied across it's terminals also will conduct ", therefore each
phase will pass the current for 240°, where for the rest of time this phase will be off and doesn't
participate in the rectification process.
Fig.3.10-a: Three-phase Full- wave rectifier
Fig.3.10-b: Three-phase voltage, and output rectified voltage
3.10-c: Phase current, and instantaneous rectified current.
Fig.3.10-d: Diode voltage
Fig.3.10-e: Output voltage , rectified current, and phase current at RL load.
The effect of circuit
inductance L on the
output voltage.
Fig.3.10- f: The output voltage with illustrated effect of the circuit inductance on this voltage.
The rectifier mathematical equations are going to be described by discussing the
following example :
Example 3.5:
Three- phase bridge rectifier with the following parameters :
VDC=750V; IDC=1000A; 50 HZ; Wye connected; the voltage reduction due to circuit inductance
is 5% of the output average voltage.
Determine : 1- the load resistance, and secondary current.
2- the diode average, rms current, and PIV.
3 - the transformer utilization.
4- the form factor and ripple factor
5- the circuit inductance.
6- the average voltage reduction, and net output voltage
7 - the average voltage reduction, and net output voltage at 120Hz frequency
Solution:
1. The average, rms voltage and current:
2
2π / 6
V DC =
π/6
∫
V LL .( ω t ) d ( ω t ) =
0
6 π/6
∫ 3 Vm sin( ω t + π / 3 ) d ( ω t ) =
π 0
3 3
π . V DC
π . 750
Vm = 1 . 654 Vm ⇒ Vm =
=
= 453 . 45 V
π
3 3
3 3
V DC
V DC
750
=
⇒ R =
=
= 0 . 75 Ω .
R
I DC
1000
2
2 π/6
6 π/6
=
V
L L .( ω t ) d ( ω t ) =
3
Vm
sin(
ω
t
+
π
/
3
)
d(ωt) =
2 π / 6 ∫0
π ∫0
=
I DC
V RMS
(
2
)
0. 5
IRMS
3 9 3
 = 1 . 6554 Vm = 1. 6554 * 453 . 45 = 750 .64 V
= Vm  +

2
4
π


V RMS 750 . 64
=
=
= 1000 . 854 A
R
0 .75
2. The diode average, rms current and PIV:
I DC
2
π
=
Im L sin
= 0 . 318 Im L
3
π
6
I DC
1000
Im L =
=
= 1048 . 21 A
3 * 0 . 318
0 . 954
2 π /6
π
I RMS
I DR =
Im L sin( ω t + ). d ( ω t ) =
= 577 . 843 A
∫
π 0
6
3
I DAV =
PIV =
3 Vm =
3 . 453 . 45 = 785 . 39 V
3. The transformer utilization:
P DC
TUF =
( VA ) rating
( VA ) rating = 3 . Vs . Is = 3 * 320 . 68 * 818 . 023 = 786 . 98 kVA .
P DC = V DC . I DC = 750 * 1000 = 750 kW
IS =
8
2π
π /6
∫ (Im
=
*
0
Vs = Vm /
∴ TUF
cos( ω t ) ) d ( ω t ) = 0 . 7804
2
L
2 = 320 . 69 V
750 kW
786 . 98 kVA
* 100 % = 95 . 3 %
4. The Form Factor, Ripple Factor and Power Factor:
Im
L
= 818 . 023 A
.
FF =
V RMS
750 . 65
=
= 1 . 00085
V DC
750
RF =
PF =
FF 2 − 1 =
1 . 0008
2
− 1 = 0 . 041
P AC
751 . 36 kW
=
= 0 . 95
( VA ) rating
786 . 98 kVA
5. The Circuit inductance:
∆ V = 6 .fs .I DC .Lc
∆ V = 5 %. V DC . = 5 %. 750 = 37 . 5 V
∆V
37 . 5
∴ Lc =
=
= 0 .125 mH
6 . fs . I DC
6 * 50 * 1000
6- the average voltage reduction, and net output voltage
V DC ( L ≠ 0 ) = V DC − ∆ V = 750 − 37 .5 = 712 .5 V
7- the average voltage reduction, and net output voltage at frequency of 120Hz
∆ V ' = 6 * 120 * 1000
*
0 . 125 / 1000 = 90 V
∴ V DC ' ( L ≠ 0 ) = V DC − ∆ V ' = 750 − 90 = 660 V
Summary:
Taking into account the obtained rectifier parameters we conclude:
1. The output average voltage is 165% of the phase magnitude .
2. High transformer utilization 95.4%, which means that the transformer must be
1/0.954=1.048 times larger that when it is used to deliver power from a pure ac
voltage, i.e. approximately there is no difference between both modes.
3. good rectification efficiency = 99%.
4. There is no dc component in the secondary current, therefore no additional losses
in the transformer core. This reason explain the high value of TUF.
5. Good form factor 100.04%, and Acceptable ripples factor 4.1% 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.
7. The presence of circuit inductance cause additional voltage reduction, as Lc
increases, the net output rectified voltage decreases.
8. As the source frequency increases the voltage reduction being significant.
3.4.3: Three-Phase Full-Wave Rectifier with R-L & E
Figure 3.11-a illustrates three-phase full- wave rectifier energized R- L load with back emf E.
depending on the values of L, E and circuit initial conditions two modes of operation are
defined:
1. continuous mode
2. Discontinuous mode, where this mode cause additional loses, reducing the circuit
efficiency, and mechanical vibration.
Figure 3.10.b illustrates the waveforms showing the effect of above mentioned parameters.
Fig.3.11-a: Three-phase Full- wave rectifier with R-L and E
Fig.3.11-b: the circuit waveforms.
The rectifier mathematical equations are going to be described by discussing the
following example :
Example 3.6:
Three- phase bridge rectifier with the following parameters :
Vab=208V; E=20V; 60 Hz; R=5Ω.
Determine : 1- the steady state load current at I1 at ωt= π /3.
2- the diode average, and rms current.
Solution:
The commutation process is repeated six time per period. Fig.3-11 illustrates the equivalent
circuit of one commutations where D1 and D6 conducts and the equilibrium voltage equation
is expressed as follows:
Fig.3-12: Equivalent circuit when D1,D6 conducts
1. The steady state current
Vab ( ω t ) = R .i + L
di
+E
dt
The phase current obtained by solving the previous first order equation is as follows:
i ( ω t ) = A 1e
Vabm =
Z =
R
2
− Rt
L
+
Vabm
E
sin( ω t − θ ) −
Z
R
3 .Vm
+ (ω L ) 2
Initial conditions : i(ωt=π/3)=I1= i(ωt=2π/3).
∴ I1 = A 1e
⇒ A1
−Rπ
3 .ω .L
+
Vabm
Z
E
Vabm

=  I1 +
−
R
Z

sin(
π
E
− θ) −
3
R
π

sin(
− θ ) e
3

Vabm
E
i( ω t ) =
sin( ω t − θ ) −
+
Z
R
∴I1= ? ….

2π
π
 sin(
− θ ) − sin(
Vabm 
3
3
I1 =
R π

Z
−

1− e L 3

E

 I1 + R

− θ )e
−
− Rπ
3 ω .L
R
−
π
(
− t)
Vabm
π

sin(
− θ )  e L 3ω .
Z
3

R π
L 3ω


− E.
 R


Now Numerical results:
ω L = 0 . 944 Ω ;
Z =
R
2
+ ( ω L ) 2 = 5 . 088 Ω
Vabm
2 . 208
=
= 57 . 813 A
Z
5 . 088
ωL
θ = tan − 1
= 10 . 7 o ;
∴ I1 = 50 . 39 A
R
− 200 ( 2 . 77 * 10 − 3 − t )
∴ i1 ( ω t ) = 57 . 81 sin( ω t − θ ) − 4 + 10 e
.
2- The average and rms current
Im =
4
: Idav = 2 π
π/6
∫ i1( ω t ) = .........
0
Idr =
4
2π
π/6
∫i
0
1
2
( ω t ) = .........