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```PALESTINE POLYTECHNIC UNIVERSITY
(PPU)
POWER
ELECTRONICS
Spring
2003 / 2004
2005/2006
Rectifier Classification
Chapter 3-A : Single Phase
Rectifiers
Chapter 3-B Three-Phase Rectifiers
Power Electronics
Chapter 3
Uncontrolled Rectifiers
Un controlled
Rectifiers
Single-Phase
Rectifiers
Three-Phase
rectifiers
Half-Wave
Full-Wave
Power Electronics
Chapter 3 : A
Single –Phase Uncontrolled Rectifiers
Single-Phase
Rectifiers
Half-Wave
“HW”
Full Wave
A: s1_1
“FW”
Bridge circuit
Center tape
Power Electronics
Chapter 3 : B
Three –Phase Uncontrolled Rectifiers
Three-Phase
Rectifiers
Half-Wave
“HW”
Full Wave
A: s1_1
“FW”
Three-Phase Half Wave Uncontrolled
I- With Resistive Load
Principle of operation :
1-Each diode must conduct for 120 dg while
the anode voltage is maximum positive
comparing with the other anode voltages .
2- Each phase voltage is connect to the load
for the time of 120 dg.
3-The source ( phase) current is unsymmetrical
Ic
because it’s flow only during the positive half
cycles .
Conclusion :
Vout
Vdiode
PIV
1- Low ripples , comparing with single-phase
rectifier
2- Relatively acceptable efficiency and TUF 74
%
3-There is a dc component in the source current
(heavy saturated transformer)
4- The output ripples are three-times the supply
frequency .
5- The diode inverse voltage is 1.731 Vm .
II- Three Phase Rectifier with R-L Load
The existing of inductance in the rectification circuit
( supply transformer & load inductance),
1- Voltage reduction in the average output voltage;
2- Current deformation of the output & phase current
3- Increasing the harmonic specter, therefore ,
increasing the harmonic losses .
The output voltage :
Vd c( L 0 )
Vd c( L0)
V
current
3  Idc Lc f
V
In every commutation interval, two diodes operate together for angle  Which called overlapping angle .
currents
III- Three Phase Rectifier with failed diode
voltage
Vdc( d1 ) failed
Vdc0 1
1
 Vx
3
The output
voltage
V
4

6
Vmco s t d  t
T
0
The
diode
voltage
The mathematical equations of HW – Three Phase Rectifier
6 - The source current:
1- The average voltage & current :

Vd c
3

3
Vm
Vm  co s(  t ) d  t 3 3
2

0
Idc
Vdc
R
2- The RMS voltage & current:
Is
3
3
2
( Im. co s t ) d  t 0 .48Im Irms

0
7 - The diode average current :


3
3
Vrms
2
( Vmco s t ) d  t 0 .84 0Vm

1
Idav
3
Im  co s(  t ) d  t 0 .33Idc

0
0
Vr ms
I rm s
8 - The diode rms current :
R

3 - The output average & AC power:
Pdc Vdc Idc
Pac
Vrms Irms

4 - The rectification efficiency:
Pdc
TUF
( VA) rating
3 Vs  Is
2
( Imco s t ) d  t
Irms
0
3
7 - The Form factor :
8- The Ripple Factor :
RF
( VA ) rating
where

3
P dc
P ac
5- The transformer utility factor:
Idr
1
FF
9- Diode PIV :
2
FF 
Vrms
1.015
Vdc
.
1 18 %
PIV Va Vb
3 Vm
Three-Phase Full Wave Uncontrolled
I- With Resistive Load
Principle of operation :
1-Each diode from anode group will conduct for
120 dg while the anode voltage is maximum
positive comparing with the other anode
voltages . And one diode from cathode group
also conduct for 120 dg, while the cathode
voltage is maximum negative .
1- Low
2Each ripples
diodes ,group
comparing
is connect
with toanother
circuits
for a
time
(4%
of ripples),
60 dg. therefore no need of filter
2- Extremely high efficiency efficiency and
3-The source ( phase) current is symmetrical,
TUF > 96%
3-There
thereforeis no
no saturation
dc component
effectin the source
current , therefore minimized losses
4- the supply voltage connected to the load is
4- The output ripples are with six-times the
line voltage .
supply frequency .
5- The diode inverse voltage is 1.731 Vm .
Conclusion
:
6- the phase rms current is 81% of the load rms
value .
7- This circuit find widespread applications in
wide range of the power specter .
II- FW Rectifier with R-L load
current
Output
voltage
Phase
current
Phase
current
The output average voltage:
Vd c( L 0 )
V
Vd c( L0)
6  Idc Lc f
V
Mathematical Modeling of FW – Three Phase Rectifier
6 - The source current:
1- The average voltage & current :


Vd c
6
6

Vab(  t ) d  t 3 3
Vm
0
Idc
Vdc
R

6
8
Is
2

ImL
3
Vm
Irms
4
6
7 - The diode average current :
6

2
( Vab(  t ) ) d  t 1 .65 5Vm
0
6
4
Idav
0
8 - The diode rms current :
3 - The output average & AC power:
Pdc Vdc Idc
Pac
Vrms Irms

4 - The rectification efficiency:
P dc
P ac
5- The transformer utility factor:
Pdc

Idr
3 Vs  Is
6
4
2
( ImLco s t ) d  t
Irms
 0factor :
7 - The 2Form
9- The Ripple Factor :
RF
( VA ) rating
( VA) rating
ImL  co s(  t ) d  t 0 .31 8ImL
2
Vr ms
R
where
Is
R
I rm s
TUF
d  t 0 .78 04ImL
R

6
2
0
2- The RMS voltage & current:
Vrms
Vab(  t )
2
FF
1 4%
10- Diode PIV :
3
FF

Vrms
.
1.0006
Vdc
PIV Va Vb
3 Vm
Single Phase Half-Wave Circuit
S1
D1
DIODE
A: v1_1
V1
-30/30V
Supply voltage
50 Hz
C1
30uF
S2
R1
200
10.00 V
-10.00 V
D2
DIODE
V2
-30/30V
30.00 V
-30.00 V
0.000ms
T1
2TO1
50 Hz
C2
30uF
R2
200
A: s1_2
Output voltage
Without C
30.00ms
60.00ms
90.00ms
30.00ms
60.00ms
90.00ms
30.00ms
60.00ms
90.00ms
30.00ms
60.00ms
90.00ms
30.00 V
20.00 V
10.00 V
0.000 V
-10.00 V
0.000ms
Conclusion :
A: s1_1
Output voltage
With C
30.00 V
20.00 V
10.00 V
0.000 V
1- High ripples , therefore large value of
capacitor
is required
2- Poor efficiency and TUF ~28%--31%
3- Dc component in the source current (
heavy saturated transformer )
4- The output ripples have the same
frequency equals the source frequency .
-10.00 V
0.000ms
A: r1[i]
with C
25.00mA
-25.00mA
-75.00mA
-125.0mA
0.000ms
Single phase Uncontrolled
Bridge rectifiers
S1
1-Electrical circuit without
filtering capacitor
D1
BRIDGE
V1
-30/30V
50 Hz
C1
30uF
A: s1_2
1-Electrical circuit with filtering
capacitor
D1
BRIDGE
V1
-30/30V
R1
200
30.00 V
50 Hz
C1
30uF
A: s1_1
30.00 V
20.00 V
20.00 V
10.00 V
10.00 V
0.000 V
0.000 V
-10.00 V
0.000ms
Conclusion :
30.00ms
60.00ms
90.00ms
R1
200
-10.00 V
0.000ms
30.00ms
60.00ms
1- Low ripples , therefore small value of capacitor
is required
2- Relatively high efficiency and TUF 81 %
3- No dc component in the source current ( no-saturation
Effect in the transformer
4- The output ripples have twice frequency with respect to the source .
90.00ms
The mathematical equations
of FW Bridge Rectifier
- The transformer utility factor:
1- The main parameters :
* rectification output parameters :
- Average output voltage & current:
2
2
T
( VA ) rating
where
Vs  Is
( VA) rating
T
Vd c
Pdc
TUF
Vm  si n(  t ) d  t
Is
Vdc
Idc
( VA) rating
R
0
2 Vs Is
I rms- for FW- center tape
Is
- The RMS voltage:
- for FW- bridge…..
Irms
2
T
2
2
Vm sin t d  t 
T
Irms
Vrms
Vrms
0
- The ripple factor:
R
2
RF
2
FF
- The harmonic factor
- The output average & AC power:
Pdc Vdc Idc
Pac
Vrms Irms
- The rectification efficiency:

HF
Is
Is1
P dc
P ac
2
1
; HF= 1.11
1
The mathematical equations
of HW Rectifier
- The transformer utility factor:
1- The main parameters :
* rectification output parameters :
- Average output voltage & current:
T
Vd c
1
2
T
Pdc
TUF
( VA ) rating
where
Vs  Is
( VA) rating
Vm  si n(  t ) d  t
apparent power
Vdc
Idc
R
0
Is
source current
Irms
- The RMS voltage:
- The form factor :
T
2
2
Vm sin t d  t
Vrms
0
1
Irms
; FF= 1.57
Vrms
- The ripple factor:
R
T
RF
- The harmonic factor
- The output average & AC power:
Pdc Vdc Idc
Pac
Vrms Irms
- The rectification efficiency:

HF
Is
Is1
P dc
P ac
2
1
2
FF
1
Single-Phase Rectifier – Center Tap
R5
500
-10.00 V
45.00ms
60 Hz
30.00ms
D5
12F120
15.00ms
5TO1CT
-20.00 V
0.000ms
V2
D4
12F120
0.000 V
10.00 V
20.00 V
1- Low ripples , therefore small value of capacitor is required
2- Relatively high efficiency and low TUF ~ 57%
3- No dc component in the source current ( no-saturation
Effect in the transformer
4- The output ripples have twice frequency with respect to the source .
5- The diode PIV voltage is twice the supply voltage
A: r5_2
Conclusion :
Mathematical Equations of FW Rectifier – Center Tap
- The transformer utility factor:
1- The main parameters :
* rectification output parameters :
- Average output voltage & current:
T
Vd c
2
2
T
Vm  si n(  t ) d  t
- The form factor :
T
2
2
Vm sin t d  t 
T
Irms
Pac
; FF= 1.11
RF
- The ripple factor:
Vrms
R
- The harmonic factor
HF
- The output average & AC power:
Pdc Vdc Idc
2 Vs Is
I rms
- for FW- center tape
2
Is
- The RMS voltage:
0
where
Vdc
Idc
R
Vrms
( VA ) rating
( VA) rating
0
2
Pdc
TUF
Is1
Vrms Irms
- The rectification efficiency:

Is
P dc
P ac
2
1
2
FF
1
+51 Volt Power Supply
V1
-170/170V
A
Q1
2N2222A
S1
+
2TO1
R1
680
C1
100uF
60 Hz
D3
DIODE
+
D1
18DB10
D4
12F120
D2
1N4757
C2
100uF
R2
50
Without stabilizer
A: r2_2
125.0 V
75.00 V
25.00 V
-25.00 V
0.000ms
10.00ms
20.00ms
30.00ms
With stabilizer
A: r2_2
125.0 V
75.00 V
25.00 V
-25.00 V
0.000ms
10.00ms
20.00ms
30.00ms
Thyristor and Triac Circuits
R2
100
V1
-220/220V
R2
100
R1
10k 20%
B
A
V1
-220/220V
R1
5k 30%
B
D1
DIODE
D1
DIODE
50 Hz
50 Hz
SCR1
2N5064
+
+
MAC210-6
A
C1
0.9uF
C1
0.9uF
Triac
voltage
A: r2_2
200.0 V
A: scr1_1
Thyristor
voltage
100.0 V
0.000 V
A: r2[i]
50.00ms
65.00ms
80.00ms
A: r2[i]
curent
1.500 A
0.500 A
-0.500 A
-1.500 A
A: d1_k
50.00ms
65.00ms
80.00ms
0.500 V
35.00ms
50.00ms
65.00ms
2.250 A
1.750 A
1.250 A
0.750 A
0.250 A
-0.250 A
20.00ms
Capacitor
voltage
A: r1_2
Capacitor
voltage
1.500 V
35.00ms
50.00ms
65.00ms
55.00ms
70.00ms
85.00ms
2.500 V
-2.500 V
-7.500 V
-0.500 V
-1.500 V
35.00ms
-50.00 V
-250.0 V
20.00ms
Triac
current
2.500 A
-2.500 A
35.00ms
50.00 V
-150.0 V
-100.0 V
-200.0 V
35.00ms
150.0 V
-12.50 V
40.00ms
50.00ms
65.00ms
80.00ms
Triac firing circuits
15/-15V
V(A+B)
A
B
50 Hz
UJT
needle
s
C
Q
A: q1_3
100
0/0V
3.00ms
MAC210-6
0/5V
B
20
5.000 V
3.000 V
1.000 V
A
-1.000 V
-3.000 V
-5.000 V
6.000ms
120 Hz
20V
8.000ms
10.00ms
voltage
S1
A: q1_2
12.50 V
7.500 V
5k
150
2.500 V
D
-2.500 V
2N2646
-7.500 V
-12.50 V
5.000ms
0.5uF
15.00ms
47
A: q1_3
25.00ms
.
12.50 V
7.500 V
2.500 V
-2.500 V
-7.500 V
-12.50 V
0.000ms
10.00ms
20.00ms
30.00ms
A: q1_3
Capacitor
voltage
A: c1_2
12.50 V
15.00 V
A: q1_2
Pulse
generator
-2.500 V
10.00ms
20.00ms
5.000 V
-5.000 V
35.00ms
7.500 V
2.500 V
-7.500 V
-12.50 V
0.000ms
25.00 V
30.00ms
voltage
12.50 V
7.500 V
-15.00 V
-25.00 V
0.000ms
12.00ms
2.500 V
-2.500 V
5.000ms
10.00ms
15.00ms
-7.500 V
-12.50 V
0.000ms
10.00ms
20.00ms
30.00ms
```