Single-Phase Half-Wave Rectifier
Waveforms
Single-Phase Half-Wave Rectifier
Performance Parameters
• Average value of the output voltage, Vdc
• Average value of the output current, Idc
• Output dc power, Pdc
– Pdc = VdcIdc
• rms value of the output voltage, Vrms
• Output ac power, Pac
– Pac = VrmsIrms
Performance Parameters (continued)
• Efficiency, η
– η = Pdc/Pac
• Effective (rms) value of the ac component
of the output voltage, Vac
– Vac = Vrms2 – Vdc2
• Form factor, FF
– FF = Vrms/Vdc
• Ripple factor, RF
– RF = Vac/Vdc
Performance Parameters (continued)
• Alternate form for ripple factor
V
R F  ( )  1  FF  1
V
rms
2
2
dc
• Transformer utilization factor, TUF
– TUF = Pdc/VsIs
– Vs, Is are rms voltage and current of the
transformer secondary
Input Voltage and Current
Performance Parameters (continued)
• Displacement angle, Φ
• Displacement Factor, DF
– DF = cos(Φ)
• Harmonic Factor, HF
 I

I I
HF  (
)  ( )  1
I
I

2
2
s
s1
1
2
s
2
s1
s1
2
1
2
Performance Parameters (continued)
• Power Factor, PF
VI
I
PF 
cos   cos 
VI
I
s
s1
s1
s
s
s
Performance Parameters (continued)
• Crest Factor, CF
CF 
I
s ( peak )
I
s
Example 3.1
• Determine η, FF, RF, TUF, PIV of the
diode, CF of the input current, input PF.
Determine the Average Voltage, Vdc
1
V 
T
1
V 
T
dc
dc
T
v
0
L
(t )dt
T
2
V
0
m
sin tdt
V
T
V 
(cos
 1)
T
2
m
dc
1
f 
T
  2 f
V 
dc
V
m

 0.318V
V
0.318V
I 

R
R
dc
dc
m
m
Determine the rms Voltage, Vrms
V
1



 v (t )dt
T

T
1
2
2
rms
0
L
1


V 
 (V sin t ) dt
T

V
V 
 0.5V
2
V
0.5V
I 

R
R
T
2
rms
0
2
m
m
rms
m
rms
rms
m
1
2
Determine Pdc, Pac, and η
(0.318V )
P 
R
(0.5V )
P 
R
(0.318V )

 40.5%
(0.5V )
2
m
dc
2
m
ac
2
m
2
m
Determine FF and RF
V
FF 
V
rms
dc
0.5V

0.318V
m
m
FF  1.57  157%
R F  FF  1
2
R F  1.57  1  1.21  121%
2
Determine the TUF
1
2
V
1


 0.707V
V 
 (V sin t ) dt 

T
2
0.5V
I I 
R
(0.318V )
P
R

T UF 
0.5V
VI
)
(0.707V )(
R
T UF  0.286
T
2
s
m
m
0
m
s
load
2
m
dc
s
m
s
m
m
Determine the PIV
• PIV is the maximum (peak) voltage that
appears across the diode when reverse
biased. Here, PIV = Vm.
-
+
-
PIV +
Determine CF
CF 
I s ( peak )
Is
Vm
I s ( peak ) 
R
0.5Vm
Is 
R
Vm
CF  R  2
0.5Vm
R
Determine PF
Pac
PF  cos  
VA
2
(0.5Vm )
R
PF 
 0.707
0.5Vm
(0.707Vm )(
)
R
Summary – Half-Wave Rectifier
• RF=121%
High
• Efficiency = 40.5 Low
• TUF = 0.286
Low
– 1/TUF = 3.496
– transformer must be 3.496 times larger than
when using a pure ac voltage source
Half-Wave Rectifier with R-L Load
Waveforms of Current and Voltage
Conduction period of D1 extends beyond ωt = π
Average Output Voltage
Vm
Vdc 
2
 

sin td (t )
0
Vm
 
Vdc 
  cos t 0
2
Vm
Vdc 
1  cos(   )
2
Vdc
I dc 
R
Increase average voltage and current by making σ = 0
Waveforms with Dm installed
Application as a Battery Charger
Diode conducts for vs > E,
starting when Vmsinα = E
Waveforms for the Battery Charger
Diode turns off when
vs < E (at β = π – α)
Charging current
io = (vs – E)/R
io = (Vmsinωt – E)/R
for α < ωt < β
Single-Phase Full-Wave Rectifier
Center-Tapped Transformer
Waveforms for the Full-Wave Rectifier
T
2
2
Vdc   Vm sin t
T 0
Vdc 
2Vm

Vdc  0.636Vm
Single-Phase Full-Wave Rectifier
PIV = 2Vm
Full-Wave Bridge Rectifier
Waveforms for the Full-Wave Bridge
Full-Wave Bridge with Waveforms
Conduction pattern
D1 – D2
D3 – D4
PIV = Vm