Single-Phase Half-Wave Rectifier
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ECE 442 Power Electronics 1
Single-Phase Half-Wave Rectifier

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Waveforms

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Single-Phase Half-Wave Rectifier

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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

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Performance Parameters (continued)
• Efficiency, η– η = Pdc/Pac
• Effective (rms) value of the ac component of the output voltage, Vac
– Vac = Vrms2 – Vdc
2
• Form factor, FF– FF = Vrms/Vdc
• Ripple factor, RF– RF = Vac/Vdc

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Performance Parameters (continued)
• Alternate form for ripple factor
• Transformer utilization factor, TUF– TUF = Pdc/VsIs
– Vs, Is are rms voltage and current of the transformer secondary
2 2( ) 1 1rms
dc
VRF FF
V

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Input Voltage and Current

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Performance Parameters (continued)
• Displacement angle, Φ
• Displacement Factor, DF– DF = cos(Φ)
• Harmonic Factor, HF
122 2
1221
2
1 1
( ) ( ) 1s s s
s s
I I IHF
I I

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Performance Parameters (continued)
• Power Factor, PF
1 1cos coss s s
s s s
V I IPF
V I I

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Performance Parameters (continued)
• Crest Factor, CF
( )s peak
s
ICF
I

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Example 3.1
• Determine η, FF, RF, TUF, PIV of the diode, CF of the input current, input PF.

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Determine the Average Voltage, Vdc
0
2
0
1( )
1sin
(cos 1)2
T
dc L
T
dc m
m
dc
V v t dtT
V V tdtTV T
VT

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1
2
0.318
0.318
m
dc m
dc m
dc
fT
f
VV V
V VI
R R

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Determine the rms Voltage, Vrms 12
2
0
12
22
0
1( )
1( sin )
0.52
0.5
T
rms L
T
rms m
m
rms m
rms m
rms
V v t dtT
V V t dtTV
V V
V VI
R R

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Determine Pdc, Pac, and η2
2
2
2
(0.318 )
(0.5 )
(0.318 )40.5%
(0.5 )
m
dc
m
ac
m
m
VP
RV
PRVV

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Determine FF and RF
2
2
0.50.318
1.57 157%
1
1.57 1 1.21 121%
rms m
dc m
V VFF
V V
FF
RF FF
RF

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Determine the TUF12
2
0
2
1( sin ) 0.707
20.5
(0.318 )
0.5(0.707 )( )
0.286
Tm
s m m
m
s load
m
dc
ms s
m
VV V t dt V
TV
I IR
VP RTUF VV I V
RTUF

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Determine the PIV
• PIV is the maximum (peak) voltage that appears across the diode when reverse biased. Here, PIV = Vm.
-
+
+- PIV

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Determine CF
( )
( )
0.5
20.5
s peak
s
ms peak
ms
m
m
ICF
I
VI
RV
IRV
RCFV
R

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Determine PF
2
cos
(0.5 )
0.7070.5
(0.707 )( )
ac
m
mm
PPF
VA
V
RPFV
VR

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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

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Half-Wave Rectifier with R-L Load

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Waveforms of Current and Voltage
Conduction period of D1 extends beyond ωt = π

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Average Output Voltage
0
0
sin ( )2
cos2
1 cos( )2
mdc
mdc
mdc
dcdc
VV td t
VV t
VV
VI
R
Increase average voltage and current by making σ = 0

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Waveforms with Dm installed

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Application as a Battery Charger
Diode conducts for vs > E, starting when Vmsinα = E

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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

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Single-Phase Full-Wave Rectifier
Center-Tapped Transformer

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Waveforms for the Full-Wave Rectifier
2
0
2sin
2
0.636
T
dc m
mdc
dc m
V V tT
VV
V V

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Single-Phase Full-Wave Rectifier
PIV = 2Vm

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Full-Wave Bridge Rectifier

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Waveforms for the Full-Wave Bridge

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Full-Wave Bridge with Waveforms
Conduction pattern
D1 – D2 D3 – D4
PIV = Vm