Single-Phase Half-Wave Rectifier

ECE 442 Power Electronics 1

Single-Phase Half-Wave Rectifier


Waveforms


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 – Vdc

2

• Form factor, FF– FF = Vrms/Vdc

• Ripple factor, RF– RF = Vac/Vdc



• 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


Input Voltage and Current



• 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



• Power Factor, PF

1 1cos coss s s

s s s

V I IPF

V I I



• Crest Factor, CF

( )s peak

s

ICF

I


Example 3.1

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


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


1

2

0.318

0.318

m

dc m

dc m

dc

fT

f

VV V

V VI

R R


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


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


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


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


Determine the PIV

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

-

+

+- PIV


Determine CF

( )

( )

0.5

20.5

s peak

s

ms peak

ms

m

m

ICF

I

VI

RV

IRV

RCFV

R


Determine PF

2

cos

(0.5 )

0.7070.5

(0.707 )( )

ac

m

mm

PPF

VA

V

RPFV

VR


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

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


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

2

0

2sin

2

0.636

T

dc m

mdc

dc m

V V tT

VV

V V


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

Single-Phase Half-Wave Rectifier

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