lec 3, parameter performance_10 sept

8/10/2019 LEC 3, Parameter Performance_10 Sept

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Diode Rectifiers and itsPerformance Parameters


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Objectives Understand the operation of diode rectifiers

Understand and examine the performance

parameters of diode rectifiers.

Examine the harmonic distortion of voltage andcurrent on the load and supply caused by dioderectifiers.


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Single-Phase Half-Wave Rectifier

( )

0

1sin

2

m

O dc m

VV V tdt

= =

( )m

O dc

V

IR=

2 2

( )

0

1sin

2 2

m

O rms m

VV V tdt

= =

( )2

mO rms

VI

R=


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DC output voltage is discontinuous and containharmonics.

Input current is not sinusoidal

The performance of this half-wave rectifier withresistive load is examined using performance

parameters

Disadvantages:


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The performance of rectifiers are evaluatedusing the following parameters

The average value of output (load) voltagegiven by Vdc

The average value of output (load) currentgivenby Idc

The output dc powergiven by Pdc = VdcIdc

The rms value of output voltagegiven by Vrms


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The rms value of output currentgiven by Irms

The output ac powergiven by Pac= VrmsIrms

The efficiencyor rectification ratioof a rectifier

given by

dc

ac

P

P=


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The output voltage consists of 2 components,an ac component and a dc component

The effectiveor (rms) value of the accomponent of output voltage is given by

2 2

ac rms dc V V V=


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The form factorwhich is a measure of theshape of the output voltage is given by

The ripple factorwhich is a measure of theripple content is given by

rms

dc

VFF V=

ac

dc

VRF

V=


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By substituting the equation for the effective

value of the ac component of the output

voltage into the ripple factor equation, we can

express the ripple factoras

221 1rms

dc

VRF FF

V

= =


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The transformer utilization factoris defined as

where Vsand Isare the rms voltage and rms

current of the transformer secondary respectively.

dc

s s

PTUF

V I=

The crest factoris a measure of the peak inputcurrent Is(peak) as compared with its rms value IS

and it is defined by

s ( peak )

s

ICF

I=


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vs is the sinusoidal input voltage

is is the instantaneous input current

is1 is the fundamental component of is


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The displacement angle () is the anglebetween fundamental components of input

current and voltage

The displacement factor (DF) or displacementpower factor (DPF) is defined as

DF = cos

The input power factor(PF) is defined as

1 1

s s s

s s s

V I IReal PowerPF cos cos

Apparent Power V I I = = =


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For a pure sinusoidal input current and voltage,input power factor is defined as the cosine of the

load angle (displacement power factor), i.e.

cosReal Powercos

Apparent Power

S S

S S

V IPF

V I

= = =


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The harmonic factor (HF) also known as totalharmonic distortion (THD) is a measure of thedistortion of a waveform. The harmonic factor of

the input current is given as

where both currents are recorded as rms values

1 21 2 2

2 2

1

21 1 1

s s s

s s

I I I

HF I I

= =


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An ideal rectifier should have:

= 100%

Vac = 0

RF = 0

TUF = 1

HF = THD = 0 PF = DPF = 1


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Half-Wave Rectifierwith Inductive Load

Load current extends beyond

the half-cycle until it becomes

zero atwt =

( )

0

1sin (1 cos )

2 2

m

O dc m

VV V tdt

= =


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During diode conduction the circuit is defined by:

sin( )m

diL Ri V t

dt+ =

which yields the load current

0 (rad)t during

{ }sin( ) sin exp( / tan )mV

i t tZ

= +


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where

and the following load current

( )2 2 2 (ohm)Z R L= +

tan / L R

=

0i =

2 (rad)t

during


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The current extinction angle is determined bythe load impedance and can be solved using the

following equation when i = 0 and t=

sin( ) sin exp( / tan ) 0 + =

This is a transcendental equation and solved byiteration techniques


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The figure below can be used to determine angle

for any load impedance angle


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Half-Wave Rectifier with InductiveLoad and Free Wheeling Diode

Inductive load current is characterized: Discontinuous current

High ripple content

Using free wheeling diode D2 will eliminate the

first draw back & second is reduced.


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Prevents negative voltageappearing across the load

Load current becomescontinuous with high inductive

load

( )

0

1sin

2

m

O dc m

VV V tdt

= =


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2sin( ) ( sin ) exp( / tan )m m

O o

V Vi t I t

Z Z = + +

0 (rad)t

The supply current (load current) is given by

during


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{1 exp ( ) / tanO D oi i I t = =

2 (rad)t

The diode current and hence load current is

given by

during


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Single-Phase Full-WaveRectifiers

Two types of full wave rectifiers exist: That formed with a center-tapped transformer and

two diodes and

That formed with or without a transformer and

four diodes, also known as a bridge rectifier.


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Full-Wave Rectifier with Centre-Tapped Transformer


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Each diode conducts on alternate half cycles ofsupply voltage producing a full-wave output

voltage across the load

The average output voltage is given by:

( )

0

22sin2

m

O dc m

V

V V tdt

= =

The peak inverse voltage of the diodes is 2Vm


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Full-Wave Bridge Rectifier


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The average output voltage is given by:

( )

0

22sin

2

m

O dc m

VV V tdt

= =

The peak inverse voltage of the diodes is only Vm


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lec 3, parameter performance_10 sept

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