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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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