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Diode rectifiers
- A rectifier is a circuit that converts an ac signal into a unidirectional signal.- A rectifier is a type of ac-dc converter.
Rectifiers are classified according to type of input supply:
Single phase rectifiersThree phase rectifiers
Single phase rectifier
Single phase half-wave rectifiersSingle phase full-wave rectifiers
Single phase half-wave rectifiers classified according to load:
1-Resistive load2 RL load
a- Without free wheeling diodeb- With free wheeling diode
3- EMF load
Performance parameters of a rectifier
Average output (load) voltage, Vdc
Average output (load) current, Idc
Output dc power Pdc=Vdc Idc
rms value of the output voltage, Vrms
rms value of the output current, Irms
Output ac power Pac=Vrms Irms
ac
dc
PP
=ηEfficiency (or rectification ratio) of a rectifier:
The output voltage can be considered as composed of 2 components: (1) the dc value and (2) the ac component or ripple.
- The ripple voltage 22dcrmsac VVV −=
- The form factor is a measure of the shape of output voltage:dc
rms
VVFF =
- The ripple factor is a measure of the ripple content:
dc
ac
VV
RF = or 11 22
−=−⎟⎟⎠
⎞⎜⎜⎝
⎛= FF
VVRF
dc
rms
ss
dc
IVP
TUF =- The transformer utilization factor is defined as
Vs and Is are the rms voltage and current of the transformer secondary respectively
- Crest factor: a measure of peak input current Is(peak) w.r.t. its rms value Is,
s
peaks
II
CF )(=
Half-wave rectifier supplying a resistive load
Figure 1
π
ππ
ωωπ
π
m
m
mdc
V
V
tdtVV
=
+−=
= ∫
))0cos(cos(2
)sin(21
0
RVI m
dc π=
2)sin(
21
0
2 mmrms
VtdtVV == ∫ ωωπ
π
RVI m
rms 2=
%53.40
2*
2
*
**
====
RVV
RVV
IVIV
PP
mm
mm
rmsrms
dcdc
ac
dc ππη
57.12
2 ====π
πm
m
dc
rmsV
V
VVFF
211.1157.11 22 =−=−== FFVVRF
dc
ac
Peak reverse (inverse) voltage (PIV) of diode D1 is Vm
Is(peak)=Vm/R and Is=0.5Vm/R. The CF of the input current is CF=Is(peak)/Is=1/0.5=2
Figure 2
Half-wave rectifier supplying an RL load
Without free-wheeling diode
RVI dcdc /=
β
β
Due to inductive load, the conduction period of diode D1extends beyond π (until the current equals zero at ωt = π+σ).
The average voltage across the inductor is zero.
)]cos(1[2
]cos[2
)(sin2 0
0σπ
πω
πωϖ
πσπ
σπ+−=−== +
+
∫ mmmdc
VtVttdVV
) (sin
:periodon During
solutionstransientstatesteadyesincorporatdtdILIRtVm +⇒+=ω
RLast
ZVI m
SSωφφω 1tan )sin( −=−=
tLR
ts eAI−
= .&
φ=∴==
+φ−ω=+==−
sinZ
VA 0t at 0I :conditions initial from
e.A )tsin(Z
VIIII
mS
tLR
mtsssoS
φσφπβτ ≅+≅⇒<<= :: , ) ( , eiloadsinductivehighlyLRFor
Disadvantages:
1. Discontinuous current
2. High ripple content (FF>1)
3. Presence of DC component in supply circuit.
With free-wheeling diode
Figure 3
- It can be noted from equation (1) that the average voltage (and load current) can be increased by making σ=0, which is possible by adding a freewheeling diode Dm as shown in Figure 3 with dashed lines.
-Diode “Dm” prevents appearing of negative voltage across the load.
-At ωt = π the current from D1 is transferred to Dm ; this process is called commutation of diodes .
- The load current i0 is discontinuous with resistive loads and continuous with highly inductive loads. R
VIVV mdc
mdc ,
ππ==
For highly inductive loads, the current doesn’t reach zero, instead, the current lies between two border values after a few cycles of transients.
]1
1)[sin( 2
maxLR
LR
m
e
eZ
VIωπ
ωπ
φ−
−
−
+=
LR
eII ωπ
maxmin .−
=&
tLR
mm eAt
ZVi
dtdILIRtV
−+−=⇒+=
→→→
. )sin( sin
,.........54,32,0 :periodon During
φωω
πππππ
EMF load
Figure 4
If output is connected to a battery, rectifier operates as battery charger.
EVm
For vs>E, diode D1conducts. The angle α at which diode starts conduction can be found as follows:
=αsinmV
E1sin −=α⇒
Diode D1 is turned off when vs<E at απβ −=
REtV
REv
i mso
−=
−=
ωsinThe charging current io can be found from: For α<ωt<β
Example: The battery voltage in figure 4 is E=12V and its capacity is 100Wh.The average charging current should be Idc=5A. The primary input voltage is Vp=120 V, 60 Hz, and the transformer has a turn ratio of n=2:1. Calculate:
(a) Conduction angle δ of the diode, (b) Current limiting resistance R,(c) Power rating PR of R(d) Charging time ho in hours, (e) Rectifier efficiency η,(f) PIV of the diode.
Solution
VVV
Vn
VVVVVE
sm
psp
85.846022
602
120 120 12
=×==
=====
o
o
o
74.16313.887.17187.17113.8180
13.885.84
12sin 1
=−=−=
=−=
== −
αβδ
β
α(a)
),2cos2(2
1
)(sin
21
EEVR
I
tdR
EtVI
mdc
mdc
πααπ
ωω
πβ
α
−+=
−= ∫
For β=π-α
(b)
Which gives
Ω=×−××+×××
=
−+=
26.4)121419.012213.8cos85.842(52
1
)2cos2(2
1
ππ
πααπ
o
mdc
R
EEVI
R
(c)
2.84.67
4.67
cos42sin2
)2(22
1
)()sin(21
2
22
2
22
2
22
==
=
⎥⎦
⎤⎢⎣
⎡−+−⎟⎟
⎠
⎞⎜⎜⎝
⎛+=
−= ∫
rms
rms
mmm
rms
mrms
I
AI
EVVEVR
I
tdR
EtVI
αααππ
ωωπ
βα
The power rating of R is,
WPR 4.28626.42.8 2 =×=
(d) The power delivered Pdc to the battery is
hP
h
PhWEIP
dco
dco
dcdc
667.160
100100100
60512
===
==×==
(e) The rectifier efficiency is
%32.174.28660
60=
+=
+=
=
Rdc
dc
PPP
powerinputtotalbatterythetodeliveredpower
η
η
(f) The peak inverse voltage PIV of the diode is
VEVPIV m 85.961285.84 =+=+=
Single-phase full-wave rectifiers
-A full-wave rectifier circuit with a center-tapped transformer is shown in figure 5.
-Each half of the transformer with its associated diode acts as a half-wave rectifier and the output of a full-wave rectifier is shown in figure 6. To get the same output as bridge rectifier, the source voltage should be doubled.
The average output voltage is: ∫ ===π
πω
π o mm
mdc VVdttVV 6366.02sin1
Figure 5
tVV mS ωsin2=
Figure 6
During the positive half cycle of the input voltage, the power is supplied to the load through diodes D1 and D2.
During the negative cycle, free-wheeling diodes D3 and D4 conduct.
..)( lno
o
VV
m
oo
VXI
00
1 2
1
Bridge rectifier:
Half-wave with free-wheeling diode
Full-wave bridge
Figure 7
Figure 8
mmm
mdc VVVtdtVV 6366.02))0cos(cos(2
)sin(1
0
==+−== ∫ ππ
πωω
π
π
RV
RVI mdc
dc6366.0
==
mm
mrms VVtdtVV 707.02
)sin(1
0
2 === ∫ ωωπ
π
RVI m
rms707.0
=
%81707.0*707.0
6366.0*6366.0
**
====
RVV
RVV
IVIV
PP
mm
mm
rmsrms
dcdc
ac
dcη
11.16366.0707.0
===m
m
dc
rms
VV
VVFF
482.0111.11 22 =−=−== FFVVRF
dc
ac
Is(peak)=Vm/R and Is=0.707Vm/R.
The CF of the input current is CF=Is(peak)/Is=1/0.707= 2
Single-phase Full-wave rectifier with RLE load
Figure 10
Figure 9
If vs=Vm sin ωt is the input voltage, the load current io can be found from:
tVERidtdiL mo
o ωsin2=++ 0≥oifor
Which has a solution of the form
REeAt
ZVi tLRm
o −+−= − )/(1)sin(2 θω
Where22 )( LRZ ω+=
RLωθ 1tan −=
Vs is the rms value of the input voltage
Case 1 continuous load current
)(
1 sin2 LR
mo e
ZV
REIA ω
π
θ ⎟⎟⎠
⎞⎜⎜⎝
⎛−+=
A1 in equation 3 can be determined from the condition: at ωt=π, io=Io.
, Substitution of A1 in equation 3 yields:
REe
ZV
REIt
ZVi
tL
Rm
om
o −⎟⎟⎠
⎞⎜⎜⎝
⎛−++−=
− )]([sin2)sin(2 ωπ
ωθθω
At steady-state: io(ωt=0)=io(ωt=π). ⇒ io(ωt=π)=Io.
RE
e
eZVI R
R
mo −
−
+=
−
−
)L (
)L (
1
1sin2
ωπ
ωπ
θ
REe
et
ZVi tLR
LRm
o −⎥⎦⎤
⎢⎣⎡
−+−= −
−)/(
)/)(/( sin1
2)sin(2 θθω ωπ
After substituting Io in equation 4 and simplification:
Case 2 discontinuous load current
The load current flows only during the period α ≤ ωt ≤ β.
Define x=E/Vm As (EMF) is constant, diodes start to conduct at ωt=α
)(sinsin 11 xVE
m
−− ==α
At ωt=α, io(ωt)=0 and equation 3 gives
)(
1 )sin(2 LR
m eZV
REA ω
α
θα ⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
Which, after substituting in equation 1, yields the load current
REe
ZV
REt
ZVi
tL
Rss
o −⎟⎟⎠
⎞⎜⎜⎝
⎛−−+−=
− )]([)sin(2)sin(2 ωα
ωθαθω
To solve for io, assume initially the current is continuous and proceed with the solution. If the assumption is not correct, the load current is considered equal to zero (discontinuous current equation), and β is found accordingly.
Fourier analysis of single-phase full-wave rectifier output:
∑∞
=++=
,..3,2,1)sincos()(
nnndco tnbtnaVtV ωω
Even function β ⇒ bn=0 & repetitive every π ⇒ n=2,4,6,8,….
∑∞
=+=
,....6,4,2cos)(
nndco tnaVtV ω
πωω
πωω
π
ππm
mmdcVtdtVtdtVV 2.sin1.sin
21
0
2
0=== ∫∫
∞=+−
−== ∫ ,.....,8,6,4,2n, )1)(1(
4.cos.sin1 2
0 nnVdttntVa m
mn πωω
π
π
..........4cos1542cos
342)( −−−= tVtVVtV mmm
o ωπ
ωππ
The second harmonic (with frequency 2f is dominant)
3 Phase half-wave rectifier
Disadvantages:
1. DC current component in supply
2. Low source power-factor
(poor utilization:1/3 cycle diode operation).
3. Peak inverse voltage of diodes = line voltage
Load
mLoadsource
sourcermsLoad
RVII
II acbacbaLoad IIIIIIII
33
332222
_
==∴
=== ++⇒++=→→→→
To get source current Ia or Ib or Ic
[ ] mmmmdc VVtVtdtVV 827.03
sin3sin3)(cos3/2
2 3/
0
3/0 ==== ∫
π π ππ
ωπ
ωωπ
RV
RVI mdc
dc827.0
==
RVI m
rms84068.0
=mmrms
mmrms
VVV
ttVtdtVV
84068.03
2sin21
323
42sin
23)(cos
3/22
21
21
3/
0
21
3/
0
22
=⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ +=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎥⎦⎤
⎢⎣⎡ +=⎥⎦
⎤⎢⎣⎡= ∫
πππ
ωωπ
ωωπ
ππ
%77.9684068.0*84068.0
827.0*827.0
**
====
RVV
RVV
IVIV
PP
mm
mm
rmsrms
dcdc
ac
dcη
0165.1827.0
84068.0===
m
m
dc
rms
VV
VVFF
1824.010165.11 22 =−=−== FFVVRF
dc
ac
6647.03P
PT.U.F.)( source
dc ===SS
dcdc
IVIVfactornutilizatiorTransforme
Three phase Bridged Rectifier
The diode behaves as a window, permitting the load to “see” the available line voltage.
Advantages:
1. Bi-directional source current;
(no source dc component)
1. Higher diodes utilization
2. Higher source power-factor
The average output voltage is found from:
mm
mdc
VV
tdtVV
654.133
cos36/2
2 60
==
= ∫
π
ωωπ
π
Where Vm the peak phase voltage.
The rms output voltage is:
mmmrms VVtdtVV 6554.14
3923cos3
6/22 2
12
16
0
22 =⎟⎟⎠
⎞⎜⎜⎝
⎛+=⎥⎦
⎤⎢⎣⎡= ∫ π
ωωπ
π
RVI mm /3=If the load is purely resistive, the peak current through a diode is:
The rms value of the diode current is:
mmmd IItdtII 5518.06
2sin21
61cos
24 2
12
1
60
22 =⎥⎦⎤
⎢⎣⎡
⎟⎠⎞
⎜⎝⎛ +=⎥⎦
⎤⎢⎣⎡= ∫
πππ
ωωπ
π
The rms value of the transformer secondary current:
mmms IItdtII 7804.06
2sin21
62cos
28 2
12
1
60
22 =⎥⎦⎤
⎢⎣⎡
⎟⎠⎞
⎜⎝⎛ +=⎥⎦
⎤⎢⎣⎡= ∫
πππ
ωωπ
π
Example 2: A three phase bridge rectifier has a purely resistive load of Rohms. Determine (a) The efficiency, (b) Form factor, (c) Ripple factor,(d) the TUF, (e) the peak inverse voltage of each diode, and (f) the peak current through a diode. The rectifier delivers Idc=60A at output voltage Vdc=280.7Vand the source frequency is 60Hz.
∫ === 6
0654.133cos3
6/22 π
πωω
π mmmdc VVtdtVV
mmmrms VVtdtVV 6554.14
3923cos3
6/22 2
12
16
0
22 =⎟⎟⎠
⎞⎜⎜⎝
⎛+=⎥⎦
⎤⎢⎣⎡= ∫ π
ωωπ
π
RVI m
rms6554.1
=
%83.996554.1*6554.1
654.1*654.1
**
====
RVV
RVV
IVIV
PP
mm
mm
rmsrms
dcdc
ac
dcη
0008.1654.1
6554.1===
m
m
dc
rms
VV
VVFF
04.010008.11 22 =−=−== FFVVRF
dc
ac
ss
dc
IVPTUF
3=
RVVIV m
mss ××××= 37804.0707.033
9542.07804.0707.033
654.1 2
=×××
=TUF
The peak line-to-neutral voltage is
VVm 7.169654.1
7.280==
VVPIV m 7.2937.16933 =×==
The average current through each diode is
mmd
md
III
tdtII
3183.06
sin2
cos24 6
0
==
= ∫π
π
ωωπ
π
The average current through each diode is
AIAI
m
d
83.623183.0/20203/60
====