ch3 half-wave rectifiers (the basics of analysis) 3-1 introduction ˙used often in low-power...
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CH3 Half-wave rectifiers (The basics of analysis)
3-1 Introduction
˙Used often in low-power applications ( average current in the supply will not be zero , and cause
problems in transformer performance.)
˙Enable the student to advance to the analysis of more complicated circuits with a minimum of effort.
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3-2 Resistive load : Fig.3-1
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R
V
R
VI
V
t)td(sinV2
1VV
m0
m
0
mavg0
Average power absorbed by the resistor:
2R
V
R
VI
2
Vt)d(t)sin(V
2
1V
R
VRIP
mrmsrms
m
0
2mrms
2rms2
rms
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3-3 Resistive-Inductive load (R-L load): Fig.3-2
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Diode on,dt
(t)idL(t)iRtsinVm
response natural(t)i
response forced(t)i(t)i(t)i(t)i
n
f
nf
The force response is the steady-state sinusoidal current.
)R
L(Tan L)(RZ
)-tsin(Z
V(t)i
1-22
mf
,
Natural response is the solution to the homogeneous differential eq.
constant) (Time R
L Ae(t)i
0dt
(t)idL(t)iR
t-
n
,
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Some numerical method is required to solve
.
angle. extinction
])e(sin)-t[sin(Z
V
esinZ
V)-tsin(
Z
V(t)i
sinZ
V)sin(-
Z
V-A
0Ae)-sin(0Z
V(0)i
Ae)-tsin(Z
V
(t)i(t)i(t)i
t/-m
t/-mm
mm
0m
t/-m
nf
0)e(sin)-sin(
0)e(sin)-[sin(Z
V)(i
t
])e(sin)-t[sin(Z
V
)e(sinZ
V)-tsin(
Z
Vt)(i
/-
/-m
t/-m
t/-mm
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Average power absorbed by the load :
0
22
0
2rms
2rmsL
t)t)d((i2
1t)t)d((i
2
1I
RIP
Average current.
0
t)t)d((i2
1I
RL )
R
L(Tan L)(RZ
2t 0
t0 )e(sinZ
V)-tsin(
Z
Vt)(i
1-22
t/-mm
,,
,
,
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3-5 R-L source load : Fig.3-5
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(t)i(t)i(t)i
Vdt
(t)idL(t)iRtsinV
ON Diode ,)V
V(sin VsinV
nf
dcm
m
dc1-dcm
,
(t)i f is determined using superposition for the two source ( dcm V ,tsinV )
t-
n
dcmf
Ae(t)i
R
V-)-tsin(
Z
V(t)i
? 0)(i
)eR
V)-sin(
Z
V(-A 0)(i
otherwise 0
t AeR
V-)-tsin(
Z
Vt)(i
dcm
t-dcm
,
,
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Average power absorbed by R, P= RI2rms
Average power absorbed by dcV , dcdc IVP
t)t)d((it)sin(V2
1
t)t)d((it)(2
1IVRIP
t)t)d((i2
1I t)t)d((i
2
1I
m
2
0dc2rmsac
2rms ,
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3-6 Inductor-Source load : Fig.3-6
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At )V
V(sint
m
dc1- , Diode conducting.
otherwise , 0
t t)-(L
Vt)cos-(cos
L
V
dVL
1-)dsin(V
L
1t)(i
L
V-tsinV
t)d(
t)(id
Vt)d(
t)(idL
Vdt
(t)idLtsinV
dcm
t
dc
t
m
dcm
dc
dcm
,
Power supplied by the AC source=Power absorbed by the dc source.
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3-7 The freewheeling diode Fig.3-7 , 2D : freewheeling diode
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The voltage across the load is a half-wave rectified sine wave and
6... 4 2n
o2m
omm
o t)cos(n1)-(n
2V-t)sin(
2
VV(t)
,,
Using superposition, ?(t)io
If L is infinitely large,
R
L
R)(
V
R
VI(t)i mo
oo ,Reducing load current harmonics.
Zero-to-peak fluctuation in load current can be estimated as being equal to the amplitude of the first ac term in the Fourier series ( )
1I
The peak-to-peak ripple is then 1o 2II
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3-8 Half-wave rectifier with a capacitor filter: Fig.3-11
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t sinVV
off diode eV
on diode tsinV(t)
m
RC)(-t
m
o
,
,
,
At , the slopes of the voltage function are equal. t
RC)(-t-
RC)(-t-
e)1
(sinV
]esin[V)(
cosV)sinV()(
RC
td
d
tttd
dm
m
m
m
)()(
11
1
sinV
cosV
sinVsinVcosV
11
)/()(m
RCTanRCTan
RCTanRCTan
RC
RCe
RC
m
m
mRCm
,
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In practical circuits where the time constant is large.
mmand VsinV2
At ONDiodet 2 ,
0)(sinsin
)sinV()2sin(V)/()2(
)/()2(
RC
RCmm
e
e
Solved numerically for ?
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on diode t )tcos(Vc
off diode t eR
sinV
)t(d
)t(dct)(i
dt
)t(dvc(t)i
R/)t(v(t)i
m
)RC/()t(m
oC
oC
oR
,22,
,2,
CRDS iiii
Since the diode is on for a short time in each cycle, the peak diode current is generally much larger than the average diode current. Peak capacitor current occurs when the diode turn on at 2t ,
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Peak diode current is
)sin
cos(V
sinVcosVI ,
Rc
Rc
m
mmpeakD
Peak-to-peak ripple voltage
)sin(VsinVVV mmmo 1
If R-C time constant is large compared to the period of the sine wave
2 ,V V ,
2 m
2
,2
......!3!2!1
1
,)(,
)2
(
]2
11[
]1[
)2(
32
,
)(2
)(2
2222
TRC
RC
x
xxxe
IrippleVC
fRC
V
RCV
RCV
eVeVVV
eVeVv
x
peakDo
mm
m
RCm
RCmmo
RCm
RCmo
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3-9 The controlled half-wave rectifier
Resistive load (R): Fig.3-13
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.angle delay)cos(
V)t(tdsinVV
,resistor load the across voltage Average
mmo
1
22
1
Power absorbed by R is R
VP rms
2
2
21
2
2
1
2
1 22
0
2
)sin(V
)t(d]tsinV[)t(d)t(V
m
morms
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R-L load : Fig.3-14
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angle ction?....condu-
angle nction?.....extiy ,numericall solved
]e)sin())[sin(Z
V()(i
otherwise , 0
t , ]e)sin()t)[sin(Z
V(
)t(i
e)]sin()Z
V([A
Ae)sin()Z
V()(i
Ae)tsin()Z
V(
)t(i)t(i)t(i
)/()(m
)/()t(m
)(m
)(m
)(t
m
nf
0
0
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Average output voltage
]cos[cos2
V)(sinV
2
1V
m
mo tdt
Average current :
)t(d)t(iI
2
1
Power absorbed by the load is RIP rms2
)t(d)t(iI rms
2
2
1
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.R-L source load : Fig.3-15
The gate signal may be applied to the SCR at any time that the ac source is larger than the dc source.
)/(dcm
)t/(-dcm
m
dc-1min
]eR
V-sin
Z
V[-A 0)(i
otherwise , 0
t ,AeR
V-)-tsin(
Z
Vt)(i
)V
V(sin
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3-11 Commutation
Effect of source inductance: Load current is constant.
Fig.3-17
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1D 2DThe interval when both and are on is called commutation time (or angle). Commutation is the process of turning off an electronic switch, which usually involves transferring the load current from one switch to another.
When both and are on, the voltage across is , and current in and source is
1D 2D SL tsinVv mLS
SL
t
0S
mm
S
t
0 SLSS
S t)cos-(1L
V0t)td(sin
L
1(0)it)d(
L
1i
)2V
XI-(1
V
cosu)(12
Vt)os(-c
2
Vt)td(sinV
2
1V
LX , )V
XI-(1cos)
V
LI-(1cosu
0cosu)-(1L
V-I(u)i
t)cos-(1L
V-Ii-Ii
m
SLm
mu
m
u mo
SSm
SL1-
m
SL1-
S
mLD2
S
mLSLD2
SL reduces average load voltage ( oV )