Upload
blanche-anthony
View
243
Download
4
Tags:
Embed Size (px)
Citation preview
Diode Circuits or Uncontrolled Rectifier
Rectification: The process of converting the alternating
voltages and currents to direct currents
The main disadvantages of half wave rectifier are :
• High ripple factor,
• Low rectification efficiency,
• Low transformer utilization factor, and,
• DC saturation of transformer secondary winding.
Performance Parameters
22dcrmsac VVV
acdc PP / rectification effeciency
dcrms VVFF /
11 22
222
FFV
V
V
VV
V
VRF
dc
rms
dc
dcrms
dc
ac
form factor
ripple factor
121
2
21
21
2
S
S
S
SSi
I
I
I
IITHD
121
2
21
21
2
S
S
S
SSv
V
V
V
VVTHD
FaactorntDisplacemeFactorDistortion
I
I
IV
IV
IV
PPF
S
S
SS
SS
SS
*
coscos
1111
Single-phase half-wave diode rectifier with resistive load.
0
sin2
1 mmdc
VtdtVV
2sin
2
1
0
22 mmrms
VtdtVV
R
V
R
VI mdcdc
R
V
R
VI mrmsrms 2
the load and diode currents R
VII mDS 2
Example 1: The rectifier shown in Fig.2.1 has a pure resistive load of R Determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) Peak inverse voltage (PIV) of diode D1.
mm
mdcVV
tdtVV ))0cos(cos(2
)sin(2
1
0 R
V
R
VI mdcdc
2)sin(
2
1
0
2 mmrms
VtVV
R
VI mrms 2
%53.40
2*
2
*
*
*
RVVR
VV
IV
IV
P
P
mm
mm
rmsrms
dcdc
ac
dc
57.12
2
m
m
dc
rms
V
V
V
VFF
211.1157.11 22 FFV
VRF
dc
ac
)d (It is clear from Fig2.2 that the PIV is
.
mV
Half Wave Diode Rectifier With R-L Load
Fig.2.3 Half Wave Diode Rectifier With R-L Load
tansinsin)(
t
m etZ
Vti
0sinsin)( tan
eZ
Vi m
)cos1(*2
sin*2
0
mmdc
Vtdt
VV
)2sin(1(5.0*2
)sin(*2
1
0
2
Vm
dwttVV mrms
Single-Phase Full-Wave Diode Rectifier Center-Tap Diode Rectifier
m
mdcV
tdtVV2
sin1
0
R
VI mdc
2
2
sin1
0
2 mmrms
VtdtVV
R
VI mrms
2
PIV of each diode = mV2
R
VII mDS 2
Example 3. The rectifier in Fig.2.8 has a purely resistive load of R Determine (a) The efficiency, (b) Form factor (c) Ripple
factor (d) TUF (e) Peak inverse voltage (PIV) of diode D1 and(f) Crest factor of transformer secondary current.
%05.81
2*
2
2*
2
*
*
R
VVR
VV
IV
IV
P
P
mm
mm
rmsrms
dcdc
ac
dc
11.1222
2
m
m
dc
rmsV
V
V
VFF
483.0111.11 22 FFV
VRF
dc
ac
The PIV is mV2
Single-Phase Full Bridge Diode Rectifier With Resistive Load
Example 4 single-phase diode bridge rectfier has a purely resistive load of R=15 ohms and, VS=300 sin 314 t and unity transformer ratio.
Determine (a) The efficiency, (b) Form factor, (c) Ripple factor, (d) The peak inverse voltage, (PIV) of each diode, , and, (e) Input power factor.
VV
tdtVV mmdc 956.190
2sin
1
0
A
R
VI mdc 7324.12
2
VV
tdtVV mmrms 132.212
2sin
12/1
0
2
%06.81rmsrms
dcdc
ac
dc
IV
IV
P
P11.1
dc
rms
V
VFF
482.011 22
222
FFV
V
V
VV
V
VRF
dc
rms
dc
dcrms
dc
ac The PIV=300V
Input power factor = 1cosRe
SS
SS
IV
IV
PowerApperant
Poweral
Full Bridge Single-phase Diode Rectifier with DC Load Current
.............,5,3,14
cos0cos2
cos2
sin*2
00
nforn
In
n
I
tnn
ItdtnIb
oo
oon
)..........9sin9
17sin
7
15sin
5
13sin
3
1(sin*
4)( ttttt
Iti o
%4615
1
13
1
11
1
9
1
7
1
5
1
3
1))((
2222222
tITHD s
2
41
oS
II
%34.4814
21
2
41))((
2
2
2
1
o
o
S
Ss I
I
I
ItITHD
Example 5 solve Example 4 if the load is 30 A pure DC
From example 4 Vdc= 190.986 V, Vrms=212.132 V AIdc 30 and rmsI = 30 A
%90rmsrms
dcdc
ac
dc
IV
IV
P
P 11.1dc
rms
V
VFF
482.011 22
222
FFV
V
V
VV
V
VRF
dc
rms
dc
dcrms
dc
ac
The PIV=Vm=300V
AI
I oS 01.27
2
30*4
2
41
Input Power factor= PowerApperant
PoweralRe
LagI
I
IV
IV
S
S
SS
SS 9.01*30
01.27cos*cos* 11
Effect Of LS On Current Commutation Of Single-Phase Diode Bridge Rectifier .
m
os
V
ILu
21cos 1 oS
oSrd ILf
ILV 4
2
4
osm
rdceinducsourcewithoutdcactualdc IfLV
VVV 42
tan
32
2 2 uII os
2
sin*2
81
u
u
II oS
32
sin2
uu
upf
Example 6 Single phase diode bridge rectifier connected to 11 kV, 50 Hz, source inductance Ls=5mH supply to feed 200 A pure DC load, find: (i) Average
DC output voltage, (ii) Power factor. And (iii) Determine the THD of the utility line current.
VVm 155562*11000
VVactualdc 9703200*005.0*50*4
15556*2
.285.015556
200*005.0*50**2*21cos
21cos 11 rad
V
ILu
m
os
917.0
3
285.
2285.0
285.0sin*2
32
sin*2
2cos*1
uu
uu
I
Ipf
S
S
AuI
I oS 85.193
3
285.0
2
200*2
32
2 22
Au
u
II oS 46.179
2
285.0sin*
285.0*2
200*8
2sin*
2
81
%84.40146.179
85.1931
22
1
S
Si I
ITHD
Three-Phase Half Wave Rectifier
mm
mdc VV
tdtVV 827.02
33sin
2
36/5
6/
R
V
R
VI mmdc
*827.0
**2
33
mmmrms VVtdtVV 8407.08
3*3
2
1sin
2
36/5
6/
2
R
VI mrms
8407.0
R
V
R
VII mmSr 4854.0
3
08407
ThePIV of the diodes is mLL VV 32
Example 7 The rectifier in Fig.2.21 is operated from 460 V 50 Hz supply at secondary side and the load
resistance is R=20. If the source inductance is negligible, determine (a) Rectification efficiency, (b)
Form factor (c) Ripple factor (d) Peak inverse voltage (PIV) of each diode.
VVVV mS 59.3752*58.265,58.2653
460
mm
dc VV
V 827.02
33
R
V
R
VI mmdc
0827
2
33
mrms VV 8407.0R
VI mrms
8407.0
%767.96rmsrms
dcdc
ac
dc
IV
IV
P
P
%657.101dc
rms
V
VFF
%28.1811 22
222
FFV
V
V
VV
V
VRF
dc
rms
dc
dcrms
dc
ac
The PIV= 3 Vm=650.54V
Three-Phase Half Wave Rectifier With DC Load Current and zero source induct
New
axis
32
13/
3/
0o
oI
tdIa
harmonicstrepleanallfor
nforn
I
nforn
I
tnn
IdwttnIa
o
o
oon
0
17,16,11,10,5,43*
,....14,13,8,7,2,13*
sincos*1 3/
3/
3/
3/
...8sin
8
17sin
7
15sin
5
14sin
4
12sin
2
1sin
3
3)( tttttt
IItI OO
s
%24.1090924.119
*21
2
3
3/1))((
2
2
2
1
O
o
S
Ss
I
I
I
ItITHD
Example 8 Solve example 7 if the load current is 100 A pure DC
VVV
V mm
dc 613.310827.02
33
AIdc 100
VVV mrms 759.3158407.0 AIrms 100
%37.98100*759.315
100*613.310
rmsrms
dcdc
ac
dc
IV
IV
P
P
%657.101dc
rms
V
VFF
%28.1811 22
222
FFV
V
V
VV
V
VRF
dc
rms
dc
dcrms
dc
ac
The PIV= 3 Vm=650.54V
Three-Phase Full Wave Rectifier With Resistive Load
LLmLLm
mdc VVVV
tdtVV 3505.1654.12333
sin33
3/2
3/
R
V
R
V
R
V
R
VI LLLLmmdc
3505.123654.133
LLmmmrms VVVtdtVV 3516.16554.14
3*9
2
3sin3
33/2
3/
2
R
VI mrms
6554.1
R
V
R
VI mmr 9667.0
3
6554.1
R
VI mS 29667.0
Example 10 The rectifier shown in Fig.2.30 is operated from 460 V 50 Hz supply and the load resistance is R=20ohms. If
the source inductance is negligible, determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) Peak inverse
voltage (PIV) of each diode .
VVV
V mm
dc 226.621654.133
AR
V
R
VI mmdc 0613.31
654.133
VVVV mmrms 752.6216554.14
3*9
2
3
AR
VI mrms 0876.31
6554.1
%83.99rmsrms
dcdc
ac
dc
IV
IV
P
P
%08.100dc
rms
V
VFF
%411 22
222
FFV
V
V
VV
V
VRF
dc
rms
dc
dcrms
dc
ac
The PIV= 3 Vm=650.54V
Three-Phase Full Wave Rectifier With DC Load Current
ttttv �t
ItI o
s
13sin13
111sin
11
17sin
7
15sin
5
1sin
32)(
%3125
1
23
1
19
1
17
1
13
1
11
1
7
1
5
1))((
22222222
tITHD s
oS II3
2 oS II
3*2
1
%01.311/3*2
3/21))((
2
2
1
S
Ss I
ItITHD
Power Factor =S
S
S
S
I
I
I
I 11 )0cos(*
LL
oS
V
ILu
21cos 1
oo
rd fLILI
V 62
6
oLLrdceinducsourcewithoutdcactualdc fLIVVVV 635.1tan
63
2 2 uII oS
2sin
621
u
u
II oS
63
sin*3
2cos
63
2
2sin
62
2cos
2
1
uu
uu
uI
u
u
Iu
I
Ipf
o
o
S
S
Example 11 Three phase diode bridge rectifier connected to tree phase 33kV, 50 Hz supply has 8 mH source inductance to feed 300A pure DC load current Find;Commutation time and commutation angle.DC output voltage.Power factor.Total harmonic distortion of line current.
oradu 61.14.2549.0
LL
oS
V
ILu
21cos 1
dLLrdceinducsourcewithoutdcactualdc fLIVVVV 635.1tan
VVdcactual 43830300*008.*50*633000*35.1
9644.0
6
2549.0
3*2549.0
2549.0sin3
63
sin*3
u
u
upf
AuI
I ds 929.239
6
2549.0
3*
300*2
63
2 22
Au
u
II oS 28.233
2
2549.0sin*
2*2549.0*
300*3432*
2sin
2
341
9644.02
2549.0cos*
929.239
28.233
2cos*1
u
I
Ipf
s
S
%05.24128.233
929.2391
22
1
S
Si I
ITHD