BFM3323-02 Rectifier (AC-DC) (1)
Transcript of BFM3323-02 Rectifier (AC-DC) (1)
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
1/51
1
Chapter 2
AC to DC CONVERSION
(RECTIFIER)
Single-phase, half wave rectifier
Uncontrolled: R load, R-L load, R-C load
Controlled
Free wheeling diode
Single-phase, full wave rectifier
Uncontrolled: R load, R-L load,
Controlled
Continuous and discontinuous current mode
Three-phase rectifier
uncontrolled
controlled
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
2/51
2
Rectifiers
DEFINITION: Converting AC (from
mains or other AC source) to DC power by
using power diodes or by controlling the
firing angles of thyristors/controllable
switches.
Basic block diagram
Input can be single or multi-phase (e.g. 3-
phase).
Output can be made fixed or variable
Applications: DC welder, DC motor drive,
Battery charger,DC power supply, HVDC
AC input DC output
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
3/51
3
Single-phase, half-wave, R-load
mm
mRMSo
mm
mavgo
VV
tdtVV
VV
tdtVVV
5.02
)sin(2
1,
(rms),tageOutput vol
318.0)sin(2
average),or(DCtageOutput vol
2
0
0
1
+
vs_
+
vo_
voio
vs
t
A K
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
4/51
4
1 2
V RL
5
Given the supply voltage v = 120 Vrms, 60 Hz. Determine:
a) The average load voltage and currentb) The load voltage in rms
c) The average power absorbed by the load, RL
d) The power factor of the circuit
Answer
a) 54 V & 10.8 Ab) 84.9 V
c) 1440 W
d) 0.707
Example
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
5/51
5
Half-wave with R-L load
tan
)(
:where
)sin()(
:isresponseforceddiagram,From
response,naturalresponse;forced:
)()()(
:Solutioneqn.aldifferentiorderFirst
)()()sin(
:KVL
1
22
R
L
LRZ
tZ
Vti
ii
tititi
td
tdiLRtitV
vvv
mf
nf
nf
m
LRs
+
vs_
+
vo
_
+
vR_
+
vL_
i
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
6/51
6
R-L load
tm
mm
m
tmnf
tn
et
Z
Vti
Z
V
Z
VA
AeZ
Vi
A
AetZ
Vtititi
RLAeti
td
tdiLRti
)sin()sin()(
as,giveniscurrenttheTherefore
)sin()sin(
)0sin()0(
:i.e,conductingstartsdiodethebeforezerois
currentinductorrealisingbysolvedbecan
)sin()()()(
Hence
;)(
:inresultswhich
0)(
)(
0,sourcewhenisresponseNatural
0
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
7/51
7
R-L waveform
:i.e,decreasingiscurrentthebecausenegativeis
:Note
dt
diLv
v
L
L
t
vo
vs,
io
vR
vL
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
8/51
8
Extinction angle
otherwise
0
0for
)sin()sin(
)(
load,L-Rithrectfier wthesummariseTo
and0betweenconductsdiodetheTherefore,
y.numericallsolvedbeonlycan
0)sin()sin(
:toreduceswhich
0)sin()sin()(
.angle,tiontheextincasknownispointThis
OFF.turnswhendiodeiszeroreachescurrent
point whenheduration)Tthatduringnegative
issourcethe(althoughradiansnlonger tha
biasedforwardinremainsdiodethat theNote
t
etZV
ti
e
eZ
Vi
tm
m
=L/R
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
9/51
9
RMS current, Power
RMSRMSs
RMSRMSs
RMS
RMS
o
IV
Ppf
IVS
S
P
S
Ppf
RI
tdtitdtiI
tdtitdtiI
.
.
i.esource,
by thesuppliedpowerapparenttheis
load.by theabsorbedpowerthetoequalwhich
source,by thesuppliedpowerrealtheiswhere
:definitionfromcomputedisFactorPower
P
:isloadby theabsorbedPower
NCALCULATIOPOWER
)(2
1
)(2
1
:iscurrentRMSThe
)(2
1)(
2
1
:iscurrent(DC)averageThe
,
,
2
o
0
22
0
2
0
2
0
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
10/51
10
1 2
V= vmsinwt
L
0.1 H
RL
100
Given Vm= 100 V , = 377 rads-1. Determine:
a) An expression for current i & extension angleb) The average current
c) The rms current
d) The power absorbed by RLe) The power factor of the circuit
Answer
a) 0.936sin(t-0.361)+0.936e-t/0.377, 201o(3.5 rad)b) 0.308 A
c) 0.474 A
d) 22.4 W
e) 0.67
Example
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
11/51
11
Half wave rectifier, R-C Load
sin
OFFisdiodewhen
ONisdiodehenw)sin(
/
m
RCt
mo
Vv
eV
tVv
+
vs_
+
vo_
iD
Vm
Vmax
vs
vo
Vmin
iD
Vo
Time constant = t = RC
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
12/51
12
Operation
Let C initially uncharged. Circuit isenergised at t=0
Diode becomes forward biased as the
source become positive
When diode is ON the output is the sameas source voltage. C charges until Vm
After t=/2, C discharges into load (R).
The source becomes less than the outputvoltage
Diode reverse biased; isolating the loadfrom source.
The output voltage decays exponentially
(with time constant RC)
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
13/51
13
Estimation of
mm
m
m
RCmm
RCtm
RCtm
mm
VV
RC
RCRC
RC
RCV
V
eRC
VV
t
eRC
V
td
eVd
tVtd
tVd
sinand
Thereforewave.sinetheofpeakthetocloseveryis
22-tan
:thenlarge,iscircuits,practicalFor
tantan
1
tan
1
1
sin
cos
1sincos
equal,areslopesthe,At
1sin
)(
sin
and
cos)(
sin
:arefunctionstheofslopeThe
11
/
/
/
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
14/51
14
Estimation of
forynumericallsolvedbemustequationThis0)(sinsin(
or
)sin()2sin(
,2tAt
)2(
)2(
RC
RCmm
e
eVV
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
15/51
15
Ripple Voltage
fRC
V
RCVV
RCe
eVeVVV
eVeVv
t
VV
VVVV
VVV
tV
mmo
RC
RCm
RCmmo
RCm
RCmo
m
mmmm
o
2
21:expansoinSeriesUsing
1
:asedapproximatisvoltagerippleThe
)2(
:is2atevaluatedtageoutput volThe
2.thenconstant,istageoutput volDC
such thatlargeisCand2,andIf
sin)2sin(
2atoccurstageoutput volMin.istageoutput volMax
2
22
2222
minmax
max
Note: Sin(A+B)=sinAcosB+cosAsinB
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
16/51
16
Capacitor Current
)2(t)(i.e
OFF,isdiodewhen
sin
)2(t)(2i.e
ON,isdiodewhen
)cos(
),(ngsubstitutiThen,
OFFisdiodewhensin
ONisdiode when)sin()(
But)(
)(
:,ofIn terms
)(
)(:asexpressedbecancapacitorin thecurrentThe
/
/
RCtm
m
c
o
RCtm
m
o
oc
oc
eR
V
tCV
ti
tv
eV
tVtv
td
tdvCti
t
td
tdvCti
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
17/51
17
Peak Diode Current
R
VCVi
R
V
R
Vi
CVCVI
iiii
mmpeakD
mmR
mmpeakc
CRDs
sincos
:iscurrentpeakdiodeThe
sin)(2sin)(2
.
:obtainedbecan)(2atcurrentResistor
cos)(2cos
Hence..)(2atoccurscurrentdiodepeakThe
:thatNote
,
,
Note: cos(A+B)= cosAcosB - sinAsinB
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
18/51
18
ExampleA half-wave rectifier has a 120V rms source at 60Hz. The
load is =500 Ohm, C=100uF. Assume and are calculatedas 48 and 93 degrees respectively. Determine (a) Expressionfor output voltage (b) peak-to peak ripple (c) capacitor
current (d) peak diode current.
(OFF)5.169
(ON))sin(7.169
(OFF)sin
(ON))sin(7.169)sin()(
:tageOutput vol(a)
;5.169)62.1sin(7.169sin
843.048
;62.193
;7.1692120
)85.18/(62.1
/
t
RCtm
m
o
m
o
o
m
e
t
eV
ttVtv
VradV
rad
rad
VV
Vm
Vmax
vs
vo
Vmin
iD
Vo
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
19/51
19
Example (cont)
A
radradu
R
VCVi
e
t
eR
V
tCV
ti
VufRC
V
RCVV
VVVVVV
VVV
mmpeakD
t
RCtm
m
c
mmo
mmmmo
o
50.4)34.026.4(
500
)62.1sin(7.169)843.0cos(7.169)100)(602(
sincos
:currentdiodePeak(d)
(OFF)A339.0
(ON)A)cos(4.6
(OFF))sin(
(ON))cos(
:currentCapacitor(c)
7.5610050060
7.1692
:ionApproximatUsing
43sin)2sin(
:Using
:(b)Ripple
,
)85.18/(62.1
)/(
minmax
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
20/51
20
Controlled half-wave
+
vo_
+
vs_
igia
2
2sin1
2]2cos(1[
4
sin2
1
voltageRMS
cos12
sin2
1
:voltageAverage
2
2
,
mm
mRMSo
mmo
Vtdt
V
tdtVV
VtdtVV
t
v
vo
ig
t
t
v s
+ VSCR -
Note power sinus: sin x= (1-cos2x)2
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
21/51
21
Controlled h/w, R-L load
eZ
VA
AeZ
Vi
i
AetZ
Vtititi
m
m
t
mnf
sin
sin0
,0:conditionInitial
sin)()()(
+
vs_
i
+
vo
_
+
vR_
+
vL_
t
vs
vo
io
+ VSCR -
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
22/51
22
Thyristor waveform
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
23/51
23
Controlled R-L load
RIP
:loadby theabsorbedpowerThe
dtti2
1I
:currentRMS
dti2
1I
:currentAverage
coscos2
V
dtsinV2
1
V
:voltageAverage
angel.conductionthecalledisAngle
esinsinZ
V0i
ynumericallsolvedbemustangleExtinction
otherwise0
tforesintsinZ
V
ti
g,simplifyinandAforngSubstituti
2
RMSo
2
RMS
o
m
mo
)(
m
t)(
m
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
24/51
24
Thyristor Triggering
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
25/51
25
Examples
1. A half wave controlled rectifier has a source of 120V
RMS at 60Hz. R = 20 ohm, L= 0.04 H, and the delay
angle is 45 degrees. Determine: (a) the expression for
current i(t), (b) average current, (c) the power absorbed
by the load.
2. Design a circuit to produce an average voltage of 40V
across a 100 ohm load from a 120V RMS, 60Hz supply.
Determine the power factor absorbed by the resistance.
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
26/51
26
Freewheeling diode (FWD)
Note that for single-phase, half wave rectifierwith R-L load, the load (output) current is
NOT continuos.
A FWD (sometimes known as commutation
diode) can be placed as shown below to make
it continuos
+
vs
_
io
+
vo
_
+
vR_
+vL_
+
vs_
+
vo
_
D1 is on, D2 is off
io
vo= vs
io
+vo
_
io
D2 is on, D1 is off
vo= 0
(a)
(b) (c)
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
27/51
27
Operation of FWD
Note that both D1 and D2 cannot be turned
on at the same time.
For a positive cycle voltage source,
D1 is on, D2 is off
The equivalent circuit is shown in Figure (b)
The voltage across the R-L load is the same as
the source voltage.
For a negative cycle voltage source,
D1 is off, D2 is on
The equivalent circuit is shown in Figure (c)
The voltage across the R-L load is zero.
However, the inductor contains energy from
positive cycle. The load current still circulates
through the R-L path.
But in contrast with the normal half waverectifier, the circuit in Figure (c) does not
consist of supply voltage in its loop.
Hence the negative part of vo as shown in the
normal half-wave disappear.
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
28/51
28
The inclusion of FWD results in continuos
load current, as shown below.
Note also the output voltage has no
negative part.
FWD- Continuous load current
iD1
io
output
Diode
currentiD2
vo
t
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
29/51
29
Why single-phase full-wave ?
To produce purely DC (less ripple) voltage
or current
Suitable for high power application
Average current in the AC source is zero,thus avoiding problem associated with
non-zero average source current especially
in transformer
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
30/51
30
Full wave rectifier
Center-tapped
D1is
+
vs
_
vo +
iD1
iD2
io
+
vs1_
+vs2_
D2
+ vD1
+ vD2
Center-tapped(CT) rectifier
requires
center-tap
transformer.
Full Bridge
(FB) does not.
CT: 2 diodes
FB: 4 diodes.
Hence, CT
experienced
only one diode
volt-drop per
half-cycle
Conduction
losses for CTis half.
Diodes ratings
for CT is twice
than FB m
mmo
m
mo
VV
tdtVV
ttV
ttVv
637.02
sin1
:voltage(DC)Average
2sin
0sin
circuits,bothFor
0
+
vs_
is
iD1
+
vo_
io
Full Bridge
D1
D2
D4
D3
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
31/51
31
Bridge waveforms
+
vs_
is
iD1
+
vo_
io
Full Bridge
D1
D2D4
D3
Vm
Vm
-Vm
-
Vm
vs
v
o
vD1
vD2
vD3 vD4
io
iD1 iD2
iD3 iD4
i
s
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
32/51
32
Center-tapped waveforms
Center-tapped
D1is
+
vs_
vo +
iD1
iD2
io
+
vs1_
+
vs2_
D2
+ vD1
+ vD2
Vm
Vm
-2Vm
-2Vm
vs
vo
vD1
vD2
io
iD1
iD2
is
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
33/51
33
Full wave bridge, R-L load
+
vs_
is
iD1
+
vo_
io
+
vR_
+
vL_
vo
vs
io
iD1 , iD2
iD3 ,iD4
is
t
2
1
4
3
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
34/51
34
Approximation with large L
,for,2
:i.e.terms,harmonicthe
alldroptopossibleisitenough,largeisIf
.increasingryrapidly vedecreasesThus
decreases.harmonicincreases,As
:currentsharmonicThe
curentDCThe
1
1
1
12
termsharmonicstheand
2
termDCthewhere
)cos()(
Series,FourierUsing
...4,2
RLR
V
R
VIti
L
nI
Vn
LjnR
V
Z
VI
R
V
I
nn
VV
VV
tnVVtv
moo
n
n
n
n
nn
o
o
mn
mo
nnoo
(average value)
(Zn increases)
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
35/51
35
R-L load approximation
vo
vs
io
iD1 , iD2
iD3 ,iD4
is
t
RIP
IIII
R
V
R
VI
RMSo
oRMSnoRMS
moo
2
2,
2
:loadthetodeliveredPower
,2
currenteApproximat
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
36/51
36
Examples
b) Given a bridge rectifier has an AC source Vm=100V at
50 Hz, and R-L load with R =100 Ohm, L= 10 mH
i) determine the average current in the load
ii) determine the first two higher order harmonics of
the load current
iii) determine the power absorbed by the load
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
37/51
37
Controlled full wave, R load
R
VP
V
tdtVV
VtdtVV
RMSo
m
mRMSo
mmo
2
2
,
:isloadRby theabsorbedpowerThe
42sin
221
sin1
VoltageRMS
cos1sin1
:voltage(DC)Average
+
vs_
is
iD1
+
vo_
ioT1
T4 T2
T3
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
38/51
38
+
vs_
is
iD1
+
vo_
io
+
vR_
+
vL_
Controlled, R-L load
vo
Discontinuous mode
io
vo
Continuous mode
+
io
T1, T2ON
T3, T4
ON
2
1
4
3
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
39/51
39
Discontinuous mode
zero.angreater thbemust)(tatcurrentoperationcontinousFor
).(isexpressioncurrentoutputtheinwhenismodecurrentusdiscontino
andcontinousbetweenboundaryThe
0)(
:conditiony withnumericallsolvedbemustandangleextinctiontheisthatNote
)(
:ensuretoneedmode,usdiscontinoFor
;tanand
)(
for
)sin()sin()(
:loadL-Rwithwavehalfcontrolledsimilar toAnalysis
1
22
)(
o
tm
i
R
L
R
LLRZ
t
etZ
Vti
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
40/51
40
Continuous mode
cos
2VdtsinV
1V
:asgivenistageoutput vol(DC)Average
R
Ltan
mode,currentcontinuousforThus
RLtan
forSolving
0,e1)sin(
),sin()sin(
:identityryTrigonometUsing
0)esin()sin(
0)i(
m
mo
1
1
)(
)(
t
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
41/51
41
Single-phase diode groups
In the top group (D1, D3), the cathodes (-) of the two
diodes are at a common potential. Therefore, the
diode with its anode (+) at the highest potential will
conduct (carry) id.
For example, when vs is ( +), D1 conducts id and D3reverses (by taking loop around vs, D1 and D3).
When vs is (-), D3 conducts, D1 reverses.
In the bottom group, the anodes of the two diodes
are at common potential. Therefore the diode with
its cathode at the lowest potential conducts id.
For example, when vs (+), D2 carry id. D4 reverses.
When vs is (-), D4 carry id. D2 reverses.
+
vs_
+
vo_
vp
vn
io
D1
D3
D4
D2
vo =vp vn
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
42/51
42
Three-phase rectifiersD1
vo =vp vn
+
vo_
vpn
vnn
ioD3
D2
D6
+ vcn -
n+ vbn -
+ van -
D5
D4
Vm
Vm
van vbn vcn
vn
vp
vo =vp - vn
D1
D6
D1
D3
D3
D2D3
D4
D5
D4
ia
ib
ic
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
43/51
43
Three-phase waveforms
Top group: diode with its anode at thehighest potential will conduct. The other
two will be reversed.
Bottom group: diode with the its cathode at
the lowest potential will conduct. The other
two will be reversed.
For example, if D1 (of the top group)
conducts, vp is connected to van.. If D6 (of thebottom group) conducts, vn connects to vbn .
All other diodes are off.
The resulting output waveform is given as:
vo=vp-vn
For peak of the output voltage is equal to
the peak of the line to line voltage vab .
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
44/51
44
Three-phase, average voltage
phase.-singleaofnhigher thamuchisrectifierphase-threea
ofcomponentvoltageDCoutputthat theNote
955.03
)cos(3
)sin(3
1
:voltageAverage
radians.3ordegrees60overaverageitsObtainsegments.sixtheofoneonlyConsiders
,,
323
,
32
3
,
LLmLLm
LLm
LLmo
VV
tV
tdtVV
vo
Vm, L-L
vo
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
45/51
45
3 Phase Current Waveform
Output current is assumed to be DC
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
46/51
46
Controlled, three-phase
Vm
van vbn vcn
T1
+
vo_
vpn
vnn
ioT3
T2
T6
+ vcn -
n
+ vbn -
+ van -
T5
T4
vo
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
47/51
47
Controlled, three-phase
waveforms
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
48/51
48
Controlled, three-phase
waveforms
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
49/51
49
Output voltage of controlled
three phase rectifier
cos3
)sin(3
1
:ascomputedbecanvoltageAverage
SCR.theofangledelaythebeletFigure,previoustheFrom
,
32
3
,
LLm
LLmo
V
tdtVV
EXAMPLE: A three-phase controlled rectifier has
an input voltage of 415V RMS at 50Hz. The load
R=10 ohm. Determine the delay angle required to
produce current of 50A.
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
50/51
50
Q1. Consider an uncontrolled full wave diode rectifier
with series R and L load. The supply voltage is given as
340 sin (314t) V.
a) By assuming the load inductance is large enough,
sketch the waveform of
i) Output voltage, Vo
ii) Diode current iD1, iD2, iD3 and iD4iii) Input supply current, i
s
b) If R-L load are 10 and 100 mH respectively, and theinstantaneous voltage across the load is given as
Where
i) Derive the average output voltage, Voii) Calculate the average load current, Ioiii) Calculate the amplitude of harmonic voltage, Vn for n =
2, 4iv) Calculate the amplitude of harmonic current, Vn for n =
2, 4
v) Calculate the load current in rms
4,2
)cos()(n
nwtVnVotvo
1
1
1
12
nn
VmVovn
-
8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)
51/51
Q2. a) Draw the circuit diagram of a single-phase
controlled full-wave rectifier using four power devices
with resistive load connected to it. Assuming the input
voltage of the rectifier is v (t) = Vmsin(t) and the delayangle is ,
i) Sketch the load voltage and current waveforms
ii) Derive the equations for the average load voltage and
current.
b) Consider using the rectifier as DC supply to a highly
inductive load (R = 2.4 and L = 0.6 H) which keeps theload current smooth and ripple-free. The rectifier is
supplied from 1414 Vrms, 50 Hz supply.
i) Sketch the load voltage waveform and show that therelationship between the average load voltage and the
delay angle is (2Vm/) cos .
ii) If the load requires a current of 450 A, calculate the
delay angle that will give this value.
iii) Prove that the load current is continuous.
iv) Determine the power absorbed by the load.
v) Explain how the power absorbed by the load is
increased by including a freewheeling diode i