Download - Chapter 1 DC Drives Part2
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onverter Control oonverter Control o
DC DrivesDC Drives
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LearningObjectives
At the end of this Chapter, you should be able to :
Analyze the operation and control of phase controlled
SCR converters for DC Motor Drive
Analyze the operation and control of DC-DC
converters for DC Motor Drive
Model and analyze the closed-loop control system forDC motor drive system
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(B) DC-DC Converter Controlled DCDrives
(A) Phase Controlled SCR ConverterDC Drives
Power ElectronicControllers for
DC DrivesThere are two types of power electroniccontrollers for DC motor control. They are:
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4
Phase Controlled SCR ConverterDC Drives
DC MotorDrives
Single-phaseDrives
Three-phaseDrives
SeparatelyExited
SelfExcited
ShuntSeries
SelfExcited
SeparatelyExcited
Series
ShuntCompound
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Average motor armature current:
In a separately-excited machine, the developed torque is proportional
to average armature current. This current Iais known as the torque
producing component of the motor current.
RMS motor armature current, ar
This is the heat producing component of the motor current.
Similar definitions for verage and !"S voltage exists.
ia# instantaneous armature current
T # time period for one cycle variation of ia.
$
%
T
&'ar
%
%
$
T
%
I
= +t
t t dti a
=
+T
a
%
%
&'T%I
t
t a dtti
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nput !o"er #actor $!#%
If supply current is a distorted sinusoid, only the fundamental
component of input current will contri(ute to the mean input power.
Input )ower *actor '!#%is an important parameter as it decides the
volt-ampere requirement of the drive system. *or the same power
demand, if the power factor is poor, more volt-amperes 'and hence
more current& are drawn from the supply.
amperesvoltinputr&m&s&
po"erinputmean
!# =
+rms supply phase voltageIrmssupply phase current
I% rmsfundamental component of the supply current
%angle (etween supply voltage and fundamental component of
supply current
+I
cos+I)*
%% =
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nput Displacement #actor $D#%
This may (e called fundamental power factor and is defined as
where % is known as the input displacement angle.
*or the same power demand, if the displacement factor is low,
more fundamental current is drawn from the supply.
'armonic #actor''#%
The input current (eing non-sinusoidal contains currents of
harmonic frequencies. The harmonic factor is defined as
(D# cos=
$%
=
(
(
))
'#
%
h
$
%
%
$
$
I
I
I
I
-* =
=
=nn
n* rms value of the nth
harmonic current
h* rms value of the net
harmonic current
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The harmonic factor indicates harmonic content of input supply
current and thus measures the distortion of the input current.The input supply current, i, can (e expressed in terms of a
*ourier series as follows( )( )
=
=
++=
++=
%nn
%nnn
&sin'nI$I
&sin'nncosI
n
t
tbtai
The d.c. component, I, and *ourier coefficients an, bnare o(tained as
=+
, +
%dti
=T
n&cos'n
T
$dttia
=
+
n +
$dttib )sin(n
$
%
n
$$
nn
$(aI
+=
= n
n
n
b
a%tan
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aria!le speed DC drives that will givea good approximation to the steady-state motoring operation are groupedunder the !road classi"cation#
D$C$ machine systems fed from an%$C$ supply
&Phase Controlled 'SC()converters*
D$C$ machine systems fed from a D$C$supply
DC-DC converters - Pulse-width-
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currentdcoutput
currentrmsInput
I
I
d
rms =
$$
r&or'ance Para'eters
oltage ratio
Po*er ratio
&'voltagedcmaximum+where,++ ddac ==
po"eroutputDC
supplytheofrating-A
!
-
d
rmsac =
+ar'onic contents a, ac co'onents in o.t.tvoltage/
C.rrent ratio
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SCR devices
currentloaddccurrentS/!
d
S/!
I
I =
currentloaddc
currentS/!peak
I
I
d
S/! =
&'voltagedcmaximum
S/!of+oltageInverse)eak
+
+
d
)I+
==
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$1
SCR 2hase-angle controlled3 drive B changing the 6ring angle/ variable DCo.t.t voltage can be obtained,
Single hase (lo* o*er) and three hase(high and ver high o*er) s.l can be .sed,
7he load c.rrent is .nidirectional/ b.t the
o.t.t voltage can reverse olarit, +ence 8-9.adrant oeration is inherentl ossible,
4-9.adrant oeration is also ossible .sing
2t*o sets3 o& controlled recti6ers (D.alconverters ,
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$4
Single-Phase ConverterControlled Separately-
Excited DC Motor DriveConverter : SCR ;.ll-Bridge Converter (8-.lse converter)
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$5
oltage and current waveforms
%rmature
current
%rmaturevoltage
a= Ea
3nput
acvoltage
3nput linecurrent = 45
3nput
linecurrent
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cos$$
0cos&cos'1$
0cos1$
sin$%
&waveformfrom'voltage,inalmotor termaveargeThe
converterthetoageinput voltrmswheresin$2et
2aw3+fromequationcircuitarmatureThe
a
aca
ac
ac
acdca
acac
g
a
aaaaa
VV
V
V
dVVV
E
VVv
edt
di
LiRev
=
++=
=
==
==+
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45%6$6$
6
$6
6
%%%6$$
6
$$6%
$$666
66
./V
V
V
V
V,VoltageInerse!ea"
cosVan#,
.
.$a% )()(
d
ac
d
PIV
acPIV
ddc
d
ac
acacddca
V
V
V
V
VVVVV
cos$$
voltage,inalmotor termverage aacVV =
$!
Per&or'anceara'eters
Mathematical Derivations
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%+
Tv
a
v
a
%
v
aaa
secradT33
!3+
secrad3
!I+
constantisfluxifspeed,veragemotor,normala*or
=
=
cos
$$
voltage,inalmotor termverage aacV
V =
%+
Tv
a
v
%
v
aa
secradT33
!
3
cos$$
secrad3
!Icos$$
constantisfluxifspeed,verage
=
=
ac
ac
V
V
$"
a=
voltageappliedtoarmatur
e$
Mathematical Derivations
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$#
Two &'a#rant ('ll controlle#) peration
*%ample:7he &ollo*ing data gives details o& asearatel e>cited d,c, 'otor .sed &or ro.lsion in an
electric train,
Ar'at.re resistance ? %,%4 oh'
Bac= e,',&, Constant ? %,5 @r'
7or9.e constant ? 4," '@A (at 'a>, 6elde>citation)
oltage alied to ar'at.re is reg.lated b a single
hase &.ll-controlled (i,e,/ t*o 9.adrant) bridge .singSCRs, 7he 6eld has a searate controller to give &.ll
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8%
7he e9.i'ent is rated &or contin.o.s oerationat &.ll seed/ at *hich 'otor seed is 8%%% r,,',and o*er o.t.t is 5%% = and the 6eld has been
*ea=ened to 5% o& 'a>i'.',a) Deter'ine the secondar voltage and A rating o&
a trans&or'er to s.l the bridge,
b) Deter'ine the 6ring dela angle *hen starting
&ro' standstill *ith a tor9.e o& =-', S=etch theo.t.t voltage *ave&or's and the cond.ctionatterns in the o*er se'icond.ctors, eglect theeEects o& sat.ration in the 'otor/ s.l
reg.lation/ voltage dros across the S,C,Rs andco''.tation delas,F7GPH ac@do
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%
rads7.$8
=
%secrad5.%7
$
%
== B
8$
$a%
V..I
I+IVspee#,f'llt
.!I
V)(.,emf-ac"
/sra#.spee#-ase
ra#/.
.spee#f'll
a0oewea"eningfiel#
spee#s,lowate%citationfiel#'ll
.0ri#ge1'a#rant)(controlle#f'llphasesinglea0y+eg'late#
."2344!.,m.p.r:)(ass'mespee#f'llt o't
476$
$497:%696
%66
46%466
5%76$%666
7$86;
$$66
6$6
f
fvaaa
a
outa
va
Bb
kA
E
rpmrpm
VkE
s
two
5ol'tion:
5 #
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5pee#
Tor1'e
(iel#
wea"ening
voltage&'Secondary+7.488&47'%%.%+%%.%+
rating&+er'Transformk+7.488)%%.%+
.k
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81
%
v
a
%
=7.5&4>47'cos
,forsolve,cos++.eqn*rom
+4&7.'%$4&I
I'k!I+voltage,rmature
?ero.toequalis@startingt
current&'starting%$4=.7
m.kA;I
l&'standstilrads
.m.kA;Twithstillstandfromstartingwhenangle,delay*iring
==
==
=+=+=
===
=
=
dadc
f
f
aaa
T
a
E
K
T
$b%
Waveforms are shown in previous slideson mathematical derivations
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84
*%ample: The following #ata gies #etails of a separately e%cite#
#.c. motor controlle# 0y a single6phase f'll conerter.
rmat're resistance 4.47 ohm
-ac" e.m.f. constant 4.8 V/rpmTor1'e constant .9 ;m/ (at ma% fiel# e%citation)
The ac s'pply oltage to the conerter is
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8
5'pply power factor:
If the motor c'rrent is constant an# ripple free, the inp't s'pply
c'rrent is a s1'are wae of amplit'#e 78 .
Th's the rms s'pply c'rrent, I = 78
5'pply olt6amps = V =
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8!
Balf6wae(5ingle61'a#rant operation) Con#'ction pattern:
Hach SCR cond.cts &or1@1radians7riggering se9.ence ? $/ 8/ 1/ $J
7riggering interval ?8@1radians
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BCB$ %66 D 9;
fordVVV
V
da sin/
)waeformfrom(oltage,terminalmotoraeargeThe
ma%
a
conerterthetooltagephasema%im'minp'trmswhere
sinoltage,phasetheet
aw?Vfrome1'ationcirc'itarmat'reThe
ma%
ma%
6
6
9966
V
Vv
edt
diLiRev g
aaaaaa
Analysis for 3-Pulse 1-Quadrant Operation
)sin.cos.(ma%
4-=;;9%
$
BB6
V
74forcos
sin/
ma%ma% E$
BB6
B$
%6 D
9;4
9;
VdVVa
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8#
Balf6wae(Two61'a#rant operation) Con#'ction pattern:
Hach SCR cond.cts &or1@1radians
7riggering se9.ence ? $/ 8/ 1/ $J
D il d A l i f 3 P l Q d O i
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1%
Detailed Analysis for 3-Pulse -Quadrant Operation
D9
;4
9;
da
ga
aaaaa
d+B$
%6+6+
+
6+
+6v
9EEe9
dt
di29i!6e6v
sin/
)waeformfrom(oltage,terminalmotoraeargeThe
conerterthetooltagephasema%im'minp'trmswhere
sinoltage,phasetheet
aw?Vfrome1'ationcirc'itarmat'reThe
ma%
a
ma%
ma%
9;4
9;
9;
4
9;
-
$
B6
B$%66 D
cosE
sin/
)waeformfrom(oltage,terminalmotoraeargeThe
ma%
ma%
a
V
dVVV
V
da
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acma%ac
ma%
ma%
ma%
ma%
ma%
thenoltage,phasermsIf
cos
)cos(
Dsin.cos.sin.cos.E
DsinsincoscossinsincoscosE
)Dcos()cos(E
VVV
VV
VV
V
V
V
a
a
66
$
BB6
B$
B6
4-=;;949=;;$
B
6
;-
;9
;
49
;
4-
$
B6
9;
99;
4-
$
B6
Detailed Analysis for 3-Pulse -Quadrant Operation
D t il d A l i f 3 P l Q d t O ti
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cos$
$B
cos$x
B$
BB
cos$
BBvoltage,inalmotor termaveargethes
$xB
voltage,phasemaximumThen
er,transformtheofconnectionstar*or
voltageline-to-2inevoltage,line2et the
22
22
maxa
22max
22
VV
VV
VV
VV
V
a
a
=
=
=
=
=
18
Detailed Analysis for 3-Pulse -Quadrant Operation
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11
!ully "ontrolled #$wo-%uadrant&
*ach 5C+ con#'cts for/7 ra#iansTriggering se1'ence =,
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15
3-phase input
volta'e
Armature
volta'eArmature
current
(nput linecurrent
#c
A l i f ) P l Q d t O ti
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1
Analysis for )-Pulse -Quadrant Operation
D
9;
9;
-dca
gaaaaaa
d+B
%6+6+
++
+B99;EE9;e9dt
di29i!6e6v
cos/
)waeformfrom(oltage,terminalmotoraeargeThe
cos:waecosineaas#efine
tage,aerageolofncalc'latioforwaeformThe
0elinesthe0etweenoltagetheet
aw?Vfrome1'ationcirc'itarmat'reThe
a
9;
9;-
+B6 Esin
A l i f ) P l Q d t O ti
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cos
cossin
Dsin)cos(cos)sin(sincoscosEsin
)Dsin()Esin(7
CC
CC
CC
CC
+B6+
;$
+B6+
;
-9
;
--
;
9
;
+B6+
9;
--9;
+6
a
a
a
Analysis for )-Pulse -Quadrant Operation
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T7PE ac8do 3rms83d %8Pdo 3SC(83d 93SC(83d
P38d:utputoltage;armonic
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D I;DGCT;C* CCGTI;:
C6DC 5C+ C;V*+T*+
oad inductance#
RKS rile c.rrent in the load '.st not e>ceedi'.' rile occ.rs at so'e $
Compute ;
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D I;DGCT;C* CCGTI;
Example# ;alf wave 'Two ,uadrant) /-phase %C-DC
oad inductance calculation# .'ber o& riles (in one ccle) ? n ? 1 ;.nda'ental s.l &re9.enc ? & ? 5%+0
Average dc voltage at = 5> do RKS rile c.rrent in the load '.st note>ceed i'.' rile occ.rs at = ?5
$
BB max
VVdo =
acmax
ac
$then
voltage,phaserms
VV
V
=
=
DC DC C t C t ll d
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DC-DC Converter ControlledDC Drives
5witche#6mo#e #ries Gsing switche# mo#e DC6DC conerter, DC oltage is
arie# 0y #'ty cycle.
$ainly 'se# for low to me#i'm power range
5ingle61'a#rant conerter (0'c"):61'a#rant
Balf60ri#ge: 61'a#rant
'll60ri#ge: 61'a#rant
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48
P+M techni,ue is a method of controlling the
voltage within a dc-dc converter$ +ith thistechni,ue> the converter output voltageinvolves a pulse width modulated wave> andthe voltage is controlled !y varying theduration of the output voltage pulses$ 7he revio.s slides sho* the voltage controlachieved b varing the hase o& the cond.ctionintervals o& SCRs/ T;.and T;1*ith resect to T;/and T;2,
The pulse width control is achieved !y phase-advancing or retarding the control signals forone or pair of switches 'Transistors> M:S and in this way the converter output
voltage can !e adBusted smoothly from
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7
!2$ peration
% control signal is compared with a repetitiveswitching-fre,uency triangular waveform inorder to generate the switching signals$
Controlling the switching duty ratios allows theavera ed D$C$ volta e out ut to !e controlled$
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losses)no(ass'ming3I
4./",34IV,/4VV,/44V.,g.e
I"II
"VT
"VVV
ratio#'ty"
fre1'encyswitchingwhere,conertertheofperio#timeT
conertertheof")(
,conertertheof"
conertertheofoltageo'tp'tV
oltage,so'rce#cV
s
##s
#Hs
ssH
#
s
s
#
s
6
6666
66
6T
66
6
6ff
%66
time-off6T-%
time-on6T
6
6
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4!
2ample:
)(
@.
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2att7.7.
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4#
appro24 trian'ular or sawtooth waveform4
.7
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.4449mB.
T")4.9(
LTVLI
:perio#theanalyKing0yChec"
mB.
ms)"()V(V
LI
LTV
LT
LI#t
#iV
oltageIn#'ctor:perio#;
(typical)rippleachieetoDesignms)(T"BKfre1'ency!2$
:#esignCho"e
Ts
666
666
F6
peak-to-peak66
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5$
2aeforms with allowances for transistor an# #io#e oltage #rops
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58
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s
sdssTsG
d
offons
ss,
T
T&k%&'$+'Tk&+$'+++
voltageofwaveformthe*rom
TTTperiod,Switching
T&k%'rtransistooftime-HffkTrtransistooftime-Hn
+==
+=
==
v
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54
2aeforms with allowances for transistor an# #io#e oltage #rops.
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offons
s
s,
dT
TTTperiod,SwitchingT&k%'rtransistooftime-Hff
kTrtransistooftime-Hn
kratio,duty&and'diodesandrstransistooftimesswitching/onsider
&and+'dropsvoltagediodeandransistor/onsider t
computedisvoltageaveragethevoltage,ofwaveformthe*rom
+=
=
=
FR tt
v
5
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s
dss$%
s
Tss$%
Gd
s
dsss$%
s
Tsss$%
Gd
T
&$+&0'T&k%'$1' T
&+$&0'+Tk$1'++
T&$+&0'T&k%'T&k%1''
T
&+$&0'+TkT1'k++
strape?iumofareaheconsider tvoltage,ofwaveformthe*rom
vtt
tt
vtt
tt
FR
FR
FR
FR
+
==
+
++
==
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The mass of an electric ehicle as shown in !(56. is 344 "g, The ehicleis going 'p a slope of 74oat a spee# of 74 "m/hr. The friction coefficient of the
s'rface at a gien weather con#ition is 4.. The acceleration #'e to graity, g =
@.8 ms6. The electric ehicle is 0eing powere# 0y a DC motor mo'nte# on the
front wheels an# the wheel #iameter is 4.3 m. re#'ction gear of 34: is 'se#.
The motor ta"es 9 V with constant e%citation, an# is controlle# 0y a two6
1'a#rant transistor !2$ #c6#c conerter. The #c motor on 0oar# the ehicler'ns at 844 rpm. The following characteristics apply to the system:
2ample:
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rmat're resistance = 4. ohms
Tor1'e constant= 4. ;m/-ac" emf constant = 4. V/ra#s6
Transistor an# #io#e switching times, on an# off = Msec an# Msec
5witching fre1'ency of transistors = 3 "BK
Transistor con#'ction oltage #rop = V
Determine:
(a) The tor1'e #eelope# 0y the motor.
(0) The oltage applie# to the motor.
(c) The #'ty ratio of the !2$ conerter.
(ns: 4.97< ;m, 7.@48 V, 4.8@)
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Solution: Data:$ass of the 0's, m = 344 "g
2heel #iameter, # = 4.3 m5pee# of the 0's, = 74 "m/hr
5lope of the hill, = 74o
riction coefficient, r= 4.
cceleration #'e to graity, g = @.8
ms6(a) The weight of ehicle is #ii#e# into two components: (i) !erpen#ic'lar to the roa# s'rface (force F), responsi0le for friction
force, Fr (ii) !arallel to the s'rface, responsi0le for p'lling the ehicle towar#s
0ottom of the slope, FlThe total force, FL= FlN Fr
ll these forces are #epen#ent on the graitational force, FgConsi#ering the force #iagram, the normal force, Fan# p'lling force, Flare
F = Fgcos
Fl= Fgsin
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The graitational force, F = m.g = 344 % @.8 = @44 ;
F = Fgcos = @44 % cos 74 = 7 ;
Fl = Fgsin= @44 % sin 74 = 34 ;The friction force, Fr= r% F= 4. % 7 =
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(c) kratio&,')
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p ' '
!irst %uadrant operation: 8otorin' mode
$wo Quadrant Operation - .e'enerative 7ra0in':
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Second %uadrant operation: .e'enerative 9ra0in' mode
$wo Quadrant Operation - .e'enerative 7ra0in':
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p ' '
$wo Quadrant Operation - .e'enerative 7ra0in':
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p ' '
$wo Quadrant Operation - .e'enerative 7ra0in':
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+hen the electric vehicle is going down the hill>the dc machine is under regenerative !raing
mode$ 3n the downward direction> the speed ofthe electric vehicle may exceed its no-loadspeed> and generate electric power that can !efed !ac to the source$
f
aabfabLb
f
afffLbab
db
ab
Lb
ab
R
VIIII
R
VIIIII
T
E
I
I
==
=+=
=
=
==
=
motor,theofcurrentTerminal
current,fieldtheiswhere,
(rakingduringspeed
torque(rakingdeveloped
(rakingduringemfinduced(rakingduringcurrentline
(rakingduringcurrentarmature
(
$wo Quadrant Operation - .e'enerative 7ra0in':
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a
abaab
db
Tv
a
v
a
T
dbabvaababa
abTdb
R
EVI
T
kk
R
k
V
k
TRkRIEV
IkT
=
=
+=+==
(
operationnormalIn the
' '
In the #ownhill operation, the loa# tor1'e changes its #irection.Th's the armat're c'rrent also reerses its #irection.
Therefore in the a0oe e1'ations T#0
is negatie.
$wo Quadrant Operation - .e'enerative 7ra0in':
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2ample: 4 V #c sh'nt motor has the following characteristics:rmat're resistance 4.799 ohms
iel# resistance 4 ohmsVoltage constant 7.@7 V sec6
Tor1'e constant 7.@7 ;m/
Determine the following #'ring regeneratie 0ra"ing when the armat're
c'rrent is
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9
orwar# c'rrent 'ses TNTan# then DN Dalternately.
+eerse c'rrent 'ses T7NTan# then D7 N Dalternately.
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Drie signals for transistors:
#elay tDis allowe# on switch6on of the transistors in or#er
to ens're that the other transistors hae f'lly cease#
con#'ction.
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9
.evision:
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.evision:6Characteristic of DC $otor
6Controlle# 0y s'pply oltage/c'rrent an#
e%cite# oltage/c'rrent ( #epen# on config'ration)
65ingle !hase Controller Conerter6Balf wae65emi conerter6'll conerter6D'al conerter
6reeweeling #io#e effect6Determine aerage oltage an# rms oltage
Controlling Techni1'es65C+ techni1'e
65ingle phase67 phase
67 p'lse conerter6< p'lse conerter
6!2$65ingle 1'a#rant6Do'0le/secon# 1'a#rant