Chapter 1 DC Drives Part2

8/9/2019 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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6/75

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


7/75!

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)*

%% =


8/75"

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


9/75#

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


10/75$%

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

=

++=

=

==

==+


17/75

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 #


22/75

5pee#

Tor1'e

(iel#

wea"ening

voltage&'Secondary+7.488&47'%%.%+%%.%+

rating&+er'Transformk+7.488)%%.%+

.k


23/75

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


28/75

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/


conerterthetooltagephasema%im'minp'trmswhere

sinoltage,phasetheet


ma%

a

ma%

ma%

9;4

9;

9;

4

9;

-

$

B6

B$%66 D

cosE

sin/


ma%

ma%

a

V

dVVV

V

da


31/75

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


D t il d A l i f 3 P l Q d t O ti


32/75

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



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11

!ully "ontrolled #$wo-%uadrant&

*ach 5C+ con#'cts for/7 ra#iansTriggering se1'ence =,


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35/75

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/


cos:waecosineaas#efine

tage,aerageolofncalc'latioforwaeformThe

0elinesthe0etweenoltagetheet


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


39/75

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:

)(

@.


48/75

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


52/75

58


53/75

s

sdssTsG

d

offons

ss,

T

T&k%&'$+'Tk&+$'+++

voltageofwaveformthe*rom

TTTperiod,Switching

T&k%'rtransistooftime-HffkTrtransistooftime-Hn

+==

+=

==

v


54/75

54

2aeforms with allowances for transistor an# #io#e oltage #rops.


55/75


56/75

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&,')


63/75

p ' '

!irst %uadrant operation: 8otorin' mode

$wo Quadrant Operation - .e'enerative 7ra0in':


64/75

Second %uadrant operation: .e'enerative 9ra0in' mode



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p ' '



66/75

p ' '



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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

(



68/75

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.



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70/75

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

Documents

Chapter 1 DC Drives Part2