DC Drives System

8/2/2019 DC Drives System

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I. D C Drive Systelns1.1 Introduction

DC motors are used extensively used in variable-speed driveswith power levels ranging from fractional HP ~ few MW.

Advantages of using DC motors:i) High starting torque;ii) Wide speed range;iii) Methods of speed control are simple;iv) Less expensive than AC drives.Disadvantages:. i) Commutators limit its very high speed applications;ii) More maintauce required;iii) Sparks.

Although the future trends is toward AC Drives, DC drives arecurrently used in many industries.

DC drives can be classified into 3 types:i) Single phase drives;ii) Three phase drives;iii) Chopper drives.


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Armature 2

~;


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3

The motor torque, Tm (s) = torque delivered to the load, TL (s) , i.e.[ f ]

where Td (s) = disturbance torque (often neglected)Load torque for rotating inertia,

2TL (s) = J s fJ(s) + !sfJ(s) [g]

From Eqs.[c],[e]-[g], the transfer function of motor-load combination:fJ(s ) 1("= =

Vj (s) s( J s+ f )(Ljs+ Rj )1(" / JLj

[h]

[il

Field

DisturbanceTd(s)

! Load Speed r---,o h ( S ) 1 ~ 1 W ( S \ I . _ _~--,t----. ~ ~ ~ : : 1Ir(s) Km Tm(s)Rf+ Lfs +Figure 2. Block diagram model of field controlled de-motor,

In terms of time constants,[jl

. 7L =J/!Typically, one finds T _ t < 7L and often T _ t may be neglected.


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ii) Armature control (voltage applied to armature terminals)Assume constant field current,

Motor torque, T I 1 , ( S ) = (](1I~If)L,(s) = 1 ( 1 1 1Ia (s)~l (s) = (R; + La s) Ia (s) + Vbrmature voltage,

where Vb = back emf : motor speed, i.e.

Armature current,Vb (s) = ](b 0) (S )

v " (s) - K"O)(S)fa (s) = = _ _ _ :; _ _ : - '- - ~ ~ _ _ :_RII+L(,sLoad torque for rotating inertia,

TL (s) = Ji()s) + f si) (s) = Tm (s) - J : l (s)From Eqs.[k],[n],[o], the transfer function of motor-load combination:

e (s ) K" ,= =v " (s) s[( Js+ f )(LlIs+ R,,) +KbK",]

For many DC motors, Ta is negligible, hence8(s) = =v , t (s ) Kill [K ", / (Ra/ +KbK:,,)]s[R(JJs+ f)+~ ,K:nJ - s(r]s+1)

RaJr-----:--"'---1- R" /+K"K, , ,here equivalent time constant,

DisturbanceTd(s)Armature Speed

+ Kill TIlI(s) - h(s) 1 (,)(s) I-- -Ra + LaS + Js +F s-KbBack emf

Figure 3. Armature controlled de-motor.

4

[k][I]

[m][n]

[0]

[p]

[q]

[r]

[s]

PositionO(s)


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1.3 Characteristics of DC Motors5

The equations describing the 1/0 relations of a separately excitedmotor shown.

+ ia, I. i f o l ,

f : v,L ,Va . Va A ,

Instantaneous field current, i _ r ;

Instantaneous armature current, i a ;

Motor back emf, known as speed voltage, eg = K; m i _ rTorque developed by motor,

where l(i = proportional constant;

Figure 4.. Equivalent circuit of sepa-rately excited de motors.

Also, ~l = J :~ +Bm + TLwhere J = moment of inertia, Kg.m2

c o = motor speed, rad/sB = viscous friction constant, N.m/rad/sK;= voltage constant, V/A-rad/sK, = K; = torque constant


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6

La = armature circuit inductance, HLf= field circuit inductance, HR;= armature circuit resistance, QRf = field circuit resistance, QTL= load torque, N.m

Under steady-state conditions, steady-state average quantities:( 1 - 1 )(1-2)

~l = Rela + Eg= RaIa + s, w1f (1-3)

r, = KtIfIa= Boi + TL

(1-4)(1-5)

Power developed, (1-6)From Eq.(1-3), speed of separately excited motor,

V , - R"IaKvVf / Rf (1-7)

Motor speed control can be varied by controllingi)ii)iii)

- voltage control- field control- Ia for fixed If

Define: Base speed, C O b - speed corresponds to rated Va' If & 1(1Note: For c o


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7

......-------------Power, PdI----------:;...


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8

From Eq.(I-9), speed of series motor,

(1-12)

For speed control, we can vary Va and/or fa .

From Eq.(I-IO), series motor can provide a high (starting) torque,commonly used in traction applications.

Note: For O J


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1.4 Operating Modes9

For DC motors, there are 5 operating modes in variable-speedapplications:

i) Motoringii) Regenerative brakingiii) Dynamic brakingiv) Pluggingv) Four quadrants

i) Motoring - Eg < Vaii) Regenerative braking - act as generator; Eg > Va ; - ia , + if ;

motor K.E. ----)-aiii) Dynamic braking - similar to ii) except Va is replaced by braking

resistance Rb, motor K.E. dissipated in Rbiv) Plugging - braking, armature terminals reversed while running,

Va & Eg same direction, i,eversed-> braking torquev) Four quadrants -

a) Forward motoring - + ( Va' Eg, t.,Ttl, O J ) , Va> Egb) Forward braking - forward direction, + Eg, - Ia , Va


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A 2 F2Separately excited motor Series motor(b) Regenerative Braking

A 2 F2Separately excited motor(a) Motoring

la A 1 F1 I f

L rR f V r

F l

L fR f

A 2 F2Separately excited motor(c) Dynamic Braking

If1

L fR f

A 2 F2Separately excited motor(d) Plugging

1 0

Series motor

la = I f +Eg

Series motor

E g+

Series motor

Figure 10. Operating modes.


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1.5 Single-Phase DrivesII

The armature circuit of a DC motor is connected to the O/P of asingle-phase controlled rectifier, ~ Va by ~ au (delay angle).

Types of DC drives:i)ii)iii)iv)

H aIf-wave-converterSemi-converterFull-converterDual-converter

Single-phaseacsupply

+

Figure (r

acsupply0-----

Controlledrectifier

0.5 KW;I5KW15KW15KW

L-quad.J-quad.2-quad.4-quad.

+'\

~ L , Single-ph( R , ac supV ,

-

aseply

B asic c ircu it amlngcl1lcnt o r a single-phase d e d riv e.

Controlledrecti fier

In SW,r-~-~ u-----~

I~(a) Armature reversal

I r

+V ,

(b) Ficici rever sal


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12

1.5.1 Single-Ph~se Half-Wave-Converter Drives

(a) Circuit

P-i-J-Ll(b) Quadrant

O~------------------~wl

o I-----------_____J-------_ wi; ~ 1 ~ '" '.

IT 2n

n

Flgurc 1 3 Single-phase half-wave converter drive.(c) Waveforms

Average armature voltage,

~I/ (1 )27f + cos ail ; for 0 _::;il _::;7fwhere Vlll = peak ac supply voltage.

(1-13)

With semi-converter field circuit, average field voltage,

V ; " (1 )~(= ff. + cos aJ ,. (1-14)

1.5.2 Single-Phase Semi-Converter Drives

l,R ,

(a)Circuj(ho l ~ ~ I "

i.

(I)) Quadrant (e) Wovclorms

Figurc 1 4 - Single-phase scm icunvcrtcr drive.


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1 3

With semi-converter armature circuit, average armature voltage,

(1-15)

With semi-converter field circuit, average field voltage,

(1-16)

1.5.3 Single-Phase Full-Converter Drives

+

i.

o f-------'--~----+- wtI.f---~------~

(a) Circuit i,~/----_-I= = : = = J _ wta.--~~-I.'--

(b) Ouadrant (e) Waveforms

Figure IS S ing le-p hase full-c onv erter driv e.

With full-wave converter in the armature circuit, average armaturevoltage,

(1-17)

With full-wave converter field circuit, average field voltage,2VVf = __II cos a1[ if ' (1-18)


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14

1.5.4 Single-Phase Dual-Converter DrivesConverter 1 Al Converter 2

L ,R ,

Figure 16 Single-phase dual-converter drive.

With 2 full-wave converters in the armature circuit, either converter 1operates +V~, or converter 2 operates -Va ; for converter 1 with aulgives average armature voltage,

(1-19)

for converter 2 . with au2 gives average armature voltage,

2 1 1 ; "-- cos au2 ; for 0::; aa2 ::; 1 C1C (1-20)

With full-wave converter field circuit, average field voltage,

(1-21)


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1 5Exam le I

The speed of a separately excited motor is controlled by a single-phase scrniconver-ter ir, Fig. 14 .a. The field current, which is also controlled by a serniconverter. is setto the maximum possible value. The ac supply voltage to the armature and fieldconverters is one-phase. 208 V, 60 Hz. The armature resistance is R" = 0.25 fl,th e field resistance is R( = 14 7 n, and the motor voltage constant is K; '= 0.7032V/A-rad/s. The load torque is T ,_ = 45 N '!l1 at 1000 rpm. The viscous friction andno-load losses arc negligible. The inductances or the armature and field circuits aresufficient enough to make the armature and field currents continuous and ripple-free. D etermine (a) the field current 1 (: (b) the delay angle of the converter in thearmature circuit. ao: and (c) the input power factor PF of the armature circuitconverter.Solution V I = 208 V. V J ) J = v '2 x 208 = 29-l.16 V. R; = 0.25 n. R _ ( = 147 n. T" =h= 4S N r m , K L " = 0 .7 03 2 V /A - rad /s. and (,) = 1000,,/30 = 104.72 rad/s.

(a) From Eq. ( 1-(6). the maximum fi eld v olta ge (anJ current) is obtained 1'01' adelay angle of aJ = 0 an d

V _ 2 V I I / _ J x 294.16(- - ---- = 187.27 V. IT ITThe field current is

I, ' -_ V i" = 187.27 __.... 1 .2 74 AR, 1 4 7(b) From E q. ( 1-4).

T ,J 4 5I" = 1\, .1 , = O.7()32 x 1.274 = 50.23 /\

rrom Eq. (T -2).E ; = I\,(I)/} = O.7().~2 ~ < IO..L 72 ,'. 1.274 - t )3.~2 V

From Eq. (1-3), th e armatu re voltage isV , = Y:U';2 + I " R t ! :co 93.82 -[- 50 .2 .i x 0.25 = l)l.K 2 -;. 12 .5() = 106.38 V

From E q. (I-15). V o / = 1 0 6 . 3 0 = (2lJ4.16/ iT) x (I + cu s (t"l and this gives the delayangle as 0'" = = X2.2" .

(c) ltthc armature current is constant and ripple-free. the output power is P " =\ '"1,, = 106.38 x 50.23 = 5343.5 W . II' the losses in the armature converter arcneglected. the power from tile supply is P " " " I ) " = 5343.5 W . 'The rms input currentof the armature converter as shown in Fig. l-l is

([ L'() _ X ' J I~0 l _

= )0._3 180 ) 37.03 Aand the input volt-ampere rating. VI = VJI( / = 208 x 37.03 7702 .24 . A ssumingnegligible harmonics, the input power factor is approximately

P I! 5343.5 4 I . )I)F = - = = 0 69 (ClQQIl1Q,VI 7702.24 . ~~ ~


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1 6Exam JIe 2 .. T he speed of a separately excited de motor is controlled by a single-phase full-wave

converter in Fig. IS a. The field circuit is also controlled by a full converter andthe field current is set to the maximum possib le value. The ac supply voltage to thearmature and field conveners is one-phase, 440 V , 60 Hz. The armature resistance isR; = 0.25 fL, the field circuit resistance is R. r = 17 5 n, and the motor voltage constantis K l' = 1.4 V / A -rad/s. The armature current corresponding to the load demand isI" = 45 A . The viscous friction and no-load losses arc negligib le. The inductances ofthe armature and field circuits arc sufficient to make the armature and field currentscontinuous and ripple-free. If the delay angle of the armature converter is 0 ' < 1 = 60and the armature current is /1 1 = 45 A , determine (u) the torque developed by themotor T,I , (b) the speed ('), and (c) the input power factor PF of the drive.Solution V , = 440 V , V,, , = V 2 x 440 = 62 2 .2 5 V . R" = 0.25 il,R , = 17 5 n, (X it =600 , and K " = 1.4 V / A - rall/s.(a) From Eq, (I -IKL th e maximum field v olta ge (and current) would be ob-tained for a delay angle or ('(, = () and

I V I x 622.2. ' i\'.= : : : : _ _ _ _ I _ I . ! =t rr ~-,-,-- J96.14 VThe Iiclt] ell rrcnt is

J~6.1417.' i = 2 .26 :\From E q. (1-4), the developed torque is

T" ::.:T: = 1\(.1,1" = 1.4 x 2 .26 x 45From Eq. ( ]: -17). th e a rm ature voltage is

142 .4 N 'm

The back emf isL ~ = \ '" - I " H " - 1l)~ U )7 - - L " x 0 .2 .'i

From E q. ( I -2 J. the speed isL .. 1~ 6.K ~ _ _ . _

w = F I . = I 4 I 1(- ::.: )~.O)rad/s or )M rpm}\.1' J . X _._)

I ~ 6 .X 2 V

(c) A ssuming loss!css converters. the total input power from the supply isP i = V " II I 1 V J IJ = I l ) X. 0 7 x 4 5 -I Jl)().14 x 2.2() = l ) E O ~ . 4 W

The input current or the armature converter Ior a highly inductive load is shown inFig. II and its rrns value is /,,, = I " = 4 S A . The rms value of the input current offield converter is l,r = Ir = 2 .2 6 A . The effective rrns supply current can be foundfrom

1 " . = (/~ + l~ .)I!~.\II .~I

and the input volt-ampere rating, V I ::.; rJ, := 440 x 45.06 = 1< ).0 2 6.4 . N e gl ec tin gthe ripples, the input power is approximately

PF = !.i = 9808.4 0 49S I .V I J 9,826.4 = . . (agging)


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1 7

Example ~If the polarity of the motor back emf in E xample 2. is reversed by reversing thepolarity of the field current, determ ine (a) the delay angle of the armature c ircuitconverter, a'i' to maintain the armature current constant at the same value of J [ J =45 A; and (b) the power fed back to 'the supply due to regenerative braking of th emotor .Solution (a) From part (b) of E xample 2 . the back emf at the lime o r polarityreversal IS E; =; 1~ 6.tl2 V and after polarity reversal L ~ = -186.82 V . FromEq. ( I -3).

\"1 = ! - ' : 1 + foRo = -186.82 + 45 x 0.25 =; -175.57 VFrom E q. (1 .-17 ).

') V ') x 622 .25V , I =; ~ cos 0."1 = cos 0"1 = - 175.57 V'iT 7 Tand this yields the delay angle 0[' tile armature converter as n , = 116.31.

(b) The rower ted back to the supply. P; = \',JII = 175.57 x 45 = 7900 .7 W .Note. The speed and back emf of the motor will decrease with time. If the

armature current is to be maintained constant at J ( / = 45 A during regeneration, thedelay angle o r the arm ature converter has to be reduced. This would require aclosed-loop control to maintain the armature current constant and to adjust thedelay angle continuously.


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1 3

1.6 Chopper DrivesWidely used in traction applications.A DC chopper is connected between a fixed-voltage DC source

and a DC motor to ~ Va' In addition to ~I control, a DC chopper canprovide regenerative braking of motors and can return energy back tothe supply. Attractive feature to transportation system with frequentstops (e.g. MTR). DC motor can be operated in one of the fourquadrant by controlling V ' ; - r and/or VI ( or currents).

Possible control modes of a DC chopper drives are:i) Power (or acceleration) controlii) Regenerative brake controliii) Rheostatic brake controliv ) Combined regenerative and rheostatic brake control

1.6.1 Power ControlTo control Va of DC motor; a one-quad. drive.

+ia1 0

Vo 0is1 0 I

Va kT TV , , ,0 kT T

VehV , - - - - I I

kT

(a) Circuit

(0) Ouadrant (c) W


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1 9

Average armature voltage, Va = k ~ ) (1-22)where k is the duty cycle.

Power supplied to motor, P v = Va I I I = k ~ ~ I a (1-23)where I a is the average armature current.

Assume lossless chopper, i.e. Pi = Po = Vs I s, average input current,(1-24)

Equivalent input resistance, Req V -I . " V -I" k (1-25)

Power flow to motor (and speed) can be controlled by ~ duty cycle, k .

Exam lie 4A de separately excited motor is powered by a de chopper (as shown in F ig. 14-18a)from a 600- V de source. T he arm ature resistance is RII = 0.05 n. The back emfconstant of the motor is Kc' = 1.52 7 V I A -rau/s. T he average arm ature current is l ; =2 50 A . The field current is I I ' = 2 .5 A .. T he arm ature current is continuous and hasnegligib le ripple. If the duty cycle of the chopper is 60% . determ ine (a) the inputpower from the sou rce. (b) the equivalen t input resistance of the chopper drive, (c )tne motor speed . and (d) the developed torque.Solution V , = 600 V . 1 ,/ = 2 50 A . and J . : . = 0 .6 . T he total armature circuit resitanceis Rill = R; = 0.05 n.

(a) From E q. ( 1-21-.).Pi = kV, .11 l = 0 .6 x 600 x 2 50 = 90 kW

(b) From E q. ( I-20"). R


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20

1.6.2 Regenerative Brake ControlMotor acts as generator and motor' & load' K.E. is returned

back to supply. A one quad. (2nd quad.) drive.ir i.is I.: J L t i. + Dm +R t 0ic isI.r---- - _ l Ieh V s 0 kT T~g ! icla- if I 1(a) Circuit kT T4,. ve nVs -- _ . . . . . . - - J II, kT T

(b) Quadrant (c) WaveformsFigure (8 Regenerative braking o r de separately excited motors.

Average chopper voltage, Veil = (J - k) ~\' (1-26)Regenerated power, Pg = fa (1- k) ~\' (1-27)Voltage generated, Eg = KvII IOJ = Vei l + Rm1a

- (1 - k) ~\' + RI1 l I II (1-28)Equivalent load resistance, Req = E !iI" ; (1 - k ) +Rill" (1-29)

By varying the duty cycle, k , R cq can be varied from Rill to(Vs/ fa +Rill) and the regenerative power can be controlled.Minimum brake speed, OJmill (1-30)

Maximum brake speed, OJ I1U1X = (1-31)


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21

Example 5A dc chopper is used in regenerative braking of a dc series motor similar to thearrangement shown in Fig. 14-19a. The de supply voltage is 600 V. The armatureresistance is R; = 0.02 n and the field resistance is Rf = 0.03 n. The back emfconstant is K; = 15.27 mV/A-rad/s. The average armature current is maintainedconstant at Ia = 250 A. The armature current is continuous and has negligible rip-ple. If the duty cycle of the chopper is 60.%, determine (a) the average voltage acrossthe chopper, Veh; (b) the power regenerated to the dc supply, Pg ; (c) the equivalentload resistance of the motor acting as a generator, Req; (d) the minimum permissiblebraking speed, wmin; (e) the maximum permissible braking speed, W ma x; arid (f) themotor speed.Solution Vs = 600 V, Ia = 250 A, K; = 0.01527 V/A-rad/s, k = 0.6. For a seriesmotor R 1 I 1 = R; + Rf = 0.02 + 0.03 = 0.05 D.(a) From Eq. ([-2(;), Veh = (l - 0.6) x 600 = 240 V.

(b) From Eq. ([.-27), Plf = 250 x 600 x (l - 0.6) = 60 kW.(c) From Eq. ([-2,9), Req = (600/250)(1 - 0.6) + 0.05 = 1.01 n.(d) From Eq. (z:-30),the minimum permissible braking speed,

Wmin = 0.~~}27 = 3.274 rad/s or 3.274 x 30 = 31.26 rpm7T(e) From Eq. (I-J/ ), the maximum permissible braking speed,

600 0.05Wma x = 0.01527 x 250 + 0.01527 = 160.445 rad/s(0 From Eq. (14-3~), E N = 240 + 0.05 x 250 = 252.5 V and the motor speed,

or 1532.14 rpm

252.5w = 0.01527 x 250 = 66.14 rad/s or 631.6 rpmNote. The motor speed would decrease with time. To maintain the arma-

ture current at the same level, the effective load resistance of the series generatorshould be adjusted by varying the duty cycle of the chopper.


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22

1.6.3 Rheostatic Brake ControlEnergy is dissipated in a rheostat, may be used as heating.

Rheostatic braking is also IUlOWIl as dynamic braking. A one quad.(2ndquad.) drive.Average braking resistor current, I b = I I I ( 1 - k) (1-32)Average braking resistor voltage, Vb = = Rb Ia ( 1 - k) (1-33)

Equivalent load resistance, Req = = ; J ; : : : : ; :, (1- If) +u;11

(1-34)

Power dissipated il l Rb, Ph ;::::;:iRb ( 1 - k) (1-35):. B y controlling k, Req can be varied from Rm to Rm +Rb .

(a) Circuitr-------~-----L~-----~t_d

. 0 -J"(b) Quadrant

r-------L------L-------~t(c) Waveforms

Figure 19 ' Rheostatic braking of de separately excited motors.


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23

1.6.4 Com billed Regenerative and Rheostatic Brake ControlMost energy efficient drive method in practical traction systems.

During regenerative brakings, if it exceeds, say 200/0, above the linevoltage, regenerative braking is removed and rheostatic braking isapplied. An almost instantaneous transfer from regenerative braking torheostatic braking if the line becomes nonreceptive.

If it is nonreceptive, thyristor TR is turn 'on' to divert the motorcurrent to the resistor Rh Thyristor TR is self-commutated whentransistor Q J is turned 'on' in the next cycle.

Figure 20 Combined regenera-tive and rheostatic braking.

+


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1.6.5 Two-Quadrant Chopper Drives24

Power control, 1st quadrant, ~l ,I a are positive. Q J & D2operate. Q J is turned 'on', Vs is connected to motor terminal. Q J isturned 'off', L, decays through D2 .

Regenerative braking control, 2nd quadrant, + Va' - Ia . Q2 & D1operate. Q2 is turned 'on', motor acts as generator & lat. Q2 is turned'off', motor acts as generator returns energy to supply through DJ

+

----l---f---"- ....ao(a)Quadrant (b)Circuit

D,

iai r

D2 : i J ~ 1Figure 21 Two-quadrant transistorized chopper drive.


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25

1.6.6 Four-Quadrant Chopper DrivesForward power control - QJ & Q2 operate. Q3 & Q4 are off. When QJ &

Q2 are both 'on', r: is connected to motor terminal & lat. When Q Jis turned 'off and Q2 is still 'on', i,decays through Q2 &D4 .

Forward regeneration - QJ' Q2 & Q3 are 'off. When Q4 is 'on', t,t andflows through Q4 & D2 When Q4 is 'off, motor acts as generator,returns energy to supply via D J &D2

Reverse power control ~Q3 & Q4 operate. QJ & Q2 are off. When Q3 &Q4 are both 'on', Iat& flows in reverse direction. When Q3 is turned'off and Q4 is still 'on', fa falls through Q4 & D2 .

Reverse regeneration - QJ' Q3 & Q 4 are 'ofr. When Q2 is 'on', Iat andflows through Q2 &D4 When Q2 is 'ofr, Ia - t . and motor returnsenergy to supply via D3 &D4

QQ02Q4 Va 1 2

4 O2(a)Quadrant (b)Circuit

Figure .22 Four-quadrant transistorized chopper drive.

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DC Drives System