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ELECTRIC MOTOR DRIVES w ww Modeling, .Ea Analysis, and Control syE ngi nee rin g.n et
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ELECTRIC MOTOR DRIVES Analysis, and Control wModeling, ww .Ea syE ngi nee rin g.n e R. Krishnan
Virginia Tech. Blacksburg. VA
t
Pn'nlin'

11,,11
Upper Saddle River, New Jersey 0745S Downloaded From : www.EasyEngineering.net
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Ubrary of Congress Cl.taJog_in·PubliQllion DaH' Krishnan. R. (Ramu) Electric motor drivt:5: modeling. analysis. and mntroll R. Knshnal!.
p.em. Includea bibliographical rckrcnces and irKkx_ ISBN 0.130910147 \. E1cctric driving. I,Tllle. TK40S8.K73 2001 62146'dc21
ww w
00050216
Vrec l>resldenl and Editorial Dircctor. ECS. MtI'C"Ul J. HOfll/tl Acquisitions Edilor. Eric: F,... Editorial Assistanl:Jr.uit:" H.(;o VICC President of Prodoction and Manufacturing. ESM: Oovid IV HIl'co"l; Executlvc Managing Edilor: V"mu O'Brien Managing Editor: Dllvid A ~ Plodut:tion Editor. Lmuhmi 8010Speciallhank!> and grato itude. They are Mr. Pnwccn ViFIY rri"C/ple of Opertllioll 183
031
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Downloaded From : www.EasyEngineering.net xvi
Contents 6.3.2 SlipEnergy R~col'ery Scheme 283 6.3.3 St~(Jd,'State Analysis 285 6.3.3.1 Range of Slip 287 6.3.3.2 Equivalent Circuit 287 6.3.3.3 Performance Characteristics 289 6.3.4 Startillg 296 6.3.5 RQ/irrg ofthe Com·trters 296 6.3.5.1 Bridge·Rectifier Ratings 297 6.3.5.2 Phase·Colltrollcd Converter 297 6.3.5.3 Filter Choke 2 of II >U1g.IC'·rhll'le cOlllrollcdhndge
ngi nee
,(;)
,.,
,, ,,
i:1ng .ne
Ill... df.:CI of source ,mp"'d~ncc on the ,'per,llt"" "f a ~m!!Ic'rhase co"',,,lkd rCCl'!'.." m slca,1 ~lale
t
source impedance. invariably reducing the load vol!age to LCro. I·!encc. the O\crall effect of source inductance is to reduce Ihe availahle dc output volla~e. Figure 3.9 contains the operational waveforms with source inductance. The source inductance can be introduced by the isolation transformer or hy inlentionally p[;1Ccd rl·,u:lurs. to reduce the rate of rise of currents in Ihe thyristors.. If the source induct'lllCC ;s 1...,,the voltage lost due 10 it is V,:::
1
l'·· .. Vmsin(w,t)d((lI,tl::: n •
Vm
[coso  cos{o + IJ.J. n
1.1.22)
where IJ. is the overlap conduclion period. By equating Ihis voltage to thc voltage drop in the source reactance. the uver· lap angle IJ. IS obtained as IJ.
= cos
where loX is the load current
,[CO'\O 
In stead~
slale. Downloaded From : www.EasyEngineering.net
Downloaded From : www.EasyEngineering.net Section 3.3
PhaseControlled Converters
S1
These characteristics are modified when the load includes a counteremf. as in the case of a dc machine.lhere is an additional feature in discontinuous operation. with the induced emf appearing across the output of the converter during zerocurrent intervals, but, if the source emf is instantaneously greater than the back emf, then the conduction starts but will not end immediately when the source emf becomes less than the induced emf, because of the energy stored in the machine inductance and in the external inductance connected in series to the armature of the machine. llierefore. conduction will be prolonged until the energy in these inductances is depleted. The threephase controlled fullbridge converter is similar in oper
K,
f
et
(3.89)
" I" ...
I
+sr,
~'".
H, FiIlU",3,J1
Smlphficd curremconlrolloop
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Downloaded From : www.EasyEngineering.net Section 3.7
where T J
:::::
Design of Controllers
79
T l + T,. The transfer (unction can be arranged simply as K,
(3.90)
+ sT;) where
T,
T~
(3.91)
I + Kn
,
K li
1
K, = H~ . 7(1::+"K;,')
(3.92)
(3.93)
ww
The resulting model of the current loop is a first·ordersystem.suitable for use in the design of a speed loop.lne gain and delay o( the current loop can also be found experimentally in a motordrive system.lnat would be more accurate (or the speedcontroller design.
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3.7.3 Speed Controller
En gin eer i
The speed loop with the firstorder approximation of the currentcontrol loop is shown in Figure 3.32.1ne loop gain function is
G H (s l ,
::: {
K,K,KbH .. } BIT,
(1 + sT,) s(1 + sT,l{l + sTmHI + sT,.)
ng. net
."","",+,,'::;''';;;::0;",,,
(3.94)
This is a fourlhorder system. To reduce the order of the system for analytical design of the speed controller. approximation serves. In the vicinity of the gain crossover frequency. the following is valid: (3.95)
w;
,
K,{I +sT,)
.
.T,
Speed controller
w_
.i.
_K_,_ 1.~T,
.2
Kt/B,
I sT..,
ro w.
Currelllioop
~ 1"~T..
Figure 3.32
RepresentaliOn of the QU1Cr speed loop ,n the de
mOlO.
dnve
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Downloaded From : www.EasyEngineering.net 80
Chapter 3
PhaseControlled DC Motor Drives
The next approximation is to build the equivalent time delay of the speed feedback filter and current loop. Their sum is very much less than the integrator time constant, T•. and hence the equivalent time delay, T•. can be considered the sum of the two delays. T j and Too. This step is very similar to the equivalent time delay introduced in the simplification of the currentloop transfer function. Hence, the approx· imate gain function of the speed loop is
(3.96) where T. '" T;
K,:
ww w.E~I asy E
+ Too
(3.97)
.
(3.98)
K,KbH .. BT
,
The closedloop transfer function of the speed to its command is I [
w.(.)
w~(s)
=
H..
+ .T.) ] 1 (>, + >,.j K K KzK, = H.. (ao + als + azs 1 + s.s+sz·+T
'T +
where
Z
3}SJ)
(3.99)
•
ngi nee r
ao = KzK/f.
(3.100)
a l = KzK.
(3.101)
"I = I
aJ
'"
T.
ing .ne t
(3.102) (3.103)
This transfer function is optimized to have a wider bandwidth and a magnitude of one over a wide frequency range by looking at its frequency resJX>nse.lts magnitude is given by
I I WaouW)
I
 = w;(jw) H.
~ + cda~ {~+ wl(ai  2a032) + m'(a~  2a,a}) + m'a}}
(3.104)
This is optimized by making the coefficients of ul and w· equal zero. to yield the following conditions:
,,~ '" 2a032
(3.105)
a~ '" 2ala}
(3.106)
Substituting these conditions in terms of the motor and controller parameters given in (3.100) into (3.103) yields (3.101)
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Design of Controllers
81
resulting in
(3.IOB) Similarly, T~
K~Kj
=
21';1'4 K,K 1
(3.109)
which, arter simplification, gives the speedcontroller gain as K
•
~
I 
(3.110)
2K 2T 4
Substituting equation (3.110) into equation (3.108) gives the time constant of Ihe speed controller as
ww
1'. = 41'.
w.E
(3.111)
Substituting for K, and 1', inlo (3.99) gives the closedloop transfer function of the speed 10 its command as
asy
w",(s) I [ w;(s) = H .. 1
I + 41'4 s
+ 41'4s + 8T~s! + 81'~sJ
En gin
]
(3.112)
It is easy to prove thaI for the openloop gain function the corner points are ]f41'. and Iff4, with the gain crossover frequency being 1121'4' In the vicinity of the gain
eer i
crossover frequency, the slope of the magnilUde response is 20 dB/decade. which is the most desirable characteristic for good dynamic behavior. Because of its symmetry at the gain crossover frequency, this transfer function is known as a symmetric optimum function. Further, this transfer function has the following features: (i) Approximate time constant of the system is 41'•. (ii)
The step response is given by
ng. n
I
w,(\) = H.. (1 + e 1rrr•  2el,.4T·cos(V3t/4T4»)
et (3.113)
with a rise time of 3.11',. a maximum overshoot of 43.4%, and a sell ling lime of 16.51'•.
(iii) Since the overshoot is high_ it can be reduced by compensating for its cause, i.e.. the zero of a pole in the speed command palh. as shown in FIgure 3.33. The resulting transfer function of Ihe speed to ils command is :;;:; =
I~J I+ 41'.s + ~~s! + 8T~Sl J
(3.114)
whose step response is w,(t) =
~ .. (I
 e 1/4T. 
~ e ,/•T'sin(V3t/4T.d)
(3.115)
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Chapter 3
PhaseControlled DC Motor Drives
, 

I
I +4T.s

l+4T.s
I
".
1+4T.S+IIT/s2+8T. V
Compenutor
.
Spc'ed·1oop lran$fer funcllOiI
Filur" J.33 Smoothmg of the o\'ershoot VIa a oompenSatOi
wilh a rise time of 7.6T•. a maximum overshoot of 8.1 %. and a settling time of 13.3T•. Even though the rise time has increased. the overshoot has been reduced 10 approximately 20% of its previous value. and the scnling time has come down by 19%. (iv) The poles of the dosedloop transfer function arc
ww
w.E asy E I
. v'3
I
s ""  2T.;  4T.
'1.
(3.116)
J4T•
The real parts of the poles are negati\·e. and there are no repeated poles at the origin. so the system is asymptotically stable. Hence. in the symmctric Opllmum design. the system stability is guaranteed. and there is no need 10 chcck for it in Ihe design process. Whether Ihis is true for Ihe original systcm without approximation will be explored in the following example. (v) Symmetrk optimum eliminates the effects due to the disturbance very rapidly compared to other optimum techniques employed in practical systems. such as linear or nlOdulus optimum. This approach indicates one of the possible meth· ods to synthesize the speed controller. Thallhe judicious choice of approxima· tion is based on the physical constants of the motor, on the converter and transducer gains. and on time delays is to he emphasized here.
ngi nee
rin
g.n et
ThaI the speed·loop transfer function is expressed in terms ofT. is siglllficant 111 thaI it dearly links the dynamic performance to the speed·feedback and currentloop time constants. That a faster current loop wilh a smaller speedfilter tl1l1o.: constant accelerates the speed response is evident from this. Expressing 1', in terms of the motor, the converter and transducer gains. and lhe time delays by using expressions (3.91) and (3.97) yields l T, ~T, + "I'.. = ~ , + T.. = T + T. +T..
I+K tt
I+K r•
(3.\17)
Since Kr, » I. T~ is found approximately after substilUting for Kr. from equation (3.93) in terms of gains and time delays as
T4
(T l + T,)T, 
Tm
I • ;::;'::;:
KIK~K,H
5 O.OOh)·
I~ :::
10 V. The lachogcncralor has Ihe Iran,fcr tunClltll1
'm e speed re ference
w.E
current renll1l1ed II1lhe molur
IS
\'0 I la~C
h as
iI
. maximum (/ fIOV'!1 . 1e md\tmUm
20 A.
asy E
Solulion til Clln\L'rlcr transfer function:
1.35 V x 2;\(1 _ .11.(" :') V/V K,'    . 1.35 10 V,.
ngi ne
V ....(mltx) = 310.5 V
The raled de \
Kb + RaB t
eer
(1 + sTrn )
ing .ne
(1 + sTd(1 + sT2 ) + HeKrKcKJ(l + sTrn )
(4.69)
and Kc and Kr are the currentcontroller and chopper gains,
respectively. The chopper gain is derived as V K ==_S r Vern
t
(4.70)
where Vs is the de link voltage and Vern is the maximum control voltage. The gain of the current controller is not chosen on the basis of the damping ratio. because the poles are 010st likely to be real ones. Lo\ver the gain of the current controller; the poles will be far removed from the zero. ~The higher the value of the gain, the closer will one pole move to the zero, leading to the approximate cancellation of the zero. The other pole will be far away from the origin and will contribute to the fast response of the current loop. Consider the workedout Example 3.4 from the previous chapter. The values of various constants are: KI
T2
==
= 0.049 He 0.0208
== 0.355
Vs == 285 V
Trn == 0.7 T] == 0.1077 Vern == lOV
Kr == 28.5 V
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Chapter 4
ChopperControlled DC Motor Drives
The zero of the closed currentloop transfer function is at 1.42. Note that this is not affected by the current controller. The closedloop poles for currentcontroller gains of 0.1,1., and 10 are given below., along with their steadystate gains, in the following table. Kc
Poles
Steadystate gain
0.1 l.0 10.0
7.18. 65.11  3.24.  203.45 1.66. 1549.0
21.33 213.3g 2580.30
It is seen that, as the gain increases, one of the poles moves closer to the zero at __1_,
ww
TO'
enabling cancellation and a better dynamic response.
In highperformance motordrive systems., it is usual to have a PI current controller instead of the simple proportional controller illustrated in this section. The PI controller provides zero steadystate current error, whereas the proportional controller will have a steadystate error. In the case of the PI current controller, the design procedure for the phasecontrolled dc tuotordrive system can be applied here without any changes. The interested reader can refer to Chapter 3 for further details on the design of the PI current controller.
w.E
asy En gin e
4.12.7 Design of Speed Controller by the SymmetricOptimum Method
eri n
The block diagram for the speedcontrolled drive system with the substitution of the currentloop transfer function is shown in Figure 4.26. Assuming that the time constant of the speed filter is negligible, the speedloop transfer function is derived from Figure 4.26 as
Wm(s) 1 ao(1 + sTs ) w;(s)  H w • ao + atS + a2s2 + a 3s3
g.n
et
(4.71 )
where ao
Ks
= KST
(4.72)
s
= 1 + HcKrKcK t + KsK s a2 = T 1 + T 2 + HcKrKcKITm
at
a3
Ks
=
T 1T2
(4.73) (4.74 ) (4.75)
HwKrKcK
= K b   t Bt
(4.76)
This is very similar to the equation derived in Chapter 3, from which the following symmetric optimum conditions are imposed:
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Downloaded From : www.EasyEngineering.net Section 4. 12
Ks(l +sTs ) sTs
.*
ClosedLoop Operation
ia(s)
la
ia
Kb/Bt
Current loop
Speed controller + Limiter W mr
Wm
(l +sT m )
i:(s)
159
Motor and load
Hw l+sTw Tachogenerator + Filter
Figure 4.26
ww
Simplified speedcontrolled dc motor drive fed from a chopper with hysteresiscurrent control
w.E
ai a~
asy E
= 2a 03 2 = 2a,33
(4.77) (4.78)
Conditions given by equations (4.77) and (4.78) result in speedcontroller time and gain constants given by
ngi
(4.79)
nee
(4.80)
rin g
For the same example, the speedcontroller gain and time constants for various gains of the current controller are given next, together with the closedloop poles and zeros and the steadystate gains of the speedloop transfer function.
Current controller gain
Speed conlroller gain
Speed controller time constant
0.1 1.0 10.0
106.7 102.9 597.0
0.045 0.0188 0.0026
Steadystate gain 1628 38,000
1.6 x 10 7
.ne t
Zero
Poles
22.22 53.2 384.6
18.3 =!= j31.  36.9  52. 7 ~ j88.9, 105.1  393 =!= j666.8.  792.4
Increasing currentcontroller gain has drastically reduced the speedloop time constant without appreciably affecting the damping ratio of the closedloop speedcontrol system. Because of the large integral gain in the speed controller, its output will saturate in time.. An antiwindup circuit is necessary to overcome the saturation in this controller design and thus to keep the speed controller responsive. The antiwindup circuit can be realized in many ways. One of the implementations is shown in Figure 4.27.
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Downloaded From : www.EasyEngineering.net 160
Chapter 4
ChopperControlled DC Motor Drives
J
ww
Dead Zone Circuit Figure 4.27
Antiwindup circuit with PI speed controller
w.E
The saturation due to the integral action alone is countered in this implementation. It is achieved by a negativefeedback control of the integralcontroller output through a deadzone circuit. This deadzone circuit produces an output only when the integral controller output exceeds a preset absolute maximum, i.e., when the controller output saturates. This feedback is subtracted from the speed error and the resulting signal constitutes the input to the integrator. When the integralcontroller output saturates, the input to the integral controller is reduced. This action results in the reduction of the integral controller's output, thus pulling the integral controller from saturation and making the controller very responsive. If there is no saturation of the integralcontroller output, then the feedback is zero (because of the deadzone circuit); hence, the antiwindup circuit is inactive in this implementation. The outputs of the integral and the proportional controllers are summed, then limited, to generate the torque reference signal. By keeping the outputs of the proportional and integral controllers separate, their individual tuning and the beneficial effect of the high proportional gain are maintained.
asy
En gin e
eri ng. net
4.13 DYNAMIC SIMULATION OF THE SPEEDCONTROLLED DC MOTOR DRIVE
A dynamic simulation is. recommended before a prototype or an actual drive system is built and integrated, to verify its capability to meet the specifications. Such a simulation needs to consider all the motordrive elements, with nonlinearities. The transferfunction approach could become invalid, because of the nonlinear current loop. This drawback is overcome with the timedomain model developed below. The speedcuntrolled drive shown in Figure 4.20, but modified to include the effects of field excitation variation and with a PWM or a hysteresis controller, is considered for the simulation.
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Dynamic Simulation of the SpeedControlled DC Motor Drive
161
4.13.1 Motor Equations The de motor equations, including its field, are . di a Va == Ral a + Lad! + KrWm Vf ==
Te

. Rfl f
T, ==
(4.81 )
die
+ L cdt
(4.82)
dW
Jilim + Btw m
(4.83 )
Te == K tf Yes
eer
ing .ne
t
Displayl Print the results
Figure 4.30
Flowchart for the dynamic simulation of the choppercontrolled de motor drive
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Downloaded From : www.EasyEngineering.net Section 4. 13
Dynamic Simulation of the SpeedControlled DC Motor Drive
165
==~___4
O..~I.L.'L..I
o"....""""~
ww w.E a
o"....""""~
2,.,....,
I
syE ngi
nee
!
!
I
oo
0.005
0.01
0.015 Time.s
Figure 4.31
rin 0.02
g.n
et
Dynamic performance of a onequadrant choppercontrolled separatelyexcited dc motor drive for a step command in speed reference. in normalized units
Figure 4.32 shows the performance of a fourquadrant drive system with hysteresiscurrent controller, for a load torque of 0.25 p.u.. The reference speed is stepcommanded, so the speed error reaches a maximum of 2 p.u., as in the firstquadrant operation described in Figure 4.31. When the speed reference goes to zero and then to negative speed, the torque command follows, but the armature current is still positive. To generate a negative torque, the armature current has to be reversed, i.e, in this case to negative value. The only way the current can be reversed is by taking it through zero value. To reduce it to zero, the armature current is made to charge the dc link by opening the transistors, thus enabling the diodes across the other pair of nonconducting transistors to conduct. This corresponds to part of the
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Chapter 4
ChopperControlled DC Motor Drives
ww
w.E asy E
ngi n
eer
o Figure 4.32
0.01
0.02
0.03
0.04
0.05 0.06 Time. s
ing 0.07
.ne t
0.08
0.09
0.1
Dynamic performance of a fourquadrant choppercontrolled separatelyexcited dc motor drive for a step command in bipolar speed reference. in normalized units
secondquadrant operation, as shown in Figure 4.7, with the diodes 0304 conducting. When the current is driven to zero, note that the speed, and hence the induced emf, are still positive. The current is reversed with the aid of the induced emf by switching T 4 only, as in to the fourthquadrant operation shown in Figure 4.10. The induced emf enables a negative current through 02' the machine armature, and T 4 . The current rises fast~ when it exceeds the current command by the hysteresis window, switch T 4 is turned off, enabling 01 to conduct along with 02' The armature current charges the dc source, and the voltage applied across the machine is + Vs during
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Section 4.14
167
this instant. If the current falls below the command by its hysteresis window, then the armature is shorted with T 4 again, to let the current build up. This priming and charging cycle continues until the speed reaches zero. The drive algorithm for the part discussed is briefly summarized as follows.
Wr
T*e
Wr
la
ia
+
+
+
+
+
If W r
w;.
If W r >
w;. itT ~ ~i. set
+
+
+
+
ww w
Condition er ~ die set Va = V s ier ::s di. reset Va = 0 (Quadrant I)
Va = Vs until i a (Quadrant I) Va
i er ::s .:ii. reset (Quadrant IV)
~()
= V. Va = 0
.Ea
To reverse the rotor speed, the negative torque generation has to be continued. When the stored energy has been depleted in the machine as it reaches standstill, energy has to be applied to the machine, with the armature receiving a negative current to generate the negative torque. This is achieved by the switching of T~ ·and T 4 . From now on, it corresponds to third and secondquadrant operations for the machine to operate in the reverse direction atthe desired speed and to bring it to zero speed, respectively. For the operation in the reverse direction, the input armature voltage conditions can be derived as in the table given above. TIle quadrants of operation are also plotted in the figure, to appreciate the correlation of speed, torque, voltage, and currents in the motor drive system.
syE
ngi
nee r
4.14 APPLICATION
ing
.ne
t
Forklift trucks are used in materialhandling applications, such as for loading, unloading, and transportation of materials and finished goods in warehouses and manufacturing plants. The forklift trucks are operated by rechargeable batterydriven dc series motordrive systems. Such electrically operated equipment is desired, to comply with zero pollution in closed workplaces and to minimize fire hazards. The dc series motor for a typical forklift [6] is rated at 10 hp, 32 Vdc. 230 A, 3600 rpm and is totally enclosed and fancooled. It operates with a duty cycle of 200/0. Fourquadrant operation is desired. to regenerate and charge the ~at teries during braking, and both directions of rotation are required, for moving in the forward and reverse directions. The torque is increased by having a gear box between the motor and the forklift drive train. A schematic diagram of the forklift control is shown in Figure 4.33. The presence of an operator means that the forklift has no automatic closedloop speed control; the operator provides the speed feedback and control by continuous variation of speed reference. Choppercontrolled dc motor drives are also used in conveyors. hoists, elevators, golf carts, people carriers in airport lobbies, and some variablespeed hand tools.
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Downloaded From : www.EasyEngineering.net 168
Chapter 4
ChopperControlled DC Motor Drives Truck drive train
1, I
I
I
I
: :
: orward Speed: reference:
I
r
Stop
Voltage sense Logic! Control! Protection
i dJverse : :I
vc
'    
 
 


 
I  __ I
  
Operator panel
ww
Figure 4.33
Limiter External torque limit
Blockdiagram control schematic of a batteryfed choppercontrolled d~ series motordrive system
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Example 4.6
asy En gin eer
A separatelyexcited choppercontrolled dc motor drive is considered for a paper winder in a windingunwinding controller. This application uses essentially constant horsepower because the tension of the strip decreases as the buildup increases on the roll~ hence, the winder can be speeded up. The maximum increase in speed is determined by the output power of the drive system. The motor ratings and parameters are as follows: 200 kW, 600 V, 2400 rpm, Ra = 0.05 fi, La = 0.005 H, Kb = 2.32 V/rad/sec, BI = 0.05 N·m/rad/sec, J = 100 K g_m2, ~ = 30 fi, Lf = 20 H, Input = 460 V :t 100/0,3 ph, 60 Hz :t 3 Hz. Design the chopper power circuit with a current capacity of 2 p.u. for short duration and dc link voltage ripple factor of 1% for dominant harmonic. (The ripple factor is the ratio between the rms ac ripple voltage and the average dc voltage.) Solution
ing
.ne t
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Application
169
+
Figure 4.34
Harmonicequivalent circuit of the de motor drive
ww
v..
=
In
w.E
_1_(3V2V .~) V2 11" 35
from equation (3.175). which when substituted for nominallinetoline voltage of 460 V. gives V in = 25.1 V
asy E V
Given ripple factor = 0.01 = ~. Vdc
where Von is the output voltage ripple across the capacitor and V de is the average de voltage. Von = 0.01 V dc = 0.01
X
ngi nee rin
1.35
X
V
= 0.01
X
1.35
X
460
= 6.21 V
Hence, the ratio of sixthharmonic output voltage to input voltage is given as Von = 6.21 = 0.247 Vin 25.1
The sixthharmonic equivalent circuit for the filter and motor load is shown in Figure 4.34. From the figure. it is seen that. to minimize the harmonic current to the load. the capacitive reactance has to be approximately 10 times smaller than the load impedance:
from which C f is evaluated as C. t
where
=
10 ;============================== (6 X 358.14)V(0.05)2 + (6 X 358.14 x 0.005)2
10
6ws VR; + (6ws La )2 Ws
g.n et =
433 ~F
corresponds to the lowest line frequency, i.e., 27r(fs
3)

= 211"(60
 3)
= 358.14 rad/sec
The load impedance is 10 times the capacitive reactance: for the purpose of finding the ratio between the capacitive voltage and the input source voltage, then. the load impedance can be treated as an open circuit. This gives rise to
nwL r S
1 nwsC,

     = 0.247
n 2w;L rC r 
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Chapter 4
ChopperControlled DC Motor Drives
from which LeC r is obtained as 1
+1
_ 0.247 LrCr 2 2 n Ws
_ 
1.09 x 10
6
Hence, the filter inductance is obtained, by using the previously calculated Cf' as
Lf
1.09 X 10 0 =
Cf
1.09 X 10 6  = 433 X 10 6
2.52 mH
(iii) Chopper Device Ratings
IT = I max = 686 A
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V o = V T = V2(V
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+ O.lV)
=
715.6 V
(iv) Device Selection IGBT.
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A single device per switch is sufficient in both these categories.
4.15 SUGGESTED READINGS
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1. P. W. Franklin, "Theory of dc motor controlled by power pulses. Part I Motor operation," IEEE Trans. on Power Apparatus and Systems. vol. PAS91, pp. 249255. Jan./Feb. 1972.
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2. P. W. Franklin, "Theory of dc motor controlled by power pulses. Part II Braking methods, commutation and additional losses:' IEEE Trans. on Power Apparatus and Systenls. vol. PAS91, pp. 256262, Jan./Feb. 1972.
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net
3. Nisit K. De, S. Sinha. and A. K. Chattophadya. "Microcomputer as a programmable controller for state feedback control of a de motor employing thyristor amplifier:' ConI Record, IEEElAS Annual Meeting. pp. 586592, Oct. 1984.
4. S. R. Doradla and N. V. P. R. Durga Prasad. "Openloop and closedloop performance of an acdc PWM converter controlled separately excited dc motor drive." Conf Record, IEEElAS Annual nleeting, pp. 411418, Oct. 1985. 5. H. lrie, T. Hirasa, and K. Taniguchi. "Speed control of dc motor driven by integrated voltage control method of chopper:' Con! Record, IEEElAS Annual Meeting, pp. 405410. Oct. 1985. 6. Report Number: DOE/BP 34906. "Adjustable Speed Drive Applications Guidebook:' 1990.
7. S. B. Dewan and A. Mirhod, "Microprocessor based optimum control for four quadrant chopper," Con! Record, IEEElAS Annual meeting, pp. 646652. Oct. 1980. 8. G. A. Perdikaris, "Computer control of a dc motor." Con! Record, IEEE lAS Annual meeting, pp. 502507, Oct. 1980. 9. N. A. Loric, A. S. Sedra. and S. B. Dewan. "A phaselocked de servo system for contouring numerical control:' Con! Record, IEEElAS Annual Meeting. pp. 12031207. Oct. 1981.
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Discussion Questions
171
4.16 DISCUSSION QUESTIONS 1. Why is the duty cycle usually changed by varying the ontime rather than the chopping frequency? 2. Very small duty cycles near zero are not possible to realize in practice. What are the constraints in such cases? 3. Similarly, a duty cycle of one is not feasible in the chopper. What is the reason? 4. Draw the gating pulses for each of the four quadrants of operation of the choppercontrolled dc motor drive.
5. Is a sinewave output current possible from a chopper circuit? 6. Is a sinewave voltage output possible from a chopper circuit? 7. The chopper can be switched at very high frequencies. What are the factors that lin1it
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highfreq uency operation?
8. A Shp dc nl0tor is to be driven from a chopper for two applications: a robot. and a winder. Discuss the power device to be selected for each of the applications.
9. Compare a fourquadrant chopper and a phasecontrolled converter fed fronl a 3phase ac supply with regard to the nurnber of power devices, speed of response. control complexity. and harmonics generated.
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10. A twoquadrant drive to operate in FM and FR is required. Which one has to be recommended between Figures 4.13(ii) and (iii) '?
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11. The averaging and the instantaneous steadystate conlputation techniques give identical results in the continuousconduction mode. Is this true for the discontinuous current mode, too?
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12. Ripple current is minimized by either increasing the chopper frequency or including an inductance in series with the armature of the dc motor. Discuss the merits and demerits of each alternative. Are the alternatives constrained by the size of the motor drive '?
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13. Discuss the control circuit design for a twoquadrant chopper circuit.
14. Discuss the impact of the choice of the current controllers on the dynamic performance of the dc motordrive systenl.
15. Design a digital controller for.a hysteresiscurrent controller.
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16. Design an analog version of the PWM current controller. Use commercially available integrated chips and operational amplifiers in the design.
17. Which one of the current controllers produces the highest switching losses and motor copper losses? 18. Is an innermost voltage loop necessary for fast response of the dc motor drive? 19. The hysteresis window. ~i. can be made a constant or a variable. Discuss the merits and demerits of both alternatives.
20. Likewise, the PWM carrier frequency can be either a constant or a variable. Discuss the merits and demerits of both alternatives.
21. The window in the hysteresis controller and the carrier frequency in the PWM controller are made to be variable quantities. What are they to be dependent upon for variation? 22. Is the parameter sensitivity of the openloop choppercontrolled dc motor drive different from that of the phasecontrolled de motor drive?
....,. 7~
Timedomain simulation can eliminate the errors in the final construction and testing of a choppercontrolled dc ITIotor drive. Discuss a nlcthod of including the frontend uncontrolled rectifier and filter in the simulation.
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Chapter 4
ChopperControlled DC Motor Drives
24. The timedomain dynamic model derived and developed in this chapter has to be very slightly modified to treat either a series or a cumulativelyexcited dc motor drive. What are the changes to be introduced? 25. The transferfunction model is true for both small and largesignal analysis of the separatelyexcited dc motor drive. A choppercontrolled dc series motor drive has to be considered for its speedcontroller design. Will the same block diagram be adequate for the task at hand?
4.17 EXERCISE PROBLEMS 1. A chopper is driving a separatelyexcited dc motor whose details are given in Example 4.1. The load torque is frictional and delivers rated torque at rated speed. Calculate and plot the duty cycle vs. speed. The maximum electromagnetic torque allowed in the motor is 2 p.u. The friction coefficient of the motor is 0.002 N·m/rad/sec. 2. Consider the de motor chopper details given in Example 4.2. Compute the torquespeed characteristics for a duty cycle of 0.4, using the averaging and instantaneous steady state computation techniques. Do the methods compare well in the discontinuous currentconduction mode? 3. Compare the switch ratings of the twoquadrant choppers shown in Figures 4.13(ii) and (iii). 4. A choppercontrolled dc series motor drive is intended for traction application. Calculate its torquespeed characteristics for various duty cycles. The motor details are as follows: 100 hp, 500 V, 1500 rpm, R a + R f = 0.01 n, La + Lse = 0.012 H, M = 0.1 H, J = 3 Kg· m2, B} = 0.1 N·m/rad/sec. The chopper has an input source voltage of 650 V and operates at 600 Hz. 5. The motor chopper given in Example 4.4 is used in a position servo application. It is required that the pulsating torque be less than 1°1 of the rated torque. At zero speed, the motor drive is producing a maximum of 3 p.u. torque. Determine (i) the increase in switching frequency and (ii) the value of series inductance to keep the pulsating torque within the specification: 6. The above problem has an outermost position loop. The position controller is of PI type. Determine the overallposition transfer function and find its bandwidth for K p = 1 and Tp = 0.2 sec, if the system is stable. Assume that the speed controller is of proportional type, with a gain of 100. The current controller is a hysteresis controller. 7. A choppercontrolled dc motor has the following parameters:
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6.3 A, 200 V, 1000 rpm, R a = 4 n, La = 0.018 H, Kb = 1.86 V/rad/sec, J = 0.1 kg . m2• B} = 0.0162 N·m/rad/sec, fe = 500 Hz, V s = 285 V Determine the following: (i) Torquespeed characteristics for duty cycles of 0.2, 0.4, 0.6, and 0.8 in the forwardmotoring mode~ (ii) The average currents at 500 rpm, using averaging and instantaneous steadystate evaluation techniques~ (Assume a suitable duty cycle.) (iii) Critical duty cycle vs. speed, with and without an external inductance of 20 mHo (Draw it.)
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Exercise Problems
173
8. Determine the speedcontroller gain and time constant for the following separatelyexcited choppercontrolled dc motor drive: 250 hp, 500 V, 1250 rpm, 920/0 Efficiency, R a = 0.0520, La = 1 mH, Kb = 3.65 V/rad/sec, J = 5 kg·m 2 , B, = 0.2 N·m/rad/sec, Vs = 648 v.
The current controller has a PWM strategy with a carrier frequency of 2 kHz. Calculate also the speed response for a speed command of 0 to 0.8 p.u., when the load torque is maintained at 0.5 p.u. {Hint: Calculate the speed controller gain and time constant~ assuming B 1 = 0.1
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CHAPTER
5
Polyphase Induction Machines
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5.1 INTRODUCTION
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A hrid introduction 10 the theory and principle of operation of induction machines is given in this chapler, llle nOI:1tions :Ire introduced. And consistent usc of them in Ihe text is adhered IO.llIC principle of operation of Ihe induction molar is devel~ oped to derive the steadyslate equivalcllI circuit and performance calculations,They are required to evaluate the steadyslalc response of variablespeed inductionmolor drives In the succeeding chaplCrs.lhe dynamic simulation is olle of the key steps in Ihe validalion of the design procc!>S of the mOlordrive systems. eliminating inadvertent design mistakes and the rc~ulting errors in the prototype construction and lesting and hence the need for dyllllmic modds of the induction machine. The dynamic model of lhe induction machine in direcd 10 obtain transient responses. smallsignal equations. and a multitude of transfer functions. all ot which arc usdul in the study of converterfed induction· motor drives. Spacephasor approes, and power output. Note thai the strayload losses have not been accounted for in the equivalent circuit of the machine. Various analytical formulae and empirical relationships are in usc, but a precise prediction of the stray losses is very difficult.
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Example 5.1 (i) Find the efficiency of 3n,induction motor operaling at full load. The machine details
arc
given in the following: 2000 hp. 2300 v, 3 phase, Star connected.  11 II "'" 1570,5 kW . P, 1491.2 % EffiCIency. '1 .. 1', 100" 1570.5 100  94.%%
(ii) The prinCIple of powerfactor imrro\'em~nl Will! capacl\or In,tol!latlO,)n ,It thc machine stator terminals is based on lhe capacitor's drawlIlg II Icadlllg reacllve curr.:nl from the supply to cancellhe lagging reactive current drawn by the inductIOn nll\dl1no::..ln order for Ihe lille power factor 10 be unity. the reacth'e component of the hne curreOl mu~t be lero. rhc reacti\'e line current IS the sum of the capacitor and IlldUCllon machille r~'actl\'': currents. Downloaded : www.EasyEngineering.net Therefore. the capilelll"'e reactive currenl (10"") has From lfl he equal III magnlHl(k hUI flPP()~IIC in
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Chapter S
Polyphase Induction Machines
direction to the machine lagging reactive current, but the machine reactive current is the imaginary part of the stator current and is given by lap + imag(I,l  0 Hence. luI'   imag(l,)  (j104.1) "" jl04.1 A
This current is controlled by the capacitors installed !:It the input. and therdore the capacitor required is leal' j 104.1 C = jw,VI» = j377. 1327.9 = 20.792 flF
5.5 STEADYSTATE PERFORMANCE
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A flowchart for the evaluation of the steadystate performance or the motor is given in Figure 5.5. The relevant equations for use are from (5.22) to (5.33). A set or sample torquevs.slip characteristics for constant input voltage is shown in Figures 5.6(a) and (b). The torquevs.slip characteristics are shown for slip \ along. the (/ and q axes. The common lerm.lhe number (If turns III Ih....... mJmg..l' c,ml:...kJ lin I::ilh~'r ..ide of Ihe equations.le,tlm!! Ihl: currcnt qualilic.... Th'" II a\I" I' a..,>umed Inix' la~gmg Ihc 1I axis by 6~. The relallOn..h;p Ocl een dqo and "hi' l:urrl.'nl.. i.. a'QI
(5.146)
These equations can be represented in equivalent circuits. Powersystem engineers and design engineers use normalized or per unit (p.u.) values for the variables. The normalization of the \ ariablcs is made via reactances rather than inductances. To facilitate such a step. a modified flux linkage is defined whose unit in volts is
ww w.E asy En gin ee
(5.147)
where wI' is the base frequency in Tad/sec. Similarly the other modified flux linkages are written as '""~
:z
l.II~
= X,i~
Xmi~
T
41", = XI>~ ...
,""", = XI,i... In
(5.149)
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'""~ "" X, i~ ~ Xmij,
Substiluting the flux linkages
(5.148)
X.ld, + Xmij,
(5.150)
g.n e
(5.151 ) (5.152)
t
lerms of Ihc modified flu..: linkages yields
~
¥
¥
~.
~
WI'
WI'
WI'
WI'
WI'
~
,~, =  . Ad' =  . A," "" 
(5.153)
WJ,
and. by substituting equation (5.153) into equal ions from (5.141) to (5.146J. the resulling equal ions III modified flux linkages are , _ R
v", 
.< .1",
We
+
WI>
0.2
uo
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0.' fl.S
u,q(1.6 IU' n.2 I
ClIO
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IIW
U.I~
11211
11 ~'
II ~Il
II:'"
7
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., ;
, U.II~
1110
U.15
020
0_25
0>
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I 0110
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0.7 0.'
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Time.~
0.20
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0.2
(II !l.0 000
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11.0

l) :~
I
0.15
,, U'0'
1
H!tl
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1.0 0.00
o I~
.
0.2 IIA
U.II
010
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OA
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005
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o.
um
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.
02 0.6
>'
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0'
0.00
231
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net II ~ll
nmc.~
till"R' 5.111 (el Fn.".,·..crdcra(lon charaC(CrtSIICS In "nchronou< rcfcrcrlC
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Chapter 5
Polyphase Induction Ma(hines 1.I 10 09 0.8 0.7 E 0.6 " 0.5 OA 0.3 0.2 0.\ 0.0 0.00
8 6
, 2 0 2
' ,
6 8 U.OO 8
0.05
D.LU
0.15
0.20
(US
,
6
, "
0 2
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0.00
0.05
0.10
0,15
0.20
0.2
2 0 2
,
0.00
0.25
0.05
0.10
0.15
0.20
11,00
,
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(1.25
0.25
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0.2 0.00 06
0.] Ul
I
0.1
2 0.00
0.0 0.00
Figure S.18
0.20
0.11
"
0
0.20
0,15
0,1
"•
U.S
0.15 Time.§
DID
0.2
0..
0.10
0.05
0.]
,
0.05

0.6
(1.25
]
e'
0.10
11.5
0.1
6
,
1115
Nv
0.3
OA
8 6
_G
0.10
0.1
2
i
0.05
0.0
0.25
0,05
0.10
, U,05
0.15
0.20
0.25
1
1
, 0.10 0.15 T,me.;;
(I
"0
(e) Freea~'Celcral;OR cllarac!cflS!'CS in syllchnHlous reference fran11111
p
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= 32AmIIrISln
2
W"
Because
>
Pa Pa Pa Pa
3phase ac supply
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Figure 6.26
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Static Scherbius drive
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simple implementation of it in a blockdiagram schematic is shown in Figure 6.26. This converter has a number of advantages:
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(i) simple line commutation; (ii) shaping of current, and hence powerfactor improvement: (iii) high efficiency; (iv) suitable and compact for MW applications; (v) capable of delivering a dc current (and hence the induction motor can be
operated as a synchronous motor).
6.3.10 Applications Slipenergyrecovery motor drives are used in large pumps and compressors in MW power ratings. It is usual to resort to an auxiliary starting means to minimize the
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Chapter 6
PhaseControlled InductionMotor Drives
Rated speed
Design point 100 ~
d ro ~
::c ;:.R 0
50
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Ot.+~
o
100
50
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0/0 Flow
Head and efficiency vs. flow of pumps
Figure 6.27
b
lOO
"'0 ~
v
:c
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System curve
50 c
?f2
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3 0 0
50
100
0/0 .
Flow
1. Throttling energy loss
2. Friction loss
3. Static head
Figure 6.28
Flow reduction by damper control
t
converter rating. Windenergy alternative power systems also use a sliprecovery scheme, but the converter rating cannot be minimized. because of the large variation in operational speed. An illustrative example of such an application can be found in reference [15] ..The application is briefly described here. The pump characteristics at rated speed are shown in Figure 6.27. The nominal operating point is at 1 p.u. flow, where the efficiency is maximum. If the flow has to be reduced, then the characteristic of the pump is changed by adding a resistance to the flow in the form of damper control, shown in Figure 6.28. The new' operating point, b, is obtained by changing the system curve to move from ca to cb by adding system resistance from a damper control. At this operating point. the input energy to the pump will not change in proportion to the flow change. resulting in poor effi
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SlipEnergy Recovery Scheme
307
System curve
a
100
1. Energy saved for 40% flow at 60% speed
"0 Cd
v
::r: '::!2
0
,\1
I
100 % speed
50
85%
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0
40 50
Figure 6.29
100 0/0 Flow Flow control by speed variation
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ciency because the input has to supply the throttling energy loss and friction energy loss. The flow can also be reduced if the pump characteristics are changed by reducing the pump speed with a variablespeedcontrolled induction motor drive, and then the operating point moves as shown in Figure 6.29. The operating point can be always attempted on the highestefficiency point by varying the speed. As the speed is reduced, the power input to the pump reduces and hence the input power to the driving motor decreases, resulting in higher energy savings compared to the damper control. Additionally, the combined efficiency of the motor and pump also improves by this method, resulting in considerable energy savings. Because the normal variation in the flow is between 60 and 100% of the rated value, the range of the variable speed control required usually is between 60 and 100 % of the rated speed. Restrictedspeed control requirement makes the slipenergyrecoverycontrolled induction motor drive an ideal choice for this kind of application in the MW power range. It is not unusual to encounter, for example. 9.5MW drive systems with a speed range of from 1135 to 1745 rpm and 11.5MW drive systems with a speed range of from 757 to 1165 rpm, olany of these in parallel or in series in many pumping stations. Such highpower drives get supplied fronl a 13.8 kV bus and are equipped with harmonic suppression circuits in the front end of the inverter and large dc reactors for smoothing in the dc link. They \vill have starting circuits with liquidcooled resistors to reduce the power capability of the converters, and the inverter will be dual threephase SCRbridge converters in series, to keep the displacement factor above 0.88. Note toat the displaceolent factor, which is the power factor considering only fundamentals, will be poor \vith a single threephase bridge converter in the system. The converter units are force cooled by air or water in many installations.
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Chapter 6
PhaseControlled InductionMotor Drives
6.4 REFERENCES 1. R.E. Bedford and V.D. Nene, UVoltage control of the three phase induction motor by thyristor switching: A timedomain analysis using the a~o Transformation," IEEE Trans. on Industry and General Applications, vol. IGA6, pp. 553562, Nov.lDec.1970. 2. T.A. Lipo, "The analysis of induction motors with voltage control by symmetrical triggered thyristors," IEEE Trans. on Power Apparatus and Systenls, vol. PAS90, pp. 515525, March/April 1971. 3. William McMurray, "A comparative study of symmetrical threephase circuits for phasecontrolled ac motor drives," IEEE Trans. on Industry Applications, vol. lA10, pp. 403411, May/June 1974. 4. B. 1. Chalmers, S. A. Hamed, and P. Schaffel, "Analysis and application of voltagecontrolled induction motors," Con! Record, IEEESecond International Conference on MachinesDesign and Applications, pp. 190194, Sept. 1985. 5. W. Shepherd, hO n the analysis of the threephase induction motor with voltage control by thyristor switching," IEEE Trans. on Industry and General Applications, vol. IGA4, no. 3, pp. 304311, May/June 1968. 6. Derek A. Paice, "Induction motor speed control by stator voltage control," IEEE Trans. on Power Apparatus and Systems, vol. PAS87, pp. 585590, Feb. 1968. 7. John Mungensat "Design and application of solid state ac motor starter:' Con! Record, IEEElAS Annual Meeting, pp. 861866, Oct. 1974. 8. R. Locke, 4' Design and application of industrial solid state contactor," Con! Record, IEEElAS Annual Meeting, pp. 517523, Oct. 1973. 9. M. S. Erlicki, "Inverter rotor drive of an induction motor," IEEE Trans. on Power Apparatus and Systems, vol. PAS84, pp. 10111016, Nov. 1965. 10. A. Lavi and R.1. Paige, "Induction motor speed control with static inverter in the rotor," IEEE Trans. Power Apparatus and Systems, vol. PAS85. pp. 7684, Jan. 1966. 11. W. Shepherd and 1. Stanway, "Slip power recovery in an induction motor by the use of thyristor inverter," IEEE Trans. on Industry and General applications~ vol. IGA5. pp. 7482, Jan./Feb. 1969. 12. T. Wakabayashi et a1.. "Commutatorless Kramer control system for large capacity induction motors for driving water service pumps~" Con! Record, IEEElAS Annual Meeting, pp. 822828, 1976. 13. A. Smith, "Static Scherbius systems of induction motor speed control," Proc. Institute of Electrical Engineers, vol. 124, pp. 557565, 1977. 14. H. W. Weiss, 4'Adjustable speed ac drive systems for pump and compressor applications," IEEE Trans. on Industry Applications~ vol. IA10, pp. 162167, Jan/Feb 1975. 15. P. C. Sen and K. H. 1. Ma, "Constant torque operation of induction motors using chopper in rotor circuit," IEEE: Trans.. on Industry Applications, vol. IA14, no. 5~ pp. 408414, Sept.lOct. 1978. 16. M. Ramamoorthy and N. S. Wani, "Dynamic model for a chopper controlled slip ring induction motor," IEEE Trans. on Electronics and Controllnstrunzentation, vol. IECI25, no. 3, pp. 26G266, Aug. 1978.
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Discussion Questions
309
6.5 DISCUSSION QUESTIONS 1. Is there a difference in gating/triggering control between the inductionmotor phase con
troller and the phasecontrolled rectifier used in the dc motor drives? 2. In the phasecontrolled induction motor drive, there could be 3 or 2. 2 or 1, 1 or 0 phases conducting, depending on the load. Envision the triggering angles and conduction patterns in these modes of operation. 3. The power switches can be connected in a number of ways to the star and deltaconnected stator of the induction motors. Enumerate the variations. 4. Is there an optimal configuration of the phase controller to be used in a deltaconnected stator? 5. A singlephase capacitorstart induction motor is to have a phase controller. Discuss the best arrangement of its connection to the motor winding. 6. Discuss the configuration of the phase controller for a capacitorstart and a capacitorrun singlephase induction motor. 7. Discuss the merits and demerits of using a phase controller as a stepdown transformer. 8. The relationship between the triggering angle and the fundamental of the output rms voltage is nonlinear in a phasecontrolled induction motor drive. How can it be made linear? Is there any advantage to making it linear? 9. The phase controller is placed in series with the rotor windings as shown in Figure 6.30.
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Discuss the following: (i) the advantages of this scheme over the statorphasecontrolled induction motor drive~
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(ii) the current and voltage ratings of the power switches and their comparison to the
statorphase controller;
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a 3Phase ac b supply
et
(}__e
0     ....
Induction motor
n
c \\'
Figure 6.30
Rotorphasecontrolled induction motor drive
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Chapter 6
PhaseControlled InductionMotor Drives
(iii) the improvement in the input power factor;
10.
11. 12.
(iv) the improvement in linecurrent waveform and the reduction of harmonics; (v) the disadvantages of this scheme; (vi) the reversibility of the motor drive~ (vii) the controlsignal generation for this scheme~ (viii) any softstart possibility; (ix) closedloop control and its requirements. Similarly to a phasecontrolled converter (used in de motor drives), whose a is limited in the inversion mode, will there be a need to limit the maximum value of a in the phasecontrolled induction motor drive? Explain. Compare the performance of the rotorphase and choppercontrolled induction motor drives. Discuss the performance, analysis, design, merits, and demerits of the rotor choppercontrolled induction motor drive shown in Figure 6.31. (Hint: References 15 and 16.) Compare the statorphase and slipenergyrecoverycontrolled induction motor drives on the basis of (i) input power factor (ii) efficiency (iii) cost (iv) control complexity (v) pulsating torque (6 th harmonic) (vi) feedback control (vii) range of speed control What is the effect of the inductor in the dc link? Suppose that the inductor in the dc link is replaced by an electrolytic capacitor. How would this affect the performance of the phasecontrolled converter and consequently the slipenergy recovery? The slipenergycontrolled drive, with a current loop controlling the dc link current, behaves like a separatelyexcited dc motor. Will a change in the load affect the flux and hence the torque constant?
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14. 15.
16.
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x
a 3phase ac b supply c
Cf Rex W
x
Induction motor Figure 6.31
Diode bridge rectifier DClink filter
Chopper
Rotor choppercontrolled induction motor drive
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Exercise Problems
311
17. Discuss the feasibility of replacing the thyristor bridge converter with a GTO or transistorbased converter to control the harmonics fed to the utility. 18. The diodebridge rectifier in the de link can be replaced by a GTO converter. By suitable control of this controlled converter, is it possible to control the current to eliminate the 6 th harmonic in the dc link and reduce the size of the dc inductor? 19. At the time of starting (assuming that there is no auxiliary starting means), what is the value of the triggering angle in the phasecontrolled converter'? 20. The following applications need a variablespeed motor drive: (i) a 20.000hp pump: (ii) a 50hp fan.
The options available are statorphase and slipenergyrecoverycontrolled induction motor drives. Which one is to be recommended for the above applications? Explain the reasons for your choice.
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21. For openloop speed control. NEMA class D induction 010tors are preferred in phasecontrolled motor drives. Is the same choice valid for openloop slipenergyrecoverycontrolledinduction motor drives?
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22. What are the modifications required in the controller of the phasecontrolled converter used in a dc motor drive for adaptation to the slipenergyrecoverycontrolled induction motor drive?
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23. Can a slipenergycontrolled induction motor be reversed in speed'! How is it done?
6.6 EXERCISE PROBLEMS
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1. Determine the range of speed control obtained with a phasecontrolled induction motor drive. The details are given below. 1 hp. 2 pole. 230 V. 3 phase. star connected. 60 Hz, R s = 1 ft R r = 6.4 fl, X m = 75 O. X 1r = 3.4 ft Xis = 3 !1. and fullload slip is 0.1495. Load is a fan requiring 0.5 p.u. torque at 3200 rpm. ex is limited to a maximum of 135 degrees.
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2. Calculate and draw the efficiency of the drive given in Example 1 as a function of rotor speed.
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3. A 1000hp induction motor is to be started with a phase controller. The current is limited to 1 p.u. Find the starting torque and the range of triggeringangle variation to run the motor from standstill to 0.9 p.u. speed at a constant load torque of 0.05 p.u. The motor details are as follows: Base kVA = 850 kVA, Base voltage = 2700 V, Rated power = 1000 hp. Rated frequency = 50 Hz, P = 4, R s = 0.041 fl, R r = 0.1663 f1. Xm = 80 O. X!r = 0.2 fl. XiS = 0.2 0, fullload slip = 0.0801 (base speed = full load speed). 4. Calculate the impact of rotor resistance increase by 80% on the speedcontrol range and efficiency of the drive given in problem 3.
5. The problen1 given in Exalnple I is driving a fan load. At 1500 rpnl. the load torque is 100%. Determine the openloop stable operatingspeed range for the triggering angles of 45° < ex < 135°. 6. Determine approximately the magnitude of the two predominant harmonics present in the phasecontrolled induction motor drive. Are these harmonics loaddependent? 7. A movie theater has been reconverted to a house onefourth of its original seating. The ventilation fans have to be accordingly derated to reduce the operational costs. Phase
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CHAPTER
7
FrequencyControlled Induction Motor Drives
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7.1 INTRODUCTION
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"Inc speed of Ihc inductilln motor is vcry near to ib ~ynchnHl(lUS "peed. and chan~· ing the synchronuus h and in vOltlUllf'. which is the I)]
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(7. I Y7).
k
where [is the 4 x 4 identity matrix. For the example under consideralioll.lhe sampling time corresponds to 2 electrical degrees. given as T,
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2
:=
f' s
(7.ltJH)
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The exponential of ATs can be evaluated from the series with!:! 10 IS value of k is given by
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k
+
asy
I
:=
360/2
:=
lcrlll~_lh:
(7.199)
180
which gives k := 179. For wave forms with halfwave symmetry.lhe final and initial values are related by other than idenlity, as is shown in the section on the sixstep inverterfed induction motor drive.
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7.4.10.4 Computation of steadystate performance.
nle currel1l vcctor is evaluated for the entire cycle from the discretized statespace equation discussed and derived above. The electromagnetic torque is compuled as
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where lhe currents are the components of the state vector X(k). Bv inverse lransformation.the phase currents are computed as
it>o(k)
:=
iAk} = i".(k) O.5i",(k)  O.866i