Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science

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1 aachuett Intitute of Technology Department of Electrical Engineering and Computer Science Electric achinery Cla Note 10: Induction achine Control and Simulation c 2003 Jame L. Kirtley Jr. 1 Introduction The inherent attribute of induction machine make them very attractive for drive application. They are rugged, economical to build and have no liding contact to wear. The difficulty with uing induction machine in ervomechanim and variable peed drive i that they are hard to control, ince their torque-peed relationhip i complex and nonlinear. With, however, modern power electronic to erve a frequency changer and digital electronic to do the required arithmetic, induction machine are eeing increaing ue in drive application. In thi chapter we develop model for control of induction motor. The derivation i quite brief for it relie on what we have already done for ynchronou machine. In thi chapter, however, we will tay in ordinary variable, kipping the per-unit normalization. 2 Volt/Hz Control Remembering that induction machine generally tend to operate at relatively low per unit lip, we might conclude that one way of building an adjutable peed drive would be to upply an induction motor with adjutable tator frequency. And thi i, indeed, poible. One thing to remember i that flux i inverely proportional to frequency, o that to maintain contant flux one mut make tator voltage proportional to frequency (hence the name contant volt/hz ). However, voltage upplie are alway limited, o that at ome frequency it i neceary to witch to contant voltage control. The analogy to DC machine i fairly direct here: below ome bae peed, the machine i controlled in contant flux ( volt/hz ) mode, while above the bae peed, flux i inverely proportional to peed. It i eay to ee that the maximum torque varie inverely to the quare of flux, or therefore to the quare of frequency. To get a firt-order picture of how an induction machine work at adjutable peed, tart with the implified equivalent network that decribe the machine, a hown in Figure 1 In Chapter 8 of thee note it i hown that torque can be calculated by finding the power diipated in the virtual reitance R 2 / and dividing by electrical peed. For a three phae machine, and auming we are dealing with RS magnitude: p 2 R 2 T e = 3 I 2 ω where ω i the electrical frequency and p i the number of pole pair. It i traightforward to find I 2 uing network technique. A an example, Figure 2 how a erie of torque/peed curve for an induction machine operated with a wide range of input frequencie, both below and above it bae frequency. The parameter of thi machine are: 1

2 I a R a X 1 X 2 I 2 < Xm < > > R 2 < Figure 1: Equivalent Circuit Number of Phae 3 Number of Pole Pair 3 RS Terminal Voltage (line-line) 230 Frequency (Hz) 60 Stator Reitance R 1.06 Ω Rotor Reitance R Ω Stator Leakage X 1.34 Ω Rotor Leakage X 2.33 Ω agnetizing Reactance X m 10.6 Ω A trategy for operating the machine i to make terminal voltage magnitude proportional to frequency for input frequencie le than the Bae Frequency, in thi cae 60 Hz, and to hold voltage contant for frequencie above the Bae Frequency. Induction otor Torque N m Speed, RP Figure 2: Induction achine Torque-Speed Curve For high frequencie the torque production fall fairly rapidly with frequency (a it turn out, 2

3 it i roughly proportional to the invere of the quare of frequency). It alo fall with very low frequency becaue of the effect of terminal reitance. We will look at thi next. 2.1 Idealized odel: No Stator Reitance I a X 1 X 2 I 2 + V < Xm < > < > R 2 Figure 3: Idealized Circuit: Ignore Armature Reitance Ignore, for the moment, R 1. An equivalent circuit i hown in Figure 3. It i fairly eay to how that, from the rotor, the combination of ource, armature leakage and magnetizing branch can be replaced by it equivalent circuit, a hown in in Figure 4. X I a 1 X 2 I 2 + V < > < < > R 2 Figure 4: Idealized Equivalent In the circuit of Figure 4, the parameter are: V X m = V Xm + X 1 X = X m X 1 If the machine i operated at variable frequency ω, but the reactance i etablihed at frequency ω B, current i: and then torque i V I = j(x 1 + X2) ω ωb + R 2 2 R 2 3p V 2 R 2 Te = 3 I 2 = ω (X 1 + X 2) 2 + ( R 2 ) 2 3

4 Now, if we note that what count i the abolute lip of the rotor, we might define a lip with repect to bae frequency: ω r ω r ω B ω B = = = B ω ω B ω ω Then, if we aume that voltage i applied proportional to frequency: and with a little manipulation, we get: V = V ω 0 ω B T e = 3p V 0 2 R 2 B ω B (X + X ) 2 + ( R ) 2 B Thi would imply that torque i, if voltage i proportional to frequency, meaning contant applied flux, dependent only on abolute lip. The torque-peed curve i a contant, dependent only on the difference between ynchronou and actual rotor peed. Thi i fine, but eventually, the notion of volt per Hz run out becaue at ome number of Hz, there are no more volt to be had. Thi i generally taken to be the bae peed for the drive. Above that peed, voltage i held contant, and torque i given by: 3p V 2 R 2 B Te = ωb (X 1 + X 2) 2 + ( R 2 B ) 2 The peak of thi torque ha a quare-invere dependence on frequency, a can be een from Figure 5. Induction otor Torque N m Speed, RP Figure 5: Idealized Torque-Speed Curve: Zero Stator Reitance 4

5 2.2 Peak Torque Capability Auming we have a mart controller, we are intereted in the actual capability of the machine. At ome voltage and frequency, torque i given by: 3 p T e = 3 I 2 2 R 2 ω = V 2 R 2 ((X 1 + X 2)( ω ωb )) 2 + (R 1 + R 2 Now, we are intereted in finding the peak value of that, which i given by the value of R 2 which maximize power tranfer to the virtual reitance. Thi i given by the matching condition: R 2 ω = R ((X 1 + X 2)( )) 2 Then maximum (breakdown) torque i given by: 3p ω V 2 R ((X 1 + X 2)( ω ω Tmax = ))2 B ((X 1 + X 2)( ω ω B )) 2 + (R + R 2 + ((X + X )( ω ω B )) ) 2 Thi i plotted in Figure 6. Jut a a check, thi wa calculated auming R 1 = 0, and the reult are plotted in figure 7. Thi plot how, a one would expect, a contant torque limit region to zero peed. ω B ) Breakdown Torque Newton eter Drive Frequency, Hz Figure 6: Torque-Capability Curve For An Induction otor 3 Field Oriented Control One of the more ueful impact of modern power electronic and control technology ha enabled u to turn induction machine into high performance ervomotor. In thi note we will develop a 5

6 300 Breakdown Torque 250 Newton eter Drive Frequency, Hz Figure 7: Idealized Torque Capability Curve: Zero Stator Reitance picture of how thi i done. Quite obviouly there are many detail which we will not touch here. The objective i to emulate the performance of a DC machine, in which (a you will recall), torque i a imple function of applied current. For a machine with one field winding, thi i imply: T = GI f I a Thi make control of uch a machine quite eay, for once the deired torque i known it i eay to tranlate that torque command into a current and the motor doe the ret. Of coure DC (commutator) machine are, at leat in large ize, expenive, not particularly efficient, have relatively high maintenance requirement becaue of the liding bruh/commutator interface, provide environmental problem becaue of parking and carbon dut and are environmentally enitive. The induction motor i impler and more rugged. Until fairly recently the induction motor ha not been widely ued in ervo application becaue it wa thought to be hard to control. A we will how, it doe take a little effort and even ome computation to do the control right, but thi i becoming increaingly affordable. 3.1 Elementary odel: We return to the elementary model of the induction motor. In ordinary variable, referred to the tator, the machine i decribed by flux-current relationhip (in the d-q reference frame): [ ] [ ] [ ] λds LS i = ds λ dr i dr [ ] [ ] [ ] λqs LS iqs = λ qr i qr 6

7 Note the machine i ymmetric (there i no aliency), and ince we are referred to the tator, the tator and rotor elf-inductance include leakage term: The voltage equation are: L S = + L Sl = + l dλ ds v ds = ωλ qs + r S i ds dλ qs v qs = + ωλ ds + r S i qs dλ dr 0 = ω λ qr + r R i dr dλ qr 0 = + ω λ dr + r R i qr Note that both rotor and tator have peed voltage term ince they are both rotating with repect to the rotating coordinate ytem. The peed of the rotating coordinate ytem i w with repect to the tator. With repect to the rotor that peed i ω = ω ω m, where ω m i the rotor mechanical peed. Note that thi analyi doe not require that the reference frame coordinate ytem peed w be contant. Torque i given by: 3.2 Simulation odel T e 3 = p (λ ds i qs λ qs i ds ) 2 A a firt tep in developing a imulation model, ee that the inverion of the flux-current relationhip i (we ue the d- axi ince the q- axi i identical): i ds = λ L 2 ds S L S 2 λ dr i dr = L S λ 2 ds λ L S L 2 dr Now, if we make the following definition (the motivation for thi hould by now be obviou): the current become: X d = ω 0 L S X kd = ω 0 X ad = ω 0 ( X d 2 ) = ω 0 L S LR i ds = ω 0 X ad ω 0 X λ ds λ dr d X k d X d i dr = X ad ω 0 X d ω 0 X λ ds X k d X λ dr d 7 S d X kd

8 The q- axi i the ame. Torque may be, with thee calculation for current, written a: 3 3 ω 0 X ad T e = p (λ ds i qs λ qs i ds ) = p (λ ds λ qr λ qs λ dr ) 2 2 X kd X d Note that the uual problem with ordinary variable hold here: the foregoing expreion wa written auming the variable are expreed a peak quantitie. If RS i ued we mut replace 3/2 by 3! With thee, the imulation model i quite traightforward. The tate equation are: where the rotor frequency (lip frequency) i: dλ ds = V ds + ωλ qs R S i ds dλ qs = V qs ωλ ds R S i qs dλ dr = ω λ qr R R i dr dλ qr = ω λ dr R S i qr dω m 1 = (T e + T m ) J ω = ω pω m For imple imulation and contant excitaion frequency, the choice of coordinate ytem i arbitrary, o we can chooe omething convenient. For example, we might chooe to fix the coordinate ytem to a ynchronouly rotating frame, o that tator frequency ω = ω 0. In thi cae, we could pick the tator voltage to lie on one axi or another. A common choice i V d = 0 and V q = V. 3.3 Control odel If we are going to turn the machine into a ervomotor, we will want to be a bit more ophiticated about our coordinate ytem. In general, the principle of field-oriented control i much like emulating the function of a DC (commutator) machine. We figure out where the flux i, then inject current to interact mot directly with the flux. A a firt tep, note that becaue the two tator flux linkage are the um of air-gap and leakage flux, λ ds = λ agd + L Sl i ds λ qs = λ agq + L Sl i qs Thi mean that we can re-write torque a: T e 3 = p (λ agd i qs λ agq i ds ) 2 8

9 Next, note that the rotor flux i, imilarly, related to air-gap flux: λ agd = λ dr l i dr λ agq = λ qr l i qr Torque now become: e 3 3 T = p (λ dr i qs λ qr i ds ) pl (i dr i qs i qr i ds ) 2 2 Now, ince the rotor current could be written a: That econd term can be written a: i dr = i ds λ dr λ qr i qr = i qs 1 i dr i qs i qr i ds = (λ dr i qs λ qr i ds ) So that torque i now: ( ) T e 3 l 3 = p 1 (λ dr i qs λ qr i ds ) = p (λ dr i qs λ qr i ds ) Field-Oriented Strategy: What i done in field-oriented control i to etablih a rotor flux in a known poition (uually thi poition i the d- axi of the tranformation) and then put a current on the orthogonal axi (where it will be mot effective in producing torque). That i, we will attempt to et λ dr = Λ 0 λ qr = 0 Then torque i produced by applying quadrature-axi current: T e 3 = p Λ 0 i qs 2 The proce i almot that imple. There are a few detail involved in figuring out where the quadrature axi i and how hard to drive the direct axi (magnetizing) current. Now, uppoe we can ucceed in putting flux on the right axi, o that λ qr = 0, then the two rotor voltage equation are: 0 = dλ dr ω λ qr + r R I dr 0 = dλ qr + ω λ dr + r R I qr 9

10 Now, ince the rotor current are: i dr = i ds λ dr λ qr i qr = i qs The voltage expreion become, accounting for the fact that there i no rotor quadrature axi flux: ( ) dλ dr λdr 0 = + r R i ds 0 = ω λ dr r R i qs Noting that the rotor time contant i we find: T R = rr dλ dr T R + λ dr = i ds ω = TR λ dr i qs The firt of thee two expreion decribe the behavior of the direct-axi flux: a one would think, it ha a imple firt-order relationhip with direct-axi tator current. The econd expreion, which decribe lip a a function of quadrature axi current and direct axi flux, actually decribe how fat to turn the rotating coordinate ytem to hold flux on the direct axi. Now, a real machine application involve phae current i a, i b and i c, and thee mut be derived from the model current i ds and i q. Thi i done with, of coure, a mathematical operation which ue a tranformation angle θ. And that angle i derived from the rotor mechanical peed and computed lip: θ = (pω m + ω ) A generally good trategy to make thi ort of ytem work i to meaure the three phae current and derive the direct- and quadrature-axi current from them. A good etimate of direct-axi flux i made by running direct-axi flux through a firt-order filter. The tricky operation involve dividing quadrature axi current by direct axi flux to get lip, but thi i now eaily done numerically (a are the trigonometric operation required for the rotating coordinate ytem tranformation). An elmentary block diagram of a (pobly) plauible cheme for thi i hown in Figure 8. In thi picture we tart with commanded value of direct- and quadrature- axi current, correponding to flux and torque, repectively. Thee are tranlated by a rotating coordinate tranformation into commanded phae current. That tranformation (imply the invere Park tranform) ue the angle q derived a part of the cheme. In ome (cheap) implementation of thi cheme the commanded current are ued rather than the meaured current to etablih the flux and lip. We have hown the commanded current i a, etc. a input to an Amplifier. Thi might be implemented a a PW current-ource, for example, and a tight loop here reult in a rather high performance ervo ytem. 10