PLC 单片机 外文翻译 外文文献 英文文献 一个逆变器在电梯门中控制两个永磁直线电动机

1、Control of two PM linear motors with asingle inverter: application to elevator doorsAbstractThis work considers the control of two PM synchronous motors using a single inverter. The standard approach to the control of a PM synchronous motor is to use a single inverter which provides independent cont

2、rol of the direct and quadrature voltages (and therefore of the direct and quadrature currents) of the motor. Here, an approach is presented that provides independent torque control of two PM synchronous motors using a single inverter. In this approach, the quadrature current of each motor is contro

3、lled while it is shown that the direct current is uncontrollable. Both parallel and series connections of the two motors to the single inverter are considered and it is shown how the singularity of the controller can be avoided in each case. The methodology is applied to the control of elevator door

4、s.1. IntroductionComposed by the order of relay control system is a realization of the first elevator control method. However, to enter the nineties, with the development of science and technology and the widespread application of computer technology, the safety of elevators, reliability of the incr

5、easingly high demand on the relay control weaknesses are becoming evident. Elevator control system relays the failure rate high, greatly reduces the reliability and safety of elevators, and escalators stopped often to take with the staff about the inconvenience and fear. And the event rather than ta

6、king the lift or squat at the end of the lift will not only cause damage to mechanical components, but also personal accident may occur. Programmable Logic Controller (PLC) is the first order logic control in accordance with the needs of developed specifically for industrial environment applications

7、 to operate the electronic digital computing device. Given its advantages, at present, the relay control the lift has been gradually replaced by PLC control. At the same time, AC variable frequency motor speed control technology, the way the lift drag speed has been a gradual transition from DC to A

8、C frequency converter. Thus, PLC control technology increases VVVF Elevator modern technology has become a hot industry.With the continuous development of urban construction, the increasing high-rise buildings, elevators and life in the national economy has a broad application. Elevator high-rise bu

9、ildings as a means of transport in the vertical run of daily life has been inextricably linked with people. In fact the lift is based on external call control signals, as well as the laws of their own, such as running, and the call is random, the lift is actually a man-machine interactive control sy

10、stem, simple to use control or logic control order can not meet the control requirements, and therefore , elevator control system uses a random control logic. Elevator control is currently generally used in two ways, first, the use of computer as a signal control unit, the completion of the lift sig

11、nal acquisition, operation and function of the set, to achieve the lift and set the automatic scheduling function to run the election, drag the control from inverter to complete; the second control mode with programmable logic controller (PLC) to replace the computer control signal sets the election

12、. From the control and performance, these two types of methods and there is no significant difference. Most of the domestic manufacturers to choose the second approach, because the smaller scale of production, their design and manufacture of high cost of computer control devices; and PLC high reliab

13、ility, convenient and flexible program design, anti-interference ability, stable and reliable operation of the characteristics of Therefore, the elevator control system is now widely used to realize programmable controller.This work considers the control of two PM synchronous motors using a single i

14、nverter. The standard approach to the control of a PM synchronous motor is to use a single inverter which provides independent control of the direct and quadrature voltages (and therefore of the direct and quadrature currents) of the motor. The quadrature current is proportional to the motor torque

15、and the direct current is used for field weakening. Here, an approach is presented that provides independent torque control of two PM synchronous motors using a single inverter. In this approach, the quadrature current of each motor is controlled while the direct current is uncontrollable. Such an a

16、pproach was also considered in the work 6.This problem was motivated by the control of elevator doors. A conventional elevator door system has the two doors mechanically connected to a single cable which forces the two doors to open and close together due to the mechanical coupling. Using position s

17、ensor feedback from the wall, the position of the doors is then controlled by a motor/inverter system that pushes/pulls on the cable. The objective here was to consider a different system where the cable system is eliminated and each of the two doors of the elevator are actuated using a linear synch

18、ronous motor. The two motors must reliably open and close the two doors of the elevator while maintaining a stiffness in the differential direction of motion on the order of 100,000N/m to have the “feel of the conventional cable driven doors. For example, in a conventional elevator door system if on

19、e door is held, the other door must stop at the same position since the doors are attached to a single cable whose stiffness is 100,000N/m. This same behavior is still desired in the new system and requires one being able to independently control each of the doors (i.e., their linear motor actuators

20、) to maintain the stiffness. However, in order to reduce costs, the question considered here is that of being able to independently control the two linear motors using a single inverter.The outline of the rest of the paper is as follows: Section 2 briefly describes the modeling of PM synchronous mot

21、ors, Section 3 develops a linear PM motor model from the rotary model, presents the door model and summarizes the standard PM synchronous motor control algorithm, Section 4 considers the control of two linear PM motors using a single inverter for both the parallel and series connection. Finally, Sec

22、tion 5 offers some conclusions.2. Modeling and control of PM synchronous motorsA linear permanent magnet motor may be modeled by considering an equivalent three-phase permanent magnet (PM) rotary synchronous motor. To do so, let x, denote the position and speed of the linear motor, m denote the mass

23、 of the linear motor, req denote the radius of the equivalent rotary motor (i.e., the linear motor travels a distance 2req for each complete revolution of the rotary motor), m is the mass of the linear motor, and F, FL denote the force produced by the linear motor and the load force on the motor. Th

24、en, for the rotary motor, it follows that the angular position is h = x/req, the angular speed is given by = /req, the moment of inertia is J = r2eq m, the torque = req F, the load torque L = req FL. A model of a three-phase PM synchronous (rotary) motor is 5.Here LS is the self-inductance of a stat

25、or winding, M is the coefficient of mutual inductance between the phases, Km is the torque/back-emf constant (so that KM = Km/req is the force/back-emf constant in the linear motor), RS the resistance of a stator winding, np is the number of pole pairs (or the number of rotor teeth for a stepper mot

26、or). If the phases were perfectly coupled, one would have M =LS. The three-phase to two-phase transformations for currents and voltages are defined bywhich transforms the original model into the equivalent modelFor a balanced three-phase system assumed here, it follows thatv0=(vs1+vs2+vs3)/=0, i0=(v

27、s1+vs2+vs3)/=0so that one obtains the two-phase equivalent model given byHere L= Ls + M(Ls),Keq =Km ia and ib are the equivalent currents in phases a and b, respectively. Letting Vbus denote the bus voltage into a three-phase inverter. The maximum voltage out of the inverter is obtained when it is r

28、un in six step mode and the peak of the fundamental of the sixstep waveform is vmax = This is taken to be the maximum limit of the phase voltage. Finally, with imax, vmax denoting the limits of the phase currents and voltages of the three-phase motor, the corresponding limits Imax,Vmax for the equiv

29、alent two-phase motor are thenThe direct-quadrature or dq transformation is defined bywhere id, iq and vd, vq are the transformed currents and voltages, respectively in the dq (for direct and quadrature) reference frame. The definition of the dq reference system assumes that the d-axis is aligned wi

30、th the rotor,s magnetic axis when = 0.Note that when =0, the d-axis is aligned the ia-axis which in turn is the same as the iS1-axis. The state-space model in the dq coordinates is (5)This model assumes that the rotor is smooth (non-salient) and that the magnetics are linear.3. Motor specificationsT

31、he motor parameters are specified for a linear motor and are converted to an equivalent rotary motor. The linear motor parameters are stator inductance LS = 6.4mH, stator resistance RS = 9.5, coefficient of mutual inductance M = 0.5LS = 3.2mH, motor mass m = 2.7 kg, force constant KM = 0.0227N/A, di

32、stance between poles dp = 0.0712/2m, np = 1 (no. of primary pole-pairs). The maximum dc bus voltage to the inverter is Vmax = 320V resulting in a peak fundamental waveform to the motor of vmax = Vmax=204V.The phase currents are limited to Imax = 10A (peak) and the maximum (linear) force put out by t

33、he motor is 320N.The radius of an equivalent rotary motor satisfies 2preq = np2dp req = 0.0227m. The torque constant of an equivalent three-phase rotary motor is found from the linear force constant by setting Km = req KM = (0.0227)(32)Nm/A = 0.726Nm/A and the moment of inertia is J =r2eqm =(0.0227)

34、22.7 =1.3910-3kgm2. The parameters LS, M, RS, np are the same as for the linear motor. Here x = 0 for the linear motor corresponds to the magnetic axis of its rotor phase a being lined up with the magnetic axis of stator phase a and similarly for the equivalent rotary motor. The corresponding equiva

35、lent two-phase parameters are then L = LS + M = 9.6mH, RS = 9.5, Keq =Km=0.9Nm/A, Imax (continuous) =imax =12.2A,Vmax=vmax=249.5V.The linear force put out by this motor isF = Keq iq / req=Km iq / req .3.1. Door modelThe door model is from the technical report of He 4 and is of the form dx/dt = Ax+bu

36、 y =Cx where A R88 , bR8 , CR88 .The values of the triple A,b,C are given in 4. Here x1 is the door position, x2 is the door speed and the input u to the door is the linear force F = Keq iq / req put out by the motor. The state variables x1, x2 are the two measured/computed state variables so that t

37、he output matrix is simplyThe mass of the door is denoted by Mc so that the total mass of the door/motor combination is Mc + m. The observability matrixhas rank 4 while the controllability matrixhas rank 5. However, A is stable. The control approach is to feed back x,(= x/req,=/req) treating the tra

38、nsfer function from input u = F to x as a double integrator. The resolution of the linear position feedback from the wall to the door control system is 0.005mm. The maximum door speed is max = 1m=s, the maximum acceleration is max = 1.2m/s2, and the jerk rate is limited to jmax = 2.4m/s3. The total

39、distance traveled by each door is 555mm. 3.2. Standard controller A straightforward way to do servo control of this motor for the standard controller in which there is one inverter for each motor is to choose the linear force aswhere the reference trajectories xref, vref, ref, iqref are as shown in

40、Section 4.1.1 and idref is typically taken to be zero 13.4. Two motors and one inverterThe interest here is to control the motor using a single inverter and the approach to independently control the quadrature current in each motor which in turn requires leaving the direct currents uncontrolled. One

41、 approach to controlling two PM synchronous motors using one inverter would be to just control the two motors identically. Specifically, as they nominally follow identical trajectories, just set vd1 = vd2, vq1 = vq2 so that 1 = 2, 1 = 2, id1 = id2, iq1 = iq2. This is a standard approach for torque/s

42、peed control of induction motor propulsion systems (light rail vehicles, subway cars, etc.) in which two traction (induction) motors are driven by a single inverter. In this case (torque control), the induction motors require only the rotor speed (to estimate the rotor fluxes) so that the average sp

43、eed of the two motors can be used in the flux estimator and for speed control. However, in the case of synchronous motors, the position of the rotor is required for control as (7), (8) show. The external disturbances L1, L2 on the individual motors are not necessarily equal and so the rotors will mi

44、salign, i.e., 1 2. In this situation, a control scheme based on putting the same input into both motors is not able to recover after misalignment occurs, that is, to realign1 = 2.Again, the objective here is to use one inverter to control two motors. (If there was an inverter for each motor, then th

45、e feedback controller given in (6) would suffice). To develop a controller using a single inverter for the two motors, two cases are considered: (1) the motors are connected in parallel (2) the motors are connected in series.4.1. Parallel connectionFirst, consider the motors connected in parallel, t

46、hat is, the applied voltage to each phase of the motors are the same. The model of the two motors in the dq coordinate system are thenandTo run these two motors off of a single inverter, one must take into account how the voltages are commanded to the motor. The same three voltages vS1, vS2, vS3, or

47、 equivalently, the same two phase equivalent voltages va, vb are commanded to both motors. The dq voltages for the two motors are given bywhere 1,2 are the angular position of motor 1 and motor 2, respectively 1.As the controller (6) indicates, this can be done by specifying the quadrature voltage v

48、q of each motor to control the torque producing current iq of each motor. To do so, Eqs. (7) and (8) show this requires choosing va, vb such thatClearly there is a singularity in the inverse at(np(2-1 )mod=0.The reference trajectories for the two linear motors are designed to maintain an angular sep

49、aration of the two motors. A control scheme for the two motors is thenThe direct voltages are determined by the quadrature voltages given by (9). Specifically, substitute (9) into1 1 = 0 is assumed to correspond with the magnetic axis of phase a of motor 1 and similarly for motor 2.so that when 2 1

50、0, vd1, vd2 and consequently, the currents id1, id2 can be large near the singularity.To avoid the singularity, the control scheme proposed here is to physically offset the angular position of the rotors of the two motors by (/2)/np. Then, as both nominally track the same trajectory (except for the

51、position offset of (/2)/np), a tight trajectory tracking control loop would keep np(2 1) close to /2 and thus keep the system away from the singularity. SimulationsA basic simulation was run using the one inverter two motor controller. The speed profile for the two motors is shown in Fig. 1 and the

52、position profiles are shown in Fig. 2. These plots show that the doors open in 4.2 s. The speed tracking error is less than 510-4m/s and therefore is not shown . The difference in position shown in Fig. 2 is due to the fact that x1(0) = 0 and x20= reqnp=0.0356m. The uncontrolled currents id1, id2 ar

53、e shown in Fig. 3. The difference in the trajectories of the two currents can be explained by the fact the initial angular position of the two motors are different by (/2)/np to avoid the singularity in the control. The consequence of not being able to control the direct current to zero results in t

54、he wasted power RSi2d because in a one-inverter/one-motor configuration, this current would typically be zero.The quadrature currents iq1, iq2 are controlled and, as shown, are on top of each other. The linear force F = Keq iq / req is required to make the move.Finally, the phase voltage vS1 is show

55、n .where it is noted that the maximum of the voltage is just under 100V.4.2. Series connectionIn the series connection, the current in each motor is the same in their corresponding phases. To analyze this situation, consider the two-phase equivalent models of the two synchronous motors:The control i

56、nput isvsa = vsa1 + vsa2vsb = vsb1 + vsb2.Using high-gain current feedback,vsa = KP (isa-ref + isa)vsb = Kp(isa-ref isa).one can force isa isa_ref, isb isb_ref fast enough that isa, isb can be considered as the inputs . Note that the high-gain feedback does not have an integrator. This is due to the

57、 fact that the currents are sinusoids and will be of high frequency at high speeds and an integrator can have trouble tracking such a fast varying signal .With 1, 2 the torque references for the two motors respectively, the following two equationsAs in the parallel connection case, this has a singul

58、arity when(np(2-1 )mod=0.As in the case of the parallel connection, the reference trajectories for the two linear motors are designed to maintain an angular separation of the two motors. A control scheme for the two motors is thenThe direct currents are determined by the quadrature currents, specifi

59、cally, substitute (13) into When2 1 02k, id1, id2 and consequently, the currents id1, id2 can be large near the singularity.Again, the control scheme is to offset the angular position of the two motors by (1/2) /np. As before, both motors nominally track the same trajectory so that the controller wo

60、uld nominally keep np(2 1) = /2 and therefore keep the system away from the singularity. SimulationsTo simulate the series connected system, the first and second equation of (10) are added to the first and second equation of (11), respectively so that with vsa = vsa2 + vsa1, vsb = vsb2 + vsb1, along