CN1404215A - Asynchronous motor optimizing excitation control method based on magnetic-field saturated non-linear motor model - Google Patents

Abstract

The present invention belongs to the field of asynchronous motor exciting speed-regulating technology and features the exciting speed-regulating method of optimizing stator-exciting current to regulate rotor magnetic linkage. It MT coordinate system, the finite element analysis to asynchronous motor magnetic field is performed to obtain the effect of saturated motor magnetic field on motor parameters and the stator and rotor magnetic linkage via solving non-linear equation. The magnetic linkage table is optimized to obtain optimized exciting table and maximum torque vs rotation speed table. By means of PI control method to obtain optimized exciting current and reference value of torque current, comparison to practical exciting current and torque current and PI regulation and voltage decoupling, final stator voltage in MT coordinate system is obtained for the control of the inverter.

Description

Optimize excitation control method based on magnetic field saturation nonlinearity motor model asynchronous machine

Technical field

Optimize the excitation control technology field that exciting method belongs to the asynchronous machine rotating speed based on magnetic field saturation nonlinearity motor model asynchronous machine

Background technology

Along with power electronic technology and development of computer, the alternating-current electric transmission just progressively replaces the main trend that DC Transmission becomes electric drive.Most widely used with asynchronous machine especially in the alternating current machine.That its advantage is is with low cost, simple in structure, reliability is high, easy to maintenance, moment of inertia is little, efficient is high, rated capacity is big etc.But alternating current machine particularly asynchronous machine belongs to multivariable, strongly coupled system, control system is very complicated and be difficult to accomplish high performance speed governing, often to participate in control, increase the complexity and the cost of system, reduce the reliability of system by various detecting instruments.

The speed regulating method of asynchronous machine can be divided at present:

1. frequency modulation and voltage modulation control.This method realizes simple, is widely used in the occasion not high to the speed governing performance requirement, and shortcoming is that load capacity is poor, and dynamic response is slow, and performance is poorer during low speed.

2. field orientation vector control.This method reaches the control performance of similar separately excited DC machine with the exciting current and the torque current component decoupling zero of asynchronous machine.This method can improve the performance of asynchronous machine speed governing greatly, is to use more control method in the present high-performance AC governing system.Shortcoming is the magnetic direction that needs to detect rotor, and the control result is subjected to the influence of parameter of electric machine variation easily.

3. direct torque control.Since adopt direct feedback bang-thereby bang control mode saved coordinate transform, simplified control structure, the influence of having avoided the parameter of electric machine to change.Shortcoming is to introduce low-speed torque ripple, has limited speed adjustable range, has reduced speed adjusting performance.

The asynchronous machine of field orientation vector control is to use wider high-performance governing system at present.Common control strategy is to keep exciting current constant, controls the torque of motor by changing torque current, and at this moment machine operation is in stable magnetic circuit saturation point, and the various inductance coefficents of motor can be considered as constant.But in tractive load occasions such as electric locomotive, electric automobile, elevating mechanisms, require the breakdown torque of motor bigger as far as possible, make drive system have torque control ability flexibly than nominal torque.Change the realization that exciting current will help this target.But at this moment motor main magnetic circuit degree of saturation will change, and inductance parameters no longer is a constant, must adopt new motor mathematical model and control strategy.

Summary of the invention

The objective of the invention is to set up and be applicable to that the saturated motor model in consideration magnetic field also can realize optimizing the asynchronous machine control method that excitation is controlled under the field orientation control.It has not only overcome the linearizing shortcoming of conventional model, and the actual conditions of motor in bigger range of operation, have truly been reflected, the asynchronous machine of optimizing excitation control is than power supply maximum current one timing of conventional vector control method at the supply motor, the output torque that can improve asynchronous machine.At motor starting, can better realize under the operating conditions such as acceleration that torque controls flexibly.

Technical scheme of the present invention is:

1. adopt and consider that the saturated motor mathematical model of motor magnetic circuit calculates motor properties.The present invention can obtain the saturated influence to the parameter of electric machine of motor-field by finite element analysis is carried out in asynchronous machine magnetic field.Structured data according to motor generates the motor end cross-sectional view with AUTOCAD, then cross section is carried out subdivision, obtain subdivision graph, generate coefficient matrix, find the solution the magnetic potential that Nonlinear System of Equations obtains each point under the different stator MT shaft currents with N-R iteration and Gaussian reduction, calculate the close distribution of magnetic of motor and the magnetic linkage of each winding according to magnetic potential, obtain the magnetic linkage of rotor M, T axle at last through heterogeneous or three-phase to the conversion of two-phase.Through to different M, the analytical calculation of motor magnetic linkage under the T shaft current can obtain a discrete exciting table, in the practical application this table is carried out interpolation and just can obtain magnetic linkage under any electric current.

Because asynchronous machine is the system of multivariable close coupling, the inductance under three phase coordinate systems in the motor model is a variations per hour, changes with rotor position angle, brings a lot of difficulties with finding the solution for machine analysis.Usually adopt the mode of coordinate transform that variations per hour is become constant.The principle of coordinate transform is that the magnetomotive force that motor model produced under different coordinates is in full accord, thereby and the purpose of conversion is the equivalence of the physical model of alternating current machine is simplified the analysis of motor for the pattern of similar direct current machine and to find the solution.The physical model of alternating current machine and Equivalent DC machine model are as shown in Figure 1.

A, B, C represent the three-phase of motor respectively among the figure, can obtain the model of motor on the MT axle by the variable with three-phase to the MT coordinate system projection of rotation.Because the MT axle is with the rotation of motor synchronous rotating speed, then the time-varying parameter of alternating current machine is converted into linear variable under the MT axle, is the DC generator model with the equivalence of alternating current machine model, thus the model equation that obtains simplifying.If ω ₁Be synchronous speed, F is the synthetic magnetomotive force that three-phase current produces.θ is the angle of M axle and stator A axle.Then pass through coordinate transform, consider that the motor mathematical model of the saturated influence of magnetic circuit is as follows:

For asynchronous machine, (1) formula is for being the motor dynamical equation of state variable with the electric current.Equation reflection be differential relationship between electric machine rotor electric current and voltage and the magnetic linkage.

Wherein:

P represents the time is carried out the differentiating operator of differential;

i _Ms, i _Ts, i _Mr, i _TrBe respectively the M of rotor, the T shaft current.

f _Ms, f _Ts, f _Tr, f _MrBe respectively the M of rotor, T axle magnetic linkage.The main purpose of FEM (finite element) calculation is calculated these magnetic linkages exactly.

u _Ms, u _TsBe the M of stator, the T shaft voltage.

R _s, R _rBe fixed rotor resistance.

ω ₁, ω ₂Be the stator and rotor current frequency. $J\frac{d{\mathrm{\ω}}_r}{\mathrm{dt}}=T_e-T_l---\left(2\right)$

(2) formula is the motor movement equation.T wherein _lBe load torque, ω _rBe rotor mechanical angular speed, T _eBe electromagnetic torque, J is the moment of inertia of motor and load.

2. on this model based, propose to optimize excitation control method.

Here the motor control method of Cai Yonging is indirect rotor field-oriented vector control.The field orientation vector control is with the direction of the space magnetic field vector reference direction as reference axis, motor stator current phasor quadrature is decomposed into excitation current component consistent with magnetic direction and the torque current component vertical with magnetic direction, by control respectively, make the alternating current function have good control performance as separately excited DC machine to these two components.In indirect rotor field-oriented vector control system, get the reference direction that the rotor field direction is a rotor M axle.Then vertical with it T axle rotor flux ψ _Tr=0 and the differential p ψ of magnetic linkage _Tr=0 and ψ _Trω ₂=0, exciting current i then _MsWith torque current i _TsThe given output torque T that has determined asynchronous machine _eBe shown below: $T_e=\frac{3}{2}{n}_p\frac{i_{\mathrm{Ts}}}{K_s}{\mathrm{\ψ}}_r---\left(3\right)$

(3) formula represents that the output torque of motor is by torque current i _TsWith rotor flux ψ _rDecision.N wherein _pBe the motor number of pole-pairs, $K_s=L_r/L_m$ , L _mAnd L _rBe the mutual inductance and the self-induction of rotor.Considering the saturated condition lower rotor part magnetic linkage ψ in magnetic field _rDetermine by following formula: ${\mathrm{\ψ}}_r-f_{\mathrm{Mr}}(i_{\mathrm{Ms}},--\frac{p{\mathrm{\ψ}}_r}{R_r})=0---\left(4\right)$

f _MrFor characterizing the saturated nonlinear function of magnetic circuit, R _rBe rotor resistance.Following formula shows that rotor flux is only by the stator excitation current i _MsDetermine, promptly according to following formula by known i _MsJust can be in the hope of rotor flux ψ _r ${\mathrm{\&omega;}}_s=\frac{R_r{i}_{\mathrm{Ts}}}{K_s{\mathrm{\&psi;}}_r}---\left(5\right)$ Following formula is slip-frequency ω _sExpression formula in indirect rotor field-oriented control system.The total current of the asynchronous machine of inverter control and the performance limitations that stator terminal voltage is subjected to inverter.Its restrictive condition is: $i_s=\sqrt{{{\mathrm{<figure-callout\; id="2"\; label="i\; Ms"\; filenames="A0214614900242.png,A0214614900292.png"\; state="\{\{state\}\}">i</figure-callout>}}_{\mathrm{Ms}}}^2+{{\mathrm{<figure-callout\; id="2"\; label="i\; Ts"\; filenames="A0214614900242.png,A0214614900292.png"\; state="\{\{state\}\}">i</figure-callout>}}_{\mathrm{Ts}}}^2}\&le;i_{s\max }$ I wherein _s, u _sBe stator current and terminal voltage, i _Smax, u _SmaxBe inverter $u_s=\sqrt{{\mathrm{<figure-callout\; id="2"\; label="u\; Ms"\; filenames="A0214614900242.png,A0214614900292.png"\; state="\{\{state\}\}">u</figure-callout>}}_{\mathrm{Ms}}^{}+{\mathrm{<figure-callout\; id="2"\; label="u\; Ts"\; filenames="A0214614900242.png,A0214614900292.png"\; state="\{\{state\}\}">u</figure-callout>}}_{\mathrm{Ts}}^{}}\&le;u_{s\max }$ Maximum possible electric current and voltage under the power supply, its value depends on the performance parameter of inverters element.

This control method makes that by adjusting the rotor flux of motor motor can be exported maximum torque under certain total current restriction.From formula (3), can see torque and change apart from electric current and long-pending being directly proportional of rotor flux.Then under the certain condition of total current size, be that variable carries out optimizing to formula (3) and calculates the result that can obtain as shown in Figure 2 to change apart from electric current or exciting current:

I among Fig. 2 _SmaxMaximum limit amplitude for stator current.

Result of calculation can obtain the exciting current of maximum torque correspondence under the different motor total currents from Fig. 2, thus the exciting table that is optimized (seeing Table 1).Conversely also can be from table as requested electromagnetic torque obtain the exciting current size of minimum total current correspondence.

Can derive by the motor equation (1) of front and to obtain following formula: $u_{\mathrm{Ms}}=R_s{i_{\mathrm{Ms}}}^{}-({\mathrm{\&omega;}}_r+{\mathrm{\&omega;}}_s)f_{\mathrm{Ts}}(i_{\mathrm{Ts}},i_{\mathrm{Ts}}/K_s)$ Can see that rotating speed is influential to the MT shaft voltage.

u _Ts＝R _si _Ts+(ω _r+ω _s)f _Ms(i _Ms，0)

Then can calculate at the breakdown torque table (see Table 2) of motor under the different rotating speeds under certain electric current and voltage limit according to the electric current and voltage restrictive condition.The concrete computational process of above-mentioned two tables is seen below appended flow chart.

The invention is characterized in: it is to optimize in the exciter control system at the asynchronous machine that is made of single-chip microcomputer, sampling plate and inverter of bus control type, changing exciting current, to realize that stator total current one is fixed electromagnetism output torque maximum and fix the field orientation vector control method of the optimization field performance of total current minimum in torque output one, and it contains successively and has the following steps:

(1) under the on all four condition of the magnetomotive force that requires motor model to produce after the coordinate transform, the variable of motor A, B, C three-phase is projected on the MT coordinate system with the rotation of the synchronous speed of motor, obtain the motor model of the saturated influence in consideration magnetic field on the MT coordinate system so that be converted into linear variable, the dc electrolysis model of alternating current machine model equivalence for simplifying at the time-varying parameter of the following alternating current machine of MT coordinate system;

(2) input:

(2.1) the breakdown torque table under the different rotating speeds;

(2.2) with the corresponding optimization excitation ammeter of minimum total current of exporting certain torque;

(2.3) magnetic linkage table that constitutes by following data item:

i _MsBe stator M shaft current, i _MrBe rotor M shaft current.

i _TsBe stator T shaft current, i _TrBe rotor T shaft current.

ψ _TSBe stator T axle magnetic linkage, ψ _TrBe rotor T axle magnetic linkage.

ψ _MSBe stator M axle magnetic linkage, ψ _MrBe rotor M axle magnetic linkage.

Stator A, B, C three-phase magnetic linkage: ψ _A, ψ _B, ψ _C

(3) according to given angular velocity omega ^*With the actual measurement angular velocity omega _rObtain torque set-point Te through the rotating speed pi regulator ^*: $T_e^{*}=(k_P+\frac{k_q}{S})({\mathrm{\&omega;}}^{*}-{\mathrm{\&omega;}}_r)$ Wherein kp is a proportionality constant, and kq is an integral constant;

(4) Te ^*Compare with the breakdown torque Temax that obtains according to rotor actual measurement rotating speed:

If: Te ^*〉=Temax then gets Te ^*=Temax;

If: Te ^*＜Temax then gets Te ^*=Te ^*

(5) according to rotor actual measurement rotating speed and Te ^*Check in i from optimizing excitation ammeter _Ms, again by i _MsCheck in rotor flux ψ from magnetic linkage table _r

(6) according to Te ^*, ψ _rObtain i from following formula _Ts: $T_e^{*}=\frac{3}{2}{n}_p\frac{i_{\mathrm{Ts}}}{K_s}{\mathrm{\&psi;}}_r{n}_p$ : the motor number of pole-pairs, $K_s=L_r/L_m$ : L _m: rotor mutual inductance L _r: the rotor self-induction.

(7) judge that the threephase stator size of current read in whether in reasonable range, if surpass amplitude limit value, then gets amplitude limit value for reading in current value;

(8) calculate the slippage angular frequency according to following formula _sWith the rotor field angular position theta ₁: ${\mathrm{\&omega;}}_s=\frac{R_r{i}_{\mathrm{Ts}}}{K_s{\mathrm{\&psi;}}_r}$ , be the rotor current angular frequency to rotor field angular speed again ₁=ω _s+ ω _rIntegration obtains the rotor field angular position theta ₁

(9) the stator phase current is become current i under the MT coordinate system _Ms, i _Ts: $\left[\begin{array}{c}{i}_M\\ i_T\end{array}\right]=\frac{2}{3}\left[\begin{array}{ccc}\cos {\mathrm{\&theta;}}_1& \cos ({\mathrm{\&theta;}}_1-\frac{2\mathrm{\&pi;}}{3})& \cos ({\mathrm{\&theta;}}_1+\frac{2\mathrm{\&pi;}}{3})\\ -\sin {\mathrm{\&theta;}}_1& -\sin ({\mathrm{\&theta;}}_1-\frac{2\mathrm{\&pi;}}{3})& -\sin ({\mathrm{\&theta;}}_1+\frac{2\mathrm{\&pi;}}{3})\end{array}\right]\left[\begin{array}{c}{i}_{\mathrm{sa}}\\ i_{\mathrm{sb}}\\ i_{\mathrm{sc}}\end{array}\right];$

(10) according to following formula i given on the stator _Ms ^*, i _Ts ^*With the actual current i that records _Ms, i _TsAfter comparing respectively, difference is obtained separately the voltage given value u of stator MT axle by exciting current pi regulator, torque current pi regulator and the compensation of process voltage decoupling _Ms ^*, u _Ts ^*: ${\hat{u}}_{\mathrm{Ms}}=L_m^M{\mathrm{pi}}_m-{\mathrm{\&omega;}}_1{\mathrm{\&psi;}}_{\mathrm{Ts}}$ $L_m^M=\frac{{\&PartialD;f}_{\mathrm{Ms}}}{{\&PartialD;i}_{\mathrm{Mr}}}$ M axle bucking voltage ${\hat{u}}_{\mathrm{Ts}}=L_{\mathrm{MT}}{\mathrm{pi}}_m+{\mathrm{\&omega;}}_1{\mathrm{\&psi;}}_{\mathrm{Ms}}$ $L_{\mathrm{MT}}=\frac{{\&PartialD;f}_{\mathrm{Ts}}}{{\&PartialD;i}_m}$ T axle bucking voltage i _m=i _Mr+ i _MsThe exciting current pi regulator is: $u_M=(k_{\mathrm{dm}}+\frac{k_{\mathrm{lm}}}{S})(i_{\mathrm{MS}}^{*}-i_{\mathrm{MS}}),$ The torque current pi regulator is: $u_T=(k_{\mathrm{dT}}+\frac{k_{\mathrm{lT}}}{S})(i_{\mathrm{TS}}^{*}-i_{\mathrm{TS}}),$ Then the given voltage of MT axle is: $u_{\mathrm{Ms}}^{*}=u_M+{\hat{u}}_{\mathrm{Ms}}$ $u_{\mathrm{Ts}}^{*}=u_T+{\hat{u}}_{\mathrm{Ts}};$

(11) obtain stator three-phase voltage set-point through 2/3 conversion again and send into pulse width modulator PWM producing six tunnel required pulses of inverter;

(12) finish.

Described magnetic linkage table is at the hypothesis cylindrical electrical machine along the axis direction Distribution of Magnetic Field evenly and ignore under the condition of the kelvin effect of rotor bar and magnetic hysteresis eddy current effect unshakable in one's determination, and the external electromagnetic field of motor is reduced to the two dimensional surface field.Use the finite element elements method at different M, calculate the electromagnetic field of motor and the magnetic linkage of each winding under the T shaft current and obtain, it contains successively and has the following steps:

(1) get with the perpendicular cross section of motor shaft as the constant nonlinear electromagnetic of analyzed two dimensional surface field, subdivision is carried out in the fan section of the radial symmetric that comprises stator and rotor;

(2) find the solution vector magnetic potential A on each summit of triangular element to finding the solution the zone triangular element that goes out of each subdivision with following electromagnetic field equation:

Wherein: first kind boundary condition Г ₁For: for stator, the outer yoke of per tooth is wide behind the subdivision;

Rotor there is not first kind boundary condition;

The second class boundary condition Г ₂For: for stator, the tooth rim of per tooth is long behind the subdivision;

For rotor, the groove girth of per tooth groove behind the subdivision;

A is a vector magnetic potential.V is a reluctivity.δ is a current density, and Ht is the magnetic field intensity of border and normal direction vertical direction.A ₀For the borderline magnetic potential value of the first kind, be known for its boundary value of special area.

(3) by each triangular element row are write above-mentioned electromagnetic field equation,, all equations find the solution the magnetic potential that obtains each summit, triangle subdivision unit thereby being synthesized a total matrix equation;

(4) suppose again under the condition of each point magnetic potential sight line distribution in the triangular element, according to each triangular apex i, j, the position coordinates of k and the magnetic potential A on this summit _iA _jA _k, determine that the magnetic potential of each point in stator, the rotor distributes;

(5) calculate A, B, C three-phase magnetic linkage ψ according to stator each point magnetic potentiometer _Aψ _Bψ _CAfter, obtain the magnetic linkage ψ on the stator MT axle more thus _MsAnd ψ _Ts: ${\mathrm{\&psi;}}_A=W_K{l}_e\left(\underset{i=1}{\overset{q}{\mathrm{\&Sigma;}}}(A_{s_i}-A_{s_{\mathrm{ai}}})\right)$

Wherein: A _SlA _SqAnd A _SalA _SaqBe respectively each groove S of stator A phase winding place _lS _qAnd S _AlS _AqMagnetic potential mean value, l _eBe effective length unshakable in one's determination, W _kBe umber of turn;

B, each corresponding magnetic potential mean value substitution following formula of C two-phase are just obtained the magnetic linkage of B, C two-phase; Thereby obtain:

θ is the angle between the positive A axle of M axle and stator.In vector control system the output torque of motor only relevant with the size of magnetic linkage and with the orientation independent of magnetic linkage, magnetic linkage is big or small definite and can not change with the variation of θ under the given condition of MT shaft current.So here can be given arbitrarily and can the magnetic linkage size that finally calculates do not exerted an influence in the θ angle.Usually be taken as 0 for simplicity.

(6) for rotor-side, establishing the pairing groove of a pair of mouse cage sliver is n and n+1, and corresponding average magnetic potential is A _nAnd A _N+1, then the magnetic linkage in this loop is ψ _n=l _e(A _N+1-A _n), corresponding, the magnetic linkage on the rotor MT axle is: Wherein, K is the rotor bar number, ₀It is the angle between two slivers;

(7) under limit in the field orientation system motor T axle magnetic linkage be zero.Then can obtain i _Ts+ K _si _Tr=0, thereby at given stator torque current i _TsCondition under obtain rotor T shaft current.By changing given stator T shaft current i _TsWith M shaft current i _MsWith rotor M shaft current i _Mr, repeat above step and obtain a discrete usefulness with reflecting stator M shaft current i _Ms, stator M axle magnetic linkage ψ _Ms, stator T shaft current i _Ts, stator T axle magnetic linkage ψ _Ts, rotor M shaft current i _Mr, rotor M axle magnetic linkage ψ _Mr, rotor T shaft current i _TsRotor T axle magnetic linkage ψ _TrBetween the magnetic linkage table of correlation; Obtain magnetic linkage table under any electric current by interpolation algorithm;

The computational process of foregoing input table is as follows:

The data item that described breakdown torque table contains is: breakdown torque, rotating speed, optimization exciting current and torque current;

The data item that described optimization excitation ammeter contains is: optimize exciting current, corresponding output torque and rotating speed;

Described breakdown torque table and optimization excitation ammeter obtain through following steps successively according to magnetic linkage table:

(1) scope of setting stator current Is:

I _S0～I _Smax(maximum is limited by inverter)

(2) differentiate: I _s＞I _Smax:

If: I _s＞I _Smax, then change step (9) over to;

If: I _s＜I _Smax, then carry out next step;

(3) scope of setting motor speed ω: ω=ω ₀＜ω _Max(maximum is limited by motor),

If: ω＞ω _Max, then returning step (2), Is increases an increment simultaneously;

If: ω＜ω _Max, then carry out next step;

(4) under stator total current and stator voltage and motor speed restriction, obtain the optional scope of stator excitation electric current: I _M0～I _Mmax, maximum is determined by following formula $u_s=\sqrt{{\mathrm{<figure-callout\; id="2"\; label="u\; Ms"\; filenames="A0214614900242.png,A0214614900292.png"\; state="\{\{state\}\}">u</figure-callout>}}_{\mathrm{Ms}}^{}+{\mathrm{<figure-callout\; id="2"\; label="u\; Ts"\; filenames="A0214614900242.png,A0214614900292.png"\; state="\{\{state\}\}">u</figure-callout>}}_{\mathrm{Ts}}^{}}\&le;K_u{u}_{\max }=u_{s\max }$ Wherein: u _sBe stator terminal voltage amplitude, K _uFor guaranteeing the voltage security coefficient of inverter trouble free service, u _MaxIt is the maximum withstand voltage that inverter can bear $u_{\mathrm{Ms}}=R_s{i}_{\mathrm{Ms}}-({\mathrm{\&omega;}}_r+{\mathrm{\&omega;}}_s)f_{\mathrm{Ts}},(i_{\mathrm{Ts}},i_{\mathrm{Ts}}/K_s)$ Wherein: ω _r=ω, ${\mathrm{\&omega;}}_s=\frac{R_r{i}_{\mathrm{Ts}}}{K_s{f}_{\mathrm{Ms}}(i_{\mathrm{Ms}},0)};$ u _Ts＝R _si _Ts+(ω _r+ω _s)f _Ms(i _Ms，0)

(5) differentiate: I _m＞I _Mmax:

If: I _m＞I _Mmax, then returning step (3), ω increases an increment simultaneously;

If: I _s＜I _Smax, then carry out next step;

(6) calculate T shaft current I according to total current _Ts

According to exciting current I _mLook into magnetic linkage table and obtain rotor magnetic linkage ψ _s, ψ _r

(7) calculate motor torque according to torque current and magnetic linkage, with the torque that calculates, stator excitation electric current and torque current, motor stator total current, motor speed deposit disk file result.dat in;

(8) I _mIncrease an increment, repeating step (5)～(7) are until I _mIncrease make I _s=I _Smax, change step (9) over to.

(9) from disk file result.dat, read in data;

(10) establish array sp[n], n is the quantity of different motor speeds in the file, and rotating speed by depositing in from low to high among the array sp, and is established k=0;

(11) establish speed=sp[k], search all motor torques with the speed correspondence hereof, and deposit this group torque in array Te[m] in, m is the number of these torques;

(12) Te is sorted by size;

(13) getting number maximum among the Te is Temax;

(14) excitation, torque current and the magnetic linkage with speed and Temax and correspondence deposits data file maxtorqu.dat in;

(15) establish j=0, T=Te[i];

(16) search all stator total currents with torque T and speed correspondence hereof, and deposit this group current value in array Is[l] in;

(17) to array Is[l] sort by size;

(18) get Is[l] in minimum value be Ismin;

(19) exciting current and the magnetic linkage with speed and Ismin, T and correspondence deposits data file optimal.dat in;

(20) do you differentiate: k＞n?:

If: k＜n, then k=k+1 returns step (11);

If: k＞n, then deposit data file maxtorque.dat and optimal.dat in disk, obtain the breakdown torque table and optimize excitation ammeter, carry out next step;

(21) finish.

The invention has the beneficial effects as follows, on the basis that does not change the asynchronous motor drive system main circuit structure, pass through to optimize the exciting current of motor, the output torque that under the constant condition of total current, can improve motor, thereby improved the performance of asynchronous motor speed-regulating system, as improved starting characteristic, start the obviously fixing control method of required time faster than traditional excitation.

Figure 13 optimizes the motor torque that produces under the excitation relatively for model machine in conventional vector control constant excitation megnet and the present invention, the given electric current of stator bigger (often being subjected to the inverter performance limitations) as seen from the figure, and the torque of then optimizing excitation increases significantly.

Figure 14 optimizes the motor torque comparison that produces under the excitation for model machine in conventional vector control constant excitation megnet and the present invention, and traditional vector control is controlled well than frequency and voltage boosting, and it is best to optimize excitation vector control starting performance.

Description of drawings

The present invention is further described with enforcement below in conjunction with accompanying drawing

Fig. 1. stator by three-phase A, B, C coordinate to MT coordinate transformation schematic diagram.

Fig. 2. stator current one is fixed torque and exciting current graph of a relation, a rated exciting current, b torque optimal exciting electric current, ismax are the given maximum of stator current.

Fig. 3. hardware system block diagram: Sn: speed probe; Si: current sensor.

Fig. 4 .CPU Task Distribution figure and control block diagram.

Fig. 5. the program flow chart of indirect rotor field-oriented vector system.

Fig. 6. magnetic linkage calculation process block diagram.

Fig. 7. the motor cross-sectional view.

Fig. 8. motor stator and rotor subdivision graph.

Fig. 9. the enlarged drawing of circle shown in Fig. 8.

Figure 10. the FB(flow block) of FEM (finite element) calculation.

Figure 11. matrix equation [K] A[]=the solution procedure block diagram of [R].

Figure 12. the computational process block diagram of table 1, table 2.

Figure 13. optimize the comparison diagram (experiment value) of excitation and rated excitation, a: exciting current, b: rated exciting current are optimized in torque.

Figure 14. the starting process under the stator current amplitude limit (experiment value), a optimizes excitation, b rated excitation, the starting of c frequency and voltage boosting.

Embodiment

Ask for an interview Fig. 3 and Fig. 4.

Motor speed at first as requested provides rotary speed setting value ω ^*, with set-point and actual measurement rotational speed omega _rSubtract each other and obtain speed error Δ ω, again speed error substitution rotating speed pi regulator is calculated torque reference value Te ^*Simultaneously according to actual speed and torque reference value Te ^*, being tabled look-up by the excitation optimal curve that has calculated obtains required optimization exciting current reference value i ^* _Ms, simultaneously the given substitution torque current of exciting current computing module is calculated the given i of torque current by formula (3) (4) ^* _TsBy excitation with torque current is given can calculate required slippage angular frequency by formula (5) _s, add the actual motor speed ω that records _rObtain the rotor current angular frequency.The rotor current angular frequency is carried out integral and calculating obtain the rotor field angular position theta ₁The motor threephase stator electric current that simultaneously detection is obtained obtains actual excitation and torque current component i through 3/2 conversion _MsAnd i _Ts, by with set-point relatively obtain current error Δ I.Current error is respectively by obtaining initial results after torque current and the exciting current pi regulator, add the offset that calculates through voltage decoupling, then obtain the stator voltage under the finally given MT coordinate system, obtain exporting to the three-phase of inverter by 2/3 conversion.Three-phase output is modulated three tunnel pulses that obtain inverter by PWM, the three-phase voltage U a/b/c that control inverter output is required.

The program flow diagram of indirect rotor field-oriented vector system is seen Fig. 5.L wherein _m ^M, L _MTBe to f by magnetic linkage table _Ms, f _TsRespective point obtains as local linearization.

In initialization, detect common storage area earlier, detect the motor speed that the user sets again.The system status parameters of reading in from external bus comprises motor current state and control system state.The validity of check input data and in addition amplitude limit be meant and judge input motor speed set-point whether between maximum (top) speed and minimum speed, otherwise be considered as invalid data.Adopt last time valid data as this secondary data to invalid data.

Whole system is made up of four parts: control unit, inverter, sensing device and motor.Control unit obtains three-phase voltage, electric current and the motor speed of the motor that obtained by sensor; The control signal (the gate pulse signal of major loop IGBT device) that inverter needs that calculates through numerical control system; Control signal is applied to through interface circuit and becomes actual controlled variable (controlled variable of voltage source inverter is the stator voltage of motor, and current source inverter is stator current) on the power electronic device.Digital control unit mainly contains three parts: arithmetic element, analog quantity input interface and control signal output interface.In addition, also have the interface of realization and computer communication, and the interface of man-machine interaction or the like.Wherein arithmetic element is the core component of control system, has adopted the structure of two CPU controls.Control unit couples together two CPU and peripheral hardware by bus.Main circuit has adopted the inverter major loop of POWERTRAN .

Below how narration uses Finite Element Method court magnetic linkage table, and its program flow chart is seen Fig. 6.If: the actual parameter of electric machine that adopts is:

Stator outer diameter (mm) 210; Diameter of stator bore (mm) 114; Rotor diameter (mm) 113

Shaft diameter (mm) 48; Silicon steel sheet length (mm) 130; Silicon steel sheet model: D21

Shaft material: H45; Number of stator slots: 24; Rotor number: 20; Rated power: 7.5kw; Rated voltage (V): 178; Specified line current (A): 31; Connected mode: Δ

The subdivision program is seen Fig. 7～9.In Fig. 8, the 1st, stator yoke portion, the 2nd, stator tooth, the 3rd, stator slot, the 4th, air gap, the 5th, rotor tooth, the 6th, rotor, the 7th, rotor yoke, the 8th, rotating shaft.

The subdivision explanation: according to the symmetry of electric machine structure, 24 of stator can be divided into symmetry are fan-shaped, and the central angle of each piece correspondence is 15 degree.Rotor then can be divided into the symmetry 20 fan-shaped, the central angle of each piece correspondence be 18 the degree.Then with stator block according to radial centre lines be divided into the symmetry two carry out subdivision.Rotor block then directly carries out subdivision.Subdivision process to stator block is: with the stator block layering, the principle of layering is thicker between the simple region layer with isocentric circular arc, and as stator yoke, and complex region such as the punishment of stator slot camber line are thin layer.Radially layer is divided into rectangle then, curved section adopts broken line to replace.With diagonal rectangle is divided into two triangles then.Shown in following subdivision graph.The rotor subdivision is similar.

After subdivision is finished like this, stator block chosen duplicates, use " serves as that axle carries out axial symmetry and duplicates with certain bar line " among the AutoCAD then thus function is duplicated the stator block upset subdivision result who obtains half motor stator part.Do to obtain rotor subdivision result partly half after same the processing for rotor portion.

Four subdivision unit as shown in Figure 9, after subdivision was finished, the data that preserve were: the sequence number P of triangle sequence number Si, triangular apex _{I, i+1, i+2}, the coordinate of triangular apex (px, py).The order of getting a little is counter clockwise direction, and the order of getting the summit as cell S i is: Pi, Pi+1, Pi+2. are stator or rotor etc. according to position, unit setup unit type simultaneously.

For stator side, A phase winding place groove is S _lS _qAnd S _AlS _Aq, wherein q is every extremely every phase groove number, if the magnetic potential mean value of each groove is A _SlA _SqAnd A _SalA _Saq, then A phase magnetic linkage is ${\mathrm{\&psi;}}_A=W_K{l}_e\left(\underset{i=1}{\overset{q}{\mathrm{\&Sigma;}}}(A_{s_i}-A_{s_{\mathrm{ai}}})\right)---\left(6\right)$

L wherein _eBe effective length unshakable in one's determination, W _kBe umber of turn.Bring the magnetic potential mean value of BC two-phase corresponding groove into magnetic linkage ψ that following formula just can calculate the BC two-phase equally _Bψ _CCan calculate the magnetic linkage of stator MT axle according to following formula:

θ angle in the following formula is the angle between M axle and the stator+A axle.

For rotor-side, establishing the pairing groove of a pair of mouse cage sliver is n and n+1, and corresponding average magnetic potential is A _nAnd A _N+1, then the magnetic linkage in this loop is

ψ _n＝l _e(A _n+1-A _n)?????????????????????????????????????(8)

Then can obtain by heterogeneous result, thereby obtain magnetic flux on the rotor MT axle to two-phase with reference to following formula.

Wherein K is the rotor bar number, ₀Be the angle between two slivers, ψ _nIt is the magnetic linkage in n loop.

The computational process of finite element

Motor is a cylinder, can suppose that cylinder is interior even along the axis direction Distribution of Magnetic Field, therefore is that calculate the two dimensional surface field with motor electromagnetic field problem reduction.Do not consider the kelvin effect and the magnetic hysteresis eddy current effect unshakable in one's determination of the rotor bar that influence relatively is less, only consider saturation effect unshakable in one's determination, the BH curve of its core material is known.Use the magnetic linkage parameter that Finite Element is calculated the electromagnetic field of motor under different M, T shaft current and obtained each winding of this moment according to result of calculation.At first get the cross section vertical as analyzed two dimensional surface field, the symmetrical sector region that comprises stator and rotor is carried out subdivision with motor shaft.Find the solution magnetic potential A on each summit, unit to finding the solution equation (10) below each subdivision triangular element is used in the zone.Here suppose in the unit that magnetic potential is a linear distribution everywhere, can obtain by three summit magnetic potential linear interpolations.By each cell columns being write equation and analyzed boundary condition, thereby the synthetic total matrix equation of all equations is found the solution the magnetic potential that obtains each summit, subdivision unit the most at last.Because the zone of finding the solution comprises stator and rotor, then can determine that the magnetic potential of rotor distributes, and obtains the magnetic linkage of rotor MT axle again according to equation (7) (9) according to the magnetic potential of unit vertex position coordinate and this point.

The motor cross section belongs to the constant nonlinear electromagnetic of two dimensional surface field, and its electromagnetic field equation and boundary condition are:

Wherein A is a vector magnetic potential.V is a reluctivity.δ is a current density, and Ht is the magnetic field intensity of border and normal direction vertical direction.A ₀Being borderline magnetic potential value, is known for its boundary value of special area.

Its variational problem of equal value is the minimization energy functional $F=\underset{\mathrm{\&Omega;}}{\&Integral;\&Integral;}({\&Integral;}_0^{\mathrm{<figure-callout\; id="2"\; label="B\; vB"\; filenames="A0214614900242.png,A0214614900292.png"\; state="\{\{state\}\}">B</figure-callout>}}\frac{{\mathrm{vB}}^{}}{}\frac{{}^2}{2}-\mathrm{A\&delta;})\mathrm{dxdy}+{\&Integral;}_{{\mathrm{\&Gamma;}}_2}{H}_t\mathrm{Adt}---\left(11\right)$

Wherein $B=\sqrt{{\left(\frac{\&PartialD;A}{\&PartialD;x}\right)}^2+{\left(\frac{\&PartialD;A}{\&PartialD;y}\right)}^2},$ $v=\frac{1}{\mathrm{\&mu;}\left(H\right)}$ 。Permeability when μ (H) is H for magnetic field intensity.

For a cell S e of triangle subdivision, its energy functional F is carried out discretization and asks its minimum value can obtain following formula:

(the discretization process of F can referring to " Theory of Electrical Moto ﹠ Electromagnetic Fields and calculating " P89～P91, Chen Pizhang etc., Science Press, 1986).If three summits of Se are numbered counterclockwise: i, j, k $\left[\begin{array}{c}\frac{\&PartialD;F}{{\&PartialD;A}_i}\\ \frac{\&PartialD;F}{{\&PartialD;A}_j}\\ \frac{\&PartialD;F}{{\&PartialD;A}_k}\end{array}\right]=\left[\begin{array}{ccc}{k}_{\mathrm{ii}}& k_{\mathrm{ij}}& k_{\mathrm{ik}}\\ k_{\mathrm{ji}}& k_{\mathrm{jj}}& k_{\mathrm{jk}}\\ k_{\mathrm{ki}}& k_{\mathrm{kj}}& k_{\mathrm{kk}}\end{array}\right]\left[\begin{array}{c}{A}_i\\ A_j\\ A_k\end{array}\right]-\left[\begin{array}{c}{f}_i\\ f_j\\ f_k\end{array}\right]---\left(12\right)$ A wherein _{I, j, k}Be an Atria summit i, j, the magnetic potential at k place. $k_{\mathrm{nn}}=\frac{{\mathrm{\&mu;}}_{\mathrm{<figure-callout\; id="4"\; label="e"\; filenames="A0214614900242.png"\; state="\{\{state\}\}">e</figure-callout>}}}{4{\mathrm{\&Delta;}}_e}({\mathrm{<figure-callout\; id="2"\; label="b\; n"\; filenames="A0214614900242.png,A0214614900292.png"\; state="\{\{state\}\}">b</figure-callout>}}_n^{}+c_n^2)n=i,j,k$

b _i＝(y _j-y _k)，b _j＝(y _k-y _i)，b _k＝(y _i-y _j)????(14)

c _i＝(x _k-x _j)，c _j＝(x _i-x _k)，c _k＝(x _j-x _i)

Wherein, δ e is the current density at place, unit, and Δ e is the leg-of-mutton area in unit, x _{Ij, k}, y _{I, j, k}Be each apex coordinate of triangular element.He is the magnetic field intensity at place, unit, μ _ePermeability for the place, unit.

Each cell columns to divided region is write top equation, synthetic then total equation:

[KIA]＝[R]?????????????????????????????????????(15)

Wherein, [K] is the summation of all k-factor matrixes, and [A] is the magnetic potential vector of being had a few, and [R] is the summation vector of all free term f.[A] that satisfies this equation is exactly the each point magnetic potential of being asked.

Figure 10 is the flow chart of FEM (finite element) calculation.

Suppose that the stator side electric current is i _A, i _BAnd i _C, suppose that the stator A phase winding axis of motor and the angle of M axle are θ this moment, the axis of first pair of sliver of rotor and the angle of M axle are γ, suppose also that simultaneously each end ring electric current of rotor-side is with position, loop Sine distribution.The pass of each phase current of stator and M, T shaft current is so

i _A＝i _Ms?cosθ-i _Ts?sinθ

i _B＝i _Ms?cos(θ-120°)-i _Ts?sin(θ-120°)(16)

i _B＝i _Ms?cos(θ+120°)-i _Ts?sin(θ+120°)

Here θ is taken as 0.

Electric close in the stator A phase groove is ${\mathrm{\&delta;}}_{\mathrm{sn}}=W_l{i}_s/S_s$ , S here _sBe stator slot sectional area, i _sBe the equivalent current in the groove, W ₁Be the number of turn.Equal the A phase current for its electric current of single layer winding, two layer winding then needs to determine according to the structure of winding the size of the equivalent current in the groove.

Rotor-side end ring electric current is i _Kn, can calculate by known rotor M, T spindle motor, as shown in the formula:

i _Kn＝i _Mr?cos[γ-(n-1) ₀]-i _Tr?sin[γ-(n-1) ₀](17)

Sliver current i then _Rn=i _Kn-i _{K (n+1)}The electric close of this sliver place groove is ${\mathrm{\&delta;}}_{\mathrm{rn}}=i_{\mathrm{rn}}/S_r$ , S _rBe the rotor area.

Be finding the solution of matrix equation [K] [A]=[R] below:

Owing to consider the saturated of motor, so that coefficient matrix K should be is nonlinear.Then adopt the N-R iterative method to calculate.

Make f (A)=[KIA], then have following being similar to:

f([A])＝f([A] ^N)+[J] ^N([A] ^N+1-[A] ^N)

N represents iteration the N time.As [A] N+1=[A] during N, just separating of equation can obtain.

Then obtain the formula of iteration form:

[J] ^N([A] ^N+1-[A] ^N)＝[R]-f([A] ^N)???????????????????????????(18)

Jacobian matrix in the iteration form $\left[J\right]=\frac{\&PartialD;\left[f\right]}{\&PartialD;\left[A\right]}$ , be equivalent to the slope of f ([A]) at the A place.For single triangular element Se, establish three summits by being numbered i counterclockwise, j, k, then the formula by the front has $f_e=\left[\begin{array}{c}{f}_{\mathrm{ie}}\\ f_{\mathrm{je}}\\ f_{\mathrm{ke}}\end{array}\right]=\left[\begin{array}{ccc}{k}_{\mathrm{iie}}& k_{\mathrm{ije}}& k_{\mathrm{ike}}\\ k_{\mathrm{jie}}& k_{\mathrm{jje}}& k_{\mathrm{jke}}\\ k_{\mathrm{kie}}& k_{\mathrm{kje}}& k_{\mathrm{kke}}\end{array}\right]\left[\begin{array}{c}{A}_i\\ A_j\\ A_k\end{array}\right]$ Therefore, ask local derviation to have to following formula ${\left[J\right]}^e=\left[\begin{array}{ccc}\frac{{\&PartialD;f}_{\mathrm{ie}}}{{\&PartialD;A}_i}& \frac{{\&PartialD;f}_{\mathrm{ie}}}{{\&PartialD;A}_j}& \frac{{\&PartialD;f}_{\mathrm{ie}}}{{\&PartialD;A}_k}\\ \frac{\&PartialD;f_{\mathrm{je}}}{{\&PartialD;A}_i}& \frac{{\&PartialD;f}_{\mathrm{je}}}{{\&PartialD;A}_j}& \frac{{\&PartialD;f}_{\mathrm{je}}}{{\&PartialD;A}_k}\\ \frac{{\&PartialD;f}_{\mathrm{ke}}}{{\&PartialD;A}_i}& \frac{{\&PartialD;f}_{\mathrm{ke}}}{{\&PartialD;A}_j}& \frac{{\&PartialD;f}_{\mathrm{ke}}}{{\&PartialD;A}_k}\end{array}\right]$ These partial differential equation of deriving can obtain $\frac{{\&PartialD;f}_{\mathrm{ie}}}{{\&PartialD;A}_i}=k_{\mathrm{iie}}+\frac{f_{\mathrm{ie}}^2}{v^{2}B{\mathrm{\&Delta;}}_e}\frac{\&PartialD;v}{\&PartialD;B}$ $\frac{{\&PartialD;f}_{\mathrm{je}}}{{\&PartialD;A}_j}=k_{\mathrm{jje}}+\frac{f_{\mathrm{je}}^2}{v^{2}B{\mathrm{\&Delta;}}_e}\frac{\&PartialD;v}{\&PartialD;B}$ $\frac{{\&PartialD;f}_{\mathrm{ke}}}{{\&PartialD;A}_k}=k_{\mathrm{kke}}+\frac{f_{\mathrm{ke}}^2}{v^{2}B{\mathrm{\&Delta;}}_e}\frac{\&PartialD;v}{\&PartialD;B}$ $\frac{{\&PartialD;f}_{\mathrm{ie}}}{{\&PartialD;A}_j}={\frac{\&PartialD;f_{\mathrm{je}}}{{\&PartialD;A}_i}=k}_{\mathrm{ije}}+\frac{f_{\mathrm{ie}}{f}_{\mathrm{je}}}{v^{2}B{\mathrm{\&Delta;}}_e}\frac{\&PartialD;v}{\&PartialD;B}$ $\frac{{\&PartialD;f}_{\mathrm{ie}}}{{\&PartialD;A}_k}={\frac{\&PartialD;f_{\mathrm{ke}}}{{\&PartialD;A}_i}=k}_{\mathrm{ike}}+\frac{f_{\mathrm{ie}}{f}_{\mathrm{ke}}}{v^{2}B{\mathrm{\&Delta;}}_e}\frac{\&PartialD;v}{\&PartialD;B}$ $\frac{{\&PartialD;f}_{\mathrm{ke}}}{{\&PartialD;A}_j}={\frac{\&PartialD;f_{\mathrm{je}}}{{\&PartialD;A}_k}=k}_{\mathrm{kje}}+\frac{f_{\mathrm{ke}}{f}_{\mathrm{je}}}{v^{2}B{\mathrm{\&Delta;}}_e}\frac{\&PartialD;v}{\&PartialD;B}$ Wherein B is a magnetic flux density. $v=\frac{1}{{\mathrm{\&mu;}}_e}$ Be the reluctivity in the unit.[J] that each unit is calculated ^eAll be superimposed in the Jacobian matrix [J], Item is superimposed to the capable j row of k of [J], then can obtain calculating required [J].

Provide calculated example below:

If the apex coordinate of a certain boundary element is:

[56.5,0], [57,0], [56.4314,2.78382], sequence number is 2.

Computational process is:

Read in apex coordinate in array pp, all information with triangular element are kept among the structural array tt simultaneously.

At first judge the position of this unit,,, calculate the current density of this unit then according to given electric current then according to the electrical degree of position judgment unit if in stator or rotor.Formula is as (11), shown in (12).If not in groove, then current density is made as 0.Here this unit is in stator yoke portion, and then current density is 0.

Calculate the coefficient of this unit:

bi＝yj-ym＝0-2.78382＝-2.78382；bj＝ym-yi＝2.78382-0＝2.78382；bm＝yi-yj＝0-0＝0；

Ci＝xm-xj＝56.4314-57＝-0.5686；Cj＝xi-xm＝56.5-56.4314＝0.0686；Cm＝xj-xi＝57-56.5＝0.5： ${\mathrm{\&Delta;}}_e=\frac{1}{2}(b_i{c}_j-b_j{c}_i)=0.6960$

Because current density is 0, and by boundary condition as can be known He be 0.Because the unit is in stator yoke, so v _e=1.0/ ((4 * PI * le-7 * 1500)=530.5165 of u0 * ur)=1.0/ $k_{\mathrm{ii}}=\frac{{\mathrm{<figure-callout\; id="4"\; label="v\; e"\; filenames="A0214614900242.png"\; state="\{\{state\}\}">v</figure-callout>}}_e}{4{\mathrm{\&Delta;}}_e}({\mathrm{<figure-callout\; id="2"\; label="b\; i"\; filenames="A0214614900242.png,A0214614900292.png"\; state="\{\{state\}\}">b</figure-callout>}}_i^{}+c_i^2)=\frac{530.5165}{4\&times;0.696}(2.78382^2+{(-0.5686)}^2)=1538.4$ $k_{\mathrm{ij}}=\frac{{\mathrm{<figure-callout\; id="4"\; label="v\; e"\; filenames="A0214614900242.png"\; state="\{\{state\}\}">v</figure-callout>}}_e}{4{\mathrm{\&Delta;}}_e}(b_i{b}_j+c_i{c}_j)=\frac{530.5165}{4\&times;0.696}(2.78382\&times;(-2.78383)+(-0.5686)\&times;0.0686)=1469.3$ Similarly can obtain other coefficients.

After each unit is similarly calculated, obtain total matrix equation noted earlier at last.This equation is found the solution the magnetic potential that can obtain final each unit to distribute.

Be kept at the magnetic linkage on Distribution of Magnetic Field in the electric machine rotor under different electric currents and the MT axle that is in the data file, the preservation form is a data form.Be example with the result under a certain electric current below:

I _Ms(stator M shaft current) 2.81095 f _Ms(stator M axle magnetic linkage) 0.381278

I _Ts(stator T shaft current) 14.9244 f _Ts(stator T axle magnetic linkage) 0.0488938

I _Mr(rotor M shaft current) 0 f _Mr(rotor M axle magnetic linkage) 0.376839

I _Tr(rotor T shaft current)-14.047 f _Tr(rotor T axle magnetic linkage) 2.67026e-007

In the practical application this table is carried out interpolation and just can obtain magnetic linkage under any electric current.

Be the part of the magnetic linkage table that calculates below

Be described as follows:

Current_iTs is a stator T shaft current; Current_iTr (') is a rotor T shaft current; Current_iMs is a stator M shaft current

Current_iMr (') is a rotor M shaft current; Link-Ts is a stator T axle magnetic linkage; Link-Ms is a stator M axle magnetic;

Link-Tr is a rotor T axle magnetic linkage; Link-Mr is a rotor M axle magnetic linkage;

Link-A is an A phase magnetic linkage; Link-B is a B phase magnetic linkage; Link-C is a C phase magnetic linkage

?Current_iTs	?9.1639	?14.6622	?18.3278
				?Current_iTr(′)	?-8.59711	?-13.7554	?-17.1942
?Current_iMs	?0	?0	?0
				?Current_iMr(′)	?0	?0	?0
?Link-Ts	?0.0480693	?0.0758845	?0.0921548
				?Link-Ms	?-1.78623e-5	?-3.03019e-5	?-3.91252e-5
?Link-Tr	?0.00564894	?0.00859067	?0.00945619
				?Link-Mr	?-1.89887e-5	?-3.22076e-5	?-4.17075e-5
?Link-A	?0.0405476	?0.0633322	?0.0752785
				?Link-B	?-0.0315718	?-0.0505208	?-0.0629877
?Link-C	?-0.0315409	?-0.0504683	?-0.0629199

Below how narration calculates the breakdown torque table and to optimize excitation ammeter according to the magnetic linkage table that has obtained.

At first set stator total current Is.The Is that sets can not be greater than current maxima Ismax, and this maximum is subjected to the inverter performance limitations.The initial value that makes Is is Is ₀, be generally 1/10 of motor rated current, then Is=Is ₀Then set rotating speed of motor ω.The maximum ω of rotating speed _MaxBe subjected to the restriction of motor performance, initial value is got 60 rev/mins.

According to following formula: $u_s=\sqrt{{\mathrm{<figure-callout\; id="2"\; label="u\; Ms"\; filenames="A0214614900242.png,A0214614900292.png"\; state="\{\{state\}\}">u</figure-callout>}}_{\mathrm{Ms}}^{}+{\mathrm{<figure-callout\; id="2"\; label="u\; Ts"\; filenames="A0214614900242.png,A0214614900292.png"\; state="\{\{state\}\}">u</figure-callout>}}_{\mathrm{Ts}}^{}}\&le;u_{s\max }$ $u_{\mathrm{Ms}}=R_s{i}_{\mathrm{Ms}}-({\mathrm{\&omega;}}_r+{\mathrm{\&omega;}}_s)f_{\mathrm{Ts}}(i_{\mathrm{Ts}},i_{\mathrm{Ts}}/K_s)$

u _Ts＝R _si _Ts+(ω _r+ω _s)f _Ms(i _Ms，0)

Can under the condition of given stator total current and rotating speed, calculate the amplitude limit value of exciting current according to magnetic linkage table.Briefly can be with top two simplified formulas for asking at i _MsEqual the inequality equal sign establishment of what time.

Try to achieve the maximum I of exciting current _MmaxAfter, the initial value of setting stator excitation electric current I m is 1/10 of a rated exciting current.Under the situation of known stator M axle exciting current and stator total current, can calculate stator T shaft current according to following formula: $i_s=\sqrt{{{i_{\mathrm{<figure-callout\; id="2"\; label="Ms"\; filenames="A0214614900242.png,A0214614900292.png"\; state="\{\{state\}\}">Ms</figure-callout>}}}^2}^{}+{{\mathrm{<figure-callout\; id="2"\; label="i\; Ts"\; filenames="A0214614900242.png,A0214614900292.png"\; state="\{\{state\}\}">i</figure-callout>}}_{\mathrm{Ts}}}^2}$

After obtaining stator M, T shaft current, can look into the rotor flux ψ that magnetic linkage table obtains motor _r, then can calculate motor torque according to following formula: $T_e=\frac{3}{2}{n}_p\frac{i_{\mathrm{Ts}}}{K_s}{\mathrm{\&psi;}}_r$

Motor torque, given stator total current, motor speed, stator M, T shaft current, rotor flux are saved in the data file.Whether judge current stator excitation electric current then greater than maximum, if less than would increase an increment, get 1/10 of rated exciting current here, recomputate stator T shaft current then and try to achieve new motor torque and deposit data file in.Repeated calculation up to Im greater than maximum.When Im greater than current maximum, whether judge given rotating speed greater than maximum, if not then increase an increment, double counting exciting current maximum is up to calculating motor torque again.Increase repeatedly motor speed up to rotating speed greater than the rotating speed maximum, judge that more whether the motor total current is greater than the stator current maximum.If do not reach maximum stator current value, then stator current is increased an increment, reset motor speed then, calculate again up to trying to achieve motor output torque.Increase repeatedly up to the maximum that reaches stator current.

After aforementioned calculation finishes, all rotating speeds that occur in calculating are kept among the array sp, the size of establishing this array is n.Get speed=sp[0], and be that all motor torques that calculate of speed are kept among the array Te with corresponding rotating speed, the size of establishing this array is m.This array is sorted by size, get its maximum of T emax, then obtain rotating speed and be the motor breakdown torque under the speed, speed, Temax and corresponding stator excitation, torque current are deposited among the data file maxtorqu.dat.Get T=Te[0], be that speed and torque are that the stator current of T deposits among the array Is with corresponding rotating speed.Is is sorted by size, and getting minimum value is Ismin, and speed, T, Ismin and corresponding stator excitation electric current are deposited among the data file optimal.dat.Make then that T is the next torque value of Te array, the step all elements in array Te above repeating are all used.Make speed point to the next element of sp array then, get operations such as torque more again, all elements in the sp array all is used to.Finally can obtain two data files.Maxtorqu.dat is the breakdown torque table, and optimal.dat is for optimizing exciting table.

By tabling look-up and can reach the effect that has improved the output torque in certain total current restriction optimizing exciting table and breakdown torque table, result of the test has also been verified this point in real system.

Table 1: optimize exciting table (exporting the exciting current of the minimum total current correspondence of certain torque)

iM1max＝29.47082468 ( A ) Te＝2.70746473 ( Nm ) Speed＝2160 ( rpm ) iM1max＝24.74302980 ( A ) Te＝2.63189971 ( Nm ) Speed＝2220 ( rpm ) iM1max＝21.09194341 ( A ) Te＝2.56003875 ( Nm ) Speed＝2280 ( rpm ) iM1max＝18.27491629 ( A ) Te＝2.49084617 ( Nm ) Speed＝2340 ( rpm ) iM1max＝16.02573427 ( A ) Te＝2.42493269 ( Nm ) Speed＝2400 ( rpm ) iM1max＝14.19040791 ( A ) Te＝2.36206414 ( Nm ) Speed＝2460 ( rpm ) iM1max＝12.70356758 ( A ) Te＝2.30202777 ( Nm ) Speed＝2520 ( rpm ) iM1max＝11.51978179 ( A ) Te＝2.24458567 ( Nm ) Speed＝2580 ( rpm ) iM1max＝10.57498681 ( A ) Te＝2.18949822 ( Nm ) Speed＝2640 ( rpm ) iM1max＝9.75600839 ( A ) Te＝2.13673122 ( Nm ) Speed＝2700 ( rpm ) iM1max＝9.15476443 ( A ) Te＝2.08613405 ( Nm ) Speed＝2760 ( rpm ) iM1max＝8.57911884 ( A ) Te＝2.03756872 ( Nm ) Speed＝2820 ( rpm ) iM1max＝8.14756482 ( A ) Te＝1.99090852 ( Nm ) Speed＝2880 ( rpm )

Subordinate list 2: the breakdown torque table that a certain rotating speed is corresponding down

( A ) ( A ) Torque＝7.64925427 ( Nm ) Speed＝5100 ( rpm ) iMs＝3.384394871 iTs＝12.194775030Torque＝7.56649265 ( Nm ) Speed＝5160 ( rpm ) iMs＝3.340630411 iTs＝12.206836409Torque＝7.48536041 ( Nm ) Speed＝5220 ( rpm ) iMs＝3.297871503 iTs＝12.218457767Torque＝7.40581435 ( Nm ) Speed＝5280 ( rpm ) iMs＝3.256083626 iTs＝12.229660097Torque＝7.32781249 ( Nm ) Speed＝5340 ( rpm ) iMs＝3.215233823 iTs＝12.240463150Torque＝7.25131404 ( Nm ) Speed＝5400 ( rpm ) iMs＝3.175290617 iTs＝12.250885526Torque＝7.17627941 ( Nm ) Speed＝5460 ( rpm ) iMs＝3.136223924 iTs＝12.260944750Torque＝7.10267014 ( Nm ) Speed＝5520 ( rpm ) iMs＝3.098004981 iTs＝12.270657350Torque＝7.03044895 ( Nm ) Speed＝5580 ( rpm ) iMs＝3.060606270 iTs＝12.280038922Torque＝6.95957961 ( Nm ) Speed＝5640 ( rpm ) iMs＝3.024001454 iTs＝12.289104193Torque＝6.89002703 ( Nm ) Speed＝5700 ( rpm ) iMs＝2.988165316 iTs＝12.297867080Torque＝6.82175712 ( Nm ) Speed＝5760 ( rpm ) iMs＝2.953073693 iTs＝12.306340741Torque＝6.75473685 ( Nm ) Speed＝5820 ( rpm ) iMs＝2.918703430 iTs＝12.314537627Torque＝6.68893419 ( Nm ) Speed＝5880 ( rpm ) iMs＝2.885032321 iTs＝12.322469524Torque＝6.62431805 ( Nm ) Speed＝5940 ( rpm ) iMs＝2.852039064 iTs＝12.330147600Torque＝6.56085831 ( Nm ) Speed＝6000 ( rpm ) iMs＝2.819703216 iTs＝12.337582439

Claims (3)

1. optimize excitation control method based on saturation nonlinearity motor model asynchronous machine, it has adopted in the field orientation vector control the method for the exciting current of asynchronous machine and torque current component decoupling zero, has also adopted the saturated motor mathematical model of following consideration magnetic circuit to calculate the relevant parameter of motor simultaneously:

Wherein:

P represents the time is carried out the differentiating operator of differential;

i _Ms, i _TsBe respectively the M of stator, the T shaft current;

i _Mr, i _TrBe respectively the M of rotor, the T shaft current;

f _Ms, f _TsBe respectively the M of stator, T axle magnetic linkage;

f _Tr, f _MrBe respectively the M of rotor, T axle magnetic linkage;

u _Ms, u _TsBe the M of stator, the T shaft voltage;

R _sBe stator resistance, R _rBe rotor resistance;

ω ₁Be stator current frequency, ω ₂Be the rotor current frequency;

M, T axle are the rotating coordinate systems with the rotation of motor synchronous rotating speed;

It is characterized in that: it is to optimize in the exciter control system at the asynchronous machine that is made of single-chip microcomputer, sampling plate and inverter of bus control type, changing exciting current, to realize that stator total current one is fixed electromagnetism output torque maximum and fix the field orientation vector control method of the optimization field performance of total current minimum in torque output one, and it contains successively and has the following steps:

(1) under the on all four condition of the magnetomotive force that requires motor model to produce after the coordinate transform, the variable of motor A, B, C three-phase projected to the motor model that obtains the saturated influence in consideration magnetic field on the MT coordinate system on the MT coordinate system with the rotation of the synchronous speed of motor so that be converted into linear variable, the dc electrolysis model of alternating current machine model equivalence for simplification at the time-varying parameter of the following alternating current machine of MT coordinate system;

(2) input:

(2.1) the breakdown torque table under the different rotating speeds;

(2.2) with the corresponding optimization excitation ammeter of minimum total current of exporting certain torque;

(2.3) magnetic linkage table that constitutes by following data item:

i _MsBe stator M shaft current, i _MrBe rotor M shaft current;

i _TsBe stator T shaft current, i _TrBe rotor T shaft current;

ψ _TSBe stator T axle magnetic linkage, ψ _TrBe rotor T axle magnetic linkage;

ψ _MSBe stator M axle magnetic linkage, ψ _MrBe rotor M axle magnetic linkage;

Stator A, B, C three-phase magnetic linkage: ψ _A, ψ _B, ψ _C

(3) according to given angular velocity omega ^*With the actual measurement angular velocity omega _rObtain torque set-point Te through the rotating speed pi regulator ^*: $T_e^{*}=(k_P+\frac{k_q}{S})({\mathrm{\&omega;}}^{*}-{\mathrm{\&omega;}}_r)$ Wherein kp is a proportionality constant, and kq is an integral constant;

(4) Te ^*Compare with the breakdown torque Temax that obtains according to rotor actual measurement rotating speed:

If: Te ^*〉=Temax then gets Te ^*=Temax;

If: Te ^*＜Temax then gets Te ^*=Te ^*

(5) according to rotor actual measurement rotating speed and Te ^*Check in i from optimizing excitation ammeter _Ms, again by i _MsCheck in rotor flux ψ from magnetic linkage table _r

(6) according to Te ^*, ψ _rObtain i from following formula _Ts: $T_e^{*}=\frac{3}{2}{n}_p\frac{i_{\mathrm{Ts}}}{K_s}{\mathrm{\&psi;}}_r{n}_p$ : the motor number of pole-pairs, $K_s=L_r/L_m$ : L _m: the rotor mutual inductance

L _r: the rotor self-induction;

(7) judge read in the threephase stator size of current whether in reasonable range, if surpass amplitude limit value, then get amplitude limit value for reading in current value;

(8) calculate the slippage angular frequency according to following formula _sWith the rotor field angular position theta ₁: ${\mathrm{\&omega;}}_s=\frac{R_r{i}_{\mathrm{Ts}}}{K_s{\mathrm{\&psi;}}_r}$ , be the rotor current angular frequency to rotor field angular speed again _l=ω _s+ ω _rIntegration obtains the rotor field angular position theta ₁

(9) the stator phase current is become current i under the MT coordinate system _Ms, i _Ts: $\left[\begin{array}{c}{i}_M\\ i_T\end{array}\right]=\frac{2}{3}\left[\begin{array}{ccc}{\cos \mathrm{\&theta;}}_1& \cos ({\mathrm{\&theta;}}_1-\frac{2\mathrm{\&pi;}}{3})& \cos ({\mathrm{\&theta;}}_1+\frac{2\mathrm{\&pi;}}{3})\\ -\sin {\mathrm{\&theta;}}_1& -\sin ({\mathrm{\&theta;}}_1-\frac{2\mathrm{\&pi;}}{3})& -\sin ({\mathrm{\&theta;}}_1+\frac{2\mathrm{\&pi;}}{3})\end{array}\right]\left[\begin{array}{c}{i}_{\mathrm{sa}}\\ i_{\mathrm{sb}}\\ i_{\mathrm{sc}}\end{array}\right];$

(10) according to following formula i given on the stator _Ms ^*, i _Ts ^*With the actual current i that records _Ms, i _TsAfter comparing respectively, difference is obtained separately the voltage given value u of stator MT axle by exciting current pi regulator, torque current pi regulator and the compensation of process voltage decoupling _Ms ^*, u _Ts ^*: ${\hat{u}}_{\mathrm{Ms}}=L_m^M{\mathrm{pi}}_m-{\mathrm{\&omega;}}_1{\mathrm{\&psi;}}_{\mathrm{Ts}}$ $L_m^M=\frac{\&PartialD;f_{\mathrm{Ms}}}{{\&PartialD;i}_{\mathrm{Mr}}}$ M axle bucking voltage ${\hat{u}}_{\mathrm{Ts}}=L_{\mathrm{MT}}{\mathrm{pi}}_m+{\mathrm{\&omega;}}_1{\mathrm{\&psi;}}_{\mathrm{Ms}}$ $L_{\mathrm{MT}}=\frac{{\&PartialD;f}_{\mathrm{Ts}}}{{\&PartialD;i}_m}$ T axle bucking voltage i _m=i _Mr+ i _MsThe exciting current pi regulator is: $u_M=(k_{\mathrm{dm}}+\frac{k_{\mathrm{lm}}}{S})(i_{\mathrm{MS}}^{*}-i_{\mathrm{MS}}),$ The torque current pi regulator is: $u_T=(k_{\mathrm{dT}}+\frac{k_{\mathrm{lT}}}{S})(i_{\mathrm{TS}}^{*}-i_{\mathrm{TS}}),$ Then the given voltage of MT axle is: $u_{\mathrm{Ms}}^{*}=u_M+{\hat{u}}_{\mathrm{Ms}}$ $u_{\mathrm{Ts}}^{*}=u_T+{\hat{u}}_{\mathrm{Ts}};$

(11) obtain stator three-phase voltage set-point through 2/3 conversion again and send into pulse width modulator PWM producing six tunnel required pulses of inverter;

(12) finish.

2. according to claim 1 based on saturation nonlinearity motor model asynchronous machine optimization excitation control method, it is characterized in that: described magnetic linkage table is at the hypothesis cylindrical electrical machine along the axis direction Distribution of Magnetic Field evenly and ignore under the condition of the kelvin effect of rotor bar and magnetic hysteresis eddy current effect unshakable in one's determination, the external electromagnetic field of motor is reduced to the two dimensional surface field, use the finite element elements method at different M, calculate the electromagnetic field of motor and the magnetic linkage of each winding under the T shaft current and obtain, it contains successively and has the following steps:

(1) get with the perpendicular cross section of motor shaft as the constant nonlinear electromagnetic of analyzed two dimensional surface field, subdivision is carried out in the fan section of the radial symmetric that comprises stator and rotor;

(2) find the solution vector magnetic potential A on each summit of triangular element to finding the solution the zone triangular element that goes out of each subdivision with following electromagnetic field equation:

(3) by each triangular element row are write above-mentioned electromagnetic field equation,, all equations find the solution the magnetic potential that obtains each summit, triangle subdivision unit thereby being synthesized a total matrix equation;

(4) suppose again under the condition of each point magnetic potential sight line distribution in the triangular element, according to each triangular apex i, j, the position coordinates of k and the magnetic potential A on this summit _iA _jA _k, determine that the magnetic potential of each point in stator, the rotor distributes;

(5) calculate A, B, C three-phase magnetic linkage ψ according to stator each point magnetic potentiometer _Aψ _Bψ _CAfter, obtain the magnetic linkage ψ on the stator MT axle more thus _MsAnd ψ _Ts: ${\mathrm{\&psi;}}_A=W_K{l}_e\left(\underset{i=1}{\overset{q}{\mathrm{\&Sigma;}}}(A_{s_i}-A_{s_{\mathrm{ai}}})\right)$

Wherein: A _Sl... A _SqAnd A _SalA _SaqBe respectively each groove S of stator A phase winding place _l... S _qAnd S _Al... S _AqMagnetic potential mean value, l _eBe effective length unshakable in one's determination, W _kBe umber of turn;

B, each corresponding magnetic potential mean value substitution following formula of C two-phase are just obtained the magnetic linkage of B, C two-phase; Thereby obtain:

θ is the angle between the positive A axle of M axle and stator, in vector control system the output torque of motor only relevant with the size of magnetic linkage and with the orientation independent of magnetic linkage, the magnetic linkage size is determined and can not changed with the variation of θ under the given condition of MT shaft current, so here can be given arbitrarily and can the magnetic linkage size that finally calculates not exerted an influence in the θ angle, is taken as 0 usually for simplicity;

(6) for rotor-side, establishing the pairing groove of a pair of mouse cage sliver is n and n+1, and corresponding average magnetic potential is A _nAnd A _N+1, then the magnetic linkage in this loop is ψ _n=l _e(A _N+1-A _n), corresponding, the magnetic linkage on the rotor MT axle is:

Wherein, K is the rotor bar number, ₀It is the angle between two slivers;

(7) under limit in the field orientation system motor T axle magnetic linkage be zero, then can obtain i _Ts+ K _si _Tr=0, thereby at given stator torque current i _TsCondition under obtain rotor T shaft current, by changing given stator T shaft current i _TsWith M shaft current i _MsWith rotor M shaft current i _Mr, repeat above step and obtain a discrete usefulness with reflecting stator M shaft current i _Ms, stator M axle magnetic linkage ψ _Ms, stator T shaft current i _Ts, stator T axle magnetic linkage ψ _Ts, rotor M shaft current i _Mr, rotor M axle magnetic linkage ψ _Mr, rotor T shaft current i _Ts, rotor T axle magnetic linkage ψ _TrBetween the magnetic linkage table of correlation; Obtain magnetic linkage table under any electric current by interpolation algorithm;

3. according to claim 1 based on saturation nonlinearity motor model asynchronous machine optimization excitation control method, it is characterized in that:

The data item that described breakdown torque table contains is: breakdown torque, rotating speed, optimization exciting current and torque current;

The data item that described optimization excitation ammeter contains is: optimize exciting current, corresponding output torque and rotating speed;

Described breakdown torque table and optimization excitation ammeter obtain through following steps successively according to magnetic linkage table:

(1) scope of setting stator current Is:

I _S0～I _Smax(maximum is limited by inverter)

(2) differentiate: I _s＞I _Smax:

If: I _s＞I _Smax, then change step (9) over to;

If: I _s＜I _Smax, then carry out next step;

(3) scope of setting motor speed ω: ω=ω ₀＜ω _Max(maximum is limited by motor),

If: ω＞ω _Max, then returning step (2), Is increases an increment simultaneously;

If: ω＜ω _Max, then carry out next step;

(4) under stator total current and stator voltage and motor speed restriction, obtain the optional scope of stator excitation electric current: I _M0～I _Mmax, maximum is determined by following formula $u_s=\sqrt{u_{\mathrm{Ms}}^2+u_{\mathrm{Ts}}^2}\&le;K_u{u}_{\max }=u_{s\max }$ Wherein: u _sBe stator terminal voltage amplitude, K _uFor guaranteeing the voltage security coefficient of inverter trouble free service, u _MaxIt is the maximum withstand voltage that inverter can bear $u_{\mathrm{Ms}}=R_s{i}_{\mathrm{Ms}}-({\mathrm{\&omega;}}_r+{\mathrm{\&omega;}}_s)f_{\mathrm{Ts}}(i_{\mathrm{Ts}},i_{\mathrm{Ts}}/K_s)$ Wherein: ω _r=ω, ${\mathrm{\&omega;}}_s=\frac{R_r{i}_{\mathrm{Ts}}}{K_s{f}_{\mathrm{Ms}}(i_{\mathrm{Ms}},0)};$ u _Ts=R _si _Ts+ (ω _r+ ω _s) f _Ms(i _Ms, 0) and (5) differentiation: I _m＞I _Mmax: if: I _m＞I _Mmax, then returning step (3), ω increases an increment simultaneously;

If: I _s＜I _Smax, then carry out next step;

(6) calculate T shaft current I according to total current _Ts

According to exciting current I _mLook into magnetic linkage table and obtain rotor magnetic linkage ψ _s, ψ _r

(7) calculate motor torque according to torque current and magnetic linkage, with the torque that calculates, stator excitation electric current and torque current, motor stator total current, motor speed deposit disk file result.dat in;

(8) Im increases an increment, and repeating step (5)～(7) are until I _mIncrease make I _s=I _Smax, change step (9) over to:

(9) from disk file result.dat, read in data;

(10) establish array sp[n], n is the quantity of different motor speeds in the file, and rotating speed by depositing in from low to high among the array sp, and is established k=0;

(11) establish speed=sp[k], search all motor torques with the speed correspondence hereof, and deposit this group torque in array Te[m] in, m is the number of these torques;

(12) Te is sorted by size;

(13) getting number maximum among the Te is Temax;

(14) excitation, torque current and the magnetic linkage with speed and Temax and correspondence deposits data file maxtorqu.dat in;

(15) establish j=0, T=Te[i];

(16) search all stator total currents with torque T and speed correspondence hereof, and deposit this group current value in array Is[l] in;

(17) to array Is[l] sort by size;

(18) get Is[l] in minimum value be Ismin;

(19) exciting current and the magnetic linkage with speed and Ismin, T and correspondence deposits data file optimal.dat in;

(20) do you differentiate: k＞n?:

If: k＜n, then k=k+1 returns step (11);

If: k＞n, then deposit data file maxtorque.dat and optimal.dat in disk, obtain the breakdown torque table and optimize excitation ammeter, carry out next step;

(21) finish.

Images